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g100num [7]
4 years ago
11

HELP!!!!!!!!!!!!!!!!!!!!!!!!!! Can U Answer All These.

Mathematics
1 answer:
mash [69]4 years ago
3 0

<u>QUESTION 1</u>

In figure A, the area of the bigger square is 625\:units^2. This means that the length of this square is the hypotenuse of the right angle triangle formed, which is 25\:units.

The area of the two smaller squares are 576\:units^2 and 49\:units^2.


The length of these two squares form the two shorter legs of the right angle triangle, which are 24 and 7 units respectively.


Therefore we can write the equation, 576+49=625.


This implies that, 24^2+7^2=25^2


We can see that the sum of the squares of the two shorter legs equals the square of the longer side which is the hypotenuse and this what the Pythagoras theorem says.


But in fig. B,

5^2+4^2\ne 6^2


Therefore the correct answer is Fig A.


<u>QUESTION 2</u>

See diagram.

We introduce the two lines as shown in the diagram.

From Pythagoras' Theorem,

|AB|^2=|AC|^2+|BC|^2


But |AC|=|7-3|=4 units.


and

|BC|=|8-2|=6 units.

This implies that,

|AB|^2=4^2+6^2


|AB|^2=16+36


|AB|^2=52


|AB|=\sqrt{52}


|AB|=2\sqrt{13}


|AB|=7.2 to the nearest tenth.



<u>QUESTION 3</u>

The numbers 3,4 and 5 form a Pythagorean triplet because 3^2+4^2=5^2.


If these numbers represent the side lengths of a triangular fence, then the triangular fence must be a right angle triangle.

The correct answer is B

Yes,because  3^2+4^2=5^2.


<u>QUESTION 4</u>

See graph.

We want to find the image of the triangle with vertices A(-5,-4), B(-2,-4), and C(-5,-1) after a reflection in the x-axis.

The mapping for this transformation is

P(x,y)\rightarrow P'(x,-y).


We just have to negate the y-coordinates.

A(-4,-5)\rightarrow A'(-4,5).


B(-4,-1)\rightarrow B'(-4,1).


C(-5,-1)\rightarrow C'(-5,1).








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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.

In addition, b must be a natural number, which means that:

1 + 2\cdot a\cdot c \ge 4

2\cdot a \cdot c \ge 3

a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

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:

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

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

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

b = 3

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