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Sergio039 [100]
3 years ago
10

A water sprinkler has a range of 5 meters as shown. The jet of water from the sprinkler sweeps out at an angle of 85° as the noz

zle turns. Solve for the total area watered by the sprinkler. Use the value π = 3.14, and round your answer to one decimal place.
a. 15.6 Square Meters
b. 17.8 Square Meters
c. 18.5 Square Meters
d. 19.2 Square Meters

Mathematics
2 answers:
kipiarov [429]3 years ago
8 0

Answer:

C. 18.5 Square Meters

Hope it helps :)

telo118 [61]3 years ago
6 0
I found the image that accompanied this problem.
We need to solve for the area of the sector.

A = n/360 * π * r²
A = 85/360 * 3.14 * 5²
A = 0.2361 * 3.14 * 25
A = 18.5347 

Answer is C. 18.5 square meters.
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3 years ago
Read 2 more answers
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

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\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

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Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

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1 year ago
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