IDK the constant and quantities...
First, find how many hours it takes for him to grade 1 essay, 0.16. Then multiply that by 35 to get 5.6.
I hope this is right -_-'
Answer:
x=1
Step-by-step explanation:
you would change it to 5x-4(-3x+5)=-3, because you put the info for the first equation into the second equation.
5x-4(-3x+5)=-3
5x+12x-20=-3
17x-20=-3
17x=17
x=1
Answer:
9x^2(5y^2 + 2x).
Step-by-step explanation:
First find the Greatest Common Factor of the 2 terms.
GCF of 18 and 45 = 9
GCF of x^2 and x^3 = x^2.
The complete GCF is therefore 9x^2.
So, dividing each term by the GCF, we obtain:
9x^2(5y^2 + 2x).
Answer: a) 46 minutes
b) 10:47
The basic knowledge for this is that 60 minutes = 1 hour
<u>a)</u> Question a asks for how long it took from a 10:34 bus from Mosley to reach Bamford. From the table you can see that the 10:34 bus reaches Bamford at 11:20. All you have to do is count from 10:34 to 11:20.
10:34 will become 11:00 at 10:60 right? Clock's generally don't show 10:60 but goes directly to 11:00 after 1 minute is passed at 10:59. So from 10:34 to 11:00, it takes 26 minutes. Remember the bus reaches at 11:20 so from 11:00 to 11:20, it takes 20 minutes. Now add them up:
26 minutes + 20 minutes = 46 minutes
Here you go! Total 46 minutes from Mosley to Bamford.
<u>b)</u> From question b, we can see that Trina did not ride the first bus or by any chance missed it because the bus left at 10:14 and she is at the station at 10:15. Now think it from your perspective! You missed the first bus and you are in a big rush. So to reach your destination as early as possible, you will obviously take the next earliest bus. The next bus is at 10:24 (Belton). So if we take the 10:24 bus in Belton, it reaches at 10:47 in Garton.
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hope it helped
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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.