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Varvara68 [4.7K]

3 years ago

7

a square is cut from rectangle the side length of the square is half of the unknown with W the area of the Shaded region is 84 s

quare inches write an equation you can use to find the width in inches

1 answer:

Brilliant_brown [7]3 years ago

6 0

Answer:

Step-by-step explanation:

Just give me the answer

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which ordered pairs in the form (x, y) are solutions to the equation 3x−4y=21 ? Select each correct answer. (−1, −6) (−3, 3) (11

Valentin [98]

Answer:

see explanation

Step-by-step explanation:

To determine which ordered pairs are solutions to the equation

Substitute the x and y values into the left side of the equation and if equal to the right side then they are a solution.

(- 1, - 6)

3(- 1) - 4(- 6) = - 3 + 24 = 21 = right side ← thus a solution

(- 3, 3)

3(- 3) - 4(3) = - 9 - 12 = - 21 ≠ 21 ← not a solution

(11, 3)

3(11) - 4(3) = 33 - 12 = 21 = right side ← thus a solution

(7, 0)

3(7) - 4(0) = 21 - 0 = 21 = right side ← thus a solution

The ordered pairs (- 1, - 6), (11, 3), (7, 0) are solutions to the equation

3 0

3 years ago

Read 2 more answers

What is the best estimate for the width of your index finger

Len [333]

I don’t know for sure by 1 cm is a good guess.

8 0

3 years ago

Carley beat the school record for the 400 meter run by 1.3 seconds. If she ran the race in 55.7 seconds, what was the record.

dybincka [34]

Answer:

57 seconds

Step-by-step explanation:

55.7 + 1.3 = 57 seconds

8 0

2 years ago

A 20% acid solution is mixed with 20 liters of a 45% acid solution. How much of a 30% acid solution will result?

vaieri [72.5K]

Answer:

50 liters

Step-by-step explanation:

If x is the volume of 20% acid, then:

x (0.20) + 20 (0.45) = (x + 20) (0.30)

0.2x + 9 = 0.3x + 6

3 = 0.1x

x = 30

30 liters of 20% acid are needed, so there will be a total of 50 liters of 30% acid.

6 0

3 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago