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GenaCL600 [577]
3 years ago
15

Which of the following numbers is a composite number that is divisible by 3?

Mathematics
1 answer:
zavuch27 [327]3 years ago
3 0
The answer is D because if you'd have to divide 261 \3 = 87
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7x - 8 = 41 what does x equal to
velikii [3]

Answer:

7

Step-by-step explanation:

41+8=49 49/7=7

4 0
2 years ago
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In a soccer tournament teams receive 6 points for winning a game 3 points for trying a game and 1 point for each goal they score
Alex777 [14]
I beleive it is 6(2) + 3 (1) + 1(5) = 20 points :)
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3 years ago
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.

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\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
bag contains 15 marbles numbered from 1 to 15. Two marbles are selected from the bag at the same time. Let event E = the two num
lbvjy [14]
The numbers DO differ by 10 if they are

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So the answer is (a) 5.
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2 years ago
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The slope of the line whose equation is given
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The slope is 9/1 hope i helped
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