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Ray Of Light [21]
3 years ago
13

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

napkin spans the plate and the plate has an 18-inch diameter, what is the area of the space on the plate that is NOT covered by the napkin? Leave your answer in terms of .
Mathematics
1 answer:
Crank3 years ago
8 0

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

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In 2004, a pine tree was 5 3/4 ft tall. If the tree grew at an average rate of 5/6 ft per year for the next 7 years, how tall wo
Nady [450]

Answer:

11 7/12 ft tall

Step-by-step explanation:

5 3/4 + (5/6*7)

139/12

11 7/12

7 0
4 years ago
A school baseball team earned $3416.90 from selling 5716 tickets to their game. If grandstand tickets sold for 65 cents each and
expeople1 [14]
If all were grandstand tickets, revenue would be 0.65*5716 = 3715.40. It was actually 298.50 less than that. Each bleacher ticket sold drops the revenue by .25, so there were 298.50/.25 = 1194 bleacher tickets sold.
3 0
3 years ago
An urn contains three red balls and four blue balls. Draw two balls at random from the urn, without replacement. Compute the exp
mezya [45]

Step-by-step explanation:

in total we have 3+4 = 7 balls.

when we draw the first ball, the probability to draw a red ball is 3/7, and a blue ball 4/7.

when we draw the second ball, we have now only 6 balls in total.

the probabilty to draw a red back now depends also on the result of the first draw.

if the first ball was already red, then we have only a chance now of 2 out of 6.

if the first ball was blue, then we have now a chance of 3 out of 6.

so, the probably to draw at least 1 red ball in 2 draws is the probability of drawing one on the first draw plus the probability of drawing one on the second :

1 - probability to see 2 blue balls

1 - 4/7 × 3/6 = 1 - 12/42 = 30/42 = 0.714285714...

the expected number of red balls in 2 draws is

1 red in first red in first red in second

but not second and second but not first

1×(3/7 × 4/6) + 2×(3/7 × 2/6) + 1×(4/7 × 3/6) = 12/42 + 12/42 + 12/42 = 36/42 = 6/7 = 0.857142857 ≈ 0.8571

7 0
2 years ago
If f (x) = 4x - 7 and G(x) = 2x + 4 , evaluate f(x) + g(x) for x = -3
avanturin [10]
Hi there!

• x = - 3

Let's first calculate f(x) and g(x) individually :-

• f(x) = 4x - 7

⇒ 4(- 3) - 7

⇒ - 12 - 7 = - 19

\boxed{\text{f(x)} = - 19}

• g(x) = 2x + 4

⇒ 2(- 3) + 4

⇒ - 6 + 4 = - 2

\boxed{\text{g(x)} = - 2}

According to th' question :-

f(x) + g(x)

⇒ - 19 + (- 2)

⇒ - 19 - 2

⇒ - 21

\boxed{\text{f(x) + g(x)} = - 21}

~ Hope it helps!
6 0
3 years ago
Calculate the value for each determinant:
Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

\begin{vmatrix}3 & -1\\ -3 & 4\end{vmatrix}=(3\times 4)-(-1)\times (-3)

               = 12 - 3

               = 9

\begin{vmatrix}8 & -1\\ -4 & -1\end{vmatrix}=8\times(-1)-(-1)\times(-4)

               = -8 - 4

               = -12

\begin{vmatrix}-12 & -1\\ 12 & 5\end{vmatrix}=(-12)\times(5)-(-1)\times(12)

                = -60 + 12

                = -48

\begin{vmatrix}2 & -1\\ 2 & -1\end{vmatrix}=2\times(-1)-(2)\times(-1)

            = -2 + 2

            = 0

\begin{vmatrix}0 & -1\\ 17 & 4\end{vmatrix}=0\times(4)-(-1)\times(17)

              = 17

\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix}=1\times(1)-(0)\times(0)

          = 1

6 0
3 years ago
Read 2 more answers
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