All critical points of f(x,y) are classified as follows:(1,2) - Local Maximum(1,-2) - Local Maximum(-1,2) - Local Maximum(-1,-2) - Saddle Point.
Given the function, f(x,y)=x³ + 3xy² − 15x + y³ − 15y.To find the critical points of the function, we differentiate it partially with respect to x and y, respectively.
∂f/∂x = 3x² + 3y² - 15 = 0 ∂f/∂y = 6xy + 3y² - 15x + 3y² - 15 = 0
On solving the above two equations, we get the critical points to be (1,2), (-1,2), (1,-2) and (-1,-2).
To classify these critical points, we use the second partial derivatives test. Let us evaluate the second-order partial derivatives of f(x,y).
∂²f/∂x² = 6x = 6 at all critical points∂²f/∂y² = 6x + 6y = 0 at all critical points∂²f/∂x∂y = 6y = 12 or -12.Thus, for (1,2), (1,-2), (-1,-2), we have ∂²f/∂x∂y = 12 which is positive.
Therefore, these points are local maxima.
For (-1,2), we have ∂²f/∂x∂y = -12 which is negative.
Therefore, this point is a saddle point.
Hence, all critical points of f(x,y) are classified as follows:(1,2) - Local Maximum(1,-2) - Local Maximum(-1,2) - Local Maximum(-1,-2) - Saddle Point.
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Which of the described new technologies is likely to have the
largest impact in GIS over the next five years? Why?
The technology that is likely to have the largest impact in GIS (Geographic Information System) over the next five years is Artificial Intelligence (AI).
1. AI has the potential to greatly enhance the efficiency and accuracy of GIS data analysis and interpretation. AI algorithms can process large volumes of data and identify patterns and relationships that may not be immediately apparent to human analysts. This can lead to more accurate and reliable GIS analyses and decision-making.
2. Machine learning, a subset of AI, can enable GIS systems to automatically learn and improve from experience without being explicitly programmed. This means that GIS software can adapt and improve its performance over time, making it more intelligent and efficient.
3. AI can also assist in automating time-consuming tasks in GIS, such as data collection, data integration, and data validation. For example, AI can analyze satellite imagery to automatically identify and classify different land cover types, saving time and effort for GIS professionals.
4. Another area where AI can have a significant impact is in predictive modeling. By analyzing historical GIS data and using AI algorithms, it is possible to predict future patterns and trends. This can be particularly useful in urban planning, transportation management, and environmental monitoring.
5. AI can also improve GIS-based decision-making by providing insights and recommendations based on complex spatial data. For instance, AI algorithms can analyze transportation networks and suggest optimal routes for emergency response or identify locations for new infrastructure development.
Overall, AI has the potential to revolutionize the field of GIS by improving data analysis, automating tasks, enhancing predictive modeling, and enabling smarter decision-making. Its ability to process and analyze large volumes of spatial data will be crucial in unlocking new insights and advancing GIS applications in the coming years.
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A medical company tested a new drug for possible side effects. The table
shows the relative frequency that a study participant experienced the side
effect.
In the medical industry, new drugs undergo thorough tests before they can be approved for use. One aspect of drug testing is the assessment of possible side effects on patients.
The medical company that conducted the tests analyzed data and used tables to show the relative frequency that a study participant experienced side effects. This approach provided the company with insights on the occurrence of side effects.The table below shows the relative frequency of a side effect that a study participant experienced.
Side Effect Number of Participants who experienced the side effect
Nausea 45
Headache 33
Dizziness 21
Vomiting 12
Fever 8
Fatigue 5
Total 124
Relative Frequency = Number of participants who experienced a side effect / Total number of participants in the study
Using this formula, we can determine the relative frequency of each side effect.
For example, the relative frequency of nausea is 45/124, which equals 0.3637 or 36.37%. This shows that out of the 124 study participants, 45 experienced nausea, which was the most frequent side effect.
The relative frequency of each of the other side effects was 0.2661 (26.61%) for headaches, 0.1693 (16.93%) for dizziness, 0.0968 (9.68%) for vomiting, 0.0645 (6.45%) for fever, and 0.0403 (4.03%) for fatigue.
The company can use these findings to decide whether to continue with the development of the drug.
For instance, the high occurrence of nausea may mean that the drug needs further development or modification before it is approved for use.
On the other hand, a lower frequency of side effects may mean that the drug can proceed to the next stage of testing.
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Use Green's Theorem to evaluate the following line integral. Assume the curve is oriented counterclockwise. A sketch is helpful. ∮(2y−3,2x^2 −4)⋅dr, where C is the boundary of the rectangle with vertices (0,0),(5,0),(5,4), and (0,4) ∮_C (2y−3,2x^2 −4)⋅dr=
The line integral [tex]\(\oint_C (2y-3, 2x^2-4) \cdot dr\)[/tex] around the boundary [tex]\(C\)[/tex] of the rectangle is equal to [tex]\(160\).[/tex]
To evaluate the line integral [tex]\(\oint_C (2y-3, 2x^2-4) \cdot dr\)[/tex] using Green's theorem, we need to compute the flux of the vector field [tex]\((2y-3, 2x^2-4)\)[/tex] across the boundary [tex]\(C\)[/tex] of the given rectangle.
First, let's sketch the rectangle with its vertices at [tex]\((0,0)\), \((5,0)\), \((5,4)\), and \((0,4)\).[/tex]
(0,4)------------------(5,4)
| |
| |
| |
| |
(0,0)------------------(5,0)
The boundary [tex]\(C\)[/tex] consists of four line segments: the top side, the right side, the bottom side, and the left side of the rectangle.
We can apply Green's theorem, which states that for a vector field [tex]\(\mathbf{F} = (P, Q)\)[/tex] and a simple closed curve [tex]\(C\)[/tex] oriented counterclockwise, the line integral [tex]\(\oint_C \mathbf{F} \cdot d\mathbf{r}\)[/tex] is equal to the double integral over the region [tex]\(D\)[/tex] enclosed by [tex]\(C\)[/tex] of the curl of [tex]\(\mathbf{F}\):[/tex]
[tex]\[\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}\right) \, dA\][/tex]
In our case, [tex]\(\mathbf{F} = (2y-3, 2x^2-4)\),[/tex] so we have [tex]\(P = 2y-3\) and \(Q = 2x^2-4\)[/tex]. We need to evaluate the double integral of the curl of [tex]\(\mathbf{F}\)[/tex] over the region enclosed by the rectangle.
