Is it (sometimes true/never true) that all three altitudes of a triangle fall outside the triangle?

Midpoint is (-2,2) point a is (-8,-1) what is the endpoint

STatiana [176]

Answer: (-2,2), (-8,-1)

Step-by-step explanation:

4 0

2 years ago

Add<br>show all steps please

choli [55]

$\bf \cfrac{3}{x-2}+\cfrac{4}{x-1}-\cfrac{1}{x}$ so, no breaks on the denominator thus, the LCD will be all three then.

$\bf \cfrac{3}{x-2}+\cfrac{4}{x-1}-\cfrac{1}{x}\implies \cfrac{[3x(x-1)]+[4x(x-2)]-[(x-2)(x-1)]}{x(x-2)(x-1)}
\\\\\\
\cfrac{[3x^2-3x]+[4x^2-8x]-[x^2-3x+2]}{x(x-2)(x-1)}
\\\\\\
\cfrac{3x^2-3x+4x^2-8x-x^2+3x-2}{x(x-2)(x-1)}
\\\\\\
\cfrac{6x^2-8x-2}{x(x-2)(x-1)}\implies \cfrac{2(x^2-4x-1)}{x(x-2)(x-1)}$

3 0

3 years ago

scZoUnD [109]

The answer to this is 21

5 0

2 years ago

Read 2 more answers

ASAP First to get it right brainliest

nlexa [21]

324

Because you multiply 9 by 36

7 0

2 years ago