There are 600 students including the seventh and eighth graders at the party.
This problem uses the concept of percentages to define the conditions that are laid in front of us.
Let the original number of students be S , and the number of seventh graders be = 0.60S
We know that percent is used to convey the mathematical term of a fraction multiplied by 100.
Total students after 20 eighth graders arrive = S + 20
And we have that
Number of seventh graders / total number of students = 58%
.60S / [ S + 20 ] = .58 we multiply both sides by S + 20
0.60S =0 .58 [ S + 20]
.60S = .58S + 11.6 we subtract 0.58S from both the sides
0.02S = 11.6 we divide both the sides by .02
S = 11/6 / 0.02 = 580
So the total number of students = 580 + 20 = 600 .
Hence there are 600 students at the party at that time.
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Answer:
Step-by-step explanation:
(1) 2x - 6y = -12
(2) x + 2y = 14
There is a -6y and a +2y. Since they have opposite signs, I'll try to eliminate the y terms. (That's my choice. There is more than one way to solve these.)
Multiply eq. (2) by 3:
3x + 6y = 42
Then add the result to eq. (1) to eliminate the y terms:
2x - 6y = -12
3x + 6y = 42
------------------
5x = 30, so x = 6
Now plug the value of x into eq. (2) and solve for y:
6 + 2y = 14
2y = 8
y = 4
Why did I use eq. (2) to solve for y? Because it's less work. I could have used eq. (1) instead:
2(6) - 6y = -12
12 - 6y = -12
-6y = -24
y = 4
More than one way to solve.
Answer:
its complemtary
Step-by-step explanation:
check this photo
Answer:
1/8
Step-by-step explanation:
-3p+1/8=-1/4
subtract 1/8
-3p=-3/8
divide by -3
p=1/8
