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Answer:
- parallel: y = 4x -6
- perpendicular: y = -1/4x +27/4
Step-by-step explanation:
If we want the new line to be written in slope-intercept form, we need to find the new value of the y-intercept. The equation of the line is ...
y = mx +b . . . . . . . for slope m and y-intercept b
Solving for b gives ...
b = y -mx . . . . . . . subtract mx from both sides.
The values of x and y come from the point we want the line to pass through. The value of m will be the same for the parallel line as for the given line: 4. For the perpendicular line, it will be the opposite reciprocal of this: -1/4.
<u>Parallel line</u>
b = 6 -4(3) = 6 -12 = -6
y = 4x -6
Perpendicular line
b = 6 -(-1/4)(3) = 6 +3/4 = 27/4
y = -1/4x +27/4
Answer:
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Answer:
Aha..our nemesis problem. :)
The area is 129.90 ft ^2.
Step-by-step explanation:
First, use a 30-60-90 triagle to find the apothem.
That is the area of 1 triangle. There are 6 of them in a hexagon, so multiply that number by 6:
21.65*6=129.90 ft ^2
Have a great day!
Answer:
True
Step-by-step explanation:
Given that a function is
We are to find the slant asymptote if any for this function
Since numerator is of degree 2 and denominator 1, let us divide and then check
Doing long division we find
![f(x)=\frac{1}{2} [x-\frac{5}{2} ]-\frac{11}{4(2x-3)}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bx-%5Cfrac%7B5%7D%7B2%7D%20%5D-%5Cfrac%7B11%7D%7B4%282x-3%29%7D)
Thus we find the asymptote y= the quotient obtained i.e
![\frac{1}{2} [x-\frac{5}{2} ]\\=\frac{x}{2} -\frac{5}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx-%5Cfrac%7B5%7D%7B2%7D%20%5D%5C%5C%3D%5Cfrac%7Bx%7D%7B2%7D%20-%5Cfrac%7B5%7D%7B4%7D)
Hence asymptote is
Statement given is true.
h(x) = -4 + 6
h(x) = 2
h(-2) = 2 because It's a constant function, when you substitute any x-values, you still get the same y-value.
Even if you substitute x = 1000, you still get y = 2.
Therefore, h(-2) = 2
-----------
If you mean h(x) = -4x+6
then h(-2) = 8+6 = 14
therefore, h(-2) = 14 for h(x) = -4x+6
