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.
3/4=6/8
1/8+1/8+1/8+1/8+1/8+1/8
=6(1/8)
=6/8
Answer: 6 days.
Answer:
<h2>the second one</h2>
Step-by-step explanation:
so much for bein a college student.
Answer:
8.5 or 8 1/2
Step-by-step explanation:
Actually, the answer is not A, if you're saying A is the first choice above. That's incorrect. You will need to use the Geometric mean for right triangles here to figure out what the value of a is. We will use this form:
. We have a value for YZ of 3; side a is XZ. That means in order to solve this we need WZ, which we can find using pythagorean's theorem. 3^2 + 4^2 = c^2 and 9 + 16 = c^2 and c = 5. Now we fill in accordingly:
. Cross-multiply to get 3XZ=25 and side XZ is
. XZ is 25/3 and YZ is 3, so 25/3 - 3 = XY. That means that XY (side a) = 16/3 or 5 1/3, choice B, or the second one down.
