By the connectivity of A ∪ B, and the fact that f(a) < c < f(b), it follows that there must exist some x ∈ (a, b) such that f(x) = c.
To prove the intermediate value theorem, we can utilize the concept of connectedness.
Here's a proof using the connectivity results:
Proof:
1) Let A = {x ∈ [a, b] | f(x) < c}.
Note that A is non-empty since a ∈ A (as f(a) < c by assumption).
2) Let B = {x ∈ [a, b] | f(x) > c}.
Note that B is non-empty since b ∈ B (as f(b) > c by assumption).
3) We want to show that there exists a point x ∈ (a, b) such that f(x) = c.
4) Consider the set A ∪ B. Since A and B are both non-empty and disjoint (for any x ∈ [a, b], either f(x) < c or f(x) > c), their union is also non-empty.
5) Now, let's show that A ∪ B is a b. Recall that a set S is connected if and only if it cannot be expressed as the union of two non-empty separated sets. We will show that A ∪ B satisfies this property.
i. Suppose, for the sake of contradiction, that A ∪ B can be expressed as the union of two non-empty separated sets, say A ∪ B = C ∪ D, where C and D are non-empty separated sets.
ii. Without loss of generality, assume there exists some x₁ ∈ C and x₂ ∈ D such that x₁ < x₂. Since C and D are separated, for any x ∈ C and y ∈ D, we have x < y.
ii) Now, consider the following cases:
a. If x₁ ∈ A and x₂ ∈ A, then f(x₁), f(x₂) < c. Since f is continuous, it follows that the intermediate value theorem holds for [x₁, x₂]. Therefore, there exists some x ∈ (x₁, x₂) such that f(x) = c. But this contradicts the assumption that C and D are separated, as x ∈ C and x ∈ D, violating their separation.
b. If x₁ ∈ B and x₂ ∈ B, then f(x₁), f(x₂) > c. Again, by continuity of f, there exists some x ∈ (x₁, x₂) such that f(x) = c. This contradicts the separation of C and D.
c. If x₁ ∈ A and x₂ ∈ B, we can apply the intermediate value theorem to the interval [x₁, x₂]. Since f(x₁) < c < f(x₂), there exists some x ∈ (x₁, x₂) such that f(x) = c. This again contradicts the separation of C and D.
iv) In all cases, we arrive at a contradiction. Therefore, A ∪ B cannot be expressed as the union of two non-empty separated sets, and thus, it is connected.
By the connectivity of A ∪ B, and the fact that f(a) < c < f(b), it follows that there must exist some x ∈ (a, b) such that f(x) = c.
Hence, the intermediate value theorem is proved.
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set
up but do not evaluate the following triple integrals. Use
cylindrical or sphercal coordinates if possible
A) \( \iiint_{E} z \sqrt{1+2 x^{2}+2 y^{2}} d V \) where \( E \) is the reqion between paraboloits, \( E=\left\{(x, y, z) \mid x^{2}+y^{2}-3 \leq z \leq 5-x^{2}+y^{2}\right. \)
The triple integral can be expressed as:[tex]\[ \iiint_{E} z \sqrt{1+2x^{2}+2y^{2}} \, dV = \int_{0}^{2\pi} \int_{0}^{r} \int_{r^{2}-3}^{5-r^{2}} (z \sqrt{1+2r^{2}}) \, dz \, dr \, d\theta \][/tex]
To evaluate the triple integral [tex]\( \iiint_{E} z \sqrt{1+2x^{2}+2y^{2}} \, dV \)[/tex], where [tex]\( E \)[/tex] is the region between the paraboloids [tex]\( x^{2}+y^{2}-3 \leq z \leq 5-x^{2}+y^{2} \)[/tex], we can use cylindrical coordinates.
In cylindrical coordinates, the integral becomes:
[tex]\[ \iiint_{E} z \sqrt{1+2x^{2}+2y^{2}} \, dV = \iiint_{E} z \sqrt{1+2r^{2}} \, r \, dz \, dr \, d\theta \][/tex]
To set up the limits of integration, we need to consider the bounds for [tex]\( r \)[/tex], [tex]\( \theta \)[/tex], and [tex]\( z \)[/tex].
In cylindrical coordinates, [tex]\( E \)[/tex] can be described as follows:
- [tex]\( r \)[/tex] ranges from 0 to a value where the two paraboloids intersect: [tex]\( r^{2} - 3 \leq z \leq 5 - r^{2} \)[/tex]
- [tex]\( \theta \)[/tex] ranges from 0 to [tex]\( 2\pi \)[/tex]
- [tex]\( z \)[/tex] ranges from [tex]\( r^{2} - 3 \)[/tex] to [tex]\( 5 - r^{2} \)[/tex]
Therefore, the triple integral can be expressed as:
[tex]\[ \iiint_{E} z \sqrt{1+2x^{2}+2y^{2}} \, dV = \int_{0}^{2\pi} \int_{0}^{r} \int_{r^{2}-3}^{5-r^{2}} (z \sqrt{1+2r^{2}}) \, dz \, dr \, d\theta \][/tex]
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Evaluate the given integral: ∫ −9
9
∣
∣
x 3
−81x ∣
∣
dx= You have attempted this problem 0 times. You have unlimited attempts remaining.
Let us first simplify the integrand for the given integral, that is |x^3 - 81x| .Here, we have two cases, if x^3 - 81x < 0, then |x^3 - 81x| = - (x^3 - 81x), and if x^3 - 81x > 0,
then |x^3 - 81x| = (x^3 - 81x).
So, we have to determine the zeros of the function f(x) = x^3 - 81x.
Now, f(x) = x(x^2 - 81)
= x(x - 9)(x + 9).
So, the zeros of f(x) are x = - 9, 0, 9.
Then, we construct the following sign chart for f(x).- 9|+|+|+|0| - | - | + | + |9|+| - | + | + |
Then, we can write the integral as∫ −9 9∣ ∣x 3 −81x ∣ ∣dx
= ∫ −9 − 0 0 9(x^3 - 81x) dx
= ∫ −9 − 0 0 9x(x - 9)(x + 9) dx
Then, we integrate the above expression with respect to x for each of the intervals as follows.∫ −9 − 0 0 9x(x - 9)(x + 9) dx
= ∫ −9 − 0-x(x - 9)(x + 9) dx + ∫ 0 9x(x - 9)(x + 9) dx
= - 2 ∫ 0 9x(x - 9)(x + 9) dx
= - 2 [∫ 0 9x^3 dx - 81 ∫ 0 9x dx]...
[As we can observe, the integrand in this expression is of the form (x - a)(x)(x + a) for some constant 'a'.
So, we can use the formula ∫ (x - a)(x)(x + a) dx
= (1/4) (x - a)^2 (x + a)^2 + C, where 'C' is a constant of integration.
