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3 years ago

The sum of two numbers is 72. One number is 8

Mathematics

1 answer:

Tpy6a [65]3 years ago

7 0

Answer:

32 and 40

Step-by-step explanation:

Let's call the smaller number x. If the smaller number is x, the equation for this problem is 2x + 8 = 72

Equation:

2x + 8 = 72

2x = 64

x = 32

So now we know the smaller number is 32, so now add 8 and we get 40.

Therefore the 2 numbers are 32 and 40

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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 plot is located at (-3, -2) Please help!!!

DaniilM [7]

The is answer E, you have to search for -3 in the x line than search for -2 in the y line

7 0

3 years ago

Read 2 more answers

She has read 1/5 of the book already what fraction of the book does she need to read now?

arlik [135]

Answer:

4/5

Step-by-step explanation:

1-1/5=4/5

she still has 4/5 to read

7 0

3 years ago

2 over 3 n =-12 what value of n makes the equation true?

galina1969 [7]

Answer:

-1/18

Step-by-step explanation:

4 0

3 years ago

Solve the inequality 3u+3(u+1)>2u+7 and write the solution in interval notation.

strojnjashka [21]

Given:

The given inequality is $3u+3(u+1)>2u+7$

We need to determine the solution of the inequality in interval notation.

Solution of the inequality:

The solution of the inequality can be determined by simplifying the inequality.

Thus, we have,

$3u+3u+3>2u+7$

$6u+3>2u+7$

Subtracting both sides by 3, we get;

$6 u>2 u+4$

Subtracting both sides by 2u, we have;

$4 u>4$

Dividing both sides by 4, we get;

$u>1$

Writing it in interval notation, we get;

$(1, \infty)$

Thus, the solution of the inequality is $(1, \infty)$

5 0

3 years ago