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3 years ago

Wassup someone talk to me

Mathematics

2 answers:

Zinaida [17]3 years ago

8 0

hi dear child, how's school?

-pilar <3

Levart [38]3 years ago

5 0

Answer:

Step-by-step explanation:

whats ur name?

I’m Valeria or Val for short

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I know you’re supposed to change the bounds and break up the integral, but for some reason, I can’t get the 44/3. Can someone ex

tatyana61 [14]

First, look for the zeroes of the integrand in the interval [0, 6] :

x² - 6x + 8 = (x - 4) (x - 2) = 0 ⇒ x = 2 and x = 4

Next, split up [0, 6] into sub-intervals starting at the zeroes we found. Then check the sign of x² - 6x + 8 for some test points in each sub-interval.

• For x in (0, 2), take x = 1. Then

x² - 6x + 8 = 1² - 6•1 + 8 = 3 > 0

so x² - 6x + 8 > 0 over this sub-interval.

• For x in (2, 4), take x = 3. Then

x² - 6x + 8 = 3² - 6•3 + 8 = -1 < 0

so x² - 6x + 8 < 0 over this sub-interval.

• For x in (4, 6), take x = 5. Then

x² - 6x + 8 = 5² - 6•5 + 8 = 3 > 0

so x² - 6x + 8 > 0 over this sub-interval.

Next, recall the definition of absolute value:

$|x| = \begin{cases}x & \text{for }x \ge0 \\ -x & \text{for }x < 0\end{cases}$

Then from our previous analysis, this definition tells us that

$|x^2 - 6x + 8| = \begin{cases}x^2 - 6x + 8 & \text{for }0$

So, in the integral, we have

$\displaystyle \int_0^6 |x^2-6x+8| \, dx = \left\{\int_0^2 - \int_2^4 + \int_4^6\right\} (x^2 - 6x + 8) \, dx$

Then

$\displaystyle \int_0^2 (x^2 - 6x + 8) \, dx = \left(\frac13 x^3 - 3x^2 + 8x\right) \bigg|_0^2 = \frac{20}3 - 0 = \frac{20}3$

$\displaystyle \int_2^4 (x^2 - 6x + 8) \, dx = \left(\frac13 x^3 - 3x^2 + 8x\right) \bigg|_2^4 = \frac{16}3 - \frac{20}3 = -\frac43$

$\displaystyle \int_4^6 (x^2 - 6x + 8) \, dx = \left(\frac13 x^3 - 3x^2 + 8x\right) \bigg|_4^6 = 12 - \frac{16}3 = \frac{20}3$

and the overall integral would be

20/3 - (-4/3) + 20/3 = 44/3

3 0

3 years ago

Marcia claims that the GCF for (2x2 + 4xy + 8xy4) is 8x2y4.

miskamm [114]

She is not correct.
The GCF for 2x^2 + 4xy + 8xy^4
is 2x
The first term does not have a 'y' in it so you cannot have a 'y' in the GCF.
The second and third terms both only have 1 'x', so you can only take out 1 'x', not 2.
The highest constant that all three terms have in common is a 2.

8 0

4 years ago

A car dealership manager has determined that the profit, in thousands of dollars, of the dealership can be determined by the exp

ss7ja [257]

Answer:

The number of cars that brings the same profit is 9 and 0

Step-by-step explanation:

Given

$x^2 - 4x = 5x$