You can write three equations in the numbers of nickels (n), dime (d), and quarters (q).
n + d + q = 23 . . . . . . . there are 23 coins total
0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters
5n +10d +25q = 250 . .the total value is $2.50
The collection includes 11 nickels, 7 dimes, and 5 quarters.
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I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.
(5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)
25q = 125 . . . . . . . simplify
q = 5
From the second equation,
d = q +2 = 7
And from the first,
n = 23 -5 -7 = 11
Answer:
a)
(i) 43, 56
(ii) -1,3
b)-27
c) -21 +4n
Step-by-step explanation:
For a,i, you can see that it decreses by 7 each time so you subtract 7 from 50 to get 43, then you subtract 7 from that to get 56. For a,ii it increases by 4 each time so you add 4 to -5 to get -1, then add 4 to that to get 3. For b, because it is the 15th term you are looking for, you can plug 15 into the equation as n so 78-7n to 78- 7*15, 78-105, -27. For c, you know that each time it increases by 4, so you know you add 4n, and then for the base, you have to find the term before -17 so you subtract 4 to get -21. Put it together and get -21+4n.
F = 18 ft.
The law of cosines states
c² = a² + b² - 2ab cos C
Using our information, we have
c² = 23² + 16² - 2(23)(16)cos 52
c² = 529 + 256 - 736cos 52
c² = 785 - 736cos 52
c² = 331.8732
Taking the square root of both sides, we have
c = √331.8732 = 18.22 ≈ 18
Answer:
1st one is 68 and the 2nd one is 25. hope that helps