+2440 Here is the predicted forecast for today and the next three days for me.
| High Temperature | Low Temperature | Brief Synopsis | |
| Wednesday | 32° | 8° | Somewhat cloudy |
| Thursday | 24° | 9° | Snowy |
| Friday | 12° | 5° | Mostly Sunny |
| Saturday | 9° | 1° | Completely Sunny |
+2440 In order to solve this problem, I would use a similar strategy as I used here https://web2.0calc.com/questions/halp_34
There is a slight difference from this problem and the one I hyperlinked, though. The coefficient of the quadratic term is not one. In order to make it one, let's factor it out without changing the value of the expression.
\(3x^2-6x-2\Rightarrow3(x^2-2x)-2\)
Now, the problem is more or less identical to the one I hyperlinked you to. Let's find the value that we can use to manipulate the expression. This magical value is found exactly the same way as completing the square.
\(x^2-2x+\left(\frac{-2}{2}\right)^2\\ x^2-2x+1\\ (x-1)^2\)
This process shows that we need 1 to be inserted inside the parentheses.
\(3(x^2-2x+1)-2-3(1)\)
Since we added one inside the parentheses, we must compensate for that by subtracting the same amount somewhere else in the expression. We need to multiply by three because of the distributive property. Simplify from here and convert into the desired form.
\(3(x^2-2x+1)-2-3(1)\\ 3(x-1)^2-2-3\\ 3(x-1)^2-5\)
Notice the parallelism.
\(\textcolor{green}{3}(x-\textcolor{blue}{1})^2+(\textcolor{red}{-5})\\ \textcolor{green}{a}(x-\textcolor{blue}{h})^2+\textcolor{red}{k}\)
Just like before, a=3, h=1, and k=-5. Therefore, the sum of these values is \(a + h + k=3+1-5=-1\)
.+2440 Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root."
| \(b^2-4ac=0\) | Plug in the given values of a, b, and c of the quadratic. |
| \((-5)^2-4(2)(k)=0\) | Now, solve for k. |
| \(25-8k=0\) | Subtract 25 from both sides. |
| \(-8k=-25\) | |
| \(k=\frac{25}{8} \) | |
+2440 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) |
+2440 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. |
| \(x=\frac{322}{990}=\frac{161}{495}\) | |
