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2 years ago

9

3x∧2±6x±x±20 sole by factorization

1 answer:

Trava [24]2 years ago

8 0

Answer:

just look at the algebra formula and simplify to get your answer. Remember to use your powers well

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Can someone pls help me find what is 1/10+12/20?

irga5000 [103]

Answer:

$\frac{7}{10}$

Step-by-step explanation:

$\frac{1}{10}$ can also be written as $\frac{2}{20}$

$\frac{2}{20} +\frac{12}{20} =\frac{14}{20}$

Simply to $\frac{7}{10}$

-OTHER METHOD-

$\frac{12}{20}$ can be simplified to $\frac{6}{10}$

$\frac{1}{10} +\frac{6}{10} =\frac{7}{10}$

7 0

3 years ago

Read 2 more answers

Plz help this is due tomorrow

kkurt [141]

7:8 Purl Stitches I believe ! :)

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3 years ago

A deposit of $35.45 was made to a checking account. Before the deposit, the account had a balance of −$12.39 .

Naddika [18.5K]

The answer would be 23.06.

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3 years ago

19. Which of the following is the measure of ÐQRS? A. 80° B. 100° C. 115° D. 90°

Ede4ka [16]

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3 years ago

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

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

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

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

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

which is about 651.918, so that the area is approximately $651.918\pi\approx\boxed{2048}$ .

Compare this to actual value of the integral, which is closer to 1967.

4 0

3 years ago