What are the three theorems we can use

Mathematics

1 answer:

Veronika [31]3 years ago

3 0

Answer:

Hope this will help you....

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Determine the function which corresponds to the given graph<br> The asymptote is x=3

stepladder [879]

The function has a horizontal asymptote at y = 3.
The function has a vertical asymptote at x = -3.
The function has a vertical intercept of 1.
The function has a zero of x = -2.

Because the function has a zero at x = -2, and a vertical asymptote at x = -3, therefore
$y= \frac{k(x+2)}{x+3}$

Write the function in the form
y = $y= \frac{k(1+ \frac{2}{x}) }{1+ \frac{3}{x} }$

As x → ∞, y → k.
Because a horizontal asymptote exists at y = 3, therefore k = 3.
A vertical intercept exists at y = 1.
The function is
$y = \frac{3(x+2)(x+1)}{(x+3)(x+1)}$

Answer: $y = \frac{3(x+2)(x+1)}{(x+3)(x+1)}$

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3 years ago

There are 60 minutes in 1 hour. Use a tape diagram to find the total number of minutes in 6 hours and 15 minutes

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2 years ago

4x − y = -7 can some one help me

Flauer [41]

Answer:

the question is not understandable

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3 years ago

Can someone help me in #66

VladimirAG [237]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)
\\ \quad \\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
$

$\bf log_4\left( \cfrac{\sqrt{x^5y^7}}{zw^4} \right)\implies log_4(\sqrt{x^5y^7})-log_4(zw^4)
\\\\\\
log_4\left[(x^5y^7)^{^\frac{1}{2}}\right]-log_4(zw^4)\implies 
\cfrac{1}{2}log_4\left[(x^5y^7)\right]-log_4(zw^4)
\\\\\\
\cfrac{1}{2}\left[ log_4(x^5)+log_4(y^7) \right] -[log_4(z)+log_4(w^4)]
\\\\\\
\cfrac{1}{2}log_4(x^5)+\cfrac{1}{2}log_4(y^7)-[log_4(z)+log_4(w^4)]$

$\bf \cfrac{1}{2}\cdot 5log_4(x)+\cfrac{1}{2}\cdot 7log_4(y)-log_4(z)-4log_4(w)
\\\\\\
\cfrac{5}{2}log_4(x)+\cfrac{7}{2}log_4(y)-log_4(z)-4log_4(w)$

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3 years ago

That's the question ??

Oksanka [162]

Step-by-step explanation:

Before we answer, the area of a Circle given it radius is

$a = \pi r {}^{2}$