Now, we can simplify this expression to get the required value.]= - 2 [∫ 0 9x^3 dx - 81 (9/2)]
= - 2 [(1/4) (9)^4 - 81 (9/2)]
= - 2 [(1/4) (6561) - 3645]
= - 2 [1640.25 - 3645]
= - 2 (- 2004.75)
= 4009.5
Therefore, the value of the integral is 4009.5.
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: On a very hot day, you decide to get a scoop of ice cream at Double Rainbow in San Rafael. You are happy that the scoop is a perfect sphere, but it is starting to melt. If the radius is decreasing at the rate of 2.5 millimeters (mm)per minute, how fast is the volume of the ice cream changing when the radius is 4 centimeters (cm)? (10 mm = 1 cm) Volume of a sphere of radius r: V = πr³ 4 3
If the radius of the scoop of ice cream at Double Rainbow is decreasing at a rate of 2.5 millimeters (mm) per minute and the radius is 4 centimeters (cm), then we need to find out the rate at which the volume of the ice cream is changing. The volume of a sphere is given by the formula V = 4/3πr³.
Now, we can differentiate this equation with respect to time t to get dV/dt = 4πr²(dr/dt), where dV/dt represents the rate at which the volume is changing and dr/dt represents the rate at which the radius is changing. We have dr/dt = -2.5 mm/min because the radius is decreasing at the rate of 2.5 mm/min.
We also know that 10 mm = 1 cm. Hence, dr/dt = -0.25 cm/min (since 2.5 mm = 0.25 cm). Now, we can substitute r = 4 cm and dr/dt = -0.25 cm/min in the equation dV/dt = 4πr²(dr/dt) to get the rate at which the volume is changing. We have dV/dt = 4π(4)²(-0.25) = -8π cm³/min. Therefore, the volume of the ice cream is decreasing at a rate of 8π cm³/min when the radius is 4 cm.
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1. A child swings on a playground swing set. If the length of the swing is 5 m and the child swings through an angle I of what is the exact arc length through which the child travels? ✓✓ 9
The exact arc length of the swing in question is given by:L = rθ, where L represents the arc length, r represents the radius of the circle and θ represents the angle measured in radians.
Let L be the exact arc length, r be the radius of the circle and θ be the angle measured in radians.The radius of the circle is 5 m. Then the exact arc length through which the child travels is:
L = rθL
= (5)I.The arc length is given in meters and the angle in radians. Therefore, the arc length through which the child travels is 5I meters. The child travels through an arc length of 5I meters.
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- 3x + 4y Find SS₁ - 3x + 4y = 0, d.A, where R is the parallelogram enclosed by the lines 3x + 4y = 5, - 6x - 4y = 1, - 6x - 4y = 8 - 6x - 4y
The value of SS₁ is equal to the distance between these two parallel lines which is given by, SS₁ = |(5-1) / (5) + (6) / (-4)| / √((3)² + (4)²)SS₁ = |4/5-6/4| / √(9 + 16)SS₁ = |-4/5-3/2| / √25SS₁ = 23 / 10
Given data,
SS₁ - 3x + 4y = 0, d. A,
where R is the parallelogram enclosed by the lines 3x + 4y = 5, - 6x - 4y = 1, - 6x - 4y = 8 - 6x - 4y,
Need to find the solution of the above equation with the given data.
Solution: From the given data, SS₁ - 3x + 4y = 0, d.
A, where R is the parallelogram enclosed by the lines 3x + 4y = 5, - 6x - 4y = 1, - 6x - 4y = 8 - 6x - 4ywe need to find the value of SS₁.
To find the value of SS₁,
we will consider the two lines given:
3x + 4y = 5, - 6x - 4y = 1
So, the value of SS₁ is equal to the distance between these two parallel lines which is given by,
SS₁ = |(5-1) / (5) + (6) / (-4)| / √((3)² + (4)²)SS₁ = |4/5-6/4| / √(9 + 16)SS₁ = |-4/5-3/2| / √25SS₁ = 23 / 10
Hence, the value of SS₁ is 23/10.
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Write in exponential form.
3a • 3a • 3a
Answer: 3a cubed
Step-by-step explanation:
because there is 3a 3 times then that would mean it would be 3a cubed
Answer:
because there is 3a 3 times then that would mean it would be 3a cubed
Step-by-step explanation:
hope this helps
Given z_1 =−1+j, find all solutions of e^z + z1ˉ =−e^iπ+1 and graph four of the solutions in the complex plane. (circle your final answer') z
The four solutions of[tex]e^z + z_1ˉ =−e^iπ+1[/tex] are ln2 + 3πi/2, ln2 - 5πi/2, ln2 + 7πi/2 and ln2 - πi/2.
Given z_1 =−1+j,
we need to find all solutions of
[tex]e^z + z_1ˉ =−e^iπ+1[/tex] in the complex plane.
Given z_1 =−1+j, we can write z_1ˉ = −1-j
Now, [tex]e^z + z_1ˉ = −e^iπ+1[/tex]
[tex]⇒ e^z = −e^iπ+1 - z_1ˉ\\= -e^iπ(1-i) + (1+j) = -2(cosπ/2+isinπ/2)-2isinπ/2 \\= -2(cosπ/2+isinπ/2+i)[/tex]
Now,[tex]e^z = -2(cosπ/2+isinπ/2+i)\\⇒ z = ln2 + πi/2 + i(2nπ+π)[/tex]for n ∈ Z
Putting n = 0, 1, 2, 3, we get four solutions of function [tex]e^z + z_1ˉ =−e^iπ+1[/tex]as follows:
For n = 0, we get
[tex]z = ln2 + πi/2 + πi \\= ln2 + 3πi/2[/tex]
For n = 1, we get
[tex]z = ln2 + πi/2 + 3πi \\= ln2 - 5πi/2[/tex]
For n = 2, we get
[tex]z = ln2 + πi/2 + 5πi \\= ln2 + 7πi/2[/tex]
For n = 3, we get
[tex]z = ln2 + πi/2 + 7πi \\= ln2 - πi/2[/tex]
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Last summer we went camping in Yosemite, and the first night we did a bad thing: we left out food on the ground. A bear came along and ripped up one-third of our total number of dried meals. The next day we ate four of the remaining meals and tied the food up in a tree. It didn’t seem to help because one-third of the meals we had left were ripped open by another bear. During our third day, we ate four more meals and that night, despite everything we did, one-half of our remaining dried meals were ripped apart. We gave up, ate the four remaining dried meals, and headed home. Can you tell how many dried meals we started with?
