#1**+1 **

\((x-h)^2+(y-k)^2=r^2\) is the center-radius form for a circle such that \((h,k)\) is the coordinates of the center of the circle on the coordinate plane and \(r\) is the radius.

1) Find the equation of the circle centered at (1,-4) with radius 3

Since \((h,k)\) of \((x-h)^2+(y-k)^2=r^2\) represents the coordinates of the center and \((1,-4)\) is already known to be the center, then we know that \(h=1\text{ and }k=-4\) . \(r\) is the radius, so r=3.

Substitute these values into the center-radius form of a circle to generate the equation of this particular circle.

\(h=1,k=-4,r=3\\ (x-h)^2+(y-k)^2=r^2\\ (x-1)^2+(y-(-4))^2=3^2\\ (x-1)^2+(y+4)^2=9\)

2) I encourage you to use the information I gave you above to try and work out the second problem.

3) Find the distance between the points \((0,8)\text{ and }(2,-5)\).

We can use the distance formula, \(d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\) . Substitute the points that we want to find the distance from.

\(d=\sqrt{(8-(-5))^2+(0-2)^2}\\ d=\sqrt{13^2+(-2)^2}\\ d=\sqrt{169+4}\\ d=\sqrt{173}\) | Substitute in the known coordinates and simplify completely. 173 has no perfect square factors, so the radical cannot be simplified any further. |

TheXSquaredFactorFeb 17, 2019

#2**+2 **

If we assume that "the graph consists of three line segments," then we can generate the equation of the lines.

Of course, I could use \(y=mx+b\), but that would require me to know the y-intercept of the first red line, and, unfortunately, the y-intercept is not clear-cut in the image.

I will use point-slope form instead; this form requires me to know the coordinates of **any ** two points on that line. \(y-y_1=m(x-x_1)\) is the form of point-slope form. \((x_1,y_1)\) is the coordinate of any one point of the line. In the image, I can pinpoint that \((-4,4)\text{ and }(-1,-1)\) both lie on the line I care about for this problem. Now that they are identified, I can now find the slope.

\(m=\frac{-1-4}{-1-(-4)}=\frac{-5}{3}\)

I will substitute in the point \((-1,-1)\) as my point.

Since we are trying to find \(f(-2)\) , x=-2.

\(y_1=-1;m=-\frac{5}{3};x=-2;x_1=-1\\ y-y_1=m(x-x_1)\\ y-(-1)=-\frac{5}{3}(-2-(-1))\) | It is time to solve for y! |

\(y+1=-\frac{5}{3}*-1\\ y+\frac{3}{3}=\frac{5}{3}\\ y=\frac{2}{3}\) | |

\(f(-2)=\frac{2}{3}\). You will see that this answer is consistent with the initially given diagram.

** **

TheXSquaredFactorFeb 15, 2019

#1**+1 **

Given three points, the simplest equation that I can think of would be a quadratic. The standard form of a quadratic is written as \(ax^2+bx+c\) , where we can tweak a,b, and c so that it intersects the points listed above.

\(f(x)=ax^2+bx+c\) | Let's substitute in the individual values of x into these equations. |

\(\boxed{1}\quad f(8)=a*8^2+b*8+c\\ \boxed{1}\quad 1=64a+8b+c\) | Substitute in x=8 to create the first equation. Simplify completely. |

\(\boxed{2}\quad f(10)=a*10^2+b*10+c\\ \boxed{2}\quad 4=100a+10b+c\) | Substitute in x=10 to create the second equation. Simplify completely. |

\(\boxed{3}\quad f(13)=a*13^2+b*13+c\\ \boxed{3}\quad 8=169a+13b+c\) | Substitute in x=13 to create the second equation. Simplify completely. |

Notice that we have now generated a three-variable system of equations. Let's solve this for all the variables. I tried my best to pick the best path possible. I have numbered all the equations so that you can refer to them easily.

\(\boxed{1}\quad 1=64a+8b+c\\ \boxed{4}\quad -1=-64-8b-c\) | Notice that all I did here was to negate all the sides. I do this so that I can eliminate the "c" values from two of the equations. |

\(\boxed{2}\quad 4=100a+10b+c\\ \boxed{4}\quad -1=-64a-8b-c\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \boxed{5}\quad 3=36a+2b\) | I added equation 2 and 4 together to generate a new equation, 5, with one fewer variable. |

\(\boxed{3}\quad 8=169a+13b+c\\ \boxed{6}\quad -8=-169a-13b-c\) | I have done the exact same process as above; I negated both sides of equation 3. This is for the same purpose; eliminate "c." |

\(\boxed{6}\quad -8=-169a-13b-c\\ \boxed{2}\quad 4=100a+10b+c\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \boxed{7}-4=-69a-3b\) | Yet again, I added equations 6 and 2 to create another equation with one fewer variable. |

We have now successfully eliminated "c" in two instances; it is now time to use these two new equations in order to isolate another variable. I will still use elimination.

\(\boxed{5}\quad 3=36a+2b\\ \boxed{8}\quad 9=108a+6b\) | In order to create equation 8, I multiplied both sides of equation 5 by 3. |

\(\boxed{7}-4=-69a-3b\\ \boxed{9}-8=-138a-6b\) | In order to create equation 9, I multiplied both sides of equation 7 by 2. |

\(\boxed{8}\quad 9=108a+6b\\ \boxed{9}\quad -8=-138a-6b\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \boxed{10}1=-30a\) | Finally! We have reduced this to one variable only! |

\(a=-\frac{1}{30}\) | We now have the value of "a," Let's use this information to help us fill out the rest. |

Let's now substitute this back into one of the equations and find "b."

