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Finger [1]
3 years ago
7

Find the distance between the points (4,-2), (4,-5)

Mathematics
1 answer:
kiruha [24]3 years ago
6 0

\text{Given that,}\\\\(x_1,y_1) = (4,-2)~~ \text{and}~~  (x_2 ,y_2) = (4,-5)\\\\\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2 -y_1)^2}\\\\~~~~~~~~~~~~=\sqrt{(4-4)^2 + (-5+2)^2}\\\\~~~~~~~~~~~~=\sqrt{0^2+(-3)^2}\\\\~~~~~~~~~~~~=\sqrt{0+9}\\\\~~~~~~~~~~~~=\sqrt 9\\\\~~~~~~~~~~~~=3

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How do I demonstrate that there is only one value of x (between 0 and 2pi) that verifies cosx=3/5 and sinx=4/5
Varvara68 [4.7K]

well, as we can see, the cos(x) as well as the sin(x) are both positive, that puts the angle in the I Quadrant only, everywhere else they differ in signs or are both negative.


\bf cos(x)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\qquad \qquad \qquad sin(x)=\cfrac{\stackrel{opposite}{4}}{\stackrel{hypotenuse}{5}}


that simply means it has an adjacent side of 3, opposite side of 4, and a hypotenuse of 5, like you see it in the picture below.


and we can get the inverse function of either function to get about 53.130102354° for ∡x.

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KATRIN_1 [288]

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Step-by-step explanation:

Let the two positive integers be x and (5 + x).

According to the given description:

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