Our goal is to find the area
area=1/2 times base times height
we know the height is 7.9 but what is the base?
use the pythagorean theorem twice
alrighty
so
remember for legs length a and b and hytponuse c in a right triangle
a²+b²=c²
we need AD and DC
so
AD²+7.9²=9.4²
AD=√25.95
DC²+7.9²=23.2²
DC=√475.83
so
AD+DC=base=(√25.95)+(√475.83)
area=1/2bh
area=1/2((√25.95)+(√475.83))(7.9)
area≈106.28518654591426812803776879893
round to 2 decimal places
area≈106.29 square units
This question is not correctly written.
Complete Question
Select all equations that can represent the question: "How many groups of 4/5 are in 1?" A ?⋅1=4/5? Times 1 is equal to 4 fifths B 1⋅4/5=?1 times 4 fifths is equal to ? C 4/5÷1=?4 fifths divided by 1 is equal to ? D ?⋅4/5=1? Times 4 fifths is equal to 1 E 1÷4/5=?1 divided by 4 fifths is equal to ?
Answer:
D ?⋅4/5=1 = ? Times 4 fifths is equal to
E 1÷4/5=? = 1 divided by 4 fifths is equal to
Step-by-step explanation:
How many groups of 4/5 are in 1?
The operation used to solve this is that Division operation.
Hence, we solve it by saying:
1 ÷ 4/5 = ?
= 1× 5/4 = ?
5/4 = ?
Cross Multiply
5 = 4 × ?
? = 5/4
The equations that can represent the question: is
Option D ?⋅4/5=1 = ? Times 4 fifths is equal to
Option E 1÷4/5=? = 1 divided by 4 fifths is equal to
SA = 2
+2

1. Find the radius.
R = 1/2D
6 x 1/2 = 3
R = 3
2. Plug them in.
SA 2(3.14)(3)(7) + 2(3.14)(3^2)
3. Solve
3 x 7 = 21
21 x 3.14 = 65.94
65.94 x 2 = 131. 88
3^2 = 9
9 x 3.14 = 28.26
28.26 x 2 = 56.52
4. Add them together.
131.88 + 56.52 = 188.4
Answer: 188.4 square millimeters
Using the quadratic equation we get:
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Factoring out 2 we get
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Factoring out the imaginary number:

So b.