The curl of [tex]\(\mathbf{F}\)[/tex] is given by:
[tex]\[\text{curl}(\mathbf{F}) = \left(\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}\right)\][/tex]
[tex]\[\text{curl}(\mathbf{F}) = \left(\frac{{\partial}}{{\partial x}}(2x^2-4) - \frac{{\partial}}{{\partial y}}(2y-3)\right)\][/tex]
[tex]\[\text{curl}(\mathbf{F}) = (4x - 0) - (0 - 2) = 4x - 2\][/tex]
Now, we can compute the double integral of [tex]\(\text{curl}(\mathbf{F})\)[/tex] over the region enclosed by the rectangle.
[tex]\[\iint_D \text{curl}(\mathbf{F}) \, dA = \int_{0}^{4} \int_{0}^{5} (4x - 2) \, dx \, dy\][/tex]
Integrating with respect to [tex]\(x\)[/tex] first, we have:
[tex]\[\int_{0}^{5} (4x - 2) \, dx = \left[2x^2 - 2x\right]_{0}^{5} = (2(5)^2 - 2(5)) - (2(0)^2 - 2(0)) = 50 - 10 = 40\][/tex]
Substituting this result into the double integral, we obtain:
[tex]\[\iint_D \text{curl}(\mathbf{F}) \, dA = \int_{0}^{4} 40 \, dy = \left[40y\right]_{0}^{4} = 40(4) - 40(0) = 160\][/tex]
Therefore, the line integral [tex]\(\oint_C (2y-3, 2x^2-4) \cdot dr\)[/tex] around the boundary [tex]\(C\)[/tex] of the rectangle is equal to [tex]\(160\).[/tex]
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2 kg of saturated water vapor at 600 kPa pressure is in a piston-cylinder arrangement. The water expands adiabatically up to a pressure of 100 kPa and it is stated that there is a work output of 700 kJ. a.) Calculate the change in entropy of water as kJ/kg.K.
b.) Is the phase change realistic? Support your answer using the T-s diagram and the second law concept for process change.
a.) The change in entropy of water can be calculated using the equation:
ΔS = Cp * ln(T2/T1) - R * ln(P2/P1)
where ΔS is the change in entropy, Cp is the specific heat capacity at constant pressure, T1 and T2 are the initial and final temperatures, and P1 and P2 are the initial and final pressures.
First, we need to determine the initial and final temperatures. From the ideal gas law, we can rearrange it to solve for temperature:
P1V1/T1 = P2V2/T2
Given that the mass of water vapor is 2 kg, we can determine the initial and final volumes using the specific volume of saturated water vapor.
Next, we need to determine the specific heat capacity at constant pressure (Cp) and the gas constant (R). For water vapor, Cp is approximately 2.09 kJ/kg.K and R is approximately 0.461 kJ/kg.K.
Substituting the values into the equation, we can calculate the change in entropy of water.
b.) To determine if the phase change is realistic, we can examine the T-s diagram and apply the second law of thermodynamics. In the T-s diagram, the phase change occurs when the water vapor undergoes an adiabatic expansion and reaches a lower pressure.
If the work output of 700 kJ is obtained during this adiabatic expansion, it suggests that the water vapor has gone through a phase change. However, the T-s diagram can help us confirm this.
On the T-s diagram, an adiabatic expansion follows a curve that is steeper than an isentropic (reversible and adiabatic) expansion. If the process shown on the T-s diagram matches an adiabatic expansion, then the phase change is realistic.
Additionally, we can apply the second law of thermodynamics, which states that the entropy of an isolated system can only increase or remain constant. If the change in entropy of the water is positive or zero, then the phase change is realistic.
By analyzing the T-s diagram and considering the second law concept for process change, we can determine if the phase change is realistic or not.
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a research firm needs to estimate within 3% the proportion of junior executives leaving large manufacturing companies within three years. a 0.95 degree of confidence is to be used. several years ago, a study revealed that 30% of junior executives left their company within three years. to update this study, how many junior executives should be surveyed? group of answer choices 897 1,085 800 782
To estimate the proportion of junior executives leaving large manufacturing companies within three years within a 3% margin of error and a 95% confidence level, we can use the formula for sample size calculation
Thus, the research firm should survey approximately 1,085 junior executives to update the study with a 3% margin of error and a 95% confidence level Where:n = required sample sizeZ = Z-value corresponding to the desired confidence level (in this case, 0.95)p = estimated proportion from the previous study (30% or 0.3)
E = margin of error (3% or 0.03)P lugging in the values, we have:
n = (1.96^2 * 0.3 * (1 - 0.3)) / 0.03^2n ≈ 1079.68 Rounding up to the nearest whole number, the required sample size is approximately 1080. Therefore, the answer closest to this value is 1,085.
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There is no demand for a certain model of a disposable camera when the unit price is $12. However, when the unit price is $8, the quantity demanded is 8000 per week. The suppliers will not market any cameras if the unit price is $2 or lower. At the $4 per camera, however, the manufacturer will market 5000 cameras per week. Both the demand and supply equations are known to be linear. a. Find the demand equation. b. Find the supply equation. c. Find the equilibrium quantity and price. (Round the quantity to the nearest whole number and the price to the nearest cent.)
