INEQUALITIES

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If 10x – 3 ≤ x + 15 ≤ 6x ; find x

Solution

10x – 3 ≤ x + 15 and  x + 15 ≤ 6x

10x – x ≤ 3+ 15 and  15 ≤ 6x - x

9x ≤ 18(divide by 9 both sides) and  15 ≤ 5x (divide by 5 both sides)

x ≤ 2 and  3 ≤ x

x ≤ 2 and  x ≥ 3 

TRY THIS...........

If 10x – 3 ≤ x + 36 ≤ 13x ;  find x

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FINDING SLOPE 2

Find the slope of a line which passes through (8, -2) and (6,-9)

Solution

x₁ = 8,  y₁ =-2,  x₂ = 6,  y₂ = -9

m = y2 –y1

       x2 – x1

m =   -9 –(-2)

          6 –8

m =   -9 + 2

           -2

m =    - 7

          -2

Hence the slope is 7/2

TRY THIS……………………

Find the slope of a line which passes through (0, -2) and (-5,-9)

MATRIX-2

ARITHMETICS 1

Evaluate 45 x 237 + 463 x 45.

Solution

= 45 x 237 + 463 x 45

= 45 x (237 + 463)

= 45 x 700

= 31500

Hence 45 x 237 + 463 x 45 = 31500

TRY THIS………….

Evaluate 127 x 516 + 484 x 127.

FINDING SLOPE 1

Find the slope of a line which passes through (10, 1) and (14, 8)

Solution

x₁ = 10,  y₁ =1,  x₂ = 14,  y₂ = 8

m = y2 –y1

       x2 – x1

m =   8 –1

        14 –10

m =   7

         4

Hence the slope is 7/4

TRY THIS………………..

Find the slope of a line which passes through (-4, -23) and (1, 8)

RADICALS 1

DIFFERENCE OF 2 SQUARES 1

Evaluate 780² – 680²

Solution

780² – 680² = (780 + 680)( 780 – 680)

                   = (1460)( 100)

                   = 146000

Hence 780² – 680² = 146000

TRY THIS……………………..

Evaluate 780² – 680²

PROBABILITY 1

The probability that Simba will beat Yanga in the national league is ⁹/13. Find the probability that it won’t be able to do so.

Solution

Let P(E)=  ⁹/13         P(E’) = ?

P(E)   + P(E’) = 1

⁹/13 + P(E’) = 1

P(E’) = 1 -  ⁹/13

P(E’) = ¹³/13 -  ⁹/13

P(E’) =  13-9

               13

P (E’) =  4

             13

Hence the probability that it won’t beat Yanga is ⁴/13.

TRY THIS…………………………………..

The probability that Mgambo JKT FC will beat Mbeya City FC in the national league is ⁹/30. Find the probability that it won’t be able to do so.

MATRIX 1

ARITHMETIC PROGRESSION 1

Find the sum of the 1st 16 terms of the following arithmetic progression: 5+11+17+…………………

Solution

d=6, A₁ = 5,n=16

S_n = n[2A₁ + (n-1)d]

       2

S₁₆ = 16[2(5) + (16-1)d]

         2

S₁₆ = 8[10 + 15d]

 

S₁₆ = 8[10 + 13(6)]       since d=6

S₁₆ = 8[10 + 78]  

         

S₁₆ = 8 x 88 

          

S₁₆ = 704

Therefore the sum of 1st 16 terms is 704.

TRY THIS............

Find the sum of the 1st 12 terms of the following arithmetic progression: 7+10+13+…………………

ALGEBRA 1

STATISTICS 1

In Civics test the following marks were recorded.

marks	10-19	20-29	30-39	40-49	50-59	60-69	70-79
No. of students	4	5	7	9	6	5	4

Calculate the mean.

Solution

In this question we are going to apply the method of assumed mean.

Here you are required to produce the frequency distribution table.

Let us take 54.5 as our assumed mean (A). Take away the assumed mean from each class mark.

Class interval	Class mark (x)	d=x-A	Frequency(f)	fx	fd
10-19	14.5	-40	4	58	-160
20-29	24.5	-30	5	122.5	-150
30-39	34.5	-20	7	241.5	-140
40-49	44.5	-10	9	400.5	-90
50-59	54.5	0	6	327	0
60-69	64.5	10	5	322.5	50
70-79	74.5	20	4	298	80
			∑f = 40	∑fx = 1770	∑fd = -410

Mean = A +  ∑fd

                       ∑f

Mean = 54.5 +  -410

                             40

Mean = 54.5 +  (-10.25)

                    

Mean = 44.25       

         

Hence Mean =  44.25

TRY THIS………..

In Civics test the following marks were recorded.

marks	10-14	15-19	20-24	25-29	30-34	35-39	40-44
No. of students	4	5	7	9	6	5	4