+2446 The gradient (usually referred to as the slope) of a line indicates its steepness. There is an equation that allows one to calculate the gradient. It is the following:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
m = slope of line
Before we can understand this equation, let's try putting it to use on some lines:
Source: http://www.coolmath.com/sites/cmat/files/images/06-lines-01.gif
Let's use this line and the points designated on the line to figure out its slope:
| \(m=\frac{y_2-y_1}{x_2-x_1}\) | Plug in the appropriate value for the y-coordinates and the x-coordinates. |
| \(m=\frac{3-(-1)}{4-(-2)}\) | Simplify the numerator and denominator by recognizing that subtracting a negative is equivalent to adding a positive. |
| \(m=\frac{3+1}{4+2}\) | |
| \(m=\frac{4}{6}\) | When finding the gradient, you should simplify it into simplest terms |
| \(m=\frac{2}{3}\) | |
It does not matter which order you subtract the y-coordinates, but make sure that you subtract them in the same order as your x-coordinates. Otherwise, your slope will be incorrect.
+2446 I am assuming that you want this equation solved using the quadratic formula. I would use that method, too. Of course, let's remind ourselves of the quadratic formula:
In a quadratic function in the form \(ax^2+bx+c=0\).
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
The question, however, asks for how much time elapsed when it (I assume a thrown ball) reached a height of 3 meters
| \(3=-5t^2+8t+1.3\) | Subtract 3 on both sides. |
| \(0=-5t^2+8t-1.7\) | |
Now that the equation is in the form of \(ax^2+bx+c=0\), let's solve for how much time, t, it took for it to reach 3 meters:
| \(0=-5t^2+8t-1.7\) | Use the quadratic formula to solve for t. | ||
| \(t = {-8 \pm \sqrt{8^2-4(-5)(-1.7)} \over 2(-5)}\) | We should first calculate the discriminant \(b^2-4ac\) to see if there are indeed solutions for t. Let's just focus on that part. | ||
| \(8^2-4(-5)(-1.7)\) | Do 8^2 first. | ||
| \(64-4(-5)(-1.7)\) | Do 4*-5 next. | ||
| \(64-(-20)(-1.7)\) | Do -20*-1.7 | ||
| \(64-34\) | |||
| \(30\) | Because \(b^2-4ac>0\), this means that there are 2 solutions. Replace \(8^2-4(-5)(-1.7)\) with its calculated value, 30. | ||
| \(t=\frac{-8\pm\sqrt{30}}{2(-5)}\) | SImplify 2*-5 in the denominator | ||
| \(t=\frac{-8\pm\sqrt{30}}{-10}\) | Break up the fraction by doing \(\frac{a\pm b}{c}=\frac{a}{c}\pm\frac{b}{c}\) | ||
| \(t=\frac{-8}{-10}\pm\frac{\sqrt{30}}{-10}\) | Simplify -8/-10 by realizing its GCF is -2. | ||
| \(t=\frac{4}{5}\pm\frac{-\sqrt{30}}{10}\) | Split your answer into two answers | ||
| |||
The question asks for the first time that the ball hits 3 meters. Of course, we do not know the exact value of either of our solutions currently. We can utilize logical thinking to figure it out, however.
If we pretend as if \(\sqrt{30}=5\), as that is an OK approximation, we can simplify both solutions into
\(t=\frac{4}{5}-\frac{1}{2}\) and \(t=\frac{4}{5}+\frac{1}{2}\)
We can infer that \(\frac{4}{5}-\frac{1}{2}\) is still positive but closer to 0 because 4/5 is greater than 1/2. Subtracting the two would make it closer to 0. Adding both would make the numbers further from 0.
Therefore, the first time it goes to the height of 3 meters is after \(\frac{4}{5}-\frac{\sqrt{30}}{10}\) that much time elapsed. \(\frac{4}{5}-\frac{\sqrt{30}}{10}\approx0.252277\).
I have supllied a link to a graph if ou would like to check it out: https://www.desmos.com/calculator/ivnsia7gdd
+2236 Hi Melody,
Ok . . . I did not know that. I was really scratching my chimp head on this.
I did note that she was totally lacking in gratitude with words or points, but such behaviors are not rare on here.
By dumb luck, I managed to snag copies of a few of the better question posts shortly after the answers posted, but before she deleted the questions. I noticed two deleted posts; I assumed they were malfunctions. I should have been aware though, because she did the same thing with her Sapchats, leaving 40+ followers without a history, except for a statement about being too lazy to post. It’s not true; no one has that many followers with no posts.
Her aberrant behavior seems to be integral to her current personality. If she does return, it’s likely her behavior will too. One of my psychology professors is always interested in anecdotal observations of anomalous behavior. I may relay this to him in the fall.
Two years ago, Nauseated noticed a forum member who started deleting his older posts—his answers to questions. He deleted six pages worth, and then quit doing it after posting the reason why. You may remember this: http://web2.0calc.com/questions/good-luck-with-this-one#r11
Naus said, “It’s only a few picoteslas. It’s not a great loss. I’m not sure if he was serious or being his trolling self—probably both. He did add if Heureka, Alan, CPhill, Rom, Bretie, or Melody did it, then it would be like having a major fire in the mathematics section of a unique library.
In the case of Miranda, I think she should have pressed ‘Ctrl-Alt’ before pressing delete. Her questions are not unique, but the solution presentations are, and those questions were the signposts for the pathways.
Maybe we can restore a few of these. I can post the questions as a guest and the original answerers can copy their answers to the new post.
GA
P.S. Have a nice day
.
+2446 If we think about the expansion of a binomial of a big power (like 7). The expansion looks like the following:
\((a+b)^n={n\choose 0}a^nb^0+{n\choose 1}a^{n-1}b^1+...+{n\choose n}a^0b^n\)
Knowing that this pattern exists, let's utilize that in the current expansion of the expression:
\((2x+3)^7={7\choose0}(2x)^7(3)^0+{7\choose1}(2x)^6(3)^1+{7\choose2}(2x)^5(3)^2+{7\choose3}(2x)^4(3)^3+{7\choose4}(2x)^3(3)^4+...\)
Why don't I care about the other terms? Well, I only care about the terms that has the x^5. If it doesn't have that, then I can ignore that term. In this case the appropriate term is \({7\choose 2}(2x)^5(3)^2\). Let's evaluate it:
| \({7\choose 2}(2x)^5(3)^2\) | First, let's evaluate the function that has the 7 choose 2. This has a formula to it. |
The formula for the choose function is the following:
\({x \choose y}=\frac{x!}{y!(x-y)!}\)
Let's apply that to \({7\choose 2}\) first:
| \(7\choose2\) | Apply the formula to figure out its true value |
| \(\frac{7!}{2!(7-2)!}\) | |
| \(\frac{7!}{2!*5!}\) | Let's expand the factorial function. |
| \(\frac{7*6*5*4*3*2*1}{2*1*5*4*3*2*1}\) | It is very obvious that there is a common factor of 5! in both the numerator and denominator, so let's factor that out. |
| \(\frac{7*6}{2}\) | Simplify the numerator completely. |
| \(\frac{42}{2}=21\) | |
Ok, now let's continue with our problem:
| \({7\choose 2}(2x)^5(3)^2\) | Replace \({7\choose 2}\) with its calculated value, 21. |
| \(21*(2x)^5*3^2\) | Distribute the power to the 5 to 2 and the x |
| \(21*2^5x^5*3^2\) | Simplify all the constants. |
| \(21*32*x^5*9\) | Do the multiplication of all the constants. |
| \(6048x^5\) | |
Therefore, the coefficient in front of the x^5 is 6048.
