ALGEBRA D2

Expand (y – 3) (y – 5)

Solution

= (y – 3) (y – 5)

= y (y – 5) – 3 (y – 5)

= y² – 5y – 3y + 15

= y² – 8y + 15 answer

TRY THIS………..

Expand (a – 8) (a – 5)

TRIGONOMETRY D2

Given that Sin Ө  = 5/9, Find Cos Ө.

Solution

Sin²Ө + Cos²Ө = 1

(5/9)² + Cos²Ө = 1

Cos²Ө = 1 - (5/9)²

Cos²Ө = 1 - ²⁵/81

Cos²Ө =⁸¹/81 - ²⁵/81

Cos²Ө = ⁵⁶/81

CosӨ =  √(⁵⁶/81)

CosӨ =  √(4 x 14)       

                9

CosӨ =  √4 x √14      

                9

CosӨ = 2 √14      answer

                9

TRY THIS………

Given that Sin Ө  = 2/7, Find Cos Ө.

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PROBABILITY D1

A number is chosen at random from 21 – 30 inclusive. Find the probability that it is a prime or a multiple of 2.

Solution

Let n(S) represent sample space

P(M) = probability of a multiple of 2

P(R) = probability of prime number

n(S) = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30} = 10

n(M) = {22, 24, 26, 28, 30} = 5

n(R) = {23, 29} = 2

P(M) = ⁵/10

P(R) = ²/10

………………………….

P(MuR) = P(M) + P(R)

P(MuR) = ⁵/10 + ²/10

P(MuR) = ⁷/10 answer

TRY THIS…………………

A number is chosen at random from 21 – 30 inclusive. Find the probability that it is a prime or a multiple of 5.

TRIGONOMETRY D1

If Sin15x = 0.5 and 0⁰ ≤ x ≤ 360⁰, solve for x.

Solution

Sin15x = 0.5 ………….. (i)

Sin30⁰= 0.5………….. (ii)

equating (i) and (ii),

15x = 30⁰

15x = 30⁰

15      15

x = 2⁰

From the given question we see that sine is positive.

From the range of 0⁰ up to 360⁰ sine is positive in the 1st and 2nd quadrants.

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

In the 1st quadrant,

x= 10⁰

In the 2nd quadrant,

SinӨ= Sin(180⁰ – Ө)

Sin30⁰= Sin(180⁰ – 30⁰) = Sin150⁰

Then,

Sin15x = 0.5 ………….. (i)

Sin150⁰= 0.5………….. (iii)

Equating (i) and (iii),

15x = 150⁰

15x = 150⁰

15       15

x = 10⁰

Hence x = 2⁰ or x = 10⁰ answer

TRY THIS…………

If Sin(2x-10) = 0.8 and 0⁰ ≤ x ≤ 360⁰, solve for x.

CONGRUENCE D2

In the figure below prove that triangle DEH is congruent to triangle GHW

solution

 

Given: figure DEHGW .        

Required to prove: ∆DEH ≡ ∆GHW

Proof: EH = HW (given)

            DH = HG (given)

<EHD = <WHG (opposite angles)

Hence ∆DEH ≡ ∆GHW by SAS rule (proved)