TRIANGLES 1

The two sides of an equilateral triangle are (4x-3) and (18+x). If the third side is e²-11, find the value of e.

Solution

All sides of an equilateral triangle are equal.

Equating the two sides

4x-3 = 18+x

4x-x -3 = 18        [Taking x on the left side]

4x-x = 18 + 3       [Taking 3 on the right side]

3x = 21       

3x = 21       [Dividing by 3 on both sides]

3      3

x = 7.

Finding the side; we use either (4x-3) or (18+x).

= 4x-3

= 4(7) – 3

= 28 – 3

= 25

Hence, each side of the triangle is 25

To find e we equate 25 with e²-11.

e²-11 = 25

e²  = 25 + 11

e²  = 36

√e²  = √36  put under root sign both sides

e = √36 

e = 6 

Hence e=6   

TRY THIS………………..

The two sides of an equilateral triangle are (3x-10) and (50-2x). If the third side is m²+147, find the value of m.

GRADIENT OR SLOPE 5

The slope of a line which passes through (-5, -12) and (c,-9) is ³/11. Find c.

Solution

x₁ = -5,  y₁ =-12,  x₂ = c,  y₂ = -9

m = y2 –y1

      x2 – x1

  3   =   -9 –(-12)

11         c –(-5)

  3   =    -9 + 12

11          c  + 5

  3   =          3   .

11           c  + 5

3(c + 5) = 3 x 11     [after cross multiplying]

3c + 15 = 33

3c = 33 - 15

3c = 18

3c = 18   

3      3

c = 6   [after dividing by 3 both sides]

Hence the slope is ³/11

TRY THIS………………..

The slope of a line which passes through (-4, -14) and (e,-7) is ³/10. Find e.

FUNCTIONS 11

Given that F(x) = 13x  +  65. Find F(-2)

Solution

F(x) = 13x  +  65

F(-2) = 13(-2)  +  65

F(-2) = -26  +  65

F(-2) = 39

Hence  F(-2) = 39

TRY THIS………………..

Given that F(x) = 17x  +  61. Find F(-2)

GRADIENT OR SLOPE 4

Find the slope of a line which passes through (-5, -12) and (6,-9)

Solution

x₁ = -5,  y₁ =-12,  x₂ = 6,  y₂ = -9

m = y2 –y1

       x2 – x1

m =   -9 –(-12)

           6 –(-5)

m =   -9 + 12

           6 + 5

m =     3

          11

Hence the slope is ³/11

TRY THIS………………..

Find the slope of a line which passes through (-15, -2) and (2,-9)

STANDARD FORM 4

Evaluate the following giving your answer in standard form.

982.7 x 10^-9

   5 x 10^-20

Solution

=  982.7 x 10^-9

       5 x 10^-20

=  982.7  x   10^-9

       5         10^-20

=  196.5 x 10^{-9 –(-20)}

=  196.5 x 10^{-9 + 20}    

=  196.5 x 10¹¹      then we change 116.54 into standard form as well

= 1.965 x 10² x 10¹¹   [when we have exponents with same base, we add 

   the powers.]

 

= 1.965 x 10¹³

= 1.97 x 10¹³ [correct to 2 d.p.]

Hence   982.7 x 10^-9   = 1.97 x 10¹³

                5 x 10^-20

TRY THIS………………..

Evaluate the following giving your answer in standard form.

0.01547 x 10^-5

   4 x 10^-30

FUNCTIONS 10

Find the inverse of { (12,6), (7,3), (-9,8), (-3,-6), (11,-7)  }

Solution

Here, the inverse is the opposite of a given expression. Therefore, the given values will be swapped. 

Hence, inverse is { (6,12), (3,7), (8,-9), (-6,-3), (-7,11)  }

TRY THIS………..

Find the inverse of { (1,-15), (10,3), (-9,8), (-7,-6), (11,-1) }

PERIMETER 1

Area of rectangle is 48m². Find its perimeter if width is 4m. 

Solution

A = L x w

48 = L x 4

48 = 4L

4      4

12 = L

Now; P =2(L + w)

= 2(12 + 4)

= 2 x 16

= 32m

Hence perimeter is 32m

TRY THIS………..

Area of rectangle is 80m². Find its perimeter if width is 4m.  

PROBABILITY 12

	A school survey found that 3 out of 10 students like pizza. If three students are chosen at random with replacement, what is the probability that all three students like pizza? Solution THIS IS AN INDEPENDENT EVENT. THE 1ST CHOICE HAS NO ANY EFFECT FOR THE 2ND OR 3RD.
	P(student 1 likes pizza)  =   3  10 P(student 2 likes pizza)  =   3  10 P(student 3 likes pizza)  =   3  10 P(student 1 and student 2 and student 3 like pizza)  =   3   ·   3   ·   3   =   27  10 10 10 1000

Hence probability that all three students like pizza is ²⁷/1000  .

TRY THIS………..

A school survey found that 3 out of 7 students like pizza. If three students are chosen at random with replacement, what is the probability that all three students like pizza?

LOGARITHMS 17

Simplify        Log0.0000001

                        Log(1/100)

Solution

=    Log0.0000001

       Log(1/100)

=    Log10^-7

       Log100^-1

=    Log10^-7

       Log(10²)^-1

=    Log10^-7

       Log10^-2

=    -7Log10

       -2Log10

=    -7

       -2

=    7/2

Hence           Log0.000001         =  ⁷/2

                      Log(1/100)

TRY THIS………..

Simplify        Log0.000001

                    Log(1/1000)

SETS 10

In a certain school, 190 students take either Chemistry or physics. 130 take chemistry and 100 take both subjects. Find those who take physics only.

Solution

In most cases, OR stands for union whereas AND/BOTH, stands for intersection.

Let Chemistry = n(C), physics = n(P).

n(C)= 130 ,

n(P)= ?

n(CuP) = 190,

n(CnP)= 100

Solution

n(CuP) = n(C) + n(P) - n(CnP)

190  =  130 + n(P)  – 100

190  =  n(P) + 30

190 - 30  =  n(P)

n(P)= 160

physics only = n(P) – n(CnP)

                   = 160 – 100

                   = 60 

Hence those taking only physics are 60.  

TRY THIS………..

In a certain school, 193 students take either Chemistry or physics. 125 take chemistry and 90 take both subjects. Find those who take physics only.