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dlinn [17]
3 years ago
15

PQ is parallel to RS. PR and PQ are perpendicular to PQ and RS. The ratio of the lengths of and is : .

Mathematics
1 answer:
Bad White [126]3 years ago
7 0
Yeah baby I’m going home now I love you so so beautiful I love you
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Please Help. Find the area. Thank you
Andru [333]

Answer:

432 yd squared

Step-by-step explanation:

This is a parallelogram. The area of a parallelogram is denoted by: A = bh, where b is the base and h is the height. Here, the base is b = 19 1/5 and the height is h = 22 1/2. Plug these in:

A = bh = (19 1/5) * (22 1/2)

To make this simpler, let's convert the numbers into decimals. 1/5 is just 0.2 so 19 1/5 is 19.2. 1/2 is just 0.5, so 22 1/2 is 22.5. Now we have:

A = 19.2 * 22.5 = 432

Thus the area is 432 yd squared.

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2 years ago
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A bag contains 5 red marbles and 4 green marbles. What is the probability of choosing a red marble then a green marble, without
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Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}

so we conclude that
q(\frac{\sqrt{19} }{2} ) = 1/4

therefore

p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right)  \right)

plug x=\sqrt{19}/2 into p( q(x) ) to get answer

p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left(  \frac{\sqrt{19} }{2}\right)^2 }{ \left(  \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow

\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow  -\dfrac{6\sqrt{19} }{361}

p(1/4) = -\dfrac{6\sqrt{19} }{361}
3 0
3 years ago
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