<span>Data:
infinite geometric series
A1
= 880
r = 1 / 4
The sum of a geometric series in sigma
notation is:
n 1 - r^n
∑ Ai = A ----------- ; where A = A1
i = 1 1-r
When | r | < 1 the infinite sum exists and is equal to</span><span><span>:
∞ A
∑ Ai = ---------- ; where A = A1
i = 1 1 - r</span>
So, in this case</span><span><span>:
∞ 880
∑ Ai = -------------- = 4 * 880 / 3 = 3520 /3 = 1173 + 1/3
i = 1 1 - (1/4)</span> </span>
Answer: 1173 and 1/3
Answer:
tan 45° = 1 = true
cos 30° = 1/2 = false
sin 45° = 1/3 = false
sin 90° = 0 = false
Step-by-step explanation:
Use the rational roots test. The possible roots are: plus/minus 6,3,2,1
Use synthetic division and you will see that 3 is a root:
3 | 1 -3 -3 11 -6
| 3 0 -9 6
____________
1 0 -3 2 0
Use rational root again, to see that possible roots are: plus/minus 2,1
Try 2:
2 | 1 0 -3 -2
| 2 4 2
_____________
1 2 1 0
The above is x^2+2x+1 which is a perfect square: (x+1)^2
So we have the final factorization: (x-3)(x-2)(x+1)^2
So the roots are: 3, 2, -1
Where -1 is a double zero.
2 (a): <span>Lena's method is correct
</span> (b) Explanation: - We can explain that she was correct by checking the equation. In order to check that, we have to put the values of x and y.
4y = 3x + 7
Substitute for y = 10, x = 11 [ Lena's answers ]
4(10) = 3(11) + 7
40 = 33 + 7
40 = 40 [ L.H.S. = R.H.S ]
Now, checking with 2nd equation,
9x + 4y -139 = 0
9(11) + 4(10) - 139 = 0
99 + 40 - 139 = 0
139 - 139 = 0
0 = 0 [ L.H.S. = R.H.S ]
As L.H.S. is equal to R.H.S. for both equation, She was correct
3) a) System of equations would be:
x + y = 1
x - y = 11
b) Solving the equation,
Take 2nd equation,
x - y = 11
x = 11 + y
Now, substitute it in 1st equation.,
x + y = 1
11 + y + y = 1
2y = 1 - 11
y = -10 / 2
y = -5
Substitute it in 2nd equation,
x - y = 11
x - (-5) = 11
x + 5 = 11
x = 11 - 5
x = 6
In short, x is equal to 6, and y is equal to -5
Hope this helps!
