+2236 Your computer program is a classic example of GIGO. Even if your program worked properly, it will never determine if a number is irrational. At best, it will give the fractional equivalent of a decimal proving it rational. It should always do that, because you are entering a finite number of decimals. You coded it to say otherwise, but that doesn’t make it true.
Computers are not magical machines Mr. BB. Computer coding isn’t the same as an incantation you’d say to a crystal ball. Computers do what you tell them to do; that’s not the same as what you want them to do.
I’m not the only one who knows you as an incompetent mathamation and coder. I’m not your first Troll, either. That honor goes to Dragonlance. He’s one brilliant kid, for sure.
Here are two of his comments:
... One more thing computers and calculators can’t cope because they are machines. They don’t think or love or hate or give a rat’sass about anything. That is a good thing because it would never do anything for you. Only humans cope. Well some humans do. You do not know much about computers or calculators do you?—Source
He has several troll posts directed toward you. He addresses and refers to you as “Anol Retentive” sometimes adding the “constipated hog” description, and as the “banker man.” One time he suggested, “You should go lock you self in a bank vault until you feel better. Make sure the time clock is set for next year.” Source
Sounds like good advice! While you are in there, you should study up on irrational numbers and computer programming.
GA
+118724 Hi Rauhan
9. Ann has some sticks that are all of the same length. She arranges them in squares and has made the following 3 rows of patterns:
Row 1
Row 2
Row 3
She notices that 4 sticks are required to make the single square in the first row,
7 sticks to make 2 squares in the second row
and in the third row she needs 10 sticks to make 3 squares.
(a) Find an expression, in terms of n, for the number of sticks required to make a similar arrangement of n squares in the nth row.
| row (r) | 1 | 2 | 3 | ... | n |
| match sticks (m) | 4 | 7 | 10 | 3n+1 |
Ann continues to make squares following the same pattern. She makes 4 squares in the 4th row and so on until she has completed 10 rows.
(b) Find the total number of sticks Ann uses in making these 10 rows.
m(10)=3*10+1=31 match sticks
Ann started with 1750 sticks.
Given that Ann continues the pattern to complete k rows but does not have sufficient sticks to complete the (k + 1)th row,
(c) show that k satisfies (3k – 100)(k + 35) < 0.
The total number of sticks in k rows is
\(4+7+10+.......(3k+1) \qquad \text{Sum of an AP}\\ =\frac{n}{2}(a+L)\\ =\frac{k}{2}(4+3k+1)\\ =\frac{k}{2}(3k+5)\\ now\\ \frac{k}{2}(3k+5)<1750\\ k(3k+5)<3500\\ 3k^2+5k<3500\\ 3k^2+5k-3500<0\\ \qquad 3*-3500=-10500\\ \qquad \text{I need 2 numbers that multiply to -10500 and add to +5}\\ \qquad \text{Those numbers are 105 and -100}\\ 3k^2+105k-100k-3500<0\\ 3k(k+35)-100(k+35)<0\\ (3k-100)(k+35)<0 \)
(d) Find the value of k.
if you graph y=(3k-100)(k+35)
it will be a concave up parabola and y will be less then 0 between the two zeros.
3k-100=0
3k=100
k=33 and a 1/3
and
k=35=0
k=-35
k cant be negative so Ann will have enough sticks for 1 to 33 rows.
She will not have enough sticks for 34 rows.
So
k=33
Any questons, just ask :)
+2446 There are a multitude of ways to find the missing angle measures in the diagram. I'll just show you my observations.
1) \(m\angle A=121^{\circ}\)
\(\angle A\) and the angle across from it form vertical angles. Thus, by the vertical angles theorem, they are congruent.
2) \(m\angle B=59^{\circ}\)
\(\angle A\) and \(\angle B\) form a linear pair, so the angles are supplementary by the linear pair theorem. If the angles are supplementary, then the sum of the measure of the angles is 180 degrees.
3) \(m\angle C=59^{\circ}\)
\(\angle B\) and \(\angle C\) together form vertical angles. As aforementioned, this means that the measure of both angles are equal.
4) \(m\angle D=55^{\circ}\)
It is given info that the figure is a parallelogram, which by definition is a quadrilateral with two pairs of opposite sides parallel. Plus, the unnamed 55 degree angle and \(\angle D\) can be classified as alternate interior angles. Since this is true, those angles are congruent.
5) \(m\angle E=4^{\circ}\)
The unnamed 121 degree angle, \(\angle D,\) and \(\angle E\) are all angles in a common triangle. The triangle sum theorem states that the sum of the measures of the interior angles of a triangle is 180 degrees. We can use this theorem to solve for the remaining angle:
| \(121+m\angle D+m\angle E=180\) | Use the substitution property of equality to substitute in the known value for the measure of the angle D. |
| \(121+55+m\angle E=180\) | Simplify the left hand side as much as possible. |
| \(176+m\angle E=180\) | Subtract 176 from both sides to isolate angle E. |
| \(m\angle E=4^{\circ}\) | |
I now want you to try to figure out the rest of the angle measures on your own now! See if you can do it.
| Statements | Reasons |
| DFGH is a kite | 1. Given |
| \(\overline{DF}\cong\overline{FG}\) | 2. Definition of a kite (A kite is a quadrilateral with 2 pairs of adjacent congruent sides) |
| \(\overline{FH}\perp\overline{DG}\) | 3. Property of a kite |
| \(m\angle DPF=90^{\circ}\\ m\angle GPF=90^{\circ}\) | 4. Perpendicular segments form right angles |
| \(\angle DPF\cong\angle GPF\) | 5. Right Angles Congruence Theorem (All rights angles are congruent) |
| \(\overline{PF}\cong\overline{PF}\) | 6. Reflexive Property of Congruence (Any geometric figure is congruent to itself) |
| \(\triangle DPF\cong\triangle GPF\) | 7. Hypotenuse Leg Triangle Congruence Theorem (Two rights triangles with corresponding hypotenuses and a select leg congruent are congruent triangles) |
| \(\overline{PD}\cong\overline{GP}\) | 8. Corresponding Parts of Congruent Triangle are Congruent (sometimes abridged to CPCTC) |
+2446 I have not seen a interminably repeating decimal be converted into a simplified fraction in the fashion Cphill described above, but I will present to you an alternate method. On further review, though, the method I have below appears to prove Cphill's method.
| \(x=0.3\overline{25}\) | Firstly, I set the repeating decimal equal to a variable. I will use the standard choice, x. |
| \(x=0.325252525...\) | A few more decimal places should be written out so that the method is clear. |
Now, multiply x by a factor of ten such that the repeating portion lines up with the first line.
| \(10x=3.25252525...\\ \hspace{5mm}x=0.325252525...\) | Notice how the repeating portion does not line up here, so this is not the correct multiple of ten. Let's multiply both sides by ten again. |
| \(100x=32.525252525...\\ -(\hspace{1mm}x=\hspace{2mm}0.325252525...)\) | Look at this! Notice how the repeating portion of both equations line up perfectly. Now, subtract the two equations from each other. |
| \(99x=32.2\) | Now, solve for x. |
| \(x=\frac{32.2}{99}\) | Apply a multiplication of 10/10 to simplify the fraction. |
| \(x=\frac{322}{990}=\frac{161}{495}\) | |
