Answer:
[-3, 3]
Step-by-step explanation:
The function/ graph is defined from -3 to 3, inclusive of those values
Answer:
the cost of running the boarding
house for 600 students is N61,000
Step-by-step explanation:
Let C represents cost
K1 represents first constant
K2 represents second constant
C= k1+k2n
3500 = k1 + 25 k2............. Eqn(1)
6000= k1 + 50 k2 .............. Eqn(2)
Subtract eqn(1) from eqn(2)
2500= 25k2
K2= 2500/25
K2= 100
To get k1 from eqn(1)
3500 = k1 + 25 k2
Substitute the value of k2
3500 = k1 + 25 (100)
3500= k1 +2500
K1= 3500- 2500
K1= 1000
The equation connecting them;
C= 1000+ 100n
The cost of running the boarding
house for 600 students is
n= 600
C= 1000+ 100(600)
C= 1000+60000
C= N 61,0000
Answer
Solution:
p = 79
n = 52
Step-by-step explanation:
Given equations:
A) 
B) 
To solve for 
and 
Naming the first equation as A and second as B.
Using elimination method to solve.
In order to eliminate we subtract equation A from B.
Subtracting A from B i.e [
(B)
- (A)
We get, 
Dividing both sides 2.26.
∴ 
We can solve for by substituting in equation A.
Subtracting both sides by 52.
∴ 
Answer:
The pizza cost 5.04 in 2000
Step-by-step explanation:
Step One: set the price of the pizza in 1980 equal to the price in 2000
$4.99 ?
=
1980 2000
Step two: Divide 1980 by 2000
1980 / 2000 = 0.99
Step Three: Divide 4.99 by 0.99
Reason: You know that 1980/0.99= 2000 and what ever you do to the denominator you have to do to the numerator and vice versa.
4.99 / 0.99 = 5.0404040
Step Four: Answer
The pizza cost 5.04 in 2000
Keep doing this until you get all of the years (2015 and 1981) pizza prices
