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3 years ago

Guys please help with this question thanks

Mathematics

1 answer:

erastova [34]3 years ago

3 0

= 16 + 49 - 3(11) - 4(10)
= 16 + 49 - 33 - 40
= -8

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allsm [11]

Answer:

Its D.

Step-by-step explanation:

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Devi has a grid in the shape of a parallelogram. She has equilateral triangles with an area of 40 square millimeters that fit on

Ludmilka [50]

The area of the parallelogram is 26 × 10 = 260 cm^2 ;
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Step-by-step explanation:

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How to find the real number root of the negative square root of 36

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3 years ago

Integrate the following w.r.t x 1) 2x^2/3. 2) (5-x)^23

Kryger [21]

Answer:

A) $\int\frac{2x^2}{3}dx=\frac{2x^3}{9}+C$

B) $\int(5-x)^{23}dx=-\frac{(5-x)^{24}}{24}+C$

Step-by-step explanation:

So we have the integral:

$\int\frac{2x^2}{3}dx$

First, remove the constant multiple:

$=\frac{2}{3}\int x^2\dx$

Use the power rule, where:

$\int x^ndx=\frac{x^{n+1}}{x+1}$

Therefore:

$\frac{2}{3}\int x^2\dx\\=\frac{2}{3}(\frac{x^{2+1}}{2+1})$

Simplify:

$=\frac{2}{3}(\frac{x^{3}}{3})$

And multiply:

$=\frac{2x^3}{9}$

And, finally, plus C:

$=\frac{2x^3}{9}+C$

We have the integral:

$\int(5-x)^{23}dx$

To solve, we can use u-substitute.

Let u equal 5-x. Then:

$u=5-x\\du=-1dx$

So:

$\int(5-x)^{23}dx\\=\int-u^{23}du$

Move the negative outside:

$=-\int u^{23}du$

Power rule:

$=-(\frac{u^{23+1}}{23+1})$

Add:

$=-(\frac{u^{24}}{24})$

Substitute back 5-x:

$=-(\frac{(5-x)^{24}}{24})$

Constant of integration:

$=-\frac{(5-x)^{24}}{24}+C$

And we're done!

8 0

3 years ago