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3 years ago

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Which identity is the result of using the pythagorean theorem to show that a triangle with side lengths x^2 - 1, 2x, and x^2 +

1 is a right triangle?
A. (x^2 - 1)^2 + (2x)^2 = -(x^2 + 1)^2
B. -(x^2 -1)^2 + (2x)^2 = (x^2 +1)^2
C. (x^2-1)^2 + (2x)^2 = (x^2 +1)^2
D. (x^2 - 1)^2 - (2x)^2 = (x^2 + 1)^2

2 answers:

vfiekz [6]3 years ago

8 0

9514 1404 393

Answer:

C. (x^2-1)^2 + (2x)^2 = (x^2 +1)^2

Step-by-step explanation:

Inexplicably, three of the four answer choices have minus signs in them. Those are all incorrect. The appropriate choice is ...

C. (x^2-1)^2 + (2x)^2 = (x^2 +1)^2

__

Expanded, this is ...

x^4 -2x^2 +1 +4x^2 = x^4 +2x^2 +1 . . . . the desired identity

ella [17]3 years ago

6 0

Answer:

C. (x^2-1)^2 + (2x)^2 = (x^2 +1)^2

Step-by-step explanation:

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3 years ago

Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6

ElenaW [278]

<h2>Answer</h2>

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

Or as ordered pairs: $(\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)$

<h2>Explanation</h2>

Lets solve our system of equations step by step

$x^2+y^2=6$ equation (1)

$x^2-y=6$ equation (2)

1. Solve for $x^2$ in equation (2)

$x^2-y=6$

$x^2=6+y$ equation (3)

2. Replace equation (3) in equation (1) and solve for $y$

$x^2+y^2=6$

$6+y+y^2=6$

$y^2+y=0$

$y(y+1)=0$

$y=0$ or $y=-1$

3. Replace the values of $y$ in equation (3) and solve for $x$

- For $y=0$

$x^2=6+y$

$x^2=6$

$x=\sqrt{6}$ or $x=-\sqrt{6}$

- For $y=-1$

$x^2=6+y$

$x^2=6-1$

$x^2=5$

$x=\sqrt{5}$ or $x=-\sqrt{5}$

So, the solutions of our system of equation are:

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

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