What is the 10th term in the pattern with the formula 5n + 100?
Answers
first term is n=1
so 10th term is n=10
5(10)+100=50+100=150
150 is 10th term
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What is the value of x in the equation 3(2x + 4) = −6?a) −3 b)1 c)12 d)19
Answers
= (x-5)(x-5)^2
= (x-5) ( x^2 -10x + 25)
= x^3 -15x^2 + 75 - 125
Hope this helped
cm 2
cm 3
cm 4
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Answer: A. cm
Step-by-step explanation: The formula for the volume of a hexagonal prism is, volume = [(3√3)/2]a2h cubic units where a is the base length and h is the height of the prism. We can also use the other formula V = 3abh, where a = apothem length, b = length of a side of the base, and h = height of the prism.
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Answer:
The end behavior of the graph of p is:
f(x) → ∞ as x → -∞ and f(x) → -∞ as x → ∞
Step-by-step explanation:
Given the polynomial function
Since the leading term of the polynomial (the term in a polynomial that contains the highest power of the variable) is -9x⁹, then the degree is 9 i.e. odd, and the leading coefficient is -9, i.e. negative.
This means that f(x) → ∞ as x → -∞ and f(x) → -∞ as x → ∞
The graph is also attached below.
Thus, the end behavior of the graph of p is:
f(x) → ∞ as x → -∞ and f(x) → -∞ as x → ∞
The graph of p(x) will start from the bottom-left and extend towards the top-right of the coordinate plane.
Use the concept of a graph defined as:
Drawing the curve that represents a function on a coordinate plane is known as graphing a function. Every point on the curve will satisfy the function equation if the curve (or graph) reflects the function.
The given polynomial is:
Since we know that,
The end behaviour of the graph of p(x) can be determined by looking at the leading term, which in this case is .
Here, the leading term has an odd degree and a negative coefficient,
the end behaviour of the graph will be as follows:
As x approaches negative infinity, the graph of p(x) will decrease without bound (goes down indefinitely).
As x approaches positive infinity, the graph of p(x) will increase without bound (goes up indefinitely).
Hence, the graph of p(x) will start from the bottom-left and extend towards the top-right of the coordinate plane.
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hope it helped you
Answers
This could also be written as, where
This would factorize to give:
So we can factorize our original expression:
We can then solve for
And also:
So our values for
As:
Our final solutions for
The solutions to the equation 2cos²(x) + cos(x) - 1 = 0 over the interval [0, 2π) are x = π/3, 5π/3, and π.
We have,
To solve the equation 2cos²(x) + cos(x) - 1 = 0 over the interval [0, 2π), we can use a substitution technique.
Let's substitute cos(x) with a variable, say, u.
The equation becomes:
2u^2 + u - 1 = 0.
Now, we can factorize the quadratic equation:
(2u - 1)(u + 1) = 0.
Setting each factor equal to zero, we have:
2u - 1 = 0 or u + 1 = 0.
Solving these equations separately, we find:
2u = 1 or u = -1.
For 2u = 1, we get u = 1/2. Taking the inverse cosine of 1/2,
We have cos(x) = 1/2.
For u = -1, we get u = -1. Taking the inverse cosine of -1, we have cos(x) = -1.
Now, we need to determine the solutions for x within the given interval [0, 2π).
For cos(x) = 1/2, the solutions within the interval are x = π/3 and x = 5π/3.
For cos(x) = -1, the solution within the interval is x = π.
Therefore,
The solutions to the equation 2cos²(x) + cos(x) - 1 = 0 over the interval [0, 2π) are x = π/3, 5π/3, and π.
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