Answer:
y=3/4x+2
Step-by-step explanation:
I found this by using desmos graphing calculator
Putting all the names in a hat and selecting six students because the rest would not have been random because you know who you picked just by looking but if u put names in a hat you don't know who you picked.
The idea is to pair up the terms, factor each sub-group, and then pull out the overall GCF to fully factor.
x^3+5x^2-6x-30
(x^3+5x^2)+(-6x-30)
x^2(x+5)+(-6x-30)
x^2(x+5)-6(x+5)
(x^2-6)(x+5)
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Answer: (x^2-6)(x+5)
For this case we have the following polynomials:
3x2
x2y + 3xy2 + 1
We have then:
For 3x2:
Classification: polynomial of one variable:
Degree: 2
For x2y + 3xy2 + 1:
Classification: polynomial of two variables
Degree: 2 + 1 = 3
Answer:
The polynomial 3x2 is of one variable with a degree of 2.
The polynomial x2y + 3xy2 + 1 is of two variables a with a degree of 3.
The correct answer is C.
You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.
f(x) = x^4 + x^3 - 2x^2
Since each term has at least x^2, we can factor it out.
f(x) = x^2(x^2 + x - 2)
Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.
f(x) = x^2(x + 2)(x - 1)
Now we can also separate the x^2 into 2 x's.
f(x) = (x)(x)(x + 2)(x - 1)
To find the zeros, we need to set them all equal to 0
x = 0
x = 0
x + 2 = 0
x = -2
x - 1 = 0
x = 1
Since there are two 0's, we know the graph just touches there. Since there are 1 of the other two numbers, we know that it crosses there.
