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Georgia [21]
2 years ago
15

This model of the backboard of a basketball goal is composed of a rectangle and a semicircle. What is the area of

Mathematics
1 answer:
irina1246 [14]2 years ago
3 0

Answer:

deal with the rectangle first A = l X w

4 X 12 = 48 squared cm

semi circle

Area of a circle πr^2

12/2 = 6, radius = 6

π6^2 = 113.097335

113.097335/2 = 56.5486677

56.5 + 48 = 104.5 squared cm

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Peaches are being sold for $2 per pound.The customer created a model to represent the total cost of peaches bought. If x represe
dalvyx [7]

Answer:B)The values of both x and y will be real numbers greater than or equal to 0.

Step-by-step explanation:

7 0
3 years ago
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Morris has 2 3/4 gallons of ice tea. He gives 3/7 of it to her friend. How many gallons of ice tea does Morris have left?
lara31 [8.8K]

Answer:

2 \frac{9}{28} gallons of ice tea is left with Morris

Step-by-step explanation:

Morris has 2 3/4 gallons of ice tea. He gives 3/7 of it to her friend.

Morris has 2 3/4 gallons of ice tea

2\frac{3}{4}= \frac{2*4+3}{4}=\frac{11}{4}

He gives 3/7 of it to her friend.

To find how many gallons of ice tea does Morris have left , we subtract 3/7 from 11/4

\frac{11}{4}- \frac{3}{7}

Make the denominators same

\frac{11*7}{4*7}- \frac{3*4}{7*4}

\frac{77}{28}- \frac{12}{28}

\frac{65}{28}

\frac{65}{28} = 2 \frac{9}{28}

7 0
2 years ago
Prove the following
fomenos

Answer:

Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
  • cot (90°-θ)  =  tan θ
  • sin (90°+θ)  =  cos θ
  • cos (90°+θ)  =  -sin θ
  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
  • sin (180°-θ)  =  sin θ
  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
  • cot (180°-θ)  =  -cot θ
  • sin (180°+θ)  =  -sin θ
  • cos (180°+θ)  =  -cos θ
  • tan (180°+θ)  =  tan θ
  • csc (180°+θ)  =  -csc θ
  • sec (180°+θ)  =  -sec θ
  • cot (180°+θ)  =  cot θ
  • sin (270°-θ)  =  -cos θ
  • cos (270°-θ)  =  -sin θ
  • tan (270°-θ)  =  cot θ
  • csc (270°-θ)  =  -sec θ
  • sec (270°-θ)  =  -csc θ
  • cot (270°-θ)  =  tan θ
  • sin (270°+θ)  =  -cos θ
  • cos (270°+θ)  =  sin θ
  • tan (270°+θ)  =  -cot θ
  • csc (270°+θ)  =  -sec θ
  • sec (270°+θ)  =  cos θ
  • cot (270°+θ)  =  -tan θ
7 0
2 years ago
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Answer:

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B) x + y = 2  then we multiply B) by -5

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A) 5x -7y = 58  we get

-12y = 48

y = -4

B) x + y = 2

x = 2 +4

x = 6



Step-by-step explanation:


6 0
3 years ago
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Answer: x > 20

Step-by-step explanation:

x + 10 > 30

    -10     -10

x > 20

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3 years ago
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