Which expression is equivalent to x^2+7x+12/x^2-x-12?

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

5 0

X^2+7x+12/x^2-x-12
x^2+4x+3x+12/x^2+3x-4x-12
x * (x+4) +3(x+4)/ x* (x+3)-4(x+3)
(x+3)(x+4)/(x-4)(x+3)

x+4/x-4

anastassius [24]3 years ago

3 0

Answer:

$\frac{(x+4)}{(x-4)}$

Step-by-step explanation:

Given : $\frac{x^2+7x+12}{x^2-x-12}$

Solution:

$\frac{x^2+7x+12}{x^2-x-12}$

Factorize the numerator and denominator.

$\frac{x^2+3x+4x+12}{x^2-4x+3x-12}$

$\frac{x(x+3)+4(x+3)}{x(x-4)+3(x-4)}$

$\frac{(x+4)(x+3)}{(x-4)(x+3)}$

$\frac{(x+4)}{(x-4)}$

So, $\frac{x^2+7x+12}{x^2-x-12}$ = $\frac{(x+4)}{(x-4)}$

Hence Option B is correct.

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A coin is to be tossed as many times as necessary to turn up one head. Thus the elements c of the sample space C are H, TH, TTH,

slavikrds [6]

Answer:

Step-by-step explanation:

As stated in the question, the probability to toss a coin and turn up heads in the first try is $\frac{1}{2}$ , in the second is $\frac{1}{4}$ , in the third is $\frac{1}{8}$ and so on. Then, P(C) is given by the next sum:

$P(C)=\sum^{\infty}_{n=1}(\frac{1}{2} )^{n}=1$

This is a geometric series with factor $\frac{1}{2}$ . Then is convergent to $\frac{1}{1-\frac{1}{2}}-1=1.$ . With this we have proved that P(C)=1.

Now, observe that

$P(H)=\frac{1}{2}, P(TH)=\frac{1}{4},P(TTH)=\frac{1}{8},P(TTTH)=\frac{1}{16},P(TTTTH)=\frac{1}{32},P(TTTTTH)=\frac{1}{64}.$

Then

$P(C1)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} =\frac{31}{32}$

$P(C2)=P(TTTTH)+P(TTTTTH)=\frac{1}{32}+\frac{1}{64} =\frac{3}{64}$

$P(C1\cap C2)=P(TTTTH)=\frac{1}{32}$

and

$P(C1\cup C2)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)+P(TTTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} +\frac{1}{64}=\frac{63}{64}$