Linear Algebra

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Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.

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Algebra is used to solve simple problems involving know and unknown variables.

25

For any fraction to equal zero, the numerator has to equal zero. So, we have to make 1-x-2=0. Simplify and get x=-1.

If you meant 1-x^2, then do the same thing: 1-x^2=0. Simplify and get x=±1

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There is no equation given in the question. However, the x-intercept can be obtained by letting y = 0 in the equation and solving for x.

problem:

The product of two whole numbers is 10,000 . If neither number contains zero, what are the numbers?????

solution:

prime factorization

10,000

^

10*1000

^ ^

2*5 100*10

^ ^

10*10 2*5

^ ^

2*5 2*5

prime factors:2*2*2*2*5*5*5*5 = 10,000

first get the value of 2^4

2*2*2*2= 16

then get the value of 5^4

5*5*5*5= 625

Now lets check...

16 * 625 = 10,000 ok

16 and 625 doesnt have any 0 ok

Another question:

If f(x)=x+1 and g(x)=(3x+1)^-2, find the domain of g o f?

1/(3x+1)^2 or 1/(3x+1)(3x+1)

since x ≠ -1/3 will only be repeated, we'll only use x ≠ -1/3

X+1≠-1/3 x≠-1 1/3 or x≠-4/3

{x|x≠-4/3}

It must be a first class in M.Sc. Applied Mathematics.

by using decimals

no.

Every equation has a solution.

I don't know you meant the ugly and large numbers, or linear single variable equations. For the former, you just need patience, practice and either a mad brain I don't have or a calculator which I do have.

For the latter, there is a way to do it. For any Linear equation, we can always simplify it to this form:

a + bx = 0 where b is NOT 0. (if it is, then a must be 0, or you've goofed somewhere and yes, it's a 0 on the right hand side RHS)

Then, bx = -a, x = -a/b

Of course that's the easy part, the hard part is to put it into a + bx = 0 form

This, I can't really give you a generic case, only that use addition and subtraction to move everything to one side. Multiplication if necessary. If you are using division somewhere, you might not be working with a linear equation.

10

No, if two lines are parallel they will not have a solution.

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Slant asymptotes are obscure and those will not be used in any decent class, but they are found by long dividing and using only the polynomial part, not the remainder over the divisor, and that is the equation for the slant asymptote. For example, ( -3x^2 + 2 ) / ( x - 1 ) when long divided out becomes -3x - 3 - ( 1 / x - 1 ). The last term drops and it becomes -3x - 3. Horizontal asymptotes are found by taking the limit of the function as x tends towards infinity and negative infinity. Vertical ones are found by finding points where the graph is undefined (where the denominator is 0).

such as travelling through an auto rishaw or by any taxi

more examples can be for increase in prises in any field

They are simply algebaric equations. They may be linear or non-linear, continuous or not, in two or more variables.

Any equation can be plotted on a coordinate grid and any point, curve, shape on a coordinate grid can be represented by an equation. However, finding the appropriate equation will not always be easy.

Because the question is tautological. You are asking how something is the same as that very samne thing!

You can have any number of free variables. If you have m variables and n linear equations then, if m > n, you will have at least (m - n) free variables.

divide top number by bottom number

Linear because neither of the variables are raised to a power.

Elimination and substitution are two methods.

It is an equation which, if plotted will be represented by a straight line (hence linear).

There are infinite number of negative non integers that have a product of 20. Any two negative fractions can make a product of 20. e.g.

-3/20 * -800/6 = 20

-3/8 * -320/6 = 20

-9/7 * -140/9 = 20

The list is endless. Only make sure you select both negative fractions to get a positive 20 and when selecting the fractions the numerator of one number is divisible by the other number's denominator such that the factors cancel to give 20.