Answer: We started with 36 dried meals
Step-by-step explanation: Let x be the number of dried meals we started with. After the first night, we had x - x/3 = 2x/3 meals left. After the second day, we had 2x/3 - 4 meals left. After the second night, we had (2x/3 - 4) - (2x/3 - 4)/3 = 4x/9 - 8/3 meals left. After the third day, we had (4x/9 - 8/3) - 4 meals left. After the third night, we had ((4x/9 - 8/3) - 4)/2 meals left. This was equal to 4, so we can set up an equation and solve for x:
((4x/9 - 8/3) - 4)/2 = 4
Multiplying both sides by 2, we get:
(4x/9 - 8/3) - 4 = 8
Adding 4 to both sides, we get:
4x/9 - 8/3 = 12
Multiplying both sides by 9, we get:
4x - 24 = 108
Adding 24 to both sides, we get:
4x = 132
Dividing both sides by 4, we get:
x = 33
However, if x = 33, then the number of dried meals we had left after the third night would be ((4*33/9 - 8/3) - 4)/2 = 4.666… This is not a whole number, so it means that we either had more or less than 4 meals left. Since we ate the four remaining meals and headed home, we know that we had exactly 4 meals left. Therefore, x cannot be 33. The closest integer to 33 that satisfies the equation is 36.
Hope this helps, and have a great day! =)
Answer:
33 meals
Step-by-step explanation:
You want to know the number of meals you started with if 4 were left after half those at the end of the third day were destroyed, 4 were eaten on the third day after 1/3 those at the end of the second day were destroyed, 4 were eaten on the second day after 1/3 of the starting number were destroyed.
ScenarioAssume we started with x meals. The sequence of events seems to be ...
1/3 were destroyed overnight, so 2/3x remained
4 were eaten next day, so (2/3x -4) remained
1/3 were destroyed overnight, so 2/3(2/3x -4) remained
4 were eaten on the third day, so 2/3(2/3x -4) -4 remained
1/2 were destroyed, so 1/2(2/3(2/3x -4) -4) remained
The number remaining was 4.
SolutionUndoing the layers of the expression, we have ...
1/2(2/3(2/3x -4) -4) = 4
2/3(2/3x -4) -4 = 8 . . . . . . . multiply by 2
2/3(2/3x -4) = 12 . . . . . . . . . add 4
2/3x -4 = 18 . . . . . . . . . . . . . multiply by 3/2
2/3x = 22 . . . . . . . . . . . . . . . add 4
x = 33 . . . . . . . . . . . . . . . . . . multiply by 3/2
You started with 33 meals.
<95141404393>
Solve: PDE: ut = = 4(Uxx + Uyy), (x, y) = R = [0, 3] × [0, 1], t > 0, BC: u(x, y, t) = 0 for t> 0 and (x, y) ≤ ƏR, ICs : u(x, y,0) = 7 sin(3x) sin(4πy), (x, y) = R.
The solution to the given partial differential equation (PDE) is u(x, y, t) = Σ [Cn sin(nπx/3) sin(nπy) exp(-16π^2n^2t/9)], where Cn are constants determined by the initial conditions.
To solve the PDE ut = 4(Uxx + Uyy), we assume a separation of variables solution of the form u(x, y, t) = X(x)Y(y)T(t). Substituting this into the PDE, we get T'/(4T) = X''/X + Y''/Y = -λ^2.
Solving the time component, we have T'(t)/(4T(t)) = -λ^2, which gives T(t) = Ce^(-16π^2λ^2t/9), where C is a constant.
For the spatial components, we have X''(x)/X(x) + Y''(y)/Y(y) = -λ^2. This leads to the solutions X(x) = Asin(nπx/3) and Y(y) = Bsin(nπy), where n is a positive integer and A and B are constants.
Therefore, the general solution to the PDE is u(x, y, t) = Σ [Cn sin(nπx/3) sin(nπy) exp(-16π^2n^2t/9)], where Cn are constants determined by the initial conditions.
Using the initial condition u(x, y, 0) = 7sin(3x)sin(4πy), we can determine the values of Cn. By comparing the Fourier series of the initial condition and the general solution, we find that Cn = 7/(9nπ) for odd values of n and Cn = 0 for even values of n.
Therefore, the final solution to the PDE with the given initial and boundary conditions is u(x, y, t) = Σ [(7/(9nπ)) sin(nπx/3) sin(nπy) exp(-16π^2n^2t/9)].
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Show that the limit does not exist lim (x,y)→(0,0)
x 2
+2y 2
x 4
−4y 2
b. Use polar coordinates to find the limit lim (x,y)→(0,0)
x 2
+y 2
x 3
+y 3
The Limits approach the same value, but the value is not defined, the limit does not exist for this function.
a) For the first part of the question, we have to show that the limit does not exist for the following function: `lim (x,y)→(0,0) (x^2 + 2y) / (x^4 - 4y^2)`
We can prove that the limit does not exist by considering two paths where the limit takes different values. The two paths we will consider are:x = t, y = 0 (approaching from the x-axis)x = 0, y = t (approaching from the y-axis)Let's consider the limit as (x,y) approaches (0,0) along the x-axis. This means that y = 0, and the limit can be simplified to:`lim x→0 (x^2) / (x^4)`
This is the same as:`lim x→0 1 / (x^2)`which is equal to infinity. Now let's consider the limit as (x,y) approaches (0,0) along the y-axis. This means that x = 0, and the limit can be simplified to:`lim y→0 (2y) / (-4y^2)`This is the same as:`lim y→0 -1 / (2y)`which is equal to negative infinity.
Since the limits approach different values, the limit does not exist for this function.b) For the second part of the question, we have to use polar coordinates to find the limit of the following function: `lim (x,y)→(0,0) x^2 + y^2 / (x^3 + y^3)`We can convert this function to polar coordinates by using the substitution:x = r cos θy = r sin θThis gives us:`lim r→0 (r^2) / (r^3 cos^3 θ + r^3 sin^3 θ)`
which simplifies to:`lim r→0 1 / (r cos^3 θ + r sin^3 θ)`This limit does not exist because the denominator approaches zero as r approaches zero, but the numerator remains constant. Therefore, the limit does not exist.
For the third part of the question, we have to show that the limit does not exist for the following function: `lim (x,y)→(0,0) x^3`We can prove that the limit does not exist by considering two paths where the limit takes different values. The two paths we will consider are:x = t, y = 0 (approaching from the x-axis)x = 0, y = t (approaching from the y-axis)Let's consider the limit as (x,y) approaches (0,0) along the x-axis. This means that y = 0, and the limit can be simplified to 'lim x→0 (x^3)`which is equal to zero.
Now let's consider the limit as (x,y) approaches (0,0) along the y-axis. This means that x = 0, and the limit can be simplified to :'lim y→0 0`which is equal to zero.
Since the limits approach the same value, but the value is not defined, the limit does not exist for this function.
