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3 years ago

Question 3 Find the equation of the line joining (-2,4) and (3,7)

Mathematics

1 answer:

otez555 [7]3 years ago

7 0

Answer:

y = (3/5)x+26/5 or 5y = 3x+26

Step-by-step explanation:

Applying,

The equation of a line in two point form

(y₂-y₁)/(x₂-x₁) = (y-y₁)/(x-x₁)............... Equation 1

From the question,

Given: y₂ = 7, y₁ = 4, x₂ = 3, x₁ = -2

Substitute these values into equation 1

(7-4)/[3-(-2)] = (y-4)/(x+2)

3/5 = (y-4)/(x+2)

5(y-4) = 3(x+2)

5y-20 = 3x+6

5y = 3x+6+20

5y = 3x+26

y = (3/5)x+26/5

Hence the equation of the line is y = (3/5)x+26/5 or 5y = 3x+26

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8 mins 56 secs

Step-by-step explanation:

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3 years ago

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3 years ago

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3 years ago

Which graph represents the solution set to the system of inequalities? { y<-1/2+2 y>-3/2x+2

Sidana [21]

The graph of the given system of linear inequalities

y < -1/2 + 2

y > -3/2x + 2

is attached below

<h3>Graph of system of linear inequalities </h3>

From the given information, we are to graph the given system of linear inequalities

The given system of linear inequalities is

y < -1/2 + 2

y > -3/2x + 2

The graph of the given system of linear inequalities

y < -1/2 + 2

y > -3/2x + 2

is shown below

Learn more on Graph of system of linear inequalities here: brainly.com/question/26965469

#SPJ1

5 0

2 years ago

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

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, 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