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Answer:

#x≈1.432" to 3 decimal places"#

Explanation:

Using the #color(blue)"law of logarithms"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(loga^n=nloga)color(white)(2/2)|)))#

#6^x=13#

Take the ln ( natural log ) of both sides.

#rArrln6^x=ln13#

#rArrxln6=ln13#

#rArrx=ln13/ln6≈1.432" to 3 decimal places"#

Answer:

#p(x)# has zeros:

#0# with multiplicity #3#

#2# with multiplicity #2#

#-2# with multiplicity #1#

#-1+sqrt(3)i# with multiplicity #1#

#-1-sqrt(3)i# with multiplicity #1#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

We find:

#p(x) = (x^3-8)(x^5-4x^3)#

#color(white)(p(x)) = (x^3-2^3)x^3(x^2-2^2)#

#color(white)(p(x)) = (x-2)(x^2+2x+4)x^3(x-2)(x+2)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)(x^2+2x+4)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)(x^2+2x+1+3)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)((x+1)^2-(sqrt(3)i)^2)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)(x+1-sqrt(3)i)(x+1+sqrt(3)i)#

Hence zeros:

#0# with multiplicity #3#

#2# with multiplicity #2#

#-2# with multiplicity #1#

#-1+sqrt(3)i# with multiplicity #1#

#-1-sqrt(3)i# with multiplicity #1#

Answer:

The vertical asymptotes are #x=sqrt8# and #x=-sqrt8#
The slant asymptote is #y=3/4x#
No horozontal asymptote.

Explanation:

Let's factorise the denominator #(4x^2-32)#

#=4(x^2-8)=4(x+sqrt8)(x-sqrt8)#

The domain of #y# is #D_y=RR- {sqrt8,-sqrt8} #

As we cannot divide by #0#,

So #x!=sqrt8# and #x!=-sqrt8#

The vertical asymptotes are #x=sqrt8# and #x=-sqrt8#

As the degree of the numerator is #># the degree of the denominator, we expect a slant asymptote.

Let's do a long division

#color(white)(aaaa)##3x^3+1##color(white)(aaaa)##∣##4x^2-32#

#color(white)(aaaa)##3x^3-24x##color(white)(aa)##∣##(3x)/4#

#color(white)(aaaaa)##0-24x+1#

So, #(3x^3+1)/(4x^2-32)=(3x)/4-(24x+1)/(4x^2+32)#

The slant asymptote is #y=3/4x#

To calculate the limits, we use the terms of highest degree.

#lim_(x->+oo)y=lim_(x->+oo)(3x^3)/(4x^2)=lim_(x->+oo)(3x)/4=+oo#

#lim_(x->-oo)y=lim_(x->-oo)(3x^3)/(4x^2)=lim_(x->-oo)(3x)/4=-oo#

There are no horizontal asymptote

#lim_(x->-sqrt8^(-))=-oo#

#lim_(x->-sqrt8^(+))=+oo#

#lim_(x->sqrt8^(-))=-oo#

#lim_(x->sqrt8^(+))=+oo#

When #x=0#, #=>#, #y=-1/32#

When #y=0#, #0>#, #x=(-1/3)^(1/3)#

graph{(y-(3x^3+1)/(4x^2-32))(y-x3/4)=0 [-28.86, 28.9, -14.43, 14.43]}

Answer:

The answer is #=30/41+(17i)/41#

Explanation:

If you want to simplify a quotient of complex numbers , multiply numerator and denominator by the conjugate of the denominator.

#z=z_1/z_2=(z_1barz_2)/(z_2barz_2)#

The conjugate of #(a+ib)# is #(a-ib)#

And #i^2=-1#

Here, #z_1=2+5i# and #z_2=5+4i#

So #(2+5i)/(5+4i)=((2+5i)(5-4i))/((5+4i)(5-4i))#

#=(10-8i+25i-20i^2)/(25-16i^2)#

#=(30+17i)/(41)#

#=30/41+(17i)/41#

Answer:

#3x^3+6x^2+8x+24 + (47)/(x-2)#

Explanation:

We set up the long division of a polynomial by a simple monomial like this:

#(x-2))bar(3x^4-4x^2+8x-1)#

It works just like the long (numerical) division most of us learned back in elementary school, except now we're dividing with variables.

