The half-life and exponential decay formulas indicates that the the half-life and amounts are;
31. 60 days
32. A(t) = 0.5·[tex]e^{(-0.0115\cdot t)}[/tex]
0.3 grams
33. A(t) = 0.5·[tex]e^{(-0.00822\cdot t)}[/tex]
84 minutes
What is the half-life of a substance?The half-life of a radioactive substance is the duration it takes for half the amount of the substance to decay.
31. The exponential growth formula indicates that we get;
0.5 = 1 × [tex]e^{(-0.0115\cdot t)}[/tex]
Where t = The time in days, which is the half life of the Iodine-125
-0.0115·t = ln(0.5)
The half life of the Iodine-125, [tex]t_{1/2}[/tex] = ln(0.5)/(-0.0115) ≈ 60.3
Therefore, the number of days for half of the Iodine-125 to decay is about 60 days
32. The amount of Iodine-125 remaining after t days, A(t) is can be represented as follows;
A(t) = 0.5 × [tex]e^{(-0.0115\cdot t)}[/tex]The amount of Iodine-125 remaining after 60 days which is the half life for the Iodine-125 is therefore;
A(60) = 0.5 × [tex]e^{(-0.0115\times 60)}[/tex] ≈ 0.3 grams33. The initial amount of the radioactive substance, A(0) = 250 g
The amount after 250 minutes, A(250) = 32 g
Therefore, we get;
A(0) = 250·[tex]e^{(k\times 0)}[/tex] = 250
A(250) = 250·[tex]e^{(k\times 250)}[/tex] = 32
32/250 = [tex]e^{(k\times 250)}[/tex]
ln(32/250) = ln([tex]e^{(k\times 250)}[/tex])
k × 250 = ln(32/250)
k = ln(32/250)/250 ≈ -0.00822
The exponential equation is; A(t) = 250·[tex]e^{(-0.00822\cdot t)}[/tex]
The half life can therefore, be found as follows;
125 = 250·[tex]e^{(-0.00822\times t)}[/tex]
[tex]e^{(-0.00822\times t)}[/tex] = 125/250 = 1/2
ln([tex]e^{(-0.00822\times t)}[/tex]) = ln(1/2)
-0.00822·t = ln(1/2)
t = ln(1/2)/(-0.00822) ≈ 84
The half life of the substance is about 84 minutes
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V7xV2
Realiza la siguiente multiplicación de raíces cuadradas
It is always a good practice to simplify the result whenever possible, but in this case, V14 is the simplest form of the product of V7 and V2.
To multiply the square roots V7 and V2, we can combine the numbers inside the square roots and simplify the result.
V7 * V2 = V(7 * 2) = V14
Multiplying the numbers under the square roots, we get 7 * 2 = 14. Therefore, the product of V7 and V2 is V14.
This means that the square root of 14 is the result of multiplying V7 and V2. However, it is important to note that V14 cannot be further simplified because 14 does not have any perfect square factors.
In summary, the product of V7 and V2 is V14. It is worth mentioning that when multiplying square roots, we can multiply the numbers inside the square roots and keep the square root symbol intact, unless the numbers inside have perfect square factors that can be simplified further.
It is always a good practice to simplify the result whenever possible, but in this case, V14 is the simplest form of the product of V7 and V2.
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What is the value of n in the equation 1/2(n-4)-3=3− (2n + 3)?
Answer:
n = 2
Step-by-step explanation:
[tex]\frac{1}{2}[/tex] (n - 4) - 3 = 3 - (2n + 3) ← distribute parenthesis on both sides
[tex]\frac{1}{2}[/tex] n - 2 - 3 = 3 - 2n - 3 ( simplify both sides )
[tex]\frac{1}{2}[/tex] n - 5 = - 2n ( multiply through by 2 to clear the fraction )
n - 10 = - 4n ( add 4n to both sides )
5n - 10 = 0 ( add 10 to both sides )
5n = 10 ( divide both sides by 5 )
n = 2
Given the function f(x) = 0.5|x - 41-3, for what values of x is f(x) = 7?
x = -24, x = 16
x= -16, x = 24
x=-1, x = 9
x = 1, x = -9
The values of x for which f(x) = 7 are x = 61 and x = 21.
To find the values of x for which f(x) = 7, we can set up the equation and solve for x.
The given function is f(x) = 0.5|x - 41| - 3.
Setting f(x) equal to 7, we have:
0.5|x - 41| - 3 = 7.
First, let's isolate the absolute value term:
0.5|x - 41| = 7 + 3.
0.5|x - 41| = 10.
To remove the absolute value, we can consider two cases:
Case: (x - 41) is positive or zero:
0.5(x - 41) = 10.
Multiplying both sides by 2 to get rid of the fraction:
x - 41 = 20.
Adding 41 to both sides:
x = 61.
So x = 61 is a solution for this case.
Case: (x - 41) is negative:
0.5(-x + 41) = 10.
Multiplying both sides by 2:
-x + 41 = 20.
