Hey there! I'm happy to help!
The median is the line cutting the triangle in this graph. In general, a median connects any of the triangle points to halfway across the side opposite to that point. This median connects R to halfway across the side across from R.
We want to find the equation of this median. We see that T is at (-2,3) and R is (3,-3). The first thing to do when looking for the equation of the line is to find the slope, or incline. Since we have two points, we can do this very easily. You simply divide the difference in the y-values by the difference in the x-values.
DIFFERENCE IN Y-VALUES
3-(-3)
3+3
6
DIFFERENCE IN X-VALUES
-2-3
-5
Now, we divide the two answers, giving us -6/5.
So, we have our slope, which gives us the equation so far y=-6/5x+b. We just need to find the b, which is our y-intercept. Well, to do this, we plug in one of our points and we can solve for b. We will use (3,-3)!
-3=-6/5(3)+b
-3=-3 3/5+b
We add 3 3/5 to both sides to isolate the b.
b=3/5
This means that this median should hit the y-axis at (0,3/5), and it looks like it does. Therefore, the equation of this median is y=-6/5x+3/5.
Now you can find the slope of a median! Have a wonderful day! :D
Try drawing the triangles: the original and the dilated one.
If the tangent of x is 4/3, and then all lengths of the original triangle are doubled, then the tangent of x is 8/6 = 4/3. Thus, B is the correct answer.
The function has neither because it goes infinitely up and infinitely down
Check the picture below. So the parabola looks more or less like so.
let's recall that the vertex is half-way between the focus point and the directrix, at "p" units away from both.
Let's notice that the focus point is below the directrix, that means the parabola is vertical, namely the squared variable is the "x", and it also means that it's opening downwards as you see in the picture, namely that "p" is negative, in this case "p" is 1 unit, and thus is -1.
![\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2\implies -4(y-5)=(x+2)^2 \\\\\\ y-5=-\cfrac{1}{4}(x+2)^2\implies y=-\cfrac{1}{4}(x+2)^2+5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%20%5Cend%7Barray%7D%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20vertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D-2%5C%5C%20k%3D5%5C%5C%20p%3D-1%20%5Cend%7Bcases%7D%5Cimplies%204%28-1%29%28y-5%29%3D%5Bx-%28-2%29%5D%5E2%5Cimplies%20-4%28y-5%29%3D%28x%2B2%29%5E2%20%5C%5C%5C%5C%5C%5C%20y-5%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%2B5)
Answer:
0.7 hours
Step-by-step explanation:
Given that Irina was able to make the same distance from work to home in 0.4 of an hour at 27 miles per hour, we can use this rate and time to find the distance she travels to and from work using the general formula:
d = rt, where d=distance, r = rate and t = time
d = 27(0.4) = 10.8 miles
Since the distance from Irina's home to work is 10.8 miles, we can again use the formula 'd = rt' to find the time it takes her to bike to work at a rate of 16 miles per hour and solving for time, 't':
10.8 = (16)t
t = 0.7 hours
