+2446 Before we can calculate the value of \(\frac{7}{10}+\frac{7}{12}\), we must create a common denominator. The LCD of both the denominators is 60.
In this step, I will convert \(\frac{7}{10}\) to a fraction with a denominator of 60.
| \(\frac{7}{10}\) | Multiply the numerator and denominator by 6/6. |
| \(\frac{7}{10}*\frac{6}{6}\) | Notice that multiplying by 6/6 is actually multiplying by 1, which means that the value of the fraction remains unchanged. |
| \(\frac{42}{60}\) | |
And now I will manipulate \(\frac{7}{12}\) as well.
| \(\frac{7}{12}*\frac{5}{5}\) | Yet again, we multiply the fraction by 1. |
| \(\frac{35}{60}\) | |
Since the denominators are the same in both fractions, now we can add them together.
| \(\frac{42}{60}+\frac{35}{60}\) | Add the numerator and preserce the denominator. |
| \(\frac{77}{60}=1\frac{17}{60}=1.28\overline{33}\) | |
+2446 Solving an inequality is similar to solving a regular equation; you are isolating the variable, in this case K.
| \(\frac{K}{3}\geq-5\) | Multiply by 3 on both sides to isolate K. |
| \(K\geq-15\) | |
The solutions, if you are wondering, are values for K that satisfies its current restriction (K is greater than or equal to -15). Examples of solutions are \(-14,-13,-12.5,0,\frac{2}{3},1.\overline{78},\pi, \sqrt{21}\).
+2446 Hello! That's quite an interesting expression you got there of \((a-7)^2-2(a-7)(a+7)+a+7\). Let's simplify it the we can.
| \((a-7)^2-2(a-7)(a+7)+a+7\) | First, let's deal with \((a-7)^2\) by knowing the expansion of a binomial squared. In general, it is \((x-y)^2=x^2-2xy+y^2\). |
| \((a-7)^2=a^2-14a+49\) | |
| \(a^2-14a+49-2(a-7)(a+7)+a+7\) | Now, let's expand \((a-7)(a+7)\) by using the following rule again of \((x+y)(x-y)=x^2-y^2\). |
| \(-2(a-7)(a+7)=-2(a^2-49)\) | Distribute the -2 to both terms in the parentheses. |
| \(-2(a^2-49)=-2a^2+98\) | |
| \(a^2-14a+49-2a^2+98+a+7\) | Let's rearrange the equation such that all the terms with the same degree are adjacent. |
| \(-2a^2+a^2-14a+a+98+49+7\) | Now, simplify. |
| \(-a^2-13a+154\) | |
Therefore, \((a-7)^2-2(a-7)(a+7)+a+7=-a^2-13a+154\).
+2446 For A, I think I know why it is not working properly. You appear to have done everything correctly, so I presume that there is a silly computational error.
1. Determine the Midpoint of Both Coordinates
The midpoint is the average of both x- and y- coordinates. In algabraic terms, the formula is \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\). Let's use it! You used this formula correctly, as I can tell. For completeness, I'll calculate it also.
| \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\) | Now, just plug into the formula. |
| \(\left(\frac{3+6}{2},\frac{3+2}{2}\right)\) | Now, simplify both sides. |
| \(\left(\frac{9}{2},\frac{5}{2}\right)\) | |
2. Determine Slope (or Gradient) of the Segment
This also has a formula, luckily, and all we have to do is plug the coordinates in. It is \(m=\frac{y_2-y_1}{x_2-x_1}\), where m is the slope of the segment. Let's just use the formual to determine the slope.
| \(m=\frac{y_2-y_1}{x_2-x_1}\) | When using this formula, it is essential to plug in the x- and y-coordinates in the same order. Otherwise, you will get the incorrect result. |
| \(m=\frac{3-2}{3-6}\) | Now, simplify the numerator and denominator. |
| \(m=-\frac{1}{3}\) | |
This is not the slope of the perpendicular bisector but rather the slope of the original 2 segments.
3. Determine The Slope of Perpendicular Bisector
A line that is perpendicular to a segment has a slope that is opposite and reciprocal to the slope of the original segment. What does that mean? Well, let's deal with the current slope as its example of \(-\frac{1}{3}\).
I find doing the reciprocal of a number first is generally easier, but the order in which you do the next steps do not matter. In general, the reciprocal of a number follows tthe following rule of \(a\Rightarrow\frac{1}{a}\). Let's apply that now with our slope. \(-\frac{1}{3}\Rightarrow-\frac{1}{\frac{1}{3}}=-3\)
Now, find the opposite. To do this, use the rule that \(a\Rightarrow-a\). \(-3\Rightarrow3\).
Therefore, \(3\) is the slope of the perpendicular bisector.
