3 years ago

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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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 A = 8, B = 2, C = 4, and D = 7, and thus A + B + C + D = 21.

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