Answer:
The sum of the geometric series is 1048575
Step-by-step explanation:
![S_n=\frac{a_1(r^n-1)}{r-1}\\ \\S_{10}=\frac{3(4^{10}-1)}{4-1}\\ \\S_{10}=\frac{3(1048576-1)}{3}\\ \\S_{10}=1048575](https://tex.z-dn.net/?f=S_n%3D%5Cfrac%7Ba_1%28r%5En-1%29%7D%7Br-1%7D%5C%5C%20%5C%5CS_%7B10%7D%3D%5Cfrac%7B3%284%5E%7B10%7D-1%29%7D%7B4-1%7D%5C%5C%20%5C%5CS_%7B10%7D%3D%5Cfrac%7B3%281048576-1%29%7D%7B3%7D%5C%5C%20%5C%5CS_%7B10%7D%3D1048575)
Answer:
f(x) = x3 - 3x + 1
f'(x) = 3x2 - 3 = 3(x2 - 1) = 3(x+1)(x-1)
Find the critical numbers: set f'(x) = 0
3(x+1)(x-1) = 0 x = 1, -1
Check the sign of f'(x) on either side of each critical number. If f'(x) is positive on an interval, then f is increasing on the interval. If f'(x) is negative on an interval, then f is decreasing on the interval.
+ l - l + sign f'(x)
-------------------------------------
-1 1
f is increasing on (-∞, -1) υ (1, ∞)
f is decreasing on (-1, 1)
For this case we have:
The arithmetic sequence formula is given by:
![a_n = a_1 + (n-1) * d](https://tex.z-dn.net/?f=a_n%20%3D%20a_1%20%2B%20%28n-1%29%20%2A%20d)
Where:
a_n: It's the last term
a_1: It's the first term
n: Position of the term
d: Common difference
We want to find
:
![a_8 = a_1 + (8-1) * d](https://tex.z-dn.net/?f=a_8%20%3D%20a_1%20%2B%20%288-1%29%20%2A%20d)
Substituting the values of a_1 and d
![a_8 = -7 + (7) (0.5)](https://tex.z-dn.net/?f=a_8%20%3D%20-7%20%2B%20%287%29%20%280.5%29)
![a_8 = -7 + 3.5a_8 = -3.5](https://tex.z-dn.net/?f=a_8%20%3D%20-7%20%2B%203.5a_8%20%3D%20-3.5)
Answer:
![a_8 = -3.5](https://tex.z-dn.net/?f=a_8%20%3D%20-3.5)
Step-by-step explanation:
1. (2^3)*(2^3)*(2^3)*(2^3)=2^12
2. (p^2)* (p^2)* (p^2)* (p^2)* (p^2)=p^10
3. (x^m)*(x^m)=x^(2m)
4. (2^3 x)* (2^3 x) = 2^6 x^2
To solve this problem, you have to know these two special factorizations:
![x^3-y^3=(x-y)(x^2+xy+y^2)\\ x^3+y^3=(x+y)(x^2-xy+y^2)](https://tex.z-dn.net/?f=%20x%5E3-y%5E3%3D%28x-y%29%28x%5E2%2Bxy%2By%5E2%29%5C%5C%20x%5E3%2By%5E3%3D%28x%2By%29%28x%5E2-xy%2By%5E2%29%20)
Knowing these tells us that if we want to rationalize the numerator. we want to use the top equation to our advantage. Let:
![\sqrt[3]{x+h}=x\\ \sqrt[3]{x}=y](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%2Bh%7D%3Dx%5C%5C%20%5Csqrt%5B3%5D%7Bx%7D%3Dy%20)
That tells us that we have:
![\frac{x-y}{h}](https://tex.z-dn.net/?f=%20%20%5Cfrac%7Bx-y%7D%7Bh%7D)
So, since we have one part of the special factorization, we need to multiply the top and the bottom by the other part, so:
![\frac{x-y}{h}*\frac{x^2+xy+y^2}{x^2+xy+y^2}=\frac{x^3-y^3}{h*(x^2+xy+y^2)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx-y%7D%7Bh%7D%2A%5Cfrac%7Bx%5E2%2Bxy%2By%5E2%7D%7Bx%5E2%2Bxy%2By%5E2%7D%3D%5Cfrac%7Bx%5E3-y%5E3%7D%7Bh%2A%28x%5E2%2Bxy%2By%5E2%29%7D%20)
So, we have:
![\frac{x+h-h}{h(\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2})}=\\ \frac{x}{\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2}}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%2Bh-h%7D%7Bh%28%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7B%28x%2Bh%29%28x%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%29%7D%3D%5C%5C%20%5Cfrac%7Bx%7D%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7B%28x%2Bh%29%28x%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%20)
That is our rational expression with a rationalized numerator.
Also, you could just mutiply by:
![\frac{1}{\sqrt[3]{x_h}-\sqrt[3]{x}} \text{ to get}\\ \frac{1}{h\sqrt[3]{x+h}-h\sqrt[3]{h}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7Bx_h%7D-%5Csqrt%5B3%5D%7Bx%7D%7D%20%5Ctext%7B%20to%20get%7D%5C%5C%20%5Cfrac%7B1%7D%7Bh%5Csqrt%5B3%5D%7Bx%2Bh%7D-h%5Csqrt%5B3%5D%7Bh%7D%7D%20)
Either way, our expression is rationalized.