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baherus [9]
3 years ago
7

Here are two circles. Their centers are A and F. What is the same about the two circles?

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
8 0

Answer:

The diameter of both is 8.

Step-by-step explanation:

The ratio is 4 of the first circle. And d=2r. So the diameter is 8.

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Tina and her friend are picking apples at Newberry Orchard. The apples cost $1.40 per pound. If they pick 2.5 pounds of apples,
sp2606 [1]

Answer:

$3.50

Step-by-step explanation:

the apples cost $1.40 per pound, so if we have 2.5 pounds, we would need to do this:

1.40 × 2.5

3.5

I hope this helps!

8 0
3 years ago
fladli sells televisions. He earns a fixed amount for each television and an additional $ 20 if the buyer gets an extended warra
ivanzaharov [21]

Answer: $80


Step-by-step explanation:

Given: The amount earned by Fadil on selling 14 televisions= $1400

Therefore, the amount he earned on selling 1 television=\frac{1400}{14}=\$100

Since, he sold televisions with extended warranty and he earns $20 additional for it.

Then, the fixed amount Fadil earns for each​ television=\$100-\$20=\$80

Hence, Fadil earns $80 as the fixed amount for each televison.

4 0
4 years ago
Help find the area of the shape
-Dominant- [34]

Answer:

44 cm²

Step-by-step explanation:

follow me for answering the question

4 0
3 years ago
1 David buys 4 kg of apples from the supermarket for £1.60.
liberstina [14]

Answer:

£ 2.8

Step-by-step explanation:

4 kg = 1.60

1 kg= 1.6/4 = £0.4

7 kg = 0.4*7 = £2.8

4 0
3 years ago
A rubber ball is dropped from the top of a hole. Exactly 20 seconds later, the sound of the rubber ball hitting bottom is heard.
gayaneshka [121]

9514 1404 393

Answer:

  4192.9 ft

Step-by-step explanation:

We assume your falling-distance formula is supposed to be ...

  s = 16t²

So, the time required to fall distance s is ...

  t = √(s/16) = (1/4)√s

The time required for sound to travel distance s is ...

  s = 1100t

  t = s/1100

Then the sum of the time for the ball to fall and the time for the sound to travel back is ...

  (1/4)√s + s/1100 = 20

  275√s = 22000 -s . . . . . multiply by 1100 and subtract s

  75625s = s^2 -44000s +484,000,000 . . . square both sides

  s^2 -119,625s +484,000,000 = 0 . . . . put in standard form

  s ≈ (1/2)(119,625 ±√12,374,140,625) = {4192.943, 115432.057}

Only the smaller of these two solutions makes any sense in this problem.

The hole is about 4192.9 feet deep.

_____

<em>Additional comment</em>

The distance equation for the falling object presumes a vacuum. The sound transmission presumes the presence of air, so the question setup is self-contradictory.

7 0
3 years ago
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