# MATH 1090-4 EXAM 2 SOLUTIONS

1. (a) Find an equation of a line parallel to the line x =
4 and passing

through the point (15,−2).

x = 15

(b) Find an equation of a line perpendicular to the line x = 4 and

passing through the point (15,−2).

y = −2

2. Find an equation of a line perpendicular to the line y = −2x + 3 and

passing through the point (5, 1).

(slopes of perpendicular lines are mutually
opposite and

reciprocal)

(point-slope equation of a line)

3. Find the intercepts and the vertex of the parabola y = 49 − (x − 3)^{2}

and determine whether the vertex is a minimum or a maximum.

x-intercepts:

y = 0

49 − (x − 3)^{2} = 0

49 = (x − 3)^{2}

x − 3 = 7 or x − 3 = −7

x = 10 or x = −4

(10, 0) and (−4, 0)

y-intercept:

x = 0

y = 49 − (−3)^{2}

y = 40

(0, 40)

Vertex is the point (3, 49) and it is a maximum since the leading coef-

ficient is negative.

4. The company will supply 50 units when the price of a product is $16

and 60 units when the price is $21. Find the supply function assuming

it is linear.

Since the supply function is linear, its graph is a line passing through

the points (50, 16) and (60, 21).

5. The demand function for a product is p = 12 − 2q where
p is the price

in dollars when q units are demanded. Form the revenue function R,

determine the level of production that will maximize the revenue and

the maximum revenue.

Since revenue=price*quantity,

R = (12 − 2q)q

R = −2q^{2} + 12q

The maximum is obtained when

The maximum revenue is

R = (12 − 2 · 3) · 3

R = 18

6. Solve the system

−4x + 6y = −10 (multiplying the first equation by -2)

y = −1 (adding the above to the second equation)

2x − 3(−1) = 5

2x = 2

x = 1

7. Nat invested $1000 at the rate 8% over 5 years compounded quarterly.

Find the compound amount at the end of 5 years.

8. The population of a town of 20000 declines 1% annually. Find the

population after 10 years.

P = 20000(0.99)^{t} where P is the population after t years.

P = 20000(0.99)^{10}

P = 18087.64

Approximately 18088.

9. Solve for x:

x = 2 (discard x = −1 because x is the base of a
logarithm, hence

it cannot be negative)

10. Solve for x:

11. Let a = log x and b = log y. Rewrite in terms of a and b.

12. Solve for x:

13. Solve for x:

x = 3 (discard x = −3 since x − 1 and x + 1 must be positive)

14. Solve for x:

15. The number of milligrams N of a radioactive substance
present after t

years is given by . After how many years will
there be

90 milligrams present?

After approximately 40 years.