Answer:
After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$, divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$. $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$,$73$, and $79$ are prime numbers greater than $67$ and less than or equal to $80$, subtract $3$ from $80$ to get the answer $80-3=\boxed{77}\Rightarrow\boxed{D}$.
Step-by-step explanation:
hope this helps
I'm assuming you mean 
, not 
, like your prompt suggests.
First, let's figure out what rule we can use. A likely noticeable one is the Power Rule, which says the following:
![\dfrac{d}{dx} [u^a] = a(u)^{a-1} du](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdx%7D%20%5Bu%5Ea%5D%20%3D%20a%28u%29%5E%7Ba-1%7D%20du)
Applying this, we can solve for the derivative:
While you can simplify the expression to your liking, I believe that this form is not overly complex and will thus leave it as is.
Thus, our answer is:
Hi there. I need points lol
This should be g(x) = 12x - 8, as multiplying 3 by 4 increases the slope.
Answer:
C. 70
Step-by-step explanation:
70/5= 14
70/7 =10
