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taurus [48]
3 years ago
10

Find the missing side . round the the nearest tenth

Mathematics
1 answer:
lys-0071 [83]3 years ago
7 0

S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}

Sin=\frac{Opposite}{Hypotenuse} Cos=\frac{Adjacent}{Hypotenuse} Tan=\frac{Opposite}{Adjacent}

7) You have the opposite. You need the adjacent.

Tan=\frac{Opposite}{Adjacent}

Tan57=\frac{12}{x}

x=\frac{12}{Tan57}

x = 7.79289... =

<h2>7.8</h2><h2 />

8) You have the hypotenuse. You need the Opposite.

Sin=\frac{x}{Hypotenuse}

Sin37=\frac{x}{13}

x = sin37 × 13

x = 7.8235... =

<h2>7.8</h2><h2 />

9) You have the hypotenuse. You need the opposite.

Sin=\frac{x}{Hypotenuse}

Sin59=\frac{x}{11}

x = sin59 × 11

x = 9.4288.... =

<h2>9.4</h2><h2 />

10) You have the hypotenuse. You need the opposite.

Sin=\frac{x}{Hypotenuse}

Sin53=\frac{x}{11}

x = sin53 × 11

x = 8.7459.... =

<h2>8.7</h2>
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Which expression is a sum of cubes?
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we know that

A polynomial in the form a^{3} +b^{3} is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> -64x^{6} y^{12} +125x^{16} y^{3}

we know that

-64=-4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case A) is not a sum of cubes

<u>case B)</u> -32x^{6} y^{12} +125x^{16} y^{3}

we know that

-32=-2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{16}=x^{15} *x=x*(x^{5})^{3} -------> is not a perfect cube

y^{3}= (y)^{3}

therefore

the case B) is not a sum of cubes

<u>case C)</u> 32x^{6} y^{12} +125x^{9} y^{3}

we know that

32=2^{5} -------> is not a perfect cube

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

therefore

the case C) is not a sum of cubes

<u>case A)</u> 64x^{6} y^{12} +125x^{9} y^{3}

we know that

64=4^{3}

x^{6}= (x^{2})^{3}

y^{12}= (y^{4})^{3}

125=5^{3}

x^{9}=(x^{3})^{3}

y^{3}= (y)^{3}

Substitute

4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}

(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}

therefore

<u>the answer is</u>

64x^{6} y^{12} +125x^{9} y^{3} is a sum of cubes

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