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Sophie [7]
3 years ago
7

Correct answer gets brainiest

Mathematics
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

x + 2x + a + 90 \degree = 360\degree \\  x + 2x + 132\degree + 90 \degree = 360\degree \\ 3x + 222\degree = 360\degree \\ 3x = 360\degree - 222\degree \\ 3x = 138\degree \\ x =  \frac{138\degree}{3}  \\ \boxed{ x = 46\degree}

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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]

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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.

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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.

3 0
3 years ago
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