Not very sure.. I'm not too good at this
12/100 of a day, which reduces to 3/25 of a day.
Hello from MrBillDoesMath!
Answer:
5
Discussion:
Consider the expansion of e^x:
e^x = 1 + x + x^2/2 + x^3/6 +...... => replace x by 5t
e^(5t) = 1 + (5t) + (5t)^2/2 + .... => subtract 1 from both sides
e^(5t) - 1 = (5t) + (5t)^2/2+.... => divide both sides by t
(e^(5t) -1)/ t = 5 + (25/2) t +....
so as t ends to 0 the quotient tends to
5 + (25/2)0 + (other terms) *0 -> 5
Thank you,
MrB
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -5 ± √((5)^2 - 4(-11)(-3)) ] / ( 2(-11) )
x = [-5 ± √(25 - (132) ) ] / ( -22 )
x = [-5 ± √(-107) ] / ( -22)
Since we conclude that √-107 is nonreal, the answer to this question is that there are no real solutions.
Answer:
Step-by-step explanation:
