This slideshare is about expressions. It is divided into expressions 1 and expressions 2. Expression 1 focuses on evaluating expressions which deals with substituting the known values in an equation to solve for the unknown. While, expressions 2 focuses on transposition of formula; change of subject of formula in a given equation. For more details, contact Youtube: https://www.youtube.com/channel/UC_FYPDg12rH5Ir3s7LLBgeg?view_as=subscriber Facebook: Adjex Academy LinkedIn: Adjex Academy Slideshare: Adjex Academy Instagram: adjex_academy E-mail: adjexacademy@gmail.com

1. EXPRESSIONS
BY ADJEKUKOR, CYNTHIA UFUOMA

2. CONTENTS
Expressions 1
Expressions 2
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3. EXPRESSIONS 1
EVALUATING EXPRESSIONS

4. EVALUATING EXPRESSIONS
 If v = (π h) / 6 (3R2 + h2)
 Determine the value of v when h=2.85 ; R= 6.24 ; π= 3.142
 Substitute the values in the expression
 V= (3.142 × 2.85)/6 (3 × 6.242 + 2.852)
 V= (3.142 × 2.85)/6 ( 3 × 38.938 + 8.123)
 V=(3.142 × 2.85)/6 (116.814 + 8.123)
 V= (3.142 × 2.85)/6 (124.937)
 V= (3.142 × 2.85 × 124.937) / 6
 V= 1118.773 / 6
 V= 186.462
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5. Evaluating Expressions
If R= (R1R2) / R1 + R2
where R1=276 ; R2=145 , what is R
Substitute the values in the expression
R = (276 × 145) / (276 + 145)
R = 40020 / 421
R= 95.6
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6. Evaluating Expressions
 If V= (πb) / 12 (D2 + Dd + d2) evaluate v to 3s.f
 Where b= 1.46; D= 0.864; d= 0.517; π= 3.142
 Substitute the values in the expression
 V= (3.142 × 1.46) / 12 (0.8642 + 0.864 × 0.517 + 0.5172)
 V= (3.142 × 1.46) / 12 (0.7465 + 0.4467 + 0.2673)
 V= (3.142 × 1.46) / 12 (1.4605)
 V= (3.142 × 1.46 × 1.4605) / 12
 V= 6.69978/12
 V= 0.558315
 V≈ 0.558 to 3 s.f
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7. Evaluating Expressions
What is the value of r in I= (nE) / (R + nr) Where
n=6; E=2.01; R=12; I=0.98
Substitute the values in the expression
0.98= (6 × 2.01) / (12 + 6r)
0.98 = 12.06 / (12 + 6r)
Cross multiply
0.98 (12 + 6r) =12.06
11.76 + 5.88r = 12.06
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8. Evaluating Expressions
Collect like terms
5.88r = 12.06 − 11.76
5.88r= 0.3
Divide both sides by 5.88
r= 0.3/5.88
r= 0.05102
r≈ 0.051
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9. EXPRESSIONS 2
TRANSPOSITION OF FORMULA

10. TRANSPOSITION OF FORMULA
 Transpose the formula v= u + at to make ‘a’ the subject of
formula
 v= u + at we want ‘a’ to stand alone
 Move ‘u’ to the other side of the equation
 v − u= at
 This can be written as v − u= a × t
 We still want ‘a’ to stand on its own, what can we do?
 Divide both sides by t
 (v − u) / t= a
 Thus, a= (v − u) / t
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11. Transposition of Formula
 Transpose d= 2 √h(2r − h) to make r the subject
 d= 2 √h(2r − h) we want ‘r’ to stand alone.
 This can be written as d= 2 × √h(2r − h)
 We can divide both sides by 2
 d/2 = √h(2r − h)
 Square both sides
 (d/2)2 = [√h(2r − h)]2
 d2 /4 = h(2r − h)
 Expand the bracket at the LHS
 d2 /4 = 2rh − h2
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12. Transposition of Formula
Move h2 to the other side
d2 /4 + h2 = 2rh
Evaluate the LHS
(d2 + 4h2 ) / 4 = 2rh
Divide both sides by 2h
(d2 + 4h2 ) / 4 ÷ 2h = 2rh / 2h
(d2 + 4h2 ) / (4 × 2h) = r
(d2 + 4h2 ) / 8h = r
r = (d2 + 4h2 ) / 8h
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13. Transposition of Formula
 Transpose v= [πh (3R2 + h2)] / 6 make R the subject of the
formula
 v= [πh (3R2 + h2)] / 6
 Cross multiply
 6v = πh (3R2 + h2)
 This can also be written as 6v = πh × (3R2 + h2)
 Divide both sides by πh
 6v / πh = 3R2 + h2
 move h2 to the LHS
 (6v / πh) − h2 = 3R2
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14. Transposition of Formula
 Evaluate the LHS
 (6v − πh3) / πh = 3R2
 Divide both sides by 3
 (6v − πh3) / πh ÷ 3 = R2
 (6v − πh3) / 3πh = R2
 Take the square root of both sides
 √[(6v − πh3) / 3πh ] = R
 R = √ [(6v − πh3) / 3πh ]
 R = √ [(6v / 3πh ) − (πh3 / 3πh) ]
 R = √ [(2v /πh ) − (h2 / 3) ]
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15. Transposition of Formula
 Transpose the formula F = S(M − m) ⁄ M + m Make m the subject of
formula
 F = S(M − m) ⁄ M + m
 Cross multiply
 F(M + m) = S(M − m)
 Expand the brackets
 FM + Fm = SM − Sm
 Collect like terms
 Fm + Sm = SM − FM
 Factorise
 m(F + S) = M(S − F)
 Divide both sides by F + S
 m= M(S − F) / (F + S)
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Expressions
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17. THANK
YOU
FOR
LISTENING
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