Answer:
v=9i+3j
Step-by-step explanation:
The given vector, v has initial point at P1 = (−5, −2) and terminal point at P2 = (4, 1).
The vector v is found by subtraction the initial point from the terminal point.
v=<4,1>-<-5,-2>
v=<4--5,1--2>
v=<9,3>
We write v as multiples of the basis vectors to obtain:
v=9i+3j
Answer: B. 2 = 3x + 10x2
Step-by-step explanation:
This is the concept of quadratic equations; We required to find the type of equation that can be solved using the model that has been used to solve the equation such that the answer is:
[-3+-sqrt(3^2+4(10)(2))]/(2(10))
The formual that was applied here was a quadratic formula given by:
x=[-b+\-sqrt(b^2-4ac)]/2a
whereby from the our substituted values above,
a=10,b=3 and c=-2
such that the quadratic equation will be:
10x^2+3x-2
6x² + 48x + 96
6(x² + 8x + 16)
6(x + 4)(x + 4) is your polynomial fully factored so I'm guessing the binomial you're looking for is (x + 4).
There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/- 
x = +/- i
They are adjacent and complementary.
Next to each other and add up to 90 degrees.