Identify the arc length of TV in terms of π. (image is attached)The answers for this question were:

≈0.67pi in., ≈1.83pi in., ≈3.67pi in., or ≈1.33pi in.
Identify the arc length of TV in terms of π. - 1

Answers

Answer 1
Answer:

The arc length of TV is ≈ 3.67π in.

What is Arc length?

Arc length is the distance between two points along a section of a curve.

Given:

∠SMV = 48 and MT= 5 in

As, ∠SMV and ∠VMT are forms linear pair. So,  

∠VMT + ∠SMV = 180∘

∠VMT= 180 -  ∠SMV

∠VMT = 180∘ − 48∘

∠VMT =132∘      

Now, length of arc will be,

= \theta/360 2πr

=132/360*2*π*5

=132/360*10π

=33/9π

≈ 3.67π in.

Hence, the arc length of TV is ≈ 3.67π in.

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Answer 2
Answer:

Answer:       ≈ 3.67π in.

Step-by-step explanation:

∠SMV and ∠VMT are supplementary. Therefore,  

m∠VMT = 180∘ − m∠SMV

It is given that m∠SMV = 48∘. Substitute the given value and simplify.

m∠VMT = 180∘ − 48∘

=132∘                                                                                                                  Arc length is the distance along an arc measured in linear units. The formula for Arc Length is L=  2πr (m∘/360∘).

The length of the radius is given as 5 in. Substitute the known values into the formula.

L= 2π (5) (132/360)

Simplify.

L= 33/9π

Round to the nearest tenth.  

L ≈ 3.67π in.

Therefore, the arc length of TV is ≈ 3.67π in.


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Answers

Step-by-step explanation:

4x + 8z = 2     Subtract 4x from both sides

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Answers

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Answers

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Answers

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Answers

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What is Significant figure?

" Significant figure are those digit of number which are important in terms of accuracy. They are single digit number from 0 to 9, which express the message of accuracy."

According to question,

Henry divides 1.060g by 1.0mL

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As per division rule of significant figure

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Here 2 is the lowest among two.

Hence, we conclude that 2 Significant figures should be include in the density value that Henry reports.

Learn more about significant figures here

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Answers