Answer:
(4×-3)(2x-1)
Step-by-step explanation:
8x²-10x+3
8x²-4x-6x+3
4x(2x-1)-3(2x-1)
(4×-3)(2x-1)
Answer:
6 units above the x-axis
Step-by-step explanation:
The pair (9,6) are a set of coordinates that indicate how far away from the origin (0,0) the point is.
The coordinates are written (x,y) where 'x' is how far to the left or right of the y-axis (vertical) the point is. The 'y' is how far above or below the x-axis (horizontal) the point is.
So in this case, the set of (9,6) means that the point is located at 9 units to the right of the origin and the y-axis and 6 units above the x-axis
Answer:if you looked at hem like they were rows, in the first row the third one would be an expression. In the second row, the second one would be an expression. This is becuase they both have an equal sign.

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.
Well, for number 15, this is what I came up with, I hope it is what your teacher is looking for:
So it starts out with 3 butterflies, and it doubles every minute:
3x2 = 6 That is for the first minute
6x2 = 12 That is for the second minute
12x2 = 24 That is for the third minute
24x2 = 48 That is for the fourth minute
So, It asks for using the power of 2 in your answer, that is what I came up with for part A.
For part B I would say something like this:
X represents the number of butterflies left.
The equation, using the information above would be this:
48-7 = X
Then simplify, which would leave you with 41 butterflies.
I hope this helps! :)