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Featured 3 months ago

Featured 1 month ago

So, we start with the two equations

..................................................................................................................

Our goal in this problem is to solve for the x and y coordinate values where the two lines intersect. We can do that in a variety of ways.

I personally find substitution is the fastest and easiest method based on the way the equations are arranged.

..................................................................................................................

We are trying to solve for values of x and y that work in BOTH equations.

You may have noticed that in the second equation, we are given the value of x.

What we need to do for substitution, is plug in

**Our original Equation**

**Plugging in the value for x**

**Using the distributive property**

**Simplifying like terms**

**Rearranging the equation by subtracting 1 from both sides**

**Solving for y by dividing both sides by 2**

Now, we have a numerical value for y.

We can plug that value into our second equation.

That leaves us with an answer of

Hope I helped!

Featured 1 month ago

Take the negative reciprocal of the given line's slope. The new slope will be

Lines that are perpendicular will have *negative reciprocal* slopes. Meaning, if one line's slope is

Why? A line's slope is equal to its rise over its runâ€”also written as

In math terms:

#Delta y_"new"=Delta x" "and" "Delta x_"new" = -Delta y#

thus

#m_"new"=(Delta y_"new")/(Delta x_"new")=(Delta x)/(-Delta y)=-(Delta x)/(Delta y)=-1/m#

*(Note: if we rotate this new line* another *90Â° (180Â° total from the beginning), this 3rd line will have a slope of #(-1)/(-1/m)#, which simplifies to #m#â€”the same slope of the first line, which is what we would expect.)*

Okay, greatâ€”so what's the slope of

#y=-2/3 x+5/3# ,

meaning that for every step of "2 down", we have a step of "3 right".

The negative reciprocal of the slope

#m_"new"=-1/m=(-1)/(- 2/3)=3/2# ,

meaning that, for a perpendicular line, a step of "3 up" comes with a step of "2 right".

Featured 1 month ago

Let's start with

Let

When only one digit repeats, one multiplies

If two digits repeat, say in

Let

Let

If in this case it is only the 6 that repeats, one only multiplies by 10, but you end up with a decimal to get rid of:

Let

Featured 3 weeks ago

column A (assumes 4 columns), row 251

We have a series of numbers that starts with:

(the question has a column label E but no numbers within it, so for this initial answer, I'm ignoring it)

The pattern is that of a "snake" - with the initial number, 2, being in column D, then the pattern increases by 2 for each number moving from D to A, then drops down to the next row and increases by 2 again as it moves from A to D.

We're looking for the column that 2008 will appear in.

We can describe the numbers in the various rows and columns this way:

And so with the pattern, we can find where 2008 will sit.

We can now divide by 8 - the remainder will tell us what column:

With the column settled, let's now find the row. We can divide 2008 by 8:

We can see that 8 is on the first row, 16 on the second, 24 on the third, and so on. 2008 is evenly divisible 251 times, and so it's on the 251st row.

~~~~~

Another way to approach finding the column is to utilize bases, specifically base 8. Let's look at the chart we have, but in base 8 numbers:

We can convert 2008 from base 10 to base 8 and use this chart to find our number

Changing from base 10, which has places such as

#1 (10^0), 10 (10^1), 100 (10^2)# , and so on, to base 8, which has #1 (8^0), 8 (8^1), 64 (8^3), and so on, involves finding the adjusted digits.

#8^3=512=> 512xxcolor(red)(3)=1536=>2008-1536=472#

#8^2=64=> 64xxcolor(green)(7)=448=>472-448=24#

#8^1=8=> 8xxcolor(brown)(3)=24=>24-24=0#

The last two digits are 30, and so that puts us in column A.

To find the row, we can take the digits, drop the final digit (because it's a 0, we're not in the middle of a series and not needing to add 1 additional row as a "final partial row", and re-evaluate the number:

Featured 1 week ago

**Proposition**

If

**Proof**

Suppose that there exists a integer solution

#x^2 + x - u = 0#

where

#m# is odd; or

#m# is even.

First, let us consider the case where

# m = 2k + 1 #

Now, since

# m^2 + m - u = 0 #

# :. (2k + 1)^2 + (2k + 1) âˆ’ u = 0 #

# :. (4k^2 + 4k + 1) + (2k + 1) âˆ’ u = 0 #

# :. 4k^2 + 6k + 2 âˆ’ u = 0 #

# :. u = 4k^2 + 6k + 2 #

# :. u = 2(2k^2 + 3k + 1) #

And we have a contradiction, as

Next, let us consider the case where

# m = 2k #

Similarly, since

# m^2 + m - u = 0 #

# :. (2k)^2 + (2k) âˆ’ u = 0 #

# :. 4k^2 + 2k âˆ’ u = 0 #

# :. u = 4k^2 + 2k #

# :. u = 2(2k^2 + k) #

And, again, we have a contradiction, as

So we have proved that there is no integer solution of the equation

Hence the proposition is proved. QED

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