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3 years ago

A cone and cylinder have the same height and their bases are congruent circles. If the

Mathematics

2 answers:

amm18123 years ago

8 0

Answer:

h=V πr2

Step-by-step explanation:

1: Find the volume of the cylinder. Volume: 120 = h=V πr2

2: Since there is no radius the answer will be that and we also finding the for the cone.

3: Volume of the cone will be h=V πr2 since we finding for the volume for both.

Hope this helps.

olganol [36]3 years ago

8 0

Answer:

H=V ||r2

Step-by-step explanation: Hope you have several wonderful day

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Lewis fills his thermos with 2 liter of water. Garret fill his thermos with 1 of water. How many more milliliter of water does l

cluponka [151]

For this case we have to:

Lewis carries 2 liters of water

Garret carries 1 liter of water

So, Lewis has 1 liter of water more than Garret.

By definition we have that 1 liter equals 1000 milliliters.

Thus, Lewis carries 1000 milliliters more water than Garret.

Answer:

1000 milliliters

7 0

3 years ago

Determine the area of a,b and c. a to b=7.5 inches a to d= 16.8 inches

Alex777 [14]

The area would be 63.The steps to how I got that anwser:
1.] Write your formula
A=bh
----
2
2.]Begin by plugging in your measurements
A=7.5 times 16.8
-------------------
2
3.]Multiply the top two numbers together
A=126
-----
2
4.]Divide to get your final anwser
A=63

3 0

4 years ago

Please help if you can pic attatched

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

$CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Subtract row 3 from row 1:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]$

Subtract row 3 from 7 times row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]$

Divide row 3 by 17:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 2 of row 3 from row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 6 of row 2 from row 1:

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

$C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]$

$C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]$