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Featured 2 months ago

See explanation.

A **vertical asymptote** occurs at

V.A. when

#2x-4=0#

#<=>2x=4#

#<=>x=2#

So the equation for the V.A. is

A **horizontal asymptote** occurs when the degree of the numerator is less than (or equal to) the degree of the denominator. ("Degree" means the highest power of *will be* a horizontal asymptote.

When the degrees are the same (like in this case), the horizontal asymptote is found at **the ratio of the leading coefficients**. Here, that happens to be

*(In the case that the denominator has a higher degree, the asymptote is always #y=0.#)*

The ** #x#-intercept** is found by letting

#0=(x+5)/(2x-4)#

#0=x+5# [multiply both sides by (2x-4) ]

#x="-5"#

So our

Similarly, the ** #y#-intercept** is found by letting

#y=((0)+5)/(2(0)-4)#

#y=5/"-4"=-5/4#

So our

With all this information, we can now draw our hyperbola:

graph{(y-(x+5)/(2x-4))(y-(x-2.0001)/(2x-4))=0 [-8.835, 11.165, -3.91, 6.09]}

Featured 1 month ago

Please read the explanation.

The standard form for a hyperbola of this type is:

where, x and y correspond to any point

Given:

We will be using algebraic steps to put the above equation into the standard form.

Divide both sides by 4 in the form of

Write the 9 as

Group the denominator of the first term into a single fraction:

The foci are located at

To help you sketch the hyperbola, the center is

There is a graph of the original equation:

Featured 1 month ago

#(x^5-15x^4+90x^3-270x^2+405x-243)/(x-3)#

#= x^4-12x^3+54x^2-108x+81#

Here's how I would do it myself.

Write the beginning of the factorisation:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(#

Then examine each power of

The first term we need to add is

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(color(blue)(x^4)#

Examining what we have so far, the

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4color(blue)(-12x^3)#

Then

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+color(blue)(54x^2)#

Then

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+54x^2color(blue)(-108x)#

Then

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+54x^2-108x+color(blue)(81))#

Finally, note that

I have used quite a few words to describe it, but most of the calculations can be done in your head without having to write down more than:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+54x^2-108x+81)#

Hence:

#(x^5-15x^4+90x^3-270x^2+405x-243)/(x-3)#

#= x^4-12x^3+54x^2-108x+81#

Featured 1 month ago

The answer is

We start by calculating the determinant

As

To determine the inverse

Now we calculate the traspose matrix of

To obtain the inverse of

Verification

Featured 6 days ago

Suppose you know that:

#log 2 ~~ 0.30103#

#log 3 ~~ 0.47712#

Then note that:

#43 = 129/3 ~~ 128/3 = 2^7/3#

So

#log 43 ~~ log(2^7/3) = 7 log 2 - log 3 ~~ 7*0.30103-0.47712 = 1.63009#

We know that the error is approximately:

#log (129/128) = log 1.0078125 = (ln 1.0078125) / (ln 10) ~~ 0.0078/2.3 = 0.0034#

So we can confidently give the approximation:

#log 43 ~~ 1.633#

A calculator tells me:

#log 43 ~~ 1.63346845558#

Featured 1 week ago

The general term of a geometric series can be represented by the formula:

#a_n = a r^(n-1)#

where

We find:

#(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a + color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a(1 - r^N)#

Dividing both ends by

#sum_(n=1)^N ar^(n-1) = (a(1 - r^N))/(1-r)#

If

#sum_(n=1)^oo ar^(n-1) = lim_(N->oo) (a(1 - r^N))/(1-r) = a/(1-r)#

In the given example,

#sum_(n=1)^oo 6*(1/2)^(n-1) = 6/(1-1/2) = 12#

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