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SCORPION-xisa [38]
3 years ago
13

The diameter of a sphere is 42 cm . find the surface area to the nearest tenth.

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
6 0
The correct answer is A
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In AVWX, V X is extended through point X to point Y, mZXVW = (3x + 13)°,
olga_2 [115]
(8+6)=ax+1/2 that’s the answer love u
3 0
3 years ago
Rosemary is driving to Chicago, and she is 84 miles away. She wants to arrive in 1.5 hours. What rate of speed will she have to
vagabundo [1.1K]

Answer:

56 mph

Step-by-step explanation:

Process of elimination. 126 mph is much too high because it's more than the distance and they will arrive too early. 65 + 65 ÷ 2 = 97.5, that's more than 84 miles. The reason we divide by 2 is because of the .5. Now all we have is 56 mph. To check, 56 + 56 ÷ 2 = 84

4 0
2 years ago
Piper has a points card for a movie theater. she receives 60 rewards points just for signing up. she earns 13.5 points for each
Arisa [49]

Answer:

10

Step-by-step explanation:

I first took away 60 from 195, as she already had 60 points. Than we were left with 135 points and since she gets 13.5 points every time she visits, I divided 135 by 13, which gives us 10.

3 0
2 years ago
Find the interior angles of a regular polygon which has 6, 10 and 20 sides​
andrew-mc [135]

Answer:

(6)=120°

(10)=72°

(20)=36°

Step-by-step explanation:

(n-2)×180°

(6-2)×180°=720°

6÷720°=120°

10÷720°=72°

20÷720°=36°

4 0
3 years ago
A gold mine is projected to produce $52,000 during its first year of operation, $50,000 the second year, $48,000 the third year,
marshall27 [118]

Answer:

its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

Step-by-step explanation:

Given that;

gold mine is projected to produce $52,000 during its first year

produce $50,000 the second year

produce $48,000 the third year

the mine is expected to produce for 20yrs; i.e n = 20

annual interest rate = 4% = 0.04%

now let P represent the present worth

we determine the present worth ;

Present worth ⇒ Cashflow(Uniform series present worth) - (2000)(uniform gradient present worth)

⇒Cashflow(P | A,i%,n) - (2000)(P | G,i%,n)

= 52000[ ((1+i)ⁿ - 1) / (i(1+i)ⁿ) ] - (2000)[ {((1+i)ⁿ - 1) / (i²(1+i)ⁿ))} - (n/i(1 + i)ⁿ) ]

= 52000[ ((1+0.04)²⁰ - 1) / (0.04(1+0.04)²⁰) ] - (2000)[ {((1+0.04)20 - 1) / ((0.04)²(1+0.04)²⁰))} - (20/0.04(1 + 0.04)²⁰) ]

= 52000[ ((1.04)²⁰ - 1) / (0.04(1.04)²⁰) ] - (2000)[ {((1.04)²⁰ - 1) / ((0.04)²(1.04)²⁰))} - (20/0.04(1.04)²⁰) ]

= 52000[ ((2.191123 - 1) / (0.04(2.191123) ] - (2000)[ {((2.191123 - 1) / ((0.0016)(2.191123))} - (20/0.04(2.191123) ]

= 52000(1.191123/0.08764) - (2000){( 1.191123/0.003506) - (20/0.87645)}

= 52000(13.59033) - (2000)(339.7582 - 228.1935)

= 52000(13.59033) - (2000)(111.5647)

= 706695.6 - 223129.4

= 483,566.2 ≈ 483,566

Therefore its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

4 0
3 years ago
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