Answer:
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.
Answer:
y = 2x - 3
Step-by-step explanation:
Slope intercept equation of a line:
y = mx + b
(0, -3) is the y-intercept, so b = -3
We have
y = mx - 3
Now we use the two points to find the slope.
slope = m = (-3 - 5)/(0 - 4) = -8/(-4) = 2
y = 2x - 3
20% decreased and it needs to increase 40,000 more to get To its normal value
What do we know about those two lines?
They are perpendicular, meaning they have the same slope.
We know the slope of both is not zero (neither is vertical).
Therefore either
1) Both slopes are positive and therefore the product is positive
2) Both slopes are negative and therefore the product is positive (minus by a minus is a plus)
For the y intercepts, we know that the line P passes through the origin.
Therefore its Y intercept is zero.
[draw it if this is not obvious and ask where does it cross the y axis]
Therefore the Y intercept of line K and line P is zero.
[anything multiplied by a zero is a zero]
So we know that the product of slopes is positive, and we know that the product of Y intercepts is zero.
So the product of slopes must be greater.
Answer A
