Answer:
i think photomath will help you with those
Step-by-step explanation:
go to your appstore and download photomath
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).
Hello,
(-25) - (25) = (-25) + (-25) = (-50)
Bye :)
Let P = number of coins of pennies (1 penny = 1 cent)
Let N = number of coins of nickels (1 nickel = 5 cents)
Let D = number of coins of dimes (1 dime = 10 cents)
Let Q = number of coins of quarters (1 quarter = 25 cents)
a) P + N + D + Q = 284 coins, but P = 173 coins, then:
173 + N + D + Q =284 coins
(1) N + D + Q = 111 coins
b) D = N + 5 OR D - N =5 coins
(2) D - N = 5 coins
c) Let's find the VALUE in CENTS of (1) that is N + D + Q = 111 coins
5N + 10D + 25 Q = 2,278 - 173 (1 PENNY)
(3) 5N + 10D + 25Q = 2105 cents
Now we have 3 equation with 3 variables:
(1) N + D + Q = 111 coins
(2) D - N = 5 coins
(3) 5N + 10D + 25Q = 2105 cents
Solving it gives:
17 coins N ( x 5 = 85 cents)
22 coins D ( x 10 = 220 cents)
72 coins D ( x 25 = 1,800 cents)
and 173 P,
proof:
that makes a total of 85+2201800+172 =2,278 c or $22.78
Answer: B.) 3/4
Step-by-step explanation:
I Got B by dividing 7 by 21 and 28 which is 3 and 4 so B is the correct answer.
