Geometry Question!! One problem. Please expain fully. Answer is know, but the work to the answer is needed. #19 only~~~

Mathematics

2 answers:

valina [46]3 years ago

8 0

#19. I also got 29,28
4x+ 2x+6= 180 cuz consecutive angles equal 180
6x+6= 180
Subtract 6
6x= 174
Divide by 6
X=29

3y-20+4y+4=180
7y-16=180
Add 16
7y= 196
Divide by 7
Y= 28

Korolek [52]3 years ago

7 0

$\bf \begin{array}{llrll}
-2x+3y=26&\implies &-2x+3y=26\\
x-y=1&\times 2\implies &2x-2y=2\\
&&-------\\
&&0x+1y=28
\end{array}
\\\\\\
\boxed{y=28}\qquad thus\implies x-y=1\implies x=1+y\implies \boxed{x=29}$

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Given the function f(x) x^2-2x/x^3 +9x^2-10x Find any holes, vertical asymptotes, and horizontal asymptotes there may be.

yulyashka [42]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ :

Holes: Since $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ reduces to $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ , then there is a hole at $x=0$ as $x$ exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).

Vertical Asymptotes: If we further reduce $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ to $f(x)=\frac{x-2}{(x+10)(x-1)}$ , then we see that there are vertical asymptotes at $x=-10$ and $x=1$