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3 years ago

6

Use the ratio go solve the percent problem

1 answer:

Drupady [299]3 years ago

5 0

Answer
40%
32 is what percent of 80? =40.

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DinElla’s test grades in her history class are 89, 94, 82, 84, and 98. What score must Ella make on her next test to have a mean

rjkz [21]

Just calculate it hahahaha

4 0

3 years ago

If h(x) = x – 7 and g(x) = x2, which expression is equivalent to (gxh)(5)??

vovikov84 [41]

We want g(h(5)).

First find h(5).

h(5) = 5 - 7

h(5) = -2

We now find g(-2).

g(x) = x^2

g(- 2) = (-2)^2

g(-2) = 4

In conclusion, g(h(5)) = 4.

7 0

3 years ago

Sketch the region enclosed by x+y2=12x+y2=12 and x+y=0x+y=0. Decide whether to integrate with respect to xx or yy, and then find

Paladinen [302]

Answer:

$A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}$

Step-by-step explanation:

For this case we have these two functions:

$x+y^2 = 12$ (1)

$x+y=0$ (2)

And as we can see we have the figure attached.

For this case we select the x axis in order to calculate the area.

If we solve y from equation (1) and (2) we got:

$y = \pm \sqrt{12-x}$

$y = -x$

Now we can solve for the intersection points:

$\sqrt{12-x} = -\sqrt{12-x}$

$12-x = -12+x$

$2x=24 , x=12$

$\sqrt{12-x} =-x$

$12-x = x^2$

$x^2 +x -12=0$

$(x+4)*(x-3) =0$

And the solutions are $x =-4, x=3$

So then we have in total 3 intersection point $x=12, x=-4, x=3$

And we can find the area between the two curves separating the total area like this:

$\int_{-4}^3 |\sqrt{12-x} - (-x)| dx +\int_{3}^{12}|-\sqrt{12-x} -\sqrt{12-x}|dx$

$\int_{-4}^3 |\sqrt{12-x} + x| dx +\int_{3}^{12}|-2\sqrt{12-x}|dx$

We can separate the integrals like this:

$\int_{-4}^3 |\sqrt{12-x} dx +\int_{-4}^3 x +2\int_{3}^{12}\sqrt{12-x} dx$

For this integral $\int_{-4}^3 |\sqrt{12-x} dx$ we can use the u substitution with $u = 12-x$ and after apply and solve the integral we got:

$\int_{-4}^3 |\sqrt{12-x} dx =\frac{74}{3}$

The other integral:

$\int_{-4}^3 x dx = \frac{3^2 -(-4)^2}{2} =-\frac{7}{2}$

And for the other integral:

$2\int_{3}^{12}\sqrt{12-x} dx$

We can use the same substitution $u = 12-x$ and after replace and solve the integral we got:

$2\int_{3}^{12}\sqrt{12-x} dx =36$

So then the final area would be given adding the 3 results as following:

$A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}$

6 0

3 years ago

3.25 3 2/5 3 3/8 from least to greatest

kirza4 [7]

Answer:

3.25, 3 2/5 , 3 3/8 is the order from least to greatest

3 0

4 years ago

The height of a rectangle is one-fourth the

Marrrta [24]

With h as height and width as w, h = 1/4*w

6 0

4 years ago