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emmasim [6.3K]
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.

Mathematics
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
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In 32 hours they will cost the same
7 0
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

6 0
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
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