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3 years ago

Need help please, keep getting different answers

Mathematics

1 answer:

natali 33 [55]3 years ago

5 0

Answer:

Step-by-step explanation:

If x is the shorter leg, then x + 5 is the longer leg, and 2x − 5 is the hypotenuse.

Using Pythagorean theorem:

c² = a² + b²

(2x − 5)² = x² + (x + 5)²

4x² − 20x + 25 = x² + x² + 10x + 25

4x² − 20x + 25 = 2x² + 10x + 25

2x² − 30x = 0

x² − 15x = 0

x (x − 15) = 0

x = 0 or 15

Since x can't be 0, x must be 15.

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3 years ago

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2 years ago

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3 years ago

A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).

maksim [4K]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}$

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2 years ago

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V125BC [204]

$5 + 2j = 9 \\ j = 2$
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2 years ago