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meriva
3 years ago
15

The perimeter of the kite is 134 cm. Find the value of x.

Mathematics
1 answer:
frez [133]3 years ago
3 0

Answer:

x = 10 cm

Step-by-step explanation:

(3x + 2) + (3x + 2) + (4x - 5) + (4x - 5) = 134 cm

combine like terms:

14x - 6 = 134

add 6 to each side of the equation:

14x = 140

divide both sides by 14:

x = 10

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Answer:

<u><em>$4.50 or $4.5</em></u>

Step-by-step explanation:

<em><u>well, this is easy</u></em>

<em><u>30% equals to 0.30 right.</u></em>

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<em><u>so your wondering. What do we do?</u></em>

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3 years ago
Can u guys help me with this problem
netineya [11]

Answer:

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Step-by-step explanation:

-0.75-(-\frac{2}{5})+0.4+(-\frac{3}{4})

The opposite of -\frac{2}{5} is \frac{2}{5}

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Convert decimal number −0.75 to fraction -\frac{75}{100}.

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-\frac{3}{4} + \frac{2}{5}+0.4-\frac{3}{4}

Least common multiple of 4 and 5 is 20. Convert -\frac{3}{4} and \frac{2}{5} to fractions with denominator 20.

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Since -\frac{15}{20} and \frac{8}{20} have the same denominator, add them by adding their numerators.

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Add -15 and 8 to get -7

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Convert decimal number 0.4 to fraction \frac{4}{10}. Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

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-\frac{7}{20}+\frac{8}{20}-\frac{3}{4}

Since -\frac{7}{20} and \frac{8}{20} have the same denominator, add them by adding their numerators.

20-7+8-\frac{3}{4}

Add -7 and 8 to get 1.

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Least common multiple of 20 and 4 is 20. Convert \frac{1}{20} and \frac{3}{4} to fractions with denominator 20.

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Since \frac{1}{20} and \frac{15}{20} have the same denominator, subtract them by subtracting their numerators.

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Subtract 15 from 1 to get -14.

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Reduce the fraction -\frac{14}{20} to lowest terms by extracting and canceling out 2.

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Hope this helps! Brainliest would be much appreciated! Have a great day! :)

8 0
2 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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kifflom [539]

we'd do the same as before on this one as well.

if we take 27.99 to be the 100%, what is 12 off of it in percentage?

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3 years ago
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