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.
Answer: x>1.2
Step-by-step explanation:
Let a number is x.
Hence,

Divide both parts of the equation by 5:

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

where
is a compact way to indicate both solutions
and
, while
are the coefficients of the quadratic equation, i.e. we consider the polynomial
.
So, in your case, we have 
Plug those values into the formula to get

So, the two solutions are


Answer:
Step-by-step explanation:
i think the answer is 1/3 from 8x 6 =-46.7 i think