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3 years ago

Multiply the polynomials (5x^2+4x-4)(4x^3-2x+6) Explain your steps!!

1 answer:

4 0

So distribute using distributive property
a(b+c)=ab+ac so

split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24

the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24

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The equations which are sums of perfect cubes are as follows;

$\rm 8x^6+27x^9+1$

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$\rm 27x^9+x^{12}$

<h3>Perfect cubes;</h3>

Perfect cubes are the numbers that are the triple product of the same number.

We have to determine

Which are sums of perfect cubes?

1. The given equation is $\rm 8x^6+27x^9+1$ .

The equation can be written as;

$\rm 8x^6+27x^9+1\\\\2^3(x^2)^3+3^3(x^3)^3+1^3$

All the terms in the expression are can be represented as perfect cubes.

2. The given equation is $\rm 81x^3+16x^6$ .

The equation can be written as;

$\rm 81x^3+16x^6\\\\9^2x^3+4^2(x^2)3$

In this, all the terms are not perfect cubes, some of them are squares.

3. The given equation is $\rm x^6+x^3$ .

The equation can be written as;

$\rm x^6+x^3\\\\(x^2)^3+x^3$

All the terms in the expression are perfect cubes.

4. The given equation is $\rm 27x^9+x^{12}$ .

The equation can be written as;

$\rm 27x^9+x^{12}\\\\3^3(x^3)^3+(x^4)^3$

All the terms in the expression are perfect cubes.

5. The given equation is $\rm 9x^3+27x^9$ .

The equation can be written as;

$\rm 9x^3+27x^9\\\\3^2x^3+3^3(x^3)^3$

In this, all the terms are not perfect cubes, some of them are squares too.

To know more about perfect cubes click the link given below.

brainly.com/question/4701925

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Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court

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Answer:

$(a)\dfrac{92}{117}$

$(b)\dfrac{8}{39}$

$(c)\dfrac{25}{117}$

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

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(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

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P(One man and one woman are appointed) $=P(MW)+P(WM)$

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(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.
Two women are appointed

P(at least one woman is appointed) $=P(MW)+P(WM)+P(WW)$

$P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}$

In Part B, $P(MW)+P(WM)=\frac{8}{39}$

Therefore:

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