Subtracting 41 from both sides:
-x = -21.
Multiplying both sides by -1 to solve for x:
x = 21.
So x = 21 is a solution for this case.
Therefore, the values of x for which f(x) = 7 are x = 61 and x = 21.
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In the problems below, f(x) = log2x and
9(x) = 10910x.
How are the graphs of fand g similar? Check all that apply.
The graphs of f(x) and g(x) are similar in terms of being increasing, passing through the point (1,0), approaching infinity as x approaches infinity, and having a vertical asymptote at x = 0.
The given functions are f(x) = log2x and g(x) = 9(x) = 10910x. Let's examine the similarities in the graphs of these functions.
Both functions are increasing: The logarithmic function f(x) = log2x and the exponential function g(x) = 10910x are both increasing functions. As x increases, the corresponding values of f(x) and g(x) also increase.
Both functions pass through the point (1,0): When x = 1, both f(x) and g(x) evaluate to 0. This means that both functions intersect the y-axis at the point (1,0).
Both functions approach infinity as x approaches infinity: As x becomes larger and larger, both f(x) and g(x) grow without bound. This indicates that the graphs of both functions have an asymptote at y = infinity.
Both functions have a vertical asymptote at x = 0: The logarithmic function f(x) = log2x has a vertical asymptote at x = 0, while the exponential function g(x) = 10910x also has a vertical asymptote at x = 0. This means that the graphs of both functions approach but never cross the y-axis.
Based on these observations, the similarities between the graphs of f(x) and g(x) are that both functions are increasing, pass through the point (1,0), approach infinity as x approaches infinity, and have a vertical asymptote at x = 0.
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The formula for the volume of a triangular pyramid can be written as V = 1/3 B•h, where B = area of the base. What is the formula for B?
A. B = πr²
B. B=I•w
C. B = ½bh
The formula for the area of the base (B) of a triangular pyramid is given by option C: B = ½bh.
In a triangular pyramid, the base is a triangle, and the area of a triangle is calculated using the formula ½bh, where b represents the length of the base of the triangle and h represents the corresponding height. Therefore, the formula for the area of the base (B) is B = ½bh.
The base of a triangular pyramid is a two-dimensional shape, and its area is determined by the length of the base and the height perpendicular to it. The formula B = ½bh is commonly used to find the area of a triangle, which applies to the base of the pyramid in this case.
The ½ factor accounts for the fact that the base is a triangle, and the product of the base length (b) and the corresponding height (h) yields the area of the triangle. Hence, option C correctly represents the formula for the base area of a triangular pyramid.
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A survey of 425 people found that 102 of them had gone to the movies in the last month.
A recent survey conducted on a sample of 425 individuals revealed that 102 of them had visited the cinema within the past month.
This data suggests that approximately 24% of the surveyed population had engaged in movie going activities during the specified period.
The findings highlight a significant proportion of the respondents who have recently enjoyed the cinematic experience.
However, it is important to note that this data is based on a limited sample size and may not be representative of the entire population.
Further research and surveys involving a larger and more diverse sample are recommended to obtain a more accurate understanding of movie attendance trends.
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if x=15-root 2 find the value of x²-5x+3
To find the value of x²-5x+3 when x=15-√2, we substitute the value of x into the expression:
x² - 5x + 3 = (15-√2)² - 5(15-√2) + 3
First, let's expand (15-√2)² using the formula for the square of a binomial:
(15-√2)² = (15)² - 2(15)(√2) + (√2)²
= 225 - 30√2 + 2
Simplifying further:
(15-√2)² = 227 - 30√2
Now we substitute this back into the expression:
x² - 5x + 3 = 227 - 30√2 - 5(15-√2) + 3
= 227 - 30√2 - 75 + 5√2 + 3
= 155 - 25√2
Therefore, the value of x²-5x+3 when x=15-√2 is 155 - 25√2.
Consider the following matrix in reduced row echelon form:
Row Echelon Form
What can you say about the solution to the associated linear system?
The reduced row echelon form of a matrix allows us to determine whether the associated linear system is consistent or inconsistent, unique or has infinitely many solutions.
In reduced row echelon form, a matrix has the following properties:
Each leading entry (the leftmost nonzero entry) of a row is equal to 1.
Each leading entry is the only nonzero entry in its column.
All entries below a leading entry are zeros.
The leading entry in each row is to the right of the leading entry in the row above it.
Based on these properties, we can make the following conclusions about the solution to the associated linear system:
Consistency: If there are no rows of the form [0 0 ... 0 | b] (where b is a nonzero constant), meaning there are no contradictory equations, the linear system is consistent. In other words, there is at least one solution.
Uniqueness: If there are no free variables (columns without leading entries), meaning each column corresponds to a pivot column, the linear system has a unique solution. This means there is only one set of values for the variables that satisfies all the equations.
Infinite solutions: If there are one or more free variables, the linear system has infinitely many solutions. This occurs when there are more variables than equations or when there are dependent equations.
