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Fynjy0 [20]
2 years ago
6

A car dealership paid $9,000 for a car. The sticker price is $10,395 for a customer to buy the car. a. What’s the markup in doll

ars? b. By what percent is the markup?
Mathematics
1 answer:
Sladkaya [172]2 years ago
7 0

Answer:

1) To determine how much the change of $9,000 to $10,395, we must subtract 9,000 from 10,395;

10,395 - 9,000 = $1,395 is the change in <u>price</u> (an increase).

Formula to find percentage of change;

Change in price/original amount x 100% (100/100)

Replace those values:

1,935 (change in price/increase)       100

_____                      x                        _____ =<u> </u><u>%0.215</u>

9,000(original amount)                      100

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Answer: 48x/64

Work (assuming ^ is multiplication):

12x^4/4x^16

12x ^ 4x = 48x

4 ^ 16 = 64

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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
Determine the constant of variation for the direct variation given.
ohaa [14]

Answer:

B. 5

Step-by-step explanation:

We have been given that R varies directly with S. When S is 16, R is 80. We are asked to find constant of variation.

We know that two directly proportional quantities are in form y=kx, where,

k = Constant of variation.

Upon substituting our given values, we will get:

R=k*S

80=k*16

\frac{80}{16}=\frac{k*16}{16}

5=k

Therefore, the constant of variation is 5 and option B is the correct choice.

6 0
3 years ago
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