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3 years ago

A prism has a volume of 64. If the dimensions of the prism are tripled, what is the volume of the new prism?

Mathematics

1 answer:

stepladder [879]3 years ago

5 0

Answer:

C 1728

Step-by-step explanation:

since a volume always goes with the cubic (power of 3) units, tripling all dimensions (in all 3 dimensions / coordinate directions) puts this factor of 3 also to the power of 3 = 27

so, the volume increases by the factor of 27.

64×27 = 1728

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A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. Each

Alecsey [184]

Answer:

There were 7 small boxes and 14 large boxes shipped.

Step-by-step explanation:

This problem may be solved by a system of equations:

I am going to say that:

x is the number of small boxes used

y is the number of large boxes used

There were twice as many large boxes shipped as small boxes shipped

This means that y = 2x

Each small box of paper weighs 45 pounds and each large box of paper weighs 80 pounds. The total weight of all boxes was 1435 pounds.

This means that 45x + 80y =1435

So we have to solve the following system:

y=2x

45x + 80y =1435

45x + 80 (2x) = 1435

205x = 1435

x= 1435/205

x=7

y=2x= 2(7) = 14

There were 7 small boxes and 14 large boxes shipped.

5 0

3 years ago

Ben and Landon took turns driving on the recent 820 mile road trip Ben averaged 60 mph while Landon average 56 mph if the trip t

AysviL [449]

Answer:

9 hours

Step-by-step explanation:

Let :

Landon hours = x

Ben hours = y

x + y = 14 - - (1)

60x + 56y = 820 - - - (2)

From (1)

y = 14 - x

60x + 56(14 - x) = 820

60x + 784 - 56x = 820

4x = 820 - 784

4x = 36

x = 36/4

x = 9

Landon drove for 9 hours

4 0

3 years ago

Solve each expression.reduce to simplest form of 5/8x4/5

Butoxors [25]

5/8 * 4/5
= 1/8 * 4/1
= 1/4 * 2/1
1/2 * 1/1

Final answer = 1/2

4 0

3 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

3 years ago

Yuki has saved £100 and spends £22 on a present. Rosetta has saved £140 and spends £28 on a present. What percentage has each gi

grigory [225]

Answer:

yuki has spent 22 percent , Rosetta spent 20 percent

step by step explanation:

1) if £100=100percent

then £22=x percent

x=(22×100)/100=22 percent

2) if £140=100 percent

then £28= x percent

x=(28×100)/140

x=2800/140

x=20 percent

6 0

3 years ago