Answer the question no.10
1 answer:
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In this case, you must synthetic division. You are given x-2. You need to set that to zero so that you now have x-2=0. You then need to get x by itself. So:
.
You then need to use that final part as the beginning of your synthetic division.
To continue, you must find the leading coefficients of every degree (remember: your degree is found by your largest exponent) beginning with your highest degree, in this case it is 3, and use the front number of the term. BE SURE TO BE AWARE OF NEGATIVES (indicated by subtraction signs)!!! Then you continue down the equation from
down to
then 
then 
(which is just the number without an "x") doing the same thing. So your setup should look similar to as follows:
2 l -3 -2 -1 -2
l__________
The numbers inside the box should be added down (the first number is added to nothing, so it is added to zero. Then it is multiplied by the outside number (2). Then that multiplied number continues into the next slot within the box which is then added to the number above it, and so the process repeats. So the process should go like this.
2 l -3 -2 -1 -2
<u>l </u><u>0
</u> -3
2 l -3 -2 -1 -2
<u /> <u>l </u><u> 0 -6 </u>
-3 -8
2 l -3 -2 -1 -2
<u>l </u><u>0 -6 -16
</u> -3 -8 -17
2 l -3 -2 -1 -2
<u>l </u><u>0 -6 -16 -34
</u> -3 -8 -17 -36
So now you are left with -3, -8, -17, and -36. This is to be expected. Now, those new numbers are your new leading coefficients. Remember your first degree? It was 3, now, you need to take your leading degree and subtract it by one. That is your new degree. So now, going left to right on the synthetic division chart, you add "x" and your degree descending as you go further to the right. So, your new equation is
.
There is a problem though, because there is still that -36. This is where that original piece that you divided the equation by comes in handy. You divided everything by x-2, so, you must add your remainder divided by x-2 to the rest of the equation.
So, the final equation is equal to:
That is your final answer.
You then need to use that final part as the beginning of your synthetic division.
To continue, you must find the leading coefficients of every degree (remember: your degree is found by your largest exponent) beginning with your highest degree, in this case it is 3, and use the front number of the term. BE SURE TO BE AWARE OF NEGATIVES (indicated by subtraction signs)!!! Then you continue down the equation from
2 l -3 -2 -1 -2
l__________
The numbers inside the box should be added down (the first number is added to nothing, so it is added to zero. Then it is multiplied by the outside number (2). Then that multiplied number continues into the next slot within the box which is then added to the number above it, and so the process repeats. So the process should go like this.
2 l -3 -2 -1 -2
<u>l </u><u>0
</u> -3
2 l -3 -2 -1 -2
<u /> <u>l </u><u> 0 -6 </u>
-3 -8
2 l -3 -2 -1 -2
<u>l </u><u>0 -6 -16
</u> -3 -8 -17
2 l -3 -2 -1 -2
<u>l </u><u>0 -6 -16 -34
</u> -3 -8 -17 -36
So now you are left with -3, -8, -17, and -36. This is to be expected. Now, those new numbers are your new leading coefficients. Remember your first degree? It was 3, now, you need to take your leading degree and subtract it by one. That is your new degree. So now, going left to right on the synthetic division chart, you add "x" and your degree descending as you go further to the right. So, your new equation is
There is a problem though, because there is still that -36. This is where that original piece that you divided the equation by comes in handy. You divided everything by x-2, so, you must add your remainder divided by x-2 to the rest of the equation.
So, the final equation is equal to:
That is your final answer.
Hypotenuse = guy wire
cos 45 = adj / hyp
cos 45 = 10 / hyp
hypotenuse = 10 / 0.42262
hypotenuse = <span> <span> <span> 23.6619185084 </span> </span> </span>
cos 45 = adj / hyp
cos 45 = 10 / hyp
hypotenuse = 10 / 0.42262
hypotenuse = <span> <span> <span> 23.6619185084 </span> </span> </span>
Answer:
- f'(1) exists: f'(1) = -2
- f'(0) DNE
Step-by-step explanation:
<h3>a)</h3>The function is continuous for x > 0 . The derivative is defined on that interval and is equal to ...
f'(x) = -2x . . . . . for x > 0
Then at x = 1, the derivative is ...
f'(1) = -2(1)
f'(1) = -2
__
<h3>b)</h3>The function has a jump discontinuity at x=0, so the derivative does not exist at that point. A condition for the existence of the derivative is that the function is continuous at the point of interest.
