16 = k/11

You’re going to need to get K by itself

So you need to multiply 11 on both sides

to remove the denominator of 11

16 x 11 = 176

k/11 x 11 = k

So now the answer would be

176 = k

Hope this helped :)

5 0

3 years ago

Suppose that a function f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence

Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

$\lim_{x \to c} \frac{x^2-cx}{(x-c)}$

We can simplify it

$\lim_{x \to c} \frac{x(x-c)}{(x-c)}$

$=\lim_{x \to c} x$

$=c$

so, we can see that limit exist and it's value defined

case-2:

$\lim_{x \to c} \frac{1}{(x-c)}$

Left limit is

$\lim_{x \to c-} \frac{1}{(x-c)}$

$=-\infty$

Right Limit is

$\lim_{x \to c+} \frac{1}{(x-c)}$

$=+\infty$

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0

2 years ago