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
In order to solve this equation, let's write it out in number form:
-3 - x = -10
In order to isolate 'x' we need to ensure that there is only 'x' on one side. We can get rid of the -3 by adding 3 to both sides:
-3 + 3 - x = -10 +3
-x = -7
Because there are negatives on both sides, you can simply cancel them out:
x = 7
3/4 that really helpssssssss
To add two functions, combine like terms. (5+3x^(3)) = 8x^3-2x. The domain is all real numbers. There is no bound on how small or large the numbers can be for cubed functions.
Yes it’s right for this problem
