Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 2 Solving Equations Exercise 2.6
Algebra 1 Common Core Solutions
Chapter 2 Solving Equations Exercise 2.6 1CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010 cm/s
A unit is defined as the standard used to measure the required system.
From above the speed of light is given and it is written in cm/s.
As, units are put at the end of any quantitative term
Hence, clearly in this case the unit is cm/s.
Chapter 2 Solving Equations Exercise 2.6 1LC
Chapter 2 Solving Equations Exercise 2.6 2CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010 cm/s
If a rocket car of the future can travel at the same speed as of the light.
Then write the rate in miles per hour.
Hence, clearly the units of measure should be in miles per hour.
Chapter 2 Solving Equations Exercise 2.6 2LC
Chapter 2 Solving Equations Exercise 2.6 3CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010 cm/s
As in the question it is asked to give the measure in miles per hour.
So, the speed of light given inshould be converted to miles per hour.
Hence, the unit that needs to be converted for solving the problem is cm/s.
Chapter 2 Solving Equations Exercise 2.6 3LC
Chapter 2 Solving Equations Exercise 2.6 4CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010 cm/s
It is asked in the question to convert the given units to miles per hour.
That is,cm should be converted to miles andshould be converted to hours.
Hence, the conversion factors used to solve the problem is:
Chapter 2 Solving Equations Exercise 2.6 4LC
Chapter 2 Solving Equations Exercise 2.6 5CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010 cm/s
It is mentioned in the question to give the answer in miles per hour.
This restricts the option of having another form of conversion.
Thus, there is no alternative way of conversion.
Hence, there are no different factors to solve the given problem.
Chapter 2 Solving Equations Exercise 2.6 5LC
Chapter 2 Solving Equations Exercise 2.6 6CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010 cm/s
In step 2 it is asked to calculate the light of speed in miles per hour.
The speed of light is given in cm/s.
So, it is clear that cm should be converted to miles andshould be converted to hours. The required unit conversion is given below:
Chapter 2 Solving Equations Exercise 2.6 6LC
Chapter 2 Solving Equations Exercise 2.6 7CB
Chapter 2 Solving Equations Exercise 2.6 7LC
Chapter 2 Solving Equations Exercise 2.6 8CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010cm/s
After converting the speed of light in miles per hour, the solution is written below:
6.7 × 108 miles/hour
The solution is coming in miles per hour as asked in the question.
Hence, the units are coming to be as expected.
Chapter 2 Solving Equations Exercise 2.6 8LC
Chapter 2 Solving Equations Exercise 2.6 9CB
Consider the statement given in the question:
Given that,
Speed of light = 3.0 × 1010cm/s
The obtained solution is:
6.7 × 108 miles/hour
This solution describes the speed of light in a mile per hour. So, this speed is also valid for the further use.
Hence, the answer do makes sense.
Chapter 2 Solving Equations Exercise 2.6 9E
Chapter 2 Solving Equations Exercise 2.6 10CB
Consider the statement given in the question:
Given that:
Density of gold = 19.3 g/cm3.
A unit is defined as the standard used to measure the required system.
From above the density of gold is given and it is written in g/cm3.
As, units are put at the end of any quantitative term
Hence, clearly in this case the unit is g/cm3.
Chapter 2 Solving Equations Exercise 2.6 10E
Chapter 2 Solving Equations Exercise 2.6 11CB
Chapter 2 Solving Equations Exercise 2.6 11E
Chapter 2 Solving Equations Exercise 2.6 12CB
Chapter 2 Solving Equations Exercise 2.6 12E
Chapter 2 Solving Equations Exercise 2.6 13CB
Chapter 2 Solving Equations Exercise 2.6 13E
Chapter 2 Solving Equations Exercise 2.6 14CB
Chapter 2 Solving Equations Exercise 2.6 14E
Chapter 2 Solving Equations Exercise 2.6 15CB
Chapter 2 Solving Equations Exercise 2.6 15E
Chapter 2 Solving Equations Exercise 2.6 16CB
Chapter 2 Solving Equations Exercise 2.6 16E
Chapter 2 Solving Equations Exercise 2.6 17CB
Chapter 2 Solving Equations Exercise 2.6 17E
Chapter 2 Solving Equations Exercise 2.6 18CB
Chapter 2 Solving Equations Exercise 2.6 18E
Chapter 2 Solving Equations Exercise 2.6 19CB
Chapter 2 Solving Equations Exercise 2.6 19E
Chapter 2 Solving Equations Exercise 2.6 20CB
Consider the statement given in the question:
Unit analysis is defined as a method for comparing the dimensions of the physical quantities occurring in a problem to find the relationship between the quantities without having to solve the problem completely.
So, by the unit analysis the given quantitative value can be solved to obtain a greater unit value like Km/h or a smaller unit value cm/s. So, without solving the in parts the solution can be obtained in a single step.
Also, it tells about the relation between the given two quantities.
