Algebra Applied Mathematics Word Problems

Question

Exam 2 Review​​​​​​​

 

2. Solve for x:  

 

3. Expand and simplify: 

 

4.  If an element has a half-life of 15 days and 184g is present initially

a) Find k.

b) How much will remain after 50 days?

c) How long until there is 1g left? 

 

5. Patricia wishes to have a rectangular-shaped garden in her backyard. She has 64 ft of fencing material with which to enclose her garden. Letting x denote the width of the garden, find a function f in the variable x giving the area of the garden.What is the domain?

 

6. The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $510 per person per day if exactly 20 people sign up for the cruise. However, if more than 20 people (up to the maximum capacity of 90) sign up for the cruise, then each fare is reduced by $5 per day for each additional passenger. Assume at least 20 people sign up for the cruise, and let x denote the number of passengers above 20.

(a) Find a function R giving the revenue per day realized from the charter.

(b) What will the revenue be if 72 people sign up?

 

7. ​Solve for x: 

 

8. ​In the pair of supply and demand equations below, x represents the quantity demanded in units of a thousand and p the unit price in dollars.

p = 74 - 4x2  and  p = x2 + 10x + 34    Find the equilibrium quantity.

 

9. ​Solve for x:      

10. Review Finance word problems from 2-27 notes.

Answer 1

a) The half-life formula for radioactive decay is:

N = N0 * (1/2)^(t/T)

where N is the amount of the radioactive element present at time t, N0 is the initial amount of the element, T is the half-life of the element, and k is the decay constant. We can use the given information to find k:

N/N0 = (1/2)^(t/T) 184/184 = (1/2)^(15/k) 1 = 2^(-15/k) k/15 = -ln(2) k = -15ln(2) ≈ -10.353

Therefore, k ≈ -10.353.

b) Using the half-life formula and the value of k from part (a), we can find the amount remaining after 50 days:

N/N0 = e^(kt) N/184 = e^(-10.353*50) N ≈ 34.39

Therefore, approximately 34.39g will remain after 50 days.

c) Using the half-life formula and the value of k from part (a), we can find the time it takes for the amount to decay to 1g:

N/N0 = (1/2)^(t/T) 1/184 = (1/2)^(t/15) t = 15ln(184/1)/ln(2) t ≈ 109.84

Therefore, it will take approximately 109.84 days for the amount to decay to 1g.

1. Let x be the width of the rectangular-shaped garden. Then, the length of the garden can be expressed as (64 - 2x)/2 (since two sides of the garden will be of length x and the remaining two sides will use up the remaining fencing material). Therefore, the area of the garden can be expressed as:

A(x) = x(64 - 2x)/2 A(x) = 32x - x^2

The domain of this function is the set of all non-negative real numbers, since the width of the garden cannot be negative and the total fencing material cannot be used up if the width of the garden is greater than 32 feet.

1. 
   

(a) Let R(x) be the revenue per day for x passengers above 20. Then:

R(x) = (510 - 5x)(20 + x) R(x) = -5x^2 + 110x + 10200

(b) To find the revenue when 72 people sign up, we can substitute x = 52 into the function R(x):

R(52) = -5(52)^2 + 110(52) + 10200 R(52) = $30,040

Therefore, the revenue will be $30,040 if 72 people sign up.

1. To find the equilibrium quantity, we need to find the quantity at which the quantity demanded and quantity supplied are equal. This occurs when the two equations are equal to each other, so we can set them equal and solve for x:

1. To find the equilibrium quantity, we need to find the quantity at which the quantity demanded and quantity supplied are equal. This occurs when the two equations are equal to each other, so we can set them equal and solve for x:

74 - 4x^2 = x^2 + 10x + 34 5x^2 - 10x - 40 = 0 x^2 - 2x - 8 = 0

Using the quadratic formula, we get:

x = (2 ± sqrt(4 + 32))/2 x = 1 ± sqrt(9) x = -2 or x = 4

Since we are dealing with a quantity of a product, we can discard the negative solution. Therefore, the equilibrium quantity is x = 4 thousand units.