Mathhemathh

My interpretation: \({d\over dx}({3\over x^2}+5x^2)\)

By sum rule: \({d\over dx}{3\over x^2}+{d\over dx}5x^2\)

By definition of negative exponents: \({d\over dx}3x^{-2}+{d\over dx}5x^2\)

By constant product rule: \(3({d\over dx}x^{-2})+5({d\over dx}x^2)\)

By power rule: \(3(-2x^{-3})+5(2x)\)

Simplify to get \(10x-{6\over x^3}\), which is your final answer. 

Thanks 

Personally, yes. 

Wow, we have the same method xD

Since the interior angle is 29x, and the exterior angle is x, then \(29x+x=180\).

Combine like terms: \(30x=180\)

Divide both sides by 30: \(x=6\).

Now, the exterior angle measures 6˚. We know that the sum of all exterior angles add up to 360˚. So... \(6s=360\), where s is the number of sides.

Divide both sides by 6: \(s=60\).

The polygon has 60 sides. 

When dividing, for example, \(a\over b\), we ask the question, "What is x if \(a=bx\)?" For \(0\over0\), this becomes \(0x=0\). Here we have a problem. This has infinitely many solutions! Therefore, 0 divided by 0 is indeterminate.

If Sara splits it between 1 kid, that kid gets 2 candy bars.

If Sara splits it between 5 kids, they each get \(2\over5\) candy bars.

If Sara splits it between 999999999999999999999999999 kids, they still each get \(2\over999999999999999999999999999\) candy bars each.

Final answer? A lot.

37 teams * 28 students per team = 1036 students. 

We have \(b=\sqrt{4ac}\)

Square both sides, \(b^2=4ac\)

Divide both sides by \(4c\) to get \(a={b^2\over4c}\) 

\(n-(-5)=n+5\). When subtracting negative integers, you add them, as \(a-(-b)=a+(-1)(-b)=a+b\).

Doing that, we get \(n+5=35\), \(n=30\).