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Xelga [282]
3 years ago
12

Graph these equations: x+2y=6, y=-1/2x+4 How many solutions does the system of equations have?

Mathematics
1 answer:
inysia [295]3 years ago
7 0

Answer:

The system has no solution

Step-by-step explanation:

we have

x+2y=6

Isolate the variable y

2y=-x+6

Divide by 2 both sides

y=-\frac{x}{2}+3 ---> equation A

y=-\frac{x}{2}+4 ---> equation B

Compare the equations

Equation A and equation B have the same slope and different y-intercept

Remember that

If two lines have the same slope, then the lines are parallel

The solution of the system is the intersection point both graphs

In this problem, the lines don't intersect, therefore, the system has no solution

using a graphing tool

The graph in the attached figure

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Answer:

C

Step-by-step explanation:

A straight diagonal line going the middle of the graph is a proportianal relationship.

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uppose we want to build a rectangular storage container with open top whose volume is $$ cubic meters. Assume that the cost of m
Ainat [17]

Answer:

a = length of the base = 2.172 m

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Step-by-step explanation:

Suppose we want to build a rectangular storage container with open top whose volume is 12 cubic meters. Assume that the cost of materials for the base is 12 dollars per square meter, and the cost of materials for the sides is 8 dollars per square meter. The height of the box is three times the width of the base. What’s the least amount of money we can spend to build such a container?

lets call a = length of the base

b = width of the base

c = height

V = a.b.c = 12

Area without the top:

Area = ab + 2bc + 2ac

Cost  = 12ab + 8.2bc + 8.2ac

Cost = 12ab + 16bc + 16ac

height = 3.width

c = 3b

Cost = 12ab + 16b.3b + 16a.3b = 12ab + 48b² + 48ab = 48b² + 60ab

abc = 12 → ab.3b = 12 → 3ab² = 12 → ab² = 4 → a = 4/b²

Cost = 48b² + 60ab = 48b² + 60b.4/b² = 48b² + 240/b

C(b) = 48b² + 240/b

C'(b) = 96b - 240/b²

Minimum cost: C'(b) = 0

96b - 240/b² = 0

(96b³ - 240)/b² = 0

96b³ - 240 = 0

96b³ = 240

b³ = 240/96

b³ = 2.5

b = 1.357m

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Answer:

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Step-by-step explanation:

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2 years ago
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ad-work [718]

\bf \begin{array}{|cc|ll} \cline{1-2} \stackrel{x}{year}&\stackrel{y}{price}\\ \cline{1-2} 0&1.12\\ 3&1.96\\ \cline{1-2} \end{array}~\hspace{10em} (\stackrel{x_1}{0}~,~\stackrel{y_1}{1.12})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1.96}) \\\\\\ \stackrel{\textit{rate of change}}{slope = m\implies} \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1.96-1.12}{3-0}\implies \cfrac{0.84}{3}\implies 0.28

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