This is as far as I could get X=-94+m<2
Answer:
Number of quarters → 15
Number of dimes → 2
Step-by-step explanation:
Let the number of dimes I have = y
And number of quarters = x
Since, I have amount in my pocket = $2
Therefore, 0.10y + 0.25x = 2
100(0.10y + 0.25x) = 100×2
25x + 10y = 200
5x + 2y = 40
2y = -5x + 40
y = -2.5x + 20 ---------(1)
Total number of coins in my pocket = 17
x + y = 17
y = -x + 17 ---------(2)
By using a graphing calculator we can graph these two lines (As attached)
Solution of the given system of equations will be the point of intersection of these lines.
Solution → (2, 15)
Number of quarters → 15
Number of dimes → 2
5:3. Simply by the common factor which is 6. 30 divided by 6 is five. 18 divided by 6 is 3. Therefore the simplest form is 5:3
First term ,a=4 , common difference =4-7=-3, n =50
sum of first 50terms= (50/2)[2×4+(50-1)(-3)]
=25×[8+49]×-3
=25×57×-3
=25× -171
= -42925
derivation of the formula for the sum of n terms
Progression, S
S=a1+a2+a3+a4+...+an
S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d] → Equation (1)
S=an+an−1+an−2+an−3+...+a1
S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d] → Equation (2)
Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)
2S=n(a1+an)
S=n/2(a1+an)
Substitute an = a1 + (n - 1)d to the above equation, we have
S=n/2{a1+[a1+(n−1)d]}
S=n/2[2a1+(n−1)d]
The answer is:
A. 8
Have a nice day :)