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Murrr4er [49]
3 years ago
6

Calculate the distance between the points F=(-2,1) and A=(4,-2) in the coordinate plane. Round your answer to the nearest hundre

dth
Mathematics
1 answer:
Anton [14]3 years ago
8 0

Answer:

The distance between F and A is approximately 6.71 units.

Step-by-step explanation:

To find the distance between any two points, we can consider using the distance formula. Recall that:

\displaystyle d  = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let F(-2, 1) be (<em>x</em>₁, <em>y</em>₁) and A(4, -2) be (<em>x</em>₂, <em>y</em>₂).

Substitute and evaluate:

\displaystyle \begin{aligned} FA &= \sqrt{((4) - (-2))^2 + ((-2) - (1)^2} \\ \\&=\sqrt{(6)^2 + (-3)^2} \\ \\&=\sqrt{36 + 9} \\ \\ &= \sqrt{45} \\ \\ &= 3\sqrt{5}\\ \\&\approx 6.71\text{ units}  \end{aligned}

In conclusion, the distance between <em>F</em> and <em>A</em> is approximately 6.71 units.

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\displaystyle \int \csc(x)\left(\sin(x)+\cot(x)\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}\left(\sin(x)+\frac{\cos(x)}{\sin(x)}\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}*\sin(x)+\frac{1}{\sin(x)}*\frac{\cos(x)}{\sin(x)}dx\\\\\displaystyle \int 1+\frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle \int 1 dx+\int \frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle x+C_1+\int \frac{1}{u^2}du \ \text{ ... where } u = \sin(x)\\\\\displaystyle x+C_1+\int u^{-2}du\\\\

Part 2

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Don't forget about the plus C constant

==========================================================

Problem 26

Fortunately, there aren't as many steps for this problem.

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