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Furkat [3]
4 years ago
13

Is it possible for a composite number to have more than one prime factorization? Is it possible for a number to have no prime fa

ctors? Why? Give an example of how prime factorization could be used in the real world.
Mathematics
1 answer:
Nitella [24]4 years ago
3 0
Prime factors are factors of a composite number that are indivisible except by the number 1 or the number itself. The answers to your questions are the following:

1. Yes, it is possible especially for very large numbers. 
2&3. No, because as mentioned previously, the default prime factors of numbers are 1 and the number itself. For example, 2 is a prime number. Its factors are 1 and 2.
4. Prime factorization are useful in fields of encryption. They make use of the basic prime numbers for the arithmetic modulus with the general equation: n=pq.
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Answer:

46 degrees

Step-by-step explanation:

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Answer: x=2/3

Step-by-step explanation:

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A rectangle has sides that measure 2 x + 11 units long and 10 x + 1 units long. What is the expression that represents the perim
HACTEHA [7]

Answer:

(24x + 24) units

Step-by-step explanation:

A rectangle is a shape that has four sides in which all angles are equal and opposite sides are equal and parallel to each other. The perimeter of a rectangle is the sum of all its sides. Since opposite sides are equal, the perimeter is given as:

Perimeter = 2(length + width)

Since the sides of the rectangle are 2 x + 11 units long and 10 x + 1 units long. Hence:

Perimeter = 2(2x + 11 + 10x + 1)

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7 0
3 years ago
If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that
Ivan

Answer:

\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• \sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

•\sf\implies cos2A =\dfrac{\sqrt3}{2}

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

\sf\implies cos2A = 1 - 2sin^2A

Therefore , here substituting the value of sinA , we have ,

\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2

Simplify the whole square ,

\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}

Add the numbers in numerator ,

\sf\implies cos2A =  1-2\times \dfrac{4-2\sqrt3}{8}

Multiply it by 2 ,

\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}

Take out 2 common from the numerator ,

\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}

Simplify ,

\sf\implies cos2A =  1 -\dfrac{ 2-\sqrt3}{2}

Subtract the numbers ,

\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}

Simplify,

\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }

Hence Proved !

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Step-by-step explanation:

I took the test

:)

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