Answer:
The ratio would be 5 to 3
Step-by-step explanation:
I know this because you would divide the amount of grapes and the amount of cashews by 3 which would equal 5 to 3
Answer:
25
Step-by-step explanation:
420 - 45 = 375
375 / 15 = 25
Answer:
No solutions.
Step-by-step explanation:
-2x - y = 1
-4x - 2y = -1 Multiply the first equation By -2:
4x + 2y = -2 Adding:
0 = -3
which is absurd so there are no solutions.
You have to find out how much time she played on Saturday, Wednesday, Friday, and Sunday. Let's look at the facts.
Sunday: No info is given.
Friday: She played the least of all four days
Saturday: Twice as much as Wednesday
Wednesday: No info is given.
We can write a problem and solve from this information. Let's use the variable W for Wednesday. Saturday is 2W. So whatever number Wednesday has, multiply it by 2. All of the total numbers should equal 10.
Since Friday was played the least amount, we know that every other day has to be at least 2 hours or above, so that Friday can be 1 hour. Sunday we have no info given.
This is a bit of guess and see for this problem. Let's say that Friday was 1 hour, Wednesday was 2, Sunday was 3, and Saturday was 4. Total that equals 10. Let's check to varify that these numbers are correct.
Friday was the least amount of hours of any day, which is why it is 1.
Wednesday had 2 hours, and Saturday had twice that many, which is 4.
Sunday has 3 hours.
In short, your answer would be: Friday 1 hour, Wednesday 2 hours, Sunday 3 hours, and Saturday 4 hours. I hope I explained it wel
Answer:
O(n2)
Step-by-step explanation:
the first iteration algorithm of the i-for loop (the outer loop), the j-for loop (the inner loop) will run 2 to
n times which is represented as
(n − 1 times).
the second iteration algorithm of the i-for loop, the j-for loop will run 3 to n times represented as
(n − 2 times).
the third to the last iteration algorithm of the i-for loop, the j-for loop will run n − 1 to n times (2 times).
And the second to the last iteration of the i-for loop, the j-for loop will run from n to n times (1 time)
For the last iteration of the i-for loop, the j-for loop will run 0 times because i + 1 > n.
Now we know that the number of times the loops are run is
1 + 2 + 3 + . . . + (n − 2) + (n − 1) = n(n − 1)/2
So we can express the number of total iterations as n(n − 1)/2.
Since we have two operations per loop (one comparison and one multiplication), we have
2 ·n(n−1)/2 = n
2 − n operations.
So f(n) = n2 − n
f(n) ≤ n2
for n > 1.
Therefore, the algorithm is O(n2) with
C = 1 and k = 1.
