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pentagon [3]
3 years ago
7

Find the domain of this problem, do not find the solution I have the solution it is x=2, I need the domain.

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

x > 1/2

Step-by-step explanation:

The lower limint of the domain of the log function, regardless of the base, is x > 0.  But that's not all.  If the right side is 2, then there's an upper limit to x.

If the solution is x = 2, then we have:


log             4.5x = 2

      2(2)-1

or...

log     9 = 2.  This is a true statement, since 9 = 3^2.

      3

----------------------------------------------------------------------------------------------------

The question here is whether or not there's an upper limit on x.

You claim that the solution is x=2.  Is that the upper limit of the domain?  Let's find out.  Let's try x = 3:

log       13.5

      5               = 2  iS THIS TRUE OR FALSE?

Use the change of base formula:

log     13.5 = log 13.5 / log 5 = 1.617.  This is actually less than 2

      5.

--------------------------------------------------------------------------------------------

Let me play the devil's advocate and hypothesize for a moment that the solution is not x = 2.  If not 2, then what is x?


(2x - 1) ^ log           (4.5x)     =    (2x -1)^2, or 4x^2 - 4x + 1

                    2x -1

Then 4.5x = 4x^2 - 4x + 1, or

4x^2 - 8.5x + 1 = 0

This is a quadratic function, with a = 4, b = - 8.5 and c = 1.  Then the discriminant is b^2 - 4(a)(c), which here is (-8.5)^2 - 4(4)(1) ), or 56.25.

This would lead to roots of

      - (-8.5) plus or minus √56.25

x = ---------------------------------------------------          

                            2(4)

x = 2 or x = 1/8    (You claim that the solution is x = 2).

Let's try once more to identify the domain:

(1) log          (4.5x) = 2  is valid only when the base, 2x-1, is greater than                

          2x-1               zero.  Setting 2x - 1 > 0, we get 2x ≥ 1. or x > 1/2.  For values of x less than 1/2, the base (2x - 1) would be 0 or smaller, which is not permissible when working with logs.

This domain, x > 1/2, applies only to the function log          y

                                                                                        2x-1

You claim that x = 2.  Does that satisfy x > 1/2, the domain of x?  Yes.  So that's promising.

Based upon these explorations, I'd claim that the domain of the log part of this equation is x > 1/2.

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The quantity demanded x of a certain brand of DVD player is 3000/week when the unit price p is $485. For each decrease in unit p
ANTONII [103]

Answer:

Demand Equation: q = 15125 - 25p.

Supply Equation: 80p - 24000 = 9q.

Equilibrium Price: $525.

Equilibrium Quantity: 2000 units.

Step-by-step explanation:

To solve this question, first, the demand equation has to be calculated. Let price be on the y-axis and quantity by on the x-axis. It is given that p = $485 when q = 3000 units. This can also be written as (q₁, p₁) = (3000, 485). It is also given that when the price goes down by $20, the quantity increases by 500 units. Therefore, (q₂, p₂) = (3500, 465). Due to the statement "For each decrease in unit price of $20 below $485, the quantity demanded increases by 500 units", the demand function will be linear i.e. a straight line. This is because the word "for each" has been used. Therefore, finding the equation of the demand function:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 485)/(q - 3000) = (465 - 485)/(3500 - 3000).

(p - 485)/(q - 3000) = (-20)/(500).

(p - 485)/(q - 3000) = -1/25.

25*(p - 485) = -1*(q - 3000).

25p - 12125 = -q + 3000.

25p + q = 15125.

q = 15125 - 25p. (Demand Equation).

Second, find the supply equation. It is given that the suppliers will not sell the product below or at the price of $300. Therefore, (q₁, p₁) = (0, 300). Also, the suppliers will sell 2000 units if the price is $525. Therefore, (q₂, p₂) = (2000, 525). Since it is mentioned that supply equation is linear, therefore:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 300)/(q - 0) = (525 - 300)/(2000 - 0).

(p - 300)/(q) = (225)/(2000).

(p - 300)/(q) = 9/80.

80*(p - 300) = 9*(q).

80p - 24000 = 9q. (Supply Equation).

Equilibrium exists when demand = supply. This means that the demand and the supply equations have to be solved simultaneously. Therefore, put demand equation in supply equation:

80p - 24000 = 9(15125 - 25p).

80p - 24000 = 136125 - 225p.

305p = 160125.

p = $525.

Put p = $525 in demand equation:

q = 15125 - 25p.

q = 15125 - 25(525).

q = 15125 - 13125.

q = 2000 units.

To summarize:

Demand Equation: q = 15125 - 25p!!!

Supply Equation: 80p - 24000 = 9q!!!

Equilibrium Price: $525!!!

Equilibrium Quantity: 2000 units!!!

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alekssr [168]

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Answer:

Step-by-step explanation:

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\frac{36}{180}= \frac{2*2*3*3}{2*2*3*3*5} =  \frac{1}{5}

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Answer:

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Step-by-step explanation:

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Answer: 9,600 for the 10% account 10,500 for the 13% why they wouldn't want to put it in the same account is because risky investment for the 13

Step-by-step explanation:

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