Question:
Find the constant of proportionality k. Then write an equation for the relationship between x and y
![\begin{array}{ccccc}x & {2} & {4} & {6} & {8} \ \\ y & {10} & {20} & {30} & {40} \ \ \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccccc%7Dx%20%26%20%7B2%7D%20%26%20%7B4%7D%20%26%20%7B6%7D%20%26%20%7B8%7D%20%5C%20%5C%5C%20y%20%26%20%7B10%7D%20%26%20%7B20%7D%20%26%20%7B30%7D%20%26%20%7B40%7D%20%5C%20%5C%20%5Cend%7Barray%7D)
Answer:
(a) ![k = 5](https://tex.z-dn.net/?f=k%20%3D%205)
(b) ![y = 5x](https://tex.z-dn.net/?f=y%20%3D%205x)
Step-by-step explanation:
Given
![\begin{array}{ccccc}x & {2} & {4} & {6} & {8} \ \\ y & {10} & {20} & {30} & {40} \ \ \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccccc%7Dx%20%26%20%7B2%7D%20%26%20%7B4%7D%20%26%20%7B6%7D%20%26%20%7B8%7D%20%5C%20%5C%5C%20y%20%26%20%7B10%7D%20%26%20%7B20%7D%20%26%20%7B30%7D%20%26%20%7B40%7D%20%5C%20%5C%20%5Cend%7Barray%7D)
Solving (a): The constant of proportionality:
Pick any two corresponding x and y values
![(x_1,y_1) = (2,10)](https://tex.z-dn.net/?f=%28x_1%2Cy_1%29%20%3D%20%282%2C10%29)
![(x_2,y_2) = (6,30)](https://tex.z-dn.net/?f=%28x_2%2Cy_2%29%20%3D%20%286%2C30%29)
The constant of proportionality k is:
![k = \frac{y_2 - y_1}{x_2 - x_1}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D)
![k = \frac{30-10}{6-2}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B30-10%7D%7B6-2%7D)
![k = \frac{20}{4}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B20%7D%7B4%7D)
![k = 5](https://tex.z-dn.net/?f=k%20%3D%205)
Solving (b): The equation
In (a), we have:
![(x_1,y_1) = (2,10)](https://tex.z-dn.net/?f=%28x_1%2Cy_1%29%20%3D%20%282%2C10%29)
k can also be expressed as:
![k = \frac{y- y_1}{x- x_1}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7By-%20y_1%7D%7Bx-%20x_1%7D)
Substitute values for x1, y1 and k
![5 = \frac{y- 10}{x- 2}](https://tex.z-dn.net/?f=5%20%3D%20%5Cfrac%7By-%2010%7D%7Bx-%202%7D)
Cross multiply:
![y - 10 = 5(x - 2)](https://tex.z-dn.net/?f=y%20-%2010%20%3D%205%28x%20-%202%29)
Open bracket
![y - 10 = 5x - 10](https://tex.z-dn.net/?f=y%20-%2010%20%3D%205x%20-%2010)
Add 10 to both sides
![y - 10 +10= 5x - 10+10](https://tex.z-dn.net/?f=y%20-%2010%20%2B10%3D%205x%20-%2010%2B10)
![y = 5x](https://tex.z-dn.net/?f=y%20%3D%205x)
Answer:
4
Step-by-step explanation:
$17
17+11 = $28
28+11 = $39
39+11 = $50
17 + 11(n-1) $$
n-1 because removes the first class of $17
9514 1404 393
Answer:
(x +6)^2 +(y -10)^2 = 225
Step-by-step explanation:
The standard form equation for a circle is ...
(x -h)^2 + (y -k)^2 = r^2
where the center is (h, k) and the radius is r.
The center of a circle is the midpoint of any diameter. The midpoint between two points is the average of their coordinates.
((-15, -2) +(3, 22))/2 = (-15+3, -2+22)/2 = (-6, 10)
The radius can be found using the distance formula, or by simply putting one of the given points in the equation for the circle to see what the constant (r^2) needs to be.
(x -(-6))^2 +(y -10)^2 = (-15-(-6))^2 +(-2-10)^2
(x +6)^2 +(y -10)^2 = 81 +144 = 225
The equation of the circle is ...
(x +6)^2 +(y -10)^2 = 225
Answer: D) 101
Step-by-step explanation:
By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out
![\int\limits^{10}_{-1} {(2x+0.5f'(x))} \, dx =\int\limits^{10}_{-1} {2x} \, dx +0.5\int\limits^{10}_{-1} {f'(x)} \, dx =x^2|^{^{10}}_{_{-1}}+0.5f(x)|^{^{10}}_{_{-1}}\\=(100-1)+0.5(f(10)-f(-1))=99+0.5(8-4))=101](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B10%7D_%7B-1%7D%20%7B%282x%2B0.5f%27%28x%29%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E%7B10%7D_%7B-1%7D%20%7B2x%7D%20%5C%2C%20dx%20%2B0.5%5Cint%5Climits%5E%7B10%7D_%7B-1%7D%20%7Bf%27%28x%29%7D%20%5C%2C%20dx%20%3Dx%5E2%7C%5E%7B%5E%7B10%7D%7D_%7B_%7B-1%7D%7D%2B0.5f%28x%29%7C%5E%7B%5E%7B10%7D%7D_%7B_%7B-1%7D%7D%5C%5C%3D%28100-1%29%2B0.5%28f%2810%29-f%28-1%29%29%3D99%2B0.5%288-4%29%29%3D101)
I used the table for values of f(x) at 10 and -1. Wouldn't be surprised if this was part of a series of questions about f because I really can't see how you could use the hypothesis that f is twice differentiable on R. Same for the other table values. I'm curious about how you found the answer. Was it a different way?