The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
Dividing we get,
Adding, we have,
Thus, the total time taken by Joshua is 
To determine the average speed we use the formula,
where 
and 
Hence, substituting the values we have,
Dividing, we get,
Thus, the average speed of Joshua during that time is 2500 m/h.
The parent function is:
y = x ^ 2
Applying the following function transformation we have:
Horizontal translations:
Suppose that h> 0
To graph y = f (x-h), move the graph of h units to the right.
We have then:
g (x) = (x-2) ^ 2
Then, we have the following function transformation:
Vertical translations
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then that the original function is:
g (x) = (x-2) ^ 2
Applying the transformation we have
f (x) = g (x) +3
f (x) = (x-2) ^ 2 + 3
Answer:
the function f(x) moves horizontally 2 units rigth.
The function f (x) is shifted vertically 3 units up.
I believe it is B. I could be wrong but I am hoping that you get it right
In a triangle, the three interior angles always add to 180°
m∠R + m∠S + m∠<span>T = 180
</span><span>(2x+10) + (2x+25) + (x-5) = 180
</span>2x+10 + 2x+25 + x-5 = 180
5x + 30 = 180
5x = 150
x = 30
