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3 years ago

PLEASE HELP PLEASE HELP

Mathematics

1 answer:

BARSIC [14]3 years ago

4 0

Answer:

<h2>Right 2 units and Up 3 units</h2>

Step-by-step explanation:

Look at the picture.

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Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

7 0

3 years ago

I need to know what the answer is please ?

vova2212 [387]

???????????? I don’t get it?

7 0

2 years ago

What is the solution of the equation?

elena-s [515]

Answer:

Step-by-step explanation:

its c=5

6y+18+3=51

6y+21=51

6y=51-21

6y=30

y=5

4 0

3 years ago

Read 2 more answers

A circular garden has a circumference of

omeli [17]

This solution to this problem is predicated on the fact that the circumference is just: $2\pir$ . A straight line going through the center of the garden would actually be the diameter, which is well known to be two times the radius of the circle, so we can say that the circumference is just:

$d\pi$

So, solving for both the radius and the diameter gives us:

$d\pi=8xy^2\pi\\ d=8xy^2\\ \\ 2\pi r=8xy^2\pi\\ r=4xy^2$

So, the length of thes traight path that goes through the center of the guardain is just $8xy^2$ , and we can use the radius for the next part of the problem.

The area of a circle is $\pi r^2$ , which means we can just plug in the radius and find our area:

$A=\pi r^2 \implies \\ A=\pi (4xy^2)^2\implies \\ A= 16\pi x^2y^4$

So, we have found our area( $16 \pi x^2 y^4$ ) and the problem is done.

6 0

3 years ago

Alina spent no more than $45 on gas for a road trip. The first gas station she used charged $3.50 per gallon and the second gas

Paraphin [41]

Answer: 3.50x + 4.00y ≤ 45

0 < y < 11.25

<u>Step-by-step explanation:</u>

$3.50x + 4.00y \leq 45\\\\\\\text{Subtract 3.50x from both sides:}\\4.00y\leq -3.50x+45\\\\\\\text{Divide everything by 4.00 (to isolate y):}\\\dfrac{4.00y}{4.00}\leq \dfrac{-3.50x}{4.00}+\dfrac{45}{4.00}\\\\\\y\leq-0.875x+11.25\\\\\\\text{Both x and y must be greater than zero, so:}\\0$

8 0

2 years ago

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