bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.
we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.
we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.
![\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%5C%5C%20%5Ctextit%7Basymptotes%7D%5Cquad%20y%3D%20k%5Cpm%20%5Ccfrac%7Bb%7D%7Ba%7D%28x-%20h%29%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
g(7-h) +3
Now g = 5 and h=2, So lets plug the values
5(7-2) +3
= 5(5) +3
= 25 + 3
= 28
Answer:
<h3>2.2</h3>
Step-by-step explanation:
- (6y-22) = (y-11)
- 6y-22 = y-11
- 6y-y = -11+22
- 5y = 11
- y = 11/5
- y = 2⅕ or 2.2
Answer:
60 mph
Step-by-step explanation:
Given;
Total distance covered d = 440 miles
Average speed in the first 4 hours v1 = 50 mph
Total time taken for the whole trip t = 8:00 to 4:00pm
t = 16:00-8:00 = 8 hours
Firstly we need to calculate the distance covered during the first 4 hours;
d1 = average speed × time = v1 × t1
t1 = 4
d1 = 50 × 4 = 200 miles
Then we need to calculate the distance covered in the second period of the journey;
d = d1 + d2
d2 = d - d1
Substituting the values;
d2 = 440 - 200
d2 = 240 miles
The time taken for the second period of travel t2;
t2 = t -t1 = 8-4
t2 = 4 hours
Average speed = distance travelled ÷ time taken
The average speed for the second part of the trip v2 is;
v2 = d2 ÷ t2
Substituting t2 and d2;
v2 = 240 miles ÷ 4 hours
v2 = 60 mph
The average speed during the second part of the trip is 60 mph
