### Video Transcript

In this video, we will learn how to
distinguish between formulas and expressions, write a formula from a given
description, and substitute numbers into a simple formula. We will begin by giving a
definition of what we mean by a formula. A formula is a fact or rule written
with mathematical symbols. It connects two or more quantities
with an equal to sign. If a formula contains two
quantities, when you know the value of one of them, you can find the value of the
other using the formula. Formulas are commonly used in
geometry to calculate perimeters, areas, and volumes. They are also useful in more
complicated areas of mathematics together with real-life problems. The first question that we will
look at involves writing a formula from some given information.

A rectangle has width π€ and length
π. A new rectangle is formed which has
the same length but double the width. Find the perimeter π and area π΄
of this new rectangle.

Letβs begin by considering the
rectangle with width π€ and length π. We know that each pair of parallel
sides in a rectangle are equal in length. Each of the angles in a rectangle
is equal to 90 degrees. We also recall that we can
calculate the perimeter of any rectangle by adding the length and width and then
multiplying by two. Distributing the parentheses or
expanding the brackets means that this is the same as two multiplied by the length
plus two multiplied by the width, two π plus two π€. The area of any rectangle can be
calculated by multiplying its length by its width. These are standard formulas that we
need to be able to recall.

We are told that our new rectangle
has the same length but double the width. This means that the length is still
equal to π, whereas the width is equal to two π€. Substituting these into our formula
for perimeter, we see that the perimeter is equal to two multiplied by π plus two
π€. Distributing the parentheses here
gives us two multiplied by π and two multiplied by two π€. This is equal to two π plus four
π€. As we want a formula for the
perimeter π, π is equal to two π plus four π€.

As already mentioned, we know that
the area of any rectangle is equal to its length multiplied by its width. For this new rectangle, this is
equal to π multiplied by two π€. This can be written as two ππ€ or
two π€π. We always put the number first, but
the letters or variables can go in either order. The formula for the area π΄ is π΄
is equal to two ππ€. If we were given specific values
for π and π€, we could then substitute these in to calculate the value of π and
π΄.

Our next question involves
substituting a value into a real-life formula.

A company produces greeting cards
with an initial cost of 2,000 Egyptian pounds and an extra cost of half an Egyptian
pound per card. The total cost is given by πΆ is
equal to a half π₯ plus 2,000, where π₯ is the number of produced cards. Find the total cost of producing
15,000 cards.

In this question, we are given a
formula for the total cost. πΆ is equal to a half π₯ plus
2,000. As π₯ is the number of cards
produced, we can substitute any value of π₯ into the formula to calculate the total
cost. In this question, we are told that
π₯ is equal to 15,000. Substituting in this value gives us
πΆ is equal to a half multiplied by 15,000 plus 2,000. One-half multiplied by 15 or a half
of 15 is 7.5. This means that a half multiplied
by 15,000 is 7,500. πΆ is equal to 7,500 plus
2,000. This is equal to 9,500. The total cost of producing 15,000
cards is 9,500 Egyptian pounds.

Our next question involves a
formula with multiple variables.

Given that the area of a trapezoid
is π΄ equals a half β multiplied by π plus π, find the value of π΄ when β equals
four centimeters, π equals three over two centimeters, and π equals five over two
centimeters.

We recall that a trapezoid is also
sometimes known as a trapezium. The formula given can be used to
calculate the area of any trapezoid. We recall that a trapezoid has a
pair of parallel sides, labeled π and π. The perpendicular heights between
them is β. We are told that our value of β is
four centimeters, π is equal to three over two centimeters, and π is equal to five
over two centimeters. We need to substitute these values
into our formula for the area. It is worth noting that three over
two is equal to 1.5 and five over two is equal to 2.5. We can calculate the value of π
plus π using either fractions or decimals. Substituting in our values, we get
a half multiplied by four multiplied by 1.5 plus 2.5. 1.5 plus 2.5 is equal to four.

We would also get this answer if we
added three over two and five over two. As the denominators are the same,
we just add the numerators, giving us eight over two or eight divided by two. This is equal to four. The area of the trapezoid is
therefore equal to a half multiplied by four multiplied by four. As multiplication is commutative,
we can multiply these numbers in any order. Four multiplied by four is 16, and
a half of 16 is eight. We would get the same answer if we
found a half of four, which is two, and then multiplied this answer by four. As our dimensions or lengths were
in centimeters, our units for area will be square centimeters. The value of π΄ is eight square
centimeters.

Our next question involves a common
formula used in a real-world context.

One measure of distance is rate
multiplied by time, π equals ππ‘. Find the distance Olivia travels if
she is moving at a rate of 60 miles per hour for 6.75 hours.

This question refers to the rate of
travel, which may also be referred to as speed. A common measure of speed or rate
is miles per hour. The formula given in the question,
π equals ππ‘, is a commonly used one. It is often given in triangular
form, as shown. The D stands for the distance. T stands for time. S stands for speed, which in our
question has been altered to π for rate. This leads us to three rearranged
formulae. Speed is equal to distance divided
by time, time is equal to distance divided by speed, and distance is equal to speed
multiplied by time. It is this version of the formula
that we are using in this question.

We are told that the rate or speed
is 60 miles per hour. And the time is 6.75 or six and
three-quarter hours. To calculate the value of π, the
distance, we need to multiply 60 by 6.75. There are lots of ways of
performing this calculation without a calculator. We could begin by splitting 60 into
six multiplied by 10 and then multiplying 6.75 by 10. This is equal to 67.5. We could then multiply 67.5 by six
using a column multiplication method. This gives us an answer of
405.0. 60 multiplied by 6.75 is 405. As our units for the rate were
miles per hour and our units for time were hours, the units for distance will be
miles. The distance that Olivia traveled
was 405 miles.

Our final question involves a more
complicated formula.

The speed π£, in feet per second,
of an ocean wave at depth π, in feet, is given by the formula π£ is equal to the
square root of 32π. Determine, to the nearest tenth,
the speed of an ocean wave at a depth of 11 feet.

In this question, we are given the
formula π£ is equal to the square root of 32π, where π£ is the speed in feet per
second and π is the depth in feet. We need to calculate the speed π£
when the depth is 11 feet. We need to substitute π equals 11
into our formula, giving us π£ is equal to the square root of 32 multiplied by
11. 32 multiplied by 10 is 320. This means that 32 multiplied by 11
is equal to 352. Our value of π£ is the square root
of 352.

352 is not a square number, so we
need to type this into our calculator and round to the nearest tenth. The square root of 352 is equal to
18.7616 and so on. Rounding to the nearest tenth is
the same as rounding to one decimal place. So our deciding number is the first
six. If the deciding number is five or
greater, we round up. π£ is therefore equal to 18.8 to
the nearest tenth. As our units for π£ were feet per
second, we can conclude that when the depth is 11 feet, the speed of the wave is
18.8 feet per second.

We will finish this video by
summarizing the key points. Formulas can be used in areas of
mathematics, such as geometry, and also in real-life situations. A formula links two or more
variables with an equal to sign. Substituting in the value of one or
more quantities or variables allows us to calculate the value of another. It is also possible to rearrange a
formula to make another variable the subject.