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I am not the brightest, but need an example on how to solve problems like these
identify the polynomials of degree one
500+(45)1/2x
289y+6-76y
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also find f(1) for f(x)=x^3x^2-1
A function gives the value of c as 2x(22/7)xr find c when r = 14cm and r = 70cm
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To solve f(x)=x³x²-1 for f(1), you simple plug in 1 for x.
Same concept for the second part, except you will need to solve the equation twice. For the first one plug in 14 for r, in the next plug in 70:
1. 2x(22/7)x(14)
2. 2x(22/7)x(70)
e...the red-headed stepchild of math.
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I still need help for the first one, find the polyomials of degree one
500+(45)1/2x
289y6-76y
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Okay I have another one for you. I do not nderstand what they want with this.
1. Formulate a three word problem from day to day life that can be translated into linear equation ione variable, two variables, and
three variables respectively.
2. Write a systm of equations having
1. a unique solution
2. An infinite number of solutions
3. No solution
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Okay if there is anyone out there who can help tonight, I would greatly appreciate it.
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2. Write a systm of equations having
1. a unique solution
2. An infinite number of solutions
3. No solution
1. A unique solution:
3x+5y=13
5x-2y=1
2. An infinite number of solutions:
2x+5y=26
8x+20y=104
3. No solution where x and y belong to Natural numbers:
x+y=5
x-y=5
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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To get a system of equations that has no real solution, make them so that if you graphed them, you'd get two lines that were parallel but didn't overlap.
Why did the vector cross the road?
It wanted to be normal.
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I have some more for you.
A new virus is released on the Internet; the administrator of a department's Local area Network (LAN) is given five minutes by a manager to estimate the impact. The administrator samples 12 of the PC's connected to the LAN, and finds that 7 are infected; use proportion to estimate the number of infected PC's if there are a total of 117 PC's connected to the LAN.
An administrator of a popular web site is told that a new server can handle 11,000 "hits" (users accessing the site) per second. the web site currently experiences a peak demand of about 85,000 hits per second; but every month the peak demand increases by 3500 hits per second. Use proportion equation to determine how many new servers the administrator should buy to address expected traffic for the next 18 months.
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1, why do intersecting lines represent a unique solutions?
2. What is the significance of the name "linear equation" to its graphical representation?
3. The solutions of line (a) are (3,3),(5,5),(15,15),(34,34),(68.678),(1234,1234).
4 The solutions of line (b) are (3,-3),(5,-5),(15,-15),(34-34),(678,-678),(1234,-1234)
a. Form the equations of both the lines.
b. What are co-ordinates of the point of intersection of line (a) and (b)
c. Write the co-ordinates of the intersections of lines (a) and (b) with the x-axis
d. Write the co-ordinates of the intersection of lines (a) and (b) with the y-axis.
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If it is possible, could someone help me with this today? I would greatly appreciate it.
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1. Intersecting lines represent a unique solution because if two lines intersect then they only overlap at one point. Therefore there is only one answer that satisfies both equations of the lines.
2. The term "linear" means that when the equation is graphed the picture will be a line.
3,4. For a you will need two points from each line: for a use (3,3) and (5,5) and plug these into the slope equation (m=(y2-y1)/(x2-x1) and in this case: (5-3)/(5-3). S the slope is 1. Now plug this into your slope intercept equation (y=mx+b). and you get y=x+b, but you still need to find b. so plug in the point (3,3) and you get 3=3+b and solve for b. b=0. So now you have everything you need for the equation (m=1, b=0). Plug this back in to the slope intercept equation and you get a final answer of: y=x. Do the same thing with two points from line b to get it's equation.
What you need to do for this one is plot each point for line a then draw the line, then do the same for b. Once you have the two pictures you can see the answers to b,c, and d.
e...the red-headed stepchild of math.
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What about the problem before this last one?
Could someone set up the equation for these next two problems?
1. Farmer Jones has horses and roosters. All together here are 88feet and 40 wings. How many horses and how many roosters
does he have?
