There are two methods to do this:
1. 2(x + 1) = 38
2*x + 2*1 = 38 (Expand the bracket)
2x + 2 = 38
2x = 36 (Subtract 2 from both sides)
x = 18 (Divide each side by 2)
2. 2(x + 1) = 38
(x + 1) = 19 (Divide each side by 2)
x = 18 (Subtract 1 from each side)
10 square units.
Triangle area: base times height divided by 2
(A triangle area is half the area of a rectangle with the same base and height)
4*5/2=10
20•9=180
180•5=900
You have to brake down the problem to it simplest form
Answer:
the correct option is option c.
Step-by-step explanation:
We have the expression: (160*243)^1/5
First, we know that 160 = 5(2^5)
And, 243 = 3^5
Then we have: (160*243)^1/5 = [5(2^5)(3^5)]^1/5 = 6(5)^1/5
So the correct option is option c.
we know that
The measurement of the external angle is the semi-difference of the arcs it comprises.
so
Step 
<u>Find the measure of the arc AJ</u>
m∠BDE=![\frac{1}{2} *[measure\ arc\ AJ-measure\ arc\ BE]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5Bmeasure%5C%20arc%5C%20AJ-measure%5C%20arc%5C%20BE%5D)
in this problem we have
m∠BDE=

substitute in the formula
=![\frac{1}{2} *[measure\ arc\ AJ-38\°]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5Bmeasure%5C%20arc%5C%20AJ-38%5C%C2%B0%5D)
=![[measure\ arc\ AJ-38\°]](https://tex.z-dn.net/?f=%5Bmeasure%5C%20arc%5C%20AJ-38%5C%C2%B0%5D)

Step 
<u>Find the measure of the arc FH</u>
m∠FGH=![\frac{1}{2} *[measure\ arc\ AJ-measure\ arc\ FH]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5Bmeasure%5C%20arc%5C%20AJ-measure%5C%20arc%5C%20FH%5D)
in this problem we have
m∠FGH=

substitute in the formula
=![\frac{1}{2} *[112\°-measure\ arc\ FH]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%5B112%5C%C2%B0-measure%5C%20arc%5C%20FH%5D)
=![[112\°-measure\ arc\ FH]](https://tex.z-dn.net/?f=%5B112%5C%C2%B0-measure%5C%20arc%5C%20FH%5D)

therefore
<u>the answer is</u>
the measure of the arc FH is 