Infinitely many solutions means that you have the same thing on both sides of the equation no matter what value of x you plug in, right?
We just need both sides to be 3x then, correct?
If a were equal to 3 and b were equal to 0, we'd have
3x = (3)x + 0
Which is essentially 3x = 3x
So that means a = 3 and b = 0 must work!
Let's say x = 5
3(5) = 3(5) + 0
15 = 15 + 0
15 = 15
That means that a = 3 and b = 0 is your final answer :)
The answer is 13 hope I helped
Answer:
F = 48
Step-by-step explanation:
F= 8/5 (c) + 40
Let c=5
F = 8/5 * 5 +40
= 8 + 40
= 48


When

, you're left with

When

or

, you're left with

Adding the two equations together gives

, or

. Subtracting them gives

,

.
Now, you have



By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that

and

. These alone tell you that you must have

and

.
So the partial fraction decomposition is
Answer:
6.73E-8 I think
Step-by-step explanation: