What is the value of k in the equation 5k-2k=12

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}$

$\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill$

5 0

3 years ago

Solve system if equations

Alexandra [31]

Answer:

(2, 9)

x = 2

y = 9

Step-by-step explanation:

To solve this system, we can either use substitution or graph it. I will be showing how to graph it.

To solve by graphing, we graph both of the equations and find the point at which the lines intersect.

[Graph]

-> See attached

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0

3 years ago

A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric f

likoan [24]

Step-by-step explanation:

Let x be the length and y be the width of the rectangular plot.

The plot is bounded on one side by a river and on the other three sides by a single-strand electric fence. It means,

x+2y = 1500

x = 1500 - 2y ....(1)

We know that the area of a rectangular plot is given by :

A = xy ....(2)

Put the value of x from equation (1) in (2)

$A=(1500-2y)y\\\\A=1500y-2y^2$ .....(3)

For largest area, differentiate above area equation wrt y.

$\dfrac{dA}{dy}=\dfrac{d}{dy}(1500y-2y^2)\\\\=1500-4y\\\\\text{Put}\ \dfrac{dA}{dy}=0\\\\1500-4y=0\\\\y=\dfrac{1500}{4}\\\\=375$

Put the value of y in equation (1).

x = 1500-2(375)

= 750 m

Put the value of y in equation (3).

$A =1500(375)-2(375)^2\\A=281250\ m^2$

Hence, the largest area is 281250 m² and its dimensions are 750 m and 375 m.

3 0

3 years ago