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4 years ago

Find an equation of a line through √3,1 / 2) parallel to the line:

Mathematics

1 answer:

vagabundo [1.1K]4 years ago

8 0

Answer:

(a) $x=\sqrt{3}$

(b) $y=\frac{1}{2}$

(c) Answer of part a is equation of line $x=\sqrt{3}$ which is parallel to y axis and in part by equation of line is $y=\frac{1}{2}$ which is parallel to x axis

Step-by-step explanation:

We have given points are $(\sqrt{3},\frac{1}{2})$

(a) Equation of line is given x = -9

We have to find the equation of line parallel to x= -9 and passing through $(\sqrt{3},\frac{1}{2})$

x = -9 line is parallel to y axis and so slope will be infinite

Equation of line is given by $y-y_1=m(x-x_1)$

So $y-\frac{1}{2}=\frac{1}{0}(x-\sqrt{3})$

$x=\sqrt{3}$

(b) Equation of line is given $y=-\sqrt{7}$

This line is parallel to x axis so slope will be zero

So equation of line will be $y-\frac{1}{2}=0(x-\sqrt{3})$

$y=\frac{1}{2}$

(c) Answer of part a is equation of line $x=\sqrt{3}$ which is parallel to y axis and in part by equation of line is $y=\frac{1}{2}$ which is parallel to x axis

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Angles ECD and CEF add to 180

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So that means we have EF parallel to CD (due to the same side interior angle theorem)

--------------

Angles BCE and ECD combine to 30+40 = 70, which is congruent to angle ABC = 70 as well.

In other words, this shows angle ABC = angle BCD. Both of these angles are alternate interior angles. Since they're congruent, they lead to AB being parallel to CD.

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So far we have

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It's like saying "P --> Q and Q --> R, therefore P --> R"

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3 years ago

Find a power series for the function, centered at c. g(x) = 4x x2 2x − 3 , c = 0

BartSMP [9]

The power series for given function $g(x)=\frac{4x}{(x-1)(x+3)}$ is $g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$

For given question,

We have been given a function g(x) = 4x / (x² + 2x - 3)

We need to find a power series for the function, centered at c, for c = 0.

First we factorize the denominator of function g(x), we have:

$\Rightarrow g(x)=\frac{4x}{(x-1)(x+3)}$

We can write g(x) as,

$\Rightarrow g(x)=\frac{1}{x-1}+\frac{3}{x+3}\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1+\frac{x}{3} }\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1-(-\frac{x}{3} )}\\$

We know that, $\frac{1}{1-x}=\sum{_{n=0}^\infty}~{x^n}$ if |x| < 1

and $\frac{1}{1-(-\frac{x}{3} )}=\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n$ if $|\frac{x}{6}| < 1$

$\Rightarrow g(x)=-\sum{_{n=0}^\infty}~x^n+\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n\\$ if |x| < 1 and if $|\frac{x}{6}| < 1$

$\Rightarrow g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$ if |x| < 1

Therefore, the power series for given function $g(x)=\frac{4x}{(x-1)(x+3)}$ is $g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$

Learn more about the power series here:

brainly.com/question/11606956

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3 years ago

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