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viva [34]
3 years ago
12

Which value is equivalent to 7 x 5 x 2 divided by 7 x 3 to the 2nd power x 5 to the 0 power divided by 2 to the negative 3rd pow

er x 2 to the negative 9 power
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

This is hard to read, but i guess that the equation is:

7*5*2/7*3^2*5^0/2^{-3}*2^{-9}

So we have some things to solve:

First, for any number:

x^0 = 1

then 5^0 = 1.

And for negative powers,

x^-n = (1/x)^n.

And we

Then we can rewrite our equation as:

\frac{7*5*2*3^2*1*2^3}{7*2^9}

now we can cancel the seven in the numerator and the denominator

\frac{5*2*3*2^3}{2^9}

now, remember the relation:

a^x*a^y = a^(x+y)

then:

2*2^3 = 2^4

and 2^4/2^9 = 2^-5 = (1^2)^5

Then our equation is:

\frac{5*3}{2^5}

15/2^5 = 0.46875

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katen-ka-za [31]

Using the vertex of a quadratic function, it is found that:

a) The revenue is maximized with 336 units.

b) The maximum revenue is of $56,448.

<h3>What is the vertex of a quadratic equation?</h3>

A quadratic equation is modeled by:

y = ax^2 + bx + c

The vertex is given by:

(x_v, y_v)

In which:

  • x_v = -\frac{b}{2a}
  • y_v = -\frac{b^2 - 4ac}{4a}

Considering the coefficient a, we have that:

  • If a < 0, the vertex is a maximum point.
  • If a > 0, the vertex is a minimum point.

The demand function is given by:

p(x) = 336 - 0.5x.

Hence, the revenue function is:

R(x) = xp(x)

R(x) = -0.5x² + 336x.

Which has coefficients a = -0.5, b = 336.

Hence, the value of x that maximizes the revenue, and the maximum revenue, are given, respectively, as follows:

  • x_v = -\frac{336}{2(-0.5)} = 336
  • y_v = -\frac{336^2 - 4(-0.5)(0)}{4(-0.5)} = 56448

More can be learned about the vertex of a quadratic function at brainly.com/question/24737967

#SPJ1

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LUCKY_DIMON [66]

Answer:

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Step-by-step explanation:

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