$by \ pythagoras \ theorem : \\(3x +1 )^2 = (3x - 1)^2 +x^2\\\\9x^2 + 6x +1 = 9x^2 + 1 -6x +x^2\\\\9x^2 + 6x +1 = 10x^2 + 1 -6x \\\\ 10x^2 + 1 -6x - 9x^2 -6x -1 = 0 \\\\x^2 -12x = 0\\x(x - 12) = 0\\x = 0, 12\\since \ length \ cannot \be \ 0, x = 12\\OJ = 3x + 1 = 3(12) +1 = 37$

$by \ pythagoras \ thm, \\(7x -12) ^2 = (5x +10)^2 +(2x)^2\\\\49x^ 2 +144 - 168x = 25x^2 +100+ 100x +2x^2\\\\49x^ 2 +144 - 168x - 25x^2 -100- 100x -2x^2 = 0\\\\22x^2 -268x -44 = 0\\11x^2 - 134x - 22 = 0\\x=12.0154$

3 0

3 years ago

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Anit [1.1K]

Answer:

thjanks

Step-by-step explanation:

thanks

8 0

3 years ago

Read 2 more answers

Which of the following are the zeros of the function y= 3(x-a)(x+b) in terms of a and b?

frosja888 [35]

The correct answer is option 2.

3 = 0

x - a = 0

x = a

x + b = 0

x = -b

We do not multiply any of these values by 3 because in a way, 3 is acting as another "zero" of the equation. Though, as shown above, it does not equal 0, it still does not apply to any of the other equations solved above.

Hope this helps!! :)

7 0

4 years ago