Lagrange multipliers:
(if
)
(if
)
(if
)
In the first octant, we assume
, so we can ignore the caveats above. Now,
so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of
.
We also need to check the boundary of the region, i.e. the intersection of
with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force
, so the point we found is the only extremum.
The answer is 30, can I get the brainliest answer
Answer:
6a) i- 2hrs 36mins ii- 3hrs 12mins
b) car A≈ 76.9km/h car B≈ 62.5km/h
c)------
7a) 35km
b) car A=75km car B=60km
c) 30km
d) car A≈36mins car B≈48mins
Step-by-step explanation:
6a) Using the graph follow the lines until they finish then go downwards until you get to the x-axis. The x-axis is going up by 12mins for each square.
b) Using the answer from a, you divide 200km by the time.
For car A 2hrs 36mins becomes 2.6 because 36mins/60mins=0.6
∴ car A: 200/2.6≈ 76.92km/h
For car B 3hrs 12mins becomes 3.2 because 12mins/60mins=0.2
∴ car B: 200/3.2≈ 62.5km/h
7a) Using the graph go down from where the line of car A finished to meet car B. The y-axis is going up by 5km for each square.
b) Starting from the x-axis at 1 hour go upwards to see where you meet the car B line (60km) and car A line(75km). (sorry if that does not really make sense).
c) Difference from car A line to car B:
155km-125km=30km
d) Going across from 50km meet car A line and go down to see it has been travelling for approx. 36mins. Then continue across to car B line, go down to see it reached 50km at approx. 48mins.
Hope this helps.
Simply plug in the value for x.
If x is equal to -18, then plug in -18 for x.
Example:
f(x) = 15 + x/5.
f(5) = 15 + 5/5.
Then solve the equation.
A function is just something that receives some value and then barfs a value out. This is an IO relationship. (Input/Output).
Answer:
9m⁸n²
Step-by-step explanation:
(3m⁴n)² Remove parentheses
= 3²m⁸n² Square the 3
= 9m⁸n²
