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2 years ago

12

I can’t figure this out. Please help

2 answers:

patriot [66]2 years ago

7 0

Answer:

itjust algebra

Step-by-step explanation:

bulgar [2K]2 years ago

6 0

Answer:

$\frac{3^{6} }{3^{4} }$ or 3²

Step-by-step explanation:

Count the 3 and put it in the exhibitor for the numerator and the denominator of the fraction. You have $\frac{3^{6} }{3^{4} }$ then put them in a line $3^{6} 3^{-4} = 3^{6-4} = 3^{2}$

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3 years ago

Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.

liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

$f(x) = \frac{x^2+4x+3}{x^2-x-2}$

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial $x^2-x-2$

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

$(x-2) (x + 1)$

We do the same with the numerator

$x^2+4x+3$

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

$(x+3)(x + 1)$

Therefore

$f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}$

Note that $\frac{(x+1)}{(x+1)}=1$ only if $x \neq -1$

So since $x = -1$ is not included in the domain the function has a discontinuity in $x = -1$

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3 years ago

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