2. Two people are traveling towards each other. They are 100 miles apart, one is traveling 30 per hour and the other is traveling 20 per hour. How long will it take for them to meet?
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I do not know what you mean by "plug"
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On question #7 How do I know that each system yields the solution type of interest, unique, infinite, and no solution respectively. I need to provide some type of reasoning to support these conclusions. thanks
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When I say plug in a number for x I mean replace every "x" in the problem with the given number.
e...the red-headed stepchild of math.
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What about the problem before this last one?
Could someone set up the equation for these next two problems?
1. Farmer Jones has horses and roosters. All together here are 88feet and 40 wings. How many horses and how many roosters
does he have?
2. Two people are traveling towards each other. They are 100 miles apart, one is traveling 30 per hour and the other is traveling 20 per hour. How long will it take for them to meet?
1 can be answered using linear equations in two variables:
Let x=the number of horses
y=the number of roosters
4x + 2y = 88 (since horses have four feet, roosters have two)
2y = 40 (roosters have 2 wings each, and I doubt Pegasus is one of the horses:D)
By simplifying the second equation, we can see that there are 20 roosters. all we need to know now is the number of horses, x.
By subtracting the 2nd equation from the first, we will get
4x + 2y = 88
- 2y = 40
4x = 48
x = 12
so, you have 20 roosters, and 12 cows.
2 also can be solved using linear equations in two variables, I think. However, I forgot how. Sorry!
"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"
Nisi Quam Primum, Nequequam
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I need somebody to answer #15 please.
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I still need help with question 15
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which is 15?
"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"
Nisi Quam Primum, Nequequam
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How do I know that each system yields the solution type of interest, unique, infinite, anno solution respectively. I need to provide some type of reasonong to support these conclusions.
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One way that's already been said is to draw each equation as a line on a graph and see how they relate to each other.
If the lines cross, then the system has a unique solution.
If the lines are exactly the same, then the system has infinite solutions.
If the lines do not cross (because they are parallel and in different places), then there are no solutions.
This makes sense if you think of any potential solution (x,y) as being a point on the graph. For (x,y) to satisfy equation I, it has to be on line I. For it to satisfy equation II, it has to be on line II. Therefore, for it to be a solution of the system, it has to be on both lines.
Clearly it can't do that if the lines never cross. Similarly, there is exactly a unique solution that can do that if the lines cross once, and there are infinite solutions that can do that if the lines are on top of each other.
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Another way of doing it that you might find easier is to work directly with the equations.
(You might not have been taught this yet - if not, sticking with graphs will probably be simpler)
You're allowed to add or take away one equation to or from the other, and you can also multiply everything in one equation by some constant.
So, if you were given:
I: x + 2y = 9
II: x + 3y = 13
Then you could add them together to get 2x + 5y = 22.
You could also multiply the first equation by 3 to get 3x + 6y = 27.
The trick is to use these tools to try to get rid of one of the variables.
In this case, you could take I away from II to get y = 4.
Sometimes you might have to multiply an equation by something before you can add or take it away from the other one.
There are three types of equations that you can get after doing this. Here are examples for each:
(i) 3x = 2.
Here, you have a variable on one side and a constant on the other. If you end up with this, then your system has a unique solution.
(ii) 3 = 3.
Here, you have a true statement that involves no variables. You could get this if you happen to cancel both variables at once. In this case, the system has infinite solutions.
(iii) 4=5.
Similar to (ii), except this time the variable-free statement is false. In this case, the system has no solutions.
Why did the vector cross the road?
It wanted to be normal.
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Please help me with this!!!!!
Write three quadratic equation, with a, b, and c (coeffients of x^2, x and the constant as:
1. integer
2. rational numbers
3. irrational numbers
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I need you all to explain something for me.
Do you find any striking difference between the graphical representation of quadratic equations and linear equation. Explain the differences.
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I need you all to explain something for me.
Do you find any striking difference between the graphical representation of quadratic equations and linear equation. Explain the differences.
The blue is quadratic, the red is linear.
Last edited by Daniel123 (2008-06-05 02:45:39)
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