Hey there!
when x = 3 and y = 9
Let's first move to x and y.
Now multiply 5 by x and y(3+9).
Now we have the number 2 left, so we have to multiply 60 by 2.
Hope this helps!
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1 answer:
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When dealing with roots, we want to factor to see if it's possible
now since we're dealing with square roots, we circle every pair of numbers we see in the final factoring
so..
now for every circled pair, we take only 1 of the 2 circled and bring it outside
anything left uncircled stays inside the root
so..
now simply multiply
now since we're dealing with square roots, we circle every pair of numbers we see in the final factoring
so..
now for every circled pair, we take only 1 of the 2 circled and bring it outside
anything left uncircled stays inside the root
so..
now simply multiply
Answers:
- Part a)
- Part b) 3
- Part c) 2.24
- Part d) 1.58
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Work Shown:
Part (a)
The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)
The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3
We apply the distance formula to say...
We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.
Saying is the same as writing d = sqrt(x^2-(x^2-3)^2)
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Part (b)
Plug in x = 0 and you should find the following
This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).
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Part (c)
Repeat for x = 1
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Part (d)
Graph the d(x) function found back in part (a)
Use the minimum function on your graphing calculator to find the lowest point such that x > 0.
See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.
The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)
This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.
