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3 years ago

Use the real zeros to factor f x^4 plus 10^3 minus 20x^2 minus 90x plus 99

Mathematics

1 answer:

Furkat [3]3 years ago

7 0

Answer:

f(x) = (x + 11) (x + 3) (x − 1) (x − 3)

Step-by-step explanation:

f(x) = x⁴ + 10x³ − 20x² − 90x + 99

f(x) is a fourth order polynomial, so it has 4 roots. Use rational root theorem to find possible rational roots.

±99, ±11, ±9, ±3, ±1

By trial and error, the zeros are -11, -3, 1, and 3.

f(x) = (x + 11) (x + 3) (x − 1) (x − 3)

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Answer:

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Step-by-step explanation:

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3 years ago

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Anettt [7]

Answer:

3√265

Step-by-step explanation:

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3 years ago

Answer the question below

lys-0071 [83]

Answer:

A) -1

Step-by-step explanation:

$\frac{3 + 2\sqrt{5} }{2\sqrt{5}-3 } \\$

We can rewrite 2√5 - 3 as -(3 + 2√5)

Therefore,

$\frac{3+ 2\sqrt{5} }{-(3 + 2\sqrt{5} )} \\$

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3 years ago

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

Trains Two trains, Train A and Train B. weigh a total of 196 tons. Train A is heavier than Train B. The

aleksklad [387]

Answer:

Train A = 128

Train B = 68

Step-by-step explanation:

We can set up a system of equations for this problem

Let A = # of tons of Train A

Let B = # of tons of Train B

A + B = 196

A = B + 60

Now, we plug in A for the first equation, using substitution

(B+60) + B = 196

2B + 60 = 196

Subtract 60 from both sides

2B = 136

Divide both sides by 2

B = 68

Plug in 68 for B in the 2nd equation

A = 68 + 60

A = 128

Checking work: 128 + 68 = 196 :D hope this helped

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2 years ago