Select the lines that represent functions.

Mathematics

2 answers:

Neko [114]3 years ago

5 0

It’s the line that starts at -5.. Box at the bottom!!

Rama09 [41]3 years ago

5 0

It’s the straight line near the bottom that’s coordinates are (0,-5) can i get brainliest? hope this helps!

You might be interested in

Answers to 17 and 19

Vesnalui [34]

#17) 8 cubic feet.
#19) 5 miles.

5 0

3 years ago

How to evaluate question number 4

ANTONII [103]

There are 2 ways that spring to mind.

One is to use the definition of the derivative at a point:

$f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

In this case, $c=0$ and $f(x)=\sqrt{5-x}$ . Then

$f'(x)=-\dfrac1{2\sqrt{5-x}}\implies f'(0)=\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\dfrac1{2\sqrt5}$

- - -

If you don't know about derivative yet (I would think you do, considering this is for a midterm in AP Calc, but I digress), the other way is to rely on algebraic manipulation. Multiply the numerator and denominator by the conjugate of the numerator to get

$\dfrac{\sqrt{5-x}-\sqrt5}x\cdot\dfrac{\sqrt{5-x}+\sqrt5}{\sqrt{5-x}+\sqrt5}=\dfrac{\left(\sqrt{5-x}\right)^2-\left(\sqrt5\right)^2}{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac x{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac1{\sqrt{5-x}+\sqrt5}$