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3 years ago

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Mathematics

1 answer:

Monica [59]3 years ago

3 0

Answer:

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Step-by-step explanation:

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19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$

4 0

2 years ago

51 divided by 2,306.2

AlladinOne [14]

Answer:

45.21960784 or 45.2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the vertices and foci of the hyperbola with equation (x-5)^2/81 -(y-1)^2/144 = 1

vesna_86 [32]

Given the equation of an hyperbola in the form:
$\frac{(x-a)^2}{b^2}- \frac{(y-c)^2}{d^2}=1$
Then the focus of the hyperbola is $(\pm\sqrt {b^2+d^2},0)$
Compute like this:
$\sqrt{81+144}=15$
So the two focus points are $(\pm15,0)$
The vertices are given by the formula: $(\pm a,0)$
We deduce the two vertices of the given hyperbola: $(\pm9,0)$
since $\sqrt{81}=9$