3/4x + 3 -2x = -1/4 + 1/2x + 5
Combine like terms:
-1 1/4x + 3 = 1/2x + 4 3/4
Subtract 3 from both sides:
-1 1/4x = 1/2x + 1 3/4
Subtract 1/x from both sides
-1 3/4x = 1 3/4
Divide both sides by-1 3/4
X = -1
The differential of

is

where the partial derivatives are

So the differential d<em>w</em> is

Answer:
30%
Step-by-step explanation:
# of bulk packs = 300
# of triple As = 90
% = 90 / 300 * 100%
% = .3 * 100%
% = 30%
This can be mathematically expressed to
250 + X = 1075
where X represents the parts you produce before the shift ends
Transpose 250 to the other side by subtracting each side by 250
Thus, it goes like this
X = 1075 - 250
X = 825
You produced 825 parts in the middle of the shift.
Answer:
17
Step-by-step explanation:
We have two congruent triangles here. Thus, b must be 17.