The reduced row echelon form provides an organized representation of the linear system that simplifies the process of determining its solution. By analyzing the structure of the matrix, we can determine the consistency, uniqueness, and the nature of the solution set.
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12
Find x.
20
X
x = [ ?
X
The calculated length x of the right triangle is 16
Finding the length x of the right triangleFrom the question, we have the following parameters that can be used in our computation:
The right triangle
The length x of the right triangle can be calculated using the following Pythagoras theorem
x² = difference of squares of the other sides
Using the above as a guide, we have the following:
x² = (20)² - (12)²
Evaluate
x² = 256
Take the square roots
x = 16
Hence, the length x of the right triangle is 16
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New flowers are being planted in an empty area of a landscaped garden. The
map shows the area that is being planted. If walkway A and walkway B are
parallel, what is the distance from F to G on walkway C?
The distance from F to G on walkway C is approximately 19.21 units.
To determine the distance from point F to point G on walkway C, we can use the information provided in the map and apply some basic geometric principles.
Looking at the map, we can see that walkway A and walkway B are parallel. Let's use this information to find the distance from F to G on walkway C.
First, let's identify the relevant points on the map. We have point F on walkway C and point G on walkway C. The map also shows two intersecting lines, representing walkways A and B, but we don't need to consider those for this particular calculation.
To find the distance from F to G on walkway C, we need to focus on the segment of walkway C that connects F and G. Based on the map, we can see that the segment of walkway C is perpendicular to both walkway A and walkway B.
Since walkway A and walkway B are parallel, and the segment of walkway C is perpendicular to both, we can conclude that the segment FG forms a right triangle with walkway A and walkway B.
In a right triangle, we can use the Pythagorean theorem to find the length of one side (FG) if we know the lengths of the other two sides.
Let's assume the length of the segment FG is represented by the variable 'd' (which we need to find).
From the map, we can see that the distance from F to walkway A is 15 units. Similarly, the distance from G to walkway B is 12 units.
Applying the Pythagorean theorem, we have:
d^2 = 15^2 + 12^2
d^2 = 225 + 144
d^2 = 369
Taking the square root of both sides, we find:
d ≈ √369
d ≈ 19.21
As a result, it takes 19.21 units to get from F to G on walkway C.
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Identify the equivalent equation to ax + by = c
Select the correct answer.
1. Y= a/b x + c/b
2. Y= ax + c
3. Y= -a/b x + c/b
4. by = ax + c
The equivalent equation to ax + by = c is by = ax + c.
The equivalent equation to ax + by = c can be found by isolating the y variable. Let's go through the options provided:
Y = a/b x + c/b: This equation represents the slope-intercept form of a linear equation (y = mx + b). It does not match the given equation, so it is not the correct answer.
Y = ax + c: This equation represents a linear equation in slope-intercept form. It does not have the y-variable coefficient (b) present, so it is not the correct answer.
Y = -a/b x + c/b: This equation represents the slope-intercept form of a linear equation. The signs of the variables are reversed compared to the given equation, so it is not the correct answer.
by = ax + c: This equation matches the given equation ax + by = c, where the y-variable is isolated on one side. Therefore, the correct answer is option 4.
In conclusion, the equivalent equation to ax + by = c is by = ax + c.
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Find the solution(s) to x2 - 16x + 64 = 0.
The solution to the quadratic equation is x = 8 only.
What are quadratic equations?The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:
[tex]\bold{ax^2 + bx + c = 0}[/tex]
The above equation is a quadratic equation, and can be solve by either formula method or factorization method or completing the square method.
We will be solving using the factorization method:
[tex]\sf x^2 - 16x + 64 = 0[/tex]
We are going to find two numbers such that its sum is equal to -16 and its product is 64
The two numbers are: -8 and -8
[tex]\sf -8 + (-8) = -16[/tex]
and [tex]\sf -8(-8)=64[/tex]
We will replace -16x by -8x and -8x
[tex]\sf x^2 - 16x + 64 = 0[/tex]
[tex]\sf x^2 - 8x - 8x + 64 = 0[/tex]
[tex]\sf (x^2 - 8x) (-8x + 64) = 0[/tex]
In the first parenthesis, x is common so we will factor out x while in the second parenthesis -8 is common and it will be factored out.
That is:
[tex]\sf x ( x- 8) -8(x - 8) = 0[/tex]
[tex]\sf (x-8)(x-8) = 0[/tex]
[tex]\sf x -8 =0[/tex]
Add 8 to both sides
[tex]\sf x -8 + 8 = 0 + 8[/tex]
[tex]\rightarrow \bold{x =8}[/tex]
Hence, the solution to the quadratic equation x² - 16x + 64 = 0 is x = 8 only.
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The complete question is
Find the solution(s) to x² - 16x + 64 = 0.
