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ad-work [718]
3 years ago
13

Marking brainliest! Please help

Mathematics
1 answer:
Ksju [112]3 years ago
7 0

Answer:

I believe it would be -2

Step-by-step explanation:

The midpoint of (0,4) and (4,-8) is (2,-2)

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PLEASE HELP HELP ASAP
Anit [1.1K]

Answer:

12.7311111111111 on and on

Step-by-step explanation:

3 0
3 years ago
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Can anybody help this question?
Oxana [17]

Answer:

the first option

Step-by-step explanation:

really, just look at the table.

is the mean value (10.4) larger than the median (13.4) ?

I hope you can see right away that it is not.

and you can see they are not the same either.

so, all the answer options mentioning mean larger than median or equal to median can be ruled out right away.

so, it is between the first two options.

now think ! how do we draw number lines ? a coordinate axis ?

the smaller numbers left, the larger numbers right. the numbers grow from left to right.

the mean value is simply the sum of all measurements divided by the number of measurements (how many median were done). if that is smaller that the median (so, the Mean is left of the Median), it means that the majority of measurements had a result smaller (to the left) than the Median. so, it is skewed-left.

6 0
2 years ago
What dose slope represent in a problem.
Angelina_Jolie [31]

Answer:

Slope is going to be the "M" or gradient of the line. It represents the change in the Y value over time as the line continues. It tells you how high or low it points depending on what the slope is.

Thanks for letting me help!!

4 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
Simplify please sorry last one(2y+1)^2-4y^2+2
Sergeu [11.5K]
4y^2+4x+1-4y^2+2
4y+1+2
4y+3
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3 years ago
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