pooja195
  • pooja195
@mathmate
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
pooja195
  • pooja195
@mathmate
pooja195
  • pooja195
@Nnesha its not spam ;)
pooja195
  • pooja195
@Nnesha its not spam ;)

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More answers

mathmate
  • mathmate
Yes!
mathmate
  • mathmate
What's your question?
pooja195
  • pooja195
O_o question? XD 2+2=?
mathmate
  • mathmate
Not allowed to give direct answers. Check the drawing: |dw:1433043776037:dw|
pooja195
  • pooja195
4 LOL
pooja195
  • pooja195
@mathmate ;-;
mathmate
  • mathmate
That means: 1. i dont like this and 2. i dont understand this OR 3. i dont want to do this Which one/ones?
pooja195
  • pooja195
3 +_+
mathmate
  • mathmate
gimme a minute!
mathmate
  • mathmate
Calculate \(\large \frac{1}{2}+\frac{2}{3}\). Show work.
pooja195
  • pooja195
7/6
mathmate
  • mathmate
Calculate 12+23. \(\huge \color{red}{Show work.}\)
pooja195
  • pooja195
Why are we doing this? .-.
mathmate
  • mathmate
Because it is the same way you do this as with rational fractions. This is a pretest!
pooja195
  • pooja195
\[12+23=35\]
pooja195
  • pooja195
idk how to show work for that .-.
pooja195
  • pooja195
2+1=3 3+2=5
mathmate
  • mathmate
ok, that's how it works: Calculate \(\large \frac{1}{2}+\frac{2}{3}\), show work.
mathmate
  • mathmate
\(\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}\)
mathmate
  • mathmate
Do you know why we multiply the first fraction by three, and the second by two?
pooja195
  • pooja195
to get a common denominator
mathmate
  • mathmate
Exactly!
mathmate
  • mathmate
Just a review of terms: 1/2 = 2/4 because they are __________________
pooja195
  • pooja195
equivalent
mathmate
  • mathmate
exactly "equivalent fractions"! That's a good start! You know your stuff.
pooja195
  • pooja195
:)
mathmate
  • mathmate
The idea of adding and subtracting rational fraction is exactly the same!
mathmate
  • mathmate
We multiply top and bottom of each term by a factor so that the denominator becomes the common denominator.
mathmate
  • mathmate
Since each term remains an equivalent fraction, the answer will not change, BUT...
mathmate
  • mathmate
we now only have to add the numerators of the equivalent fractions, just like the numerical example.
mathmate
  • mathmate
\(\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}=\frac{3}{6}+\frac{4}{6}=\frac{7}{6}\)
mathmate
  • mathmate
Sorry it took a while!
pooja195
  • pooja195
this looks easy :)
mathmate
  • mathmate
The next one will be equally easy! gimme a minute.
mathmate
  • mathmate
Calculate \(\large \frac{1}{x}+\frac{2}{x^2}\) show work!
pooja195
  • pooja195
\[\huge~\frac{ 1*x }{ x*x }+\frac{ 2 }{ x^2}=\frac{ 1x+2 }{ x^2 }\]
mathmate
  • mathmate
Excellent! Just remember that 1x is usually written as x, except when showing work.
mathmate
  • mathmate
Now: Calculate \(\large \frac{2}{(x-1)}-\frac{1}{(x-1)^2}\) show work!
pooja195
  • pooja195
\[\frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }=\frac{ 1(x-1) }{ (x-1) }=1\]
mathmate
  • mathmate
The denominator is still meant to be (x-1)^2 ? You need to do the math on the numerators. Give it another shot!
mathmate
  • mathmate
T_T not allowed!
pooja195
  • pooja195
;-;
pooja195
  • pooja195
i dont get it
mathmate
  • mathmate
Good! I'll explain!
mathmate
  • mathmate
\(\large \frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }\) is perfect!
mathmate
  • mathmate
At this point, both denominators are identical, so we can do the math just on the numerators \(\large \frac{ 2*(x-1) -1 }{ (x-1)^2 }\)
mathmate
  • mathmate
following so far?
pooja195
  • pooja195
yes
mathmate
  • mathmate
Now can you do the math (on the numerator) and finish?
pooja195
  • pooja195
\[\frac{ x(x-1) }{ (x-1)^2 }\]
mathmate
  • mathmate
You need to distribute the first term, and add like terms, 2*(x-1)-1 = 2x-2 -1 =2x-3
mathmate
  • mathmate
following?
