Discover How to Solve Literal Equations: Key Techniques

Stellar Classification
Class	Temperature (K)	Apparent Color
O	$\geq 30000$	blue
B	$10000 - 30000$	blue white
A	$7500 - 10000$	white
F	$6000 - 7500$	yellow white
G	$5000 - 6000$	yellow
K	$3500 - 5000$	light orange
M	$2000 - 3500$	orange red

Stellar Classification

Class

Temperature (K)

Apparent Color

\geq 30000

blue

10000 - 30000

blue white

7500 - 10000

white

6000 - 7500

yellow white

5000 - 6000

yellow

3500 - 5000

light orange

2000 - 3500

orange red

Time (s)	Distance Traveled (m)
$10$	$25$
$20$	$50$
$30$	$75$
$40$	$100$
$50$	$125$
$60$	$150$

Time (s)

Distance Traveled (m)

10

25

20

50

30

75

40

100

50

125

60

150

Time (s)	Speed $\times$ Time	Distance Traveled (m)
$10$	$2.5 (10) = 25$	$25$
$20$	$2.5 (20) = 50$	$50$
$30$	$2.5 (30) = 75$	$75$
$40$	$2.5 (40) = 100$	$100$
$50$	$2.5 (50) = 125$	$125$
$60$	$2.5 (60) = 150$	$150$

Time (s)

Speed

\times

Time

Distance Traveled (m)

10

2.5 (10) = 25

25

20

2.5 (20) = 50

50

30

2.5 (30) = 75

75

40

2.5 (40) = 100

100

50

2.5 (50) = 125

125

60

2.5 (60) = 150

150

The mechanical energy of an object is the sum of its potential and kinetic energy. The potential energy

E_{p}

and the kinetic energy

E_{k}

of an object are found by the following formulas.

E_{p} = m g h and E_{k} = \frac{m v ^{2}}{2}

In these formulas,

m

is the mass of the object,

g

is the acceleration of gravity,

h

is the height above the ground, and

v

is the velocity.

Which of the following expressions is obtained when the potential energy formula is solved for

m ?

A . E_{p} g h B . \frac{E _{p}}{g h} C . \frac{g E _{p}}{h}

Which of the following expressions is obtained when the kinetic energy formula is solved for

m ?

A . \frac{2 E _{k}}{v ^{2}} B . \frac{E _{k} v ^{2}}{2} C . v E_{k}

Let

E_{p} = E_{k} .

Which of the following expressions is obtained if the equation is solved for

v ?

A . \frac{m}{2 g h} C . 2 g h B . \frac{g h}{2 m} D 2 g h

We see that m is multiplied by g and h. We can isolate the variable m by first dividing the equation by g and then by h.

Therefore, the answer is B.

We first multiply the equation by 2 to get rid of the fraction on the right hand side. Then, we divide both sides by v^2.

This corresponds to A.

We equate the expressions of the different energies and solve for v.

We can exclude the negative solution because in this context the sign of the expression only indicates the direction. v=sqrt(2gh) The answer is D.

There is a relationship between the speed of light

c,

the electric constant

ε_{0},

and the magnetic constant

μ_{0} .

c = \frac{1}{μ _{0} ε _{0}}

Which of the following equations is formed when the equation is solved for

μ_{0} ?

A . C . μ_{0} = \frac{1}{c ε _{0}} μ_{0} = \frac{1}{c ^{2} ε _{0}} B . μ_{0} = c^{2} ε_{0} D . μ_{0} = c ε_{0}

We will start by multiplying both sides by sqrt()μ_0 ε_0. Then we will raise both sides by 2 to get rid of the radical sign.

This equation matches with C.

The square shown has side lengths $a$ and diagonal length $d .$

Write a formula for

a

that contains

d .

Simplify the answer.

Notice that two sides of the square and the diagonal forms a right triangle. Since the whole figure is a square, the legs have length a. The diagonal is the hypotenuse of the right triangle.

For any right triangle, we can use the Pythagorean Theorem to relate its side lengths. Let's use it to see relationship between a and d. a^2+a^2=d^2 Now we will solve this equation for a.

Because a is a length, it is always positive. Therefore, we can ignore the negative root. a=d/sqrt(2)

The gravitational force between any two objects in the universe is calculated by the following formula.

F = \frac{G m _{1} m _{2}}{r ^{2}}

Here,

m_{1}

and

m_{2}

are the masses of objects,

G

is the universal gravitational constant and

r

is the distance between the centers of the objects. Which of the following is equal to

r ?

A . C . \frac{G m _{1} m _{2}}{F} \frac{G m _{1} m _{2}}{F} B . \frac{F}{G m _{1} m _{2}} D . \frac{G m _{1}}{m _{2} F}

We need to find an expression equivalent to r. To do so, we will use the Properties of Equality. Let's first multiply both sides by r^2, then isolate r.

Since distances are always positive, the negative solution is irrelevant in this context and thus we can remove it. r=sqrt(Gm_1m_2/F) The expression on the right-hand side corresponds to option C.

Rewriting and Solving Literal Equations

Catch-Up and Review

Rewriting the Formula for the Perimeter of a Triangle

Finding the Amount of Time Spent Jogging

Equivalent Equations

Determining Equivalent Equations

Hint

Solution

Equation I

Equation II

Equation III

Equation IV

Literal Equation

Solving Literal Equations

Answer

Hint

Solution

Practice Solving Literal Equations

Rewriting a Formula for Volume

Hint

Solution

Finding the Amount of Time Spent Jogging

Answer

Hint

Solution

Rewriting and Solving Literal Equations

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