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Art [367]
3 years ago
8

I need to find x but I don't know how

Mathematics
2 answers:
-BARSIC- [3]3 years ago
6 0
Use tan theta = opp ÷ adjecent
Tan (x)= 3÷4
X = tan ^-1 (3÷4)
vagabundo [1.1K]3 years ago
5 0
If you're looking for the angle it is a 30, 60, 90, but if you're looking for the length of the hypotenuse it's 5
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Rationalize the denominator of $\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, whe
musickatia [10]

Rationalizing the denominator involves exploiting the well-known difference of squares formula,

a^2-b^2=(a-b)(a+b)

We have

(\sqrt{16}-\sqrt2)(\sqrt{16}+\sqrt2)=(\sqrt{16})^2-(\sqrt2)^2=16-2=14

so that

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{32}(\sqrt{16}+\sqrt2)}{14}

Rewrite 16 and 32 as powers of 2, then simplify:

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{2^5}(\sqrt{2^4}+\sqrt2)}{14}

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{2^2\sqrt2(2^2+\sqrt2)}{14}

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{4\sqrt2(4+\sqrt2)}{14}

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+4(\sqrt2)^2}{14}

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+8}{14}

\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{8\sqrt2+4}7

So we have <em>A</em> = 8, <em>B</em> = 2, <em>C</em> = 4, and <em>D</em> = 7, and thus <em>A</em> + <em>B</em> + <em>C</em> + <em>D</em> = 21.

3 0
3 years ago
Irwin rode his bicycle from one end of a trail to another and back at a speed of 20 miles per hour. If the trail is 4 miles long
pentagon [3]

Answer: I believe it’s C. 2.5

Step-by-step explanation:

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4 years ago
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Answer:

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Step-by-step explanation:

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The difference between h and 9 divided by 3.
Nadusha1986 [10]

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A

Step-by-step explanation:

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-2(3x - 1) = -6x -1 does it have one Solution, no Solution, or infinite Solutions
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None

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3 years ago
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