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2 years ago

At what point on the curve y = 19 2x is the tangent line perpendicular to the line 20x 4y = 1?

Mathematics

1 answer:

Gennadij [26K]2 years ago

7 0

The points are; (7/2, -1/2).

<h3>What is the given point?</h3>

Now the tangent line is given as;20x 4y = 1. When we rewrite it in the slope intercept form, we have the equation as; y = 1 - 20x/4 or y = 1/4 - 5x.

Then to obtain the slope of the curve we have; y = 19 - 2x

dy/dx = 2

Using the relation;

m1m2 = -1

m2 = -1/2

Hence;

y = 1/4 - 5(-1/2)

y = 1/4 + 5/2

y = 7/2

Thus the points are; (7/2, -1/2)

Learn more about normal curve:brainly.com/question/10664419

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3 years ago

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4=1 times 4
2 times 2
they don't add to 2
set up equation
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first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
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use quadratic formula which is
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a=1
b=-2
c=4
x= $\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}$
x= $\frac{ 2+/-\sqrt{4-16} }{2}$
x= $\frac{ 2+/-\sqrt{-12} }{2}$
we have $\sqrt{-12}$ and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x= $\frac{ 2+/-2\sqrt{-3} }{2}$
i= $\sqrt{-1}$
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