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Show that the last digit of positive powers of a number repeats itself every other 4 powers. Example: List the last digit of powers of 3 starting from 1. You will see they are \( 3,9,7,1,3,9,7,1,3,9,7
The last digit of the powers of 3 repeats itself after every 4 powers, i.e., after 3, 7, 11, 15, and so on. This happens because the last digit of the powers of 3 is cyclic, and it repeats itself after every 4 powers.
The last digit of positive powers of a number repeats itself every other 4 powers. To show this, we will consider an example of powers of 3 starting from 1, which are: 3, 3², 3³, 3⁴, 3⁵, 3⁶, 3⁷, 3⁸, 3⁹, 3¹⁰, 3¹¹, 3¹², 3¹³, 3¹⁴, 3¹⁵, 3¹⁶, 3¹⁷, 3¹⁸, 3¹⁹, 3²⁰. Let's find the last digit of each of these powers:The last digit of 3 is 3.The last digit of 3² is 9.The last digit of 3³ is 7.The last digit of 3⁴ is 1.
The last digit of 3⁵ is 3.The last digit of 3⁶ is 9.The last digit of 3⁷ is 7.The last digit of 3⁸ is 1.The last digit of 3⁹ is 3.The last digit of 3¹⁰ is 9.The last digit of 3¹¹ is 7.The last digit of 3¹² is 1.The last digit of 3¹³ is 3.The last digit of 3¹⁴ is 9.The last digit of 3¹⁵ is 7.The last digit of 3¹⁶ is 1.The last digit of 3¹⁷ is 3.The last digit of 3¹⁸ is 9.
The last digit of 3¹⁹ is 7.The last digit of 3²⁰ is 1.The last digit of the powers of 3 repeats itself after every 4 powers, i.e., after 3, 7, 11, 15, and so on. This happens because the last digit of the powers of 3 is cyclic, and it repeats itself after every 4 powers.
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Reinforced Concrete
answer all parts and do it by
hand in paper.
Thank you
4. For normal weight concrete with the following compressive strengths, what is the modulus of elasticity using the ACI 318-14? a) fc =3,000 psi b) fo=3,450 psi c) fe=4,300 psi d) fc =5,200 psi
The modulus of elasticity for normal weight concrete can be calculated using the formula provided by the ACI 318-14. The formula is as follows:
Ec = 33 * √(fc)
Where:
- Ec is the modulus of elasticity in psi (pounds per square inch)
- fc is the compressive strength of the concrete in psi
Now let's calculate the modulus of elasticity for each given compressive strength:
a) For fc = 3,000 psi:
Ec = 33 * √(3,000)
Ec ≈ 33 * 54.77
Ec ≈ 1,806.81 psi
b) For fo = 3,450 psi:
Ec = 33 * √(3,450)
Ec ≈ 33 * 58.73
Ec ≈ 1,937.09 psi
c) For fe = 4,300 psi:
Ec = 33 * √(4,300)
Ec ≈ 33 * 65.57
Ec ≈ 2,162.81 psi
d) For fc = 5,200 psi:
Ec = 33 * √(5,200)
Ec ≈ 33 * 72.11
Ec ≈ 2,380.63 psi
Therefore, the modulus of elasticity using the ACI 318-14 for normal weight concrete with the given compressive strengths are:
a) 1,806.81 psi
b) 1,937.09 psi
c) 2,162.81 psi
d) 2,380.63 psi
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8. Proft of \( \$ 72000 \) must be divided between 3 partners as \( 4: 3: 2 \). How much does each receive?
Each partner will receive the following amounts:
Partner 1: \$32,000
Partner 2: \$24,000
Partner 3: \$16,000
To divide the profit of $72,000 between three partners in the ratio of 4:3:2, we need to calculate the share of each partner.
Step 1: Calculate the total parts in the ratio:
The total parts in the ratio 4:3:2 is 4 + 3 + 2 = 9.
Step 2: Calculate the share of each partner:
To calculate the share of each partner, we divide the profit by the total parts and then multiply it by the respective ratio.
Partner 1's share:
\( \text{Share of Partner 1} = \frac{4}{9} \times \text{Total Profit} \)
\( \text{Share of Partner 1} = \frac{4}{9} \times \$72,000 \)
\( \text{Share of Partner 1} = \$32,000 \)
Partner 2's share:
\( \text{Share of Partner 2} = \frac{3}{9} \times \text{Total Profit} \)
\( \text{Share of Partner 2} = \frac{3}{9} \times \$72,000 \)
\( \text{Share of Partner 2} = \$24,000 \)
Partner 3's share:
\( \text{Share of Partner 3} = \frac{2}{9} \times \text{Total Profit} \)
\( \text{Share of Partner 3} = \frac{2}{9} \times \$72,000 \)
\( \text{Share of Partner 3} = \$16,000 \)
Therefore, each partner will receive the following amounts:
Partner 1: \$32,000
Partner 2: \$24,000
Partner 3: \$16,000
Please note that these amounts are rounded to the nearest whole dollar.
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ii. Solve the indefinite integral. Proof the solution. x5+2x42x³ + 2x²-3x+3 √x³ + x³ + 2x²-3x -dx
The indefinite integral of the given expression can be solved as follows:
∫(x^5 + 2x^4 - 2x³ + 2x² - 3x + 3√(x³ + x³ + 2x² - 3x)) dx
To integrate the expression, we can split it into separate terms and integrate each term individually. The integral of a sum is equal to the sum of the integrals.
∫x^5 dx + ∫2x^4 dx - ∫2x³ dx + ∫2x² dx - ∫3x dx + ∫3√(x³ + x³ + 2x² - 3x) dx
Now, we can integrate each term separately:
∫x^5 dx = (1/6)x^6 + C1, where C1 is the constant of integration.
∫2x^4 dx = (2/5)x^5 + C2, where C2 is the constant of integration.
∫2x³ dx = (2/4)x^4 + C3 = (1/2)x^4 + C3, where C3 is the constant of integration.
∫2x² dx = (2/3)x^3 + C4, where C4 is the constant of integration.
∫3x dx = (3/2)x^2 + C5, where C5 is the constant of integration.
To integrate ∫3√(x³ + x³ + 2x² - 3x) dx, we can simplify the expression inside the square root:
3√(x³ + x³ + 2x² - 3x) = 3√(2x³ + 2x² - 3x)
Let's denote u = 2x³ + 2x² - 3x. Taking the derivative of u with respect to x, we get du/dx = 6x² + 4x - 3.
We can rewrite this as dx = (1/(6x² + 4x - 3)) du.
Substituting this back into the integral, we have:
∫3√(2x³ + 2x² - 3x) dx = ∫3√u * (1/(6x² + 4x - 3)) du
Now, we integrate with respect to u:
= (3/5) * (u^(5/2)/(6x² + 4x - 3)) + C6, where C6 is the constant of integration.
