Y=-x^2+8<br> y= =3/4x

For every $5 that Micah saves, his parents give him $10.

Fittoniya [83]

Answer:

1:2

Step-by-step explanation:

5 0

2 years ago

The Wagner school store is having a sale! At the sale this week, a life sized stuffed animal of Wagner’s mascot is being sold fo

mash [69]

Answer:

$330

Step-by-step explanation:

Please see the attached picture for the full solution.

☆ There is no need to round the answer off to the nearest cents as the answer is a whole number.

8 0

2 years ago

2 questions Please helpppppp( 20 points. ) PICTURE BELOW <br> Helpp ASAP

SOVA2 [1]

Answer:

5) The third side must be more than 4 and less than 14.

6) The third side must be more than 3 and less than 13.

Step-by-step explanation:

The third side must be more than the difference and less than the sum of the two numbers.

5) 9-5=4

9+5=14

So the third side must be more than 4 and less than 14.

6) 8-5=3

8+5=13

So the third side must be more than 3 and less than 13.

6 0

2 years ago

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,

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

5 0

3 years ago

AA, B BB, and C CC are collinear, and B BB is between A AA and C CC. The ratio of A B ABA, B to B C BCB, C is 3 : 1 3:13, colon,

Katena32 [7]

Answer:

C = (1, -1)

Step-by-step explanation:

For that 3:1 division, the calculated value of point B is ...

B = (3C +1A)/(3+1)

Solving for the value of C, we find ...

4B = 3C + A . . . . multiply by 4

4B -A = 3C . . . . . subtract A; next divide by 3.

C = (4B -A)/3 = ((4(-1)-(-7))/3, (4(0)+-3)/3) . . . . substitute given values

C = (3/3, -3/3) . . . . simplify a bit

C = (1, -1)

7 0

2 years ago

Y=-x^2+8 y= =3/4x– 3