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3 years ago

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What is 7(f+3)+9=5(f-2)-3f

1 answer:

muminat3 years ago

8 0

Answer:

f = -8

Step-by-step explanation:

7(f+3)+9=5(f-2)-3f

7f + 21 + 9 = 5f - 10 - 3f

7f + 30 = 2f - 10

5f = -40

f = -8

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A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to $180^{o}$ . The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to $180^{o}$ .

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

So that;

$7^{2}$ = $6^{2}$ + $(6.5)^{2}$ - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

= 29.25

Cos C = $\frac{29.25}{78}$

= 0.375

C = $Cos^{-1}$ 0.375

= 67.9757

C = $68^{o}$

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{6.5}{Sin B}$ = $\frac{7}{Sin 68}$

SIn B = $\frac{6.5 *Sin 68}{7}$

= 0.861

B = $Sin^{-1}$ 0.861

= 59.43

B = $59.4^{o}$

Thus to determine the value of A, we have;

A + B + C = $180^{o}$

A + $59.4^{o}$ + $68^{o}$ = $180^{o}$

A = $180^{o}$ - 127.4

= 52.6

A = $52.6^{o}$

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

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