3n^2+7n-6n-14+4n^2+6n= 7n^2+7n-14
Hope that helps and best of wishes!
ur not dum
Answer:
Paul
Step-by-step explanation:
Paul earn more than monica. Though he is getting less interest but because of his higher initial amount he is getting more return.
Computing the return of both
Monica return is 100*3.4%= $3.4 in a year
Paul return is 200*%2.2= $4.4 in a year.
Answer:
In a paragraph proof, statements and their justifications are written in sentences in a logical order.
A two-column proof consists of a list statements and the reasons the statements are true.
A paragraph proof is a two-column proof in sentence form.
Step-by-step explanation:
- In a paragraph proof, statements and their justifications are written in sentences in a logical order.
- A two-column proof consists of a list statements and the reasons the statements are true.
- A paragraph proof is a two-column proof in sentence form.
A paragraph proof is only a two-column proof written in sentences. However, since it is easier to leave steps out when writing a paragraph proof.
A two-column geometric proof consists of a list of statements, and the reasons that we know those statements are true. The statements are listed in a column on the left, and the reasons for which the statements can be made are listed in the right column
Answer:c
Step-by-step explanation:
when you set (x-6) and (x-1) to zero you get 6 and 1
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
