1427 = (F) (+ F + 60) (2F - 50) (+ 3F)
<em>Each pair of brackets represents one of the classes - F meaning Freshmen Class. It has been split into brackets for demonstrational purposes - nothing is being multiplied.</em>
F is an unknown number and each other class size is based off of that so we put it in algebraic terms in order to work it out.
1 - Freshmen Class
2 - Sophomore Class (Freshmen Class + 60 more students)
3 - Junior Class ( Twice the size of the Freshmen Class - 50 students)
4 - Senior Class (Three times the size of the Freshmen Class)
All this can be simplified to 7F + 10 = 1427
1427 - 10 = 1417
1417/7 = 202.428....
Is there a mistake in the question?
Following the question with 202 as the answer - the number 1424 is reached
If increased to 203 - 1431 is reached.
The answer shouldn't include half a person.
Answer:
n = 12.3
Step-by-step explanation:
I'm assuming we're looking for n, to find it, we need to isolate it.
Our equation will be solved through PEMDAS :
Parenthesis
Exponent
Multiply/Divide
Add/Subtract
n / 10^2 = 0.123
Get rid of the exponent by applying it to 10,
10^2 = 100
n/100 = 0.123
Now to get rid of the division, multiply the denominator (100) to both sides.
n/100 * 100 = 0.123 * 100
n = 12.3
Answer:
Area of shaded part ABCEF = 66 sq.cm
Step-by-step explanation:
AB = 8cm
CD = 8cm
Let DE = x cm
CE = 3x cm
CD = CE + DE = 8cm
x + 3x = 8
4x = 8
x = 8/4 = 2 cm
DE = 2cm
CE = 3 * 2 = 6 cm
Area of triangle ADE = 1/2 * base * height
= 1/2 * DE * AD
= 1/2 * 2 * 11 = 11 sq. cm
Area of triangle AEF = Area of triangle ADE = 11 sq. cm
Area of Rectangle ABCD = l * b = 8 * 11 = 88 sq.cm
Area of shaded part ABCEF = Area of Rectangle ABCD - (Area of triangle AEF + Area of triangle ADE)
= 88 - ( 11 + 11 ) = 88 -22 = 66 sq.cm
