D.dilation changes the size is your answer. Hope this helps!
Answer:
$105
Step-by-step explanation:
30% of 150 is 45.
150-45 = 105
Hope that helps
5x means that 5 should be multiplied by the value of x
Every time a variable is touching a number it means it’s multiplying
5x + 8 = ?
If x = 9, you would multiply 5 and 9 which is 45. You then add 45 and 8
45 + 8 = 53
Therefore 5x + 8 = 53
Hope this helps you! :)
Duration, time taken, the completion time
Answer:

Step-by-step explanation:
The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

We first compute the n-th derivative of
, note that

Now, if we compute the n-th derivative at 0 we get

and so the Maclaurin series for f(x)=ln(1+2x) is given by