The demand equation is y = -2000x + 24000 and the supply equation is y = 2500x + 0. The equilibrium quantity is approximately 5 units and the equilibrium price is $16.67.
a. The demand equation is a linear equation that expresses the relationship between the quantity demanded of a good and its price. The given information shows that there is no demand for the certain model of a disposable camera when the unit price is $12. However, when the unit price is $8, the quantity demanded is 8000 per week. Therefore, we can use the two points (12, 0) and (8, 8000) to find the demand equation using the slope-intercept form:
y = mx + b where y is the quantity demanded and x is the price, m is the slope, and b is the y-intercept. The slope of the line can be calculated as:
m = (y2 - y1) / (x2 - x1)
= (8000 - 0) / (8 - 12)
= -2000
The y-intercept can be found by substituting the values of one of the points in the equation and solving for b. For example, using the point (8, 8000):
8000 = -2000(8) + b
=> b = 24000
Therefore, the demand equation is:
y = -2000x + 24000
b. The supply equation is also a linear equation that expresses the relationship between the quantity supplied of a good and its price. The given information shows that the suppliers will not market any cameras if the unit price is $2 or lower. At the $4 per camera, however, the manufacturer will market 5000 cameras per week. Therefore, we can use the two points (2, 0) and (4, 5000) to find the supply equation using the slope-intercept form:
y = mx + b where y is the quantity supplied and x is the price, m is the slope, and b is the y-intercept. The slope of the line can be calculated as:
m = (y2 - y1) / (x2 - x1)
= (5000 - 0) / (4 - 2)
= 2500
The y-intercept can be found by substituting the values of one of the points in the equation and solving for b. For example, using the point (4, 5000):
5000 = 2500(4) + b
=> b = 0
Therefore, the supply equation is:
y = 2500x + 0c.
The equilibrium quantity and price are the values at which the quantity demanded equals the quantity supplied, i.e., the point at which the demand curve intersects the supply curve. To find the equilibrium quantity and price, we can set the demand equation equal to the supply equation and solve for x:
y = -2000x + 24000= 2500x + 0
=> 4500x = 24000
=> x = 5.33
Therefore, the equilibrium quantity is approximately 5 units (since it must be a whole number) and the equilibrium price is $16.67 (since it must be rounded to the nearest cent).Thus, the demand equation is y = -2000x + 24000 and the supply equation is y = 2500x + 0. The equilibrium quantity is approximately 5 units and the equilibrium price is $16.67.
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In a class of 120 students 41 liked swimming,47 liked tennis and 42 liked football.14 students liked both swimming and tennis,15 liked swimming and football and 19 liked tennis and football,while 8 students liked all three sports.find the number of students that liked at least one sport
Answer:
114 students liked at least one sport
Step-by-step explanation:
Total number of students = 120
Number of students who liked at least one sport =
(Number of students who liked swimming) +
(Number of students who liked tennis) +
(Number of students who liked football) -
(Number of students who liked swimming and tennis) -
(Number of students who liked swimming and football) -
(Number of students who liked tennis and football) +
(2 * Number of students who liked all three sports)
= 41 + 47 + 42 - 14 - 15 - 19 + (2 * 8)
= 114
Therefore, therefore 114 students liked at least one sport.
A bag contains 10 balls. 6 of the balls are blue balls while the balances are red balls. Two balls are taken at random from the bag, one after another without replacement. Find the probability that (i) both balls are blue balls. (ii) the balls taken are of different colour
(i) Probability of getting both balls are blue balls:We know that there are a total of 10 balls, of which 6 are blue balls. Also, we need to take out 2 balls one by one without replacement, therefore, for the first ball the probability of getting a blue ball will be 6/10, as there are 6 blue balls out of 10.
Therefore the probability of getting a red ball will be 4/10. Now we have only 5 blue balls left and a total of 9 balls are there in the bag. So for the second ball, the probability of getting a blue ball will be 5/9 as there are only 5 blue balls left. Now the probability of getting both balls blue will be:=(6/10) x (5/9)=1/3.
(ii) Probability of getting both balls of different colour:
We can approach this question in the same manner as we did for the first question. The only difference is that we have to find the probability of taking out two balls of different colours, i.e. one blue and one red, one after the other without replacement. Therefore, the probability of getting the first ball red will be 4/10, and the probability of getting a blue ball will be 6/10. Now we have 5 blue balls and 4 red balls left in the bag for the second ball.
Therefore, the probability of getting a blue ball will be 5/9, and the probability of getting a red ball will be 4/9.Now we need to find the probability of taking out two balls of different colours, i.e. one blue and one red ball. There are two possibilities, either the first ball is blue, and the second ball is red or vice versa. Therefore, the probability of getting two different colour balls will be:=(6/10) x (4/9) + (4/10) x (5/9)=24/90+20/90=44/90=22/45.
Thus, the probability of getting two blue balls is 1/3, and the probability of getting two balls of different colours is 22/45.
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Suppose that \( f \) is continuous and that \( \int_{-2}^{2} t(x) d z=0 \) and \( \int_{-2}^{5} t(x) d x=7 \). Find \( -\int_{2}^{5} d e(x) d x \). \( -28 \) \( -7 \) 28
The correct option is 4.
Given f is continuous function.
[tex]\( \int_{-2}^{2} t(x) d z=0[/tex]
[tex]\int_{-2}^{5} t(x) d x=7[/tex]
To find
[tex]-\int_{2}^{5} 4 e(x) d x \).[/tex]
[tex]\( \int_{-2}^{5} f(x) d x = \( \int_{-2}^{2} f(x) d x+ \( \int_{2}^{5} f(x) d x[/tex]
[tex]\( -\int_{-2}^{5} f(x) d x=0 = \( \int_{-2}^{2} f(x) d x- \( \int_{-2}^{5} f(x) d x= 0 - 7= 7[/tex]
[tex]\( -\int_{2}^{5} 4e(x) d x \)= -4\times7=-28[/tex]
Therefore, [tex]\( -\int_{2}^{5} 4 e(x) d x \). = 28[/tex]
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Complete Question:
Suppose that f is continuous and that [tex]\( \int_{-2}^{2} t(x) d z=0[/tex] and[tex]\( \int_{-2}^{5} t(x) d x=7 \).[/tex] Find[tex]\( -\int_{2}^{5} 4 d(x) d x \).[/tex]
[tex]\( -28 \), \-4,\( -7 \), 28[/tex]
Which table shows no correlation?
The table that shows no correlation is the third table, counting from the tom.
