Answer: Downhill:10mph Uphill:5mph
Step-by-step explanation:
We are looking for Dennis’s downhill speed.
Let
r=
Dennis’s downhill speed.
His uphill speed is
5
miles per hour slower.
Let
r−5=
Dennis’s uphill speed.
Enter the rates into the chart. The distance is the same in both directions,
20
miles.
Since
D=rt
, we solve for
t
and get
t=
D
r
.
We divide the distance by the rate in each row and place the expression in the time column.
Rate
×
Time
=
Distance
Downhill
r
20
r
20
Uphill
r−5
20
r−5
20
Write a word sentence about the time.
The total time traveled was
6
hours.
Translate the sentence to get the equation.
20
r
+
20
r−5
=6
Solve.
20(r−5)+20(r)
40r−100
0
0
0
=
=
=
=
=
6(r)(r−5)
6
r
2
−30r
6
r
2
−70r+100
2(3
r
2
−35r+50)
2(3r−5)(r−10)
Use the Zero Product Property.
(r−10)=0
r=10
(3r−5)=0
r=
5
3
The solution
5
3
is unreasonable because
5
3
−5=−
10
3
and his uphill speed cannot be negative. So, Dennis's downhill speed is
10
mph and his uphill speed is
10−5=5
mph.
Check. Is
10
mph a reasonable speed for biking downhill? Yes.
Downhill:
10 mph
5 mph⋅
20 miles
5 mph
=20 miles
Uphill:
10−5=5 mph
(10−5) mph⋅
20 miles
10−5 mph
=20 miles
The total time traveled was
6
hours.
Dennis’ downhill speed was
10
mph and his uphill speed was
5
mph.
Answer:
6000
Step-by-step explanation:
Answer:
It's easy a point of course !!!!!
First you plot in the y-intercept of the equation. To find the y-intercept, substitute 0 into x. -3m will cancel our giving you y=5. x=0, y=5, the first ordered pair is (0,5). Now after you plot in the y-intercept, use your slope, which is -3, to graph the points of the equation. Starting from (0,5), move down 3 spaces on the y-axis (because it’s -3) and you’ll end up at (0,2). Next move over 1 ( all slopes with just a whole number moves on the x-axis 1 since the whole number divided by 1 doesn’t change the slope number) to the right because it’s a negative linear equation so it’ll go downward. After moving right, you’ll get (1,2). Do a couple more points starting from (1,2) then the 3rd point ABD and so on to get 3 or more points to be able to draw a linear line.
