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Answer:

See explanation.

Explanation:

A vertical asymptote occurs at #x#-values that make the denominator 0. To find the vertical asymptote(s) (V.A.'s), set your denominator equal to zero and solve for #x:#

V.A. when #2x-4=0#
#<=>2x=4#
#<=>x=2#

So the equation for the V.A. is #x=2.#

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 #x.#) Since both sides of the fraction have a degree of 1, there will be a horizontal asymptote.

When the degrees are the same (like in this case), the horizontal asymptote is found at #y=#the ratio of the leading coefficients. Here, that happens to be #y=1/2# (from #(color(red)1x+5)/(color(red)2x-4)#).

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

The #x#-intercept is found by letting #y=0# and solving for #x:#

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

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

#x="-5"#

So our #x#-intercept is at #("-5",0).#

Similarly, the #y#-intercept is found by letting #x=0# and solving for #y:#

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

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

So our #y#-intercept is at #(0,-5/4).#

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]}

Answer:

Please read the explanation.

Explanation:

The standard form for a hyperbola of this type is:

#(y - k)^2/a^2 - (x - h)^2/b^2 = 1#

where, x and y correspond to any point #(x,y)# on the hyperbola, h and k correspond to the center point, #(h,k)#, "a" is the vertical distance from the center to the vertices, and #sqrt(a^2 + b^2)# is the vertical distance from the center to the foci.

Given: #9y^2 - x^2 = 4#

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

Divide both sides by 4 in the form of #2^2#:

#9y^2/2^2 - x^2/2^2 = 1#

Write the 9 as #3^2# and the denominator of the divisor:

#y^2/(2^2/3^2) - x^2/2^2 = 1#

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

#y^2/((2/3)^2) - x^2/2^2 = 1#

#sqrt(a^2 + b^2) = sqrt((2/3)^2 + 2^2) = (2sqrt(10))/3#

The foci are located at #(0, -(2sqrt(10))/3) and (0, (2sqrt(10))/3)#

To help you sketch the hyperbola, the center is #(0,0)# and the vertices are at #(0,-2/3) and (0,2/3)#

There is a graph of the original equation:

Desmos.com

Answer:

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

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

Explanation:

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 #x# in turn in descending order to work out what the next term needs to be.

The first term we need to add is #x^4# in order to get #x^5# when multiplied:

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

Examining what we have so far, the #(-3)*x^4# will result in #-3x^4#, when what we want for the next product is #-15x^4#. So we want an extra #-12x^4#. We can get this by writing #-12x^3# as our next term:

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

Then #(-3)(-12x^3) = 36x^3#, but we want #90x^3#, so we need an extra #54x^3#. We can get this by making our next term #54x^2#:

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

Then #(-3)(54x^2) = -162x^2#, but we want #-270x^2#, so we need an extra #-108x^2#. We can get this by making our next term #-108x#:

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

Then #(-3)(-108x) = 324x#, but we want #405x#, so we need an extra #81x#. We can get this by making our next term #81#:

#(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 #(-3)(81) = -243#, just as we want. So the factorisation is exact.

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#

Answer:

The answer is #=((-2,-3,-1),(-3,-3,-1),(-2,-4,-1))#

Explanation:

#A=((1,-1,0),(1,0,-1),(-6,2,3))#

We start by calculating the determinant

#detA= | (1,-1,0), (1,0,-1), (-6,2,3) | #

#=1| (0,-1), (2,3) |+1| (1,-1), (-6,3) |+0| (1,0), (-6,2) |#

#=(2)+(-3)=-1#

As #detA=-1#, the matrix is invertible

To determine the inverse #A^(-1)#, we start by calculating the matrix of cofactors

#C=((| (0,-1), (2,3) |,-| (1,-1), (-6,3) |,| (1,0), (-6,2) |),(-| (-1,0), (2,3) |,| (1,0), (-6,3) |,-| (1,-1), (-6,2) |),(| (-1,0), (0,-1) |,-| (1,0), (1,-1) |,| (1,-1), (1,0) |))#

#=((2,3,2),(3,3,4),(1,1,1))#

Now we calculate the traspose matrix of #C#

#C^T=((2,3,1),(3,3,1),(2,4,1))#

To obtain the inverse of #A#, we divide #C^T# by the determinant

#A^-1=C^T/det(A)#

#=((-2,-3,-1),(-3,-3,-1),(-2,-4,-1))#

Verification

#A*A^-1=((1,-1,0),(1,0,-1),(-6,2,3))*((-2,-3,-1),(-3,-3,-1),(-2,-4,-1))#

#=((1,0,0),(0,1,0),(0,0,1))=I#

Answer:

#log 43 ~~ 1.633#

Explanation:

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#

Answer:

#12#

Explanation:

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

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

where #a# is the initial term and #r# the common ratio.

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 #(1-r)# we find:

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

If #abs(r) < 1# then #lim_(N->oo) r^N = 0# and we find:

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

In the given example, #a=6# and #r=1/2#. So #abs(r) < 1# and:

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

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