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andrey2020 [161]
3 years ago
5

Plz help me fastttt it surface area of a rectangular prism

Mathematics
1 answer:
horrorfan [7]3 years ago
4 0

Answer:

72.2 square meter

Step-by-step explanation:

(3.8)(3.8)= 14.44

(14.44)(5)=72.2

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Natalee wanted to demonstrate the volume of a square pyramid. To do this, she took a hollowed out pyramid that has a base of 25
swat32

Answer:

The answer is below

Step-by-step explanation:

Natalee transferred water from a square pyramid to a cube. To calculate how many times she will need to dump the water from the pyramid into the cube to completely fill the cube, we divide the volume of the cube by the volume of the square pyramid. Hence:

Number of times = volume of cube / volume of pyramid

The perimeter of the pyramid base = 25 in, hence the length of one side of the bae = 25 / 4 = 6.25 in

Volume of square pyramid  = base² × (height / 3) = (6.25 in)² * (5 in / 3) = 65.1 in³

Volume of cube = 125 in³

Number of times = 125 in³ / 65.1 in³ = 1.92

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3 years ago
Can someone help me with this pls i’ll give brainliest
Rom4ik [11]

Answer: C

Step-by-step explanation:

1 1/8 = 1.125

3/4 = .75

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3 years ago
Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI
cricket20 [7]

Answer:

The system has infinitely many solutions

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

  1. Switch any two rows
  2. Multiply a row by a nonzero constant
  3. Add one row to another

To solve the following system

\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}

Step 1: Transform the augmented matrix to the reduced row echelon form

\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

which corresponds to the system

\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}

The system has infinitely many solutions.

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

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