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3 years ago

What is the slope of the line with equation y-3=-1/2(x-2)?

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

6 0

Answer:

slope = - $\frac{1}{2}$

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

y - 3 = - $\frac{1}{2}$ (x - 2) ← is in point- slope form

with slope m = - $\frac{1}{2}$

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$a)200x+250y \leq 44000\\ x+y \leq 150,\ where\ x\ is\ the\ number\ of\ children,\ and\ y\ is\ the\ number\ of\ adults.\\ c)The\ number\ of\ children\ and\ adults\ respectively\ maybe\ (20,100),\\ (100,20),(40,40),(80,60).$

Step-by-step explanation:

$We\ are\ given\ that,\\Cost\ of\ the\ Ticket\ To\ the\ Play\ for\ Adults= 250\ php\\Cost\ of\ the\ Ticket\ To\ the\ Play\ for\ Children=200\ php\\Maximum\ number\ of\ Adults\ and\ Children\ permitted\ to\ watch\ the\ play=150\ heads\\Maximum\ amount\ to\ be\ collected\ after\ the\ sales=44,000\ php$

$a)Let\ the\ number\ of\ Children\ who\ attended\ the\ play\ be\ x.\\Let\ the\ number\ of\ Adults\ who\ attended\ the\ play\ be\ y.\\Hence,\\(Total\ Cost\ of\ the\ ticket\ for\ Children)+(Total\ Cost\ of\ the\ ticket\ for\\ Adults) \leq 44,000.\\(No.\ of\ Children\ who\ attended\ the\ play)+(No.\ of\ Adults\ who\ attended\ the\ play) \leq 150.\\Hence,\\200x+250y \leq 44000\\x+y \leq 150$ $Hence,\\We've\ found\ the\ inequalities\ that\ best\ represent\ our\ conditions, and\ paired\\ them\ into\ a\ system\ of\ inequalities.$

$b) [Kindly\ refer\ the\ graph\ attachment]\\\\We've\ drawn\ the\ graph\ fixing\ the\ scale\ of\ 1\ unit\ to\ be\ 20\ heads (Adults\ or\ Children).\\The\ red\ slope\ and\ the\ red\ area\ shaded\ below\ it,\ represent\ Inequality\ 1.\\The\ blue\ slope\ and\ the\ blue\ area\ shaded\ below\ it,\ represent\ Inequality\ 2.\\As\ we\ already\ know\ that,\\The\ overlapped\ region\ contains\ the\ solutions\ to\ the\ system\ of\ Inequalities.$

$c)Considering\ any\ four\ points\ on\ the\ overlapped\ region,\ would\ work\ great.\\Also,\\We\ need\ to\ be\ a\ bit\ cautious\ here\ and\ consider\ solutions\ only\ from\ the\\ First\ Quadrant,\ as\ the\ physical\ quantity-Counting\ Numbers\ can\\ only\ be\ Whole\ Numbers.$

$Now,\\Lets\ consider\ points\ (20,100), (100,20),(40,40),(80,60).\\Lets\ verify\ for\ each\ of\ them\ separately:\\$

$I]Substituting\ x=20,\ y=100\ in\ Inequality\ 1:\\200x+250y \leq 44000\\200*20+250*100 \leq 44000\\4000+25000 \leq 44000\\29000 \leq 44000\\This\ is\ TRUE.\\\\Substituting\ x=20,\ y=100\ in\ Inequality\ 2:\\x+y \leq 150\\20+100 \leq 150\\120 \leq 150\\This\ is\ TRUE\ too.$

$II]\ Substituting\ x=100,\ y=20\ in\ Inequality\ 1:\\200x+250y \leq 44000\\200*100+250*20 \leq 44000\\20000+5000 \leq 44000\\25000 \leq 44000\\This\ is\ TRUE.\\\\Substituting\ x=100,\ y=20\ in\ Inequality\ 2:\\x+y \leq 150\\100+20 \leq 150\\120 \leq 150\\This\ is\ TRUE\ too.$

$III]\ Substituting\ x=40,\ y=40\ in\ Inequality\ 1:\\200x+250y \leq 44000\\200*40+250*40 \leq 44000\\8000+10000 \leq 44000\\18000 \leq 44000\\This\ is\ TRUE.\\\\Substituting\ x=40,\ y=40\ in\ Inequality\ 2:\\x+y \leq 150\\40+40 \leq 150\\80 \leq 150\\This\ is\ TRUE\ too.$

$IV]\ Substituting\ x=80,\ y=60\ in\ Inequality\ 1:\\200x+250y \leq 44000\\200*80+250*60 \leq 44000\\16000+15000 \leq 44000\\31000 \leq 44000\\This\ is\ TRUE.\\\\Substituting\ x=80,\ y=60\ in\ Inequality\ 2:\\x+y \leq 150\\80+60 \leq 150\\140 \leq 150\\This\ is\ TRUE\ too.$

$Hence,\ Verified.$

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3 years ago

What is the slope of the line with equation y-3=-1/2(x-2)?​

What is the slope of the line with equation y-3=-1/2(x-2)?