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3 years ago

What's the slope-intercept form that passes through the points (0, -1) and (1, 5)?

Mathematics

1 answer:

Rzqust [24]3 years ago

5 0

Answer:

Step-by-step explanation:

$\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}$

(0, -1) ⇒ x₁ = 0, y₁ = -1

(1, 5) ⇒ x₂ = 1, y₂ = 5

So the slope:

$\bold{m=\dfrac{5+1}{1-0}=\dfrac{6}{1}=6}$

The slope-intercept form of the equation of line is y = mx + b, where m is the slope and b is the y-intercept of the line.

(0, -1) ⇒ x₀ = 0, y₀ = -1 ⇒ b = -1

Therefore:

y = 6x - 1 ← the slope-intercept form of the equation

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3 years ago

Simplify the radical expression 4 square root 81x^20y^8

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The simplified expression is $3x^5 y^2$

Step-by-step explanation:

The radical expression to simplify is

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We notice the following:

$81=3^4$

And using the property

$(x^a)^b=x^{a\cdot b}$

we can write:

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$y^8 = (y^2)^4$

So the argument of the radical can be written as

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And now by using the property

$\sqrt[n]{a^n}=a$

We find:

$\sqrt[4]{(3x^5y^2)^4}=3x^5 y^2$

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4 years ago

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Compute the second partial derivatives ∂2f ∂x2 , ∂2f ∂x ∂y , ∂2f ∂y ∂x , ∂2f ∂y2 for the following function. f(x, y) = 2xy (x2 +

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Answer with step-by-step explanation:

We are given that a function

$f(x,y)=2xy(x^2+y^2)^2$

Differentiate partially w.r.t x

Then, we get

$\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)$

Differentiate again w.r.t x

$\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3$

Differentiate function w.r.t y

$\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y$

$\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)$

Again differentiate w.r.t y

$\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3$

Differentiate partially w.r.t y

$\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4$ $\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta x\delat y}=10x^4+36x^2y^2+10y^4$

Hence, if f(x,y) is of class $C^2$ (is twice continuously differentiable), then the mixed partial derivatives are equal.

i.e $\frac{\delta^2f}{\delta y\delta x}=\frac{\delta^2f}{\delta x\delta y}$

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4 years ago