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Answer:

#x = +5#

Explanation:

#" "#
# If" " color(red)( a/b = c/d)" " then" " color(red)( axxd = bxxc )#
#" "#
# (x-11)/6 = -1#
#" "#
#rArr (x-11)/6 = (-1)/1#
#" "#
#rArrcolor(red)((x - 11)xx1 = -6 xx 1)" " #Applying above method
#" "#
#rArr x-11 = -6#
# " "#
#rArrx = -6 + 11#
#" "#
#rArrx = +5#

Answer:

#(-0.5,2.5)#

Explanation:

So, we start with the two equations

#2x+7=6#
#y-3=x#?
..................................................................................................................

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.

#x=y-3#

What we need to do for substitution, is plug in #(y-3)# for x in the first equation. That looks like this.

#2x+7=6" "# Our original Equation

#2(y-3)+7=6" "# Plugging in the value for x

#2y-6+7=6" "# Using the distributive property

#2y+1=6" "# Simplifying like terms

#2y=5# Rearranging the equation by subtracting 1 from both sides

#y=2.5" "# 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.

#x=y-3#

#x=2.5-3#

#x=-0.5#

That leaves us with an answer of #(-0.5,2.5)#

Hope I helped!

Answer:

Take the negative reciprocal of the given line's slope. The new slope will be #3/2#.

Explanation:

Lines that are perpendicular will have negative reciprocal slopes. Meaning, if one line's slope is #m#, then a perpendicular line will have a slope of #-1/m#.

Why? A line's slope is equal to its rise over its run—also written as #m=[Delta y]/[Delta x]#. If we rotate that line 90° counterclockwise (making it perpendicular to its old self), the run (to the right) becomes a rise (up), and the rise (up) becomes a backwards run (to the left):

enter image source here
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 90° (180° total from the beginning), this 3rd line will have a slope of #(-1)/(-1/m)#, which simplifies to #m#—the same slope of the first line, which is what we would expect.)

Okay, great—so what's the slope of #2x+3y=5#? If we rearrange this into slope-intercept form, we get

#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=-2/3# is

#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".

Answer:

#0.dot0dot9=1/11#, #0.dot8 = 8/9#, #0.dot0dot6=2/33#

Explanation:

Let's start with #0. dot 8#, where the 8 repeats.
Let #x=0. dot 8#
When only one digit repeats, one multiplies #x# with 10:
#10x=8.dot 8#
#10x-x=8.dot8-0.dot8#
#9x=8#
#x=8/9#

If two digits repeat, say in #0.dot0dot9# one multiplies with 100.
Let #x=0.dot0dot9#
#100x=9.dot0dot9#
#100x-x=9.dot0dot9-0.dot0dot9#
#99x=9#
#x=9/99#
#x=1/11#

Let #x=0.dot0dot6#
#100x=6.dot0dot6#
#100x-x=6.dot0dot6-0.dot0dot6#
#99x=6#
#x=6/99# (You could have seen this from knowing that #0.dot0dot9=9/99#)
#x=2/33#

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 #x=0.0 dot6#
#10x=0.6 dot 6#
#10x-x=0.6dot6 -0.0dot6#
#9x=0.6#
#z={0.6}/9#
#=6/90#
#=1/15#

Answer:

column A (assumes 4 columns), row 251

Explanation:

We have a series of numbers that starts with:

#((A, B, C, D),(8, 6,4,2),(10,12,14,16),(24,22,20,18),(26,28,30,32),(vdots, vdots, vdots, vdots))#

(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:

#((A, B, C, D),(2+(2)3, 2+(2)2,2+(2)1,2+(2)0),(2+(2)4,2+(2)5,2+(2)6,2+(2)7),(2+(2)11, 2+(2)10,2+(2)9,2+(2)8),(2+(2)12,2+(2)13,2+(2)14,2+(2)15),(vdots, vdots, vdots, vdots))#

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

#2008=2+2(n)#

#2n=2006#

#n=2006/2=1003#

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

#1003/8=125 3/8=>"column A"#

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

#2008/8=251#

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:

#((A, B, C, D),(10, 6,4,2),(12,14,16,20),(30,26,24,22),(32,34,36,40),(vdots, vdots, vdots, vdots))#

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#

#2008_10=color(red)(3)color(white)(00)color(green)7color(white)(00)color(brown)3color(white)(00)0#
#color(white)(00000000)bar(8^3)color(white)(0)bar(8^2)color(white)(0)bar(8^1)color(white)(0)bar(8^0)#

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:

#3(8^2)+7(8^1)+3(8^0)=3(64)+7(8)+3(1)=251#

Proposition
If #u# is an odd integer, then the equation #x^2 + x - u = 0# has no solution that is an integer.

Proof
Suppose that there exists a integer solution #m# of the equation:

#x^2 + x - u = 0#

where #u# is an odd integer. We must examine the two possible cases:

#m# is odd; or
#m# is even.

First, let us consider the case where #m# is odd, then there exists an integer #k# such that:

# m = 2k + 1 #

Now, since #m# is a root of our equation, it must be that:

# m^2 + m - u = 0 #
# :. (2k + 1)^2 + (2k + 1) − u = 0 #
# :. (4k^2 + 4k + 1) + (2k + 1) − u = 0 #
# :. 4k^2 + 6k + 2 − u = 0 #
# :. u = 4k^2 + 6k + 2 #
# :. u = 2(2k^2 + 3k + 1) #

And we have a contradiction, as #2(2k^2 + 3k + 1)# is even, but #u# is odd.

Next, let us consider the case where #m# is even, then there exists an integer #k# such that:

# m = 2k #

Similarly, since #m# is a root of our equation, it must be that:

# m^2 + m - u = 0 #
# :. (2k)^2 + (2k) − u = 0 #
# :. 4k^2 + 2k − u = 0 #
# :. u = 4k^2 + 2k #
# :. u = 2(2k^2 + k) #

And, again, we have a contradiction, as #2(2k^2 + k)# is even, but #u# is odd.

So we have proved that there is no integer solution of the equation #x^2 + x - u = 0# where #u# is an odd integer.

Hence the proposition is proved. QED

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