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Mathematics

Grade 9

Numbers

Module 2

Easier algebra with exponents

Easier algebra with exponents

CLASS WORK

  • Do you remember how exponents work? Write down the meaning of “three to the power seven”. What is the base? What is the exponent? Can you explain clearly what a power is?
  • In this section you will find many numerical examples; use your calculator to work through them to develop confidence in the methods.

1 DEFINITION

2 3 = 2 × 2 × 2 and a 4 = a × a × a × a and b × b × b = b 3

also

(a+b) 3 = (a+b) × (a+b) × (a+b) and 2 3 4 = 2 3 × 2 3 × 2 3 × 2 3 size 12{ left ( { {2} over {3} } right ) rSup { size 8{4} } = left ( { {2} over {3} } right ) times left ( { {2} over {3} } right ) times left ( { {2} over {3} } right ) times left ( { {2} over {3} } right )} {}

1.1 Write the following expressions in expanded form:

4 3 ; (p+2) 5 ; a 1 ; (0,5) 7 ; b 2 × b 3 ;

1.2 Write these expressions as powers:

7 × 7 × 7 × 7

y × y × y × y × y

–2 × –2 × –2

(x+y) × (x+y) × (x+y) × (x+y)

1.3 Answer without calculating: Is (–7) 6 the same as –7 6 ?

  • Now use your calculator to check whether they are the same.
  • Compare the following pairs, but first guess the answer before using your calculator to see how good your estimate was.

–5 2 and (–5) 2 –12 5 and (–12) 5 –1 3 and (–1) 3

  • By now you should have a good idea how brackets influence your calculations – write it down carefully to help you remember to use it when the problems become harder.
  • The definition is:

a r = a × a × a × a × . . . (There must be r a’s, and r must be a natural number)

  • It is good time to start memorising the most useful powers:

2 2 = 4; 2 3 = 8; 2 4 = 16; etc. 3 2 = 9; 3 3 = 27; 3 4 = 81; etc. 4 2 = 16; 4 3 = 64; etc.

Most problems with exponents have to be done without a calculator!

2 MULTIPLICATION

  • Do you remember that g 3 × g 8 = g 11 ? Important words: multiply ; same base

2.1 Simplify: (don’t use expanded form)

7 7 × 7 7

(–2) 4 × (–2) 13

( ½ ) 1 × ( ½ ) 2 × ( ½ ) 3

(a+b) a × (a+b) b

  • We multiply powers with the same base according to this rule:

a x × a y = a x+y also a x + y = a x × a y = a y × a x size 12{a rSup { size 8{x+y} } =a rSup { size 8{x} } times a rSup { size 8{y} } =a rSup { size 8{y} } times a rSup { size 8{x} } } {} , e.g. 8 14 = 8 4 × 8 10 size 12{8 rSup { size 8{"14"} } =8 rSup { size 8{4} } times 8 rSup { size 8{"10"} } } {}

3 DIVISION

  • 4 6 4 2 = 4 6 2 = 4 4 size 12{ { {4 rSup { size 8{6} } } over {4 rSup { size 8{2} } } } =4 rSup { size 8{6 - 2} } =4 rSup { size 8{4} } } {} is how it works. Important words: divide ; same base

3.1 Try these: a 6 a y size 12{ { {a rSup { size 8{6} } } over {a rSup { size 8{y} } } } } {} ; 3 23 3 21 size 12{ { {3 rSup { size 8{"23"} } } over {3 rSup { size 8{"21"} } } } } {} ; a + b p a + b 12 size 12{ { { left (a+b right ) rSup { size 8{p} } } over { left (a+b right ) rSup { size 8{"12"} } } } } {} ; a 7 a 7 size 12{ { {a rSup { size 8{7} } } over {a rSup { size 8{7} } } } } {}

  • The rule for dividing powers is: a x a y = a x y size 12{ { {a rSup { size 8{x} } } over {a rSup { size 8{y} } } } =a rSup { size 8{x - y} } } {} .

Also a x y = a x a y size 12{a rSup { size 8{x - y} } = { {a rSup { size 8{x} } } over {a rSup { size 8{y} } } } } {} , e.g. a 7 = a 20 a 13 size 12{a rSup { size 8{7} } = { {a rSup { size 8{"20"} } } over {a rSup { size 8{"13"} } } } } {}

4 RAISING A POWER TO A POWER

  • e.g. 3 2 4 size 12{ left (3 rSup { size 8{2} } right ) rSup { size 8{4} } } {} = 3 2 × 4 size 12{3 rSup { size 8{2 times 4} } } {} = 3 8 size 12{3 rSup { size 8{8} } } {} .

4.1 Do the following:

  • This is the rule: a x y = a xy size 12{ left (a rSup { size 8{x} } right ) rSup { size 8{y} } =a rSup { size 8{ ital "xy"} } } {} also a xy = a x y = a y x size 12{a rSup { size 8{ ital "xy"} } = left (a rSup { size 8{x} } right ) rSup { size 8{y} } = left (a rSup { size 8{y} } right ) rSup { size 8{x} } } {} , e.g. 6 18 = 6 6 3 size 12{6 rSup { size 8{"18"} } = left (6 rSup { size 8{6} } right ) rSup { size 8{3} } } {}

5 THE POWER OF A PRODUCT

  • This is how it works:

(2a) 3 = (2a) × (2a) × (2a) = 2 × a × 2 × a × 2 × a = 2 × 2 × 2 × a × a × a = 8a 3

