Answer:
Friday's game generates $1,664 while Sunday's, game generates $1,768 for a total of $3432.00.
Step-by-step explanation:
First, identify what you know:
1) A baseball team has home games on Friday and Sunday
2) Both games, combined, earned $3432.00
3) Friday's game generated $104.00 less than Sunday's game.
We can create a formula to solve our problem!
$3,432 = x + (x - 104)
$3,432 + 104 = x + x
$3,536 = 2x
$3,536/2 = x
x = $1,768
Sunday's game earned $1,768! If Friday's game earned $104 less, then simply subtract that from Sunday's earning!
$1,768 - $104 = $1,664
Friday's game cost $1,664, and Sunday's game earned $1,768!
Hope this helps! :)
Answer:
8
Step-by-step explanation:
Reduce the fraction to 1/2 then multiply it by 8 which is 8:) this this helps! Plzz mark brainliest
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.