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beks73 [17]
3 years ago
10

Plz help :)))))))))))

Mathematics
1 answer:
dusya [7]3 years ago
5 0

Given:

b1 = 17 ft

b2 = 20 ft

h = 13 ft

r = 17/2 = 8.5 ft


Area of the Semi-circle = 1/2 π r²

= 1/2 (22/7) (8.5²)

= (11/7) (72.25)

= 113.535 ft²


Area of the trapezium = (h) (b1 + b2) / 2

= (13)(17 + 20)/2

= 240.5 ft²


Total area of the amphitheater stage = Area of Semi-circle + Area of the trazezium = 113.535 + 240.5 = 354.035 ft²

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Jada says she can find 87-59 by taking away 60 and adding 1 so it is the same as 27+1 or 28. Explain why jadas method to calcula
astraxan [27]

Answer:

The correct answer is 28.

Step-by-step explanation:

Actually Jada makes problem simpler by adding and subtracting 1

that is

87 ₋ 59 ₋1 ₊ 1

so there would be no effect on our final answer because we have added 1 and also subtracted 1

Now,

87 - 59 -1 + 1

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3 0
3 years ago
Rajan needs 1 liter of 15% hydrochloric acid solution for an experiment. He checks his storage cabinet, but he only has a 5-lite
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Answer:

  1. <em>0.75</em><em> liters of 5% hydrochloric acid solution </em>
  2. <em>0.25</em><em> liters of 45% hydrochloric acid solution.</em>

Step-by-step explanation:

Let us assume that, x liters of the 5% hydrochloric acid and y liters of the 45% hydrochloric acid solutions are combined.

As Rajan need total of 1 liter of solution, so

x+y=1

i.e x=1-y --------------------1

As Rajan needs 5% hydrochloric acid and 45% hydrochloric acid to make a 1 liter batch of 15% hydrochloric acid, hence acid content of the mixture of two acids will be same as of the final one, so

0.05x+0.45y=1\times 0.15

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Putting value of x from equation 1 in equation 2,

\Rightarrow 5(1-y)+45y=15

\Rightarrow 5-5y+45y=15

\Rightarrow 5+40y=15

\Rightarrow 40y=15-5=10

\Rightarrow y=0.25

Putting the value of y in equation 1,

x=1-0.25=0.75

Therefore, Rajan must use 0.75 liters of 5% hydrochloric acid solution and 0.25 liters  of 45% hydrochloric acid solution.

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