pooja195
  • pooja195
Yes
mathmate
  • mathmate
So the final answer is....
pooja195
  • pooja195
\[\huge\frac{ x(x-1) }{ 2x-3 }\]
pooja195
  • pooja195
>_<
pooja195
  • pooja195
wrong spot...
pooja195
  • pooja195
\[\frac{ 2x-3 }{ (x-1)^2 }\]
mathmate
  • mathmate
Whew! Yes, excellent!
mathmate
  • mathmate
One thing we did without saying it is...nothing. We did "nothing" to factorize the numerator, because it was "obvious" there are no factors.
mathmate
  • mathmate
In general, after the arithmetic on the numerator, we need to factor the numerator, can you tell me why?
pooja195
  • pooja195
:/ so its easier to multiply or is it to break down the numbers? :/
mathmate
  • mathmate
So that we can cancel (with condition) IF there are common factors between the numerator and denominator, just like what we did before in the simplification. Does that make sense?
pooja195
  • pooja195
yes
mathmate
  • mathmate
Are there any points to clarify, or are we good?
pooja195
  • pooja195
We're good.
mathmate
  • mathmate
ok, now, try this: Calculate \(\large \frac{3}{x}-\frac{2}{x-1}\) show work!
pooja195
  • pooja195
ummm O_O
mathmate
  • mathmate
Look at the denominators.
mathmate
  • mathmate
there are no common factors! So you just "cross multiply", which is a short cut when there are no common factors. like \(\large \frac{1}{2}+\frac{3}{5}= \frac{1*5}{2*5}+\frac{2*3}{x*5}=\frac{1*5+2*3}{2*5}=\frac{11}{10}\)
mathmate
  • mathmate
* 2*5 in denominator
pooja195
  • pooja195
O_o
pooja195
  • pooja195
i thought u can only do that if there is an equal sighn :/
mathmate
  • mathmate
It's not a real cross multiplication, but it works in a similar way.
mathmate
  • mathmate
The pattern helps when there are no common factors. You can use that in the new problem.
mathmate
  • mathmate
Calculate \(\large \frac{3}{x}-\frac{2}{x-1}\), show work! Remember that when there is no common factors in the denominators, the common denominator is just the product, namely x(x-1)
pooja195
  • pooja195
:?
mathmate
  • mathmate
\(\large \frac{3}{x}-\frac{2}{x-1}=\frac{3(x-1)}{x(x-1)}-\frac{2x}{(x-1)x}=\frac{3(x-1)-2x}{x(x-1)}\)
pooja195
  • pooja195
hmm ok makes a bit more sense
mathmate
  • mathmate
Good! you can finish it?
pooja195
  • pooja195
\[\frac{ x-1 }{ x(x-1) }\]
mathmate
  • mathmate
brb
mathmate
  • mathmate
IF it was x(x-1)/(x-1), you would have simplified it to x, x\(\ne\)1. But 3(x−1)−2x=3x-3 -2x (distribute 3(x-1) to 3x-3 first, then add -2x) =x-3
mathmate
  • mathmate
With x-3 as numerator, what would the answer be then?
pooja195
  • pooja195
>_< idk :(
mathmate
  • mathmate
the common denominator does not change, so still x(x-1) Answer would then be: \(\huge \frac{ x-3 }{ x(x-1) }\) nothing to further simplify!
mathmate
  • mathmate
Here's one for you to think about when I'm off for a few minutes. Calculate \(\large \frac{2}{(x+1)}+\frac{1}{(x-1)}\) , show work! Do remember to work out the math in the numerator, and the common denominator (product of x+1 and x-1) does not change.
mathmate
  • mathmate
@pooja195
pooja195
  • pooja195
\[\frac{ (x-1)*2 }{ (x-1)*(x+1) }+\frac{ (x+1)*1 }{ (x+1)*x-1) }=\frac{ 3(x+1)(x-1) }{ (x-1)(x+1) } \]
mathmate
  • mathmate
The first part is very good!
mathmate
  • mathmate
Recall that since the denominator is the same, and does not change, we need to ADD the numerators (and not multiply).
