+2446 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.
+2446 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 \(f\left(x\right)=-\frac{1}{30}x^2+\frac{21}{10}x-\frac{41}{3}\)
.+2446 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. |
+2446 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.
+2446 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 \(\{60^{\circ},120^{\circ},240^{\circ},300^{\circ}\}\)
.