It takes Dwight 1 1/3 hours to run the sunshine trail. Mike 3 1/5 hours to walk the same trail. How many times as long does it t
1 answer:
For this case we convert the mixed numbers to fractions:
Dwight:
Mike:
It is observed, that in fact, Mike takes more time to travel the road.
We subtract to know how much more time it takes Mike:
So, Mike takes hours more than Dwight to walk the road.
Answer:
Mike takeshours longer than Dwight to walk the road.
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Please refer to my image where it shows my work as I’m explaining.
Okay, for system 1:
1. I am using the elimination method to solve. So I check if all the terms are lined up and if any are the same. I found that 2X are common in both equations.
2. The goal is to “eliminate” the term hence the name. So I can choose to add or subtract. I chose subtraction because 2 - 2 equals 0 which is our goal. Solve for the rest of the terms. This will lead to getting y =4. Refer to image for the work.
3. Last step to to find the X value. We do this by picking any of the given equations,then substitute y with 4 and solve to eventually get x = 10. Refer to image for the work.
FOR SYSTEM 2:
1. Again, I am using the elimination method to solve. I noticed that NONE of the terms are in common so I will have to intervene. You can chose any term to create a match with but I chose Y since it was the one I could use the smallest number to multiply with. When multiplying, DONT just multiply Y, multiply ALL the terms in the equation or else everything will crash.
2. Now that I have terms in common I can choose to add or subtract. I chose subtraction because 2-2 equals zero which is what we want. Solve look at image for my process which lead to X = -8
3. Last step is to find the value of Y. Chose any of the given equations in system 2 then substitute x with -8. Refer to image to see process. It lead to y = 20
To check the validity of the answers, substitute the x and y values into both equations both side of the equal side should have the same number. Hope that helped!
Okay, for system 1:
1. I am using the elimination method to solve. So I check if all the terms are lined up and if any are the same. I found that 2X are common in both equations.
2. The goal is to “eliminate” the term hence the name. So I can choose to add or subtract. I chose subtraction because 2 - 2 equals 0 which is our goal. Solve for the rest of the terms. This will lead to getting y =4. Refer to image for the work.
3. Last step to to find the X value. We do this by picking any of the given equations,then substitute y with 4 and solve to eventually get x = 10. Refer to image for the work.
FOR SYSTEM 2:
1. Again, I am using the elimination method to solve. I noticed that NONE of the terms are in common so I will have to intervene. You can chose any term to create a match with but I chose Y since it was the one I could use the smallest number to multiply with. When multiplying, DONT just multiply Y, multiply ALL the terms in the equation or else everything will crash.
2. Now that I have terms in common I can choose to add or subtract. I chose subtraction because 2-2 equals zero which is what we want. Solve look at image for my process which lead to X = -8
3. Last step is to find the value of Y. Chose any of the given equations in system 2 then substitute x with -8. Refer to image to see process. It lead to y = 20
To check the validity of the answers, substitute the x and y values into both equations both side of the equal side should have the same number. Hope that helped!
There's no if about it,
has a zero
so 
is a factor. That's the special case of the Remainder Theorem; since we'll get a remainder of zero when we divide 
by 
At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover
too, so 
is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.
So far we have
where
is the zero we haven't guessed yet. Again we could divide by 
but just looking at the constant term we must have
so
We check
We usually talk about the zeros of a function and the roots of an equation; here we have a function whose zeros are
has a zero
At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover
So far we have
where
so
We check
We usually talk about the zeros of a function and the roots of an equation; here we have a function