Question

Can someone please help me, i need the help?

Answer 1

Answer:

$f(x)\div g(x)=x-10$

Step-by-step explanation:

$\frac{x^2-19x+90}{x-9}$

$\frac{(x-10)(x-9)}{x-9}$

$x-10$

Answer 2

Answer:

$f(x)\div g(x)=-10$

Step-by-step explanation:

Solve $f(x)$

$f(x)=x^2-19x+90$

$x^2-19x+90$

$\mathrm{Break\:the\:expression\:into\:groups}$

$\left(x^2-9x\right)+\left(-10x+90\right)$

$\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2-9x$

$=x(x-9)$

$\mathrm{Factor\:out\:-10\;from}\:\:-10x+90$

$=-10(x-9)$

$=x\left(x-9\right)-10\left(x-9\right)$

$\mathrm{Factor\:out\:common\:term\:}x-9$

$\bold{\left(x-9\right)\left(x-10\right)}$

Solve $g(x)$

$g(x)=x-9$

$\bold{x-9}$

$(x-9\right)\left(x-10)}\div \:x-9$

Now, solve:

$(x-9\right)\left(x-10)}\div \:x-9$

$=\frac{(x-9)\left(x-10)}{:x-9}$

$= -10$

Therefore, $f(x)\div g(x)=-10$