15/100=0.15
13/0.15= 86 2/3%
Hope this helps :)
Answer:
x=15
Step-by-step explanation:
divide 12/20 by 4 to get 3/5 then multiply by 3 to get your value of x
If 3 squared is 9 then 9 times -2 is -18 plus 7 is -11
Answer:
2. Cost of six books
3. Cost of one book
Step-by-step explanation:
I don't know about the rest though.
Take the derivative with respect to t

the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero

divide by w

we add sin(wt) to both sides

divide both sides by cos(wt)

OR

(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

since 2npi is just the period of cos

substituting our second soultion we get

since 2npi is the period

so the maximum value =

minimum value =