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3 years ago

Here is a system of equations.

Mathematics

2 answers:

goblinko [34]3 years ago

8 0

Answer:

It should be "one solution"

Step-by-step explanation:

After graphing the equations, the two lines only intersect at one point which makes it "one solution." Hope this helps.

solong [7]3 years ago

7 0

The answer Is one solution

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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 .

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.

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:

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

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

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

Example: $a = 2$ , $c = 2$

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

$b = 3$

4 0

3 years ago

Solve the equation. Round the answer to the nearest tenth.<br> 5*6^3n=20

solniwko [45]

Step-by-step explanation:

5×6'3n=20

30'3n=20

3n=20-30

3n/3= -10/3

n= -3

7 0

3 years ago

Pleaseeeeeee help me 7th grade math

vampirchik [111]

Answer:

Step-by-step explanation:

Np bro, I've dealt with these all through 7th XD

8 0

3 years ago

Read 2 more answers

The average life of a lightbulb is 2500 h, with a standard deviation of 200 h. Suppose that 10,000 bulbs are purchased to equip

Anna35 [415]

0 lightbulbs would be expected to burn out in 1800 hours

4 0

3 years ago

The question is in the picture please help

ahrayia [7]

Answer:

Claim 2

Step-by-step explanation:

The Inscribed Angle Theorem* tells you ...

... ∠RPQ = 1/2·∠ROQ

The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.

... 2·∠RPQ = ∠ROQ

The symmetric property of equality says you can rearrange this to ...

... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ

_____

* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.

You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.

4 0

3 years ago