Step-by-step explanation:
The system of linear equations are , (i) y = 2x and (ii) y = x + 1 . We will find some coordinates and then we will plot its graph. The point where both Graphs will meet will be the solution of the graph .
<u>•</u><u> </u><u>Findin</u><u>g</u><u> </u><u>coordinates</u><u> </u><u>of </u><u>Equ</u><u>ⁿ</u><u> </u><u>(</u><u>i</u><u>)</u><u> </u><u>:</u><u>-</u>
Step 1 : <u>Put</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>0</u><u> </u><u>:</u><u>-</u>

Step 2: <u>Put</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>


<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>
<u>•</u><u> </u><u>Finding</u><u> </u><u>coord</u><u>inates</u><u> </u><u>of</u><u> </u><u>equⁿ</u><u> </u><u>(</u><u>ii</u><u>)</u><u> </u><u>:</u><u>-</u>
Step 1 : Put x = 0 :-

Step 2: Put x = 1 :-


Now plot these points on the graph Taking appropriate scale . Refer to the attachment in graph . From graph the Solution is (1,2) .
Answer:
30
Step-by-step explanation:
you divide 20 by 4 then multiply that by 6
Answer:9 times 10 equal 90 $
Step-by-step explanation:
Check the picture below.
so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2} \\\\\\ AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AA%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B8%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B%5B6-%28-4%29%5D%5E2%2B%5B8-%28-2%29%5D%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B%286%2B4%29%5E2%2B%288%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B10%5E2%2B10%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B10%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BAC%3D10%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2} \\\\\\ BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2} \\\\\\ BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AB%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B6%7D%29%5Cqquad%20%0AD%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%5Cqquad%20%5Cqquad%20BD%3D%5Csqrt%7B%5B4-%28-2%29%5D%5E2%2B%5B0-6%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B%284%2B2%29%5E2%2B%28-6%29%5E2%7D%5Cimplies%20BD%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B6%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BBD%3D6%5Csqrt%7B2%7D%7D)
that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.
namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,
Answer:
Congruence between pentagons
Step-by-step explanation:
The relationship occurs because having two congruent Pentagons and generating a segment within them (or outside), congruence is extrapolated to the triangles generated within them. Thus, if there is congruence among the pentagons, it will exist between the formed triangles. In other words since the two pentagons are congruent, the corresponding angle pair is congruent. also the two corresponding side pairs are also congruent.
In the attached image for example, the ABCDE and KLMNO pentagons are congruent, so all of their internal division lines are also congruent (AC and KM)