\(a=-\frac{1}{30}\\ \boxed{5}\quad 3=36a+2b\\ 3=36*-\frac{1}{30}+2b\) | I have substituted in the value of "a" into equation 5. I can now solve for b. I will multiply both sides of this equation by 30 to eliminate the presence of those fractions. |

\(90=-36+60b\\ 15=-6+10b\\ 21=10b\\ b=\frac{21}{10}\) | Notice how I decided to divide the equation by 6 immediately to simplify it into more manageable numbers. While this is not strictly necessary, I think this is always a good strategy to use when equations get out of hand. Look at that! We have the value of "b!" |

Equation 2 looks like the easiest one to substitute back into so that is the one that I will do.

\(a=-\frac{1}{30}; b=\frac{21}{10}\\ \boxed{2}\quad 4=100a+10b+c\\ 4=100*-\frac{1}{30}+10*\frac{21}{10}+c\) | You might be able to see why I preferred equation 2. The substitution appears to be the path of least resistance. |

\(4=-\frac{10}{3}+21+c\\ -17=-\frac{10}{3}+c\\ 51=10-3c\\ 41=-3c\\ c=-\frac{41}{3}\) | We have finally solved for the final unknown. |

The simplest equation I could come up with is \(f\left(x\right)=-\frac{1}{30}x^2+\frac{21}{10}x-\frac{41}{3}\)

.TheXSquaredFactorFeb 15, 2019

#1**+1 **

Hello, APatel!

A perfect square is the product of a whole number multiplied by itself. Below is a list of the first few perfect squares, in order.

x | x^2 |

1 | 1 |

2 | 4 |

3 | 9 |

4 | 16 |

5 | 25 |

6 | 36 |

7 | 49 |

8 | 64 |

9 | 81 |

Let *x *equal a positive integer. Let *x^2 *equal any perfect square.

Let *x+1 *equal the subsequent positive integer. Let *(x+1)^2 *equal the next perfect square.

Notice how this is very similar to how I generated consecutive perfect squares in the table; I incremented the "x" column by 1 and squared the result. Also, notice that the question asks for the positive difference of the perfect squares. \((x+1)^2>x^2\) because *(x+1)^2 *represents the next perfect square. Therefore, it is possible to generate the following equation for this problem:

\((x+1)^2-x^2=59\) | Expand the binomial. |

\(x^2+2x+1-x^2=59\) | Look at that! The quadratic terms cancel out, and we are left with quite a basic equation. |

\(2x+1=59\) | Now, subtract one on both sides. |

\(2x=58\) | Divide by 2 on both sides. |

\(x=29\) | |

We are not done yet! Remember, our goal is to find the greater of the perfect squares. Of course, in this problem, I let *(x+1)^2 *represent the larger one.

\(x=29\\ (x+1)^2\) | Substitute 29 into this expression and simplify accordingly. |

\((29+1)^2\) | |

\(30^2\) | |

\(900\) | This is the larger of the 2 perfect squares. |

TheXSquaredFactorFeb 7, 2019

#1**+3 **

The average hourly wage h(x) of workers in an industry is modeled by the function where \(h(x)=\frac{16.24x}{0.062x+39.42}\) x represents the number of years since 1970.

a) *x *represents the number of years since 1970.

b) *h(x)* represents [t]he average hourly wage ... of workers

c) This question asks for *h(x) *given an input *x*.

Since *x *represents the number of years since 1970, we will have to find the number of years that have elapsed by using subtraction.

\(x=1993-1970=23\text{ years}\)

Now, let's find *h(x)*:

\(x=23\\ h(x)=\frac{16.24x}{0.062x+39.42}\) | This is the model of the hourly wage. |

\(h(23)=\frac{16.24*23}{0.062*23+39.42}\\\ h(23)=9.14459188...\approx\$9\) | I inputted the number of years and rounded appropriately to the nearest dollar |

d)

Since [t]he average hourly wage ... of workers is assumed to be $25, *h**(x)* =25, and we are solving for *x:*

\(h(x)=25\\ h(x)=\frac{16.24x}{0.062x+39.42}\) | Substitute in 25 for h(x) |

\(25=\frac{16.24x}{0.062x+39.42}\\ 25(0.062x+39.42)=16.24x\\ 1.55x+985.5=16.24x\\ 985.5=14.69x\\ x=\frac{985.5}{14.69}\approx 67.09\text{ years} \) | Solve for x by multiplying by the LCD, 0.062x+39.42 |

67 years after 1970 certainly lands in 2037, but the extra 0.09 years would fall into the following year, so the hourly wage will hit $25 per hour in 2038.

TheXSquaredFactorFeb 3, 2019

#2**+1 **

60° and 240° are not the only solutions in the interval 0 < x < 360. 120º and 300º are both perfectly valid.

Let's isolate the trigonometric term and see where to go from there.

\(4\sin^2x=3\) | Divide by 4 on both sides. |

\(\sin^2x=\frac{3}{4}\) | Take the square root of both sides. |

\(|\sin x|=\sqrt{\frac{3}{4}}\) \(|\sin x|=\frac{\sqrt{3}}{2}\) | Use the definition of the absolute value to split this into two separate equations. |

\(\sin x=\frac{\sqrt{3}}{2}\\ \sin x=\frac{-\sqrt{3}}{2}\) | |

Here, you locate which angle yields an answer of \(\pm\frac{\sqrt{3}}{2}\) , which is \(\{60^{\circ},120^{\circ},240^{\circ},300^{\circ}\}\)

.TheXSquaredFactorFeb 3, 2019