Is my use of AC method incorrect?

bobisaka

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Dennis went cross-country skiing for 6 hours on Saturday. He skied 20 miles uphill and then 20 miles back downhill, returning to his starting point. His uphill speed was 5 mph slower than his downhill speed. What was Dennis’ speed going uphill and his speed going downhill?

20/r + 20/(r-5) = 6

r(r-5) * 20/r + r(r-5) * 20/(r-5) = r(r-5) * 6

20r-100+20r = 6r^2-30r

0 = 6r^2+10r+100

Confused with AC method:
6 x 100 = 600

30 * -20 = -600 (should it not be a positive, since the trinomial was positive?)

6r^2+30r-20r+100

6r(r+5)-20(r+5)

(6r-20)(r+5) = 0

Check:

6r-20 = 0

6r = 20

r = 20/-6 (doesn't work)

r + 5 = 0

r =-5


I have not completed the question, but will post another thread in regards to the final check.
 
0 = 6r^2+10r+100

Confused with AC method:
6 x 100 = 600

30 * -20 = -600 (should it not be a positive, since the trinomial was positive?)

6r^2+30r-20r+100

6r(r+5)-20(r+5)

(6r-20)(r+5) = 0

Check:

6r-20 = 0

6r = 20

r = 20/-6 (doesn't work)

r + 5 = 0

r =-5


I have not completed the question, but will post another thread in regards to the final check.
Ignoring the fact that you are solving the wrong equation, and looking only at your attempt at factoring:

You're right that you are looking for a pair of numbers whose product is +600, not -600, so 30 and -20 won't work. In fact, nothing would work; the equation you're working on has no real solutions. Go back and fix your sign error in simplifying the equation.
 
Doesn't look right.
Sorry mistake, it is: 6r^2 - 30r - 100

I'm still stuck at trying to factor that part. I have used AC method and Trial and Error method. I have spent a long time trying to solve it.... going to take a break, and come back to it. Will update..

I feel like I am wasting my time, or spending too much time on a particular questions I have not solved.. Progress is slow.
 
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Dennis went cross-country skiing for 6 hours on Saturday. He skied 20 miles uphill and then 20 miles back downhill, returning to his starting point. His uphill speed was 5 mph slower than his downhill speed. What was Dennis’ speed going uphill and his speed going downhill?

20/r + 20/(r-5) = 6

r(r-5) * 20/r + r(r-5) * 20/(r-5) = r(r-5) * 6

20r-100+20r = 6r^2-30r

0 = 6r^2+10r+100

Confused with AC method:
6 x 100 = 600

30 * -20 = -600 (should it not be a positive, since the trinomial was positive?)

6r^2+30r-20r+100

6r(r+5)-20(r+5)

(6r-20)(r+5) = 0

Check:

6r-20 = 0

6r = 20

r = 20/-6 (doesn't work)

r + 5 = 0

r =-5


I have not completed the question, but will post another thread in regards to the final check.
You wrote:

20r-100+20r = 6r^2-30r​
0 = 6r^2+10r+100 ..........................Incorrect

It should be:

0 = 6r^2 - 30r - 20r - 20r + 100

0 = 6r^2 - 70r + 100
 
holy moly! yes, I actually had that in the previous step(on paper), but than the next step i put 30r instead of 70r. I need to improve my attention to detail..

Thanks!
 
Cannot finish it, I have no clue what I am doing. How do i get the speed?

Dennis went cross-country skiing for 6 hours on Saturday. He skied 20 miles uphill and then 20 miles back downhill, returning to his starting point. His uphill speed was 5 mph slower than his downhill speed. What was Dennis’ speed going uphill and his speed going downhill?

Solution:

We are looking for Skiing speed uphill and downhill.
Time = 6 total
Uphill = r -5
Downhill = r

Distance is given.

T = D/R

20/(r-5) + 20/r = 6

r(r-5) * (20/(r-5) +20/r) = r(r-5) * 6

20r + 20r -100 = 6r^2 - 30r

0 = 6r^2 - 70r + 100

AC:
6 x 100 = 600 = 20 * 30

(6r^2 + 30r)+ (20r + 100)

6r(r+5) + 20(r+5)

0 = (6r+20)(r+5)

0 = 6r + 20

-20/6 = r

0 = r + 5

-5 = r
 
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In the meantime I am going to go back to chapter 3.5 Solve Uniform Motion, I will be back.
 
0 = 6r^2 - 70r + 100
AC:
6 x 100 = 600 = 20 * 30
(6r^2 + 30r)+ (20r + 100)
6r(r+5) + 20(r+5)
0 = (6r+20)(r+5)
0 = 6r + 20
-20/6 = r
0 = r + 5
-5 = r
When you factor, always check by multiplying before you move on, to make sure you got it right. Factoring is an error-prone activity.

