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4 years ago

The quotient of (x4 – 3x2 + 4x – 3) and a polynomial is (x2 + x – 3). What is the polynomial?

Mathematics

2 answers:

Black_prince [1.1K]4 years ago

7 0

Answer:

$x^2-x+1$

Step-by-step explanation:

We are given that

Dividend= $x^4-3x^2+4x-3$

Quotient= $x^2+x-3$

We have to find the divisor.

Let the polynomial = $p(x)$

$\frac{x^4-3x^2+4x-3}{p(x)}=x^2+x-3$

$p(x)=\frac{x^4-3x^2+4x-3}{x^2+x-3}$

Hence, the polynomial $=p(x)=x^2-x+1$

Answer: $x^2-x+1$

suter [353]4 years ago

4 0

The right answer for the question that is being asked and shown above is that: "x2 – x + 1."The quotient of (x4 – 3x2 + 4x – 3) and a polynomial is (x2 + x – 3). The polynomial is this one: <span>x2 – x + 1</span>

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zzz [600]

Answer:

C. {-5,-4, -3, 1, 2, 5}

Step by step explanation:

We have been given a graph and we are asked to find the domain of the relation represented in graph.

We can see that our graph is a series of unconnected points. Our function represents integer values. So we can see that our graph represents a discrete function.

Since we know that domain of a discrete function is set of inputs values consisting of only certain values in an interval. .

The set of first value from each of the given points would made domain of our function. Upon looking at our graph we can see that domain of our function is -5,-4, -3, 1, 2 and 5.

Therefore, option C is the correct choice.

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3 years ago

Doug is in line at the store. He has three items that cost $4.50, $3.05, and $1.50. Explain how Doug can add the cost of th

Angelina_Jolie [31]

Answer:

He can add through calculater or by doing addition in his mind

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3 years ago

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What is -x divided by 4?

exis [7]

The answer is just -x/4 because you can divide a variable

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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

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suter [353]

Its either 3 or 2
the third way matches only if there's another reference line
I'll go with 3

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3 years ago

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