Worksheet: Reforming Equations for a Specific Variable

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In this worksheet, we will practice rewriting or re-forming an equation for a specific variable.

Q1:

Solve 𝑦=2π‘₯+13π‘₯+4 with an expression for π‘₯ in terms of 𝑦.

  • A π‘₯ = 4 𝑦 + 1 3 𝑦 βˆ’ 2
  • B π‘₯ = 4 𝑦 βˆ’ 1 2 βˆ’ 3 𝑦
  • C π‘₯ = 4 𝑦 βˆ’ 1 3 𝑦 βˆ’ 2
  • D π‘₯ = 4 𝑦 + 1 2 βˆ’ 3 𝑦
  • E π‘₯ = 4 𝑦 βˆ’ 1 2 βˆ’ 𝑦

Q2:

Rearrange 𝑦=3π‘₯βˆ’5 to make π‘₯ the subject.

  • A π‘₯ = 𝑦 + 5 3
  • B π‘₯ = 3 𝑦 + 1 5
  • C π‘₯ = 𝑦 + 5
  • D π‘₯ = 𝑦 βˆ’ 5
  • E π‘₯ = 𝑦 βˆ’ 5 3

Q3:

In 1897, Amos Dolbear derived a formula linking the number of cricket chirps and the temperature. The law states that the temperature 𝑇, in degrees Celsius, is related to the number of cricket chirps in a minute 𝑁 by the formula 𝑇=10+π‘βˆ’407.

Rearrange the formula to make 𝑁 the subject.

  • A 𝑁 = ( 7 𝑇 + 1 0 ) βˆ’ 4 0
  • B 𝑁 = ( 7 𝑇 βˆ’ 1 0 ) + 4 0
  • C 𝑁 = 7 ( 𝑇 + 1 0 ) βˆ’ 4 0
  • D 𝑁 = 7 ( 𝑇 + 1 0 ) + 4 0
  • E 𝑁 = 7 ( 𝑇 βˆ’ 1 0 ) + 4 0

Given that the temperature on a particular day is 22∘C, estimate the number of cricket chirps that you would expect to hear in a minute.

Q4:

The final velocity of an object with uniform acceleration can be calculated using the formula 𝑣=𝑒+π‘Žπ‘‘. Make 𝑑 the subject.

  • A 𝑣 π‘Ž βˆ’ 𝑒
  • B 𝑣 π‘Ž + 𝑒
  • C π‘Ž ( 𝑣 βˆ’ 𝑒 )
  • D 𝑣 + 𝑒 π‘Ž
  • E 𝑣 βˆ’ 𝑒 π‘Ž

Q5:

The displacement, 𝑠, of an object traveling with uniform acceleration, π‘Ž, can be calculated using the formula 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨, where 𝑒 is the initial velocity and 𝑑 is the time.

Make π‘Ž the subject.

  • A π‘Ž = 𝑠 βˆ’ 𝑒 𝑑 2 𝑑 
  • B π‘Ž = 2 𝑠 βˆ’ 𝑒 𝑑 𝑑 
  • C π‘Ž = 2 ( 𝑠 + 𝑒 𝑑 ) 2 𝑑 
  • D π‘Ž = 2 ( 𝑠 βˆ’ 𝑒 𝑑 ) 𝑑 
  • E π‘Ž = 2 𝑠 + 𝑒 𝑑 𝑑 

Calculate the acceleration of an object that has started from rest and traveled 200 m in 8 s.

Q6:

The variables π‘₯ and 𝑦 where π‘₯ is non-negative are related by the formula 𝑦=6(7βˆ’π‘₯)11. Rewrite the formula with π‘₯ as the subject.

  • A π‘₯ = ο„ž 7 βˆ’ 6 1 1 𝑦
  • B π‘₯ = ο„ž 7 + 1 1 𝑦 6
  • C π‘₯ = ο„ž 7 βˆ’ 1 1 𝑦 6
  • D π‘₯ = ο„ž 7 βˆ’ 1 1 6 𝑦
  • E π‘₯ = ο„ž 7 + 1 1 6 𝑦

Q7:

The variables π‘Ž and 𝑏 are related by the formula π‘Ž(𝑏+5)=π‘˜(π‘βˆ’19π‘Ž). Make 𝑏 the subject.

