X + 3 = 9-4
First simplify the right side.
9-4 = 5
Now you have x + 3 = 5
Now subtract 3 from both sides:
X = 2
Answer: x = 2
3x+5y=9 3x+5y=9
-1(3x+2y=3) -3x-2y=-3
3y=6
y=2
3x+5(2)=9
3x+10=9
-10 -10
3x=-1
x=-1/3 (-1/3,2)
Each week, a cook purchased 12 LBS. of Butter:
During the Last year: (12 Months):
Cook Paid:
Little: $23.04
Much: $29.40, For Butter he or she purchased in a week:
Question: is: what is the Difference between, the Greatest price per pound, and the least price per pound of butter the cook paid within the last year?
EQUATION:
Least Paid / 12 =====> 23.04 /12
Most Paid / 12 ======> 29.40 / 12
Divide:
23.04 / 12 = 1.92 / LB
29.40 / 12 = 2.45 / LB
Now Subtract:
2.45 - 1.92
Answer ======> 0.53 is the difference, between the greatest price per round, and least price per round of butter the cook would have paid within the last year.
Hope that helps!!! : )
Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:
diagram pls
then I can answer this question