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OLga [1]
3 years ago
9

ASAP!!! Which side of the original parallelogram corresponds to the y-centimeter side of the enlarged parallelogram? Explain how

you know.
A small parallelogram has a left side length of 10 centimeters and a top side length of 15 centimeters. A larger parallelogram has a left side length of x centimeters and top side length of y centimeters.
Not drawn to scale

The ____-centimeter side corresponds to the y-centimeter side.
A.10 cm because both are the shorter sides of the parallelogram
B.10 cm because both are the longer sides of the parallelogram
C.15 cm because both are the shorter sides of the parallelogram
D.15 cm because both are the longer sides of the parallelogram
Mathematics
2 answers:
alina1380 [7]3 years ago
5 0

Answer is D

I took the test

garri49 [273]3 years ago
5 0

Answer:

d

Step-by-step explanation:

15 cm because both are the longer sides of the parallelogram

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Answer:

The two roots of the quadratic equation are

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Step-by-step explanation:

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Divide both sides by 9:

x^{2} - \frac{x}{3} - \frac{2}{9}=0

Add \frac{2}{9} to both sides to get rid of the constant on the LHS

x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}

This simplifies to

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

\left(x - \frac{1}{6}\right)^2 = x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}

So

\left(x - \frac{1}{6}\right)^2=\frac{1}{4}

Taking square roots on both sides

\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

So the two roots or solutions of the equation are

x - \frac{1}{6}=-\sqrt{\frac{1}{4}}  and x - \frac{1}{6}=\sqrt{\frac{1}{4}}

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So the two roots are

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and

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sesenic [268]
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