Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
Answer:
yes, I can help you do it.
Answer:
4a^2b-8ab^2
Step-by-step explanation:
The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle (x,y):

Plugging in points A and B into distance formula, then setting them equal to each other gives:

Right away we can cancel out the x terms leaving:

Expand Left side and Solve for y:


Plug in points B and C as before:

Here we can cancel the y-terms.
Expand and solve for x:



Therefore the center of the circle is the point (6,3)
If the price was $1 and it increased by 15% it would now be $1.15 therefore the multiplier is 1.15 as that increases a number by 15%