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2 years ago

What is the perimeter of this pentagon?

Mathematics

1 answer:

aleksandrvk [35]2 years ago

3 0

Answer:

5.715 centimeters or 27 Inches

Step-by-step explanation:

You just have to add all the fractions together and then convert your final answer into cm, in, ft, etc...

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What is the image of (7, -3) after a reflection over the x-axis?

telo118 [61]

Answer:(7,3)

Step-by-step explanation:

5 0

2 years ago

3. Solve the system using elimination (not substitution or matrices). negative 2 x plus y minus 2 z equals negative 8A N D7 x pl

riadik2000 [5.3K]

Elimination Method

$\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}$

If we multiply the equation 3 by (-1) we obtain this:

$\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}$

If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z

1. Using the 3rd equation we can obtain X(Y,Z)

$\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\ \end{gathered}$

2. We can replace this value of X in the 1st and 2nd equations

$\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}$

3. If we simplify:

$\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}$

4. We can obtain Y from this two equations:

$\begin{gathered} Y=-\frac{-12Z+58}{9} \\ \end{gathered}$

5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)

$\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}$

6. If we simplify, we obtain:

$X=\frac{-3Z+7}{9}$

7. In conclusion, we obtain that

(X,Y,Z) =

$(\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)$

8 0

6 months ago

One angle of an isosceles triangle measures 90°. Which other angles could be in that isosceles triangle? Choose all that apply.

bonufazy [111]

Answer:

45°

Step-by-step explanation:

(180-90)/2 = 90/2 = 45°

7 0

2 years ago

Read 2 more answers

What is the area of the triangle

vagabundo [1.1K]

Answer:

150

Step-by-step explanation:

A=B*H/2

A=15*20/2

A=300/2

A=150

Ignore the hypotenuse btw

3 0

2 years ago

Read 2 more answers

Check answer please

Cerrena [4.2K]

The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

$\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}$

$\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}$

$\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}$

$\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}$

Hope it helps.

5 0

2 years ago