mathmate
  • mathmate
So it would read: \(\LARGE \frac{ (x-1)*2 + (x+1)*1}{ (x-1)(x+1) }\) \(=\LARGE \frac{ 2X-2 + X+1}{ (x-1)(x+1) }\) \(=\LARGE \frac{ 3X-1 }{ (x-1)(x+1) }\), OR \(=\LARGE \frac{ 3X-1 }{ x^2-1 }\)
pooja195
  • pooja195
O_O
mathmate
  • mathmate
Your steps are correct. It's just the last step of expanding and adding that there was a correction to make. Are you ok with the correction, or confused?
pooja195
  • pooja195
x_x
mathmate
  • mathmate
Well, if it's the part of expansion perhaps you could use some help. (x-1)*2 + (x+1)*1 is the same as 2(x-1) + (x+1) .... a coefficient of 1 is understood. Then we distribute, basically a half FOIL 2*x + 2*(-1) + x+1 ..... parentheses after a plus sign can be removed without change that makes 2x -2 +x +1 Now add like terms 2x+x -2+1 =3x-1 this is the numerator, since the denominator does not change, we have as the answer: =\(\Large \frac{ 3X-1 }{ x^2-1 }\)
pooja195
  • pooja195
seems easier... :)
mathmate
  • mathmate
Good! Would you like to do one while I'm offline, probably in less than 5 minutes?
pooja195
  • pooja195
Sure i gotta go too xD
mathmate
  • mathmate
\(\large \frac{3}{x-1}-\frac{2x}{x^2-1}\)
pooja195
  • pooja195
huh? O_o
pooja195
  • pooja195
http://prntscr.com/7bnxzi
mathmate
  • mathmate
Simplify \(\large \frac{3}{x-1}-\frac{2x}{x^2-1}\). Show work!
pooja195
  • pooja195
im not sure how to do this .-.
mathmate
  • mathmate
I would factor the second denominator, x^2-1 into (x+1)(x-1) using difference of 2 squares (note 1=1^2). This gives \(\large \frac{3}{x-1}-\frac{2x}{(x+1)(x-1)}\) the proceed as before.
pooja195
  • pooja195
\[\frac{3*x+1 }{ x-1*x+1 }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 6x(x+1) }{ (x^2-1 }\]
mathmate
  • mathmate
A minor correction by adding parentheses: (using PEMBAS) \(\Large\frac{3*(x+1) }{ (x-1)*(x+1) }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 3x+3-2x }{ (x^2-1) }=?\)
mathmate
  • mathmate
The first step was perfect. Then you would distribute(expand) and add/subtract as needed. The final step I leave to you, if possible.
pooja195
  • pooja195
Ew factoring
pooja195
  • pooja195
no wait
mathmate
  • mathmate
in this case, there is nothing to factor! 3x-3-2x = 3x-2x-3=x-3 so answer is \(\Large \frac{x-3}{x^2-1}\)
pooja195
  • pooja195
i was gonna say that XD
mathmate
  • mathmate
of course! I knew that! xD
pooja195
  • pooja195
lol u are learning all mai tricks >.>
mathmate
  • mathmate
Well, this is either called learning, or contamination, depending on the point of view! xD
pooja195
  • pooja195
contamination >.> because now i cant use it :P
mathmate
  • mathmate
Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam! Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!
pooja195
  • pooja195
xD
mathmate
  • mathmate
Just another contamination!
pooja195
  • pooja195
LOL
mathmate
  • mathmate
Ok, would you feel sad if I say...
pooja195
  • pooja195
Are we done for the day? :o
pooja195
  • pooja195
Not at all
mathmate
  • mathmate
nope, but we're done with 11.2 to 11.6 ?
pooja195
  • pooja195
We are done? :o YAY NO MORE MATH
mathmate
  • mathmate
Not too fast, read the whole line, please! lol
mathmate
  • mathmate
So now we're onto Rational equations! Interesting stuff!
pooja195
  • pooja195
T_T
mathmate
  • mathmate
Solve for x: \(\large x+1=\frac{72}{x}\)
pooja195
  • pooja195
i dont think i have learned this .-.
mathmate
  • mathmate
If not, now you will!
pooja195
  • pooja195
O_O
pooja195
  • pooja195
wait im starting to remeber things >.>
mathmate
  • mathmate
Any idea where to start?
pooja195
  • pooja195
\[\huge~\frac{ x+1 }{ 1}=\frac{ 72 }{ x }\]
mathmate
  • mathmate
Excellent! Memories are coming back!
mathmate
  • mathmate
next?
