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Elan Coil [88]
3 years ago
6

How do you do this question?

Mathematics
1 answer:
Kitty [74]3 years ago
4 0

Answer:

C) One vertical but no horizontal

Step-by-step explanation:

Use implicit differentiation to find dy/dx.

xy² = 2 + xy

x (2y dy/dx) + y² = x dy/dx + y

(2xy − x) dy/dx = y − y²

dy/dx = (y − y²) / (2xy − x)

When the tangent lines are horizontal, dy/dx = 0.

0 = (y − y²) / (2xy − x)

0 = y − y²

0 = y (1 − y)

y = 0 or 1

When the tangent lines are vertical, dy/dx is undefined, so the denominator is 0.

0 = 2xy − x

0 = x (2y − 1)

x = 0, y = 1/2

There are two possible points where there can be a horizontal tangent line, and two possible points where there can be a vertical tangent line.  Plug all back into the original equation to see which are actual points.

If y = 0:

x(0)² = 2 + x(0)

0 = 2

No solution.

If y = 1:

x(1)² = 2 + x(1)

x = 2 + x

0 = 2

No solution.

If x = 0:

(0)y² = 2 + (0)y

0 = 2

No solution.

If y = 1/2:

x(1/2)² = 2 + x(1/2)

x/4 = 2 + x/2

-x/4 = 2

x = -8

Therefore, there is one vertical line and no horizontal lines.

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Pedro is building a playground in the shape of a right triangle. He wants to know the area of the playground to help him decide
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Step-by-step explanation:

The right triangle has base as 6 yards and height as 3 1/3 of yards.

When we match this triangle with another right triangle to make  the rectangle is another right triangle with sides 6 yards and 3 1/3 yards.

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3 years ago
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just olya [345]

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

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Explanation:

I hope this helped!

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8 0
2 years ago
Find a function where f(0)=2 and f(1)=2
xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

f(x) = 2 is a valid solution.

Want something more fun? Why not a parabola? f(x)= ax^2+bx+c.

At this point you have three parameters to play with, and from the fact that f(0)=2 we can already fix one of them, in particular c=2. At this point I would recommend picking an easy value for one of the two, let's say a= 1 (or even a=-1, it will just flip everything upside down) and find out b accordingly:f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1

Our function becomes

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Notice that it works even by switching sign in the first two terms: f(x) = -x^2+x+2

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: f(x) = A cos (kx)

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need A= 2, and at that point the first condition is guaranteed; using the second to find k we get 2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi

f(x) = 2cos(2\pi x)

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

f(x)= 2+sin(2\pi x)

Sky is the limit.

8 0
3 years ago
A red light flashes every 14 minutes. A blue light flashes every 24 minutes. when will the two lights flash together again if th
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Answer: 10:48AM 

4 0
3 years ago
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