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4 years ago

Explain why you usually need to use to find the volume of a cylinder.

2 answers:

8 0

The volume of a cylinder is the amount of space that will fit inside it. You can use the formula for the volume of a cylinder to find that amount

gregori [183]4 years ago

3 0

Height, diameter and I think radius.

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Please help me please

baherus [9]

Answer:

The answer is C

4 0

3 years ago

Does anyone know how to solve this? I don't even know how to start it ;-; please help, thanks

Maru [420]

These shapes are geometric, which means they're equal. XW would be the same as XY, which is 18. The same goes towards WZ. WZ = ZY, so it'll be 29. Remember, since it's geometric, each side has to equal each other

8 0

3 years ago

A store diplays will have 6 rows,12 contaners in each row.if 39 contaners has been set UPS so farsa explain how to find the numb

torisob [31]

Multiply:
6•12=72

Then, you subtract 39 from 72:
72-39=33

33 containers would still need to be set up.

4 0

4 years ago

Is 10 the solution of the equation 4x - 8 = 31?

Tom [10]

4x-8 = 31

4x = 31 + 8

4x = 39

x = 9.75

6 0

3 years ago

Read 2 more answers

Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$

Hence,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1$

Substitute it into the second equation:

$23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0$

Substitute them into the first equation:

$3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2$

The solution is $x=2,\ y=0,\ z=-1.$

3 0

4 years ago