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2 years ago

6

Find the measure of b.

1 answer:

Alika [10]2 years ago

5 0

Answer:

b = 55°

Step-by-step explanation:

The angle at the centre is twice the angle on the circle subtended by the same arc, then

a = 0.5 × 110° = 55°

Angles on the circumference subtended by the same arc are congruent, so

b = a = 55°

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Answer and Step-by-step explanation:

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kicyunya [14]

Parallel have same slope
Perpendicular have opposite reciprocal slopes. Ex. 2/3 is slope of one line so the perpendicular line would have slope of -3/2.

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Y= -4x/5 +4/5

Second line: 10x-8y=-1

10x+1=8y
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3 years ago

Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

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