Factors of 35: 1, 5, 7, 35
Of those, only 5 and 7 are prime.
7>5
Final answer: 7
Answer:
$30, $75, $45
Step-by-step explanation:
We can set up an equation, where x is an unknown value:
2x+5x+3x=150
10x = 150
x=15
Now that we know what x is, we can plug x in to solve for 2x, 5x, and 3x:
2x=2*15=30
5x=5*15=75
3x=3*15=45
Therefore, the answer is $30, $75, $45
Answer:
(a - 3b)(a + 3b)
Step-by-step explanation:
a² - 9b²
(a)² - (3b)²
(a - 3b)(a + 3b)
I suspect you meant
"How many numbers between 1 and 100 (inclusive) are divisible by 10 or 7?"
• Count the multiples of 10:
⌊100/10⌋ = ⌊10⌋ = 10
• Count the multiples of 7:
⌊100/7⌋ ≈ ⌊14.2857⌋ = 14
• Count the multiples of the LCM of 7 and 10. These numbers are coprime, so LCM(7, 10) = 7•10 = 70, and
⌊100/70⌋ ≈ ⌊1.42857⌋ = 1
(where ⌊<em>x</em>⌋ denotes the "floor" of <em>x</em>, meaning the largest integer that is smaller than <em>x</em>)
Then using the inclusion/exclusion principle, there are
10 + 14 - 1 = 23
numbers in the range 1-100 that are divisible by 10 or 7. In other words, add up the multiples of both 10 and 7, then subtract the common multiples, which are multiples of the LCM.
This one is A. they are complementary
