Please help me i really need it fast!!

Mathematics

1 answer:

rusak2 [61]4 years ago

8 0

Answer:

They do not form a right triangle

Step-by-step explanation:

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What is the perimeter of a triangle with vertices of (5,1) (-3,3) (-7,-3)

timurjin [86]

$Vertices:\\
A(5,1)\\B(-3,3)\\C(-7,-3)\\\\ You\ must\ find\ length\ of\ AB,\ BC,\ AC.\\\\
A(5,1)\ \ \ \ \ \ B(-3,3)\\\\Distance\ between\ A \ and\ B:\\AB=\sqrt{(x_B-x_A)^2+(y_A-y_B)^2}\\\\AB=\sqrt{(-3-5)^2+(3-1)^2}\\AB=\sqrt{8^2+2^2}\\AB=\sqrt{64+4}\\AB=\sqrt{68}
$ $
 B(-3,3)\ \ \ \ C(-7,-3)\\\\Distance\ between\ B \ and\ C:\\BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\\\BC=\sqrt{(-7-(-3))^2+(-3-3)^2}\\BC=\sqrt{(-4)^2+(-6)^2}\\BC=\sqrt{16+36}\\BC=\sqrt{52}$ $A(5,1)\ \ \ \ \ \ C(-7,-3)\\\\Distance\ between\ A \ and\ C:\\AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}\\\\AC=\sqrt{(-7-5)^2+(-3-1)^2}\\AC=\sqrt{(-12)^2+(-4)^2}\\AC=\sqrt{144+16}\\AC=\sqrt{160}$ $Perimeter\ of\ triangle=\ AB+AC+BC=\\\sqrt{68}+\sqrt{52}+\sqrt{160}=2\sqrt{17}+2\sqrt{13}+2\sqrt{40}$

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3 years ago

HELP PLEASE!!!!! Solve each system.

Ede4ka [16]

$15.\ \text{Elimination method}\\\underline{+\left\{\begin{array}{ccc}y=-3x+9\\y=3x+3\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2y=12\qquad\text{divide both sides by 2}\\.\qquad \boxed{y=6}\\\\\text{put the value of y to the second equation}\\\\6=3x+3\qquad\text{subtract 3 from both sides}\\3=3x\qquad\text{divide both sides by 3}\\1=x\to\boxed{x=1}\\\\Answer:\ x=1,\ y=6\to(1,\ 6)$

$16.\ \text{Substitution method}\\\left\{\begin{array}{ccc}y=11x+4\\y=7x-16\end{array}\right\Rightarrow11x+4=7x-16\\\\11x+4=7x-16\qquad\text{subtract 4 from both sides}\\11x=7x-20\qquad\text{subtract 7x from both sides}\\4x=-20\qquad\text{divide both sides by 4}\\\boxed{x=-5}\\\\\text{Put the value of x to the first equation}\\\\y=11(-5)+4\\y=-55+4\\\boxed{y=-51}\\\\Answer:\ x=-5,\ y=-51\to(-5,\ -51)$