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sertanlavr [38]

3 years ago

7

a small bridge sit atop for cube shaped powers that all have the same volume. The combined volume of the four pillars is 503 how

many inches long are the size of the powers explain your answer

1 answer:

Natasha_Volkova [10]3 years ago

6 0

Answer:

60 inches long are the sides of the pillars.

Step-by-step explanation:

Given : A small bridge sits atop four cube shaped pillars that all have the same volume. the combined volume of the four pillars is 500 ft cubed.

To find : How many inches long are the sides of the pillars?

Solution :

Refer the attached picture below for Clarence of question.

The volume of the cube is $V=a^3$

Where, a is the side.

The combined volume of the four pillars is 500 ft cubed.

The volume of each cube is given by,

$V=\frac{500}{4}=125\ ft^3$

Substitute in the formula to get the side,

$125=a^3$

$a=\sqrt[3]{125}$

$a=5\ ft$

We know, 1 feet = 12 inches

So, 5 feet = $5\times 12=60$ inches

Therefore, 60 inches long are the sides of the pillars.

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Which of the following linear systems has a solution <br> (5,0)<br> SOMEONE PLS HELPPP

Korvikt [17]

Answer:

D

Step-by-step explanation:

out of the choices

A) 2x-y=-2

3x+2y=-10

B) x+4y=3

2x+5y=3

C) 2x+3y=5

x-4y=-14

D) 3x+4y=15

x+y=5

only the system in D has a solution of x=5 and y=0

solve by substitution x+y=5 so x=5-y

plug into the other equation so 3(5-y)+4y=15

distribute 15-3y+4y=15 then minus 15 on both sides and combine like terms

so y=0 then put that value back into the other equation

so x+y=5 equals x+0=5 so x=5 this means the solution to the system of equations is the point (5,0)

Hope this helps! :)

3 0

3 years ago

Read 2 more answers

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

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When Mya looked out at the store's parking lot, she noticed that for every 4 cars, there were 6 trucks. If there's a total of 22

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Answer:

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Step-by-step explanation:

This is a ratio problem, the ratio of cars to trucks

for every 4 cars, there are 6 trucks

represented as a ratio we have 4:6

1. how many of them are cars

applying the part to whole strategy we have

4+6 = 10

let cars be x

$\frac{4}{10}= \frac{x}{220} \\\\x= \frac{220*4}{10} \\\\x= 880/10\\\\x= 88 cars$

2. how many of them are trucks?

let trucks be y

$\frac{6}{10}= \frac{y}{220} \\\\x= \frac{220*6}{10} \\\\y= 1320/10 \\\\y= 132 trucks$

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