All categories

2 years ago

What is the completely factored form of x3 – 64x?

Mathematics

2 answers:

WITCHER [35]2 years ago

6 0

Answer:

$x(x - 8)(x + 8)$

Step-by-step explanation:

${x}^{3} - 64x \\ x( {x}^{2} - 64) \\ x( {x}^{2} - {8}^{2} ) \\ = x(x - 8)(x + 8)$

hope this helps

brainliest appreciated

good luck! have a nice day!

Arada [10]2 years ago

4 0

Answer:

x( x-8)(x+8)

Step-by-step explanation:

x^3 - 64x

Factor out an x

x( x^2 -64)

This is the difference of squares x^2 -8^2 which is a^2 - b^2 = (a-b)(a+b)

x( x-8)(x+8)

You might be interested in

∠1 is complementary ∠2 and ∠1=62°. Find m∠2.

NeX [460]

Answer:

Step-by-step explanation:

A complimentary angle means that both angles (together) will equal a right angle (90°)

∠1 + ∠2 = 90°

90 - 62 = 28

8 0

2 years ago

Sin 90 +cos 50 -tan1

Feliz [49]

Answer:

1.62533254

Step-by-step explanation:

7 0

2 years ago

I need points 3+4%34

kkurt [141]

Answer:

I do too :(

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Simplify the rational expression. State any restrictions on the variable.

ExtremeBDS [4]

we are given

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}$

Firstly, we will factor numerators and denominators

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}$

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

$n^2-6\neq 0$

$n\neq -+ \sqrt{6}$

we can simplify it

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}$

we know that denominator can not be zero

$n^2-3\neq 0$

$n\neq -+ \sqrt{3}$

so, option-B.......Answer

3 0

3 years ago

An adult ticket costs $5.25 and a child ticket costs $3.05. If $498.60 were collected for 100 tickets, how many child tickets we

lisabon 2012 [21]

Answer:

work is shown and pictured

6 0

2 years ago