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3 years ago

Given f(x) = 3 and g(x) = cos(x). What is lim [F(x) - g(x)]?

Mathematics

1 answer:

Flura [38]3 years ago

6 0

Answer:

Step-by-step explanation:

$\lim_{n \to \ -\pi } (f(x) - g(x))\\\\ = \lim_{n \to \ -\pi } (3 - cos x)\\\\ = 3 - cos( - \pi) = 3 - cos(\pi)\\\\ = 3 - (-1) = 3 + 1 = 4$

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As Sales Manager for ISeeYou Productions, you are planning to review the prices you charge clients for television advertisement

aksik [14]

The development fee p can be modeled by the equation p=-400q+9600 where p is the price the company charges and q is the number of contracts they get. The total revenue R ISeeYou obtains by signing q contracts is found by using the function: $R(q)=-400q^{2}+9600q$ . The cost ISeeYou Productions can be found by using the following function: C(q)=800q+16000. The monthlhy profit of Express ISeeYou Productions if found with the next function: $P(q)=-400p^{2}+8800q-16000$ . ISeeYou should sign 2 or 20 contracts to break even.

a) In order to solve part a of the problem, we will suppose the price will have a linear relation with the amount of contracts they signed, so we need to find two points to build this function.

We know that when signing 6 contracts, the company charges $7,200, so our first point will be:

(6, 7200)

we also know that the company signs 14 contracts when they charge $4,000. So our second point is:

(14, 4000)

so we can use these points to find the slope of the line by using the

slope formula:

$m=\frac{p_{2}-p_{1}}{q_{2}-q_{1}}$

$m=\frac{4000-7200}{14-6}$

so the slope is:

m=-400

We can now use one of those points and the slope to find the function for the price, so we get:

$y-y_{1}=m(x-x_{1})$

$p-p_{1}=m(q-q_{1})$

if we used the point: (6, 7200) we get:

$p-7200=-400(q-6)$

which can now be simplified:

p=-400q+2400+7200

so the price function is:

p=-400q+9600

b) In order to find the revenue equation we need to multiply the price equation by the amount of contracts they signed in a month, so we get:

R(q)=pq

$R(q)=(-400q+9600)q$

$R(q)=-400q^{2}+9600q$

c) The cost can be found by adding the fixed cost and the variable cost, so we get:

C(q)=800q+16000

d) The profit is found by subtracting the montly cost from the monthly revenue, so we get:

P(q)=R(q)-C(q)

$P(q)=-400q^{2}+9600q-(800q+16000)$

so we can now simplify our function to get:

$P(q)=-400q^{2}+9600q-800q-16000$

$P(q)=-400q^{2}+8800q-16000$

e) In order to find the number of contracts for the company to break even, we need to find the q contracts that will give a profit of $0 so we set the function of the previous part of the problem equal to zero so we get:

$P(q)=-400q^{2}+8800q-16000$

$-400q^{2}+8800q-16000=0$