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Featured 1 month ago

Note that

#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)#

Featured 4 weeks ago

Slope of the line perpendicular to

See a solution process below:

The equation in the problem is in slope-intercept form. The slope-intercept form of a linear equation is:

Where

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

Let's call the slope of a perpendicular line:

The slope of a perpendicular line is:

Substituting gives:

Featured 3 weeks ago

So

Let's expand the parentheses:

Then, let's simplify the equation:

Now, let's solve for

Featured 3 days ago

Axis of symmetry:

Vertex (minimum point):

X-intercepts:

Y-intercept:

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

Given:

where:

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

**Vertex:** the maximum or minimum point of the parabola. If

We have the

Simplify.

All terms must have a common denominator of

**Vertex:**

**Approximate vertex:**

Substitute

Plug in the known values.

Simplify.

Prime factorize

Simplify.

**Roots:** values for

Approximate values for

**X-intercepts:** values of

Approximate

**Y-Intercept:** value of

Y-intercept:

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]}

Featured 3 days ago

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

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

For example, to find the decimal representation of

In this simple example, the remainder

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

Here's a much longer example to calculate the exact decimal representation of

Notice that the remainder

#114/268 = 0.4bar(253731343283582089552238805970149)#

In this particular case, we could have identified the common factor

If the simplest form of the fraction has a denominator whose only prime factors are

Featured 3 days ago

It depends...

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

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

if and only if there is a one to one function

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

#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

#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

**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

#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

**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

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