A. x = 8 only
B. x = 8 and x = -8
C. x = 4 and x = 16
D. x = -2 and x = 32
Two different functions are shown. Function A: Function B: How do the x-intercepts of the two functions compare? The x-intercept in function B is one-third as large as the x-intercept in function A. The x-intercept in function B is three times as large as the x-intercept in function A. The distance between the x-intercepts in function A is half the distance between the x-intercepts of function B. The distance between the x-intercepts in function A is twice the distance between the x-intercepts of function B.
The x-intercept in function B is one-third as large as the x-intercept in function A.
The x-intercepts of two functions, A and B, are compared in terms of their relative sizes and distances. According to the given information, the x-intercept in function B is one-third as large as the x-intercept in function A. This means that the value of the x-coordinate at which function B intersects the x-axis is one-third of the value of the x-coordinate at which function A intersects the x-axis.
Conversely, we can say that the x-intercept in function A is three times as large as the x-intercept in function B. The x-coordinate at which function A intersects the x-axis is three times the value of the x-coordinate at which function B intersects the x-axis.
However, the information does not provide any direct comparison of the distances between the x-intercepts of the two functions. Therefore, we cannot determine whether the distance between the x-intercepts in function A is half or twice the distance between the x-intercepts of function B based on the given information.
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Let x = 43. Rewrite the equation in the box, replacing 4
with a (substitute).
The equation in the box with x replacing [tex]4^{\frac{1}{5} }[/tex] in the equation can be presented as follows;
What is an equation?An equation is a statement that indicates that two expressions are equivalent, by joining them with an '=' sign.
The details in the diagram indicates; x = [tex]4^{\frac{1}{5} }[/tex]
Required, to rewrite the equation in the box by plugging in x to substitute [tex]4^{\frac{1}{5} }[/tex] in the equation
The equation in the box can be presented as follows;
[tex]\underline{(4^{\frac{1}{5} })^5 = 4}[/tex]
Plugging in x = [tex]4^{\frac{1}{5} }[/tex] in the above equation, we get;
[tex]\underline{(4^{\frac{1}{5} })^5 =x^5 = 4}[/tex]
Therefore, the equation in the box becomes;
x⁵ = 4
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100 Points! Geometry question. Photo attached. Please show as much work as possible. Thank you!
Answer:
[tex]BC=5.1[/tex]
[tex]B=23^{\circ}[/tex]
[tex]C=116^{\circ}[/tex]
Step-by-step explanation:
The diagram shows triangle ABC, with two side measures and the included angle.
To find the measure of the third side, we can use the Law of Cosines.
[tex]\boxed{\begin{minipage}{6 cm}\underline{Law of Cosines} \\\\$c^2=a^2+b^2-2ab \cos C$\\\\where:\\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides.\\ \phantom{ww}$\bullet$ $C$ is the angle opposite side $c$. \\\end{minipage}}[/tex]
In this case, A is the angle, and BC is the side opposite angle A, so:
[tex]BC^2=AB^2+AC^2-2(AB)(AC) \cos A[/tex]
Substitute the given side lengths and angle in the formula, and solve for BC:
[tex]BC^2=7^2+3^2-2(7)(3) \cos 41^{\circ}[/tex]
[tex]BC^2=49+9-2(7)(3) \cos 41^{\circ}[/tex]
[tex]BC^2=49+9-42\cos 41^{\circ}[/tex]
[tex]BC^2=58-42\cos 41^{\circ}[/tex]
[tex]BC=\sqrt{58-42\cos 41^{\circ}}[/tex]
[tex]BC=5.12856682...[/tex]
[tex]BC=5.1\; \sf (nearest\;tenth)[/tex]
Now we have the length of all three sides of the triangle and one of the interior angles, we can use the Law of Sines to find the measures of angles B and C.
[tex]\boxed{\begin{minipage}{7.6 cm}\underline{Law of Sines} \\\\$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c} $\\\\\\where:\\ \phantom{ww}$\bullet$ $A, B$ and $C$ are the angles. \\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides opposite the angles.\\\end{minipage}}[/tex]
In this case, side BC is opposite angle A, side AC is opposite angle B, and side AB is opposite angle C. Therefore:
[tex]\dfrac{\sin A}{BC}=\dfrac{\sin B}{AC}=\dfrac{\sin C}{AB}[/tex]
Substitute the values of the sides and angle A into the formula and solve for the remaining angles.
