Answer:
Correct option: first one
Step-by-step explanation:
The equation f(x) = 5x + 1 is a function, because each value of x gives only one value of y.
Now let's find the inverse by switching f(x) by x and x by f'(x), then isolating f'(x):
![x = 5f'(x) + 1](https://tex.z-dn.net/?f=x%20%3D%205f%27%28x%29%20%2B%201)
![5f'(x) = x - 1](https://tex.z-dn.net/?f=5f%27%28x%29%20%3D%20x%20-%201)
![f'(x) = \frac{x - 1}{5}](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cfrac%7Bx%20-%201%7D%7B5%7D)
The inverse f'(x) is also a function, because each value of x gives only one value of x.
So we have that both the equation and its inverse are function, therefore the correct answer is the first option.
Answer:
y= -2x -8
Step-by-step explanation:
I will be writing the equation of the perpendicular bisector in the slope-intercept form which is y=mx +c, where m is the gradient and c is the y-intercept.
A perpendicular bisector is a line that cuts through the other line perpendicularly (at 90°) and into 2 equal parts (and thus passes through the midpoint of the line).
Let's find the gradient of the given line.
![\boxed{gradient = \frac{y1 -y 2}{x1 - x2} }](https://tex.z-dn.net/?f=%5Cboxed%7Bgradient%20%3D%20%20%5Cfrac%7By1%20-y%202%7D%7Bx1%20-%20x2%7D%20%7D)
Gradient of given line
![= \frac{1 - ( - 5)}{3 - ( - 9)}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B1%20-%20%28%20-%205%29%7D%7B3%20-%20%28%20-%209%29%7D%20)
![= \frac{1 + 5}{3 + 9}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B1%20%2B%205%7D%7B3%20%2B%209%7D%20)
![= \frac{6}{12}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B6%7D%7B12%7D%20)
![= \frac{1}{2}](https://tex.z-dn.net/?f=%20%3D%20%20%20%5Cfrac%7B1%7D%7B2%7D%20)
The product of the gradients of 2 perpendicular lines is -1.
(½)(gradient of perpendicular bisector)= -1
Gradient of perpendicular bisector
= -1 ÷(½)
= -1(2)
= -2
Substitute m= -2 into the equation:
y= -2x +c
To find the value of c, we need to substitute a pair of coordinates that the line passes through into the equation. Since the perpendicular bisector passes through the midpoint of the given line, let's find the coordinates of the midpoint.
![\boxed{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2}) }](https://tex.z-dn.net/?f=%5Cboxed%7Bmidpoint%20%3D%20%28%20%5Cfrac%7Bx1%20%2B%20x2%7D%7B2%7D%20%2C%20%5Cfrac%7By1%20%2B%20y2%7D%7B2%7D%29%20%20%7D)
Midpoint of given line
![= ( \frac{3 - 9}{2} , \frac{1 - 5}{2} )](https://tex.z-dn.net/?f=%20%3D%20%28%20%5Cfrac%7B3%20%20-%20%209%7D%7B2%7D%20%2C%20%5Cfrac%7B1%20-%205%7D%7B2%7D%20%29)
![= ( \frac{ - 6}{2} , \frac{ - 4}{2} )](https://tex.z-dn.net/?f=%20%3D%20%28%20%5Cfrac%7B%20-%206%7D%7B2%7D%20%20%2C%20%5Cfrac%7B%20-%204%7D%7B2%7D%20%29)
![= ( - 3 , - 2)](https://tex.z-dn.net/?f=%20%3D%20%28%20-%203%20%2C%20-%202%29)
Substituting (-3, -2) into the equation:
-2= -2(-3) +c
-2= 6 +c
c= -2 -6 <em>(</em><em>-</em><em>6</em><em> </em><em>on both</em><em> </em><em>sides</em><em>)</em>
c= -8
Thus, the equation of the perpendicular bisector is y= -2x -8.
Answer:
A. x² +y² -6x -16y +48 = 0
Step-by-step explanation:
The standard-form equation for a circle centered at (h, k) with radius r is ...
(x -h)² +(y -k)² = r²
For your circle, this is ...
(x -3)² +(y -8)² = 5²
To put this in general form, you subtract the constant on the right, and eliminate parentheses:
x² -6x +9 +y² -16x +64 -25 = 0
x² +y² -6x -16y +48 = 0 . . . . . rearrange to descending powers of x, y
(103*102*1.1)-4.9 or at least that is the order I would do it in. The answer to the problem is11551.7 if you need it. :)