80 x 5 = 400
32 x 5 = 160
so you have about 160 <span>defective bulbs</span>
<span>The nonpermissible replacement for the variable x in this expression is x = 2
Substituting x = 2 yields:
(2)^2/(2(2)-4)
4/0, which is undefined.
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Answer:
THE ANSWER IS 17/22
Step-by-step explanation:
STOP DELETING MY ANSWER!!!
Kono Dio Da!!!
Answer:
Part c: Contained within the explanation
Part b: gcd(1200,560)=80
Part a: q=-6 r=1
Step-by-step explanation:
I will start with c and work my way up:
Part c:
Proof:
We want to shoe that bL=a+c for some integer L given:
bM=a for some integer M and bK=c for some integer K.
If a=bM and c=bK,
then a+c=bM+bK.
a+c=bM+bK
a+c=b(M+K) by factoring using distributive property
Now we have what we wanted to prove since integers are closed under addition. M+K is an integer since M and K are integers.
So L=M+K in bL=a+c.
We have shown b|(a+c) given b|a and b|c.
//
Part b:
We are going to use Euclidean's Algorithm.
Start with bigger number and see how much smaller number goes into it:
1200=2(560)+80
560=80(7)
This implies the remainder before the remainder is 0 is the greatest common factor of 1200 and 560. So the greatest common factor of 1200 and 560 is 80.
Part a:
Find q and r such that:
-65=q(11)+r
We want to find q and r such that they satisfy the division algorithm.
r is suppose to be a positive integer less than 11.
So q=-6 gives:
-65=(-6)(11)+r
-65=-66+r
So r=1 since r=-65+66.
So q=-6 while r=1.