Which table shows no correlation?A table will show no correlation if we can't find any rule that relates the changes in oe of the variables with the changes in the other variable.
First table:
As x increases, y decreases until the point (6, -3), then increases to (8, -2), then it decreases and so on.
Second table.
Like the first one, but with more variation, it first decreases, then it increases until (10, 0), it decreases again to (14, -1), then it increases again.
Third table.
It increases steadily until the last point, where there is a sudden change.
Fourth table:
As x increases, y decreases steadly.
While in table 1 and 2 we can't see a prior any relation, we can see a semi periodic behavior in the increases-decreases, and there are no jumps in values of y.
For the third table the behavior is more random, and we can see two jumps in the y-values, on from -4 to 6 (10 units in total) and other from 10 to -16.
So this is the correct option.
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Determine The Standard And General Equation Of A Plane Containing The Points (1,−1,2), (−3,4,−1) And (3,−2,5).
The general equation of the plane containing the points (1, -1, 2), (-3, 4, -1), and (3, -2, 5) is:
7x + 18y + 6z + 1 = 0.
To determine the standard and general equation of a plane containing the points (1, -1, 2), (-3, 4, -1), and (3, -2, 5), we can use the fact that three non-collinear points uniquely determine a plane.
Step 1: Find two vectors in the plane
Let's choose two vectors that lie on the plane. We can find them by subtracting the coordinates of the given points.
Vector v1 = (−3, 4, −1) - (1, -1, 2) = (-4, 5, -3)
Vector v2 = (3, -2, 5) - (1, -1, 2) = (2, -1, 3)
Step 2: Find the normal vector of the plane
The normal vector of the plane is perpendicular to both v1 and v2. We can find it by taking the cross product of v1 and v2.
Normal vector n = v1 x v2 = (-4, 5, -3) x (2, -1, 3)
To compute the cross product, we can use the determinant of a 3x3 matrix:
n = (5*(-3) - (-1)*(-4), (-4)*3 - (-3)*2, (-4)*(-1) - 5*2)
= (-7, -18, -6)
Step 3: Write the standard equation of the plane
The standard equation of a plane is of the form Ax + By + Cz + D = 0, where (A, B, C) is the normal vector of the plane.
Using the normal vector n = (-7, -18, -6) and one of the given points (1, -1, 2), we can substitute the values into the equation:
-7x - 18y - 6z + D = 0
To find D, we can substitute the coordinates of any of the given points into the equation. Let's use (1, -1, 2):
-7(1) - 18(-1) - 6(2) + D = 0
-7 + 18 - 12 + D = 0
-D = 1
So, D = -1.
The standard equation of the plane is:
-7x - 18y - 6z - 1 = 0
Step 4: Write the general equation of the plane
To obtain the general equation of the plane, we can multiply the equation by -1 to make the constant term positive:
7x + 18y + 6z + 1 = 0
Therefore, the general equation of the plane containing the points (1, -1, 2), (-3, 4, -1), and (3, -2, 5) is:
7x + 18y + 6z + 1 = 0.
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Consider the function f(t)=−3t3+t−6 on the interval [−1,1]. Use the Extreme Value Theorem to determine the absolute extrema. Absolute maximum of when t is Absolute minimum of when t is
f(t) = -3t³ + t - 6 on the interval [-1, 1].The extreme value theorem states that if f is a continuous function on the interval [a, b], then f has both an absolute maximum and an absolute minimum on the interval [a, b]
So, we are to determine the absolute extrema of the given function on the interval [-1, 1]. To find the absolute extrema, we need to follow these steps: Calculate the critical points of the function f on the interval [-1, 1].Evaluate the function at the critical points and at the endpoints of the interval. Compare the values obtained in steps 1 and 2 to find the absolute maximum and minimum, if they exist.
Firstly, we calculate the critical points of the function f on the interval [-1, 1].To find the critical points, we differentiate f(t) with respect to t and set the derivative equal to zero. f(t) = -3t³ + t - 6f'(t) = -9t² + 1Equate f'(t) = 0, we have-9t² + 1 = 0 ⇒ 9t² = 1 ⇒ t² = 1/9 ⇒ t = ±1/3.So, the critical points of the function f on the interval [-1, 1] are -1, -1/3, 1/3 and 1.Now, we evaluate the function at the critical points and at the endpoints of the interval:For t = -1, f(-1) = -3(-1)³ + (-1) - 6 = -2.For t = -1/3, f(-1/3) = -3(-1/3)³ + (-1/3) - 6 = -61/27.For t = 1/3, f(1/3) = -3(1/3)³ + (1/3) - 6 = -569/27.For t = 1, f(1) = -3(1)³ + (1) - 6 = -8.So, we have f(-1) = -2 < f(-1/3) = -61/27 < f(1/3) = -569/27 < f(1) = -8.Hence, the absolute maximum of the function f is -2, which occurs at t = -1, and the absolute minimum of the function f is -569/27, which occurs at t = 1/3.
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If A Bacteria Doubles Its Size Every Four Hours, In How Many Hours Will It Triple? Round To The Nearest Tenth Of An Hour.
The bacteria would take approximately 6.6 hours to triple in size, if it doubles its size every four hours.
A bacteria that doubles its size every four hours takes approximately 6.6 hours to triple in size. As an initial point, when a bacteria doubles its size, it increases by two. To find the number of times the bacteria size has doubled to triple, we will need to know the number of times it has doubled its size.
We can, therefore, obtain this by calculating the logarithm of the ratio of the final size to the initial size. Let's represent the initial size of the bacteria as X. The number of times it doubles its size as Y. When the bacteria triples in size, it will have grown by three times its initial size:
Final size = 3X
If the bacteria doubles its size every four hours, its growth rate is 2 per four hours. Therefore, we can represent its growth rate in terms of the number of times it doubles its size as follows:
Growth rate = 2 ^ Y.
This means that the bacteria doubles in size for every Y number of times, it will have grown by a factor of 2. Therefore, if it triples in size, it doubles its size twice and then grows by half its size of the initial size (1/2). Therefore, we can represent the final size of the bacteria as follows:
Final size = (2^2) (1/2) X
= 2X.
