Please help! And thank you for your time.

Mathematics

1 answer:

Ulleksa [173]4 years ago

6 0

B. You can get this by factoring the bottom of the first expression and then cancelling the terms. Then multiply by the inverse of the second.

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T(6, 12) is the midpoint of CD The coordinates of D are (6, 15). What are the coordinates of C?

vovikov84 [41]

Answer:

(6,9)

Step-by-step explanation:

Let the coordinates of C be (x,y).

$\frac{6+x}{2}=6 \implies x=6 \\ \\ \frac{15+y}{2}=12 \implies y=15 \\ \\ \therefore C=(6,9)$

7 0

1 year ago

Does anyone know the answer to these two questions? thank you.

tamaranim1 [39]

Problem 24

Part 1

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

Part 2

$\displaystyle x+C_1+\frac{1}{1+(-2)}u^{-2+1}+C_2\\\\\displaystyle x+C_1+\frac{1}{-1}u^{-1}+C_2\\\\\displaystyle x+C_1-u^{-1}+C_2\\\\\displaystyle x+C_1-\frac{1}{u}+C_2\\\\\displaystyle x+C_1-\frac{1}{\sin(x)}+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C_1+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C\\\\\displaystyle x-\csc(x)+C\\\\$