Yooo plz help me w dis asap!!!

Factor Completely.<br><br> x^4-16x <br><br> Group of answer choices

Verdich [7]

Answer:

x(x^3-16)

Step-by-step explanation:

The only thing they both have in common that you can factor out is an x.

7 0

3 years ago

Chad ran 1/2 mile in 3/4 of an hour. At this rate of speed, how far would he run in 2 hours?

Rudik [331]

Explanation is in the file

tinyurl.com/wtjfavyw

4 0

3 years ago

Describe and correct the error in the solution, make sure to show your work

ASHA 777 [7]

Hope i helped!!! good luck.

5 0

3 years ago

Daniel's savings account balance is 20 times the amount of Henrys savings account balance. The total amount of money contained i

Juli2301 [7.4K]

Answer: Daniel has $22 in his savings account balance.

Step-by-step explanation:

Let Daniel's savings account balance be x

and Henry's savings account balance be y, then as given Daniel's savings account balnce is 20 times that of Henry's, We get $x=20y$ ...........(1)

And also total amount of both savings account is $462 so we get $x+y=462$ ......(2)

Now by substituting values of x from (1) into (2) we get,

$20y + y=462\\21y=462\\y=22$

And $x=$440$

So, Daniel's savings account balance is $440

and Henry's savings account balance is $22

5 0

3 years ago

Read 2 more answers

Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2.

insens350 [35]

Answer:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

Step-by-step explanation:

Given

$p(x) = \frac{x^2-2x-3}{x+2}$ -- Missing from the question

Required

The behavior of the function around its vertical asymptote at $x = -2$

$p(x) = \frac{x^2-2x-3}{x+2}$

Expand the numerator

$p(x) = \frac{x^2 + x -3x - 3}{x+2}$

Factorize

$p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}$

Factor out x + 1

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

$x -> -2^{-}$ Say x = -3

$p(x) = \frac{(x -3)(x + 1)}{x+2}$

$p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12$

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

$x->-2^{-}, p(x)->-\infty$

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: $x>-2$

Say x = -2.1

$p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1$

We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

$x->-2^{+}, p(x)->-\infty$

So, the behavior is:

$x->-2^{-}, p(x)->-\infty$ and as $x->-2^{+}, p(x)->-\infty$

6 0

3 years ago