Here is the region of integration of the integral integral from negative 6 to 6 integral from x squared to 36 integral from 0 to 36 minus y dz dy dx. rewrite the integral as an equivalent integral in the following orders. a. dy dz dx by. dy dx dz c. dx dy dz d. dx dz dy e. dz dx dy

a) ∫_{-6}^{6} ∫_{0}^{36} ∫_{x²}^{36} (-y) dy dz dx

b)  ∫_{0}^{36} ∫_{-6}^{6} ∫_{x²}^{36} (-y) dy dx dz

c)  ∫_{0}^{36} ∫_{x²}^{36} ∫_{-6}^{6} (-y) dx dy dz

e) ∫_{x²}^{36} ∫_{-6}^{6} ∫_{0}^{36} (-y) dz dx dy

Step-by-step explanation:

We write the equivalent integrals for given integral,

we get:

a) ∫_{-6}^{6} ∫_{0}^{36} ∫_{x²}^{36} (-y) dy dz dx

b)  ∫_{0}^{36} ∫_{-6}^{6} ∫_{x²}^{36} (-y) dy dx dz

c)  ∫_{0}^{36} ∫_{x²}^{36} ∫_{-6}^{6} (-y) dx dy dz

e) ∫_{x²}^{36} ∫_{-6}^{6} ∫_{0}^{36} (-y) dz dx dy

We changed places of integration, and changed boundaries for certain integrals.

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d

step-by-step explanation:

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1