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3 years ago

Please answer and show work.

Mathematics

1 answer:

andrew11 [14]3 years ago

4 0

Simplifying
9 + -2x = 35

Solving
9 + -2x = 35

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-9' to each side of the equation.
9 + -9 + -2x = 35 + -9

Combine like terms: 9 + -9 = 0
0 + -2x = 35 + -9
-2x = 35 + -9

Combine like terms: 35 + -9 = 26
-2x = 26

Divide each side by '-2'.
x = -13

Simplifying
x = -13

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The figure of the prism is attached below.

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So, the Area of 3 rectangles will be = (4 x 6) + (3 x 6) + (5 x 6) = 72 square units

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Since the two lines are parallel and it is cut by a transversal,
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Answer:

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

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

$Hence,\ Verified.$

3 0

3 years ago