Help with my math test (thanks!)

Andyisvenom

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Oct 22, 2013
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Currently in class and idk what to do i cant fail this can someone help me out

 

HallsofIvy

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Jan 27, 2012
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This is very strange! A math test where the instructions are "find someone else to do the test for you"? I don't recall having any such tests when I was in school!

If, as you appear to be saying, you have no idea at all how to do these problems (I don't know what problems they are as you did not even properly upload them) why should you pass? The point of any class is to learn, not pass by cheating.

What's really offensive is that you seem to think think there should be people here waiting to help you cheat!
 

Subhotosh Khan

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Currently in class and idk what to do i cant fail this can someone help me out

http://s7.postimg.org/pl9pjmhtn/image.jpg[\img][/QUOTE]

I do commend you for your honesty (while trying to cheat)!!

You could have pretended to be grandpa trying to help grandson and all that.....

If you were skillful enough to upload your problems - somebody may have taken that hook, line and sinker .............
 

stapel

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The test page in question contains the following exercises:


Solve the following, using \(\displaystyle a\, =\, 3,\, b\, =\, 4, c\, =\, \frac{1}{3},\, d\, =\, \frac{1}{2},\, m\, =\, 6,\, \) and \(\displaystyle n\, =\, \frac{1}{4}\).

17. \(\displaystyle \displaystyle{\frac{\sqrt{b}\, +\, \sqrt{2d}}{2}\, -\, \frac{\sqrt{3c}\, +\, \sqrt{8d}}{4}}\)

18. \(\displaystyle \displaystyle{\frac{2\sqrt{a^2 b^2}}{3}\, +\, \frac{3\sqrt{2\, +\, d^2}}{4}\, -\, a\sqrt{n}}\)

Solve the following, using \(\displaystyle a\, =\, 2,\, b\, =\, \frac{1}{3}, x\, =\, -2,\, y\, =\, -1,\, m\, =\, 3,\, \) and \(\displaystyle n\, =\, \frac{1}{2}\).

16. \(\displaystyle \displaystyle{\frac{3a}{x}\, +\, \frac{2y}{m}\, +\, \frac{3n}{y}\, -\, \frac{m}{n}\, +\, 2\left(x^3\, -\, y^2\, +\, 4\right)}\)

Solve each system of equation[sic]. Choose a different method every time you solve different set[sic].

16. \(\displaystyle \displaystyle{\begin{cases}x\, =\, -\frac{3y\, +\, 3}{4}\\y\, =\, -\frac{1\, +\, 5x}{4}\end{cases}}\)

17. \(\displaystyle \displaystyle{\begin{cases}\frac{x\, +\, y}{6}\, =\, \frac{x\, -\, y}{12}\\\frac{2x}{3}\, =\, y\, +\, 3\end{cases}}\)

18. \(\displaystyle \displaystyle{\begin{cases}3x\, -\, \frac{y\, -\, 3}{5}\, =\, 6\\3y\, -\, \frac{x\, -\, 2}{7}\, =\, 9\end{cases}}\)


Yes, that's the numbering in the image. ;)
 

HallsofIvy

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Jan 27, 2012
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I would say that most of those are basic arithmetic problems except that, by the numbering of the problems, these are clealy in some extremely complex numeration system.
 
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