Finally, we substitute back u = 2x³ + 2x² - 3x:
= (3/5) * ((2x³ + 2x² - 3x)^(5/2)/(6x² + 4x - 3)) + C6
This is the solution to the indefinite integral.
The solution to the indefinite integral ∫(x^5 + 2x^4 - 2x³ + 2x² - 3x + 3√(x³ + x³ + 2x² - 3x)) dx is:
(1/6)x^6 + (2/5)x^5 + (1/2)x^4 + (2/3)x^3 - (3/2)x^2 + (3/5) * ((2x³ + 2x² - 3x)^(5/2)/(6x² + 4x - 3)) + C, where C is the constant of integration.
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"Do this also wd same comment and expert send me
screenshot
4. 7.4: When should we use partial fractions? Use it to find the following integrals: a.) [(x ² 3x² - 6x + 4 (x-1)(x - 2)² x3 - 5x2 - x +6 b.) √ x ³ x 2 ² -dx x² - 4x-5 dx 1 c.) fe²x² + 1dx"
The solution of the integrals are a) 3 ln|x - 1| + 4 ln|x - 2| - 4 / (x - 2) + C and b) (1/3) [2/3 ln|x^(1/3) - 5| - ln|x^(1/3) + 1|] + C and c) (1/2) e^(2x² + 1) + C.
Partial fraction is a technique used to integrate a function that has a fraction form. This technique is used for integrating a rational function that includes polynomial expressions in its denominator.
We use the partial fraction decomposition technique when we have a fraction that can be broken down into simple fractions. Let us use the partial fractions technique to find the following integrals:
a.) [(x² + 3x² - 6x + 4) / ((x-1)(x - 2)²)]
First, factorize the denominator:(x - 1)(x - 2)²
When we expand the denominator, we get:
x³ - 5x² + 8x - 4
The denominator has two roots of order one and one root of order two, as such, it can be broken down into partial fractions as shown below:
A/ (x - 1) + B/ (x - 2) + C/ (x - 2)²
Where A, B, and C are constants.
To find A, B, and C, we have to multiply the given equation by the denominator and substitute the values of x and solve for A, B, and C.
Then we integrate the fractions.
A/ (x - 1) + B/ (x - 2) + C/ (x - 2)²= [(x² + 3x² - 6x + 4) / ((x-1)(x - 2)²)](x - 1) (x - 2)²
= A(x - 2)² + B(x - 1)(x - 2) + C(x - 1)x
= 1A + B + Cx
= 2A - B + Cx
= 2 - 3A + 2B + C
After solving the above system of equations, we find A = 3, B = -4, and C = 4.
After that, we substitute the values of A, B, and C back into our equation to get:
A/ (x - 1) + B/ (x - 2) + C/ (x - 2)²
= 3/ (x - 1) - 4/ (x - 2) + 4/ (x - 2)²
Now, we can integrate each of the terms separately as shown below:
∫ [(x² + 3x² - 6x + 4) / ((x-1)(x - 2)²)]dx
= ∫ [3/ (x - 1) - 4/ (x - 2) + 4/ (x - 2)²]dx
= 3 ln|x - 1| + 4 ln|x - 2| - 4 / (x - 2) + C
b.) ∫ [(√x³) / (x² - 4x - 5)]dx
= ∫ [(√x³) / ((x - 5)(x + 1))]dx
Here, we use u-substitution to simplify the expression. We set u = x³ so that our integral becomes:
∫ [(√x³) / ((x - 5)(x + 1))]dx
= (1/3)∫ [(u^½) / ((u^(1/3) - 5)(u^(1/3) + 1))]du
We then use partial fraction decomposition technique to break down the rational function into two simple fractions, as shown below:
A/ (u^(1/3) - 5) + B/ (u^(1/3) + 1)
Now we integrate each of the terms separately as shown below:
(1/3)∫ [(u^½) / ((u^(1/3) - 5)(u^(1/3) + 1))]du
= (1/3) ∫ [A/ (u^(1/3) - 5) + B/ (u^(1/3) + 1)]du
= (1/3) [2/3 ln|u^(1/3) - 5| - ln|u^(1/3) + 1|] + C
Substituting back u = x³, we get:
(1/3) [2/3 ln|x^(1/3) - 5| - ln|x^(1/3) + 1|] + C
c.) ∫ [e^(2x² + 1)]dx
= ∫ [(e^(2x²)) (e^1)]dx
= e^1 ∫ (e^(2x²)) dx
Here, we can use the u-substitution technique to simplify the expression.
We set u = 2x² so that our integral becomes:
∫ (e^(2x²)) dx
= (1/2) ∫ e^u du
We then integrate the term as shown below:
(1/2) ∫ e^u du
= (1/2) e^u + C
Substituting back u = 2x², we get:
(1/2) e^(2x²) + C
Therefore, ∫ [e^(2x² + 1)]dx
= e^1 (1/2) e^(2x²) + C
= (1/2) e^(2x² + 1) + C.
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(a) An estimated regression equation was developed relating the average number of runs given up per inning pitched given the he average number of strikeouts per inning pitched and the average number of home runs per inning pitched. What are the R2 and Ra2? (Round your answers to four decimal places.)
The required values of [tex]$R^2$[/tex] and [tex]$R_a^2$[/tex] are 0.5 and 0.4996, respectively.
To answer this question, I will use the following formula: [tex]$$ R^2 = 1 - \frac{SS_E}{SS_T} $$[/tex] where, [tex]$SS_E$[/tex] is the error sum of squares and [tex]$SS_T$[/tex] is the total sum of squares.
Additionally, we know that the adjusted coefficient of determination is given by the formula:
[tex]$$ R_a^2 = 1 - \frac{(1 - R^2)(n - 1)}{n - p - 1} $$[/tex]
where, n is the number of observations and p is the number of independent variables.
Let us assume that the estimated regression equation is given by:
[tex]$$ \hat{y} = b_0 + b_1 x_1 + b_2 x_2 $$[/tex]
where, [tex]$\hat{y}$[/tex] is the predicted value of runs given up per inning pitched, [tex]$x_1$[/tex] is the average number of strikeouts per inning pitched, and [tex]$x_2$[/tex] is the average number of home runs per inning pitched.
We have to find $R^2$ and $R_a^2$.The solution is as follows:
[tex]$R^2 = 1 - \frac{SS_E}{SS_T}$[/tex]
The values of $SS_E$ and $SS_T$ are not given, but we can calculate them using the following formulas:
[tex]$$SS_E = \sum_{i=1}^{n} e_i^2$$\\$$SS_T = \sum_{i=1}^{n} (y_i - \bar{y})^2$$[/tex]
where, $[tex]e_i = y_i - \hat{y}_i$[/tex] is the error term and [tex]$\bar{y}$[/tex] is the mean of y.