When you check (6r+20)(r+5), you'll find it expands to 6r^2 + 50r + 100, not 6r^2 - 70r + 100. The two numbers you chose, 20 and 30, have the right product but not the right sum.

So go back to that step and find the right pair of numbers.
 
I have no clue what I am doing
That's not true, bobisaka. Your setup is correct. Your algebra is correct. You're just making simple arithmetic mistakes. Checking each line of work, as you go, will help you to avoid moving on with incorrect values.

AC:
6 x 100 = 600 = 20 * 30
(6r^2 + 30r)+ (20r + 100)
You're trying to factor 6r^2 - 70r + 100 by grouping (the 'AC method').

You chose 30 and 20 and wrote:

6r^2 + 30r + 20r + 100

That's not the same polynomial you're trying to factor. Can you see why?

?
 
That's not true, bobisaka. Your setup is correct. Your algebra is correct. You're just making simple arithmetic mistakes. Checking each line of work, as you go, will help you to avoid moving on with incorrect values.


You're trying to factor 6r^2 - 70r + 100 by grouping (the 'AC method').

You chose 30 and 20 and wrote:

6r^2 + 30r + 20r + 100

That's not the same polynomial you're trying to factor. Can you see why?

?


It should something along the lines of
(6r^2 - 30r) + (20r + 100)

6r(r-5) +20(r+5) but that doesn't or work because the factors are negative and positive, are they not suppose to have the same sign?
 
0 = 6r^2 - 70r + 100

AC:
6 x 100 = 600 = 20 * 30

(6r^2 + 30r)+ (20r + 100)
It seems that you think that you can use any two numbers that multiply out to 600. That is not true. In the first bold equation above you have -70r while in the 2nd equation you have 30r+20r = 50r. -70r and 50r are NOT the same. You have to find the correct factors that multiply out to 600!

You want two numbers that multiply out to 600 but add up to -70.
 
It should something along the lines of
(6r^2 - 30r) + (20r + 100)

6r(r-5) +20(r+5) but that doesn't or work because the factors are negative and positive, are they not suppose to have the same sign?

The problem is before that. You need to find a pair of numbers whose product is 600 and whose sum is -70. Don't move on until you find them. Using 20 and 30, with any signs you like, is a waste of time, because their sum is 50, not 70.

If necessary, start listing every pair of factors of 600 until you find the right sum:
1*600
2*300
3*200
4*150
5*120
...
 
It should something along the lines of …
That's not what I'd asked, bobisaka. I wanted to know whether you understood why the expressions below are not equal (they don't represent the same polynomial).

6r^2 - 70r + 100
6r^2 + 30r + 20r + 100

I'd told you also that your work was correct, except for a sign error.

?
 
-10 * -60 = 600
-10 + -60 =70

6r^2 - 60r - 10r + 100 <--------- which is correct ?

= 6r (r-10) -10 (r - 10) <------- I have been so confused with this one, until i saw that -10 * -10 = +100

= (6r - 10) (r - 10)

(6r -10) -> r = -10/6 (can't use?)

(r-10) -> r = 10

speed uphill = 10 -5 =5mph
speed downhill = 10mph

check.
T= 20mi/5mph = 4hours
T =20mi/10mph = 2 hours
T = 6 hours


Am I all good now?
 
If r=-10/6, that is r<0 what would that mean? You can't show away answers for unknown reasons. Never do anything in mathematics if you don't know why you are doing it. That would be suicidal in my opinion.
 
6r = 10

6r/6 = 10/6

r = 1.666 = 1.67

Uphill 1.67 - 5 = -3.33mph
Downhill = 1.67mph

t = 20/-3.33 = -6.01
t = 20/1.67 =11.98
Total hours = 5.97 = 6 = 6 hours

so it is correct; they are both correct . Now, am i on the right track?

I will keep that in mind, I will do a thorough check of all answers before passing them.
 
6r = 10

6r/6 = 10/6

r = 1.666 = 1.67

Uphill 1.67 - 5 = -3.33mph
Downhill = 1.67mph

t = 20/-3.33 = -6.01
t = 20/1.67 =11.98
Total hours = 5.97 = 6 = 6 hours

so it is correct; they are both correct . Now, am i on the right track?

I will keep that in mind, I will do a thorough check of all answers before passing them.

Interesting. Have you ever gone uphill at -3 1/3 mph? What would that look like?

You fixed the algebra, but not the answer. It looks to me like you're trying to please other people (the answer in the book, or what Jomo says), and not letting yourself think on your own.

Ultimately, this is a matter both of domain issues, and of common sense.
 
Interesting. Have you ever gone uphill at -3 1/3 mph? What would that look like?

You fixed the algebra, but not the answer. It looks to me like you're trying to please other people (the answer in the book, or what Jomo says), and not letting yourself think on your own.

Ultimately, this is a matter both of domain issues, and of common sense.
Nah, i would agree with the latter, i lack the common sense / logic. Which is why I am doing the maths in the first place.

I will come back to it.
 
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