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:
sadly, there is no indicated angle. you have to attach a photo/screenshot of your problem for anyone to help. There is limited information given from your question.
Step-by-step explanation:
Answer:
(3,2)
Step-by-step explanation:
Y=-2x+8 equation 1
y=x-1 equation 2
x-1=-2x+8 substi y of second equation into y of first equation.
3x=9
x=3
solve for y by putting the x value into either equation
y=x-1
y=3-1
y=2
Answer:
4, 12, 8, 10
Step-by-step explanation:
Ratio:
Hydrogen : Oxygen
2:1
1) 8/2 = 4 Oxygen atoms
2) 6*2 = 12 Hydrogen atoms
3) 16/2 = 8 Oxygen atoms
4) 20/2 = 10 Oxygen atoms
