(x^2-3x+2)-1(3x^2-5x-1)=
x^2-3x+2-3x^2+5x+1=
x^2-3x^2-3x+5x+2+1=
-2x^2+2x+3
Assuming we need to find i such that
1 ≤ i ≤ n and t[i]=i.
If we need to find only the first occurrence, we can do:
for i:1 to n {
if t[i]=i then return(i)
}
If exhaustive search is required, then put the results (values of i) in an array or a linked list, return the number of values found, and the array (or linked list).
Answer:
the amount left to spend is $740.77
Step-by-step explanation:
The computation of the amount left to spend is as follows:
= Each month payment - last month payment bills
= $2,185.76 - $73.49 - $897.19 - $474.31
= $740.77
Hence, the amount left to spend is $740.77
We simply deduct three amount bills from the each month payment
Answer:
The team has 92 wins and 38 losses.
Step-by-step explanation:
In a season 130 games are played by a baseball team.
Let us assume that the team wins x games and losses y games.
Therefore, we can write, x + y=130....... (1)
It is given that if the team won sixteen more games twice as many games as it lost.
So, we can conclude that x =2y+16 ......(2)
Hence, solving equations (1) and (2) we get, (2y+16) + y =130
⇒ 3y = 114
⇒ y =38
And from equation (2), x = 2×38 + 16 = 92
Therefore, the team has 92 wins and 38 losses. (Answer)
