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3 years ago

Indicate the equation of the given line in standard form, writing the answer in the equation box below.

Mathematics

1 answer:

yawa3891 [41]3 years ago

5 0

9514 1404 393

Answer:

x -3y = 4

Step-by-step explanation:

A plot of the points reveals the longer diagonal to be BD. (Also, point C is on that line.)

We can write the standard form equation of the line through (x1, y1) and (x2, y2) as ...

(y2 -y1)x -(x2 -x1)y = (y2 -y1)x1 -(x2 -x1)y1

For points (-2, -2) and (16, 4), this equation is ...

(4 -(-2))x -(16 -(-2))y = (4 -(-2))(-2) -(16 -(-2))(-2)

6x -18y = -12 +36

Dividing by the greatest common factor (6), we have the standard-form equation ...

x -3y = 4

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What is an equation of the line in point-slope form? How can the point-slope form be written in function notation?

harina [27]

The equation of a line in point slope form = $y - y_1 = m(x - x_1)$

To put $y - y_1 = m(x - x_1)$ into function notation, we need to get y by its self.
Lets say we had a point (2,2) and a slope of 3. Ok, now that we have our point and slope, lets turn point slope form into function notation.

$y - y_1 = m(x - x_1)$
$y - 2 = 3(x - 2)$
$y - 2 = 3x - 6$
$y = 3x - 6 + 2$
$y = 3x - 8$
Change the y to f(x)
$f(x) = 3x - 8$ <-----Function notation.

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3 years ago

Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

$=\sec^2(x-y)-y'\sec^2(x-y)$

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

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3 years ago

A marching band sold 350 tins of popcorn for $8 each. Each tin cost the

photoshop1234 [79]

Answer:

The band made $1,400

Step-by-step explanation:

8-4=4

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2 years ago

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Answer:

Step-by-step explanation:

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2 years ago

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alexgriva [62]

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