She washed 30 plates from 7:30 to 7:35.....so she washed 30 plates in 5 minutes.....30/5 = 6 plates per minute
so if she started washing plates at 7:15 and ended at 7:38...how many plates did she wash....
from 7:15 to 7:38 is 23 minutes...and if she can wash 6 plates per minute, then in 23 minutes, she can wash (23 * 6) = 138 plates <==
the equation...
y = -6x + b...with b being the number of plates she started with and x being the number of minutes and y being the plates she has left to wash...I am not 100% sure on this equation...I am so sorry
Answer:
The greatest number of 15 inches pieces that can be cut from 5 rolls of length 9 feet is: 35
Step-by-step explanation:
Given
Total length of one roll of ribbon = 9 feet
As the pieces have to be cut into inches, we will convert the measurement in feet to inches
As there are 12 inches in one feet, 9 feet will be equal to:
9*12 = 108 inches
Now first of all, we have to see how many 15 inches pieces can be cut from one role
So,

So the seamstress can cut 7 15-inch long pieces from a roll.
Now given that he has to cut from 5 rolls, the total number of 15-inch pieces will be:

Hence,
The greatest number of 15 inches pieces that can be cut from 5 rolls of length 9 feet is: 35
Answer:
Step-by-step explanation:
first fit:
115 -> 300
500-> 600
358 -> 750
200 -> 350
375 -> not able to allocate
Best fit:
115 -> 125
500 -> 600
358 -> 750
200 -> 200
375 -> not able to allocate
worst fit:
115 -> 750
500 -> 600
358 -> not able to allocate
200 -> 350
375 -> not able to allocate
You can solve the median by putting all the numbers in order. Then cross off one at a time at the start, then cross off the one at the end, then keep going until you get to one.
You can solve the mode by looking for the most frequent number. I remembered that by looking at mode and seeing MO and remembering most often.
I hope this helps :-)
The equation of the tangent line at x=1 can be written in point-slope form as
... L(x) = f'(1)(x -1) +f(1)
The derivative is ...
... f'(x) = 4x^3 +4x
so the slope of the tangent line is f'(1) = 4+4 = 8.
The value of the function at x=1 is
... f(1) = 1^4 +2·1^2 = 3
So, your linearization is ...
... L(x) = 8(x -1) +3
or
... L(x) = 8x -5