Using algebraic equations, the number of points Yael has is calculated as: 29 points.
<h3>How to Use Algebraic Equations to Solve Word Problems?</h3>
In the given world problem, we known the following:
E = Eric
S = Shenna
Y = Yael
Their total points would be: E + S + Y = 68
S = 2(E) (Shenna has twice the points of Erik)
Y = S + 3 (Yael has three points more than Shenna)
We would therefore have the following algebraic equation:
E + 2E + (S + 3) = 68
Substitute S with 2E
E + 2E + 2E + 3 = 68
Solve for E
5E + 3 = 68
5E = 68 - 3
5E = 65
5E/5 = 65/5
E = 13
Eric's points is: 13
Yael's points = Y = S + 3
Y = S + 3 = 2E + 3
Plug in the value of E
Y = 2(13) + 3
Y = 26 + 3
Y = 29
Yael has 29 points.
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Let the number be x.
sqrt(x/4) = 6
x/4 = 6^2 = 36
x = 36 * 4 = 144.
44*1 , 43+1, 44+0 , 42+2, there are numerous ways a problem could equal 44
Answer:
a) -7/9
b) 16 / (n² + 15n + 56)
c) 1
Step-by-step explanation:
When n = 1, there is only one term in the series, so a₁ = s₁.
a₁ = (1 − 8) / (1 + 8)
a₁ = -7/9
The sum of the first n terms is equal to the sum of the first n−1 terms plus the nth term.
sₙ = sₙ₋₁ + aₙ
(n − 8) / (n + 8) = (n − 1 − 8) / (n − 1 + 8) + aₙ
(n − 8) / (n + 8) = (n − 9) / (n + 7) + aₙ
aₙ = (n − 8) / (n + 8) − (n − 9) / (n + 7)
If you wish, you can simplify by finding the common denominator.
aₙ = [(n − 8) (n + 7) − (n − 9) (n + 8)] / [(n + 8) (n + 7)]
aₙ = [n² − n − 56 − (n² − n − 72)] / (n² + 15n + 56)
aₙ = 16 / (n² + 15n + 56)
The infinite sum is:
∑₁°° aₙ = lim(n→∞) sₙ
∑₁°° aₙ = lim(n→∞) (n − 8) / (n + 8)
∑₁°° aₙ = 1
