Answer:
A‘‘ (1,6)
Step-by-step explanation:
after a reflection about a horizontal line the x coordinates remain the same
the middle point of the segment AA’ lies on the line so it has y = -1
so we have
(y+y’)/2 = -1
y’ = -2 -y
y‘ = -2 -2 = -4
so A‘ (1,-4)
we can use the same reasoning to find the reflection about y = 1
x‘ = x
(y + y’)/2 = 1
y‘ = 2+4
y’ = 6
A‘‘(1,6)
Answer:
No, it will take Lin 82.25 min to finish her book
40 pages take 70 min which is over an hour so she can finish in a hour. She needs an additional 22.25 mins
Step-by-step explanation:
Answer:
1:18
Step-by-step explanation
18*2=36
18*3=54
so the propotion is 1 to 18, for every 1 hour 18 dollars are added
<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>
