There could be five in each class.
Answer: X = 18
Step-by-step explanation:
Crossmultiply
Answer:
<u>First question answer:</u> The limit is 69
<u>Second question answer:</u> The limit is 5
Step-by-step explanation:
For the first limit, plug in
in the expression
, that's the answer for linear equations and limits.
So we have:

The answer is 69
For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value
into the simplified expression to get the correct answer. Shown below:

<em>Now putting 1 in
gives us the limit:</em>

So the answer is 5
Ab = (7x-6) → b = (7x-6)/a (1)
bc = (12-2x) → b = (12-2x)/c (2)
Since (1) = (2) → (7x-6)/a = (12-2x)/c OR a/c = (7x-6)/(12-2x) (3)
Multiply both numerators of (3) by the SQUARE of their respective denominators;
(a*c²)/c = (7x-6)(12-2x)²/(12-2x)
Now simplify:
ac = (7x-6)(12x-2x) or ac = -14x² + 96x - 72
Answer:
use the cymath calculator
Step-by-step explanation: