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olya-2409 [2.1K]
3 years ago
13

the greatest common factor of 24 and 36 in the least common multiple of a pair of numbers which two could they be

Mathematics
1 answer:
stealth61 [152]3 years ago
7 0
Two and Nine are the pair of the numbers
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Rubber balls are packaged in a box with dimensions 8.5 cm, 5 cm, 5 cm.
Gre4nikov [31]

Answer:

42.9 cm^3

Step-by-step explanation:

Volume of the box= (base x width x height) = 8.5 x 5 x 5 = 212.5 cm^3

Radius of a ball = 1.5 cm

Volume of a ball = 4/3 * radius^3 * pi = 4/3 x 1.5^3 x pi = 14,137167 cm^3

Volume of the balls = 14,137167 x 12 = 169,646003 cm^3

Final answer = 212.5 - 169,646003 = 42.9 cm^3

7 0
3 years ago
Can anyone please help me.
vovangra [49]

Answer:

I think it was 3/5

I'm not sure

5 0
3 years ago
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the table shows the highest of four students arrange the students in order from shortest to tallest Kim 56.03 + x + 56.3 + 56.14
saul85 [17]
56.03 56.14 56.3 57.1
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3 years ago
Liza comes to the swimming pool once every 6 days. Jenny comes once every 4 days and Olga comes once every 10 days. This Monday
a_sh-v [17]

Answer:

Step-by-step explanation: 60 days

7 0
3 years ago
The length of a rectangular poster is 2 more inches than three times its width. The area of the poster is 33 square inches. Solv
laiz [17]

Answer:

Step-by-step explanation:

Area of the rectangle = 33 inches sq

Area of a rectangle (formula) = l*b

let the width be x

length = 2 inches + 3x = 2+3x

Area of the rectangle = l*b = 2+3x + x

33 inches sq = 2+3x + x

33 in sq = 2+4x

33/2 = 4x

16.5/4 = x

x = 4.125

width= <u>4.125 inches</u>

length = 2+3x = 2+3(4.125) = <u>14.375</u>

<u></u>

(unless your poster is 32 sq inches and you made a mistake, just replace 33 with 32 and solve it)

7 0
2 years ago
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