Question

Please Help :) ~

Answer 1

whats the question?

Answer 2

sorry, i dont know how to do that

Answer 3

ok neither do i lol

Answer 4

lol. do u know if u can help me with this?

Compare and Contrast. Below are two sets of points. Find the slopes between the two points and compare them. Choose the statement that is true about the values of the slopes.

Set #1 Set #2
(2, 4) and (1, 1) (4, 6) and (2, 7)
The slope of Set #1 is smaller than the slope of Set #2.

The slope of Set #1 is larger than the slope of Set #2.

The slope of Set #1 is the same as the slope of Set #2.

None of the statements above describes slopes of sets shown.

Answer 5

oh, nope sorry i suck in math

Answer 6

$$
\cfrac{n}{m^a} \implies \cfrac{n}{1} \times \cfrac{1}{m^a}
\implies \cfrac{n}{1} \times m^{-a}
\implies nm^{-a}
$$

Answer 7

so if you just take a peek at \(\large \frac{1}{8} y^2 \)
if you multiply them, what would they give you?

Answer 8

could i turn one eight as a decimal (.125)

Answer 9

I don't think it's what's required in this case, because you're dealing with exponentials

Answer 10

just bear in mind that
$$
\cfrac{1}{anything^{\color{red}{exp}}}
\implies anything^{\color{red}{-exp}}
$$

Answer 11

cross multiply ?

Answer 12

yes

Answer 13

\[\frac{ 1 }{ 64 }\]

Answer 14

hmmm

Answer 15

hold on, ahem, I guess we'll stick with 1/8 :/

Answer 16

ok, sorry lol

Answer 17

you have a common factor, see it?

Answer 18

y ?

Answer 19

well, \(\large y^2\)

Answer 20

$$
125x^3y^2-\cfrac{1}{8}y^2\\
y^2\pmatrix{125x^3-\cfrac{1}{8}}
$$

Answer 21

Ah, ok

Answer 22

so, if you keep an eye on 125, 125 = \(5^3\)
and if you take a peek at 8, 8 = \(2^3\)
and well, 1 is really \(1^3\)

Answer 23

ok.

Answer 24

$$
125x^3y^2-\cfrac{1}{8}y^2\\
y^2\pmatrix{125x^3-\cfrac{1}{8}}\\
y^2\pmatrix{(\color{red}{5}x)^3-\pmatrix{\cfrac{\color{red}{1}}{\color{red}{8}}}^3}\\
$$

Answer 25

what is your opinion from B.)

can you check it ?

Answer 26

Me or jdoe?

Answer 27

who are the asker

Answer 28

Im the asker.

Answer 29

so, what you really have is \(\large (a^3 - b^3) \) type of binomial

Answer 30

ok @jdoe0001 but can you prove it now here with calculi ?

Answer 31

.... ahemm, not sure :/

Answer 32

@jhonyy9 do you think the answer is B ?

Answer 33

dunno, what do you think?

\(
(a^3 - b^3) = (a^2-b^2)(a^2+ab+b^2)
\)

Answer 34

I was thinking B.

Answer 35

b) would suggest \(a^2-ab+b^2\)

Answer 36

:/

Answer 37

which is good for \( (a^3+b^3)\)

Answer 38

Bleh, no clue :/

Answer 39

@jdoe0001 can you say me why a^3 + b^3 ? and why not minus ?

Answer 40

well, the original expression factored out with a "-" :)

Answer 41

ok but there not is factored out - y

Answer 42

\(
y^2\pmatrix{(\color{red}{5x})^3-\pmatrix{\cfrac{\color{blue}{1}}{\color{blue}{8}}}^3}\\
(\color{red}{a}^3 - \color{blue}{b}^3) = (\color{red}{a}^2-\color{blue}{b}^2)(\color{red}{a}^2+\color{red}{a}\color{blue}{b}+\color{blue}{b}^2)
\)
so, expand away, see what you get :)

Answer 43

not is right

check it please

Answer 44

ohh yea, you're right,, my bad

Answer 45

Soo, it is B ? 0_o

Answer 46

\(
y^2\pmatrix{(\color{red}{5x})^3-\pmatrix{\cfrac{\color{blue}{1}}{\color{blue}{8}}}^3}\\
(\color{red}{a}^3 - \color{blue}{b}^3) = (\color{red}{a}^2-\color{blue}{b})(\color{red}{a}+\color{red}{a}\color{blue}{b}+\color{blue}{b}^2)
\)

Answer 47

darn, missed one :/

Answer 48

Its ok. :)

Answer 49

this is wrong again too

Answer 50

yes one sec

Answer 51

Ugh, 7 more 4 me yippie -_-

Answer 52

\(
y^2\pmatrix{(\color{red}{5x})^3-\pmatrix{\cfrac{\color{blue}{1}}{\color{blue}{8}}}^3}\\
(\color{red}{a}^3 - \color{blue}{b}^3) = (\color{red}{a}-\color{blue}{b})(\color{red}{a}^2+\color{red}{a}\color{blue}{b}+\color{blue}{b}^2)
\)

Answer 53

hjeheh

Answer 54

@jdoe0001 comen please you know the right way but there not is right 1/8

Answer 55

ahemm, ohh ...shoot... well,, I surely am glad you're here to pick up my typos @jhonyy9 heheh
I'd have been doing some more correcting later :), thanks

Answer 56

heheeh

Answer 57

\(
y^2\pmatrix{(\color{red}{5x})^3-\pmatrix{\cfrac{\color{blue}{1}}{\color{blue}{2}}}^3}\\
(\color{red}{a}^3 - \color{blue}{b}^3) = (\color{red}{a}-\color{blue}{b})(\color{red}{a}^2+\color{red}{a}\color{blue}{b}+\color{blue}{b}^2)
\)

Answer 58

anyhow @PatrickJordon , tis 2 thus 8 = 2^3 :)

Answer 59

@jdoe0001 okey tnx :)

Answer 60

so, from the expansion of (a^3 - b^3), it produces a 2nd binomial with a + all around, so that rules out a) and b)

Answer 61

I was bout to ask if that was gonna happen lol xD

Answer 62

a times b, or 5x times 1/2, is not "5x" as c) is suggesting

Answer 63

Okey!

Answer 64

anyhow there :)