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3 years ago

Factor the trinomial X2-12x +16

Mathematics

1 answer:

wlad13 [49]3 years ago

6 0

2*&/45/+=67364/3/4/32:5,6544 is hothead answer

You might be interested in

What does increase mean

Darya [45]

Increase- to be more significant in value, to be of more worth, to have a larger quantity.

Basically, if I said I added 2 to 5, that would be an increase.

2+5=7 7 is bigger than 5.

If I subtracted 2 from 5, that would make a decrease in value.

5-2=3 3 is smaller than 5.

I hope this helps!
~kaikers

8 0

3 years ago

A random sample selected from an infinite population is a sample selected such that each element selected comes from the same __

bulgar [2K]

Answer: population; independently

Step-by-step explanation:

A random sample selected from an infinite population is a sample selected such that each element selected comes from the same *population* and each element is selected *independently*.

3 0

3 years ago

Suppose y=f(x)+k. what effect does k have on the parent function?

Aleksandr [31]

It moves it up and down

6 0

3 years ago

Is the following relation a function? (4,2), (1,1), (0,0), (1, -1), (4, -2). Give the domain and range.

Vlada [557]

(x,y)
to be a function
for every x, ther must be only 1 y to corespond to it
basically, x must NEVER repeat with a different y
list them

ah, we see (1,1) an d(1,-1)
also (4,2) and (4,-2)
double offense
1 repeats with 1 and -1
4 repeats with 2 and -2

not a function

domain is all x values
range is all y values
if they repeat, don't list them

domain=(0,1,4)
range=(-2,-1,0,1,2)

8 0

3 years ago

For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

pav-90 [236]

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$

$\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1$

All this to say that f(x) is differentiable everywhere on its domain, except at the point x = -1.

4 0

3 years ago