heartSTORM907

yeah!! that's so helpful!! thank you so much i get it now :D

thank you!!! that was really helpful!

To find the largest real number $c$ such that $1$ is in the range of $f(x) = x^2 - 5x + c - 3x + 8$, we first simplify the function to: $$f(x) = x^2 - 8x + (c + 8)$$ For $1$ to be in the range of $f(x)$, there must be some value(s) of $x$ for which $f(x) = 1$. This can be written as: $$x^2 - 8x + (c + 7) = 0$$ For this equation to have at least one real root, the discriminant of the quadratic equation must be non-negative: $$\Delta = b^2 - 4ac$$ $$\Delta = (-8)^2 - 4 * 1 * (c + 7)$$ $$\Delta = 64 - 4c - 28$$ $$\Delta = 36 - 4c$$ For $\Delta$ to be non-negative, $36 - 4c \geq 0$, or $c \leq 9$. Therefore, the largest real number $c$ for which $1$ is in the range of $f(x)$ is c = 9.

try punching it into a calculator! :)

$5c+cd = 5(\frac{1}{5}) + (\frac{1}{5})(15)$

OH MY GOD i almost completely forgot about that part.... i kept thinking i was doing something wrong lol

thank you so so so much!!!!

thank you so so so so much!!!!! i think i understand now :)