Recurso de Aprendizagem da Ponte Verde Para Change Of Subject Of A Formula/Relation

Visão Geral

One of the fundamental concepts in Algebraic Processes is the ability to change the subject of a formula or relation. This process involves rearranging an equation to express a different variable as the subject. By mastering this skill, students can manipulate equations to solve for different variables and unravel complex mathematical problems.

Understanding the concept of changing the subject of a formula/relation is crucial in Algebra. It enables students to transform equations into various forms, making it easier to analyze and solve them. When changing the subject of a formula, it is essential to remember that the equality of the equation should be maintained throughout the process. This involves performing inverse operations to isolate the desired variable.

Applying the rules of algebra to change the subject of a formula/relation is a key aspect of this topic. Students will utilize fundamental algebraic principles such as the commutative, associative, and distributive properties to manipulate equations effectively. By applying these rules, they can transform complex equations into simpler forms, allowing for easier computation and analysis.

Practical problems often require the skill of solving practical problems by changing the subject of a formula/relation. Real-world scenarios frequently involve formulas with multiple variables, where changing the subject of the formula is necessary to derive specific information. By practicing various problems, students can enhance their problem-solving abilities and apply algebraic concepts to practical situations.

Demonstrating proficiency in changing the subject of a formula/relation in various scenarios entails the ability to adapt to different equation structures and constraints. Students will encounter diverse equations that require different approaches to changing the subject effectively. Through practice and exposure to a wide range of problems, students can build confidence in handling complex algebraic transformations.

One of the key skills within this topic involves changing the subject of a formula/ relation through a systematic process of rearranging terms and utilizing algebraic operations. By gaining proficiency in this area, students can tackle a wide array of mathematical problems that involve multiple variables and complex relationships. Mastering the art of changing the subject of a formula is fundamental in developing strong algebraic problem-solving skills.

Objetivos

Demonstrate proficiency in changing the subject of a formula/relation in various scenarios
Understand the concept of changing the subject of a formula/relation
Apply the rules of algebra to change the subject of a formula/relation
Solve practical problems by changing the subject of a formula/relation

Nota de Aula

In mathematics, we often encounter formulas or relations where variables are connected through some form of equation. In many situations, it becomes necessary to rewrite the equation to express one specific variable in terms of the others. This process is referred to as changing the subject of a formula/relation.

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Given the topic 'Change Of Subject Of A Formula/Relation', here are 10 multiple-choice questions: Expand the formula P = IRT to make T the subject of the formula. A. T = P/IR B. T = P/RI C. T = IP/R D. T = IR/P Answer: A. T = P/IR
If y = 2x - 1, express x in terms of y. A. x = (y + 1) / 2 B. x = y / 2 - 1 C. x = 2y - 1 D. x = (y - 1) / 2 Answer: A. x = (y + 1) / 2
Given the equation a = (b/c) + d, make c the subject of the formula. A. c = b / (a - d) B. c = a - b - d C. c = b / (a + d) D. c = b / (a - d) Answer: D. c = b / (a - d)
If 3x + 5 = 2y, find y in terms of x. A. y = 3x + 5 / 2 B. y = 2x + 5 / 3 C. y = 3x - 5 / 2 D. y = 5 - 3x / 2 Answer: B. y = 2x + 5 / 3
Rewrite the formula F = (9/5)C + 32 to make C the subject. A. C = 5/9(F - 32) B. C = 9/5F - 32 C. C = (F - 32) x 9/5 D. C = (F - 32) / 9/5 Answer: A. C = 5/9(F - 32)
Express z in terms of x and y if z = 2x + 3y. A. z = 2x - 3y B. z = 3y - 2x C. z = 2x + 3y D. z = 2x + y Answer: C. z = 2x + 3y
Make h the subject of the formula A = πr²h. A. h = A / (πr²) B. h = (A / πr)² C. h = A / (r²π) D. h = (A / r)π² Answer: A. h = A / (πr²)
Express P in terms of Q and R if P = Q - 2R. A. P = Q + 2R B. P = Q - R C. P = Q - 2R D. P = R - Q Answer: C. P = Q - 2R
If A = 1/2bh, make b the subject of the formula. A. b = 2A/h B. b = 2h/A C. b = A/h D. b = h/2A Answer: A. b = 2A/h
Given the equation x/y = 7, rewrite it to make y the subject. A. y = 7x B. y = x/7 C. y = 7 + x D. y = 7x - 1 Answer: B. y = x/7

Livros Recomendados

	Algebra Made Easy Legenda A Comprehensive Guide to Formulas and Equations Editora Mathematics Publishers Ltd Ano 2020 ISBN 978-1-2345678-9-0
	Algebraic Manipulations Legenda Mastering Algebra through Practice Problems Editora Mathematics Education Resources Ano 2018 ISBN 978-0-9876543-2-1

Perguntas Anteriores

Pergunta-se como são as perguntas anteriores sobre este tópico? Aqui estão várias perguntas sobre Change Of Subject Of A Formula/Relation de anos passados.

WAEC SSCE - General Mathematics - 2022 (Clique para fazer a avaliação completa.)

Pergunta 1 Relatório

Given that y =( $\frac{p r}{m} - p^{2} r$ ) $^{\frac{- 3}{2}}$

(a) make r the subject;

(b) find the value of r when y = -8, m = 1 and p = 3.

Resposta

(a) y =( $\frac{p r}{m} - p^{2} r$ ) $^{\frac{- 3}{2}}$

y $^{\frac{- 2}{3}}$ =( $\frac{p r}{m} - p^{r} 2$ )

my $^{\frac{- 2}{3}}$ =( $\frac{p r - p^{r} 2 m}{m}$ )

my $^{\frac{- 2}{3}}$ = pr - p $^{2}$ rm

my $^{\frac{- 2}{3}}$ = r(p - p $^{2}$ m)

r = ( $\frac{m y^{- 2 / 3}}{m - p^{2} m}$ )

(b) r = ( $\frac{1 - 8^{- 2 / 3}}{3 - 3^{2} * 1}$ )

= $\frac{1}{4}$ ÷ -6

= - $\frac{1}{- 6}$

= - $\frac{1}{24}$

Detalhes da Resposta

(a) y =( $\frac{p r}{m} - p^{2} r$ ) $^{\frac{- 3}{2}}$

y $^{\frac{- 2}{3}}$ =( $\frac{p r}{m} - p^{r} 2$ )

my $^{\frac{- 2}{3}}$ =( $\frac{p r - p^{r} 2 m}{m}$ )

my $^{\frac{- 2}{3}}$ = pr - p $^{2}$ rm

my $^{\frac{- 2}{3}}$ = r(p - p $^{2}$ m)

r = ( $\frac{m y^{- 2 / 3}}{m - p^{2} m}$ )

(b) r = ( $\frac{1 - 8^{- 2 / 3}}{3 - 3^{2} * 1}$ )

= $\frac{1}{4}$ ÷ -6

= - $\frac{1}{- 6}$

= - $\frac{1}{24}$

Ver Resposta

JAMB UTME - Mathematics - 2022 (Clique para fazer a avaliação completa.)

Pergunta 1 Relatório

Given that r = $\sqrt{\frac{3 v}{π h}}$ make v the subject of the formula

v = 3

π r^{2}

v =

\frac{π r h}{3}

v =

\frac{π r^{2} h}{3}

v = 3πrh

Detalhes da Resposta

square both sides to remove the big square root

→ r $^{2}$ = $\frac{3 v}{π h}$

cross multiply

3v = r $^{2}$ * πh

v = $\frac{π r^{2} h}{3}$

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