Hence, the unit analysis is helpful in solving a problem.
Chapter 2 Solving Equations Exercise 2.6 20E
Chapter 2 Solving Equations Exercise 2.6 21CB
Consider the statement given in the question:
Measuring the object means that the correct length of the object has been taken.
Approximating the length means taking a near about value of the given object.
As the carpenter wants to make an entertainment center, now, some space should be left on the either side of the television. For this the exact measurement will be better than the approximation as approximation can lead the measure to false prediction. That is, the measurement can be smaller or greater than the exact measurement.
Hence, exact measurement is better than the approximation.
Chapter 2 Solving Equations Exercise 2.6 21E
Chapter 2 Solving Equations Exercise 2.6 22CB
Consider the statement given in the question:
The unitary method is a technique in elementary algebra for solving a class of problem in variation. It consists of altering one of the variables to a single unit and then performing the operation necessary to alter it to the desired value.
It is given that forgallon the distance covered is 25 miles.
So, for 13.5 gallons of gas the distance covered will be:
13.5 × 25 = 337.5 miles
Hence, the distance covered is more than 85 miles.
Thus, the brother is correct.
Chapter 2 Solving Equations Exercise 2.6 22E
Chapter 2 Solving Equations Exercise 2.6 23CB
Chapter 2 Solving Equations Exercise 2.6 23E
Chapter 2 Solving Equations Exercise 2.6 24CB
Chapter 2 Solving Equations Exercise 2.6 24E
Chapter 2 Solving Equations Exercise 2.6 25E
Chapter 2 Solving Equations Exercise 2.6 26E
Chapter 2 Solving Equations Exercise 2.6 27E
Chapter 2 Solving Equations Exercise 2.6 28E
Chapter 2 Solving Equations Exercise 2.6 29E
Chapter 2 Solving Equations Exercise 2.6 30E
Chapter 2 Solving Equations Exercise 2.6 31E
Chapter 2 Solving Equations Exercise 2.6 32E
Chapter 2 Solving Equations Exercise 2.6 33E
Chapter 2 Solving Equations Exercise 2.6 34E
Chapter 2 Solving Equations Exercise 2.6 35E
Chapter 2 Solving Equations Exercise 2.6 36E
Chapter 2 Solving Equations Exercise 2.6 37E
Chapter 2 Solving Equations Exercise 2.6 38E
a) Consider the unit 3 min 20s.
Note that 1 minute is equal and so the number of units of seconds is greater than that of minutes.
Hence, in conversion, the new unit’s seconds will be greater than the original units 3min 20s.
b) Consider the unit 23cm.
Here the original unit is centimeter.
Note that 1 centimeter is equal 0.39 inches and so the number of units of inches is less than that of centimeter.
Hence, in conversion, the new unit’s inches will be lesser than the original units 23cm.
c) Consider the unit kilometers per hour.
Here the original unit is kilometer.
Note that 1 kilometer is equal 0.62 miles and so the number of units of miles is less than that of kilometer.
Hence, in conversion, the new unit’s miles per hour will be lesser than the older unit’s kilometers per hour.
Chapter 2 Solving Equations Exercise 2.6 39E
Chapter 2 Solving Equations Exercise 2.6 40E
Chapter 2 Solving Equations Exercise 2.6 41E
Measurement refers to the dimensions of the given solid shape and estimations refers to approximating the dimensions of the given solid shape.
Measuring the televisions approximately will be more convenient as compared to approximating it as the approximation can be lesser or larger in terms of the dimensions then the space of area to be left on either side can vary. It can become less than 1 foot.
Hence, measuring the dimensions correctly will be more convenient.
Chapter 2 Solving Equations Exercise 2.6 42E
A traveler changed $300 into euros for the purpose of a tour to Germany.
Unfortunately, the trip was cancelled.
By that time, the traveler has euros currency.
After 3 months the traveler changed euros back to dollars.
The conversion factor of dollars and euros vary on each day.
It follows that the traveler did not get back exactly same $300 .
Chapter 2 Solving Equations Exercise 2.6 43E
Chapter 2 Solving Equations Exercise 2.6 44E
Chapter 2 Solving Equations Exercise 2.6 45E
Chapter 2 Solving Equations Exercise 2.6 46E
Chapter 2 Solving Equations Exercise 2.6 47E
Chapter 2 Solving Equations Exercise 2.6 48E
Chapter 2 Solving Equations Exercise 2.6 49E
Chapter 2 Solving Equations Exercise 2.6 50E
Chapter 2 Solving Equations Exercise 2.6 51E
Chapter 2 Solving Equations Exercise 2.6 52E
Chapter 2 Solving Equations Exercise 2.6 53E
Chapter 2 Solving Equations Exercise 2.6 54E
Chapter 2 Solving Equations Exercise 2.6 55E
Chapter 2 Solving Equations Exercise 2.6 56E
Chapter 2 Solving Equations Exercise 2.6 57E
