Hey there!
If we start out with the same amount of boys and girls at the beginning, then we must add the same number to both groups to maintain that same amount. In this case, since there's already 4 in both groups, you would need to split 2 in half and add the remainder to both groups.
2 ÷ 2 = 1
boys: 4 + 1 = 5
girls: 4 + 1 = 5
So, 5 girls and 5 boys are running.
Hope this helped you out! :-)
A + b = 547
b= a +1
a + a+1 = 547
2a= 547-1
2a= 546
a= 273
b= 273+1= 274
a= 273 b= 274
we have 5^4 . (5^-2)^3 = 5^4 . 5^-6 = 5^ (4-6) = 5^-2
ok done. Thank to me :>
Answer:
2 1/20
Step-by-step explanation:
It was trick math problem and I was working it out, so mad I didn't pay attention to the numbers more closely. :)
There are two 3/8 but one is negative so they cancel each other out
So you have now (-4/5) + 5/4
Find the least common multiple they have.
In this case 20
-4/5 = x/20 = 16/20
5/4 = x/20 = 25/20
So now add them both together
16/20 + 25/20 = 41/20
Find the least common factor for both the denominator and numerator of 41/20
41 is a prime number so
41/20 is an improper fraction
Divide 41 by 20 to get 2 1/20
Answer: that well be (1,-2) or (5,-6)
Step-by-step explanation: