10m15n10 because you just multiply everything by 5
Answer:
Remainder= 5, and the binomial
is not a factor of the given polynomial.
Step-by-step explanation:
Given polynomial is
, we have to divide this with a binomial [tex}(x-1)[/tex] using remainder theorem.
Remainder theorem says if
is a factor then remiander would be 
Therefore for 

Thus the remainder is 5 and since it is not 0 , so the binomial
is not a factor of the given polynomial.
Answer:
23 1/13
Step-by-step explanation:
You have done a pretty good job of writing the problem, negative 300 divided by negative thirteen. It can be translated directly to your favorite calculator (see attachment) for a solution.
If you want to perform the division by hand, the particular method of writing the problem depends on the method of division you want to use. (Several styles are taught these days). Numerous web sites and videos explain <em>long division</em> in all its detail. The second attachment shows an example where a decimal fraction result is obtained. The decimal fraction is an infinite repeating decimal with a 6-digit repeat.
For starters, you would generally convert both numbers to positive numbers, since the result of 300/13 is the same as the result of -300/-13 and positive numbers are easier to deal with.
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<em>Comment on symbols</em>
(The symbol ÷ generally means the same thing as the symbol /. Both mean "divided by". In some cases, the symbol ÷ is given the meaning "everything to the left of it divided by everything to the right of it." This is often the case when it is used as part of a compound fraction: 3/5÷4/3, for example. The preferred representation of such a division is (3/5)/(4/3), with parentheses clearly identifying numerators and denominators.)
Clearly, |S| = 50.
Count the multiples of 2 between 1 and 50:
⌊50/2⌋ = ⌊25⌋ = 25
(where ⌊x⌋ denotes the "floor of x", or the largest integer that is smaller than or equal to x; in other words, round <u>down</u> to the nearest integer)
Count the multiples of 3 between 1 and 50:
⌊50/3⌋ ≈ ⌊16.667⌋ = 16
Since LCM(2, 3) = 6, the sets of multiples of 2 and multiples of 3 have some overlap. Count the multiples of 6 between 1 and 50:
⌊50/6⌋ ≈ ⌊8.333⌋ = 8
Then by the inclusion/exclusion principle, we remove from S
25 + 16 - 8 = 33
elements, so that the new set S contains 50 - 33 = 17 elements.
Answer:
y=5
Step-by-step explanation:
4(y-6)=-4
4y-24=-4
+24 +24
4y=20
y=5