So, to find the number of times the bacteria has doubled its size, we can calculate the logarithm of the ratio of the final size to the initial size:
3X/X = 2^Y(3X/X)
= (2^Y)3
= 2^YY
= log base 2 of 3
≈ 1.585
Therefore, the number of hours required for the bacteria to triple in size is the number of times it doubles its size (1.585) multiplied by the number of hours required to double in size (4 hours):
= 1.585 * 4 hours
= 6.34 hours (rounded to the nearest tenth)
Therefore, the bacteria would take approximately 6.6 hours to triple in size if it doubles its size every four hours.
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At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, as requested. x 5
y 5
=32, normal at (2,1) A. y=2x−3 B. y=− 2
1
x+2 c. y=−2x+5 D. y= 16
1
x
Therefore, the line that is normal to the curve at the point (2,1) is represented by the equation y = x - 1.
To find the slope of the curve and the line that is normal to the curve at the point (2,1) on the curve represented by the equation 5x+5y=32, we need to manipulate the equation and use some calculus.
First, let's rearrange the given equation to express y in terms of x:
5x + 5y = 32
5y = 32 - 5x
y = (32 - 5x) / 5
y = 6.4 - x
Now, let's find the derivative of y with respect to x:
dy/dx = -1
The slope of the curve at any point is -1.
To find the equation of the line that is normal to the curve at the point (2,1), we need to find the negative reciprocal of the slope (-1). The negative reciprocal of -1 is 1.
Using the point-slope form of a line, where the slope is 1 and the point is (2,1), we can find the equation of the normal line:
y - y₁ = m(x - x₁)
y - 1 = 1(x - 2)
y - 1 = x - 2
y = x - 1
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"Area of ellipse
x2/9+y2/36=1"
The area of the ellipse x²/9 + y²/36 = 1 is 18π square units.
Given an equation of an ellipse, x²/9 + y²/36 = 1
We know that the equation of an ellipse is given as: (x²/a²) + (y²/b²) = 1
The area of an ellipse is given as: A = π × a × b
Where a and b are the lengths of the major and minor axes, respectively
Comparing the given equation with the standard equation, we have a = 3, b = 6
Hence, the area of the given ellipse is: A = π × 3 × 6 = 18π square units
Therefore, the area of the ellipse x²/9 + y²/36 = 1 is 18π square units.
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Determine the surface void ratio classification if the voids measured in an as cast wall surface. the approximate circular voids had the following counts of diameter in inches:
Size of void, Inches No. of Voids
3/32 19
1/8 17
5/32 15
3/16 13
7/32 10
1/4 14
9/32 7
5/16 6
11/32 2
3/8 1
We can calculate the percentage of voids for each diameter and classify them accordingly. The total number of voids is the sum of the counts for each diameter, so the void ratio in this case is 104. The percentage void is 18.27%.
The surface void ratio classification of the as cast wall surface can be determined based on the counts of voids of different diameters. The voids were measured in inches, and the counts of voids for each diameter were recorded. By analyzing this data, the surface void ratio classification can be determined.
To determine the surface void ratio classification, we can calculate the percentage of voids for each diameter and classify them accordingly. The total number of voids is the sum of the counts for each diameter, which in this case is 104.
First, we calculate the percentage of voids for each diameter by dividing the count of voids of that size by the total number of voids (104) and multiplying by 100. For example, for a void size of [tex]\frac{3}{32}[/tex], there are 19 voids, so the percentage of voids is [tex]\frac{19}{104} * 100[/tex] = 18.27%.
Next, we categorize the surface void ratio based on the percentage of voids. The classification may vary depending on the specific criteria used, but typically, the following classifications are considered:
Low void ratio: Less than 5%, Medium void ratio: 5% to 15%, and High void ratio: More than 15%
By examining the percentages of voids for each diameter, we can determine the classification for this as cast wall surface. It is important to note that this classification is specific to the surface of the wall and may not reflect the overall quality or performance of the wall structure.
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Now Answer The Following Question: Compute The Integral Of F(X,Y,Z)=Z Over The Region W Within The Cylinder X2+Y2≤4 Where
The integral of f(x, y, z) = z over the region W within the cylinder x^2 + y^2 ≤ 4 where 0 ≤ z ≤ 5 is equal to 25π/2.
To compute the integral of f(x, y, z) = z over the region W within the cylinder x^2 + y^2 ≤ 4 where 0 ≤ z ≤ 5, we need to set up the triple integral.
The integral can be expressed as:
∫∫∫W z dV
Since the region W is within the cylinder x^2 + y^2 ≤ 4, we can use cylindrical coordinates to simplify the integral.
In cylindrical coordinates, the region W can be defined as 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π, where r represents the radial distance and θ represents the angle.
Therefore, the integral becomes:
∫∫∫W z dV = ∫∫∫W z r dr dθ dz
The limits of integration for each variable are as follows:
z: 0 to 5
r: 0 to 2
θ: 0 to 2π
The integral can now be evaluated using these limits:
∫∫∫W z r dr dθ dz = ∫[0, 2π] ∫[0, 2] ∫[0, 5] z r dz dr dθ
Integrating with respect to z first, then r, and finally θ, we get:
∫[0, 2π] ∫[0, 2] ∫[0, 5] z r dz dr dθ
= ∫[0, 2π] ∫[0, 2] (r/2) * (5^2) dr dθ
= ∫[0, 2π] (25/2) * (r^2/2) ∣[0, 2] dθ
= ∫[0, 2π] (25/4) * 4 dθ
= (25/4) * (2π)
Simplifying further, we get:
(25/4) * (2π) = 50π/4 = 25π/2
Therefore, the integral of f(x, y, z) = z over the region W within the cylinder x^2 + y^2 ≤ 4 where 0 ≤ z ≤ 5 is equal to 25π/2.
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In January of 2008, a survey of 150
macroeconomists found 89 who believed that the
recession had already begun. A survey of 120
purchasing agents found 87 who believed the
recession had begun.
