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dlinn [17]
3 years ago
10

The profits in hundreds of dollars, P(c), that a company can make from a product is modeled by a function of the price, c, they

charge for the product: P(c) = –20c2 + 320c + 5,120. What is the maximum profit the company can make from the product?
Mathematics
2 answers:
postnew [5]3 years ago
7 0

Answer:

The correct answer is B. 640,000

Step-by-step explanation:


Jobisdone [24]3 years ago
4 0
The maximum profit is the y-coordinate of the vertex of the parabola represented by the equation.

P(c)=-20c^2+320c+5120 \\ \\
a=-20 \\
b=320 \\ \\
\hbox{the vertex } (h,k): \\
h=\frac{-b}{2a}=\frac{-320}{2 \times (-20)}=\frac{-320}{-40}=8 \\
k=f(h)=f(8)=-20 \times 8^2+320 \times 8+5120= \\
=-1280+2560+5120=6400

The maximum value is 6400, but the profit is given in hundreds of dollars, so multiply the value by 100.

The maximum profit the company can make is $640,000.
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20 points + brainliest = 30 points!
riadik2000 [5.3K]

Answer:

see below

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7 0
3 years ago
Read 2 more answers
If f ( x ) f(x) is an exponential function where f ( − 3.5 ) = 25 f(−3.5)=25 and f ( 6 ) = 33 f(6)=33, then find the value of f
gavmur [86]

Given:

f(x) is an exponential function.

f(-3.5)=25, f(6)=33

To find:

The value of f(6.5).

Solution:

Let the exponential function is

f(x)=ab^x         ...(i)

Where, a is the initial value and b is the growth factor.

We have, f(-3.5)=25. So, put x=-3.5 and f(x)=25 in (i).

25=ab^{-3.5}         ...(ii)

We have, f(6)=33. So, put x=6 and f(x)=33 in (i).

33=ab^{6}         ...(iii)

On dividing (iii) by (ii), we get

\dfrac{33}{25}=\dfrac{ab^{6}}{ab^{-3.5}}

1.32=b^{9.5}

(1.32)^{\frac{1}{9.5}}=b

1.0296556=b

b\approx 1.03

Putting b=1.03 in (iii), we get

33=a(1.03)^{6}

33=a(1.194)

\dfrac{33}{1.194}=a

a\approx 27.63

Putting a=27.63 and b=1.03 in (i), we get

f(x)=27.63(1.03)^x

Therefore, the required exponential function is f(x)=27.63(1.03)^x.

4 0
3 years ago
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