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3 years ago

A figure is made up of 5 identical squares, the area of the figure is 405 squares inches, what is the perimeter

1 answer:

6 0

The side length of each square is √(405/5) = 9 inches, but the perimeter of the figure depends on how the squares are arranged.

# # #
# # . . . . . . . . will have a perimeter of 90 inches

# # # # # . . . will have a perimeter of 108 inches⊕

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Find the LCM of the pair of polynomials.

svet-max [94.6K]

$2x^{2}-8x+8$

Factor the polynomial.

2( $x^{2}-4x+4$ )

2(x-2) $^{2}$

Factor the other Polynomial.

3 $x^{2}+27x-30$

$3(x^{2}+9x-10)$

$3(x+10)(x-1)$

This is your answer.

$6(x-2)^{2}(x+10)(x-1)$

7 0

3 years ago

Find the value of x. 16.2 0.03 38.5 34.8

ella [17]

Hi there!

$\large\boxed{x = 38.5}}$

To solve, we can use right triangle trig.

We are given the value of ∠A, and side "x" is its adjacent side. We are also given its opposite side, so:

tan (A) = O / A

tan (33) = 25 / x

Solve:

x · tan(33) = 25

x = 38.49 ≈ 38.5

7 0

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.