\frac{d}{dx}\left(\frac{1+x^4+x^6}{x^2+x+1}\right)=\frac{4x^7+5x^6+8x^5+3x^4+4x^3-2x-1}{\left(x^2+x+1\right)^2}
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Answer: (-2, 0) and (0, -2)
Step-by-step explanation:
This system is:
y + x = -2
y = (x + 1)^2 - 3
To solve this we first need to isolate one of the variables in one fo the equations, in the second equation we have already isolated the variable y, so we can just replace it in the first equation:
(x + 1)^2 - 3 + x = -2
Now we can solve this for x.
x^2 + 2*x + 1 - 3 = -2
x^2 + 2*x + 1 -3 + 2 = 0
x^2 + 2*x + 0 = 0
The solutions of this equation are given by the Bhaskara's formula, then the solutions are:
The two solutions are:
x = (-2 - 2)/2 = -2
In this case, we replace this value of x in the first equation and get:
y - 2 = -2
y = -2 + 2 = 4
This solution is x = -2, y = 0, or (-2, 0)
The other solution for x is:
x = (-2 + 2)/2 = 0
If we replace this in the first equation we get:
y + 0 = -2
y = -2
This solution is x = 0, y = -2, or (0, -2)
Answer:
8 4/20
0.2
-8 1/10
Step-by-step explanation:
Answer:
- C) Simon is correct because even though the input values are opposite in the reflected function, any real number can be an input.
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The exponential growth function has:
- Domain - all real numbers,
- Range - all real numbers excluding zero.
When the function is reflected across the y-axis, we'll have no change to domain or range from what is described above.
Alissa is correct, the input values change to opposite, however the domain stays same - all real numbers.
It means Simon is correct with his statement.
The matching answer choice is C.
Answer:
Step-by-step explanation:
Prove that for any natural number 3^(n+4)-3n is divisible by 16.
(I'm going to assume that you mean 3^(n+4)-3^n.)
1. We can break up 3^(n+4)-3n into 3^n * 3^4-3^n (by the rule a^b*a^c = a^b+c).
2. Solve to get 3^n * 81 - 3^n
3. Factor out the 3^n, and you'll get 3^n(81-1), and simplify: 3^n(80)
You may notice that 80 is divisible by 16.
4. Rewrite what we got from the last step as: 3^n*5(16).
Hope this helped you!
