The arc length of TV is ≈ 3.67π in.
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,
= /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.
Learn more about Arc length here:
#SPJ2
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.
Step-by-step explanation:
4x + 8z = 2 Subtract 4x from both sides
8z = 2 - 4x Now, divide both sides by 8
z = (2-4x) / 8 or (1-2x) / 4
2 Significant figures should be include in the density value that Henry reports.
" 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
1.060 has 4 significant figure
1.0 has 2 significant figure
As per division rule of significant figure
When we divide two more numbers, count the significant figure of the given number. Least is the significant figure of the concluded one.
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
#SPJ2