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2 years ago

In ΔOPQ, q = 6.5 inches, o = 8.6 inches and ∠P=55°. Find ∠Q, to the nearest 10th of an degree.

Mathematics

2 answers:

Nikitich [7]2 years ago

8 0

Answer:

∠Q = 35.5°

Step-by-step explanation:

We are given;

q = 6.5 inches

o = 8.6 inches

∠P = 55°

Let's first use cosine rule to find p.

p² = q² + o² - 2qo cos P

Plugging in the relevant values;

p² = 6.5² + 8.6² - (2 × 6.5 × 8.6 × cos 55)

p² = 42.25 + 73.96 - 32.0629

p² = 84.1471

p = √84.1471

p = 9.17 inches

Using sine rule, we can find ∠Q;

p/sin P = q/sin Q

sin Q = (q•sinP)/p

sin Q = (6.5 × sin 55)/9.17

sin Q = 0.5806

Q = sin^(-1) 0.5806

Q ≈ 35.5°

Vlad1618 [11]2 years ago

3 0

Answer:

47.5

Step-by-step explanation:

From delta math

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<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

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$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

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