So sorry but can you answer this question gonna pull my hair out
1 answer:
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Answer:
20 meters
Step-by-step explanation:
The track is circular so it means that after Patrick raced the entire track he is back at the starting point. In other words, every 440 meters he is back to the beginning.
So we would have that, if he races round the track twice, he would run 440(2) = 880 meters and he would be back at the starting point.
The problem asks us how far is he from the starting point at the 900 meter mark. If at 880 meters he is at the starting point, then at 900 meters he would be meters from the starting point.
Answer:
a) 5+13k where k is integer
b) 20+13k where k is integer
c)12+13k where k is integer
Step-by-step explanation:
(a)
.
8x-1=13k
Subtract 13k on both sides:
8x-13k-1=0
Add 1 on both sides:
8x-13k=1
I'm going to use Euclidean Algorithm.
13=8(1)+5
8=5(1)+3
5=3(1)+2
3=2(1)+1
Now backwards through the equations:
3-2=1
3-(5-3)=1
3-5+3=1
(8-5)-5+(8-5)=1
2(8)-3(5)=1
2(8)-3(13-8)=1
5(8)-3(13)=1
So compare this to:
8x-13k=1
We see that x is 5 while k is 3.
Anyways 5 is a solution or 5+13k is a solution where k is an integer.
b)
8x-4=13k
Subtract 13k on both sides:
8x-13k-4=0
Add 4 on both sides:
8x-13k=4
We got this from above:
5(8)-3(13)=1
If we multiply both sides by 4 we get:
8(20)-13(12)=4
So x=20 and 20+13k is also a solution where k is an integer.
c)
[tex]99x \equiv 5 (mod 13)[/tex
99x-5=13k
Subtract 13k on both sides:
99x-13k-5=0
Add 5 on both sides:
99x-13k=5
Using Euclidean Algorithm:
99=13(7)+8
13=8(1)+5
Go back through the equations:
13-8=5
13-(99-13(7))=5
8(13)-99=5
99(-1)+8(13)=5
Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.