Help me plzzzzz!!!!!

Mathematics

1 answer:

elena55 [62]3 years ago

5 0

1 is on “greater than”
2 is on “less then”
3 is on “between”
4 is on “between”
5 is on “between”
6 is on “less then”
7 is on “greater then”

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BRANILIEST OFFER!!what is the median of this data set?<br> 25, 8, 10, 35,45, 5, 40, 30. 20.

ivann1987 [24]

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5 8 10 20 25 30 35 40 45
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3 years ago

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Find the coordinates of the image after a translation of the figure below using (x, y) (x + 3, y + 2).

san4es73 [151]

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2 years ago

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Evaluate the integral ∫2−1|x−1|dx

defon

I think you might be referring to the definite integral,

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx$

Recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

Then $|x-1|=x-1$ if $x\ge1$ , and $|x-1|=1-x$ is $x$ . So spliting up the integral at <em>x</em> = 1, we have

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx$

The rest is simple:

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}$