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4 years ago

Select the statement that is true about squares from the choices below.

1 answer:

7 0

The answer is (C). All sides and angles of the square is equal. Its diagonals bisect each other and meet at 90°. If two lines are said to meet at 90°, these lines are perpendicular to each other. Therefore, the diagonals of a square is perpendicular to each other

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The Astronomical Telescope shop plans to introduce a new model based on the following information: Rent and utilities per period

dimaraw [331]

Using the profit concept, it is found that the break-even point is of 61.6 units.

Profit is given by revenue subtracted by cost.
The break-even point is when the revenue is the same as the cost, which means that the profit is zero.

In this problem:

Variable cost per unit x of $217, fixed cost of $5,420, thus, the cost equation is given by:

$C(x) = 217x + 5420$

Selling price per unit of $305, thus, the revenue equation is given by:

$R(x) = 305x$

The break-even point is the <u>value of x</u> for which:

$R(x) = C(x)$

Then

$305x = 217x + 5420$

$88x = 5420$

$x = \frac{5420}{88}$

$x = 61.6$

The break-even point is of 61.6 units.

A similar problem is given at brainly.com/question/4001746

5 0

3 years ago

Kim buys six pet gerbils. Every 2 weeks, the number of gerbils doubles. Kim's friend Jared also purchased 6 gerbills. Every two

Ber [7]

The answer is four weeks. Do you want an explanation?

7 0

3 years ago

The high school soccer booster club sells tickets to the varsity matches for $4 for students and $8 for adults. The booster club

tangare [24]

Answer:(B)

Step-by-step explanation:

it is because for the people your final price was 200 you don't know how much money the people donated

8 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

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