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labwork [276]
3 years ago
15

UP

Mathematics
1 answer:
Anestetic [448]3 years ago
6 0

Answer:

................................................................................................................................................................................................

Step-by-step explanation:

A cylinder has a height of 14 centimeters and a radius of 17 centimeters. What is its volume? Use ​ ≈ 3.14 and round your answer to the nearest hundredth.

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You’re answer would be B
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Which is equivalent to √/10 * ?<br> O (10) 4x<br> O (10)³x<br> O (10) 4x<br> O (V10)³x
cricket20 [7]

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Step-by-step explanation:

D is the correct answer

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The area of a rectangle is 6x^3+4x^2−8x. The width of the rectangle is 2x. Find the length of the rectangle.
Vitek1552 [10]

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1776

Step-by-step explanation:

6 0
3 years ago
Determine whether the number is a perfect square P.S there is a lot
mihalych1998 [28]

Answer:

okay so to figure out a perfect square a number has to be outlined by itself. 8 is not. 4 is because 2 times itself is 4. 81 is because 9 time itself is 81. 44 is not. 49 is because 7 times itself is 49. 125 is not. 150 is not. 144 is because 12 times itself is 144.

6 0
3 years ago
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
hram777 [196]

Write the system in augmented-matrix form:

\left[\begin{array}{ccc|c}2&2&4&16\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]

Multiply through row 1 by 1/2:

\left[\begin{array}{ccc|c}1&1&2&8\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]

Add -1(row 1) to row 3, and add -5(row 1) to row 2:

\left[\begin{array}{ccc|c}1&1&2&8\\0&-7&-7&-41\\0&1&-5&-17\end{array}\right]

Swap rows 2 and 3:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&-7&-7&-41\end{array}\right]

Add -7(row 2) to row 3:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&-42&-160\end{array}\right]

Multiply through row 3 by -1/42:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&1&\frac{80}{21}\end{array}\right]

Add 5(row 3) to row 2:

\left[\begin{array}{ccc|c}1&1&2&8\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]

Add -1(row 2) and -2(row3) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-\frac53\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]

So the solution to the system is

x=-\dfrac53,y=\dfrac{43}{21},z=\dfrac{80}{21}

7 0
4 years ago
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