# repating decimels etc.

• Nov 15th 2005, 11:34 AM
rpatel
i am struggling with these questions
hello i have attached the questions which i am struggling with.

there are three questions which i have attached into one image.

please help me. i have done most questions but am struggling with these. i looked in text books but couldn't solve the problem still.
• Nov 15th 2005, 02:03 PM
The Pondermatic
Problem #1:

Are you sure you wrote down the problem correctly? I've tried to answer it but I couldn't.

12 - (x - 7) ^ 2
12 - (x - 7) * (x - 7) <---- Use FOIL to multiply this out.
12 - (x^2 - 7x - 7x + 49)
12 - (x^2 - 14x + 49)
12 - x^2 + 14x - 49
-x^2 + 14x - 37 <---- Multiply everything by -(1/7).
(1/7)x^2 - 2x + 37/7

Everything works except for 37/7. Either I'm wrong, you copied the problem incorrectly, or the textbook you got the question from made a mistake.

Problem #2

There is a very neat trick my algebra teacher showed me.

0.51... = x
51.51... = 100x <---- Multiply both sides by 100.
51.51... - 0.51... = 100x - x <---- Subtract x (or 0.51...) from both sides.
51 = 99x <---- The last step gets rid of the icky repeating decimal!
51/99 = x <---- Solve for x by dividing both sides by 99.
(51 / 3)/(99 / 3) = x <---- Simplify.
17/33 = x <---- Simplify.
Thus, 0.51... = 17/33

Problem #3

This is a hard, complex problem. You need to know these things:
how to find slope ((y1-y2)/(x1-x2))
how to find a parallel line (the slopes are the same)
how to write an equation for a graph (y = mx + b)

First, take the first line, PQ, and find its slope. The formula is: slope = (y1-y2)/(x1-x2). This means you need to take the y coordinate of one of the points, subtract it from the y coordinate of the other point, and divide the difference by the difference of subtracting the first point's x coordinate from the other x coordinate.

For example, the slope of line PQ (containing points P(-1,11) and Q(4,-4) is:
(y1-y2)/(x1-x2)
(11-(-4))/((-1)-4)
(15)/(-5)
5/(-3)

Before I continue, it is incredibly important that you understand what I just found. The slope of a line is how steep it is. The number in the numerator (5) is how many units up and the number in the denominator (-3) is how many units right you must move to find another point on a line. For example, by moving up 5 units and right -3 (or left 3) units from P(-1,11), I can find that the point (-4,16) is on line PQ. You can also change the fraction to something equivalent to find points farther or closer away (i.e., move up -2.5 (or down 2.5) units and right 1.5 units, since each is -1/2 times the numerator and denominator).

Now, I'll explain the equation that you need to find. It is easiest to write this one in slope-intercept form, or y = mx + b. This is an extremely important equation that you must understand. If you do, you can graph any line if you know the equation for it. In it, y is the y coordinate of a point, x is the x coordinate of the same point, m is the slope of the line, and b is the y coordinate when x = 0.

Something else that is important is how the slopes of parallel lines are related. This is easy. Any lines that are parallel to each other have the same slope. If you have line a, and its slope is 123,456,789,003.14159, and lines b, c, d, and e are parallel to it, then lines b, c, d, and e all have a slope of 123,456,789,003.14159 as well.

Since we know that line B is parallel to line PQ and the slope of PQ, we can find half of the equation that you need. Since m is the slope, and the slopes of B and PQ are both 5/(-3), m must equal 5/(-3). Our your equation so far is y = (5/(-3)) * x + b.

Because we are given that point R is at (0,-7) and b is the y coordinate when x is 0, we can quickly find b: -7. That's it.

We now have our equation and the answer to the question: y = (5/(-3)) * x - 7. You are probably looking at the x and y and wondering why we don't have any values for them. The answer is that in any equation for a graph, you don't need to find what x and y stand for since those are what you want the equation to find for you. It's similar to how area = length * width finds you thre area of a rectangle any lentgh and width, even though you don't have conrete, constant value for any of the variables. Using the equation, you can make up any value for x or y, solve for the other variable, and put them together to find that the point (x,y) is on the line that the equation describes.
• Nov 15th 2005, 02:14 PM
CaptainBlack
Question 1:

Show that:

$12\ +\ (x-7)^2\ =\ \frac{1}{7}x^2\ -\ 2x\ +\ 19$

There must be something wrong here; in that the RHS does not equal
the LHS. To see this just substitute x=0 into both sides.

Maybe it should have asked you to solve the equation?
• Nov 15th 2005, 02:58 PM
CaptainBlack
Question 2:

Prove that the recurring decimal $0.\overline{51}$ can be
written as 17/33.

Someone else has already posted a solution for this, but this
shows a different method, which may be of interest.

By definition:

$0.\overline{51}\ =\ \frac{51}{100}\ +\ \frac{51}{100^2} \ + \ \frac{51}{100^3}\ +\ ...$
$=\ \frac{51}{100}\ \sum_{n=0}^\infty\ \frac{1}{100^n}$

but the summation is a geometric series and so may be summed
to give:

$0.\overline{51}\ =\ \frac{51}{100}\cdot\frac{1}{1-\frac{1}{100}}\$
$=\ \frac{51}{100} \cdot \frac{100}{99}$

$=\ \frac{17}{33}$
• Nov 15th 2005, 08:30 PM
CaptainBlack
Question 3:

That the line through B is parallel to PQ means that the slope of B
is equal to the slope of PQ. The slope of PQ is:

$m\ =\ (11+4)/(-1-4)\ =\ -3$

Equation of line B:

$y\ =\ m.x\ +\ c\ =\ -3x\ +\ c$

and as on B y=-7 when x=0:

$y\ =\ -3x\ -\ 7$
.
• Nov 16th 2005, 11:10 AM
rpatel
thank you everyone. i will have to look into question 1 as there might of been a mis print.

thank you :D

regards