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CHAPTER3.5T
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à 3.5ïWorking with Formulas.
äïPlease substitute the values of the given variables into
êêthe formula and solve for the remaining unknown.
âêêê1
êêèGivenïA = ─ ∙ b ∙ hï,ïb = 4,ïh = 2
êêêê2
êêêê1
êêêèA = ─ ∙ 4 ∙ 2ï───>ïA = 4
êêêê2
éS
Given any equation such as, P = 2L + 2W, if you are given numerical
values for all but one letter, it is possible to solve for the remain-
ing letter by using the addition and multiplication properties of
equality.
Suppose in this example P is given to be 28 and W is given to be 4.ïYou
must substitute ç numbers in for the appropriate letters and solve
for the last remaining letter.
ïP = 2L + 2Wë┌>ï28 - 8 = 2L + 8 - 8ë1ë 1
28 = 2L + 2∙4è │è20 = 2L + 0êï┌>ï─∙20 = ─∙2 ∙ Lï──> 10 = L
28 = 2L + 8ï────┘è20 = 2Lï───────────┘è2ë 2
Thus the value for L is 10 under the conditions that P is 28 and W is 4.
This formula represents the perimeter of a rectangle.
1
êè GivenïP = 2L + 2W,ëP = 32 and L = 6,ïfind W.
êA)ï10ê B)ï12êïC)ï14êD)ïå of ç
üë P = 2L + 2Wêêè1ê 1
êè 32 = 2 ∙ 6 + 2 ∙ Wë┌─>è─ ∙ 20 = ─ ∙ (2W)
êè 32 = 12 + 2Wêè │ë2ê 2
êè 32 - 12 = 12 - 12 + 2W │
êè 20 = 0 + 2Wêë│ë10 = Wè The width is 10.
êè 20 = 2Wï──────────────┘
Ç A
2
êïGiven the formula for the circumference of a circle, C = 2πr,
êêêfind C if π = 3.14ïand r = 6.
êA)ï28.2ë B)ï37.68ë C)ï39.51è D)ïå of ç
ü
êêêêïC = 2πr
êêêêïC = 2(3.14)(6)
êêêêïC = 37.68
Ç B
3
#êèGiven the formula for the volume of a cylinder, V = πrìh,
êêêfind h when V = 36 and r = 2.
êA)ï8êïB)ï4.26êC)ï2.87ëD)ïå of ç
ü
#êïV = πrìhêêê1êê1
êêêêè┌─>ï───── ∙ (36) = ───── ∙ (12.56h)
#êï36 = (3.14)(2)ì∙hë│è 12.56êè12.56
êï36 = (3.14)(4)∙hë │
êï36 = 12.56 ∙ hï──────┘è 2.87 = h
Ç C
4ëGiven the formula for the area of a trapezoid,
êê 1
êè A = ─∙h(a + c),ïfind A when h = 6, a = 16, and c = 14.
êê 2
êA)ï37.5ë B)ï90êïC)ï106ë D)ïå of ç
üêêï1êêêë1
êêè A = ─∙h(a + c)ê ┌─>ïA = ─∙(6)(30)
êêê 2êêè│ê 2
êêêêêë│
êêê 1êêè│
êêè A = ─∙(6)(16 + 14)ï──┘ëA = 90
êêê 2
Ç B
5
êè Given that interest equals principle times rate times time,
êè I = p∙r∙t,ïfind P when I = 300, r = .06, and t = 6.
êA)ï1133ë B)ï12,000ëC)ï833.33èD)ïå of ç
üêêêè I = p∙r∙t
êêêê300 = p∙(.06)(6)
êêêê 300 = p∙(.36)
êê1êêè ┌ï1ï┐
êë ───∙(300) = p∙(.36)∙│ ─── │è ─────>è 833.33 = p
êë .36êêè└ .36 ┘
Ç C
äïPlease solve the given formula for the variable that is
êêindicated.
â
êêêè SolveïD = r∙tïfor r
êêêêëD
êêêêë─ = r
êêêêët
éS
Given any formula such asïL = a + n ∙ d,ïyou can solve for any letter
in the formula in terms of the other letters by using the addition and
multiplication properties of equality.ïThe formula is presently solved
for "L" in terms of "a", "n", and "d".
Let's solve this formula for "n" in terms of "l", "a", and "d".
êêL = a + n ∙ dêêêè1ê 1
êêL - a = a - a + n ∙ dè┌─>ï(L - a)∙─ = (nd)∙─
êêL - a = 0 + ndêè│êëdê d
êêL - a = ndï────────────┘ëL - a
êêêêêêï─────ï= n
êïL - aêêêêëd
Thus n = ─────.
êè dë Note that it doesn't matter which side "n" is on.
6êè Solveè P = a + 2b + cïfor c
êë A)ïP - a - 2b = cêïB)ïP + a - 2b = c
êêëP
êë C) ──────ï= cêë D)ïå of ç
êêïa + 2b
üêêêïP = a + 2b + c
êêêê P - a = a - a + 2b + c
êêêê P - a = 0 + 2b + c
êêêê P - a - 2b = 2b - 2b + c
êêêê P - a - 2b = 0 + c
êêêê P - a - 2b = c
Ç A
7
êêêSolveè A = P + Prtïfor t
êë AêêA - Pêè A - Pêè å
ë A)ï────────ë B)ï─────êC)ï─────ê D)ïof
êèP(1 - r)êërêê Prêëç
üë A = P + Prtêêï1êë1
êëA - P = P - P + Prtè┌>ï──∙(A - P) = ──∙(Prt)
êëA - P = 0 + Prtê│èPrêè Pr
êëA - P = Prtï─────────┘
êêêêêèA - P
êêêêêè─────ï=ït
êêêêêëPr
Ç C
8
êêêSolveè y = Ax + Bèfor x
êèA - yêë y - Bêè B - Aêè å
ë A)ï─────êïB)ï─────êC)ï─────ê D)ïof
êëBêêèAêê yêë ç
üêêêy = Ax + Bêë1
êêêë y - B = Ax + B - B
êêêë y - B = Ax
êêêë y - B
êêêë ───── = x
êêêê A
Ç B