$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ R(\stackrel{x_1}{-7}~,~\stackrel{y_1}{5})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{5}) ~\hfill RS=\sqrt{(~~ 17- (-7)~~)^2 + (~~ 5- 5~~)^2} \\\\\\ ~\hfill RS=\sqrt{( 24)^2 + ( 0)^2}\implies \boxed{RS=24} \\\\\\ S(\stackrel{x_1}{17}~,~\stackrel{y_1}{5})\qquad T(\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill ST=\sqrt{(~~ 5- 17~~)^2 + (~~ 0- 5~~)^2}$

$~\hfill ST=\sqrt{( -12)^2 + ( -5)^2}\implies \boxed{ST=13} \\\\\\ T(\stackrel{x_1}{5}~,~\stackrel{y_1}{0})\qquad R(\stackrel{x_2}{-7}~,~\stackrel{y_2}{5}) ~\hfill TR=\sqrt{(~~ -7- 5~~)^2 + (~~ 5- 0~~)^2} \\\\\\ ~\hfill TR=\sqrt{( -12)^2 + (5)^2}\implies \boxed{TR=13} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{\LARGE perimeter}}{24~~ + ~~13~~ + ~~13\implies \text{\LARGE 50}}~\hfill$

3 0

1 year ago

Plsssssssssssssss help me nobody knows how to do it and its due tonight

inessss [21]

I believe it’s 1.5558 x 10 to the 4th power

5 0

2 years ago

Read 2 more answers

Please help me solve

arsen [322]

Answer:

18 square centimeters

Step-by-step explanation:

Notice that if e is the midpoint of the side CB, and angle $\angle x = 45^o$ , then this rectangle is in fact two squares of side 3 cm put together. therefore, side CD has a length of 6 cm, and as a result, the area of the figure is given by the product base times height = 6 x 3 = 18 $cm^2$

4 0

3 years ago

Read 2 more answers

I need help !!30 PTS!!<br>Pls DO help on ALL 3 on the screenshot

GenaCL600 [577]

1. (16,25)-(24,5) the interval it falls from is 25 mph to 5 mph or it decreased by 20 mph.

2. From D (24,5) to E (28,45). The interval of time from point D to point E is 5 mph to 45 mph. or it increased by 40 mph.

3. From B (6,25) to C (16,25) and the constant rate is 25 MPH.

I'm not 100% sure about the everything.

7 0

2 years ago