If you are interested, here is another method to obtain the slope of the perpendicular bisector. If you take the original slope of the segment and multiply it by the slope of the perpendicular bisector, you will get -1. Let's see this in action!
| \(-\frac{1}{3}m=-1\) | Multiply by -3 on both sides. |
| \(m=3\) | |
Look at that! you get the same answer.
4. Use Point-Slope Form
There are 2 methods again to get to the final equation of the slope, but I generally like point-slope form for some reason, but I will demonstrate both to you, so that you can choose your method. Point-slope form is the following:
\(y-y_1=m(x-x_1)\) where \((x_1,y_1)\) is a point on said segment or line and m is the slope of said segment or line. We know what segment is on this perpendicular bisector; we already calculated it. The perpendicular bisector is a segment that both intersects a segment at its midpoint and at a perpendicular angle. We have already calculated the midpoint of BC, which is \(\left(\frac{9}{2},\frac{5}{2}\right)\). We also know the slope, which is 3. Since we have all that information, let's plug it in!
| \(y-y_1=m(x-x_1)\) | As aforementioned, plug all the information we know into the equation. |
| \(y-\frac{5}{2}=3\left(x-\frac{9}{2}\right)\) | Now, we are transforming this equation to slope-intercept form. To do that, let's distribute the 3 to both sides. |
| \(y-\frac{5}{2}=3x-\frac{27}{2}\) | We know that we must do some manipulation with the fraction. |
| \(y-\frac{5}{2}=3x+\frac{-27}{2}\) | Now, add \(\frac{5}{2}\) |
| \(y=3x-11\) |
This is the equation of the perpendicular bisector. I'm not sure where you went wrong, unfortunately, but it seems to me that it is probably some silly computational error. That's all.
Another method of doing this is using the slope-intercept form from the beginning. Let's see how that would work:
| \(y=mx+b\) | We know what the slope is; it is 3, so let's plug that in. |
| \(y=3x+b\) | Now, plug in a point that we know is on the perpendicular bisector. We know that \(\left(\frac{9}{2},\frac{5}{2}\right)\) is a point, so plug it in. |
| \(\frac{5}{2}=\frac{3}{1}*\frac{9}{2}+b\) | Now, solve for b. |
| \(\frac{5}{2}=\frac{27}{2}+b\) | Subtract 27/2 on both sides. |
| \(b=-\frac{22}{2}=-11\) | Therefore, the equation becomes the following. |
| \(y=3x-11\) | This should look very familiar to what we determined above. |
Hopefully, this helped to find the segment that is the perpendicular bisector.
Now, it is time to do B. Thankfully, this will not be as time-consuming as generating this repsonse has taken me hours. The point you want me to identify is called the circumcenter of the triangle. it is the point located at the center of a circle such that circle contains all the points in a triangle. The circumcenter of a triangle can be found by finding the intersection of 2 of the perpendicular bisectors! Let's do that. We already know 2 lines and their respective equations. They are \(y=3x-11\) and \(y+2x=4\). Let's set up a system of equations.
{ \(y=3x-11\)
{ \(y+2x=4\)
I will use substitution in this case as we already know what y equals.
| \(3x-11+2x=4\) | Combine the like terms on the left-hand side. |
| \(5x-11=4\) | Add 11 to both sides. |
| \(5x=15\) | Divide by 5 on both sides to get the corresponding x-coordinate. |
| \(x=3\) | |
Now, plug in x into one of the equations. I think that equation 1 will be easier, so I will use that!
| \(y=3x-11\) | Plug in for x, which is 3. |
| \(y=3*3-11=9-11=-2\) | |
Therefore, the circumcenter of the triangle is \((3,-2)\).
If you would like, click here to view a graph of the triangle, its respective lines that are perpendicular bisectors and the circumcenter.
+2446 \(35*35=1225\)
I believe you want an easier method to work out problems like these where you square a number. Here is an idea for you.
\(35^2=?\)
When you square a number, this is a trick you can use! Round the number to the nearest multiple of 10. In this case, that is 30. Now, go up by the distance you went down. In this example, you went down by 5 to get to 30. Now, go up 5 to get to 40. Now multiply them together.
This should be much simpler now. \(30*40=1200\). All you need to do now is to add the square of the distance. In this case, it is \(5^2=25\).
\(1200+25=1225\)
That's much easier, don't you think? In fact, this is usually so easy that you may be able to do it mentally.
But why does it work? Well, that is actually easier to explan than you may think. Let's call the number we want to square A. Let's call the distance we want to go up and down d. Therefore, we get the following equation:
| \(A^2=(A-d)(A+d)+d^2\) | Now, expand the multiplcation of 2 binomials. |
| \(A^2=A^2-d^2+d^2\) | Now, simplify. |
| \(A^2=A^2\) | |
This means that no matter what distance you pick, you will eventually make it the original number squared. That's pretty cool!
