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4 years ago

)Figure XYZ is reflected about the y-axis to obtain figure X’Y’Z’:

2 answers:

8 0

B. Figure XYZ is congruent to figure X'Y'Z'
All you are doing is rotating the figure so therefore you are not changing the shape or size.
Hope this helps you.

stellarik [79]4 years ago

8 0

Answer:

Step-by-step explanation:

edge 2020

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NO LINKS NO FILES. After you add 3 to both sides of the equation 27 = m/6 - 3, what should you do next?

a_sh-v [17]

Answer:

Multiply each side by 6

Step-by-step explanation:

27 = m/6 - 3

Add 3 to each side

27+3 = m/6 - 3+3

30 = m/6

Multiply each side by 6

30*6 = m/6*6

180 = m

3 0

3 years ago

F.O.I.L

user100 [1]

If you are talking about what is in between the two sets of parentheses then the answer would be multiplication hope that helped

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3 years ago

Read 2 more answers

If a soccer team made 37 penalty shots out of 100, what fraction and what percentage of penalty shots did the team miss? choose

Pie

The correct answer is letter c

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3 years ago

Read 2 more answers

In a basketball skills competition, Corson successfully made 5 of 7 shots and Charleigh made 3 of 5

xxTIMURxx [149]

Answer:

Corson had a better performance at the skills competition.

Step-by-step explanation:

In order to determine which player had a better performance at the skills competition, we need to find the percentage performance of both the players.

Therefore,

Corson's percentage performance

$=\frac{5}{7}\times 100= 71.43%$

Charleigh's percentage performance

$=\frac{3}{5}\times 100= 60%$

$\because$ Corson's percentage performance > Charleigh's percentage performance

So, Corson had a better performance at the skills competition.

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3 years ago

Jane’s buying clothes. She can get 5 shirts & 4 sweaters for $229 or she can get 2 shirts & 3 sweaters for $128. How muc

makkiz [27]

Cost of 1 shirt is $\$25$ and cost of one sweater is $\$26$

<u>Solution: </u>

Given that Jane’s buying clothes.

She can get 5 shirts & 4 sweaters for $\$229$

She can get 2 shirts & 3 sweaters for $\$128$

Need to determine cost of 1 shirt and 1 sweater.

Let assume cost of 1 shirt = $\$x$

And assume cost of 1 sweater = $\$y$

Let us create equations from give conditions

First condition is Cost of 5 shirt + cost of 4 sweater = $\$229$

$\begin{array}{l}{\text { cost of 5 shirts }=5 \times \text { cost of 1 shirt }=5 \times x=5 x} \\\\ {\text { cost of 4 sweater }=4 \times \text { cost of 1 sweater }=4 \times y=4 y}\end{array}$

So using first condition equation which we get is as follows

$5 x+4 y=229 \rightarrow (1)$

Second condition is Cost of 5 shirt + cost of 4 sweater = $\$128$

$\begin{array}{c}{\text { cost of } 2 \text { shirts }=2 \times \text { cost of } 1 \text { shirt }=2 \times x=2 x} \\\\ {\text { cost of } 3 \text { sweater }=3 \times \text { cost of } 1 \text { sweater }=3 \times y=3 y}\end{array}$

So using second condition equation which we get is as follows

$2 \mathrm{x}+3 \mathrm{y}=128\quad\rightarrow (2)$

Now we have two equations to be solved

$\begin{array}{l}{5 x+4 y=229 \rightarrow (1)} \\\\ {2 x+3 y=128 \rightarrow (2)}\end{array}$

On multiplying equation (1) by 2 and equation (2) by 5 to make coefficients of x equal in both equations, we get

$\begin{array}{l}{2 x(5 x+4 y)=2\times 229} \\\\ {10 x+8 y=458 \rightarrow (3)} \\\\ {5 x(2 x+3 y)=5 \times 128} \\\\ {10 x+15 y=640 \rightarrow (4)}\end{array}$

On subtracting (3) from (4), we get

$\begin{array}{l}{(10 \mathrm{x}-10 \mathrm{x})+(15 \mathrm{y}-8 \mathrm{y})=640-458} \\\\ {\Rightarrow 7 \mathrm{y}=182} \\\\ {\Rightarrow \mathrm{y}=\frac{182}{7}=26}\end{array}$

On substituting y = 26 in equation (1) we get

$\begin{array}{l}{\Rightarrow 5 \mathrm{x}+(4 \times 26)=229} \\\\ {\Rightarrow 5 \mathrm{x}=229-104} \\\\ {\Rightarrow 5 \mathrm{x}=125} \\\\ {\Rightarrow \mathrm{x}=\frac{125}{5}=25}\end{array}$

Cost of 1 shirt = $\$x = \$25$

Cost of 1 sweater = $\$y = \$26$

7 0

4 years ago