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Sorry, I don't understand.
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From 1 to 100 have: (100 - 1) : 1 + 1 = 100 (terms).
Let A be the set of natural numbers from 1 to 100 which are multiples of 3 and 7.
=> A is the set of natural numbers from 1 to 100 which are multiples of 3*7=21.
=> A = {21;42;63;84}
We see: set A has 4 elements.
=>The number of natural numbers ranging from 1 to 100 which are not multiples of 3 and 7 are:
100 - 4 = 96 (numbers)
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60s*60'*24h*365*100 = 3153600000s
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We have: a < b < c =>\(\dfrac{1}{a}>\dfrac{1}{b}>\dfrac{1}{c}\).
=> \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1=\dfrac{3}{3}< \dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}=\dfrac{3}{a}\)
=> a < 3. Since a > 0 (positive integer).
=> a = 1 or a = 2.
*If a = 1 => \(\dfrac{1}{a}=1\).
=> \(\dfrac{1}{b}+\dfrac{1}{c}=1-1=0\) (false)
*If a = 2 => \(\dfrac{1}{a}=\dfrac{1}{2}\).
=> \(\dfrac{1}{b}+\dfrac{1}{c}=1-\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{2}{4}< \dfrac{1}{b}+\dfrac{1}{b}=\dfrac{2}{b}\).
=> b < 4. Since b > 2 (a < b).
=> b = 3.
=> \(\dfrac{1}{c}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\) => c = 6
Hence a + b + c = 2 + 3 + 6 = 11.
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It's incorrect.Because...
We have: 23 = 8 divided by 7 remaining 1
=> (23)4 divided by 7 remaining 14
=> 212 divided by 7 remaining 1.
Hence 212 is not divisible by 7.
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Let the number of student who took the test (the number of mark) be x.
We have: 81x - 100 = 80(x - 1)
=> 81x - 100 = 80x - 80
=> 81x - 80x = -80 + 100
=> x = 20
Hence there are 20 student including Kate in the class who took the test.
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We have: A team of 4 people has just finished \(\dfrac{1}{2}\) of a project in 10 days.
Let the number of people are needed to be done the rest of the project in 5 days be x.
Due the number of people and the project are two inversely proportional variable, so we have:
\(\dfrac{4}{x}=\dfrac{5}{10}=\dfrac{1}{2}\)
So \(x=\dfrac{4}{\dfrac{1}{2}}=8\)
Hence 8-4=4 more people are needed.
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Before they meet 1 hour, the distance between each other is:
70*1+40*1=110 (miles)
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