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svlad2 [7]
2 years ago
15

What is 73 divided by 8 equal in remainder???

Mathematics
2 answers:
galina1969 [7]2 years ago
7 0

8 \sqrt{73}

the quotient is 9remainder 1

9r.1

checking :

9×8=72+1=73

hope this help

olasank [31]2 years ago
4 0

Answer:

9 remainder 1

Step-by-step explanation:

Not sure but uh... 73/8=9 remainder 1.

Is this what you are looking for?

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3 years ago
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.,

<em>When</em><em> </em><em>x</em><em> </em><em>approaches</em><em> </em><em>some</em><em> </em><em>constant</em><em> </em><em>value</em><em> </em><em>b</em><em>u</em><em>t</em><em> </em><em>th</em><em>e</em><em> </em><em>curve</em><em> </em><em>moves</em><em> </em><em>towards</em><em> </em><em>infinity</em><em>.</em><em> </em>

  • 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:

<u>Step I</u>

(Factorise the numerator and denominator)

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

<em>x</em><em>²</em><em> </em><em>-</em><em> </em><em>36</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>facto</em><em>rised</em><em> </em><em>into</em><em> </em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>6</em><em>)</em><em>(</em><em>x</em><em> </em><em>-</em><em> </em><em>6</em><em>)</em>

<em>and</em><em>,</em><em> </em><em>ofcourse</em><em>,</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>write</em><em> </em><em>6</em><em>x</em><em> </em><em>as</em><em> </em><em>6</em><em>(</em><em>x</em><em> </em><em>-</em><em> </em><em>0</em><em>)</em><em> </em>

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

<u>Step</u><u> </u><u>II</u>

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

<em>There aren't any common factors in the numerator and denominator in this case.</em>

<u>Step</u><u> </u><u>III</u>

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

<em>If</em><em> </em><em>we</em><em> </em><em>put</em><em> </em>x = 6

<em>denominator</em><em> </em><em>becomes</em><em> </em><em>0</em>

Also,

<em>If</em><em> </em><em>we</em><em> </em><em>substitute</em><em> </em><em>x</em><em> </em><em>with</em><em> </em> -6

<em>denominator</em><em> </em><em>becomes</em><em> </em><em>0</em><em>.</em><em> </em>

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

Therefore,

<u>Vertical</u><u> </u><u>Asymptotes</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>function</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u><u>are</u><u>:</u>

\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 }  }

<em>dividing</em><em> </em><em>and</em><em> </em><em>multiplying</em><em> </em><em>by</em><em> </em><em>x</em><em>²</em><em> </em><em>(</em><em>Yep</em><em>!</em><em> </em><em>so</em><em> </em><em>if</em><em> </em><em>x</em><em> </em><em>becomes</em><em> </em><em>infinity</em><em> </em><em>1</em><em>/</em><em> </em><em>x</em><em> </em><em>and</em><em> </em><em>1</em><em>/</em><em> </em><em>x</em><em>²</em><em> </em><em>all</em><em> </em><em>such</em><em> </em><em>terms</em><em> </em><em>become</em><em> </em><em>0</em><em>,</em><em> </em><em>'</em><em>cause</em><em> </em><em>1</em><em>/</em><em> </em><em>∞</em><em> </em><em>is</em><em> </em><em>0</em><em>)</em><em> </em>

\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 <u>Vertical Asymptotes</u><u>,</u>Equate the denominator to 0. And
  • For finding <u>Zeroes</u><u>,</u> Equate the numerator to 0]

__________________

[That's what it's graph looks like. ]

3 0
3 years ago
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igomit [66]
You might have to check it but the answer might be 1776
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S_A_V [24]

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9+x

Step-by-step explanation:

The area of a rectangle is the width multiplied by the length. Working backwards, then, you can find the width or the length by dividing the area by the other one. In this case, the length of the rectangle is the area, or 81-x^2, divided by the width, 9-x. 81-x^2 factored out is (9-x)(9+x), so dividing this by 9-x yields a length of 9+x. Hope this helps!

8 0
3 years ago
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elena-14-01-66 [18.8K]
12*13=156 hope this helped 
8 0
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