Which situation is best represented by the following equation?

2 answers:

6 0

D) Eric was paid $458 last week. He was paid a $40 bonus and $11 for each hour he worked. What is h, the number of hours Eric worked last week?

The total amount he was paid is what the addends add up to, which is $458. This eliminates options A and B. Furthermore, it can't be C because it says "$40 for each hour he worked". In the equation, it says 11h, not 40h. The remaining option is option D.

Bezzdna [24]3 years ago

6 0

Answer:

the answer is D

Step-by-step explanation:

You might be interested in

Use the given transformation to evaluate the integral. (15x + 15y) dA R , where R is the parallelogram with vertices (−1, 4), (1

MA_775_DIABLO [31]

Answer:

$\int_R 15x+15y dA = \frac{8}{16875}$

Step-by-step explanation:

Recall the following: x = 15u+15v, y = -60u+15v. So, x-y = 75u. Then u = (x-y)/75. 4x+y = 75v. Then v = (4x+y)/75.

We will see how this transformation maps the region R to a new region in the u-v domain. To do so, we will see where the transformation maps the vertices of the region.

(-1,4) -> ((-1-4)/75,(4(-1)+4)/75) = (-1/15, 0)

(1,-4)->(1/15,0)

(3,-2)->(1/15,2/15)

(1,6)->(-1/15,2/15)

That is, the new region in the u-v domain is a rectangle where $\frac{-1}{15}\leq u \leq \frac{1}{15}, 0\leq v \leq \frac{2}{15}$ .

We will calculate the jacobian of the change variables. That is

$\left |\begin{matrix} \frac{du}{dx}& \frac{du}{dy}\\ \frac{dv}{dx}& \frac{dv}{dy}\end{matrix}\right|$ (we are calculating the determinant of this matrix). The matrix is

$\left |\begin{matrix} \frac{1}{75}& \frac{-1}{75}\\ \frac{4}{75}& \frac{1}{75}\end{matrix}\right|=(\frac{1}{75^2})(1+4) = \frac{1}{15\cdot 75}$ (the in-between calculations are omitted).

We will, finally, do the calculations.

Recall that

$15x+15y = 15(15u+15v) + 15(-60u+15v) = (15^2-15\cdot 60 )u+2\cdot 15^2v = 15^2(-3)u+2\cdot 15^2 v$

We will use the change of variables theorem. So,

$\int_R 15x+15y dA = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}} 15^2(-3)u+2\cdot 15^2 v \cdot (\frac{1}{15^2\cdot 5}) dv du = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu$

This si because we are expressing the original integral in the new variables. We must multiply by the jacobian to guarantee that the change of variables doesn't affect the value of the integral. Then,

$\int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu = \int_{\frac{-1}{15}}^{\frac{1}{15}}\frac{-3}{5}u\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}}du = \frac{-3}{5}\left.\frac{u^2}{2}\right|_{\frac{-1}{15}}^{\frac{1}{15}}\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}} = \frac{8}{16875}$