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solmaris [256]
3 years ago
7

A pair of corresponding side of two similar pentagons have length of 12cm and 28cm

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
5 0
The ratio of their areas = 12^2 / 28^2
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Find two mixed numbers so that the sum is 15 3/10 and the difference is 8 5/10
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11 9/10 and 3 2/5

\left \{ {{ 11\frac{9}{10}+3\frac{2}{5}=15\frac{3}{10}} \atop { 11\frac{9}{10}-3\frac{2}{5}=8\frac{5}{10}}} \right.

To solve for this, you can create a series of two equations
\left \{ {{y+x=15 \frac{3}{10} } \atop {y-x= 8\frac{5}{10} }} \right.

After that, you solve one of the equations for x and plug it into the other equation.
\left \{ {{x=15 \frac{3}{10}-y } \atop {y-x= 8\frac{5}{10} }} \right.

{2y-15 \frac{3}{10}= 8\frac{5}{10}

Then, solve for variable y
{2y=23 \frac{8}{10}
{y=11\frac{9}{10}

Then, you plug in this value for y into one of the original equations

11\frac{9}{10}+x=15 \frac{3}{10}

x=15 \frac{3}{10} -11\frac{9}{10}

x=3 \frac{2}{5}
8 0
3 years ago
Solve for x, the figure is a parallelogram. will give brainliest, pls show work !!
MakcuM [25]

Answer:

x = 7

Step-by-step explanation:

In a parallelogram the opposite angles are congruent, then

14x + 8 = 106 ( subtract 8 from both sides )

14x = 98 ( divide both sides by 14 )

x = 7

6 0
2 years ago
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At Mike's Moving Supplies, cardboard boxes are sold in packs of 5. Depending on the size, the
QveST [7]
Should be like 200 mins
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3 years ago
Caroline rolls a fair dice 90 times.
Darya [45]

Answer:

Answer: 45

Step-by-step explanation:

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3 years ago
Which statement is true about the function f(x) = square root of x
BlackZzzverrR [31]

Answer:


Step-by-step explanation:

Correct choice is A


Step-by-step explanation:


Consider parent function  The domain of this function is  and the range of this function is


Thus, the function  has the domain  and the range


State the domain and the range of all given functions:


A. The domain:


B. The range:


C. The domain:


D. The range:  




Read more on Brainly.com - brainly.com/question/10648085#readmore

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