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Simplify the following radical expressions :
Question 1 :
√3 + √12
Question 2 :
√75 + √3
Question 3 :
√18 + √98
Question 4 :
√5 + √20 - √125
Question 5 :
√5 + 3√7 - 4√5 - 5√7
Question 6 :
3√3 + 4√3 - √2
Question 7 :
2(√5 - √3) + 3(√3 - √5)
Question 8 :
3√16 + 3√54
Question 9 :
√25 + 53√64
Question 10 :
83√686 - 53√250
Question 11 :
√(12w) + √(27w)
Question 12 :
√45y3 + √25y3
Question 13 :
3√8x3y6 + √9x2y4
Question 14 :
√4p2q4 - 3√125p3q6
Question 15 :
A square has sides each measuring 2√7 feet. Determine the area of the square.
Question 16 :
The period of a pendulum is the time required for it to make one complete swing back and forth. The formula of the period P of the pendulum is P = 2 π√(l/32), where l is the length of the pendulum in feet. If a pendulum in a clock tower is 8 feet long, find the period. Use 3.14 for π.
Question 17 :
A rectangle has width 3√5 centimeters and length 4√10 centimeters. Find the area of the rectangle.
Question 18 :
A rectangle has length √(a/8) meters and width √(a/2) meters. What is the area of the rectangle ?
Question 19 :
The formula for the area A of the square with side length s is A = s2. Solve this equation for s, find the side length of a square having an area of 72 square units.
Question 20 :
Find the area of the rectangle
1) 3√3
2) 6√3
3) 10√2
4) -2√5
5) -3√5 - 2√7
6) 7√3 - √2
7) -√5 + √3
8) 53√2
9) 25
10) 313√2
11) 5√3w
12) 5y√5y
13) 5xy2
14) - 3pq2
15) area of the square is 28 square feet.
16) the length of period is 3.14 feet.
17) 60 √2 cm2
18) area of the rectangle is a/4 square meter.
19) the side length of the square is 6 √2 units.
20) the area of the rectangle is 18√30 - 38√2 square units.
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.
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.
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
Problem 4 :
If x = 81 b2 and b > 0, then √x = ?
a) -9b b) 9b c) 3b√27 d) 27b√3
Problem 5 :
Find the perimeter and the area of a square whose sides measure 4 + 3√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 ?
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
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.
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
1) a) v = √(2E/m)
b) 13.38 per second.
2) a) 3 √2d
b) 2√6d
c) 44 miles per hour, 51 miles per hour.
3) 64 a3
4) 9b
5) the perimeter of the square is 16 + 12√6 feet.
6) 14.05 more volts are needed.
7) 2(13√7 + √5)
8) the width of the rectangle is 2 + √3 centimeter.
9) The second object was dropped from 256 feet.
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