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butalik [34]
3 years ago
6

Can someone please help me factor and write the equations for these two problems ?

Mathematics
1 answer:
prisoha [69]3 years ago
5 0

Answer:

Step-by-step explanation:

Asymptotes 3

g(x) = \frac{3x}{x^2-3x-10}

Factors of denominator will be,

x² - 3x - 10 = x² - 5x + 2x - 10

                 = x(x - 5) + 2(x - 5)

                 = (x + 2)(x - 5)

Therefore, factored form of g(x) will be,

g(x) = \frac{3x}{(x + 2)(x - 5)}

Asymptotes 4

h(x) = \frac{(x-5)}{x^{2} + 14x + 40}

      = \frac{x-5}{x^{2}+10x+4x+40}

      = \frac{x-5}{x(x+10)+4(x+10)}

      = \frac{x-5}{(x+10)(x+4)}

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The sum of squares of two consecutive positive even integers is 52 more than their product. Find these numbers.
natulia [17]

Answer:

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The second set is ( - 8 and - 6) which also works.

Step-by-step explanation:

Equation

(x)^2 + (x + 2)^2 = (x)(x + 2) + 52            Remove the brackets on both sides

Solution

x^2 + x^2 + 4x + 4 = x^2 + 2x + 52        Collect the like terms on the left

2x^2+ 4x+ 4 = x^2 + 2x + 52                  Subtract right side from left

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Answer

Try the one you know works.

x - 6 = 0

x = 6

Therefore the two integers are 6 and 8

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So 6 and 8 is one set of  consecutive even numbers that works.    

========================

What about the other set.

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x = - 8

x and x + 2

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(- 8 )^2 + (- 6)^2 = 100

(-8)(-6) + 52 = 100

Both sets of consecutive numbers work.


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