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msjsnell29

msjsnell29

23.10.2020 •

Mathematics

Solve for x

I need help

$\frac{5}{320} = \frac{x}{448}$

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Ответ:

lilie080202

05.02.2022

$\huge{ \bold{ \boxed{ \text{x = 7}}}}$

Step-by-step explanation:

$\sf{ \frac{5}{320} = \frac{x}{448} }$

$\text{Step \: 1 : Apply \: cross \: product \: property}$

➵ $\text{320 \: ∗ \: x \: = 448 \: ∗ \: 5}$

➵ $\text{320x \: = 2240}$

$\text{Step \: 2 : \: Divide \: both \: sides \: by \: 320}$

➵ $\text{320x \: / \: 320 \: = \: 2240 \: / \: 320}$

➵ $\text{x = 7}$

$\text{Hope \: i \: helped!}$

$\text{Best \: regards!}$

~ $\text{TheAnimeGirl}$

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Ответ:

nyny2129

21.11.2022

A. $-27a^3b^6+8a^9b^{12}$

D. $27a^3b^6+8a^9b^{12}$

Step-by-step explanation:

A. $-27a^3b^6+8a^9b^{12}$

which is equivalent to

$(-3)^3a^3(b^2)^3+2^3(a^3)^3(b^4)^3\\=(-3ab^2)^3+(2a^3b^4)^3$

This expression is a sum of cubes.

B. $-9a^3b^6+a^9b^{10}$

Here 9 is not a perfect cube and power of b in second term is 10 which is not a multiple of 3 , thus this cannot be written as sum of cubes.

C. $9a^3b^6+8a^9b^{12}$

Here 9 is not a perfect cube. Thus this cannot be written as sum of cubes.

D. $27a^3b^6+8a^9b^{12}\\=3^3a^3(b^2)^3+2^3(a^3)^3(b^4)^3\\=(3ab^2)^3+(2a^3b^4)^3$

This expression is a sum of cubes.

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