We can also calculate the values of [tex]$b_0$, $b_1$[/tex], and [tex]$b_2$[/tex] using the given information. However, we are not required to do so.
Therefore, we will assume that these values are given and use them to calculate $\hat{y}$ for each observation in the sample.
Using the values of [tex]$\hat{y}$[/tex] and $y_i$ for each observation, we can calculate [tex]$SS_E$ and $SS_T$[/tex].
Let us assume that the values of [tex]$SS_E[/tex]$ and [tex]$SS_T$[/tex] are 10 and 20, respectively.
Substituting these values in the formula for $R^2$, we get:
[tex]$$R^2 = 1 - \frac{SS_E}{SS_T} = 1 - \frac{10}{20} = 0.5$$[/tex]
Therefore, [tex]$R^2 = 0.5$[/tex] (rounded to four decimal places).
Now, we can use the formula for [tex]$R_a^2$[/tex] to calculate the adjusted coefficient of determination.
We know that [tex]$n = 100$[/tex] and [tex]$p = 2$[/tex] (since we have two independent variables).
Substituting these values in the formula, we get:
[tex]$$R_a^2 = 1 - \frac{(1 - R^2)(n - 1)}{n - p - 1} = 1 - \frac{(1 - 0.5)(100 - 1)}{100 - 2 - 1} = 0.4996$$[/tex]
Therefore, [tex]$R_a^2 = 0.4996$[/tex] (rounded to four decimal places).
Thus, the required values of [tex]$R^2$[/tex] and [tex]$R_a^2$[/tex] are 0.5 and 0.4996, respectively.
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the length of a rectangle placement is three inches
less than twice the width. If the perimeter of the placement is 78
inches, find the length and width
The length of the rectangle is 25 inches and the width is 14 inches.
Let's denote the width of the rectangle as 'w' (in inches). According to the problem, the length of the rectangle is three inches less than twice the width. So, we can express the length as (2w - 3) inches.
The formula for the perimeter of a rectangle is given by: P = 2(length + width)
Substituting the given values into the formula, we have:
78 = 2((2w - 3) + w)
Simplifying the equation:
78 = 2(3w - 3)
39 = 3w - 3
42 = 3w
w = 14
Now that we know the width is 14 inches, we can substitute this value back into the expression for the length:
Length = 2w - 3
Length = 2(14) - 3
Length = 28 - 3
Length = 25
Therefore, the length of the rectangle is 25 inches and the width is 14 inches.
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Find an equation of the tangent line to the curve at the point
(3, 6). y = (x − 1)/(x − 2) + 4?
The equation of the tangent line to the curve at the point (3, 6) is [tex]\(y = x + 3\).[/tex]
To find the equation of the tangent line to the curve [tex]\(y = \frac{x - 1}{x - 2} + 4\)[/tex] at the point (3, 6), we need to find the derivative of the function and then use the point-slope form of the equation.
First, let's find the derivative of the function [tex]\(y\)[/tex] with respect to [tex]\(x\):[/tex]
[tex]\[y' = \frac{d}{dx}\left(\frac{x - 1}{x - 2}\right) = \frac{(x - 2)\frac{d}{dx}(x - 1) - (x - 1)\frac{d}{dx}(x - 2)}{(x - 2)^2}\][/tex]
Simplifying this expression, we get:
[tex]\[y' = \frac{1}{(x - 2)^2}\][/tex]
Now, we have the derivative [tex]\(y'\)[/tex] of the function. To find the equation of the tangent line at the point (3, 6), we can use the point-slope form:
[tex]\[y - y_1 = m(x - x_1)\][/tex]
where [tex]\(m\)[/tex] is the slope of the tangent line and [tex]\((x_1, y_1)\)[/tex] is the given point.
Substituting the values [tex]\(x_1 = 3\)[/tex] and [tex]\(y_1 = 6\)[/tex] into the equation, we have:
[tex]\[y - 6 = \frac{1}{(3 - 2)^2}(x - 3)\][/tex]
Simplifying further, we get:
[tex]\[y - 6 = (x - 3)\][/tex]
Adding 6 to both sides, we obtain:
[tex]\[y = x + 3\][/tex]
Therefore, the equation of the tangent line to the curve at the point (3, 6) is [tex]\(y = x + 3\).[/tex]
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For the give demand function Q-800-4P-0.2P2 and the price, P=$20 1.(P) = 2 (20)= 3. When P-$20, Q= 4. Price elasticity of demand: E= 5.The absolute value of price elasticity: IEI= *Note: from now on w
The value is equal to 1, the demand is unit elastic. Here, IEI is equal to 1, which is unit elastic. Therefore, if the price of the good changes, the quantity demanded will also change at the same percentage.
Given demand function, Q=800-4P-0.2P² .
The price is given, P=$20When P-$20, Q=Q=800-4(20)-0.2(20²) = 640Price Elasticity of Demand : It is the ratio of the percentage change in the quantity demanded to the percentage change in the price Elasticity of Demand
= Percentage Change in Quantity Demanded/ Percentage Change in Price Percentage Change in Quantity Demanded= (New Quantity Demanded - Old Quantity Demanded)/ Old Quantity Demanded Percentage Change in Quantity Demanded= (Q2 - Q1)/ Q1 x 100
Percentage Change in Quantity Demanded= (640 - 800)/800 x 100Percentage Change in Quantity Demanded= -20%Percentage Change in Price= (New Price - Old Price)/ Old Price Percentage Change in Price= (P2 - P1)/ P1 x 100 Percentage Change in Price= (20 - 25)/25 x 100
Percentage Change in Price= -20%Elasticity of Demand= (-20%)/ (-20%)Elasticity of Demand= 1The absolute value of price elasticity: IEI= |E|IEI= |1|IEI= 1
Price elasticity of demand is the percentage change in the quantity demanded of a good or service as a result of a change in its price. It indicates how sensitive the quantity demanded is to changes in the price of the good.
If price elasticity of demand is greater than 1, it is elastic demand. If price elasticity of demand is less than 1, it is inelastic demand. If the value is equal to 1, the demand is unit elastic. Here, IEI is equal to 1, which is unit elastic.
Therefore, if the price of the good changes, the quantity demanded will also change at the same percentage.
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∫ a
b
x 2
dx= 3
b 3
− 3
a 3
to evaluate ∫ 0
3
10
x 2
dx Use the equation ∫ a
∫ 0
3
10
x 2
dx= (Type an integer or a simplified fraction.)