At the 90% confidence level, can one conclude
that the purchasing agents were more pessimistic
about the economy than the macroeconomists
were?
The test statistic (1.990) exceeded the critical value (1.645), leading us to reject the null hypothesis. Thus, we can infer that, at the 90% confidence level, the purchasing agents exhibited greater pessimism towards the economy compared to the macroeconomists.
Based on the given information, we can analyze whether the purchasing agents were more pessimistic about the economy compared to the macroeconomists. To determine this, we need to conduct a hypothesis test at the 90% confidence level.
Let's define the hypotheses:
- Null Hypothesis (H₀): The proportion of purchasing agents who believed the recession had begun is equal to or less than the proportion of macroeconomists who believed the recession had begun.
- Alternative Hypothesis (H₁): The proportion of purchasing agents who believed the recession had begun is greater than the proportion of macroeconomists who believed the recession had begun.
Next, we need to calculate the test statistic. We'll use the two-proportion z-test formula:
z = (p₁ - p₂) / √[(p_cap(1 - p_cap) / n₁) + (p_cap(1 - p_cap) / n₂)]
where:
- p₁ and p₂ are the sample proportions of purchasing agents and macroeconomists, respectively.
- p_cap is the pooled proportion.
- n₁ and n₂ are the sample sizes of purchasing agents and macroeconomists, respectively.
Calculating the proportions:
p₁ = 87/120 = 0.725
p₂ = 89/150 = 0.593
Calculating the pooled proportion:
p_cap = (x₁ + x₂) / (n₁ + n₂) = (87 + 89) / (120 + 150) ≈ 0.645
Calculating the test statistic:
z = (0.725 - 0.593) / √[(0.645(1 - 0.645) / 120) + (0.645(1 - 0.645) / 150)] ≈ 1.990
Looking up the critical value for a one-tailed z-test at the 90% confidence level, we find it to be approximately 1.645.
Since the calculated test statistic (1.990) is greater than the critical value (1.645), we reject the null hypothesis. Therefore, we can conclude that at the 90% confidence level, the purchasing agents were more pessimistic about the economy than the macroeconomists were.
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TV and UW are diagonals of rhombus TUVW. UW=12, TX=8, and m
Applying the properties of a rhombus, the measures required are determined as: WX = 6; TV = 16; m<UVX = 33°; m<TXU = 90°.
What are the Properties of a Rhombus?Some of the properties of a rhombus are:
1. Diagonals bisect each other at 90°
2. All its sides have the same length, while its opposite sides are parallel top each other.
Thus, we have the following using the properties of a rhombus:
WX = 1/2(UW)
WX = 1/2(12)
WX = 6
TV = 2(TX)
TV = 2(8)
TV = 16
m<UVX = m<WVX
m<UVX = 33°
m<TXU = 90°
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Given that
y
= 6 cm and
θ
= 55°, work out
x
rounded to 1 DP.
x rounded to 1 decimal place is approximately 3.4 cm.
To work out the value of x, we can use the trigonometric function cosine (cos).
The cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse in a right triangle.
In this case, the length of the adjacent side is
x, and the length of the hypotenuse is 6 cm.
The given angle θ is 55°.
Using the cosine function, we have:
[tex]cos(\theta ) =\frac{adjacent }{hypotenuse}[/tex]
[tex]cos(55^{\circ}) =\frac{x}{6}[/tex]
To solve for x, we can rearrange the equation:
[tex]x = 6 \times cos(55^{\circ})[/tex]
Now we can calculate x using the given values:
[tex]x \approx 6 \times cos(55^{\circ})[/tex]
[tex]x \approx 6 \times 0.5736[/tex]
[tex]x \approx 3.4416[/tex]
Therefore, x rounded to 1 decimal place is approximately 3.4 cm.
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In the following questions do not perform any calculations. You may, if you like, include sketches: c. Explain why for the cylindrical surface x² + y² = 1, -15 zs1 if xds=0; ii.ff, (i+j). ds = 0; iii.ff, (i+j). ds = 4x clue: in part (iii) consider the form of the unit normal to this surface in caartesian coordinates
This is obtained by rotating ds by 90 degrees anticlockwise. Now, we have (x, y, 0) × (y, -x, 0) = (0, 0, -xy² - x²y)So, the unit normal in Cartesian coordinates is (0, 0, -xy² - x²y)/sqrt(x² + y² + (xy² + x²y)²)
Given cylindrical surface is x² + y²
= 1.a)
To prove that z
= -15
is a tangent plane to the surface when x
= 0,
consider the equation of the tangent plane at the point
(x0, y0, z0).f(x, y, z)
= g(x, y, z), where f(x, y, z)
= x² + y² - 1 and g(x, y, z)
= z.
If the tangent plane touches the surface, then the normal to the plane must be perpendicular to the surface. So the gradient of f(x, y, z) and g(x, y, z) should be perpendicular at the point
(x0, y0, z0).(∂f/∂x, ∂f/∂y, ∂f/∂z)
= (2x, 2y, 0)(∂g/∂x, ∂g/∂y, ∂g/∂z)
= (0, 0, 1)
Hence, the condition for the tangent plane at
(x0, y0, z0) is 2x0 * 0 + 2y0 * 0 + 0 * 1
= 0.
This gives x0
= y0
= 0. Hence the tangent plane is z
= z0
which is equal to -15 when x
= 0.b) Here, i + j is the unit vector in the direction of positive z-axis. Thus, the dot product of i + j with the differential element ds gives the z-component of ds which is 0. So, (i + j).ds
= 0.c) For the unit normal to the cylindrical surface, consider the gradient of the surface.∇f = (2x, 2y, 0)Therefore, the unit normal is
(2x, 2y, 0)/2
= (x, y, 0)Let (x, y, z)
be a point on the surface. The unit normal is also given by the cross product of two vectors tangent to the surface. We already know that ds
= (dx, dy, 0)
and we need to find another tangent vector. Notice that
x² + y² = 1
defines a circular cylinder whose cross-sections perpendicular to the x-axis are all circles of radius 1. Thus, another tangent vector at the point (x, y, z) can be chosen as (y, -x, 0). This is obtained by rotating ds by 90 degrees anticlockwise. Now, we have
(x, y, 0) × (y, -x, 0)
= (0, 0, -xy² - x²y)
So, the unit normal in Cartesian coordinates is
(0, 0, -xy² - x²y)/square root
(x² + y² + (xy² + x²y)²).
