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2 years ago

7x ( 3 x 4) = ( 7 x 3) x4

Mathematics

1 answer:

WARRIOR [948]2 years ago

4 0

Commutative, because no matter how you group the equation you will still get to same answer

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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

Which is the best estimate for the sum 2/9 +10/11 ?

irakobra [83]

Answer:

Step-by-step explanation:

6 0

2 years ago

Kinda confused on what the sum would be?!? HELP NOWW PLEASE

valentinak56 [21]

That is b. aka 2
thanks

4 0

3 years ago

Don’t troll I need a answer!

Andre45 [30]

A)(2x+48x)-7 (6x•9x)+32

6 0

2 years ago

Read 2 more answers

The point (-3, 4) is on the line x+4y=13.

lys-0071 [83]

Answer:

yes

Step-by-step explanation:

To determine if the point lies on the line substitute the coordinates into the left side of the equation and if equal to the right side then the point lies on the line

- 3 + (4 × 4) = - 3 + 16 = 13 = right side

Hence (- 3, 4) lies on the line x + 4y = 13

5 0

2 years ago