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NeX [460]
3 years ago
12

Area of a circular ring: A = 4

bsmiddle" class="latex-formula">pw
Solve for p. Find p when A = 22 cm^2 and w = 2 cm

Perimeter of a track: P = 2\pir + 2x
Solve for r. Finf r when P = 440 yrds and x = 110 yrds
Mathematics
1 answer:
OleMash [197]3 years ago
8 0
1a)      A = 4πpw
      /4πw = /4πw
  A / 4πw = p

1b) A = 4πpw
     22 = 4πp(2)
       p = 11/4π (≈0.87)

2a)        P = 2πr + 2x
      P - 2x = 2πr
        /2π      /2π
 P-2x / 2π = r

2b)    P = 2πr + 2x
      440 = 2πr + 2(110)
          r = 110/π (≈35.014)
      

      
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1) Slope =-1, y-intercept = 0
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1. C       2.D       3.D        4.A

Step-by-step explanation:

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3 years ago
What is the area of this ^2
Ad libitum [116K]
<h2>Question:- Area of the given diagram</h2>

<h2>Answer :- </h2>

Divide the diagram as shown in attachment.

Now first calculate the area of the rectangle having side 1 and 12

So area = length ×breadth

Area = 1× 12 = 12 unit²

Now find the Area of the remaining triangle with sides 12,13,5 units

So

Area =  \frac{1}{2}  \times base  \times height \:

Now put the value as -> base= 5 and height= 12

Area =  \frac{1}{2}  \times base  \times height \:  \\ Area =  \frac{1}{2}  \times 5 \times 12 \\ Area =  \frac{1}{ \cancel2}  \times 5 \times  \cancel{12}^{ \:  \: 6}  \\ Area = 5 \times 6 \\ Area = 30 \: units^{2}

Now add both areas for your answer

Area{ \tiny{1}} = 12 {units}^{2}  \\ Area{ \tiny2} = 30 {units}^{2}  \\ Area{ \tiny{total}} = Area{ \tiny{1}} + Area{ \tiny{2}} \\ Area{ \tiny{total}} = 12 + 30 = 42 {units}^{2}

8 0
2 years ago
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