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3 years ago

Use the substitution method to solve the system of equations. Choose the correct ordered pair. 3x – y = 7 2x – 2y = 2

Mathematics

2 answers:

quester [9]3 years ago

5 0

Y=2, and x=3 so it is c

makkiz [27]3 years ago

3 0

Answer:

The solution set is (3,2)

C is correct

Step-by-step explanation:

Given: System of equation.

$3x-y=7$

$2x-2y=2$

We will solve x and y using substitution method.

$3x-y=7$

$y=3x-7------------------(1)$

Substitute into 2x-2y=2

$2x-2(3x-7)=2$

$2x-6x+14=2$

$-4x=-12$

$x=3$

Put x=3 into eq(1)

$y=3(3)-7$

$y=2$

Hence, The solution set is (3,2)

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Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti

spayn [35]

Answer:

(a) $D(x)=-2,500x+60,000$

(b) $R(x)=60,000x-2500x^2$

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

$\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500$

Therefore, we have:

$y=-2500x+b$

At point (12,30000)

$30000=-2500(12)+b\\b=30000+30000\\b=60000$

Therefore:

$D(x)=-2,500x+60,000$

(b)Revenue

$R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2$

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

$R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12$

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

$R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0$

Therefore:

Optimal ticket price:$12
Maximum Revenue:$360,000

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