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ZanzabumX [31]
3 years ago
12

2/5x = -3/15 + 1/x x=0 x=15 x=5 x=3

Mathematics
1 answer:
Taya2010 [7]3 years ago
7 0

Answer:

x = 3

Step-by-step explanation:

\dfrac{2}{5x}=-\dfrac{3}{15}+\dfrac{1}{x}\\\\\dfrac{1}{5}=\dfrac{5-2}{5x}\qquad\text{add $\frac{3}{15}-\frac{2}{5x}$}\\\\x=3\qquad\text{multiply by 5x}

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Step-by-step explanation:

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Divide nine by three then subtract two double the results
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Step-by-step explanation:

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A line contains points M(1, 3) and N(5, 0). What is the slope of MN? -4/3 -3/4 3/4 4/3
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A company’s profit in dollars is modeled by the equation p(x) = –0.5x2 + 100x, where x represents the number of units sold.
serg [7]

Answers:

a. The company needs to sell 100 units to reach the maxmum profit.

b. The company's maximum profit is $5,000.

Solution:

a. How many unit does the company need to sell to reach the maximum profit?

p(x)=-0.5x^2+100x

This is a quadratic equation, and its graph is a parabola vertical (because the veriable "x" is square). Comparing with the general form:

p(x)=ax^2+bx+c; a=-0.5, b=100, c=0

a=-0.5<0 (negative), then the parabola opens downward, and it has a maximimun value (maximum profit) at its vertex.

We can find the abscissa of the vertex (units that the company needs to sell to reach the maximum profit using the following formula:

x=-b/(2a)

Replacing b by 100 and a by -0.5 in the formula above:

x=-100/[2(-0.5)]

x=-100/(-1)

x=100

The company needs to sell 100 units to reach the maximum profit.


b. What's the company's maximum profit?  

To determine the maximum profit we substitute the value of "x" obrained in part "a" in the quadratic equation:

x=100→p(100)=-0.5(100)^2+100(100)

p(100)=-0.5(10,000)+10,000

p(100)=-5,000+10,000

p(100)=5,000

The company's maximum profit is $5,000.

6 0
3 years ago
Drag the tiles to the correct boxes to complete the palrs.
Brut [27]

Answer:

(f+g)(2)  = 4

(f-g)(4) = 8

(f ÷g)(2) = 7

(f x g)(1) = 0

Step-by-step explanation:

We are given these following functions:

f(x) = 2x + 3

g(x) = x - 1

(f+g)(2)

(f+g)(x) = f(x) + g(x) = 2x + 3 + x - 1 = 3x - 2

At x = 2

(f+g)(2) = 3(2) - 2 = 6 - 2 = 4

Then

(f+g)(2)  = 4

(f-g)(4)

(f-g)(x) = f(x) - g(x) = 2x + 3 - (x - 1) = 2x + 3 - x + 1 = x + 4

At x = 4

(f-g)(4) = 4 + 4 = 8

Then

(f-g)(4) = 8

(f ÷g)(2)

(f \div g)(x) = \frac{f(x)}{g(x)} = \frac{2x+3}{x-1}

At x = 2

(f \div g)(2) = \frac{7}{1} = 7

Then

(f ÷g)(2) = 7

(f x g)(1)

(f \times g)(x) = f(x)g(x) = (2x+3)(x-1) = 2x^2 -2x + 3x - 3 = 2x^2 + x - 3

Then

(f \times g)(1) = 2(1)^2 + 1 - 3 = 3 + 1 - 3 = 0

So

(f x g)(1) = 0

4 0
3 years ago
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