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Answer:

#(sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007)) = 10/99 = 0.bar(10)#

Explanation:

Note that #sqrt(10) = 10^(1/2)# and if #a, b, c > 0# then:

#a^b*a^c = a^(b+c)" "# and #" "(a^b)^c = a^(bc)#

So we find:

#(sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007)) = 10^(1009/2)/(10^(1011/2)-10^(1007/2))#

#color(white)((sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007))) = 10^(1007/2+1)/(10^(1007/2+2)-10^(1007/2+0))#

#color(white)((sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007))) = color(red)(cancel(color(black)(10^(1007/2))))/color(red)(cancel(color(black)(10^(1007/2))))*(10/(100-1))#

#color(white)((sqrt(10)^1009)/(sqrt(10)^(1011)-sqrt(10)^(1007))) = 10/99 = 0.bar(10)#

Answer:

Slope of the line perpendicular to #y# is #-3/2#
See a solution process below:

Explanation:

The equation in the problem is in slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y = color(red)(2/3)x - color(blue)(6)#

Therefore, the slope of the line represented by the equation in the problem is:

#color(red)(m = 2/3)#

Let's call the slope of a perpendicular line: #m_p#

The slope of a perpendicular line is:

#m_p = -1/m#

Substituting gives:

#m_p = -1/(2/3) = -3/2#

Answer:

#p = frac(2 m + 13)(m + 4)#

Explanation:

#p - 2# is a root of #f(x) = x^(2) - m x + 9 - p^(2)#.

So #(p - 2)^(2) - m (p - 2) + 9 - p^(2) = 0#.

Let's expand the parentheses:

#Rightarrow p^(2) - 4 p + 4 - m p + 2 m + 9 - p^(2) = 0#

Then, let's simplify the equation:

#Rightarrow p^(2) - p^(2) - 4 p - m p + 2 m + 9 + 4 = 0#

#Rightarrow - p (4 + m) + 2 m + 13 = 0#

Now, let's solve for #p#:

#Rightarrow - p (4 + m) = - (2 m + 13)#

#Rightarrow - p = - frac(2 m + 13)(4 + m)#

#therefore p = frac(2 m + 13)(m + 4)#

Answer:

Axis of symmetry: #x=9/2#

Vertex (minimum point): #(9/2,-55/2)#

X-intercepts: #((9+sqrt55)/2,0)# and #((9-sqrt55)/2,0)#

Y-intercept: #(0,13)#

Refer to the explanation for the process and approximate values for vertex, x-intercepts, and y-intercept.

Explanation:

Given:

#y=2x^2-18x+13# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=2#, #b=-18#, and #c=13#

To graph a quadratic function, you need to have at least the vertex and x-intercepts. The y-intercept is helpful, also.

Axis of Symmetry: vertical line #(x,+-oo)# that divides the parabola into two equal halves. The variable for the line is #x=(-b)/(2a)#.

#x=(-(-18))/(2*2)#

#x=18/4#

#x=9/2# #larr# axis of symmetry and #x#-value for the vertex

Vertex: the maximum or minimum point of the parabola. If #a>0#, the vertex is the minimum point and the parabola will open upward. If #a<0#, the vertex is the maximum point and the parabola will open downward.

We have the #x#-value of the vertex. To determine the #y#-value, substitute #9/2# for #x# in the equation and solve for #y#.

#y=2(9/2)^2-18(9/2)+13#

Simplify.

#y=2(81/4)-162/2+13#

All terms must have a common denominator of #4#. Multiply fractions without a denominator of #4# by a multiplier equal to #1# that will produce an equivalent fraction with a denominator of #4#. For example, #color(magenta)3/color(magenta)3=1# Recall that any whole number, #n#, is understood to have a denominator of #1#. So #13=13/1#

#y=162/4-162/2xxcolor(red)2/color(red)2+13/1xxcolor(blue)4/color(blue)4#

#y=162/4-324/4+52/4#

#y=((162-324+52))/4#

#y=(-110)/4#

#y=-55/2#

Vertex: #(9/2,-55/2)# #larr# minimum point of the parabola

Approximate vertex: #(4.5,-27.5)#

Substitute #0# for #y# and use the quadratic formula to find the roots and the x-intercepts.

#0=2x^2-18x+13#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values.

#x=(-(-18)+-sqrt((-18)^2-4*2*13))/(2*2)#

Simplify.

#x=(18+-sqrt(324-104))/4#

#x=(18+-sqrt(220))/4#

Prime factorize #220#.

#x=(18+-sqrt((2xx2)xx5xx11))/4#

#x=(18+-2sqrt(55))/4#

Simplify.

#x=(9+-sqrt55)/2#

Roots: values for #x#

#x=(9+sqrt55)/2#,#(9-sqrt55)/2#

Approximate values for #x#.

