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3 years ago

Please help me on this. you need to match the terms to the correct number.

Mathematics

1 answer:

Gnom [1K]3 years ago

5 0

Answer:

64=8

49=7

36=6

25=5

100=10

16=4

9=3

20=4.5

4=2

1=1

67=8.2

81=9

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Which expression is equivalent to (a^-3 b/a^-5 b3)^-3? Assume a=0, b=0.

prohojiy [21]

$Use:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(ab)^n=a^nb^n$

$\left(\dfrac{a^{-3}b}{a^{-5}b^3}\right)^{-3}=\left(a^{-3-(-5)}b^{1-3}\right)^{-3}=\left(a^{-3+5}b^{-2}\right)^{-3}=\left(a^2b^{-2}\right)^{-3}\\\\=(a^2)^{-3}(b^{-2})^{-3}=a^{2(-3)}b^{-2(-3)}=a^{-6}b^6=\dfrac{b^6}{a^6}$

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3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.