anonymous
  • anonymous
(Please help!) One way to describe the concentration of an acid is as a percent by volume. For example, in 40 mL of a 30% acid solution, the volume of pure acid is 40 x 30/100 or 12 mL, and the volume of water is 40-12 or 28 mL. If 5 mL of pure acid is mixed with 20 mL of water to give 25 mL of acid solution, the concentration of the solution is given by 5/25 x 100% = 20 %. If water is mixed with 50 mL of 40% acid solution, write an equation that describe the acid concentration, C(x), as a function of the volume of water added, x.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The answer is C(x) = (20/50 + x)*100% and I have to graph C(x) versus x. But how do I graph it on the graphing calculator?
anonymous
  • anonymous
The answer on the back of the book is: |dw:1332552677856:dw|
Hero
  • Hero
If that's the answer, then just graph this : C(x) = (20/50 + x) forget about the 100% because 100% = 1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
but on my graphing calculator it's a diagonal line.. o.o
anonymous
  • anonymous
it's still a diagonal squiggly line.. o.o
Hero
  • Hero
Are you sure you have posted the proper function?
anonymous
  • anonymous
ohh.. I plot that on my graph and yes I posted the proper function D:
Hero
  • Hero
Then why are there negative numbers associated with it? Something has to be wrong
anonymous
  • anonymous
Okay so the function of the volume of water added, x is: \[C(x) = 20/50 + x \times 100\]
anonymous
  • anonymous
*100%
Hero
  • Hero
Your function isn't clear enough because it could be interpreted in two different ways
anonymous
  • anonymous
oh boy. D: So with one of the two ways, what function will have the same graph as i drew above?
Hero
  • Hero
\[C(x) = \frac{20}{50} + x \times 100\] or \[C(x) = \frac{20}{50+x} \times 100\]
Hero
  • Hero
And I forgot the percent sign.
anonymous
  • anonymous
ohh.. it shows from the back of my book the second function..
Hero
  • Hero
idk, there is something about the function you must be misinterpreting
anonymous
  • anonymous
in the question above, what function would you get?
Hero
  • Hero
Okay, well in that case, it is : \[C(x) = \frac{2000}{50+x}\]
Hero
  • Hero
Graph that and tell me what you get
anonymous
  • anonymous
I get a diagonal line.. o.o
Hero
  • Hero
The something is wrong with your calc or you're just not graphing correctly
anonymous
  • anonymous
how do I graph it correctly?
Hero
  • Hero
Enter this exactly: 2000/(50+x)
Mertsj
  • Mertsj
If you have 50 ml of 40% acid, you have a constant amount of acid and that is 20 ml So the acid fraction as you add water is 20/(50+x)
Hero
  • Hero
I didn't even bother to calculate the acid fraction because she said the answer was in the back
Hero
  • Hero
If you get a diagonal this time, check to see if you are set to radian (not degree) mode
anonymous
  • anonymous
There's nothing shown on the graph.. o.o
Hero
  • Hero
You're probably in degree mode. switch back to radians
anonymous
  • anonymous
but It's on radians :(
Hero
  • Hero
What kind of calculator do you have ?
Mertsj
  • Mertsj
radians and degrees have nothing to do with this function. The y intercept is 40. Scroll up on your graph
anonymous
  • anonymous
TI-83+
Hero
  • Hero
You have to re-set the domain and range
Hero
  • Hero
Mertsj, be quiet, I'm just trying to help
anonymous
  • anonymous
reset the domain and range? so the y-int has to 40?
anonymous
  • anonymous
i am confused.
Hero
  • Hero
Sometimes, students have their calcs set incorrectly
Hero
  • Hero
I was just eliminating possibilities
anonymous
  • anonymous
oh. So how do i reset it?
Hero
  • Hero
I don't have a ti-83, so I don't remember. There is a way to re-set it so that you can actually see the function
anonymous
  • anonymous
ohh.. how did Mertsj know that the y-int had to be 40?
Mertsj
  • Mertsj
By letting x = 0
Mertsj
  • Mertsj
Also, try graphing 20/(50+x). All you are doing when you multiply by 100 is you are making y 100 times greater.
Mertsj
  • Mertsj
If you graph y=20/(50+x) the y intercept will be .4
anonymous
  • anonymous
ohh.. when x=0?
Mertsj
  • Mertsj
And that might be in your window
Mertsj
  • Mertsj
yes
anonymous
  • anonymous
why x=0?
Mertsj
  • Mertsj
To find the y intercept. To get some idea of where to look for the graph that you are having trouble seeing.
Mertsj
  • Mertsj
Can you set the range of x values?
anonymous
  • anonymous
yes I can
Mertsj
  • Mertsj
Set it from -40 to 40 and set the y values from 0 to 7
Mertsj
  • Mertsj
And then graph y = 20/(50+x)
anonymous
  • anonymous
Okay, I graphed it.
Mertsj
  • Mertsj
Cool. When you do something like that and can't see the graph it is almost always because of an improper setting of x and y values. That's why finding a point or two on the graph can give you guidance about where to set those values.
anonymous
  • anonymous
ohh.. what do you mean by finding a point or two on the graph?
anonymous
  • anonymous
so you set x=0 for the equation to find the y-int?
Mertsj
  • Mertsj
Also if you can graph on an iphone of ipad where you can stretch and squeeze the graph that sometimes helps.
anonymous
  • anonymous
oh, we can only use graphing cals (mine is the TI-83+)
anonymous
  • anonymous
but thank you so much!
Mertsj
  • Mertsj
Choose an x value and find the corresponding y value and then you know a point that is on the graph. For example, in this one, when I let x = 0, I found that y was .4 so i knew where to look for the graph.
Mertsj
  • Mertsj
Ok. Then use the trick of finding a couple of points on the graph.
Mertsj
  • Mertsj
You are welcome. (even though I was told to be quiet)
Hero
  • Hero
No need to rub it in.
anonymous
  • anonymous
By the way, Mertsj, why would the y values be 0 to 7?

Looking for something else?

Not the answer you are looking for? Search for more explanations.