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Leviafan [203]
3 years ago
13

Type the correct answer for each.

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
If point \left(x,\ \frac{\sqrt{7}}{3}\right) is on the unit ircle, then:

x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\  \\ \Rightarrow x^2+ \frac{7}{9} =1 \\  \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9}  \\  \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}

Since, the point is in the second quadrant, x is negative.

Thus, (x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)

Part A:

3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\  \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\  \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\  \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta

Therefore, -6\tan\theta=3\sqrt{7}.



Part B:

- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} }  \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x}  \\  \\ =\tan\theta

Therefore, \tan\theta=- \frac{ \sqrt{7} }{2}
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The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
2 years ago
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Answer:

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Step-by-step explanation:

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