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3 years ago

Give one example of a polynomial of degree 4 that has a zero at x=5 with multiplicity 2

Mathematics

1 answer:

mr Goodwill [35]3 years ago

4 0

9514 1404 393

Answer:

y = (x -5)^2(x^2 +1)

Step-by-step explanation:

The zero at x=5 means that (x-5) is a factor. The multiplicity of 2 means that factor occurs twice (is squared). The remaining 2nd degree factor can be anything. In this equation, we have elected to make it have complex zeros.

y = (x -5)^2(x^2 +1)

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A car travels 420 miles in 6 hours. What is the unit rate?

Semmy [17]

Answer:

Step-by-step explanation:

because 70 times 6 equals 420

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3 years ago

Bob Nale is the owner of Nale's Texaco GasTown. Bob would like to estimate the mean number of litres (L) of gasoline sold to his

Fudgin [204]

Answer:

The point estimate of the population mean is $\= x = 56$

The 80% confidence level is $50.57 < \mu < 61.43$

There is 80% confidence that the true population mean lies within the confidence interval.

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The standard deviation is $\sigma = 18 \ L$

The sample mean is $\= x = 56$

Generally the point estimate of the population mean is equivalent to the sample mean whose value is $\= x = 56$

Given that the confidence interval is 80% then the level of significance is mathematically represented as

$\alpha = 100 - 80$

$\alpha = 20 \%$

$\alpha = 0.20$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

The value is $Z_{\frac{ \alpha }{2} } = 1.28$

Generally the margin of error is mathematically evaluated as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.28 * \frac{18 }{\sqrt{18} }$

=> $E = 5.43$

Generally the 80% confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

=> $56 - 5.43 < \mu < 56 + 5.43$

=> $50.57 < \mu < 61.43$

The interpretation is that there is 80% confidence that the true population mean lies within the limit

3 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

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3 years ago

Solve the system: {-2(a-b)+16=3(b+7), 6a-(a-5)=-8-(b+1)

givi [52]

Answer:

a=−3 and b=1

Step-by-step explanation:

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3 years ago

What is the ratio of the length of DE to the length of BC?

laila [671]

Answer: OD, 1/5

Step-by-step explanation:

Well what I did was take DE and seen how many time it could fit into BC. BC would take up a total of 5 DE's. So since we already have one which is DE then we would have 1/5.

HOPE THIS HELPS! ^_^

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2 years ago