[tex]\dfrac{\sin 41^{\circ}}{5.12856682...}=\dfrac{\sin B}{3}=\dfrac{\sin C}{7}[/tex]
Therefore:
[tex]\dfrac{\sin B}{3}=\dfrac{\sin 41^{\circ}}{5.12856682...}[/tex]
[tex]\sin B=\dfrac{3\sin 41^{\circ}}{5.12856682...}[/tex]
[tex]B=\sin^{-1}\left(\dfrac{3\sin 41^{\circ}}{5.12856682...}\right)[/tex]
[tex]B=22.5672442...^{\circ}[/tex]
[tex]B=23^{\circ}[/tex]
From the diagram, we can see that angle C is obtuse (it measures more than 90° but less than 180°). Therefore, we need to use sin(180° - C):
[tex]\dfrac{\sin (180^{\circ}-C)}{7}=\dfrac{\sin 41^{\circ}}{5.12856682...}[/tex]
[tex]\sin (180^{\circ}-C)=\dfrac{7\sin 41^{\circ}}{5.12856682...}[/tex]
[tex]180^{\circ}-C=\sin^{-1}\left(\dfrac{7\sin 41^{\circ}}{5.12856682...}\right)[/tex]
[tex]180^{\circ}-C=63.5672442...^{\circ}[/tex]
[tex]C=180^{\circ}-63.5672442...^{\circ}[/tex]
[tex]C=116.432755...^{\circ}[/tex]
[tex]C=116^{\circ}[/tex]
[tex]\hrulefill[/tex]
Additional notes:
I have used the exact measure of side BC in my calculations for angles B and C. However, the results will be the same (when rounded to the nearest degree), if you use the rounded measure of BC in your angle calculations.
What is the range of the exponential function shown below?
f(x)=11•(1/3)^x
A. y<0
B. y>0
C. All real numbers except 11
D. All real numbers
The range of the exponential function is B. y > 0.
The range of the exponential function f(x) = 11 × (1/3)ˣ can be determined by analyzing the behavior of the function as x approaches positive and negative infinity.
As x approaches positive infinity, (1/3) becomes smaller and smaller, tending towards zero.
f(x) approaches 11 × 0, which is equal to 0.
As a result, the function approaches 0 as x goes to infinity.
On the other hand, as x approaches negative infinity, (1/3)ˣ becomes larger and larger.
Since 1/3 is between 0 and 1, raising it to a negative power causes it to grow exponentially.
f(x) approaches infinity as x goes to negative infinity.
Combining these two behaviors, we can conclude that the range of the function f(x) = 11 × (1/3)ˣ is all positive real numbers, excluding zero.
In other words, the range is y > 0.
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1. x^6-2x^5+x^4/2x^2
2. Sec^3x+e^xsecx+1/sec x
3. cot ^2 x
4. x^2-2x^3+7/cube root x
5. y= x^1/2-x^2+2x
(1) The integral of the function is (1/10)x⁵ - (1/8)x⁴ + (1/6)x³ + C.
(2) The integral is (1/4)(sec x)⁴ + eˣ(sec x) + (1/2)(sec x)² + C.
(3) The integral of cot²x dx is -1/sin(x) - sin(x) + C.
(4)The integral of the function [tex]\frac{3}{8} x^{\frac{8}{3} } - \frac{6}{13} x^{\frac{13}{3} } + 21x^{\frac{2}{3}} + C.[/tex]
(5) The shaded area under the curve is 7.22 sq units.
What is the integral of the functions?(1) The integral of (x⁶ - 2x⁵ + x⁴) / 2x² is determined as follows;
(x⁶ - 2x⁵ + x⁴) / 2x² = (x⁴(x² - 2x + 1)) / 2x²
= (x⁴(x - 1)²) / 2x²
= (x²(x - 1)²) / 2
∫(x²(x - 1)²) / 2 dx
= (1/2) ∫x²(x - 1)² dx
= (1/2) ∫x²(x² - 2x + 1) dx
= (1/2) ∫(x⁴ - 2x³ + x²) dx
= (1/2)(1/5)x⁵ - (1/2)(1/4)x⁴ + (1/2) (1/3)x³ + C
Simplifying further:
= (1/10)x⁵ - (1/8)x⁴ + (1/6)x³ + C
(2) The integral of (sec³x + eˣsecˣ + 1) / (sec x) dx, is calculated as follows;
(sec³x + eˣsecˣ + 1) / (sec x) = (sec³x + eˣsecˣ + 1)(sec x / sec x)
= (sec⁴x + eˣsec²x + sec x) / sec x
Note; sec x as 1/cos x
= sec⁴x/cos x + eˣsec²x/cos x + sec x/cos x
= sec³x/cos x + eˣsec x + sec x/cos x
Integrate by substitution method.
u = sec x
du = sec x tan x dx.
∫(sec³x + eˣsec x + sec x/cos x) dx
= ∫(u³ + eˣu + u) du
= (1/4)u⁴ + eˣu + (1/2)u² + C
Substitute u back in terms of sec x;
= (1/4)(sec x)⁴ + eˣ(sec x) + (1/2)(sec x)² + C
(3) The integral of cot²x dx;
cot²(x) = (cos²(x))/(sin²(x))
Let u = sin(x)
du = cos(x) dx
= ∫(1-u²)/u² du
= ∫(1/u²) - 1 du
= ∫u⁻² - 1 du
= -1/u - u + C
= -1/sin(x) - sin(x) + C
(4) The integral of the function is;
∫(x² - 2x³ + 7)/∛x dx = ∫x²/∛x dx - ∫2x³/∛x dx + ∫7/∛x dx
= [tex]\frac{3}{8} x^{\frac{8}{3} } - \frac{6}{13} x^{\frac{13}{3} } + 21x^{\frac{2}{3}} + C.[/tex]
(5) The shaded area under the curve is calculated as follows;
the given function;
[tex]y = x^{1/2} - x^{2} + 2x[/tex]
∫y = A = [tex]\frac{2}{3} x^{3/2} - \frac{1}{3} x^3 \ + x^2[/tex]
the limits = 2 and 0
A = [tex]\frac{2}{3} (2)^{3/2} - \frac{1}{3} (2) ^3 \ + (2)^2[/tex]
A = 1.89 - 2.67 + 8
A = 7.22 sq units
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Find the 23rd term of an arithmetic sequence where the fourteenth term is 37 and the fifth term is 82.