pooja195
  • pooja195
\[ \huge~72*1=x(x+1)\] \[\huge~72=x^2+1x\]
mathmate
  • mathmate
Excellent again! next?
pooja195
  • pooja195
\[\huge~x^2+1x-72=0\]
mathmate
  • mathmate
Beautiful! Pray continue!
pooja195
  • pooja195
|dw:1433116854133:dw| \[\huge~(x+9)(x-8)\]
mathmate
  • mathmate
You're almost done. Equate each factor to zero to find the answers.
pooja195
  • pooja195
x=-9 x=8
mathmate
  • mathmate
100% Excellent job!
pooja195
  • pooja195
I think i have this section lets not do anymore :D ?
mathmate
  • mathmate
Just two more! Solve for x: \(\large \frac{10}{x+4}=\frac{15}{4(x+1)}\) This is easier than FOIL.
pooja195
  • pooja195
NO T_T
mathmate
  • mathmate
There is no factoring involved, you'd do it in 10 seconds, + 30 minutes of typing.
pooja195
  • pooja195
\[\huge~40(x+1)=15(x+4)\]
mathmate
  • mathmate
Too fast.... but everything is right!
pooja195
  • pooja195
40x+40=15x+60 40x=15x+20 25x=20 x=20/25
mathmate
  • mathmate
Excellent, and the answer is.....
pooja195
  • pooja195
x=20/25
mathmate
  • mathmate
... or 4/5
pooja195
  • pooja195
OR 20/25.
mathmate
  • mathmate
I think most teachers would tolerate this (not simplifying), although I don't. I would give 95% instead of 100%. So it depends on your teacher!
pooja195
  • pooja195
smh
mathmate
  • mathmate
You're in trouble when I learn more of these! Well, you need to give extra credit to those who do a perfect job, that's my reasoning!
pooja195
  • pooja195
xD
mathmate
  • mathmate
Now the last one, you'd be sad when it's done!
pooja195
  • pooja195
WAIT
pooja195
  • pooja195
After this no more math??? :O
pooja195
  • pooja195
T_T <---tears of joy
mathmate
  • mathmate
Yeah, after this last problem, no more math (until your quiz) unless you have questions.
pooja195
  • pooja195
i have 2 questions on my hw thing i dont get at all : /
mathmate
  • mathmate
We'll look at that after this question. Perhaps by that time you will know how to do. Solve \(\large 3x=\frac{5x+6}{2x+3}\). Show work.
pooja195
  • pooja195
\[\huge~\frac{ 3x }{ 1 }=\frac{ 5x+6 }{ 2x+3 }\] \[\huge3x(2x+3)=1(5x+6)\]
mathmate
  • mathmate
Excellent, so far so good. You're living up to your reputations!
pooja195
  • pooja195
\[6x^2+9x=5x+6\] \[6x^2+4x-6=0\] |dw:1433118120725:dw|
mathmate
  • mathmate
So far so good! Pray continue!
mathmate
  • mathmate
Actually, to make you life simpler, divide the equation by 2 to get 3x^2+2x-3=0 This will give smaller numbers, so easier to work with.
mathmate
  • mathmate
|dw:1433118297396:dw|
pooja195
  • pooja195
None.
mathmate
  • mathmate
Excellent. What's the next step?
pooja195
  • pooja195
idk
mathmate
  • mathmate
Quadratic formula!
pooja195
  • pooja195
OMG NO T_T
pooja195
  • pooja195
So much work
pooja195
  • pooja195
lets just skip this one ;-; lets do one that doesnt have this
mathmate
  • mathmate
The reason I choose this one is because in the real world (including exams), answers don't always come out in integers.
mathmate
  • mathmate
Just to make sure you're prepared for the quadratic formula. You could type the discriminant faster than lightning, why not the quad. formula?
pooja195
  • pooja195
What would the equation be?
pooja195
  • pooja195
:/
mathmate
  • mathmate
\(6x^2+4x-6=0\), or half of that: 3x^2+2x-3 Either one will do, doesn't make a differce with the quad. formula. Makes a big diff. with factoring.
pooja195
  • pooja195
a=3 b=2 c=-3 \[\huge~x=\frac{ -2 \pm \sqrt{(-2)^2-4(3)(-3)} }{ 2(3) } \]
mathmate
  • mathmate
Correct! Now pull out your calculator, or do it mentally, to find the final solution.
