If we let p and t be the masses of the paper and textbook, respectively, the equations that would best represent the given in this item are:
(1) 20p + 9t = 44.4
(2) (20 + 5)p + (9 + 1)t = 51
The values of p and t from the equation are 0.6 and 3.6, respectively. Thus, each paperback weighs 0.6 pounds and each textbook weighs 3.6 pounds.
Answer:
who mows the lawn for 7 dollars :/
Step-by-step explanation:
Answer: she combined the constants incorrectly in the last step.... i just got it right on the test
Answer:
Below.
Step-by-step explanation:
I won't do all of these for you but I'll show you the general method.
First write each number as prime factors.
For example number 7:
LCM of 24 and 34.
24 = 2 * 2 * 2 * 3
34 = 2 * 17
The LCM is the multiple of all these factors EXCEPT if there is a duplicate number you only use it once.
There is one duplicate here - the 2 ( in bold) so we only use this once.
So the LCM = 2 * 2 * 2 * 3 * 17 = 408.
Number 1:
13, 25
13 = 13
25 = 5 * 5
There are no duplicates so the LCM = 13 * 5 * 5 = 325.
Number 18:
15, 84
15 = 3 * 5
84 = 2 * 2 * 3 * 7
Number 3 is common to both sets so it is only used once:
LCM = 2 * 2 * 3 * 5 * 7 = 420.
Number 40:
18, 48
18 = 2 * 3 * 3
48 = 2 * 2 * 2 * 2 * 3
There are 2 sets of duplicates here, 2 and 3 .
LCM = 2 * 2 * 2* 2 * 3 * 3 = 144.
Answers:
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Step-by-step explanation:
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Dividing both sides by n · R:
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Multiplying both sides by 1000:
Dividing both sides by tc:
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