The 23rd term of the arithmetic sequence is -8.
To find the 23rd term of an arithmetic sequence, we need to know the common difference (d) of the sequence.
Given that the 14th term is 37 and the 5th term is 82, we can use this information to find the common difference.
The formula for the nth term of an arithmetic sequence is:
Term n = a + (n - 1) * d,
where a is the first term and d is the common difference.
Using the information provided, we can set up two equations based on the given terms:
Equation 1: a + (14 - 1) * d = 37, (14th term is 37)
Equation 2: a + (5 - 1) * d = 82. (5th term is 82)
Simplifying the equations, we have:
Equation 1: a + 13d = 37,
Equation 2: a + 4d = 82.
We can solve this system of equations to find the values of a and d. Subtracting Equation 2 from Equation 1, we have:
(a + 13d) - (a + 4d) = 37 - 82,
13d - 4d = -45,
9d = -45,
d = -5.
Now that we have the common difference, we can find the first term (a) using Equation 2:
a + 4d = 82,
a + 4(-5) = 82,
a - 20 = 82,
a = 82 + 20,
a = 102.
Now, we can use the formula for the nth term to find the 23rd term:
Term 23 = a + (23 - 1) * d,
Term 23 = 102 + 22 * (-5),
Term 23 = 102 - 110,
Term 23 = -8
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Peter is going to visit 5 cities this summer. He will choose from 7 different cities, and the order in which he visits the cities does not matter. How many different city combinations are possible for the summer traveling?
There are 21 different city combinations possible for Peter's summer traveling.
To determine the number of different city combinations possible for Peter's summer traveling, we need to calculate the number of combinations of choosing 5 cities out of the 7 available cities.
Since the order in which he visits the cities does not matter, we are dealing with combinations rather than permutations.
The formula to calculate the number of combinations is given by the binomial coefficient, also known as "n choose k," denoted as C(n, k) or (nCk).
In this case, we have 7 cities to choose from, and Peter will visit 5 cities. Thus, we can calculate the number of city combinations as follows:
C(7, 5) = 7! / (5! * (7-5)!)
= 7! / (5! * 2!)
= (7 * 6 * 5!) / (5! * 2)
= (7 * 6) / 2
= 42 / 2
= 21
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When Ibuprofen is given for fever to
children 6 months of age up to 2 years, the
usual dose is 5 milligrams (mg) per kilogram
(kg) of body weight when the fever is under
102.5 degrees Fahrenheit. How much
medicine would be usual dose for a 18
month old weighing 21 pounds?
milligrams
Round your answer to the nearest milligram.
Answer: The usual dose for an 18-month-old weighing 21 pounds is 48 mg of ibuprofen.
Step-by-step explanation: To find the usual dose of ibuprofen for a child, we need to follow these steps:
Convert the child’s weight from pounds to kilograms. One pound is equal to 0.4536 kilograms, so we multiply 21 by 0.4536 to get 9.5256 kilograms.Multiply the child’s weight in kilograms by the dose per kilogram. The dose per kilogram is 5 mg when the fever is under 102.5 degrees Fahrenheit, so we multiply 9.5256 by 5 to get 47.628 mg.Round the result to the nearest milligram. To round a number to the nearest milligram, we look at the digit after the decimal point. If it is 5 or more, we add one to the digit before the decimal point and drop the rest. If it is less than 5, we keep the digit before the decimal point and drop the rest. In this case, the digit after the decimal point is 6, which is more than 5, so we add one to the digit before the decimal point and drop the rest. The result is 48 mg.Therefore, the usual dose for an 18-month-old weighing 21 pounds is 48 mg of ibuprofen. Hope this helps, and have a great day! =)
A tank is full of water. Find the work (in ft-lb) required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. (Round your answer to the nearest whole number.) 3 ft6 ft12 ft A frustum of a cone with a spout is given. The smaller radius is 3 ft, the larger radius is 6 ft, and the height is 12 ft.
The work required to pump the water out of the spout is approximately 64,307,077 ft-lb
To find the work required to pump the water out of the spout, we need to calculate the weight of the water in the tank and then convert it to work using the formula: work = force × distance.
First, let's calculate the volume of water in the tank. The frustum of a cone can be represented by the formula: V = (1/3)πh(r1² + r2² + r1r2), where r1 and r2 are the radii of the two bases and h is the height.
