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Fofino [41]
3 years ago
15

Rearranging literal equations for z

Mathematics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

z = \frac{3n - 3}{6 - 2n}

Step-by-step explanation:

\frac{1}{(n-1)^{\frac{1}{3} } }  = (\frac{2z + 3}{4z} )^{\frac{1}{3} }

cubing on both sides we get

\frac{1}{n-1} = \frac{2z + 3}{4z}

4z = (2z+3)(n-1)

4z = 2nz + 3n - 2z - 3

4z - 2nz + 2z = 3n - 3

z(6 - 2n) = 3n - 3

z = \frac{3n - 3}{6 - 2n}

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Answer:

A = 26

Step-by-step explanation:

sum of students = classA + classB + classC

let's say classA = A, classB = B, and classC = C

A + B + C = 66

class A has five more students than class B, so A = 5 more than B so A = 5+B

class C has 2 less students than class B, so C = 2 less than class B = B -2, so C = B-2

A + B + C = 66

A = 5+B

C = B-2

substitute 5+B for A and B-2 for C in the first equation to limit this to one variable (B)

(5+B) + B + (B-2) = 66

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subtract 3 from both sides to isolate the variable and its coefficient

3B = 63

divide both sides by 3 to solve for B
B = 21

A = 5 + B = 5 + 21 = 26

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