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6 months ago

If x3 = 125, find x²

Mathematics

1 answer:

Burka [1]6 months ago

5 0

The value of $x^{2}$ is 25 when we have (cube of x = 125) $x^{3}$ = 125.

According to the question,

We have the following expression:

$x^{3} = 125$

Now, we will do the cube roots on both the sides:

$\sqrt[3]{x^{3} } = \sqrt[3]{125}$

(We can find the cube roots of 125 by finding the Least Common Multiple or LCM.)

So, we will have the following expression:

x = 5

Now, to find the value of $x^{2}$ , square on both the sides:

$x^{2} = 5^{2}$

$x^{2}$ = 25

(More to know: when we square a number we multiply a number two times and when we find cube of a number then we multiply a number three times.)

Hence, the value of $x^{2}$ is 25.

To know more about cube here

brainly.com/question/28134860

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We split up the domain of integration into 10 subintervals,

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where the left and right endpoints for the $i$ -th subinterval are, respectively,

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with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$