First we check: how many times does our leading #x# term in the divisor, in this case just #x# (coefficient of #1#), go into our leading #x# term in the dividend, #3x^4#? We would have to multiply #x# by #3x^3# to get #3x^4#. Just like in long division with integers, we put this above the bar.

#color(white)(SPACE)3x^3#
#(x-2))bar(3x^4-4x^2+8x-1)#

Now, we multiply #3x^3# by the divisor and subtract that from #3x^4#.

#color(white)(SPACE)3x^3#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^3(x-2)#

#=>#

#color(white)(SPACE)3x^3#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#

We can subtract #3x^4# from #3x^4# to get #0#. We see that we have no #x^3# term in the dividend, so we treat this as though we're adding #6x^3# to #0#.

Now, we check: how many times does our leading coefficient #x# in the divisor go into #6x^3#? We would have to multiply #x# by #6x^2# to get #6x^3#. Similarly to above:

#=>#

#color(white)(SPACE)3x^3+6x^2#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACE)-6x^2(x-2)#

#=>#

#color(white)(SPACE)3x^3+6x^2#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACES)-6x^3+12x^2#

This gets rid of our #x^3# term, but now we have an #x^2# term. We also have an #x^2# term in the dividend, so we can add this to the current remainder.

#=>#

#color(white)(SPACE)3x^3+6x^2#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPA)+12x^2#

#=>#

#color(white)(SPACE)3x^3+6x^2#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#

Next we check: how many times does our leading coefficient #x# in the divisor go into #8x^2#? We would have to multiply #x# by #8x# to get #8x^2#.

#=>#

#color(white)(SPACE)3x^3+6x^2+8x#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESP)-8x(x-2)#

#=>#

#color(white)(SPACE)3x^3+6x^2+8x#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESP)-8x^2+16x#

#=>#

#color(white)(SPACE)3x^3+6x^2+8x#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESPACES)+16x#

Now we add #16x# to the #8x# term in the dividend.

#=>#

#color(white)(SPACE)3x^3+6x^2+8x#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESPACESPA)24x#

Next we check: how many times does our leading coefficient #x# in the divisor go into #24x#? We would have to multiply #x# by #24# to get #24x#.

#=>#

#color(white)(SPACE)3x^3+6x^2+8x+24#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESPACESPA)24x#
#color(white)(SPACESPACE)-24(x-2)#

#=>#

#color(white)(SPACE)3x^3+6x^2+8x+24#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESPACESPA)24x#
#color(white)(SPACESPACE)-24x+48#

#=>#

#color(white)(SPACE)3x^3+6x^2+8x+24#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESPACESPA)24x#
#color(white)(SPACESPACESPACE)48#

Now we add #48# to the #-1# in our dividend.

#=>#

#color(white)(SPACE)3x^3+6x^2+8x+24#
#(x-2))bar(3x^4-4x^2+8x-1)#
#color(white)(SPA)-3x^4+6x^3#
#color(white)(SPACESPA)6x^3#
#color(white)(SPACESPAC)8x^2#
#color(white)(SPACESPACESPA)24x#
#color(white)(SPACESPACESPACE)47#

And of course, #x# goes into #47# zero times. This leaves us with a remainder.

The final answer is therefore #3x^3+6x^2+8x+24 + (47)/(x-2)#

Answer:

A reflection about the #x#-axis.

Explanation:

For any function #f#, its output at any given input #x# is just #f(x)#.

When we take that output #f(x)# and make it negative (i.e. #-f(x)#), we're just flipping the sign of the output—positives become negatives, and vice versa.

Let's say that when #x=3,# we have #f(x)=5#.
Thus, when #x=3,# we have #"-"f(x)="-"5#.

All that happens is the sign of the output value changes—points that were once above the #x#-axis are now below, with no shift left or right.

This is simply stated as a reflection where the mirror is the #x#-axis, also called a reflection about the #x#-axis.

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