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asambeis [7]
3 years ago
8

Classify the system of equations 1/3x+y+2=0 1/2x+y-5=0 A. intersecting B. parallel C. coincident

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

A. intersecting


Step-by-step explanation:

write the equations in the from of y = mx + c

Where m = slope

            c = y-intercept.

1/3x + y + 2 = 0

1/3x + y = -2

y = -1/3x - 2

1/2x + y - 5 = 0

1/2x + y = 5

y = -1/2x + 5

The equations are not parallel

-1/3x - 2 = -1/2x + 5

1/2x - 1/3x = 5 + 2

1/6x = 7

x = 7× 6/1

   = 42

y =  -1/3×42 -2

  = -14 - 2

 = -16

The equations intersect at point (42,-16)

sergiy2304 [10]3 years ago
4 0

Answer: A. intersecting


Step-by-step explanation:

1. To solve this problem and classify the system of equations shown above, you can graph it has you can see in the graph attached.

2. As you can see the lines intersect each other, this means that the system of equation has one solution and the lines are intersecting.

 3. Therefore, the answer is the option A: Intersecting.



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A store has been selling 100 Blu-ray disc players a week at $600 each. A market survey indicates that for each $40 rebate offere
Sav [38]

Answer:

<em>The rebate should be $220</em>

Step-by-step explanation:

<u>Demand Curve</u>

It's the relationship between price (P) and quantity (Q) demanded a certain product or service.

(a) We need to find the function that relates both magnitudes assuming a linear equation. The equation of a line can be found with the point-point formula:

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Two sets of data are given: 100 Blu-ray disc players are sold a week at $600 each. The ordered pair for this condition is (P,Q)=(600,100).

The other point comes from the market survey: The number of units sold will increase by 80 (100+80=180) when the price goes down $40 (600-40=560). The new point is (P,Q)=(560,180)

We set up the equation of the demand

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Rearranging

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Or

80P+40Q=48800

Simplifying

2P+Q=1120

(b) The revenue function is Q times the price

R=Q.P

Solving the equation of the demand for P

\displaystyle P=\frac{1120-Q}{2}

Thus, the revenue is

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(c) To find the optimum value of the revenue, we take the derivative of R and equate to 0

\displaystyle R'=\frac{1120-2Q}{2}=560-Q=0

Solving

Q=560 units a week

For which the revenue is

\displaystyle R=\frac{1120(560)-560^2}{2}

R=\$156,800

And the price is

\displaystyle P=\frac{1120-560}{2}=280

P=\$280

The rebate should be $600-$280=$220

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