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3 years ago

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The volume of the large cube is 125 cubic inches. The volume of the small cube is 27 inches. What is the difference between the

length of one side of the large cube and the length of one side of the small cube?

1 answer:

slega [8]3 years ago

4 0

Answer:

<h3><em>2cm</em></h3>

Step-by-step explanation:

Step one:

given data

for the large cube

the volume= 125cm^3

we know that volume = l*l*l

volume=l^3

we want to find l, so we have to cube both sides

125=l^3

l=∛125

l=5cm

Step two:

given data

for the large cube

the volume= 27cm^3

we know that volume = l*l*l

volume=l^3

we want to find l, so we have to cube both sides

27=l^3

l=∛27

l=3cm

<h3><em>The difference between the length of one side of the large cube and the length of one side of the small cube</em></h3><h3><em>=5-3= 2cm</em></h3>

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Read 2 more answers

Find an m > 0 such that the the equation x^4−(3m+2)x^2+m^2=0 has four real solutions that form an arithmetic sequence.

Aleonysh [2.5K]

Answer:

The value of m is 6.

Step-by-step explanation:

Here, the given equation,

$x^4-(3m+2)x^2+m^2=0$

$x^4+0x^3-(3m+2)x^2+0x+m^2=0$

Let the roots of the equation are a-3b, a-b, a+b and a + 3b, ( they must be form an AP )

Thus, we can write,

$a-3b+a-b+a+b+a+3b=\frac{\text{coefficient of }x^3}{\text{coefficient of }x^4}$

$=\frac{0}{1}=0$

$\implies a=0----(1)$

$(-3b)(-b)+(-b)(b)+(b)(3b)+(3b)(-3b)+(-b)(3b)+(-3b)(b)=\frac{\text{coefficient of }x^2}{\text{coefficient of }x^4}}$

$=\frac{-3m-2}{1}$

$3b^2-b^2+3b^2-9b^2-3b^2-3b^2=-3m-2$

$-10b^2=-3m-2$

$\implies b^2=\frac{3m+2}{10}-----(2)$

$(-3b)(-b)(b)(3b)=\frac{\text{Constant term}}{\text{coefficient of}x^4}= m^2$

$9b^4=m^2$

$9(\frac{3m+2}{10})^2=m^2$

$9(\frac{9m^2+4+12m}{100})=m^2$

$81m^2+36+108m=100m^2$

$-19m^2+108m+36=0$

$19m^2-108m-36=0$

$19m^2-114m+6m-36=0$

$19m(m-6)+6(m-6)=0$

$(19m+6)(m-6)=0$

$\implies m=-\frac{6}{19}\text{ or }m=6$

But m > 0,

Hence, the value of m is 6.

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3 years ago