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alexandr1967 [171]
3 years ago
7

Simplify (5x - 3y)³ - (5x + 3y)³​

Mathematics
1 answer:
lianna [129]3 years ago
5 0

Answer:

−6y((5x−3y)^2 +(5x−3y)(5x+3y)+(5x+3y)^2)

Step-by-step explanation:

Use Difference of Cubes : a^3 - b^3 = (a-b)*(a^2+ab+b^2)

= (5x - 3y - (5x+3y))*((5x-3y)^2 + (5x-3y)*(5x+3y)+(5x+3y)^2)

Simplify brackets:

= (5x−3y−5x−3y)((5x−3y)​^2+(5x−3y)(5x+3y)+(5x+3y)^2)

Collect like temrs:

= ((5x−5x)+(−3y−3y))((5x−3y)^2​+(5x−3y)(5x+3y)+(5x+3y)^2)

Simplify:

= −6y((5x−3y)^2 +(5x−3y)(5x+3y)+(5x+3y)^2)

Simpliest form it can get!

Hope this helped :3

​​

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After a dilation with a center of (0, 0), a point was mapped as (4, –6) → (12, y). A student determined y to be –2. Evaluate the
SpyIntel [72]

Answer:

B. The student incorrectly calculated the scale factor to be –2

Step-by-step explanation:

Given that :

After a dilation with a center of (0, 0), a point was mapped as (4, –6) → (12, y).

The student determined y to be -2

If a figure dilated with a center of (0, 0) and scale factor k, then

(x , y) → (kx , ky)

(4, -6) → (12, y)

k = \dfrac{x'}{x}

k = \dfrac{12}{4}

k = 3

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Now; the y-coordinate can now be calculated as;

ky = (3 × -6)

ky = -18

Therefore; the value of y = -18 and the student incorrectly calculated the scale factor to be -2.

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3 years ago
In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
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Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

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The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

             = 1 - P (X = 0) - P (X = 1)

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Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

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Answer:

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