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2 years ago

Part 4. NO LINKS!! Find the area of the shaded area shown. SHOW WORK PLEASE !! #21

Mathematics

2 answers:

SIZIF [17.4K]2 years ago

5 0

Area of square

32²
1024

Radius of each circle

32/4/2
8/2
4

Area of 16 circles

16πr²
16π(4)²
16π(16)
256π
803.8cm²

Area of shaded region

1024-803.8
220.2cm²

Julli [10]2 years ago

3 0

<h2>Area of Shaded</h2>

Find the area of the shaded area shown.

<h3>Solution:</h3>

A = s²
A = 32cm
A = 32cm x 32cm
A = 1024
A = 32/4/2
A = 32 ÷ 2
A = 8
A = 8 ÷ 2
A = 4

So in the inscribed circle there are 16.

A = 16 radius ²
A = 16 x 4r²
A = 4 x 4
A = 16
A = 16π x 16
A = 256π
A = 256 x 3.14
A = 803.84
A = 1024cm - 803.84
A = 220.16cm²

#ProblemSolve

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Step-by-step explanation:

39-15=24

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3 years ago

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The Smith family went to eat at Buffalo Wild Wings. The bill was $43.86. They gave the server a 20% tip. How much did they pay a

Marysya12 [62]

<h2>Answer:</h2>

$52.63

<h2>Step-by-step explanation:</h2>

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2 years ago

PLEASE PLEASE PLEASE PLEASE HELP!! even if it's only one! actually answer it!

Alex

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6 0

3 years ago

If EF = 4x-15, FG=3x-7, and EG =20, find the value of x, EF and FG. The drawing is not to scale.

Andrej [43]

Step-by-step explanation:

EF = 4x - 15

FG = 3x - 7

EG = EF + FG = 20

so,

4x - 15 + 3x - 7 = 20

7x - 22 = 20

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4 0

2 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

Part 4. NO LINKS!! Find the area of the shaded area shown. SHOW WORK PLEASE !! #21​

Part 4. NO LINKS!! Find the area of the shaded area shown. SHOW WORK PLEASE !! #21