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3 years ago

How do you write 2y=3(-x+5) in standard form ?

Mathematics

1 answer:

Drupady [299]3 years ago

7 0

2y=3(-x+5)

Distributed property

2y= -3x +15

Move x term to left side to get

3x+2y=15

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Two equivalent equations of the line in point slope form for (-4,8) and (0,5)

34kurt

Answer:

Equation of line 1 is 3 X - 4 Y = 20

Equation of line 2 is 3 X + 4 Y = 20

Step-by-step explanation:

Given co ordinates of points as,

( -4 , 8) and (0 , 5)

From the given two points we can determine the slop of a line

I. e slop (m) = $\frac{(y2 - y1)}{(x2 - x1)}$

Or, m = $\frac{(5 - 8)}{(0 + 4)}$

So, m = $\frac{(-3)}{(4)}$

Now equations of line can be written as ,

Y - y1 = m ( X - x1)

At points ( -4 , 8)

Y - 8 = $\frac{(-3)}{(4)}$ (X + 4)

So , Equation of line 1 is 3 X - 4 Y = 20

Again with points ( 0 , 5)

Y - 5 = $\frac{(-3)}{(4)}$ ( X - 0)

So, Equation of line 2 is 3 X + 4 Y = 20

Hence Equation of line 1 is 3 X - 4 Y = 20 and Equation of line 2 is 3 X + 4 Y = 20 Answer

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3 years ago

Think about the SSS Congruence Postulate. Why do you think it is a postulate and not a theorem?

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I don't know I am sorry I could not help U

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3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago