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| 10 Theory slides |
| 13 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here is some recommended reading before getting started with this lesson.
Consider △ABC with side lengths a, b, and c.
Maya plans to jog 150 meters at a constant speed of 2.5 meters per second.
Equivalent equations have the same solution(s). Use the Properties of Equality and inverse operations to solve for the variable t.
Each equation will be solved for t, and then the solutions will be compared.
LHS−7=RHS−7
Commutative Property of Addition
Subtract terms
LHS/(-1)=RHS/(-1)
-b-a=ba
Put minus sign in front of fraction
1a=a
LHS/5=RHS/5
Calculate quotient
ca⋅b=ca⋅b
aa=1
Identity Property of Multiplication
Rearrange equation
While at the planetarium on Saturday, Dominika learns about Annie Jump Cannon, known as the census taker of the sky,
who introduced the Harvard Spectral Classification. In its simplest form, this system classifies stars according to their surface temperatures.
Stellar Classification | ||
---|---|---|
Class | Temperature (K) | Apparent Color |
O | ≥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 |
LHS−273=RHS−273
LHS⋅59=RHS⋅59
LHS+32=RHS+32
Rearrange equation
K=3500
Subtract term
ca⋅b=ca⋅b
Calculate quotient
Add terms
Round to nearest integer
K=5000
Subtract term
ca⋅b=ca⋅b
Calculate quotient
Add terms
Round to nearest integer
Use the Properties of Equality to solve the given literal equation for the indicated variable. Some equations may require dividing both sides of the equation by a variable. Since division by zero is not defined, for these equations assume that the variable is not equal to 0.
On Sunday Dominika spends time on her hobby, which is candle making. She uses 170 cubic centimeters of wax to produce a cylindrical candle of radius 3 centimeters.
LHS/πr2=RHS/πr2
Cancel out common factors
Simplify quotient
Rearrange equation
Time (s) | Distance Traveled (m) |
---|---|
10 | 25 |
20 | 50 |
30 | 75 |
40 | 100 |
50 | 125 |
60 | 150 |
It takes Maya 60 seconds to run 150 meters.
Time (s) | Speed × 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 |
Maya will run 150 meters in 60 seconds at a speed of 2.5 meters per second.
LHS/v=RHS/v
Cancel out common factors
Simplify quotient
Rearrange equation
A formula is given by S=B(1+p).
To isolate B, we will divide both sides of the formula by (1+p).
We will start by dividing both sides of the given formula by B. Then, we will use the inverse operations to isolate p.
Isolating R means that it should stand alone on one side of the equation. We can write the law in that form by dividing both sides by nT.
Therefore, R is equal to PVnT. This corresponds to C.
This time we will divide the formula by P to isolate V.
The volume V can be expressed as nRTP. This corresponds to A.
The formula V=πr2h is used to calculate the volume of a cylinder with radius r and height h.
We need to solve the formula for h. Since h is multiplied by both π and r^2, we will divide the equation by π r^2.
The formula for h thus becomes h = Vπ r^2.
We will use the formula from the previous part and substitute the given values.
The Coca-Cola jar has a height about 11 centimeters.
Solve the following formulas for b.
In the given equation, a is being added to b. Since the inverse operation of addition is subtraction, we can isolate the variable b by applying the Subtraction Property of Equality.
In this case, we have b and another variable g on the left-hand side of the equation. We can get rid of g by adding g to both sides.
Therefore, b is equal to the sum of t and g.
On the left side, b is multiplied by 7. To get rid of 7, we need to divide the equation by 7.
Now we have a y in the denominator of the fraction by. To isolate b, we multiply both sides of the equation by y.
An athletic facility is building an indoor track. The track is composed of three parts, a rectangle and two semicircles.
The perimeter of the given figure is made up of two sides of a rectangle and two semicircles. The two sides each have a length of x, and the two semicircles make up a circle with a radius r. To find their length, let's substitute the radius into the formula for the circumference of a circle. C = 2 π r Now that we have the circumference of the circle, we can write the perimeter P of the entire figure, which is a sum of all the sides. P= x + x+ 2 π r Let's simplify the expression by adding like terms. P=x + x + 2 π r ⇕ P=2x+2 π r
In order to solve the formula for x, we need to isolate x on one of the sides.
The perimeter is given as 562 meters and the radius of the half circles is given as 45 meters. We can find x by substituting these values into the formula that we found in Part B.
When rounded to the nearest integer, the straight path x is about 140 meters long.
Let's begin with recalling the formula for the distance traveled. d=r t In this formula d represents the distance, r is the rate, and t stands for time. In our exercise we are given that the truck travels 108 kilometers per hour and we want to determine the time the truck needs to travel 25 miles. Let's first isolate t. d=r t ⇒ t = d/r We can now substitute 108 for r and 162 for d into the formula.
The truck needs 1.5 hours to travel 162 kilometers. To convert it into minutes, we will multiply the time by the conversion factor 60 min1 h.
The truck travels 162 kilometers in 90 minutes.