Question

three number are in ratio 3:4:5 the sum of their cube is 7,290,000 find the numbers

Answer 1

Answer:45 , 60 , 75

Step-by-step explanation:

→ Given the three numbers which are in ratio 3:4:5 and their sum of cubes is 729000.

→ let the three numbers are 3x,4x and 5x

→ As their sum of cubes is 729000

→ (3x)³ + (4x)³ + (5x)³ = 729000

→ 27x³ + 64x³ + 125x³ = 729000

→ 216x³ = 729000

→ x³ = 3375

→ Taking cube root on both sides, we get

→ x = 15

→ 3x = 3 * 15 = 45

→ 4x = 4 * 15 = 60

→ 5x = 5 * 15 = 75

→ Hence, the three numbers are 45,60 and 75.

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Answer 2

Answer:

Required first number is 96.96 and second number is 129.28 and third number is 161.6.

Step-by-step explanation:

Given,three number are in ratio 3:4:5

Let first number be 3x and second number be 4x and third number be 5x.

It is also given,the sum of their cube is 7,290,000.

According to question,

${(3x)}^{3} + {(4x)}^{3} + {(5x)}^{3} = 7290000 \\ 27 {x}^{3} + 64 {x}^{3} + 125 {x}^{3} = 7290000 \\ 216 {x}^{3} = 7290000 \\ {x}^{3} = \frac{7290000}{216} \\ x = {( \frac{7290000}{216}) }^{ \frac{1}{3} } \\ x = 32.32$

So, required first number is

$(3 \times 32.32) = 96.96$

and second number is

$(4 \times 32.32) = 129.28$

and third number is

$(5 \times 32.32) = 161.6$

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