I’m not sure i understand your question, could you send a picture of the question ??? i want to help you out ☝
Answer:
Total = 2,400 * (1 + (.083/4))^4*5
Total = 2,400 * (1.02075)^20
Total = 2,400 * 1.5079528829
Total = 3,619.09
Step-by-step explanation:
Answer:
w=3/5 or w=-5
Step-by-step explanation:
5w^2+22w=15
5w^2+22w-15=0 (quadratic equation)
a=5, b=22, c=-15
w1,2 =(-b+-sqrt(b^2-4ac))/2a
w1,2 =(-22+-sqrt(22^2-4*5*(-15))/2*5
w1,2 =(-22+-sqrt(484+300))/10
w1,2=(-22+-sqrt(784))/10
w1,2=(-22+-28)/10
w1=(-22+28)/10, w2=(-22-28)/10
w1=6/10, w2=-50/10
w1=3/5, w2=-5
Let lowest integer be n then
4(n + 4) = 3n + 2(n + 2) + 4
4n + 16 = 5n + 8
8 = n
so three integers are 8,10 and 12.

The surface element is

and the integral is


###
To compute the last integral, you can integrate by parts:



For this integral, consider a substitution of




and the result above follows.