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2 years ago

Given: Circle k (O), OA and OC -radii, AP and CP - tangents, m∠AOC=150° Find: m∠APC

Mathematics

1 answer:

viva [34]2 years ago

3 0

Answer:

m∠APC = 30°

Step-by-step explanation:

To solve for the above question, we would be making use of circle theorems

Looking at the circle, we can see that

m∠APC is an Angle outside the circle

m∠AOC is a smaller arc in the circle.

Therefore, Circle Theorem states that for a circle with two tangents,

The Angle outside the circle + The smaller arc = 180°

Hence,

m∠AOC + m∠APC = 180°

m∠APC = 180° - m∠AOC

m∠APC = 180° - 150°

m∠APC = 30°

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Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

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Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

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$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

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Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

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We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

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