If you're interested in determining which player's game performances show the least variability, you'd want to identify the smallest and largest number of points (minimum and max).
Answer:
1. f(x) is the total cost
2. x is the hours spent on the internet
3. the base rate is 3.50
4. hourly charge is .75
5. 4.63
Step-by-step explanation:
1. f(x) is the total cost or the amount of money he has to pay
2. x is the hours spent on the internet, it is the time spent on the internet
3. the base rate is 3.50, or the minimum amount just to get connected
4. hourly charge is .75
5. Let x = 1.5 for the hour and a half
f(1.5) = .75(1.5) + 3.5
1.125+3.5
4.625
To the nearest cent
4.63
Answer:
16.9 units
Step-by-step explanation:
Sometimes the easiest way to work these problems is to get a little help from technology. The GeoGebra program/app can tell you the length of a "polyline", but it takes an extra segment to complete the perimeter. It shows the perimeter to be ...
14.87 + 2 = 16.87 ≈ 16.9 . . . units
_____
The distance formula can be used to find the lengths of individual segments. It tells you ...
d = √((Δx)² +(Δy)²)
where Δx and Δy are the differences between x- and y-coordinates of the segment end points.
If the segments are labeled A, B, C, D, E in order, the distances are ...
AB = √(5²+1²) = √26 ≈ 5.099
BC = √(1²+3²) = √10 ≈ 3.162
CD = Δx = 3
DE = √(3²+2²) = √13 ≈ 3.606
EA = Δy = 2
Then the perimeter is ...
P = AB +BC +CD +DE +EA = 5.099 +3.162 +3 +3.606 +2 = 16.867
P ≈ 16.9
x=2 is only solution while x=1 is extraneous solution
Option C is correct.
Step-by-step explanation:
We need to solve the equation
and find values of x.
Solving:
Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)
Multiply the entire equation with x(x-1)
![\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-1%7D%2B%5Cfrac%7B2%7D%7Bx%7D%3D%5Cfrac%7Bx%7D%7Bx-1%7D%5C%5C%5Cfrac%7B1%7D%7Bx-1%7D%2Ax%28x-1%29%2B%5Cfrac%7B2%7D%7Bx%7D%2Ax%28x-1%29%3D%5Cfrac%7Bx%7D%7Bx-1%7D%2Ax%28x-1%29%5C%5CCancelling%5C%2C%5C%2Cout%5C%2C%5C%2Cthe%5C%2C%5C%2Csame%5C%2C%5C%2Cterms%3A%5C%5Cx%2B2%28x-1%29%3Dx%5E2%5C%5Cx%2B2x-2%3Dx%5E2%5C%5C3x-2%3Dx%5E2%5C%5Cx%5E2-3x%2B2%3D0)
Now, factoring the term:
![x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2](https://tex.z-dn.net/?f=x%5E2-2x-x%2B2%3D0%5C%5Cx%28x-2%29-1%28x-2%29%3D0%5C%5C%28x-1%29%28x-2%29%3D0%5C%5Cx-1%3D0%5C%2C%5C%2Cand%5C%2C%5C%2C%20x-2%3D0%5C%5Cx%3D1%5C%2C%5C%2Cand%5C%2C%5C%2C%20x%3D2)
The values of x are x=1 and x=2
Checking for extraneous roots:
Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.
If we put x=1 in the equation,
the denominator becomes zero i.e
which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.
If we put x=2 in the equation,
![\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-1%7D%2B%5Cfrac%7B2%7D%7Bx%7D%3D%5Cfrac%7Bx%7D%7Bx-1%7D)
![\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2-1%7D%2B%5Cfrac%7B2%7D%7B2%7D%3D%5Cfrac%7B2%7D%7B2-1%7D%5C%5C%5Cfrac%7B1%7D%7B1%7D%2B1%3D%5Cfrac%7B2%7D%7B1%7D%5C%5C1%2B1%3D2%5C%5C2%3D2)
So, x=2 is only solution while x=1 is extraneous solution
Option C is correct.
Keywords: Solving Equations and checking extraneous solution
Learn more about Solving Equations and checking extraneous solution at:
#learnwithBrainly
Answer:
(1,-4)
Step-by-step explanation:
First manipulate the second equation so that y is by itself
x-y=5
-y=-x+5
y=x-5
then substitute that y in for the y in the above equation and solve for y
y=-2x-2
x-5=-2x-2
3x-5=-2
3x=3
x=1
Now that you have the x-coordinate, find the y-coordinate by plugging in the x-value in either of the equations
y=-2x-2
y=-2(1)-2
y=-2-2
y=-4
I see that you're using deltamath, and I think for the graph questions, you could first place a point for the y-intercept of one of the equations, and then place another point that is on the graph of the equation. Once you plotted both lines, just see the point where they meet and that is where the two systems of equations meet.