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3 years ago

Suppose your demand function is given by D(q) = - q? - 2q + 597, where q is thousands of units sold and

Mathematics

1 answer:

Dahasolnce [82]3 years ago

8 0

Part A

q = 9 represents selling 9000 units, since q is thousands of units sold

Plug this into the D(q) function

D(q) = -q^2 - 2q + 597

D(9) = -(9)^2 - 2(9) + 597

D(9) = 498

The price per unit should be $498

You have the correct answer.

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Part B

Plug in D(q) = 477. Then solve for x.

D(q) = -q^2 - 2q + 597

477 = -q^2 - 2q + 597

-q^2 - 2q + 597 = 477

-q^2 - 2q + 597-477 = 0

-q^2 - 2q + 120 = 0

q^2 + 2q - 120 = 0

(q+12)(q-10) = 0

q+12 = 0 or q-10 = 0

q = -12 or q = 10

Ignore negative q values. It is not possible to have negative demand.

So if the unit price is $477, then you can expect to sell 10,000 units.

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Part C

Plug D(q) = 0 and solve for q

-q^2 - 2q + 597 = 0

q^2 + 2q - 597 = 0

1q^2 + 2q + (-597) = 0

1x^2 + 2x + (-597) = 0

We have an equation in the form ax^2+bx+c = 0 with a = 1, b = 2, c = -597

Use the quadratic formula to solve for x

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(2)\pm\sqrt{(2)^2-4(1)(-597)}}{2(1)}\\\\x = \frac{-2\pm\sqrt{2392}}{2}\\\\x \approx \frac{-2\pm48.90807704}{2}\\\\x \approx \frac{-2+48.90807704}{2} \text{ or } x \approx \frac{-2-48.90807704}{2}\\\\x \approx \frac{46.90807704}{2} \text{ or } x \approx \frac{-50.90807704}{2}\\\\x \approx 23.45403852 \text{ or } x \approx -25.45403852\\\\$

We ignore any negative solution. The only practical solution is roughly x = 23.454, so q = 23.454 is when D(q) is equal to 0

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Write these numbers in order from least to greatest <br> 0.65, 2/3, 3/5 , 0.5

svp [43]

Answer:

Step-by-step explanation:

Convert the fractions to the decimals (you can use the calculator):

2/3 = 2 : 3 = 0.6666...

3/5 = 3 : 5 = 0/6

We have

0.65

0.666...

0.6

0.5

The order from least to greatest:

0.5

0.6

0.65

0.666..

$\begin{array}{c|c|c|c|c}0&.&5\\0&.&6\\0&.&6&5\\0&.&6&6&6...\end{array}$

8 0

3 years ago

The top of a 15-foot ladder leans against a wall, and the bottom of the ladder slides away from the wall at a rate of 4 ft/sec.

Andrej [43]

The ladder, the ground and the wall form a right triangle; the ladder length (L) is the longest side of this triangle. L^2 = h^2 + x^2, where h represents the height of the point on the wall where the ladder touches the wall, and x represents the distance of the base of the ladder from the wall.

We need dh/dt, which will be negative because the top of the ladder is sliding down the wall.

Starting with h^2 + x^2 = L^2, we differentiate (and subst. known values such as x = 5 feet and 4 ft/sec to find dh/dt. Note that since the ladder length does not change, dL/dt = 0. This leaves us with

dh dx
2h ---- + 2x ----- = 0.
dt dt

Since x^2 + h^2 = 15^2 = 225, h^2 = 225 - (5 ft)^2 = 200, or
200 ft^2 = h^2. Then h = + sqrt(200 ft^2)

Substituting this into the differential equation, above:

2[sqrt(200)] (dh/dt) + 2 (5) (4 ft/sec) = 0. Solve this for the desired quantity, dh/dt:

[sqrt(200)] (dh/dt) + (5)(4) = 0, or

dh/dt = -20 / sqrt(200) = (-1.41 ft / sec) (answer)

This result is negative because the top of the ladder is moving downward.

4 0

4 years ago

What times what equals 98 and adds up to -21

lesya [120]

-7 & -14 because

(-7) + (-14) = -21
(-7)*(-14)=98

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3 years ago

What is 1/6 divided by 3

ira [324]

1/6 divided by 3 is 1/18

1/6÷3/1 then do reciprocal so it is 1/6×1/3 equals 1/18

6 0

3 years ago

Compute the sum:

Nady [450]

You could use perturbation method to calculate this sum. Let's start from:

$S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}$

On the other hand, we have:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}$

So from (1) and (2) we have:

$\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}$

Now, let's try to calculate sum $\sum\limits_{k=0}^{n}k\cdot k!$ , but this time we use perturbation method.

$S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\
$

but:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}$

When we join both equation there will be:

$\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1$

So the answer is:

$\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}$

Sorry for my bad english, but i hope it won't be a big problem :)

8 0

4 years ago