Given r1 = 3 ft, r2 = 6 ft, and h = 12 ft, we can calculate the volume:
V = (1/3)π(12)(9 + 36 + 18) = 270π ft³
Now, we can calculate the weight of the water using the density of water:
Weight = density × volume = 62.5 lb/ft³ × 270π ft³ ≈ 53125π lb
Next, we convert the weight to force by multiplying it by the acceleration due to gravity (32.2 ft/s²):
Force = Weight × acceleration due to gravity = 53125π lb × 32.2 ft/s² ≈ 1709125π lb·ft/s²
Finally, we can calculate the work by multiplying the force by the distance. Since the water is being pumped out of the spout, the distance is equal to the height of the frustum, which is 12 ft:
Work = Force × distance = 1709125π lb·ft/s² × 12 ft ≈ 20509500π lb·ft ≈ 64307077 lb·ft
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solve
3x + y = 10
5x - 2y - 2 = 0
The solution to the system of equations is x = 18/11 and y = 56/11.
To solve the system of equations:
3x + y = 10 ...(1)
5x - 2y - 2 = 0 ...(2)
We can use the method of substitution or elimination to find the values of x and y that satisfy both equations.
Let's use the method of elimination:
Multiply equation (1) by 2 to make the coefficients of y in both equations opposite:
6x + 2y = 20 ...(3)
Now, add equations (2) and (3):
(5x - 2y - 2) + (6x + 2y) = 0 + 20
11x = 18
Divide both sides by 11:
x = 18/11
Substitute the value of x into equation (1):
3(18/11) + y = 10
54/11 + y = 10
y = 110/11 - 54/11
y = 56/11
Therefore, the solution to the system of equations is x = 18/11 and y = 56/11.
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50 Points! Multiple choice geometry question. Photo attached. Thank you!
The volume of the oblique cone in this problem is given as follows:
D. 100.5 in³.
How to obtain the volume?The volume of a cone of radius r and height h is given by the equation presented as follows:
V = πr²h/3.
Applying the Pythagorean Theorem, the diameter of the cone is obtained as follows:
d² + 6² = 10²
d² = 100 - 36
d² = 64
d = 8.
The radius is half the diameter, hence it's measure is given as follows:
r = 4 in.
The height of the cone is given as follows:
h = 6 in.
Hence the volume is given as follows:
V = π x 4² x 8/3
V = 100.5 in³ (rounded down).
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What is the domain of the function y=2 x-6
Answer: The domain of the expression is all real numbers except where the expression is undefined.
Step-by-step explanation:
The Conjecture: Summarize the conjecture each person has made. (2 points: 1
point for each row of the chart)
Person
Student
Teacher
Conjecture
Analyze the Conjecture:
1. Do you agree with the student or the teacher? Does rationalizing the denominator
just make it easier for the teacher to grade, or will the rationalized form be easier to
use? (1 point)
Rationalizing the denominator is not solely for the teacher's convenience but ultimately makes mathematical expressions easier to work .
The student's conjecture is that rationalizing the denominator is a purely cosmetic operation that only makes it easier for the teacher to grade. The teacher's conjecture, on the other hand, is that the rationalized form of the denominator is actually easier to use.
Upon analyzing these conjectures, I tend to agree more with the teacher's viewpoint. Rationalizing the denominator serves a practical purpose beyond grading convenience. When a denominator is rationalized, it eliminates radicals or complex expressions from the denominator, which can simplify calculations and make them more manageable.
In many mathematical operations and applications, working with rationalized denominators can lead to simpler and more elegant solutions. It allows for easier addition, subtraction, multiplication, and division of fractions, and it can also facilitate simplification, cancellation, and comparisons.
Additionally, rationalizing the denominator is often necessary in certain contexts, such as when solving equations or expressing mathematical relationships in a standardized form. It can help in identifying patterns, formulating general rules, and making mathematical expressions more accessible and comprehensible.
While it is true that rationalizing the denominator may add extra steps or seem more cumbersome initially, the long-term benefits of having a rationalized form outweigh the short-term inconvenience. Therefore, rationalizing the denominator is not solely for the teacher's convenience but ultimately makes mathematical expressions easier to work with and enhances the understanding and applicability of mathematical concepts.
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Answer? Please someone ASAP!
The measure of angle Q is equal to 70 degrees.
What is a parallelogram?In Mathematics and Geometry, a parallelogram is a geometrical figure (shape) and it can be defined as a type of quadrilateral and two-dimensional geometrical figure that has two (2) equal and parallel opposite sides.
Generally speaking, the the sum of all the interior angles of a parallelogram is equal to 360 degrees and as such, we have the following:
(6x+4) + (6x+4) + 10x + 10x = 360
32x + 8 =360
32x =360-8
32x=352
x = 11
Now, we can determine the measure of angle Q as follows;
Q = 6x + 4
Q = 6(11) + 4
Q = 66 + 4
Q = 70 degrees.
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PLEASE HELP IMMEDIATELY!!!
Find the cosine of < Q.
