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sammy [17]
2 years ago
5

What is the following product? startroot 30 endroot times startroot 10 endroot

Mathematics
1 answer:
aleksandrvk [35]2 years ago
7 0

Answer:

Hello there! 10 square of 3

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Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i
Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

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So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333

So the correct answer is C, the area between the two functions is 4/3 units.

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3 years ago
Y-2=-4<br><br> Help please tyy
kolezko [41]

Answer:

y = -2

Step-by-step explanation:

y - 2 = -4

add 2 to both sides

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kakasveta [241]

From what I see, I believe the answer is [A] about 65 degrees

[B] would be too low

[C] would be close but no right

[D] would just be flat out straight

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