Question

I need help. I`m trying to finish my FLVS and the equation is 8/x + 2/9 = 1/3. I know the x = 72 but can someone show me how to solve it?

Answer 1

\[\frac{8}{x} + \frac{2}{9} = \frac{1}{3}\]\[\frac{8}{x} \times \frac{9}{9}+ \frac{2}{9} \times \frac{x}{x}= \frac{1}{3} \times \frac{3x}{3x}\]\[\frac{72 + 2x}{9x} = \frac{3x}{9x}\]Drop the denominators.\[72 + 2x = 3x\]\[72 = x\]

Answer 2

Btw, what math are you doing on flvs?

Answer 3

Algebra 2 honors!

Answer 4

Oh. I literally JUST took the final exam yesterday! lol

Answer 5

Haha I got another question. Can you help?

Answer 6

Sure.

Answer 7

It`s just like the one i asked but instead of us given the equations it just says x=64. So what are the equations?

Answer 8

I don't see how that works. Is it an essay problem?

Answer 9

Well it`s for the collaboration activity with candy.

Answer 10

Oh ok. THat one. You have to basically set up a rational expression using the candy. I did that project alone...let me recheck the details.

Answer 11

Okay thanks (:

Answer 12

Are you trying to finish tonight? The deadline is tomorrow.

Answer 13

Yes I`m trying to finish tonight.

Answer 14

ALright. Yeah. all work has to be in by midnight tonight.

Answer 15

Yeah I know. I just have

Answer 16

*i just have to do the collaboration then two other things and I`m done.

Answer 17

Alright. Did they give you the information on how much candy already?

Answer 18

Or did you make it up?

Answer 19

Made it up. Lol. If 64 is too hard you can change it. :p

Answer 20

Alright. Did you set up rational expressions or rather numbers you got from the candy?

Answer 21

Like \[\frac{8}{64} + \frac{32}{64} = \frac{40}{64}\]Then you simplified everything except the \[\frac{8}{64}\]and put x in for the 64 in what I just put above.
\[\frac{8}{x} + \frac{1}{2} = \frac{5}{8}\]

Answer 22

Something like that?

Answer 23

Yeah so the x would equal 64 right?

Answer 24

Yeah. You're supposed to give it to another student and solve for x.

Answer 25

I know my friend is here with me and thats the equation I have to solve. So how would I solve it?

Answer 26

\[\frac{8}{x} + \frac{1}{2} = \frac{5}{8}\]\[\frac{8}{x} \times \frac{8}{8} + \frac{1}{2} \times \frac{4x}{4x}= \frac{5}{8} \times \frac{x}{x}\]\[\frac{64}{8x} + \frac{4x}{8x} = \frac{5x}{8x}\]You can drop the denominators now that they have a common denominator. \[64 + 4x = 5x\]

Answer 27

Can you're friend solve it now?

Answer 28

*your

Answer 29

No. lol.

Answer 30

Do you understand what I was doing when I multiplied stuff?

Answer 31

No I don`t. I feel stupid. Lol.

Answer 32

Ok. What I'm doing there is finding a common denominator. The common denominator is 8x. What would I have to multiply x by to get 8x?

Answer 33

8.

Answer 34

ALright. However, if you multiplied the denominator, you have to do the same to the numerator. Do you see why I multiplied by \[\frac{8}{8}\]? I'm multiplying in both the numerator and the denominator by the same amount.

Answer 35

I got it now. Thank you (:
Did you do the Activity for Module 6?

Answer 36

Yup. I did the rectangular prism because it has less decimals and less stuff with π in it.

Answer 37

Just read the directions and it'll tell you what to do. I can't really help you much until you have your polynomial set up.

Answer 38

Well I`ll try to set it up really quick if you can still help me after?

Answer 39

Sure.

Answer 40

Thank you so much (:
Did you take Honors or general?

Answer 41

Honors

Answer 42

I did Module 10 too.

Answer 43

Yeah me too. Was the final hard?

Answer 44

Um...there's a practice final exam to see how hard it is. It's about 53 questions with like 7 essay questions.

Answer 45

Literally, anything that's not on the practice is not on the real final.

Answer 46

ohh okay. if i have any questions I`m going to ask you! Lol.

Answer 47

lol. Just hope that I'm online when that happens. :)

Answer 48

Well I`m about to take it as soon as I finish this activity.

Answer 49

Don't you have another activity?

Answer 50

The activity with the box from module 6.

Answer 51

I`m kind of confused on this one too.
The length is 8 inches
The width is 2 inches
And the height is 10 inches

Answer 52

Ok. THose are the measurements. What did it say to do next?

Answer 53

To find the volume. So i did that and it =160

Answer 54

Ok. After that?

Answer 55

Now suppose you knew the volume of this object and the relation of the length to the width and height, but did not know the length. Rewriting the equation with one variable would result in a polynomial equation that you could solve to find the length.

Rewrite the formula using the variable x for the length. Substitute the value of the volume found in step 2 for V and express the width and height of the object in terms of x plus or minus a constant. For example, if the height measurement is 4 inches longer than the length, then the expression for the height will be (x + 4).

Simplify the equation and write it in standard form.
Find the solutions to this equation algebraically using the Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes' Rule of Signs, and the Factor Theorem.

(Hint: If the numbers are large, graph the function first using GeoGebra to help you find one of the zeros. Use that zero to find the depressed equation which can be solved by factoring or the quadratic formula.)

Substitute 0 for the function notation and, using graphing technology, graph the function.

Answer 56

Alright. Where are you confused here?

Answer 57

You know the volume right? 160 right? The formula for volume is lwh. It says you however supposedly DON'T know the length, but do know the relations. The height is 2 inches bigger than the length. The width is 6 inches smaller than the length.
160 = x(x - 6)(x + 2)

Answer 58

So now what?

Answer 59

Multiply x(x - 6)(x + 2) out.

Answer 60

So (x^2-6x)(x^2+2x) or is there more to it?

Answer 61

160 = x(x - 6)(x + 2)
160 = x(x^2 - 4x - 12)
160 = x^3 - 4x^2 - 12x
0 = x^3 - 4x^2 - 12x - 160
Can you solve now using synthetic division, rational root theorem, and Descartes Rule of Signs.

Answer 62

I know how to do synthetic division but what number do i use in the little box?