What is the classification of AABC with vertices A(0, 0), B(4, 3), and C(4, -3) by

Mathematics

1 answer:

lbvjy [14]11 months ago

6 0

Check the picture below.

well, let's take a peek at the distances of each side.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{4}~,~\stackrel{y_2}{3}) ~\hfill AB=\sqrt{[ 4- 0]^2 + [ 3- 0]^2} \\\\\\ ~\hfill AB=\sqrt{25}\implies \boxed{AB=5} \\\\\\ B(\stackrel{x_1}{4}~,~\stackrel{y_1}{3})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-3}) ~\hfill BC=\sqrt{[ 4- 4]^2 + [ -3- 3]^2} \\\\\\ ~\hfill BC=\sqrt{36}\implies \boxed{BC=6}$

$C(\stackrel{x_1}{4}~,~\stackrel{y_1}{-3})\qquad A(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) ~\hfill CA=\sqrt{[ 0- 4]^2 + [ 0- (-3)]^2} \\\\\\ ~\hfill CA=\sqrt{25}\implies \boxed{CA=5} \\\\\\ \textit{so it's an \underline{isosceles triangle}}$

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Arthur drops a ball from a height of 81 feet above the ground. Its height, h, is given by the equation h = –16t2 + 81, where t i

Lorico [155]

Answer:

Step-by-step explanation:

We are given the position function and need to find the value of t when h<17.

Create an inequality that represents this situation:

$-16t^2+81$ The "less than" sign makes this very specifically a conjunction problem as opposed to a disjunction. That's important to the solution. But we'll get there.

The simplest way to solve this is to subtract 81 from both sides:

$-16t^2$ then divide both sides by -16:

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