pooja195
  • pooja195
\[\huge~x=\frac{ -2\pm~\sqrt{40} }{ 6 }\]
pooja195
  • pooja195
\[\huge~x=\frac{ -2 \pm \sqrt{4\times10} }{ 6 }\]
pooja195
  • pooja195
\[\huge~x=\frac{ -2 \pm 2\sqrt{10} }{ 6 }\]
mathmate
  • mathmate
Yes, yours is correct (again 95%), unless ... \(\Large~x=\frac{ -2 \pm \sqrt{40} }{ 2(3) }\) \(\Large~x=\frac{ -1 \pm \sqrt{10} }{ 3 }\)
pooja195
  • pooja195
T_T
mathmate
  • mathmate
lol, don't worry about the 5%, but I give you credit for coming that close, especially the part going from sqrt(40) to sqrt(10), excellent job!
pooja195
  • pooja195
:)
mathmate
  • mathmate
Well, in case you have nightmares with + and - signs poking at you, you can wake up and try this. It's not as hard as it seems. \(\Large \frac{3c}{c^2-4}+\frac{1}{c-2}=\frac{2}{c+2}\) You will need to simplify the left-hand side first, then do the equation solving.
pooja195
  • pooja195
O_O where did u get that?
mathmate
  • mathmate
Does it look too difficult? BTW, the answer is -3, so you don't have to cheat! lol
pooja195
  • pooja195
noo it looked fimiliar
mathmate
  • mathmate
Did you get it from your teacher? Well, I'll show you a gold mine. If you can't sleep, go to http://www.regentsprep.org/regents/math/algtrig/ate11/RationalEqPract.htm and practise all those problems!
pooja195
  • pooja195
omg no way T_T
mathmate
  • mathmate
With this one, do the simplification on the left-hand side and then solve. Do not follow the solution provided.
pooja195
  • pooja195
hmmmmk
mathmate
  • mathmate
You said you had two hw questions!
pooja195
  • pooja195
nvm :)
pooja195
  • pooja195
Can we practice more on everything once yr back? :)
mathmate
  • mathmate
sure!
pooja195
  • pooja195
Ready. :)
pooja195
  • pooja195
@mathmate
mathmate
  • mathmate
Start with 11.2?
mathmate
  • mathmate
or you have specific questions?
pooja195
  • pooja195
No lets just go with 11.2 etc
mathmate
  • mathmate
|dw:1433125010350:dw| name the variation.
pooja195
  • pooja195
inverse variation
mathmate
  • mathmate
Good! x 1 2 3 4 5 y 2 4 6 8 10 what is x when y=14
pooja195
  • pooja195
7? :/
mathmate
  • mathmate
Yep, good!
mathmate
  • mathmate
x 1 5 10 y 20 4 2 What is y when x=40
pooja195
  • pooja195
80
mathmate
  • mathmate
Hmmm....
pooja195
  • pooja195
10?
mathmate
  • mathmate
We see that y is decreasing as x increases. See if you can find the rule!
pooja195
  • pooja195
0
mathmate
  • mathmate
no, that's too small. The rule for direct variation is y=kx, where k is a constant. In the first example, x 1 2 3 4 5 y 2 4 6 8 10 we see that the rule is y=2x, so it is a direct variation.
mathmate
  • mathmate
The rule for inverse variation is y=k/x. For the second example, we have x 1 5 10 y 20 4 2 so the rule is y=20/x (20=20/1, 4=20/5, 2=20/10, ...) so what is y when x=40?
pooja195
  • pooja195
2
mathmate
  • mathmate
Another way to write the rule is xy=k For example 2, k=20, so 1*20=20 4*5=20...
mathmate
  • mathmate
According to the rule, y=20/40 = 1/2, so when x=40, y=1/2 or use the alternate rule, xy=20 40y=20, or y=1/2
mathmate
  • mathmate
x 2 7 12 y 6 21 36 Name the variation, and find x when y=120
pooja195
  • pooja195
i dont like there >.>
pooja195
  • pooja195
these
mathmate
  • mathmate
You've done those before?
mathmate
  • mathmate
I know we did not go over it, but never too late! :(
mathmate
  • mathmate
I'll show a few examples first. Direct variation: rule y=kx (when x is greater, y is greater) Inverse variation: rule xy=k (when x increases, y decreases, that's why inverse) so far so good?
mathmate
  • mathmate
|dw:1433125957505:dw|
mathmate
  • mathmate
Notice that inverse variation is not a straight line. It is a curve.
