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german
3 years ago
11

PLS PLS PLS PLS HELP 15 POINTS

Mathematics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

560 students like the new menu :)

Step-by-step explanation:

700 times .8

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Pre image ABCD was dilated to produce image A’B’C’D why is the scale factor from the pre image? Enter your answer in the box as
shepuryov [24]

The scale factor of the dilation from ABCD to A′B′C′D′ is 3.

Step-by-step explanation:

Step 1:

In the pre-image ABCD, the length of one of the sides is given as 14 units.

For the other shape A′B′C′D′, the same side as the previous shape is given as 8 units.

Step 2:

To determine the scale factor, we divide the measurement after scaling by the same measurement before scaling.

In this case, it is the given length of the sides CD and C′D′.

So the scale factor = \frac{C^{1} D^{1} }{CD} = \frac{8}{14} = \frac{4}{7} .

So the shape ABCD is dilated by a scale factor of \frac{4}{7} to produce the shape A′B′C′D′.

7 0
3 years ago
20 POINTS! HELP!
ziro4ka [17]
Im not entirely sure but i thinks its c

6 0
3 years ago
Read 2 more answers
4. Mario was asked to find the GCF of 15, 30, and 45. What number did he find? A. 15 C. 45 B. 30 D. 60
SVETLANKA909090 [29]
4. is A 15
5. is C 12
6. is B 4, 8, 12, 16

hope this helps :)
7 0
3 years ago
Read 2 more answers
What is 3(2x-6)-11=4(x-3)+6
navik [9.2K]

Answer:

x=11.5

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2
evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}

Step # 1 : Swap the first and second matrix rows,

\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}

Step # 2 : Cancel leading coefficient in row 2 through R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1,

\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0
3 years ago
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