Question

man i need help!!!!!!!!!!!!!!!!!!!!!! help me with mi math

Answer 1

What exactly do you need help with?

Answer 2

question??

Answer 3

Divide (x2 - 9) by (x + 3).

Answer 4

x2 - 9 = (x-3)(x+3)

Answer 5

so (x-3)(x+3)/(x + 3) = x-3

Answer 6

We could use a few methods here.
Synthetic division (the divisor polynomial is degree 1 and has leading coefficient 1), long division, or just factoring the dividend polynomial and trying to remove a common factor that way.

Answer 7

Find the product of 3c and (4c - 5).

Answer 8

The product is
(3c)(4c - 5)
we could just use distributive property here, taking 3c to 4c and -5 for the answer.

Answer 9

so whats the answer to it i still dont understand

Answer 10

nevermind i got it thanks for helping me

Answer 11

Distributive Property is... a(b + c) = ab + ac
so, in this case
(3c)(4c - 5) = (3c)(4c) + (3c)(-5) Then, we can reorder and simplify
= (3)(4) (c*c) + (3)(-5) c
= 12c^2 - 15c

okay, no problem. :)

Answer 12

Multiply (2x + 5)(3x - 4). i know its easy but your just so nice

Answer 13

Divide and simplify:


x3



x2 + 2

Answer 14

The usual technique is "FOIL" but I prefer using distribution twice
FOIL: First Outer Inner Last
(2x + 5)(3x - 4)

(2x)(3x) + (2x)(-4) + (5)(3x) + 5(-4)
first outer inner last

Distributive:
(2x + 5)(3x - 4) treating (3x - 4) like the "a" in that distributive property
(2x + 5)a
2x(a) + 5(a)
(2x)(3x - 4) + 5(3x - 4)
(2x)(3x) + (2x)(-4) + (5)(3x) + 5(-4)

Both ways work... from there, we can simplify by multiplying together the coefficients and adding like terms...

6x^2 - 8x + 15x - 20
6x^2 + 7x - 20

Answer 15

Is it
\[ \frac{x^3}{x^2 + 2} \]
?

Answer 16

yes bt i got another question 4exponets9 over4 expontents3. n since where here can i get to kno my tutors name?

Answer 17

there aren't any common factors between x^3 and x^2 + 2, so we end up with a remainder if we divide. I'm not sure if it still wants us to divide it

As for next problem, we could just expand it out and cross out the common terms between the numerator and denominator (since 4/4 = 1, multiplying by 1 will not change the number)
\[ \frac{4^9}{4^3} = \frac{\cancel{4\times4\times4}\times4\times4\times4\times4\times4\times4}{\cancel{4\times 4\times 4}} \]
We would have 6 fours multiplied together as the remaining numbers

Answer 18

whats your name?

Answer 19

sorry, I prefer to go by this alias here. I do not like to give out my personal information on-line. :P