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LiRa [457]
3 years ago
7

The leg of a right triangle is 3 units and the hypotenuse is 7 units. What is the length, in units, of the other leg of the tria

ngle?
4
Square root of 40
Square root of 58
10
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
4 0
A^2 + b^2 = c^2....where a and b are the legs, c is the hypotenuse
3^2 + b^2 = 7^2
9 + b^2 = 49
b^2 = 49 - 9
b^2 = 40.....take sqrt of both sides, eliminating the ^2
b = square root of 40 <==
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For this case we must indicate an expression equivalent to:

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Then, applying the property to the given expression we have:

17 (x + 5) = 17 * x + 17 * 5 = 17x + 85

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ANswer:

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
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Answer:

Where's the graph? I need the graph to answer

3 0
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Liem is 6 feet 2 inches, Eli is 5 feet 9 inches, Faith is 6 feet, and Simon is 5 feet 4 inches. In yards, what is the total of t
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7 0
3 years ago
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