When x=-1:

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.
From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4
and solve for x'.
Hope that helps!
Answer:
The answer is C.)
Step-by-step explanation:
I hope its help Please let me know.
Answer:
go on the y-axis and find line 8. draw a line from one side of the graph to the other going through 8. then shade everything BELOW the line. #markasbrainliest
Answer:
You can tell whether or not something is positive or negative by looking at the specific number for example if you look at this number -5 you know its a negative number because it has a - sign.Although when its positive it does not have any kind of format of signs according to my calculations.
Hope this helps!
Please give me 5 stars if it did i would very much appreciate it !!
Answer:
or 6.32
Step-by-step explanation:
We can find the distance between two points by using this formula

where the x and y values are derived from the given points
The points given are ( -2 , 5 ) and ( 4 , 3 )
So we plug in the x and y values of those coordinates into the distance formula

or 6.32