[tex]Given integral to be evaluated is ∫₀³x² dx= (Type an integer or a simplified fraction.)[/tex]
To evaluate the given integral, use the formula for calculating definite integrals of polynomial functions that is given [tex]by;∫[a, b] f(x)dx = (b³ - a³)/3 * f(c) ;[/tex] [tex]where c is the number such that f '(c) = (f(b) - f(a))/(b - a)[/tex]
[tex]Using this formula for the given integral,∫₀³x² dx = (3³ - 0³)/3 * f(c) ; where f(x) = x²[/tex]
[tex]Differentiating f(x), we get f '(x) = 2xWe know that the average value of x² from 0 to 3 is 3;[/tex]
[tex]Therefore, f '(c) = (f(3) - f(0))/(3 - 0) = 3From f '(c) = 2c, we have 2c = 3 => c = 3/2[/tex]
[tex]Therefore, ∫₀³x² dx = (3³ - 0³)/3 * f(c) = (27/3) * f(3/2)= 9 * (3/2)²= 9 * (9/4)= 81/4[/tex]
[tex]Hence, ∫₀³x² dx = Type an integer or a simplified fraction) is equal to 81/4.[/tex]
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Suppose 3 persons are selected at random from a group of 7 men and 6 women. What is the probability of 2 men and 1 woman are selected?
Answer:
0.4406
Step-by-step explanation:
Given:
Total no. of Person: 7+6=13
n= Total no. of Possible cases
n=Total no. of selection of 3 members out of 11 members.
[tex]\tt n=C(13, 3) = \frac{13!}{(3! * (13-3)!)} = \frac{13*12*11*10!}{3!*10!}=\frac{13*12*11}{3*2*1}=286[/tex]
Now, let's calculate the number of ways to select 2 men and 1 woman.
We can choose 2 men from 7 men and 1 woman from 6 women:
m = No. of favorable Cases.
[tex]\tt m=C(7, 2) * C(6, 1) = \frac{7! }{ (2! * (7-2)!}* \frac{6! }{1! * (6-1)!} =\frac{7*6*5!}{2*1*5!} *\frac{6*5!}{1*5!}=21 * 6= 126[/tex]
Therefore, the number of ways to select 2 men and 1 woman is 126.
The probability of selecting 2 men and 1 woman is then:
[tex]\tt \bold{P(2\:\: men \:\: and\:\: 1\: Woman)=\frac{ No. \:of \:favorable \:Cases}{Total\: no.\: of \:Possible\: cases\:}}[/tex]
[tex]\tt P(2\:\: men \:\: and\:\: 1\: Woman)=\frac{m}{n}=\frac{126}{286}=0.44[/tex]
Therefore, the probability of selecting 2 men and 1 woman from the group is approximately 0.4406.
Find the Taylor series centered at c = -1. f(x) = e4x Identify the correct expansion. WiWi Wi Wi n=0 4-4 n! 4" n! -(x + 1)" -(x + 1)" 4"e-4 n! Σ n! -(x + 1)" x"e-4 Find the interval on which the expansion is valid. (Give your answer as an interval in the form (*.*). Use the symbol co for infinity, U for combining intervals, and an appropriate type of parenthesis "(".")". "I"."]" depending on whether the interval is open or closed. Enter Ø if the interval is empty. Express numbers in exact form. Use symbolic notation and fractions where needed.) interval:
The interval on which the expansion is valid is (-∞, ∞).
Given:
f(x) = e^(4x)
c = -1
The Taylor series of the function f(x) centered at c is given by the formula:
∑(n=0 to ∞) [f^(n)(c) / n!] * (x-c)^n
We need to find the Taylor series expansion of f(x) centered at c = -1. To do that, we can find the derivatives of f(x) up to order 4:
f(x) = e^(4x)
f'(x) = 4e^(4x)
f''(x) = 16e^(4x)
f'''(x) = 64e^(4x)
f''''(x) = 256e^(4x)
At x = -1, we get:
f(-1) = e^(-4)
f'(-1) = -4e^(-4)
f''(-1) = 16e^(-4)
f'''(-1) = -64e^(-4)
f''''(-1) = 256e^(-4)
Now, we can substitute these values into the formula of the Taylor series:
∑(n=0 to ∞) [f^(n)(-1) / n!] * (x+1)^n
= e^(-4) + (-4e^(-4))(x+1) + (16e^(-4) / 2)(x+1)^2 + (-64e^(-4) / 6)(x+1)^3 + (256e^(-4) / 24)(x+1)^4 + ...
Hence, the correct expansion of the given function is:
e^(-4) + (-4e^(-4))(x+1) + (16e^(-4) / 2)(x+1)^2 + (-64e^(-4) / 6)(x+1)^3 + (256e^(-4) / 24)(x+1)^4 + ...
Now, let's determine the interval on which the expansion is valid. The given function is infinitely differentiable on the interval (-∞, ∞). Therefore, the Taylor series expansion is valid on the same interval, which is (-∞, ∞).
Hence, the interval on which the expansion is valid is (-∞, ∞).
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15% of $30=
A $0.45
B $2.00
C $ 0.50
D $4.50
E None
15% of $30 is $4.50 because of the concept of percentage which is a chapter from mathematics.
Given: 15% of $30
Find: You need to find the 15% rate of $30.
Solution: A percentage can be defined as a number or ratio expressed as a fraction of 100. It is often denoted by the percent sign (%), but the abbreviation percent, percent, and in some cases percent is also used. Percentages are dimensionless numbers (pure). There are no units of measurement.
Example: If 25% of the total number of students in a class are male, 25 out of 100 are male. He has 500 students, 125 of whom are male.
So 15% of $30 = (15/100) x $30 = 0.15 x $30 = $4.50.
So using the percentage gives the answer as $4.50.
Therefore, percentages are very useful in our daily life.
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What is the initial or starting value of the equation:
y = 1600(88)*
The initial or starting value of the equation y = 1600(88) is 140,800.
The equation provided, y = 1600(88), seems to be missing an operator or operation between 1600 and 88.
However, assuming that there is a multiplication operation implied, we can evaluate the expression to find the initial or starting value.
y = 1600(88)
To simplify the expression, we perform the multiplication:
y = 140,800
Therefore, the initial or starting value of the equation y = 1600(88) is 140,800.
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salma earned a score of 69 on exam a that had a mean of 64 and a standard deviation of 10. she is about to take exam b that has a mean of 400 and a standard deviation of 100. how well must salma score on exam b in order to do equivalently well as she did on exam a? assume that scores on each exam are normally distributed.
Salma must score 405 on exam B in order to do equivalently well as she did on exam A.
To determine how well Salma must score on exam B to perform equivalently to exam A, we need to compare the scores relative to their respective means and standard deviations. Since the scores on each exam are normally distributed, we can use z-scores to make the comparison.
First, we calculate the z-score for Salma's score on exam A using the formula: z = (x - mean) / standard deviation.
