Hi,
Solving:
Answer: y = -128
Hope this helps.
If you have any questions about the answer, comment below and I'll reply.
r3t40
Let the weightage of Ease of Use be x
Ease of Use = x
<span>Compatibility is 5 times more than ease of use:
</span>Compatibility = 5x
<span>Reputation is 3 times more important than compatibility:
</span>Reputation = 3(5x)
Reputation = 15x
<span>Cost is 2 times more important than reputation:
</span>Cost = 2(15x)
Cost = 30x
So the weightage are:
Ease of Use : 1
Compatibility : 5
Reputation :15
Cost : 30
Answer:
4000
Step-by-step explanation:
Answer:
(a) b = (4/7)c
(b) Bill: 40 shingles/hour; Chip: 70 shingles/hour
Step-by-step explanation:
Let b and c represent Bill's and Chip's rates in shingles per hour, respectively. Then we have ...
7b = 4c
c - b = 30 . . . . shingles per hour difference in rates
(a) Bill's rate in terms of Chip's rate can be found by dividing the first equation by 7
b = (4/7)c . . . . . Bill's rate is 4/7 of Chip's rate
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(b) To find the rates, we can multiply the second equation by 7 and substitute using the first equation:
7c -7b = 210
7c -4c = 210
c = 210/3 = 70
b = (4/7)(70) = 40
Bill's rate is 40 shingles per hour; Chip's rate is 70 shingles per hour.
Answer:
Step-by-step explanation:
Bob ran the first part of a 12 km race at a speed of 8 kmph. He ran the second part of the
race at 10 kmph. If his total time for the entire race was 1.74 hours, how far did he run in
the first part of the race?
First part :
Total Distance = 12 km
First part Speed = 8kmph
Let distance covered on first part = x
Second part speed = 10kmhr
(8*x)