#x=8.21,##0.792#

X-intercepts: values of #x# when #y=0#

#x#-intercepts: #((9+sqrt55)/2,0)# and #((9-sqrt55)/2,0)#

Approximate #x#-intercepts: #(8.21,0)# and #(0.792,0)#

Y-Intercept: value of #y# when #x=0#

#y=2(0)^2-18(0)+13#

#y=13#

Y-intercept: #(0,13)#

Plot the vertex and x-intercepts and sketch a parabola through the points. Do not connect the dots.

graph{y=2x^2-18x+13 [-13.95, 18.07, -40.31, -24.29]}

Answer:

Long divide the numerator by the denominator until the remainder repeats or you have as many digits as you want.

Explanation:

One way which is guaranteed to work is to long divide the numerator by the denominator.

Since the running remainder is a non-negative integer that is always less than the divisor, it will eventually repeat, at some time after you have exhausted the digits from the dividend.

So the decimal will always repeat or terminate (i.e. a repeating #0# remainder).

For example, to find the decimal representation of #1/7#, divide #1# by #7# ...

enter image source here

In this simple example, the remainder #1# recurs, so we can deduce that the digits #142857# will repeat from that point onwards and

#1/7 = 0.142857142857... = 0.bar(142857)#

Here's a much longer example to calculate the exact decimal representation of #114/268# ...

enter image source here

Notice that the remainder #68# repeats, so we find:

#114/268 = 0.4bar(253731343283582089552238805970149)#

In this particular case, we could have identified the common factor #2# of the numerator and denominator before we started the long division and instead divided #57# by #134#, but it would give the same result.

If the simplest form of the fraction has a denominator whose only prime factors are #2# and/or #5# then the long division will terminate, so the decimal representation will terminate.

Answer:

It depends...

Explanation:

It depends what you mean by "infinity" and by addition if you have such a thing.

Cardinal numbers

The cardinal number of a set of objects is the number of objects in it.

So:

#abs({ 0, 1, 2, 3 }) = abs({"red", "green", "blue", "yellow"}) = 4#

We can express inequality of cardinal numbers in terms of mappings from one set to another.

So, if #A# and #B# are two sets, then:

#abs(A) <= abs(B)#

if and only if there is a one to one function #f:A->B#

For example, we could define:

#{ (f(0) = "red"), (f(1) = "green"), (f(2) = "blue"), (f(3) = "yellow") :}#

So we can tell that:

#abs({ 0, 1, 2, 3 }) <= abs({"red", "green", "blue", "yellow"})#

Similarly, we can find:

#abs({"red", "green", "blue", "yellow"}) <= abs({ 0, 1, 2, 3 })#

and deduce:

#abs({ 0, 1, 2, 3 }) = abs({"red", "green", "blue", "yellow"})#

We can then define the natural numbers (including #0#) as cardinal numbers, where:

#n = abs({ m : m < n })#

So:

#0 = abs(O/}#

#1 = abs({ 0 })#

#2 = abs({ 0, 1 })#

etc.

If we define the natural numbers in this way then if #A# and #B# are disjoint sets, we can define addition as:

#abs(A) + abs(B) = abs(A uu B)#

What happens if you extend this to infinite sets, for example the set of all natural numbers?

Let us write:

#omega = abs( { 0, 1, 2, 3, 4,...} )#

Then we find:

#omega = abs( { 0, 1, 2, 3, 4,... } ) = abs( { 1, 2, 3, 4, 5,...} )#

and:

#omega + 1 = abs( { 1, 2, 3, 4, 5,...} uu { 0 } ) = abs({0, 1, 2, 3, 4, 5,...}) = omega#

So in cardinal addition, we have:

#omega + 1 = omega#

There are other transfinite (i.e. infinite) cardinals larger than #omega#

Ordinal numbers

Ordinal numbers are similar to cardinal numbers, but different. They identify distinct well orderings of sets.

In ordinal arithmetic:

#1 + omega = omega#

but:

#omega + 1 > omega#

Why?

In ordinal arithmetic #omega# is the ordinal number of the natural numbers, which have a natural well ordering:

#0, 1, 2, 3, 4,...#

If we add another number to the beginning of this sequence, then we get a sequence isomorphic to the first:

#k, 0, 1, 2, 3, 4,...#
#uarruarruarruarruarr#
#darrdarrdarrdarrdarr#
#0, 1, 2, 3, 4, 5,...#

but if we add another item to the end of the infinite list, then we get a distinct well ordering:

#0, 1, 2, 3, 4,... k#

So #omega+1 != omega#

Other systems

There are other methods of adding transfinite quantities to ordinary numbers and still getting some kind of addition, etc. to work.

The most extreme is probably Conway's Surreal Numbers.

In that system, all of the numbers #omega - 1#, #omega# and #omega+1# are distinct.

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