Lesson Video: Writing and Evaluating Formulas | Nagwa Lesson Video: Writing and Evaluating Formulas | Nagwa

# Lesson Video: Writing and Evaluating Formulas Mathematics

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.

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### 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.

Letβs 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 ππ€ 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 ππ€. 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 πΆ 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. πΆ 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 πΆ 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. πΆ 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 ππ‘. 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 ππ‘, 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.

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