PRACTICE ON OPERATIONS WITH RADICALS

Problem 1 :

The formula for the kinetic energy of a moving object is

E = (1/2) mv²

where E is the kinetic energy in joules, m is the mass in kilograms and v is the velocity in meters per second.

a) solve the equation for v.

b) Find the velocity of an object whose mass is 0.6 kilograms and whose kinetic energy is 54 joules.

Solution :

E = (1/2) mv²

a) Solving for v,

2E = mv²

2E/m = v²

v = √(2E/m)

b) To find the velocity, we have to apply m = 0.6 kilogram and E = 54 joules.

v = √[(2 x 54)/0.6]

= √[(2 x 2 x 3 x 3 x 3)/0.6]

= (2 x 3) √[3/0.6]

= 6 √5

= 6(2.23)

= 13.38 per second.

Problem 2 :

Police officers can use the formula s = √(30 fd) to determine the speed s that a car was travelling in miles per hour by measuring the distance d is feet of its skid marks. In this formula f is the coefficient of friction for the type and condition of the road.

a)  Write a simplified expression for the speed if f = 0.6 for a wet asphalt road.

b)  What is the simplified expression for the speed if f = 0.8 for a dry asphalt road ?

c) an officer measures skid marks that are 110 feet long. Determine the speed of the car for both wet road conditions and for dry road conditions.

Solution :

s = √(30 fd)

a) f = 0.6

Applying the value of f, we get

s = √(30 x 0.6 d)

= √18d

= √(2 x 3 x 3 x d)

= 3 √2d

b) f = 0.8

Applying the value of f, we get

s = √(30 x 0.8 d)

= √24d

= √(2 x 2 x 2 x 3 x d)

= 2 √(2 x 3 x d)

= 2√6d

c)  f = 0.6 and d = 110 feet

s = √(30 fd)

= √(30 x 0.6 x 110)

= √1980

= 44.49

44 miles per hour.

f = 0.8 and d = 110 feet

s = √(30 fd)

= √(30 x 0.8 x 110)

= √2640

= 51.38

51 miles per hour.

Problem 3 :

If the cube has a surface area of 96 a², what is the volume ?

a)  32 a³     b)  48 a³    c)  64 a³     d) 96 a³

Solution :

Surface area of cube = 96 a²

Let x be the side length of cube.

6 x² = 96 a²

x² = 96 a²/6

x² = 16a²

x = √16a²

= 4a

So, side length of cube is 4a

Volume of cube = x³

= (4a)³

= 64 a³

So, option c is correct.

Problem 4 :

If x = 81 b² and b > 0, then  √x = ?

a)  -9b     b)  9b    c)  3b√27     d) 27b√3

Solution :

x = 81 b²

Taking square on both sides, we get

√x = √81 b²

√x = √(9 x 9 x b x b)

√x = 9b

So, option b is correct.

Problem 5 :

Find the perimeter and the area of a square whose sides measure 4 + 3√6 feet.

Solution :

Side length of the square = 4 + 3√6 feet.

Perimeter of square = 4 x side length

= 4 (4 + 3√6)

Using distributive property, we get

= 16 + 12√6

So, the perimeter of the square is 16 + 12√6 feet.

Problem 6 :

The voltage V required for a circuit is given by V = √PR, where P is the power in watts and R is the resistance in ohms. How many more volts are needed to light a 100 watt bulb that a 75 watt bulb if the resistance for both is 110 ohms ?

Solution :

V = √PR

P = power in watt

P = 100, R = 110 and P = 75 and R = 110

= √(100 x 110) -  √(75 x 110)

= √11000 - √8250

= 104.88 - 90.82

V = 14.05

14.05 more volts are needed.

Problem 7 :

Find the perimeter of a rectangle whose length is 8√7 + 4√5 inches and whose width is 5√7 - 3√5 inches.

Solution :

Perimeter of rectangle = 2(length + width)

Length =  8√7 + 4√5 inches

Width = 5√7 - 3√5 inches.

= 2(8√7 + 4√5 + 5√7 - 3√5)

= 2(8√7 + 5√7 + 4√5 - 3√5)

= 2(13√7 + √5)

Problem 8 :

The perimeter of the rectangle is 2√3 + 4√11 + 6 centimeters and its length is 2√11 + 1 centimeters. Find the width.

Solution :

Perimeter of the rectangle = 2√3 + 4√11 + 6

Length = 2√11 + 1

Width = ?

2(2√11 + 1 + width) = 2√3 + 4√11 + 6

(2√11 + 1 + width) = √3 + 2√11 + 3

Width = √3 + 2√11 + 3 - (2√11 + 1)

= √3 + 2√11 + 3 - 2√11 - 1

= 2 + √3

So, the width of the rectangle is 2 + √3 centimeter.

Problem 9 :

Two objects are dropped simultaneously. The first object reaches the ground in 2.5 seconds, and the second object reaches the ground 1.5 seconds later. From what heights were the two objects dropped ?

t = √h / 4

Solution :

t = √h / 4

t = 2.5 seconds

2.5 = √h / 4

2.5(4) = √h

10 = √h

Squaring on both sides

 √h² = 10²

h = 100

The time it took the second object to fall was 2.5+1.5 seconds or 4 seconds.

4 = √h / 4

4(4) = √h

16 = √h

 √h² = 16²

h = 256

The second object was dropped from 256 feet.

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