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Vika [28.1K]
2 years ago
6

Find the area and perimeter of the shaded region.

Mathematics
2 answers:
morpeh [17]2 years ago
6 0
Perimeter- 20
Area- 50-100
defon2 years ago
4 0

Answer:

Area: 50\pi -100

Perimeter: 20\pi

Step-by-step explanation:

If we take half of one of these "pedals" we can see that it is simply 1/4 of a circle with radius 5, subtracted by a triangle. Let's calculate this half-pedal.

1/4(25 \pi) - 1/2(5* 5)

That means 4 pedals is equal to:

8(1/4(25\pi) - 1/2 (25))

50\pi - 100

So.. The area of the shaded region is 50\pi -100

Perimeter is even simpler. the half-pedal is just 1/4 of the circumference of the circle. The circumference is just 10\pi, which means our half pedal is:

1/4(10\pi )

Multiplying by 8, our perimeter is just 20\pi.

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The following lists show the free throw percentages of 16 players for the Detroit Pistons and Miami Heat during the 2012-2013 NB
Alina [70]

Answer:

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05

8 0
3 years ago
URGENT!!! 30 POINTS!!!<br><br><br>Expand log(1/2)(3x^2/2) using properties and rules for logarithms.
Korvikt [17]

\log_\frac{1}{2}\left(\dfrac{3x^2}{2}\right)=\log_\frac{1}{2}3x^2-\log_\frac{1}{2}2=\log_\frac{1}{2}3+\log_\frac{1}{2}x^2-\log_\frac{1}{2}2\\\\=\log_\frac{1}{2}(3)+2\log_\frac{1}{2}(x)-\log_\frac{1}{2}\left(\dfrac{1}{2}\right)^{-1}\\\\=\log_\frac{1}{2}(3)+2\log_\frac{1}{2}(x)-\left[-\log_\frac{1}{2}\left(\dfrac{1}{2}\right)\right]\\\\=\log_\frac{1}{2}(3)+2\log_\frac{1}{2}(x)+\log_\frac{1}{2}\left(\dfrac{1}{2}\right)\\\\=\boxed{\log_\frac{1}{2}(3)+2\log_\frac{1}{2}(x)+1}

Used:\\\\\log_a\dfrac{b}{c}=\log_ab-\log_ac\\\\\log_a(bc)=\log_ab+\log_ac\\\\\log_ab^n=n\log_ab\\\\\log_aa=1

6 0
2 years ago
What is the solution?<br> 45-[5+8y-3(y+3)]= -3(3y-5)-[5(y-1)-2y+6
monitta

Answer:

-5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

45−(5+8y−3(y+3))=−3(3y−5)−(5(y−1)−2y+6)

45+−1(5+8y−3(y+3))=−3(3y−5)+−1(5(y−1)−2y+6)(Distribute the Negative Sign)

45+(−1)(5)+−1(8y)+−1(−3(y+3))=−3(3y−5)+−1(5(y−1))+−1(−2y)+(−1)(6)

45+−5+−8y+3y+9=−3(3y−5)+−5y+5+2y+−6

45+−5+−8y+3y+9=(−3)(3y)+(−3)(−5)+−5y+5+2y+−6(Distribute)

45+−5+−8y+3y+9=−9y+15+−5y+5+2y+−6

(−8y+3y)+(45+−5+9)=(−9y+−5y+2y)+(15+5+−6)(Combine Like Terms)

−5y+49=−12y+14

−5y+49=−12y+14

Step 2: Add 12y to both sides.

−5y+49+12y=−12y+14+12y

7y+49=14

Step 3: Subtract 49 from both sides.

7y+49−49=14−49

7y=−35

Step 4: Divide both sides by 7.

7y

7

=

−35

7

y=−5

8 0
3 years ago
QUESTION 20
atroni [7]

.Answer:

The answer is below

Step-by-step explanation:

The patient loses 1 kg every week for 5 weeks.

1 kg = 2.2 lbs

Therefore the patient loses 2.2 lbs every week for 5 weeks.

a) The weight of the patient after 5 weeks = 245 lbs. - (5 weeks)(2.2 lbs per week)

The weight of the patient after 5 weeks = 245 lbs. - 11 lbs. = 234 lbs.

b) The weight of the patient after 5 weeks = 245 lbs. - 11 lbs. = 234 lbs.

1 kg = 2.2 lbs.

234 lbs. = 234 lbs. * 1 kg per 2.2 lbs. = 106.36 kg

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3 years ago
24 students in a class took an algebra test. if 18 students passed, what percent passed.
tia_tia [17]

Answer:

75%

Step-by-step explanation:

18/24=0.75

8 0
3 years ago
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