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AveGali [126]
3 years ago
6

Solve the equation 4\pi =w-6\pi

Mathematics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

4/pi=w-6/pi

<u>4</u><u> </u><u> </u><u> </u>=<u> </u><u>w</u><u>-</u><u>6</u>

<u>Pi</u><u> </u><u> </u><u> </u><u> </u><u> </u><u>pi</u>

Cross mutiply

Pi(w-6)=4(pi)

Wpi - 6pi =4pi

Collect like terms

Wpi=4pi + 6pi

Wpi =10pi

<u>Divide</u><u> </u><u>both</u><u> </u><u>sides</u><u> </u><u>by</u><u> </u><u>pi</u>

<u>Wpi</u><u> </u>= <u>10pi</u>

Pi pi

W = 10

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Answer:

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Step-by-step explanation:

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Find the zeros of x^2+4x=-10 using the quadratic formula
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A square napkin is folded in half on the diagonal and placed on the diameter of a round plate (see diagram below). If the folded
Crank

Answer:

A=81(\pi-1)\ in^2

Step-by-step explanation:

step 1

Find the area of the plate

The area of a circle is given by the formula

A=\pi r^{2}

we have

r=18/2=9\ in ---> the radius is half the diameter

substitute

A=\pi (9)^{2}\\A=81\pi\ in^2

step 2

Find the area of the square napkin folded (is a half of the area of the square napkin)

we know that

The diagonal of the square is the same that the diameter of the plate

Applying Pythagorean theorem

D^2=2b^2

where

b is the length side of the square

we have

D=18\ in

substitute

18^2=2b^2

solve for b^2

b^2=162\ in^2 -----> is the area of the square

Divide by 2

162/2=81\ in^2

step 3

Find the area of the space on the plate that is NOT covered by the napkin

we know that

The  area of the space on the plate that is NOT covered by the napkin, is equal to subtract the area of the square napkin folded (is a half of the area of the square napkin) from the area of the plate

so

A=(81\pi-81)\ in^2

simplify

A=81(\pi-1)\ in^2

8 0
3 years ago
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