Find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)

we just need the slope

take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1

the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation

solve for where y=x-1 and y=x-x² intersect

set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero

x-1=0
x=1
we got this point already

x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2

it intersects at (-1,-2)

7 0

3 years ago

Solve for x and y:<br> 3/x -1/y =13/10<br> 1/x + 2/y =9/10

harkovskaia [24]

Answer:

(x, y) = (2, 5)

Step-by-step explanation:

I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...

3x' -y' = 13/10

x' +2y' = 9/10

Adding twice the first equation to the second, we get ...

2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)

7x' = 35/10 . . . . . . simplify

x' = 5/10 = 1/2 . . . . divide by 7

Using the first equation to find y', we have ...

y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5

So, the solution is ...

x = 1/x' = 1/(1/2) = 2

y = 1/y' = 1/(1/5) = 5

(x, y) = (2, 5)

_____

The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.

3 0

3 years ago

Kenneth bought a new roll of tape. There were 59.6 yards of tape on the roll. Then Kenneth used 4 yards of the tape to make a co

barxatty [35]

Answer:

55.6 yards

Step-by-step explanation:

8 0

3 years ago

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