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kaheart [24]
4 years ago
10

How am I supposed to integrate this?

Mathematics
1 answer:
svlad2 [7]4 years ago
4 0
6x^2-4x+1=6\left(x-\dfrac13\right)^2+\dfrac13\ge0

which means the parabola lies above the x-axis over its entire domain. This means the area is given by

\displaystyle\int_{-1}^2(6x^2-4x+1)\,\mathrm dx=2x^3-x^2+x\bigg|_{x=-1}^{x=2}=10-(-5)=15
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A) Find the value of 8^1/3<br> b) Find the value of 8^2/3<br> c) Find the value of 16^3/4
Valentin [98]

Answer:

a)

\frac{ {8}^{1} }{3}  \\  =  \frac{8}{3}  \\  = 2.67

b)

\frac{ {8}^{2} }{3 } \\  =  \frac{64}{3}  \\  = 21.33

c)

{ \frac{16}{4} }^{3}  \\  =  \frac{4096}{4}  \\  = 1024

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3 years ago
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Flura [38]
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3 years ago
Diagram not drawn to scale<br> please helppp
pychu [463]

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144?

Step-by-step explanation:

8 0
3 years ago
Please help. Find the value for x to the nearest tenth. See picture.
kogti [31]
To me the answer is 6.4 and welcome get a good grade
5 0
3 years ago
I need help I’m struggling
andrew-mc [135]
All you need for a point to be left unchanged is it being on the line of reflection. (Generally speaking)

The first one is wrong cause you can have a line of reflection anywhere. If the line is horizontal ( x axis) then the slope would be zero. Whereas a line on y axis would be an undefined slope. There are also diagonal lines of reflections which have slopes

I am pretty sure the second one would be correct because, as previously stated, all the line needs to stay in the same place is it being on the line of reflection.

Again, lines of reflection do not have a set slope, a diagonal line of reflection going like / on the chart through the axis could have a slope of one, but it is not needed for the point to remain in the same place

The point does not necessarily need to be on the origin, it can be anywhere on the line of reflection, and if the line of reflection does not pass through the origin, a point on the origin would be moved in a reflection.

I hope this makes sense and helps you out a bit... have a good day
8 0
3 years ago
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