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2 years ago

What is the summary for thisssss? 3/7 + 1/5=

Mathematics

1 answer:

patriot [66]2 years ago

3 0

Answer:

22/35

Step-by-step explanation:

find the common denominator, which is 35.

multiply the numerator and the denominator of 3/7 by 5

- that would equal 15/35

multiply the numerator and the denominator of 1/5 by 7

- that would equal 7/35

now add 15/35 + 7/35 = 22/35

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Show all work to identify the asymptotes and zero of the function f(x) = 6x / x^2 - 36

eduard

Answer:

Zero of the function f(x) is at x = 0

Vertical Asymptotes at x = ±6

Horizontal Asymptotes at y = 0

Step-by-step explanation:

<h3>Vertical Asymptotes </h3>

For a given function f(x):

Vertical Asymptotes are obtained at those values of x, where the function f(x) tends to infinity, I.e.,

When x approaches some constant value but the curve moves towards infinity.

If f(x) is a fraction, it'll tend to infinity when it's denominator becomes zero.

Vertical Asymptotes of the given function can be obtained by walking thru the following steps:

Step I

(Factorise the numerator and denominator)

$\mathsf{ f(x) = \frac{6x}{ {x}^{2} - 36 } }$

x² - 36 can be factorised into (x + 6)(x - 6)

and, ofcourse, we can write 6x as 6(x - 0)

$\mathsf{ f(x) = \frac{6(x - 0)}{ (x + 6)(x - 6) } }$

Step II

(Reduce the fraction to its simplest form by canceling out the common factors)

There aren't any common factors in the numerator and denominator in this case.

Step III

(Look for the values of x which cause the denominator to be zero)

If we put x = 6

denominator becomes 0

Also,

If we substitute x with -6

denominator becomes 0.

The two values of x indicate the two Vertical Asymptotes of the function f(x).

Therefore,

Vertical Asymptotes of the given function f(x) are:

$\boxed{ \mathsf{x = \pm6}}$

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

<h3 /><h3>Horizontal Asymptotes:</h3>

Horizontal Asymptotes are obtained When x tends to infinity and y approaches some constant value.

I'll be using the concept of limits for this.

$\mathsf{y = \frac{6x}{ {x}^{2} - 36 } }$

dividing and multiplying by x² (Yep! so if x becomes infinity 1/ x and 1/ x² all such terms become 0, 'cause 1/ ∞ is 0)

$\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6x}{ {x}^{2} } }{ \frac{ {x}^{2} - 36 }{ {x}^{2} } } ) }$

$\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6}{ x } }{ 1- \frac{36 }{ {x}^{2} } } ) }$

Substitute x with ∞, you get zero/ 1

$\implies \boxed{\mathsf{y = 0}}$

So, the horizontal Asymptote of the function is y = 0, that is the x axis

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

<h3>Zeroes of a function:</h3>

The values of x that reduces f(x) to zero are called the zeroes of f(x).

Here, only x = 0 acts as the zero of the function.

[NOTE:

For finding Vertical Asymptotes,Equate the denominator to 0. And
For finding Zeroes, Equate the numerator to 0]

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