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Sliva [168]
3 years ago
14

4 && 5 struggling.... little difficult

Mathematics
1 answer:
zlopas [31]3 years ago
3 0
I can’t answer 5 because I don’t know what was in Example two, but I can tell you how to graph the equation y = -200x + 2400!

This equation is in slope intercept form, which means that the number next to the variable is the rate of change and the number added onto the first term is the initial value. Since your initial value is 2400, your first point will be at (0, 2400). Each time x increases by 1, y will decrease by 200 (because the rate of change is -200). So your second point will be (1, 2200) and so on.
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Step-by-step explanation:

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lbvjy [14]

~\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}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}

BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}

DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89

8 0
3 years ago
What is the area of the figure below?
erastovalidia [21]

Answer:

please mark my answer brainliest

Step-by-step explanation:

the answer will be...9in square +25in square +24 in square.....=58 in square.....that is ur answer...find individual areas then sum up

6 0
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