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3 years ago

The endpoints of EFare E(xE , yE) and F(xF , yF). What are the coordinates of the midpoint of EF?

Mathematics

2 answers:

jekas [21]3 years ago

3 0

The coordinates of the midpoint of EF is ((xE + xF)/2, (yE + yF)/2)
The third option is the correct answer.

algol133 years ago

3 0

Answer:

(C)

Step-by-step explanation:

It is given that the end points of EF are $E(x_{E},y_{E})$ and $F(x_{F},y_{F})$ .

Let the mid point of EF be M, then the coordinates of M can be found by using the mid point formula,

M= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

M= $(\frac{x_{E}+x_{F}}{2},\frac{y_{E}+y_{F}}{2})$

Thus, option C is correct.

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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:

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

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