Answer:
A bar graph shows amounts as bars of different sizes and, sometimes, of different colors. Longer bars represent larger numbers.
Distance<-------------300--------------><-------------------x----------------->
⊕---------------------------------⊕--------------------------------------⊕ Express Freight Overtaking point
Speed 55 speed 30
Let x be the distance travelled by Freight until the point of overtaking
Time = Distance/Speed, then:
Time (freight) = Distance (fright)/Speed (freight) And
Time (Express) = Distance (Express)/Speed (Express).
Now plug in the relative number:
Time (freight) = x/30 and Time (express) = (x+300)/55. But note that both times are equal, then:
x/30 = (x+300)/55 OR (x÷30 = (x+300)÷55) [Answer C]
Now let's find the distance x. Cross multiplication;
55x= 30(300+x) →55x = 9,000 + 30x
55x - 30x = 9,000 → 25x = 9,000 and x = 9,000/25 = 360 miles
Time needed for Freight to travel 360 m = 360/30= 12 Hours
Time needed for Express to travel (300+360) m = 660/55= 12 Hours
As you see, it's the same time of 12 hours
Don't forget that Distance = Speed x time or time = Distance/Speed, etc.
1.25
hope it helped and sorry if it's wrong i wish you luck
The requirement is that every element in the domain must be connected to one - and one only - element in the codomain.
A classic visualization consists of two sets, filled with dots. Each dot in the domain must be the start of an arrow, pointing to a dot in the codomain.
So, the two things can't can't happen is that you don't have any arrow starting from a point in the domain, i.e. the function is not defined for that element, or that multiple arrows start from the same points.
But as long as an arrow start from each element in the domain, you have a function. It may happen that two different arrow point to the same element in the codomain - that's ok, the relation is still a function, but it's not injective; or it can happen that some points in the codomain aren't pointed by any arrow - you still have a function, except it's not surjective.
