The rate at which the ice changes is -3/8 lb per hr
<h3>What is the rate the ice changes?</h3>
The given parameters are:
Changes =1 3/4 lb to 1/4 lb
Time = 1/4 hr.
The rate the ice changes is calculated as:
Rate = Change/Time
So, we have
Rate = (1/4 lb - 1 3/4 lb)/(1/4 hr)
Evaluate the difference
Rate = (-1 1/2 lb)/(1/4 hr)
Evaluate the quotient
Rate = -3/8 lb per hr
Hence, the rate at which the ice changes is -3/8 lb per hr
Read more about rates at:
brainly.com/question/19493296
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18 kg of 15% copper and 72 kg of 60% copper should be combined by the metalworker to create 90 kg of 51% copper alloy.
<u>Step-by-step explanation:</u>
Let x = kg of 15% copper alloy
Let y = kg of 60% copper alloy
Since we need to create 90 kg of alloy we know:
x + y = 90
51% of 90 kg = 45.9 kg of copper
So we're interested in creating 45.9 kg of copper
We need some amount of 15% copper and some amount of 60% copper to create 45.9 kg of copper:
0.15x + 0.60y = 45.9
but
x + y = 90
x= 90 - y
substituting that value in for x
0.15(90 - y) + 0.60y = 45.9
13.5 - 0.15y + 0.60y = 45.9
0.45y = 32.4
y = 72
Substituting this y value to solve for x gives:
x + y = 90
x= 90-72
x=18
Therefore, in order to create 90kg of 51% alloy, we'd need 18 kg of 15% copper and 72 kg of 60% copper.
The answer is 1/8, hope this helps you :)
First you will divide 525 by 100. Then you multiply your answer by 4. That will equal 21
Answer:
20/7 days (just less than 3 days)
Step-by-step explanation:
Recall that (1 job) = (rate)(time), so time = (1 job) / (rate).
Set up and solve the following equation:
1 job
------------------------------- = time required for 2 pumps working together
1 job 1 job
---------- + -------------
4 days 10 days
This comes out to:
1 job
------------------------------------------- = time required
10 job-days 4 job-days
------------------ + -----------------
40 days 40 days
or:
1 job
-------------------- = (40/14) days, or 20/7 days (just less than 3 days)
14 job·days
-----------------
40 days
