Answer:hello
Step-by-step explanation:
<h3>Solution :</h3>
<u>Let x be the number</u>
By the Pythagorean Theorem, we are looking for
, or <u />
.
Count the number of multiples of 3, 4, and 12 in the range 1-2005:
⌊2005/3⌋ ≈ ⌊668.333⌋ = 668
⌊2005/4⌋ = ⌊501.25⌋ = 501
⌊2005/12⌋ ≈ ⌊167.083⌋ = 167
(⌊<em>x</em>⌋ means the "floor" of <em>x</em>, i.e. the largest integer smaller than <em>x</em>, so ⌊<em>a</em>/<em>b</em>⌋ is what you get when you divide <em>a</em> by <em>b</em> and ignore the remainder)
Then using the inclusion/exclusion principle, there are
668 + 501 - 2•167 = 835
numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 <em>and</em> 4 that occur twice. Subtracting again removes them altogether.
Answer:
X^2 + 11x + 30
Step-by-step explanation:
x*x= x^2
x*6= 6x
x*5= 5x
6*5= 30
x^2 +6x+5x+30
Combine like terms
x^2 + 11x + 30
