In an equation, constant can change.

Mathematics

1 answer:

valina [46]3 years ago

8 0

I think the answer is false

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

maksim [4K]

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

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

$\stackrel{\textit{\large perimeter of ABC}}{4\sqrt{2}+\sqrt{13}+\sqrt{53}~~\approx~~ 16.54}$

3 0

2 years ago

I need desperate help <br>

Lady_Fox [76]

Answer:

Step-by-step explanation:

y equals 1. x equals 15. 15 over 15 is 1.

3 0

3 years ago

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Erin swims one half mile in one third hour. Write an equation for the distance y Erin can swim in x hours

cestrela7 [59]

3/2 x = y is the answer

7 0

3 years ago

Solve for x. Round to 3 decimal places

iVinArrow [24]

Answer:

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\Domain:\ x+2>0\to x>-2\\\\\log_3(x+2)-8=-2\qquad\text{add 8 to both sides}\\\\\log_3(x+2)=6\iff x+2=3^6\\\\x+2=729\qquad\text{subtract 2 from both sides}\\\\x=727\in D$