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.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

its 50 miles away from the starting point