A rectangular plot of ground is 5 meters longer than it is wide, It’s area is 20,000 square meters

1 answer:

8 0

I'm assuming you're looking for the dimensions of the plot. I'm going with that. ;) If the length of the plot is 5 meters longer than the width, then L = w + 5. The area for a rectangle is L*w, and we have an area value of 20,000 so our formula is 20000=(w+5)(w) and $20,000=w^2+5w$ . We will bring the 20,000 over by subtraction and set the polynomial equal to 0 to factor and solve for w. $w^2+5w-20,000=0$ Solving for w we get values of w=138.9 and -143.9. Of course the 2 things in math that will never EVER be negative are time and distance/length, so -143.9 is out. Our width is 138.9 and the length is 138.9 + 5 so the length is 143.9. And there you go! Hope that's what you needed!

You might be interested in

Aaliyah is a salesperson who sells computers at an electronics store. She makes a base

hjlf

Answer:

$P=1.25x+70$

Step-by-step explanation:

So we know that Aaliyah earns $1.25 commission for every computer she sells. No matter what, she will make $70 per day, so that would be our constant. The $1.25 would be our independent variable since it can change depending on the amount of computers x she sells. Therefore, our equation is:

$P=1.25x+70$

P represents the total pay, 1.25x represents the total commission for x computers, and the 70 is the base pay.

And we're done!

6 0

2 years ago

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}$ $A(5,1)\ \ \ \ \ \ C(-7,-3)\\\\Distance\ between\ A \ and\ C:\\AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}\\\\AC=\sqrt{(-7-5)^2+(-3-1)^2}\\AC=\sqrt{(-12)^2+(-4)^2}\\AC=\sqrt{144+16}\\AC=\sqrt{160}$ $Perimeter\ of\ triangle=\ AB+AC+BC=\\\sqrt{68}+\sqrt{52}+\sqrt{160}=2\sqrt{17}+2\sqrt{13}+2\sqrt{40}$