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3 years ago

Given the flag ABCDE, determine the single rule (x,y) that creates the

Mathematics

1 answer:

cricket20 [7]3 years ago

6 0

Answer:

(y, -x)

Step-by-step explanation:

Rotation of 90°

(Clockwise)

(x, y) ---> (y, -x)

Rotation of 90°

(CounterClockwise)

(x, y) --> (-y, x)

Rotation of 180°

(Both Clockwise and Counterclockwise)

(x, y) ---> (-x, -y)

Rotation of 270°

(Clockwise)

(x, y) ---> (-y, x)

Rotation of 270°

(CounterClockwise)

(x, y) ---> (y, -x)

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Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

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Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

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$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

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$=x^4+12x^3+54x^2+108x+81$

Therefore,

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IRISSAK [1]

Answer:

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Step-by-step explanation:

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