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3 years ago

7

Gary used candle molds, as shown, to make candles that were perfect cylinders and spheres:What is the approximate difference in

the amount of wax needed to make a candle from each of these molds? Use π = 3.14.

1 answer:

Nikolay [14]3 years ago

6 0

The correct answer is 16.7

Hope this helped :)

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painter A charges $376 plus $12 per and painter B charges $280 plus $15 per hour. in how many hours will they cost the same

sladkih [1.3K]

In 32 hours they will cost the same

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2 years ago

Read 2 more answers

How do i simplify <br><br>16v. 10<br>----- . ------<br>6v. 7

viva [34]

Answer:

Step-by-step explanation:

16v/6v=16/6=8/3

10/7=1 3/7

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3 years ago

According to a soccer coach, 75% of soccer players have had at least one sprained ankle. An athletic trainer would like to inves

Greeley [361]

Answer:

H0 : μ = 0.75

H1 : μ > 0.75

Step-by-step explanation:

Given :

Sample size, n = 125

x = 99

Phat = x / n = 99 / 125 = 0.792

Population proportion, P = 0.75

The hypothesis :

Null hypothesis :

H0 : μ = 0.75

Alternative hypothesis ;

Egates the null hypothesis ; since the sample proportion is greater than the the population proportion or claim "; then we use the greater than sign.

H1 : μ > 0.75

5 0

2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

WILL MARK BRAINLIEST<br>solve for x: m(2m-x)=3(x+6) if m≠3

Lera25 [3.4K]

Answer:

x=2m−6

Step-by-step explanation:

4 0

3 years ago