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2 years ago

What is the slope? Please help

Mathematics

2 answers:

ss7ja [257]2 years ago

8 0

Answer:

-1/2

Step-by-step explanation:

slope = rise/run = (difference in y)/(difference in x)

Look for two points that are easy to read since they fall on grid lines intersections.

For example, pick point (0, 3) as the first point.

Pick point (2, 2) as the second point.

We start at the first point (0, 3) and need to go to point 2 by using only vertical and horizontal moves.

Start at (2, 2). Go down 1 unit. That is a rise or a change in y of -1.

Now go right 2 units. That is a run or a change in x of 2.

slope = rise/run

slope = -1/2

Answer: -1/2

tresset_1 [31]2 years ago

7 0

Answer:

Slope is -1/2

step by step explanation

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If 2y^2+2=x^2, then find d^2y/dx^2 at the point (-2, -1) in simplest form.

horrorfan [7]

Answer:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}$

Step-by-step explanation:

We have the equation:

$2y^2+2=x^2$

And we want to find d²y/dx² at the point (-2, -1).

So, let's take the derivative of both sides with respect to x:

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

On the left, let's implicitly differentiate:

$4y\frac{dy}{dx}=\frac{d}{dx}[x^2]$

Differentiate normally on the left:

$4y\frac{dy}{dx}=2x$

Solve for the first derivative. Divide both sides by 4y:

$\frac{dy}{dx}=\frac{x}{2y}$

Now, let's take the derivative of both sides again:

$\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{x}{2y}]$

We will need to use the quotient rule:

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

So:

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

Differentiate:

$\frac{d^2y}{dx^2}=\frac{(1)(2y)-x(2\frac{dy}{dx})}{4y^2}$

Simplify:

$\frac{d^2y}{dx^2}=\frac{2y-2x\frac{dy}{dx}}{4y^2}$

Substitute x/2y for dy/dx. This yields:

$\frac{d^2y}{dx^2}=\frac{2y-2x\frac{x}{2y}}{4y^2}$

Simplify:

$\frac{d^2y}{dx^2}=\frac{2y-\frac{2x^2}{2y}}{4y^2}$

Simplify. Multiply both the numerator and denominator by 2y. So:

$\frac{d^2y}{dx^2}=\frac{4y^2-2x^2}{8y^3}$

Reduce. Therefore, our second derivative is:

$\frac{d^2y}{dx^2}=\frac{2y^2-x^2}{4y^3}$

We want to find the second derivative at the point (-2, -1).

So, let's substitute -2 for x and -1 for y. This yields:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(-1)^2-(-2)^2}{4(-1)^3}$

Evaluate:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(1)-(4)}{4(-1)}$

Multiply:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2-4}{-4}$

Subtract:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{-2}{-4}$

Reduce. So, our answer is:

$\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}$

And we're done!

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4 years ago

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qaws [65]

Step-by-step explanation:

One sec I'll answer this right now

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5 0

3 years ago

Solve the equation.

-BARSIC- [3]

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3 years ago

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torisob [31]

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4 years ago

Read 2 more answers

A car travels 2 1/3 miles in 3 1/2 minutes at a constant speed. Write an equation to represent the car travels in miles and minu

Flura [38]

Answer:

The speed of car in miles per minute is $\frac{2}{3}$ miles per minute and

The speed of car in mile per hour is 40 miles per hour,

Step-by-step explanation:

Given as :

The distance cover by car = 2 $\frac{1}{3}$ = $\frac{7}{3}$ miles

The Time taken by car to cover distance = 3 $\frac{1}{2}$ min = $\frac{7}{2}$ min

The speed o car in miles per minute = S miles/min

So, Speed = $\dfrac{\textrm Distance}{\textrm Time}$

Or, S = $\frac{\frac{7}{3}}{\frac{7}{2}}$ mie per min

∴ S = $\frac{2}{3}$ miles per minute

Now, Let The speed of car in miles per hour = x miles/h

So, Time in hour = $\frac{7}{2}$ × $\frac{1}{60}$ = $\frac{7}{120}$ hours

So , Speed = $\dfrac{\textrm Distance}{\textrm Time}$

Or, x = $\frac{\frac{7}{3}}{\frac{7}{120}}$ miles per hours

Or, x = 40 miles per hours

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4 years ago