This is an excellent practice for the solution of quadratic equations.
1*36=36 => (1,36)
2*18=36 => (2,18)
3*12=36 => (3,12)
4*9=36 => (4,9)
6*6=36 => (6,6)
9*4=36 => (9,4)
12*3=36 => (12,3)
18*2=36 => (18,2)
36*1=36 => (36,1)
We can see that the sum decreases until the two factors are close (or equal) and then increases again.
The pair of integers with a sum of 20 is therefore (2,18) or (18,2).
Answer: 4 years
Step-by-step explanation:
A(0) has to be amount at start. Assume that's 5mg
Then A(t) = 5×(0.5)^(0.25t) = 5×2^(-t/4),
(also known as 5 exp(-λ t) with λ = ln(2)/4, incidentally).
We need to such that A(t) = 2.5mg, or 2^(-t/4) is 1/2, which happens when -t/4 is -1, or t is 4.
Answer:
Step-by-step explanation:
-1/2
Answer:
7 and 6 respectively
Step-by-step explanation:
Firstly, we have to solve the equations simultaneously.
2x + 7p = 56
3x - 11p = -45
Multiply equation I by 3 and ii by 2
6x +21p = 168
6x - 22p = -90
Subtract the second from first to yield:
43p = 258
p = 6
Insert this in equation 1 where we have 2x + 7p = 56
2x + 7(6) =56
2x + 42 = 56
2x = 14 and x = 7
The equilibrium price is 6 and the equilibrium quantity is 7