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solution.
3) (18 points) Graph a) r = 2cose Table
The equation[tex]r = 2cos(θ)[/tex] is a polar equation for a curve that is a circle with radius 2 centered at (1, 0) in Cartesian coordinates. To graph this equation, we can create a table of values and then plot the points to get a sense of the curve.
Table of values for [tex]r = 2cos(θ):θr (radius)00 (initial side) x-axis20.8 (approx) 40.3 (approx) 60-260-20.3 (approx) -40.8 (approx) -60Plotting[/tex] the points on a polar graph, we get: Graph of[tex]r = 2cos(θ):[asy]size(150)[/tex]; [tex]draw((0,-2)--(0,2)[/tex], [tex]black+1bp[/tex], End [tex]Arrow(5))[/tex]; [tex]draw((-2,0)--(2,0), black+1bp,[/tex]
[tex]End Arrow(5)); for(int i=0;i < =360;i+=30)[/tex]
[tex]{ draw((0,0)--dir(i), red); } draw(circle((1,0),2),[/tex]
[tex]red+1bp); label("$x$",(2,0),SE);[/tex]
[tex]label("$y$",(0,2),NE); for(int i=0;i < =360;i+=30){[/tex][tex]label("$"+string(i)+"^\circ$",dir(i),dir(i)); }[/tex]
[tex]label("$r = 2\cos(\theta)$",(-1.5,-2), red);[/asy][/tex]
Therefore, the graph of [tex]r = 2cos(θ)[/tex] is a circle with radius 2 centered at (1, 0) in Cartesian coordinates.
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Find the volume of the solid that lies under the hyperbolic paraboloid \( z=3 y^{2}-x^{2}+6 \) and above the rectangle \( R=[-1,1] \times[1,2] \). Answer:
The volume of the solid that lies under the hyperbolic paraboloid is 10/3 cubic units.
To find the volume of the solid that lies under the hyperbolic paraboloid
z=3y²-x²+6 and above the rectangle R= [-1, 1]×[1, 2]
we need to integrate the height of the solid over the given rectangle.
The volume can be calculated using a double integral:
[tex]V=\int\int_R (3y^2-x^2+6)dA[/tex]
where dA represents the differential area element.
Let's proceed with the integration:
[tex]V=\int _{-1}^1\int _1^2\:\left(3y^2-x^2+6\right)dydx[/tex]
First, we integrate with respect to y:
[tex]V=\int _{-1}^1\:\left(y^3-x^2y+6y\right)^2_1dydx[/tex]
[tex]V=\int _{-1}^1\:\left(-5x^2+6\right)dx[/tex]
Integrating, we get:
[tex]V=\left[-\frac{5}{3}x^3+26x\right]^1_{-1}[/tex]
V=10/3
Hence, the volume of the solid that lies under the hyperbolic paraboloid is 10/3 cubic units.
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A local sandwich shop makes an Italian sandwich that contains ham, salami, and pepperoni meats. Let X represent the
weight of ham, Y represent the weight of salami, and Z represent the weight of pepperoni for each Italian sandwich
made. The mean of X is 2 ounces, the mean of Y is 1.25 ounces, and the mean of Z is 1.75 ounces. What is the mean
of the sum, S=X+Y+Z?
Ou, = 1.67 ounces
O, = 3.0 ounces
Op, = 3.25 ounces
-
OP,= 5.0 ounces
The mean of the sum of X+Y+Z is 5 ounces. The Option D.
What is the mean of X+Y+Z? Show your workings.To find the mean of X+Y+Z, we need to add the means of X, Y and Z. Since the mean of X is 2 ounces, the mean of Y is 1.25 ounces and the mean of Z is 1.75 ounces, we will calculate mean of X+Y+Z.:
Mean(X+Y+Z) = Mean(X) + Mean(Y) + Mean(Z)
= 2 ounces + 1.25 ounces + 1.75 ounces
= 5 ounces
Therefore, the mean of X+Y+Z is 5 ounces.
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In a right triangle, one angle measures b ∘
, where cosb ∘
= 10
6
. What is the
In a right triangle, one angle measures b°: The sin(90° - b°) is equal to 6/10.
In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. Since one angle measures b° and the cosine of this angle is given as 10/6, we can use the Pythagorean identity to find the length of the other side.
Let's assume that the side opposite the angle b° is represented by the length 'x' and the hypotenuse is represented by the length 'h'. According to the given information, we know that cos(b°) = 10/6, which is equal to the adjacent side (x) divided by the hypotenuse (h).
Using the Pythagorean identity, we have:
cos(b°) = x/h
(10/6) = x/h
Simplifying the equation, we find:
6x = 10h
x = (10/6)h
Now, let's consider the angle 90° - b°. The sine of this angle is equal to the ratio of the side opposite this angle to the hypotenuse. Since we have the value of x (the side opposite b°) in terms of h, we can substitute it into the equation:
sin(90° - b°) = x/h = (10/6)h/h = 10/6
Therefore, sin(90° - b°) is equal to 6/10.
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Complete question:
In a right triangle, one angle measures b° , where cosb° = 10/6 . What is the sin(90° −b° )?
Suppose the quantity demanded weekly q (in units of a thousand) of a product is related to its unit price p (in dollars/unit) by the equation 100q 2 + 9p 2 = 3600 What is the rate of change of the quantity demanded when the unit price p = $14 and the selling price is dropping at the rate of $.15/unit/week?
The rate of change of the quantity demanded when the unit price p = $14 and the selling price drops at the rate of $.15/unit/week is approximately 0.189.