  • A 𝑏 = 5 π‘Ž + 1 9 π‘˜ π‘Ž π‘˜ βˆ’ π‘Ž
  • B 𝑏 = 5 + 1 9 π‘Ž π‘˜ βˆ’ π‘Ž
  • C 𝑏 = 5 π‘Ž βˆ’ 1 9 π‘˜ π‘Ž π‘˜ βˆ’ π‘Ž
  • D 𝑏 = 5 π‘Ž βˆ’ 1 9 π‘˜ π‘Ž π‘˜ + π‘Ž
  • E 𝑏 = 5 + 1 9 π‘Ž π‘˜ + π‘Ž

Q8:

Albert Einstein’s famous formula 𝐸=π‘šπ‘οŠ¨, where the constant 𝑐 is the speed of light, relates the energy, 𝐸, contained in matter and its mass, π‘š. Rearrange the formula to make 𝑐 the subject.

  • A 𝑐 = ο„ž π‘š 𝐸
  • B 𝑐 = ο„ž 𝐸 π‘š
  • C 𝑐 = 𝐸 2 π‘š
  • D 𝑐 = 𝐸 π‘š
  • E 𝑐 = π‘š 𝐸

Q9:

The variables 𝑑 and π‘š are related by the formula 𝑑=π‘šπ‘š+1. Make π‘š the subject.

  • A π‘š = 𝑑 𝑑 βˆ’ 1
  • B π‘š = 𝑑 + 1 𝑑
  • C π‘š = βˆ’ 𝑑 βˆ’ 1 𝑑
  • D π‘š = βˆ’ 𝑑 𝑑 βˆ’ 1
  • E π‘š = 𝑑 𝑑 + 1

Q10:

The circumference 𝐢 of a circle can be estimated using the formula 𝐢=447π‘Ÿ, where π‘Ÿ is the radius. Find an estimate of the radius of a circle with 𝐢=67.1. Round your answer to the nearest tenth.

Q11:

The circumference of a circle as a function of its radius is given by 𝐢(π‘Ÿ)=2πœ‹π‘Ÿ. Express the radius of a circle as a function of its circumference, denoting it by π‘Ÿ(𝐢), and then find π‘Ÿ(36πœ‹).

  • A π‘Ÿ ( 𝐢 ) = 𝐢 πœ‹ , 18
  • B π‘Ÿ ( 𝐢 ) = 𝐢 2 πœ‹ , 72
  • C π‘Ÿ ( 𝐢 ) = 𝐢 2 πœ‹ , 36
  • D π‘Ÿ ( 𝐢 ) = 𝐢 2 πœ‹ , 18
  • E π‘Ÿ ( 𝐢 ) = 𝐢 πœ‹ , 36

Q12:

The volume, 𝑉, of a cylinder with radius π‘Ÿ and height β„Ž is 𝑉=πœ‹π‘Ÿβ„ŽοŠ¨. Given that a cylinder has a height of 6 meters, write an equation for the radius of the cylinder as a function of 𝑉, and then use this to find the radius of the cylinder if its volume is 300 cubic meters. Give your answer to two decimal places.

  • A π‘Ÿ = √ 𝑉 6 πœ‹ , 0.92 meters
  • B π‘Ÿ = 𝑉 √ 6 πœ‹ , 69.10 meters
  • C π‘Ÿ = ο„ž 𝑉 6 πœ‹ , 3.99 meters
  • D π‘Ÿ = 𝑉 6 πœ‹ , 15.92 meters
  • E π‘Ÿ = 1 πœ‹ ο„ž 𝑉 6 , 2.25 meters

Q13:

The volume of a right circular cone with radius π‘Ÿ and height β„Ž is 𝑉=13πœ‹π‘Ÿβ„ŽοŠ¨. First, write an equation for the radius of a cone with a height of 12 inches as a function of 𝑉. Then, use this to find the radius of the cone to the nearest whole number given that its volume is 50 cubic inches.

