See the graph here: https://www.desmos.com/calculator/kzr4vscuj0
The first and last lines are parallel
The second line will intersect these at one point each
Always easy to solve this type of problem
1/20 + 1/30 = [ 30 + 20 ] / 600 = 50/ 600 = 1 / 12
Take the reciprocal of the answer ⇒ 12 min together for 1 lb
So....36 min for 3 lb
thank u so much the solution was not what i thought of at all !!1
Just add the fxns together x^3 - x + x^3 + 2x^2 -10 =.............. I think YOU can do THAT !
F rate 1 lb / 20 min
T rate 1 lb /30 min
3 / (1/20 + 1/30) = 36 minutes
It would take 60 minutes for Mr. F. to eat 3 pounds of cereal, and it would take 90 minutes for Mr. T. To eat 3 pounds of cereal, so if they were eating together it would take 90 minutes for them to finish.
thanks, so the intervals to answer the question are [-8,-4] [0,8] ?
3 to 8 =
[ - log(3) + 5 ] - [ -log(8) + 5] log(8) - log(3) log(8/3)
_______________________ = ____________ = _______ =
3 - 8 -5 -5
≈0.4259687322722811
___________________ ≈ -0.085 = -0.09
- 5
\(\frac{1}{2}\times 3^3+3.4\times 4=\frac{1}{2}\times 27 +3.4\times 4= \frac{27}{2}+13.6=27.1\)
You are very welcome!
:P
You submitted 'If ........... when b = 3' Something is missing in the ........ position ....soemthing like ' a = 45'
Convention dictates the 'i' = iitial 'f' = final in a lot of chemical, physics etc equations
I left nothing out of this this is why im confused in he first plase plz respond
(f/g)(x)=f(x)/g(x)
x^2-16/x^2-6x+8
(x-4)(x+4)/(x-4)(x-2)
(x+4)/(x-2) = (f/g)(x)
Not sure but here's a link:
https://en.wikipedia.org/wiki/P_versus_NP_problem
This should help :
The mean = 22
Three SDs occur at 13 and 31
Hi must be greater than Hf ? Do you mean products must be less than products? So C, "The enthalpy of the reactants is greater than that of the products" ?
The ΔH is a positive change in endothermic reactions, and negative in heat-releasing exothermic processes.
ΔH is change in enthalpy = Hf-Hi For this to be NEGATIVE Hi must be greater than Hf so 'C'
im sorry i forgot to post this on offtopic