We are given the equation for the quantity demanded weekly q (in units of a thousand) of a product which is related to its unit price p (in dollars/unit) as
100q² + 9p² = 3600
We need to find the rate of change of the quantity demanded when the unit price p = $14 and the selling price drops at the rate of $.15/unit/week. The given equation is
100q² + 9p² = 3600
Let's differentiate both sides concerning time t.
d/dt (100q² + 9p²) = d/dt (3600)
=200q (dq/dt) + 18p (dp/dt) = 0
Divide both sides by 2 to get the rate of change of
q: dq/dt = - (9p/100)(dp/dt)
Substitute p = $14 and up/dt = -$.15 to get the rate of change of
q: dq/dt = - (9 x 14/100) x (-0.15)
= 0.189
The rate of change of the quantity demanded when the unit price p = $14 and the selling price is dropping at the rate of $.15/unit/week is approximately 0.189.
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4 If f is continuous and f(x) dx 2 S² 0 Answer: f(2x) dx. Preview My Answers f(x) dx = 11, evaluate Submit Answers You have attempted this problem 0 times. If f is continuous and 3 5.³ xf(x²) dx.
The answer to the question is:4) If f is continuous and f(x) dx 2 S² 0 Answer: f(2x) dx is given by f(2x)5) f(x) dx = 11, evaluate cannot be evaluated since there is not enough information provided to solve the integral or determine the value of f(x).6) If f is continuous and 3 5.³ xf(x²) dx is given by ∫(3/2)(5³/2)f(t) dt [from t=0 to t=25].
The given expressions are;4) If f is continuous and f(x) dx 2 S² 0 Answer: f(2x) dx.5) f(x) dx
= 11, evaluate.6) If f is continuous and 3 5.³ xf(x²) dx.Answers4) If f is continuous and f(x) dx 2 S² 0 Answer: f(2x) dx.This problem is related to the change of variable theorem. Let t
=2x, then x
=t/2 and dx/dt
=1/2. As S2 is evaluated in terms of x, replace x in terms of t, i.e., 2x
=t.The given integral is;S2
= ∫(f(x) dx) [from xhttps://brainly.com/question/31523914
=0 to x
=2]Now, changing the variable from x to t, the integral becomes;S2
= ∫f(t/2) [dx/dt dt] [from t
=0 to t
=4]S2 = ∫(1/2)f(t/2) dt [from t
=0 to t
=4]S2
= [1/2]∫f(t/2) dt [from
t=0 to t
=4]S2
= [1/2]∫f(x) dx [from x
=0 to x
=4]S2
= [1/2]S4 5) f(x) dx
= 11, evaluate.The given integral is;∫f(x) dx
= 11We cannot determine the value of f(x) using this information. There is not enough information provided to solve the integral or determine the value of f(x).6) If f is continuous and ∫3[5³ x f(x²) dx].The given integral is;∫3[5³ x f(x²) dx]Now, let t
=x². Then x
=√t and dx/dt
=1/2√t. The limits of integration must also be changed to reflect the change in variable from x to t.The integral now becomes;∫(3/2)[(5³/2)∫f(t) dt] [from t
=0 to t
=25]∫(3/2)(5³/2)f(t) dt [from t
=0 to t=25]
.The answer to the question is:4) If f is continuous and f(x) dx 2 S² 0 Answer: f(2x) dx is given by f(2x)5) f(x) dx
= 11, evaluate cannot be evaluated since there is not enough information provided to solve the integral or determine the value of f(x).6) If f is continuous and 3 5.³ xf(x²) dx is given by ∫(3/2)(5³/2)f(t) dt [from t
=0 to t
=25].
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Evaluate the integral (Hint: use u-substitution) S dx 1 √x (1+√x)
The solution of the integral using the u-substitution technique is ln|1+√x| - ln|1-√x| + C, where C is the constant of integration.
Evaluate the integral S dx 1 √x (1+√x) using u-substitution technique.u-substitution:
u-substitution is an important method of integration that involves substitution of an expression with a new variable known as the u-variable.
The general formula for u-substitution is given as follows:
∫f(g(x))g'(x)dx = ∫f(u)du
∫dx / √x (1+√x)
As given, ∫dx / √x (1+√x)
We notice that we can make a substitution of the form u = 1+√x, and so we compute du/dx = (1/2) (x^(-1/2)) and therefore 2 du = (x^(-1/2)) dx.
This means we can substitute the following into our integral:
∫dx / √x (1+√x)
= 2 ∫du/(u^2-1)
= ∫ (1/ (u-1) - 1/(u+1)) du
= ln|u-1| - ln|u+1| + C
= ln|1+√x| - ln|1-√x| + C
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1500-200 x (200+ 50)
Answer:
-48,500
Step-by-step explanation:
1500 - 200 × (200 + 50) =
= 1500 - 200 × 250
= 1500 - 50,000
= -48,500
Answer:
-48500Step-by-step explanation:
1500-200 x (200+ 50) = (remember PEMDAS)
1500 - 200 x 250 =
1500 - 50000 =
-48500
Find sin2x,cos2x, and tan2x if tanx=−8/15 and x terminates in quadrant IV.
The trigonometric equations are solved and the angle is in the fourth quadrant.
a) sin 2x = 240/289
b) cos 2x = 240/289
c) tan 2x = -240/161
Given data:
The measure of the angle tan 2x = -8/15
In the fourth quadrant going anti-clockwise, only cos is positive
So, from the trigonometric relation:
The hypotenuse of the triangle is [tex]H = 15^2+8^2[/tex]
H = 17 units
So, the value of sin x = -8/17
The value of cos x = 15/17
Now, sin 2x = 2 sinx cos x
On simplifying the equation:
sin 2x = 2 ( -8/17 ) ( 15/17 )
sin 2x = 240/289
The value of [tex]cos 2x = cos^2x-sin^2x[/tex]
[tex]cos2x=\frac{225}{289}-\frac{64}{289}[/tex]
[tex]cos2x=\frac{161}{289}[/tex]
Now, the value of tan 2x = sin2x / cos2x
So, tan 2x = -240/161
The sign is negative for tan angle in the fourth quadrant.
Hence, the trigonometric relation is solved.
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