mathmate
  • mathmate
ok so far?
pooja195
  • pooja195
Yea
mathmate
  • mathmate
Ok, now, to find the rule, this is what you do.
mathmate
  • mathmate
First find if the function is increasing, or decreasing. ok
mathmate
  • mathmate
do you know how to tell if a function is increasing?
pooja195
  • pooja195
If the line is going up?
pooja195
  • pooja195
|dw:1433126223857:dw|
mathmate
  • mathmate
Yes. An excellent example. If it is a straight line, we know it is not inverse.
pooja195
  • pooja195
Right :)
mathmate
  • mathmate
Is it automatically direct variation?
pooja195
  • pooja195
yes? :/
mathmate
  • mathmate
Unfortunately, there are other variations. The one you drew is called a partial variation. It is a straight line that does not pass through the origin. A straight line that passes through the origin AND increasing, then it's direct!
pooja195
  • pooja195
Lets skip this.
mathmate
  • mathmate
|dw:1433126603203:dw|
mathmate
  • mathmate
It has to do with the rule. A partial is y=mx+b where b\(\ne\)0 A direct is when y=kx (i.e. b=0) For example, x 2 5 8 y 6 12 18 is partial because y=2x+2 while x 2 5 8 y 4 10 16 is direct because y=2x
pooja195
  • pooja195
hmmmm makes a bit more sense :3
mathmate
  • mathmate
Good! We'll get back to the last problem. x 2 7 12 y 6 21 36 Name the variation, state the rule, and find x when y=120
pooja195
  • pooja195
T_T 120=k/x
mathmate
  • mathmate
Can you start with the variation? (is it increasing? If it is, it is not inverse)
pooja195
  • pooja195
increasing .-.
mathmate
  • mathmate
so it cannot be inverse. The rule you gave is for inverse.
pooja195
  • pooja195
so y=mx+b?
mathmate
  • mathmate
When you give a rule, you have to check that it works for at least two points.
pooja195
  • pooja195
mm ;-; <---
mathmate
  • mathmate
and you need to find k. Let's say we decide that it is direct variation, then y=kx substitute (2,6) we have 6=2k, or k=3 Let's try it for another point: (12,36), so that is 36=12k, k=3 again. So we have found the rule y=3x. If y=120, then 120=3x, what is x?
pooja195
  • pooja195
40
mathmate
  • mathmate
Exactly. So the rule is: 1. examine and assume a distribution. 2. substitute one of the points in the rule (of the assumed distribution) 3. check with another point, to see if the rule is still valid. 4. Give the answer.
pooja195
  • pooja195
mmm k
mathmate
  • mathmate
Now, try x 2 4 6 y 6 3 2 State 1. the variation 2. the rule 3. value of x when y=24
pooja195
  • pooja195
1) PArtial 2) y=mx+b 3) Lets skip it ;-;
mathmate
  • mathmate
Well, it's decreasing
mathmate
  • mathmate
so it is not direct. Is it a straight line, let's check the slope.
mathmate
  • mathmate
x 2 4 6 y 6 3 2 slope = (y2-y1)/(x2-x1) between 2 and 4, slope = (3-6)/(4-2) = -3/2 = -1.5 between 4 and 6, slope = (2-3)/(6-4) = 1/2 Since the slope is not constant, it is not a straight line.
mathmate
  • mathmate
So it's not partial. Try inverse: x 2 4 6 y 6 3 2 2*6=12 4*3=12 6*2=12 so k=12 for inverse!
mathmate
  • mathmate
Can you finish answering?
pooja195
  • pooja195
y=12x :/
mathmate
  • mathmate
Variation: inverse Rule: either xy=12, or y=12/x Now when y=24, what would x be?
pooja195
  • pooja195
2
mathmate
  • mathmate
Hin: Rule: either xy=12, or y=12/x
pooja195
  • pooja195
weve spent about an hour on this .-. y=2? ??
mathmate
  • mathmate
xy=12, y=24 so x=12/y=12/24=1/2 lol Yes, almost an hour.
mathmate
  • mathmate
and it was a good thing, because now you won't miss the easy questions.
pooja195
  • pooja195
@mathmate
mathmate
  • mathmate
k, where do we go from here? (it's laggy)
mathmate
  • mathmate
Want to start a new post?
pooja195
  • pooja195
ok

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