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taurus [48]
2 years ago
10

Find the missing side . round the the nearest tenth

Mathematics
1 answer:
lys-0071 [83]2 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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Answer:

28/9

Step-by-step explanation:

If the roots are J and K, then:

3 (x − J) (x − K) = 0

3 (x² − (J+K)x + JK) = 0

So if we factor out the leading coefficient:

3x² − 2x − 4 = 0

3(x² − 2/3x − 4/3) = 0

The coefficient of the second term is the sum of the roots:

J + K = 2/3

And the constant is the product of the roots:

JK = -4/3

If we take the sum of the roots and square it:

(J + K)² = (2/3)²

J² + 2JK + K² = 4/9

And subtract twice the product:

J² + K² = 4/9 − 2JK

J² + K² = 4/9 − 2(-4/3)

J² + K² = 4/9 + 8/3

J² + k² = 28/9

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Please Help me with number two and three
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Step-by-step explanation:

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