1. Twelve increased by a number a
2. Thirty one less than a number r
3. The product of nineteen and a number t
4. b divided by a three
5. Three is less than a number c
6. Two is added by the number x which is divided by eight
7. Six is less than a number y which is divided by eleven
8. Twenty one times a number d increased by thirteen
Answer:
total number of students is 25
AIZA AND GLAIZA came 60% of the 25 students that is 15th position
A) therefore students which came below them were= 10
B) Students who has scored greater than them were=> 15-2=13 (as they niether got less than 12 nor more than 12.)
This is a sum and difference problem, with the equations
R-N=35
R+N=635
The solution of which is
R=(635+35)/2=335
N=(635-35)/2=300
In general, the sum and difference problem can be solved (most of the time mentally) using
Larger number = (sum + difference)/2
Smaller number = (sum - difference)/2
Answer:
We can set up a system of equations.
x + y = 111
0.25x + 0.10y = 18.30
x + y = 111
Subtract 'y' to both sides:
x = -y + 111
Plug in '-y + 111' for 'x' in the 2nd equation:
0.25(-y + 111) + 0.10y = 18.30
Distribute 0.25 into the parenthesis:
-0.25y + 27.75 + 0.10y = 18.30
Combine like terms:
-0.15y + 27.75 = 18.30
Subtract 27.75 to both sides:
-0.15y = -9.45
Divide -0.15 to both sides:
y = 63
Plug this back into any of the two equations to find the 'x' value.
x + y = 111
x + 63 = 111
Subtract 63 to both sides:
x = 48
So there are 48 quarters and 63 dimes.
Step-by-step explanation:
Hope This Helps!
Answer:
The value of f(2) is larger than the value of f(3).
Step-by-step explanation:
