2/5, 1/8a commondenominator:

Mathematics

1 answer:

Goshia [24]3 years ago

3 0

Answer:

Step-by-step explanation:

5(8)

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5x-y=1/4 intercept form

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5x - y = 1/4; this equation is written in slope - intercept form

Slope - Intercept form formula: y = mx + b

-Move '5x' to the left side of the equation
-Move 'y' to the right side of the equation

y = 5x - 1/4

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Step-by-step explanation:

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aliya0001 [1]

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Step-by-step explanation:

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3 years ago

Find an equation of the tangent line to the bullet-nose curve y=|x|/sqrt(2−x^2) at the point (1,1) I think that square root is w

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$\bf y=\cfrac{|x|}{\sqrt{2-x^2}}\qquad \boxed{|x|=\pm\sqrt{x^2}}\qquad y=\cfrac{\sqrt{x^2}}{\sqrt{2-x^2}}\\\\
-------------------------------\\\\
\cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x\cdot \sqrt{2-x^2}~~-~~\sqrt{x^2}\cdot \frac{1}{2}(2-x^2)^{-\frac{1}{2}}\cdot -2x}{(\sqrt{2-x^2})^2}}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{x\sqrt{2-x^2}}{\sqrt{x^2}}~~+~~\frac{x\sqrt{x^2}}{\sqrt{2-x^2}}}{2-x^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{x(2-x^2)~~+~~x(x^2)}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{2x\underline{-x^3+x^3}}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}$

$\bf \cfrac{dy}{dx}=\cfrac{2x}{\sqrt{x^2}\sqrt{2-x^2}(2-x^2)} \implies \boxed{\cfrac{dy}{dx}=\cfrac{2x}{|x|\sqrt{2-x^2}(2-x^2)}}
\\\\\\
\left. \cfrac{dy}{dx} \right|_{1,1}\implies \cfrac{2(1)}{|1|\cdot \sqrt{2-1^2}(2-1^2)}\implies 2\\\\
-------------------------------\\\\
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=2(x-1)\implies y-1=2x-2
\\\\\\
y=2x-1$

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3 years ago

2/5, 1/8a commondenominator:​

2/5, 1/8a commondenominator: