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2 years ago

Solve for the value of x in the diagram below. Show your work and explain the steps you used to solve.

Mathematics

1 answer:

Mama L [17]2 years ago

6 0

Answer: x = 2

Step-by-Step Solution:

Given:

OH = WO = LW

OH = 2x + 1

WO = 3x - 1

LW = 5

To Find:

The value of 'x'

Solution:

We know, OH = WO = LW

So, equate any two of the values given :-

2x + 1 = 5

2x = 5 - 1

2x = 4

x = 4/2

x = 2

Hence, x = 2.

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

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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.