Wouldn't it be 6 cm, 8 cm, 10 cm?
The linear function which represents the line given by the point-slope equation is (B)
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<h3>
What is a linear function?</h3>
- The word linear function in mathematics refers to two distinct but related concepts.
- A linear function in calculus and related fields is a function whose graph is a straight line, that is, a polynomial function of degree zero or one.
To find the linear function which represents the line given by the point-slope equation:
Given: 
Distribute the right side:

Adds 8 on both sides:

Convert to function notation:

Therefore, the linear function which represents the line given by the point-slope equation is (B)
.
Know more about linear functions here:
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The complete question is given below:
Which linear function represents the line given by the point-slope equation y – 8 = y minus 8 equals start fraction one-half end fraction left-parenthesis x minus 4 right-parenthesis. (x – 4)?
A) F(x) = f(x) equals StartFraction one-half EndFraction x plus 4.X + 4
B) f(x) = f(x) equals StartFraction one-half EndFraction x plus 6.
C) X + 6 f(x) = f(x) equals StartFraction one-half EndFraction x minus 10.X –10
D) f(x) = f(x) equals StartFraction one-half EndFraction x minus 12.X – 12
The answer is: not likely. There isn’t much of a chance !
Answer:
(x+1)²/6 - (y+3)²/12 = 1
Step-by-step explanation:
The standard form of writing the equation of an hyperbola is expressed as;
(x-h)²/a² - (y-b)²/b² = 1 where (h,k) is the centre of the hyperbola.
Given the equation:
6x²-3y²+12x-18y-3 = 0
We are to convert it to the standard form of writing the equation of a hyperbola.
Collecting the like terms will give;
(6x²+12x)-(3y²+18y)-3 = 0
Divide through by 3
(2x²+4x)-(y²+6y)-1 = 0
Completing the square of the equation in parenthesis and adding the constants to the other side of the equation:
(2x²+4x)-(y²+6y+(6/2)²)-1 = 0+(6/2)²
(2x²+4x)-(y²+6y+9)-1 = 9
2x²+4x -{(y+3)²} = 9+1
2x²+4x - (y+3)² = 10
Divide through by 2
x²+2x -(y+3)²/2 = 5
(x²+2x+(2/2)²)-(y+3)²/2 = 5+(2/2)²
(x²+2x+1) - (y+3)²/2 = 5+1
(x+1)²-(y+3)²/2 = 6
Divide through by 6
(x+1)²/6 - (y+3)²/12 = 1
The resulting equation is the required standard form of a hyperbola with centre at (-1, -3)