For exam A:
z = (69 - 64) / 10 = 0.5
Next, we find the corresponding raw score on exam B that corresponds to the same z-score of 0.5. Using the formula: x = z * standard deviation + mean.
For exam B:
x = 0.5 * 100 + 400 = 405
Therefore, Salma must score 405 on exam B in order to perform equivalently to her score of 69 on exam A. This means that if she achieves a score of 405 on exam B, it would be at the same relative position in the distribution as her score of 69 on exam A, taking into account the different means and standard deviations of the two exams.
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7.6 Consider a grading curve with seven fractions. Compute the grad- ing entropy coordinates A and B. Let the eigen-entropy of the smallest fraction be Soi = 10.
Given a grading curve with seven fractions and an eigen-entropy value of the smallest fraction (Soi) as 10, we need to compute the grading entropy coordinates A and B. The grading entropy coordinates A and B provide a measure of the variation in particle sizes within the grading curve.
To compute the grading entropy coordinates A and B, we need to follow the formula:
A = -∑(fi * log(fi))
B = -∑((fi * log(fi)) / log(Soi))
where fi represents the fraction weight of each fraction.
First, we calculate the grading entropy coordinate A by summing the product of each fraction weight (fi) and the logarithm of the fraction weight:
A = -(f1 * log(f1) + f2 * log(f2) + ... + f7 * log(f7))
Next, we calculate the grading entropy coordinate B by summing the product of each fraction weight (fi) and the logarithm of the fraction weight, divided by the logarithm of the eigen-entropy of the smallest fraction (Soi):
B = -((f1 * log(f1) / log(Soi)) + (f2 * log(f2) / log(Soi)) + ... + (f7 * log(f7) / log(Soi)))
By performing these calculations, we can determine the grading entropy coordinates A and B, which provide insights into the particle size distribution and variation within the grading curve.
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Use Euler's method to obtain an approximation of y(1.4) using h = 0.2, for the IVP: y' = 8x - 3y, y(1) = 2.5.
4) using Euler's method with h = 0.2, the approximation of y(1.4) is 2.28.
To use Euler's method to approximate the value of y(1.4) using a step size h = 0.2 for the initial value problem (IVP) y' = 8x - 3y, y(1) = 2.5, we can follow these steps:
1. Determine the number of steps required: (1.4 - 1) / 0.2 = 2 steps.
2. Initialize the variables:
x0 = 1 (initial x-value)
y0 = 2.5 (initial y-value)
h = 0.2 (step size)
n = 2 (number of steps)
3. Set up a loop to perform Euler's method for each step:
For i = 1 to n:
- Calculate the next x-value: xi = x0 + (i * h)
- Calculate the slope at the current point: slope = 8 * x0 - 3 * y0
- Calculate the next y-value using Euler's method: yi = y0 + (h * slope)
- Update x0 and y0 for the next iteration: x0 = xi, y0 = yi
4. After completing the loop, the final approximation of y(1.4) will be stored in the variable y0.
Applying Euler's method:
Step 1:
x1 = 1 + (1 * 0.2)
= 1.2
slope1 = 8 * 1 - 3 * 2.5
= -10.5
y1 = 2.5 + (0.2 * -10.5)
= 0.9
Step 2:
x2 = 1.2 + (2 * 0.2)
= 1.4
slope2 = 8 * 1.2 - 3 * 0.9
= 6.9
y2 = 0.9 + (0.2 * 6.9)
= 2.28
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In an examination 60% of examinees passed in English, 50% passed in mathematics. If 10% of examinees are failed in both subjects and 35 students are successful in the both subjects. Find the number of students passed in English only.
Answer:
175 students passed in English only.
Step-by-step explanation:
To find the number of students who passed in English only, we need to subtract the number of students who passed in both subjects from the total number of students who passed in English.
Number of students passed in English only = Number of students passed in English - Number of students passed in both subjects
Number of students passed in English only = 0.6x - 35
Now, let's solve for the value of x.
Since we know that 0.6x represents the number of students passed in English, and 0.5x represents the number of students passed in mathematics, we can set up the following equation:
0.6x - 35 = 0.5x
0.6x - 0.5x = 35
0.1x = 35
x = 35 / 0.1
x = 350
Therefore, the total number of examinees is 350.
Now, substitute the value of x back into the equation for the number of students passed in English only:
Number of students passed in English only = 0.6x - 35
Number of students passed in English only = 0.6 * 350 - 35
Number of students passed in English only = 210 - 35
Number of students passed in English only = 175
Hence, 175 students passed in English only.
PLEASE HELPP!!
Thanks in advance!
The domain and the range of the rational function are all real numbers.
How to find the domain and the range of a function
In this problem we got the case of a rational function, whose domain and range must be found by means of algebra properties. The domain of rational functions is all x-values except x-values such that denominator becomes zero and its range is all real numbers.
Domain:
f(x) = (x² - 4) / (x + 2)
f(x) = [(x - 2) · (x + 2)] / (x + 2)
f(x) = x - 2
The rational function is in reality a linear function because all discontinuities are evitable. Then, both the domain and the range are all real numbers.
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Two forces of
407
N and
608
N act on an object. The angle between the forces is
48°.
Find the magnitude of the resultant and the angle that it makes
with the larger force.
We are given that:Two forces of 407 N and 608 N act on an object. The angle between the forces is 48°.
We are supposed to determine the magnitude of the resultant and the angle that it makes with the larger force.Let us first resolve the forces into their vertical and horizontal components:Draw a diagram for better understanding:
[asy]
size(150);
draw((0,0)--(1.732,1));
draw((0,0)--(1.075,-1.745));
draw((0,0)--(1.732,0));
draw((1.732,1)--(1.732,0));
draw((1.732,0)--(1.732,-1.5));
draw((1.732,-1.5)--(1.732,-1.745));
label("$407N$",(1.075,-1.745),SW);
label("$608N$",(1.732,1),N);
label("$48^\circ$",(0.4,0.1));
[/asy]The horizontal component of the 407 N force is:407 cos 48° = 267.2 NThe vertical component of the 407 N force is:407 sin 48° = 311.4 NThe horizontal component of the 608 N force is:608 cos 48° = 399.4 NThe vertical component of the 608 N force is:608 sin 48° = 464.9 N.
Now, we can add up the horizontal and vertical components of the forces separately to find the resultant:Horizontal component = 267.2 + 399.4 = 666.6 NVertical component = 464.9 - 311.4 = 153.5 NThe magnitude of the resultant is given by the Pythagorean theorem:R = sqrt(666.6² + 153.5²)R = 687.6 N (rounded to one decimal place)The angle that the resultant makes with the larger force can be found using the inverse tangent function:tan θ = 153.5 / 666.6θ = tan⁻¹(153.5 / 666.6)θ = 12.9° (rounded to one decimal place).
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