Let they be x.
Cost of textbooks + laptops cant greater than total amount.
(cost of 1 textbooks * number of textbooks) + (cost of 1 laptop * number of laptops ≤ 6500
(30*116) + (x*439) ≤ 6500
3480 + 439x ≤ 6500
439x ≤ 3020
x ≤ 6.8(approx)
Last natural number is 6, says, they can buy 6 laptops at max.
Answer:
Infinite many solutions
Meaning there is more than one solution
Hope this helps!
Let me know if this is right.
Answer:
Step-by-step explanation:
Given:
Carbon-14
t1/2 = 5700 years
t1 = 11400 years
t2 = 17100 years
N(t) = No(1/2)^(t/t1/2)
Where,
N(t) = amount of carbon left at time, t
No = initial amount of carbon
t = time taken
t1/2 = half life
Assume No = 1
N(11400) = 1 × (1/2)^(11400/5700)
= (1/2)^2
= 1/4
N(17100) = 1 × (1/2)^(17100/5700)
= (1/2)^3
= 1/8
From the fractions got from N(11400) and N(17100), 1/4 and 1/8 we can see a decrease in the initial amount of carbon by 1/2.
B.
t1/2 = 80000 years
N(t) = No × (1/2)^(t/80000)
Assume No = 1
N(t) = 1 × (1/2)^(t/80000)
Answer:
t = 20.772
Step-by-step explanation:
r = R/100
r = 6/100
r = 0.06 per year,
Then, solve the equation for t
t = ln(A/P) / n[ln(1 + r/n)]
t = ln(5,200.00/1,500.00) / ( 12 × [ln(1 + 0.06/12)] )
t = ln(5,200.00/1,500.00) / ( 12 × [ln(1 + 0.005)] )
t = 20.772 years
