The equivalence

means that n-5 is a multiple of 12.
that is
n-5=12k, for some integer k
and so
n=12k+5
for k=-1, n=-12+5=-7
for k= 0, n=0+5=5 (the first positive integer n, is for k=0)
we solve 5000=12k+5 to find the last k
12k=5000-5=4995
k=4995/12=416.25
so check k = 415, 416, 417 to be sure we have the right k:
n=12k+5=12*415+5=4985
n=12k+5=12*416+5=4997
n=12k+5=12*417+5=5009
The last k which produces n<5000 is 416
For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,
thus there are 417 integers n satisfying the congruence.
Answer: 417
Answer:
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Answer:
30 minutes.
Step-by-step explanation:
55 - 25 = 30
Hope this helped, have a great day/night!
3 sevenths plus 2 sevenths equals 5 sevenths
3/7 + 2/7 = 5/7 because you're adding 3 and 2 which gives you 5.
Answer:
4
Step-by-step explanation:
<h3><u>some relevant limit laws</u></h3>
lim C = C where c is a constant.
lim( f(x) + g(x)) =lim f(x) + lim g(x)
lim( f(x)g(x)) =lim f(x) * lim g(x)
lim( cg(x)) =clim g(x)
lim( f(x)/g(x)) =lim f(x) / lim g(x) if lim g(x) is not equal to zero.
lim( f(x))^2 = (lim f(x) )^2
lim square root( f(x)) = square root(lim f(x) )
![\lim_{n \to 3} g(x) = 9\\\\\lim_{n \to 3} f(x) = 6\\\\ \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10} \\\\ = \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10}\\\\= \sqrt[3]{lim_{n \to 3}f(x) \times lim_{n \to 3}g(x) + 10}\\\\= \sqrt[3]{6 \times 9 + 10}\\\\= \sqrt[3]{64}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%203%7D%20g%28x%29%20%20%3D%209%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%203%7D%20f%28x%29%20%20%3D%206%5C%5C%5C%5C%20%5Clim_%7Bn%20%5Cto%203%7D%20%5Csqrt%5B3%5D%7Bf%28x%29g%28x%29%20%2B%2010%7D%20%5C%5C%5C%5C%20%3D%20%5Clim_%7Bn%20%5Cto%203%7D%20%5Csqrt%5B3%5D%7Bf%28x%29g%28x%29%20%2B%2010%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B3%5D%7Blim_%7Bn%20%5Cto%203%7Df%28x%29%20%5Ctimes%20lim_%7Bn%20%5Cto%203%7Dg%28x%29%20%2B%2010%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B3%5D%7B6%20%5Ctimes%209%20%2B%2010%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B3%5D%7B64%7D)
= 4