Answer:
10 units
Step-by-step explanation:
Given data
P = (3, 1) and Q = (-3, -7)
x1=3
y1=1
x2= -3
y2= -7
The expression for the distance between two coordinate is
d=√((x_2-x_1)²+(y_2-y_1)²)
Substitute
d=√((-3-3)²+(-7-1)²)
d=√((-6)²+(-8)²)
d=√36+64
d=√100
d=10 units
Hence the distance is 10 units
The product of given polynomials is: 
Step-by-step explanation:
Given polynomials are:

In order to multiply both polynomials, we will use the distributive property
So,

Combining like terms

Hence,
The product of given polynomials is: 
Keywords: Polynomials, variables
Learn more about polynomials at:
#LearnwithBrainly
Answer:
Assuming this is for a Conic Section. There is no property as it does not match a conic section.
Answer:
a) y = 2x +4
b) y = 1/2x +4
c) y = -2x +11
Step-by-step explanation:
The given equations are in slope-intercept for, so we can read the slope directly from the equation. It is the x-coefficient.
We can then write an equation of a parallel line using the point-slope form of the equation of a line:
y -k = m(x -h) . . . . for a line with slope m through point (h, k)
If you like, you can rearrange this to "slope-intercept" form. Add k and simplify.
y = mx +(k -mh)
__
a) m = 2, (h, k) = (3, 10)
y = 2x +(10 -2·3)
y = 2x +4
__
b) m = 1/2, (h, k) = (0, 4)
y = 1/2x +(4 -(1/2)·0)
y = 1/2x +4
__
c) m = -2, (h, k) = (4, 3)
y = -2x +(3 -(-2)(4))
y = -2x +11
The value of f(0) is 1
and
The roots of f(x) = 0 are -2 and -1
<h3>Graph of a function</h3>
From the question, we are to determine the value of f(0) and the roots of f(x) = 0
The value of f(0) is the value of f(x) or the value of y when x = 0
In the given graph, we can observe that the value of y when x = 0 is 2
∴ f(0) = 2
The roots of f(x) = 0 correspond to the values of x when f(x) = 0 or when y = 0
In the graph, we can observe that the values of x when f(x) = 0 are -2 and -1
∴ The roots of f(x) = 0 are -2 and -1
Hence, the value of f(0) is 1
and
The roots of f(x) = 0 are -2 and -1
Learn more on Graph of a function here: brainly.com/question/21265554
#SPJ1