  • A π‘Ÿ = πœ‹ 𝑉 3 6 , 9 inches
  • B π‘Ÿ = ο„ž 𝑉 3 6 πœ‹ , 1 inch
  • C π‘Ÿ = ο„ž πœ‹ 𝑉 3 6 , 3 inches
  • D π‘Ÿ = ο„ž 𝑉 4 πœ‹ , 2 inches
  • E π‘Ÿ = 𝑉 4 πœ‹ , 4 inches

Q14:

The volume, 𝑉, of a sphere with radius length π‘Ÿ is given by 𝑉=43πœ‹π‘ŸοŠ©. Find the radius length of a sphere with a volume of 4.851Γ—10 cm3. (Take πœ‹=227.)

Q15:

Use the formula 𝐴=12π‘β„Ž to determine the height, β„Ž, of a triangle given that its area, 𝐴, is 4.5 and its base, 𝑏, is 2.

  • A 3 3 5
  • B 4 1 2
  • C 1 4 5
  • D 2 1 4
  • E3

Q16:

A room’s temperature ranges from 25∘C to 30∘C. Determine its temperature range in degrees Fahrenheit, using the formula πΉβˆ’32=1.8𝐢, where 𝐹 is the temperature in degrees Fahrenheit, and 𝐢 is the temperature in degrees Celsius.

  • A 6 2 ∘ F to 86∘F
  • B 8 6 ∘ F to 347∘F
  • C 6 2 ∘ F to 347∘F
  • D 6 2 ∘ F to 77∘F
  • E 7 7 ∘ F to 86∘F

Q17:

The surface area, 𝐴, of a cylinder in terms of its radius, π‘Ÿ, and height, β„Ž, is given by 𝐴=2πœ‹π‘Ÿ+2πœ‹π‘Ÿβ„ŽοŠ¨. Express the radius, π‘Ÿ, of a cylinder with a height of 4 feet as a function of 𝐴. Find, to the nearest foot, the radius of such a cylinder whose surface area is 200 square feet.

  • A π‘Ÿ = ο„ž 𝐴 + 8 πœ‹ 2 πœ‹ + 2 , 6 feet
  • B π‘Ÿ = ο„ž 𝐴 + 8 πœ‹ 2 πœ‹ βˆ’ 2 , 4 feet
  • C π‘Ÿ = ο„ž 𝐴 βˆ’ 4 πœ‹ 2 πœ‹ + 2 , 7 feet
  • D π‘Ÿ = ο„ž 𝐴 + 8 πœ‹ 2 πœ‹ βˆ’ 2 , 6 feet
  • E π‘Ÿ = ο„ž 𝐴 + 8 πœ‹ 2 πœ‹ + 2 , 8 feet

Q18:

The surface area, 𝐴, of a sphere in terms of its radius, π‘Ÿ, is given by 𝐴(π‘Ÿ)=4πœ‹π‘ŸοŠ¨. Express π‘Ÿ as a function of 𝐴 and find, to the nearest tenth of an inch, the radius of a sphere whose surface area is 1,000 square inches.

  • A π‘Ÿ = ο„ž 4 πœ‹ 𝐴 , 0.1 inches
  • B π‘Ÿ = 𝐴 4 πœ‹ , 79.6 inches
  • C π‘Ÿ = 4 πœ‹ √ 𝐴 , 0.4 inches
  • D π‘Ÿ = √ 𝐴 4 πœ‹ , 2.5 inches
  • E π‘Ÿ = ο„ž 𝐴 4 πœ‹ , 8.9 inches

Q19:

The volume, 𝑉, of a right circular cone with radius length π‘Ÿ is given by 𝑉=13πœ‹π‘Ÿβ„ŽοŠ¨. Find the height of a right circular cone with volume 4,312 cm3 and base diameter length 28 cm. ο€Όπœ‹=227.Take

Q20:

The formula for the circumference of a circle is 𝐢=2πœ‹π‘Ÿ. Rearrange this formula to make πœ‹ the subject.

