What is the minum of two colors needed?

Mathematics

1 answer:

Alexus [3.1K]2 years ago

8 0

4 colour needed for this…i think

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Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$

6 0

2 years ago

Is y equals 3x + 2 proportional number or non proportional

vodomira [7]

Answer:

Non-proportional

Step-by-step explanation:

because it has a y-intercept of 2 in order for it to be proportional its y-intercept should be 0.

3 0

3 years ago

Conplete the square to solve the quadratic equation. x<br><img src="https://tex.z-dn.net/?f=1%20%7Bx%7D%5E%7B2%7D%20%20-%206x%20

o-na [289]

Answer:

$\large\boxed{x=4\pm2\sqrt6}$

Step-by-step explanation:

$x^2-6x+12=2x+20\qquad\text{subtract}\ 2x\ \text{and}\ 12\ \text{from both sides}\\\\x^2-8x=8\\\\x^2-2(x)(4)=8\qquad\text{add}\ 4^2=16\ \text{to both sides}\\\\x^2-2(x)(4)+4^2=8+4^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-4)^2=24\iff x-4=\pm\sqrt{24}\\\\x-4=\pm\sqrt{4\cdot6}\\\\x-4=\pm\sqrt4\cdot\sqrt6\\\\x-4=\pm2\sqrt6\qquad\text{add 4 to both sides}\\\\x=4\pm2\sqrt6$