  • It is usually done in two steps, like this: (2a) 3 = 2 3 × a 3 = 8a 3

5.1 Do these yourself: (4x) 2 ; (ab) 6 ; (3 × 2) 4 ; ( ½ x) 2 ; (a 2 b 3 ) 2

  • It must be clear to you that the exponent belongs to each factor in the brackets.
  • The rule: (ab) x = a x b x also a p × b p = ab b size 12{a rSup { size 8{p} } times b rSup { size 8{p} } = left ( ital "ab" right ) rSup { size 8{b} } } {} e.g. 14 3 = 2 × 7 3 = 2 3 7 3 size 12{"14" rSup { size 8{3} } = left (2 times 7 right ) rSup { size 8{3} } =2 rSup { size 8{3} } 7 rSup { size 8{3} } } {} and 3 2 × 4 2 = 3 × 4 2 = 12 2 size 12{3 rSup { size 8{2} } times 4 rSup { size 8{2} } = left (3 times 4 right ) rSup { size 8{2} } ="12" rSup { size 8{2} } } {}

6 A POWER OF A FRACTION

  • This is much the same as the power of a product. a b 3 = a 3 b 3 size 12{ left ( { {a} over {b} } right ) rSup { size 8{3} } = { {a rSup { size 8{3} } } over {b rSup { size 8{3} } } } } {}

6.1 Do these, but be careful: 2 3 p size 12{ left ( { {2} over {3} } right ) rSup { size 8{p} } } {} 2 2 3 size 12{ left ( { { left ( - 2 right )} over {2} } right ) rSup { size 8{3} } } {} x 2 y 3 2 size 12{ left ( { {x rSup { size 8{2} } } over {y rSup { size 8{3} } } } right ) rSup { size 8{2} } } {} a x b y 2 size 12{ left ( { {a rSup { size 8{ - x} } } over {b rSup { size 8{ - y} } } } right ) rSup { size 8{ - 2} } } {}

  • Again, the exponent belongs to both the numerator and the denominator.
  • The rule: a b m = a m b m size 12{ left ( { {a} over {b} } right ) rSup { size 8{m} } = { {a rSup { size 8{m} } } over {b rSup { size 8{m} } } } } {} and a m b m = a b m size 12{ { {a rSup { size 8{m} } } over {b rSup { size 8{m} } } } = left ( { {a} over {b} } right ) rSup { size 8{m} } } {} e.g. 2 3 3 = 2 3 3 3 = 8 27 size 12{ left ( { {2} over {3} } right ) rSup { size 8{3} } = { {2 rSup { size 8{3} } } over {3 rSup { size 8{3} } } } = { {8} over {"27"} } } {} and a 2x b x = a 2 x b x = a 2 b x size 12{ { {a rSup { size 8{2x} } } over {b rSup { size 8{x} } } } = { { left (a rSup { size 8{2} } right ) rSup { size 8{x} } } over {b rSup { size 8{x} } } } = left ( { {a rSup { size 8{2} } } over {b} } right ) rSup { size 8{x} } } {}

end of CLASS WORK

TUTORIAL

  • Apply the rules together to simplify these expressions without a calculator.

1. a 5 × a 7 a × a 8 size 12{ { {a rSup { size 8{5} } times a rSup { size 8{7} } } over {a times a rSup { size 8{8} } } } } {} 2. x 3 × y 4 × x 2 y 5 x 4 y 8 size 12{ { {x rSup { size 8{3} } times y rSup { size 8{4} } times x rSup { size 8{2} } y rSup { size 8{5} } } over {x rSup { size 8{4} } y rSup { size 8{8} } } } } {}

3. a 2 b 3 c 2 × ac 2 2 × bc 2 size 12{ left (a rSup { size 8{2} } b rSup { size 8{3} } c right ) rSup { size 8{2} } times left ( ital "ac" rSup { size 8{2} } right ) rSup { size 8{2} } times left ( ital "bc" right ) rSup { size 8{2} } } {} 4. a 3 × b 2 × a 3 a × b 5 b 4 × ab 3 size 12{a rSup { size 8{3} } times b rSup { size 8{2} } times { {a rSup { size 8{3} } } over {a} } times { {b rSup { size 8{5} } } over {b rSup { size 8{4} } } } times left ( ital "ab" right ) rSup { size 8{3} } } {}

5. 2 xy × 2x 2 y 4 2 × x 2 y 3 2 xy 3 size 12{ left (2 ital "xy" right ) times left (2x rSup { size 8{2} } y rSup { size 8{4} } right ) rSup { size 8{2} } times left ( { { left (x rSup { size 8{2} } y right ) rSup { size 8{3} } } over { left (2 ital "xy" right ) rSup { size 8{3} } } } right )} {} 6. 2 3 × 2 2 × 2 7 8 × 4 × 8 × 2 × 8 size 12{ { {2 rSup { size 8{3} } times 2 rSup { size 8{2} } times 2 rSup { size 8{7} } } over {8 times 4 times 8 times 2 times 8} } } {}

end of TUTORIAL

Some more rules

CLASS WORK

1 Consider this case: a 5 a 3 = a 5 3 = a 2 size 12{ { {a rSup { size 8{5} } } over {a rSup { size 8{3} } } } =a rSup { size 8{5 - 3} } =a rSup { size 8{2} } } {}

Questions & Answers

what does the ideal gas law states
Joy Reply
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
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Felix Reply
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ALIYU
field is a region of space under the influence of some physical properties
Collete
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WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
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Adah
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
Nassze Reply
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Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
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Source:  OpenStax, Mathematics grade 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11056/1.1
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