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pishuonlain [190]
3 years ago
13

Find the value of the following expression:

Mathematics
1 answer:
Phoenix [80]3 years ago
8 0

(3^8 ⋅ 2^-5 ⋅ 9^0)^-2 ⋅ (2^ -2 / 3^3) ^4 ⋅ 3^28 =

(6561 * 0.03125 * 1)^2 * (0.00925)^4 * 22876792454961 =

42037.81348 * 0.00000000732094 * 22876792454961 =

7040477235.56798349

 round answer as needed

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Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.
vesna_86 [32]

Factorize the denominator:

\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}

If x\neq\pm2, we can cancel the factors of x^2-4, which makes x=-2 and x=2 removable discontinuities that appear as holes in the plot of g(x).

We're then left with

\dfrac1{x+1}

which is undefined when x=-1, so this is the site of a vertical asymptote.

As x gets arbitrarily large in magnitude, we find

\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0

since the degree of the denominator (3) is greater than the degree of the numerator (2). So y=0 is a horizontal asymptote.

Intercepts occur where g(x)=0 (x-intercepts) and the value of g(x) when x=0 (y-intercept). There are no x-intercepts because \dfrac1{x+1} is never 0. On the other hand,

g(0)=\dfrac{0-4}{0+0-0-4}=1

so there is one y-intercept at (0, 1).

The domain of g(x) is the set of values that x can take on for which g(x) exists. We've already shown that x can't be -2, 2, or -1, so the domain is the set

\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}

8 0
3 years ago
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kolbaska11 [484]

Answer:

5x^2 -6x +1

Step-by-step explanation:

(3х^2 – 2) + (2х^2 – бх + 3).

Combine like terms

(3х^2  + 2х^2 – бх + 3-2)

5x^2 -6x +1

4 0
3 years ago
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Romashka [77]

Yes this is the right answer

6 0
3 years ago
QUESTION 9
Ratling [72]
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Is the expression 3x+2 a binomial?
Wewaii [24]

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