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3 years ago

‼️‼️Help anyone ‼️‼️‼️

Mathematics

1 answer:

Debora [2.8K]3 years ago

4 0

Method 1
You can use The equilateral triangle (look at the picture)
$\dfrac{a\sqrt3}{2}=1\ \ \ \cdot2\\\\a\sqrt3=2\ \ \ |\cdot\sqrt3\\\\3a=2\sqrt3\ \ \ |:3\\\\a=\dfrac{2\sqrt3}{3}\\\\\boxed{|AC|=\dfrac{2\sqrt3}{3}}$

Method 2
You can use the trigonometric function
$\sin60^o=\dfrac{1}{|AC|}\\\\\sin60^o=\dfrac{\sqrt3}{2}\\\\\dfrac{1}{|AC|}=\dfrac{\sqrt3}{2}\ \ \ |\cross\ multiply\\\\|AC|\sqrt3=2\ \ \ |\cdot\sqrt3\\\\3|AC|=2\sqrt3\ \ \ \ |:3\\\\\boxed{|AC|=\dfrac{2\sqrt3}{3}}$

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A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?

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Step-by-step explanation:

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Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

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Next, you use the general equation of line:

$y-y_o=m(x-x_o)$

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$y-0.721=(-1)(x-0.235)\\\\y=-x+0.956$

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

$0=-x+0.956\\\\x=0.956$

hence, the x-intercept of the tangent line is 0.956

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