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Nana76 [90]
3 years ago
13

Suppose that a number written as a decimal has an infinite number of non-repeating digits after the decimal point. What is this

number called unnatural irrational natural rational
Mathematics
1 answer:
enyata [817]3 years ago
4 0

Answer:

irrational numbers

Step-by-step explanation:

These types of numbers are known in mathematics as irrational numbers. This is because there is no way to truly rationalize these numbers as we cannot truly rationalize the meaning of infinity. These numbers keep going endlessly and there doesn't exist an end. One example of an irrational number is the value of Pi which we know as a simplified 3.14 but in reality, the value of Pi is endless. The first thousand places of Pi can be seen below

3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989

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Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

\frac{3}{x + 4}  - \frac{1}{x + 3}  = \frac{x + 9}{(x^2 + 7x + 12)}

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

\frac{3*(x + 4)}{x + 4}  - \frac{(x + 4)}{x + 3}  = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}

3  - \frac{(x + 4)}{x + 3}  = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}

Now we can multiply both sides by (x + 3)

3*(x + 3)  - \frac{(x + 4)*(x+3)}{x + 3}  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

3*(x + 3)  - (x + 4)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

(2*x + 5)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

Now we can multiply both sides by (x^2 + 7*x + 12)

(2*x + 5)*(x^2 + 7x + 12)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)

(2*x + 5)*(x^2 + 7x + 12)  = (x + 9)*(x + 4)*(x+3)

Now we need to solve this:

we will get

2*x^3 + 19*x^2 + 59*x + 60 =  (x^2 + 13*x + 3)*(x + 3)

2*x^3 + 19*x^2 + 59*x + 60 =  x^3 + 16*x^2 + 42*x + 9

Then we get:

2*x^3 + 19*x^2 + 59*x + 60 - (  x^3 + 16*x^2 + 42*x + 9) = 0

x^3 + 3x^2 + 17*x + 51 = 0

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

The factors of -51 are:

-3 and -17

Let's try -3

p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0

Then x = -3 is one solution of the equation.

But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0

The other two solutions are when the other term is equal to zero.

Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

6 0
3 years ago
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