S
√42
√91
R
Q
Write your answer in simplified, rationalized form. Do not round.
cos (Q)=
Cosine of < Q
cos(Q) = (√42) / (√91)
To find the cosine of angle Q, we need to determine the ratio of the adjacent side to the hypotenuse in a right triangle.
Let's consider a right triangle with sides S, √42, and √91, where angle Q is the angle between the side S and the hypotenuse.
- The side adjacent to angle Q is S.
- The hypotenuse of the triangle is √91.
cos(Q) = adjacent side / hypotenuse
cos(Q) = S / √91
To simplify the expression, we multiply both the numerator and denominator by √91:
cos(Q) = (S * √91) / (√91 * √91)
cos(Q) = (S * √91) / 91
In the given problem, the values of S, √42, and √91 were provided. Substitute the corresponding value of S into the expression above to obtain the simplified, rationalized form of the cosine of angle Q.
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Simplify the radical expression: √99
Answer: 3√11.
Step-by-step explanation:
To simplify the radical expression √99, we can look for perfect square factors of 99.
The prime factorization of 99 is 3 * 3 * 11.
We can rewrite √99 as √(3 * 3 * 11).
Taking out the perfect square factors, we get:
√(3 * 3 * 11) = 3√11.
Therefore, the simplified radical expression for √99 is 3√11.
X^2 + x -72 rewrite the giving expression
Answer:
(x + 9)(x - 8)
Step-by-step explanation:
Factorising the expression
x² + x - 72
consider the factors of the constant term (- 72) which sum to give the coefficient of the x- term (+ 1)
the factors are + 9 and - 8 , since
9 × - 8 = - 72 and 9 - 8 = + 1 , then
x² + x - 72 = (x + 9)(x - 8) ← in factored form
Differentiate the function. Then find an equation of the tangent line at the indicated point on the graph of the function.
Answer:
The derivative of the function:
[tex]f'(x)=-\dfrac{1}{2\sqrt{5-x} }[/tex]
The tangent line of the function at the given point:
[tex]y=-\dfrac{1}{6}x-\dfrac{25}{3}[/tex]
Step-by-step explanation:
Find the equation of the tangent line of the given function using the given point.
[tex]f(x)=6+\sqrt{5-x}; \ (x,y)=(-4,9)[/tex]
[tex]\hrulefill[/tex]
To find the equation of the tangent line to a function at a given point, follow these step-by-step instructions:
Step 1: Identify the point of tangency
Determine the x-coordinate of the point of tangency. Let's call it x₀.Find the corresponding y-coordinate of the point of tangency. Let's call it y₀.Step 2: Find the derivative of the function
Calculate the derivative of the given function. Let's denote it as f'(x).Step 3: Substitute the x-coordinate into the derivative
Replace the variable x in the derivative function f'(x) with the x-coordinate of the point of tangency (x₀).Evaluate the derivative at x₀ to find the slope of the tangent line. Let's denote it as m.Step 4: Write the equation of the tangent line
Use the point-slope form of a line: y - y₀ = m(x - x₀).Substitute the values of m, x₀, and y₀ into the equation.Simplify and rearrange the equation to obtain the final form.Step 5: Optional - Simplify the equation
If necessary, simplify the equation by performing any algebraic manipulations.Step 6: Optional - Verify the equation
Check the obtained equation by plugging in other points along the tangent line and ensuring they satisfy the equation.[tex]\hrulefill[/tex]
Step 1:
[tex](x_0,y_0) \rightarrow (-4,9)[/tex]
Step 2:
[tex]f(x)=6+\sqrt{5-x}\\\\\\\Longrightarrow f(x)=6+(5-x)^{1/2}\\\\\\\Longrightarrow f'(x)=\dfrac{1}{2} (5-x)^{1/2-1} \cdot -1\\\\\\\therefore \boxed{f'(x)=-\dfrac{1}{2\sqrt{5-x} } }[/tex]
Step 3:
[tex]f'(x)=-\dfrac{1}{2\sqrt{5-x} } ; \ (-4,9)\\\\\\\Longrightarrow f'(-4)=-\dfrac{1}{2\sqrt{5-(-4)} } \\\\\\\Longrightarrow f'(-4)=-\dfrac{1}{2\sqrt{9}} \\\\\\\Longrightarrow f'(-4)=-\dfrac{1}{2(3)} \\\\\\\therefore \boxed{ f'(-4)=m=-\dfrac{1}{6} }[/tex]
Step 4 and 5:
[tex]y-y_0=m(x-x_0)\\\\\\\Longrightarrow y-9=-\dfrac{1}{6}(x-(-4)) \\\\\\\Longrightarrow y-9=-\dfrac{1}{6}(x+4) \\\\\\\Longrightarrow y-9=-\dfrac{1}{6}x-\dfrac{2}{3}\\\\ \\\therefore \boxed{\boxed{ y=-\dfrac{1}{6}x-\dfrac{25}{3}}}[/tex]
Thus, the problem is solved.