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stiks02 [169]
3 years ago
5

What would the area of an actual pool be in the scale from the drawing is 3:1000

Mathematics
1 answer:
elena55 [62]3 years ago
8 0

Answer:

If this is the complete question them the answer is 1250 m²

Step-by-step explanation:

the pool in the drawing is 15x7.5 and the scale is 3/1000 so if the size in the drawing  is 3 then the actual is 1000.

x=15*1000/3=5000cm which would be 50m

x=7.5*1000/3=2500cm which is 25m

the actual area of the pool is 50*25=1250 m²

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Answer:

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Step-by-step explanation:

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3 years ago
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(16+10)-1+13*5
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4 years ago
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What are the answers ?
Anika [276]
If c = 8 and d = -5:

a) c - 3 = 8 - 3
= 5

b) 15 - c = 15 - 8
= 7

c) 3(c + d) = 3(8 + (-5))
= 3*3
= 9

d) 2c - 4d = 2(8) - 4(-5)
= 16 + 20
= 36

e) d - c^2 = -5 - (8)^2
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f) 2d^2 + 5d = 2(-5)^2 + 5(-5)
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5 0
3 years ago
Which of the following is a solution of z^5 = 1 + √3 i?
nordsb [41]

Answer:

Option 2 is right

Step-by-step explanation:

Given that

z^5=1+\sqrt{3} i

We can write this in polar form with modulus and radius

|z^5|= \sqrt{1+3} =2\\tan of Angle t =\sqrt{3} \\

Hence angle = 60 degrees and

|z^5|= 2(cos60+isin60)

Since we have got 5 roots for z, we can write 60, 420, 780, etc. with periods of 360

Using Demoivre theorem we get 5th root would be

5th root of 2 multiplied by 1/5 th of 60, 420, 780,....

z= \sqrt[5]{2} (cos12+isin12)\\z=\sqrt[5]{2} (cos84+isin84)\\\\z=\sqrt[5]{2} (cos156+isin156)\\\\z=\sqrt[5]{2} (cos228+isin228)\\\\z=\sqrt[5]{2} (cos300+isin300)\\

Out of these only 2nd option suits our answer

Hence answer is Option 2.

8 0
3 years ago
Each day for 12 consecutive days, Ricardo jogged for 1 2/3 hours at an average rate of 6 1/4 miles per hour. Find the total dist
kirill115 [55]

Answer:

Your answer is 125 miles

Step-by-step explanation:

1 \frac{2}{3} ×6 \frac{1}{4} = 10 \frac{5}{12}

10 \frac{5}{12} × 12 = 125

8 0
3 years ago
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