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3 years ago

18. Find the measure of each numbered angle.

Mathematics

1 answer:

DiKsa [7]3 years ago

7 0

Answer:

m∠1 = 50°, m∠2 = 90°, m∠3 = 40°, and m∠4 = 50°

Step-by-step explanation:

Angles on straight line are equal;

$m \angle1 + 130 = 180$

$m \angle1= 180 - 130 \\ m \angle1= 50$

Vertical angles are equal,

$m \angle4= m \angle \: 1= 50 \degree$

Complementary angles add up to 90°.

$m \angle3 + m \angle4 = 90 \degree \\ 40 + m \angle3 = 90 \degree. \\ m \angle3 = 90 \degree - 50 \degree \\ m \angle3 = 40 \degree$

Angles on straight line adds up to 180°

$m \angle1 + m \angle2 + m \angle3 = 180 \degree$

$40 + m \angle2 + 50 = 180 \\ m \angle2 = 180 - 50 - 40 = 90$

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

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

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

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

Step-by-step explanation:

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

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

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

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

2) $2^{-1}.2^{-4}$

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

Using this rule:

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

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

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

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

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

4) $\frac{2}{2^{-4}}$

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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