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3 years ago

ASAPPP

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

6 0

Answer:

Hello

Rational Number

Rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q.

Irrational Number

Irrational numbers are numbers that cannot be expressed as the ratio of two integers.

Hope it helps you...

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For f(x)=2x+1 and g(x)=x^2-7, find (f+g)(x) A. 2x^2-15 B.X^2+2x-6 C.2x^3-6 D.x^2+2x+8

Advocard [28]

Answer:

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) , thus

f(x) + g(x)

= 2x + 1 + x² - 7 ← collect like terms

= x² + 2x - 6 → C

5 0

3 years ago

Y varies directly as x if y is 2 when x is 8 then the constant of variation is 1/4

sp2606 [1]

Answer:

a. True.

Step-by-step explanation:

y = kx

y = 2 when x = 8 gives:

2 = 8 * k

k = 2/8 = 1/4.

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3 years ago

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vaieri [72.5K]

i think it is 487.5. i may be wrong tho. hope this helps!

3 0

3 years ago

Read 2 more answers

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

Shelby will take a total of 6 tests. On the first 5 tests, her scores were: 80, 85, 73, 78, 90. If she wants a mean grade of 80,

Andrew [12]

Answer:

Step-by-step explanation:

4 0

3 years ago