Answer:
As consequence of the Taylor theorem with integral remainder we have that
If we ask that 
has continuous 
th derivative we can apply the mean value theorem for integrals. Then, there exists 
between 
and 
such that
Hence,
Thus,
and the Taylor theorem with Lagrange remainder is
.
Step-by-step explanation:
Let's simplify step-by-step.
o25n−3(n−6)
Distribute:
=o25n+(−3)(n)+(−3)(−6)
=no25+−3n+18
Answer:
=no25−3n+18
(Pls mark as brainliest) (:
-3x^2 - 10x + 5 = 0
Quadratic formula:
-b +/- sqrt[b^2 - 4(a)(c)]/2(a)
So. . . 10 +/- sqrt[(10^2) - 4(-3)(5)]/2(-3)
10 +/- sqrt[100+60]/ -6
10 +/- sqrt[160]/-6
Answer: 10x + 50
Step-by-step explanation:
lets say that x is equal to the amount of weeks that she runs. She starts off the first week with 50
Then, since she is adding 10 every week, you would do 10x, or 10 times the amount of weeks that she runs.
Since you originally started with 50, you have to add the 50 to the 10x.
So you get...
<h3>
10x + 50</h3>
For example
on week 2, if you plug in 2 to x you get
10(2) +50
20 + 50
70 laps after the second week
