<span>y = sqrt(25-x^2) at point (3,4)
The derivative gives us the slope at 3 to be:
-2x
------------ at x=3: -3/4
2sqrt(25-x^2)
</span><span>so we have a vector that is parallel to the slope of the tangent line is: <4,-3>
</span>
<span>the mag = 5 so; unit tangent = <4/5 , -3/5>
</span>
<span>since perpendicular lines have a -1 product between slopes we get the normal to be...
<3/5,4/5>
</span>
<span>It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector.</span>
Answer:
down on left, down on left
Step-by-step explanation:
<em>Look at the first term or the one with the highest degree.</em>
Even Degree and positive coefficient: <em>*</em>up left* and *up right*
Even degree and negative coefficient: *down left* and *down right*
Odd Degree and positive coefficient: *down left* and *up right*
Odd Degree and negative coefficient: *up left* and *down right*
Answer:
yes pick .98 l
Step-by-step explanation: .
AnswerI think 11
Step-by-step explanation:48+24=66 and that divided by 6 is 11 sorry if it’s wrong I think we start form 48+24 though. :)
Answer:
Solutions are (6,3) and (-2,3).
Step-by-step explanation:
From the first equation given we can write:
substituting for 
in the second equation given we get
∴ y=3
Putting y=3 in the first equation we get
x-2 = ±4
Hence x=6 or x=-2.