  • A πœ‹ = 2 𝐢 π‘Ÿ
  • B πœ‹ = 𝐢 2 π‘Ÿ
  • C πœ‹ = 2 𝐢 π‘Ÿ
  • D πœ‹ = 𝐢 π‘Ÿ
  • E πœ‹ = 2 π‘Ÿ 𝐢

Q21:

Using the formulae for the circumference and area of a circle, eliminate the variable π‘Ÿ to find a formula that allows you to calculate the circumference of a circle from its area.

  • A 𝐢 = 4 πœ‹ √ 𝐴
  • B 𝐢 = √ 𝐴 4 πœ‹
  • C 𝐢 = 2 πœ‹ √ 𝐴
  • D 𝐢 = πœ‹ ο„ž 𝐴 2 πœ‹
  • E 𝐢 = 2 πœ‹ ο„ž 𝐴 πœ‹

Q22:

Using the formulae for the circumference and area of a circle, eliminate the variable π‘Ÿ to find a formula that allows you to calculate the area of a circle from its circumference.

  • A 𝐴 = ο€½ 𝐢 2 πœ‹  
  • B 𝐴 = ο€½ 𝐢 4 πœ‹  
  • C 𝐴 = 𝐢 4 πœ‹
  • D 𝐴 = 𝐢 4 πœ‹ 
  • E 𝐴 = 𝐢 2 πœ‹ 

Q23:

The picture shows the design of a logo which is formed from two semicircles with a common center.

Work out the perimeter of the logo, giving your answer in terms of πœ‹.

  • A 1 6 πœ‹ βˆ’ 8
  • B 1 6 πœ‹ + 8
  • C 8 πœ‹ + 4
  • D 8 πœ‹ βˆ’ 1 6
  • E 1 6 πœ‹

Work out the area of the logo, giving your answer in terms of πœ‹.

  • A 1 1 πœ‹
  • B 7 πœ‹
  • C 1 7 πœ‹
  • D 1 3 πœ‹
  • E 1 0 πœ‹

Q24:

The volume 𝑉 of a right circular cone in terms of its height β„Ž and base radius π‘Ÿ is 𝑉=13πœ‹π‘Ÿβ„ŽοŠ¨. Give a formula for the radius π‘Ÿ in terms of 𝑉 and β„Ž.

  • A π‘Ÿ = ο„ž 𝑉 β„Ž
  • B π‘Ÿ = ο„ž πœ‹ 𝑉 3 β„Ž
  • C π‘Ÿ = ο„ž 𝑉 3 πœ‹ β„Ž
  • D π‘Ÿ = ο„ž 3 𝑉 πœ‹ β„Ž
  • E π‘Ÿ = 3 𝑉 πœ‹ β„Ž

Q25:

A container holds 100 mL of a solution that is 25 mL acid. If 𝑛 mL of a solution that is 60% acid is added, the function 𝐢 gives the concentration, 50%, as a function of the number of milliliters added, 𝐢=25+0.6𝑛100+𝑛. Express 𝑛 as a function of 𝐢 and determine the number of milliliters needed to have a solution that is 60% acid.

  • A 𝑛 = 1 0 0 𝐢 βˆ’ 2 5 𝐢 , 50 mL
  • B 𝑛 = 2 5 + 1 0 0 𝐢 𝐢 + 0 . 6 , 68 mL
  • C 𝑛 = 1 0 0 𝐢 βˆ’ 2 5 0 . 6 βˆ’ 𝐢 , 250 mL
  • D 𝑛 = 1 0 0 𝐢 + 2 5 0 . 6 βˆ’ 𝐢 , 750 mL
  • E 𝑛 = 1 0 0 𝐢 βˆ’ 2 5 𝐢 + 0 . 6 , 23 mL

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