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3 years ago

What is the slope of the line that passes through the points (1,1) and (-1,5)

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

7 0

(1,1)(-1,5)

your answer would be m=-2

hope this helped.

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K = {(x, y) | x - y = 5}, and the domain of x = 0, 2, 4

abruzzese [7]

Answer:

The range = {-5 , -3 , -1}

Step-by-step explanation:

Verified

4 0

3 years ago

Use implicit differentiation to find the slope of the tangent line at the given point:

Salsk061 [2.6K]

Answer:

$\frac{dy}{dx}=0$

Step-by-step explanation:

So we have the equation:

$(x^2+y^2)^2=4x^2y$

And we want to find the slope of the tangent line at the point (1,1).

So, let's implicitly differentiate. Take the derivative of both sides:

$\frac{d}{dx}[(x^2+y^2)^2]=\frac{d}{dx}[4x^2y]$

Let's do each side individually.

Left:

We can use the chain rule:

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

Let's let v(x) be x²+y². So, u(x) is x². Thus, the u'(x) is 2x. Therefore:

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

We can differentiate x like normal. However, for y, we must differentiate implicitly. pretend y is y(x). This gives us:

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

Differentiate:

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

Therefore, our left side is:

$2(x^2+y^2)(2x+2y\frac{dy}{dx})$

Right:

We have:

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

Let's move the 4 outside:

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

Use the product rule:

$=4(\frac{d}{dx}[x^2]y+x^2\frac{d}{dx}[y])$

Differentiate:

$=4(2xy+x^2\frac{dy}{dx})$

Therefore, our entire equation is:

$2(x^2+y^2)(2x+2y\frac{dy}{dx})=4(2xy+x^2\frac{dy}{dx})$

So, to find the derivative at (1,1), substitute 1 for x and 1 for y.

$2((1)^2+(1)^2)(2(1)+2(1)\frac{dy}{dx})=4(2(1)(1)+(1)^2\frac{dy}{dx})$

Evaluate.

$2((1)+(1))(2+2\frac{dy}{dx})=4(2+\frac{dy}{dx})$

Simplify. Also, let's distribute the right:

$2(2)(2+2\frac{dy}{dx})=8+4\frac{dy}{dx}$

Multiply.

$4(2+2\frac{dy}{dx})=8+4\frac{dy}{dx}$

Distribute the left:

$8+8\frac{dy}{dx}=8+4\frac{dy}{dx}$

Subtract 8 from both sides:

$8\frac{dy}{dx}=4\frac{dy}{dx}$

Subtract 4(dy/dx) from both sides:

$4\frac{dy}{dx}=0$

Divide both sides by 4:

$\frac{dy}{dx}=0$

Therefore, the slope at the point (1,1) is 0.

And we're done!

We can verify this using the graph. The slope of the line tangent to the point (1,1) seems like it would be horizontal, giving us a slope of 0.

Edit: Typo

5 0

3 years ago

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Solve the system of equations.<br> y-10x=5<br> y=x^2+7x+5

vesna_86 [32]

Answer:

x = 0,3

Step-by-step explanation:

y-10x=5 move -10x to the other side to get y by itself

y=10x+5 then you would substitute this equation into the other equation where you see y

10x+5=x^2+7x+5 move everything to one side

0=x^2+7x+5-10x-5 combine like terms

0=x^2-3x factor

0 = x(x-3) set each equal to zero

x=0

x-3 = 0 add 3 to both sides

x = 3

the answer is x=0 and x=3

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3 years ago

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25 fewer than twice a number?

julia-pushkina [17]

Answer:

2x - 25

Step-by-step explanation:

Since we don't know what the number is, just call it x. Twice x can be written as 2x and 25 fewer than 2x is 2x - 25.

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4 years ago

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Help me guys it's important

motikmotik

Answer:

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3 years ago

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