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# Practice Problems with Circular Trigonometric Functions Video
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• 0:01 The Unit Circle
• 4:13 Example 1
• 5:03 Example 2
• 5:36 Example 3
• 6:57 Lesson Summary
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will be able to use the unit circle to help you find your answer to trig problems without making too many calculations. Watch and learn how you can find your angle in radians on the unit circle.
## The Unit Circle
In this video lesson, we are dealing with circular trigonometric functions. These are your trig functions calculated using a circle. What kind of circle? We will use a special circle called a unit circle. We call this special circle a unit circle because it has a radius of 1. If we plot this on our Cartesian coordinate graph, the center of our circle will be at the origin. The circle crosses the x axis on the right side at (1, 0) and on the left side at (-1, 0). The circle also crosses the y axis on the top at (0, 1) and on the bottom at (0, -1).
Using the unit circle to help solve our trig functions means that all of our points will be somewhere on the unit circle. Where do the angles come into play with the unit circle? This is a good question. If we take a point, we can draw a line from the center of our circle to the point. We will have a radius of 1. The angle formed by this line and our positive x axis is the angle that we use for our calculations. The real neat part about the unit circle is that our x value then becomes cos (theta) and our y value is sin (theta).
How does this work? Remember our definitions of our cosine and sine functions using a right triangle? Cosine is adjacent over hypotenuse. Looking at our unit circle and the triangle that is formed by the radius, the x value and the y value of our point, we see that our adjacent is x and our hypotenuse is 1. So adjacent over hypotenuse is then x/1 or x. We have cos (theta) = x/1 = x.
As for the sine function, we have sin (theta) equals opposite over hypotenuse. Our opposite is y and our hypotenuse is 1. So sin (theta) = y/1 = y. Pretty neat, huh? This helps us out a lot because we can actually fill in our unit circle with various points that are easy to calculate.
Just look at all these points! If you can remember these, then you are well on your way to becoming a circular trig function pro! How can you remember all of these points? They look a bit complicated, don't they? They do, but if you see the pattern that is there, then you won't have as much trouble.
Look carefully at the numerators. Now look at what the points are doing. As you go from point to point, you see that your numbers are actually increasing or decreasing by 1. You still have your square root, but the number inside is going from 1 to 2 to 3 and back again. It is the same for both the x and y values.
As you go higher up on the y axis, the numerator for the y value goes from 1 to the square root of 2 to the square root of 3. To help you with this, you can also think of the 1 as the square root of 1 so that you have all square roots. You can see that the numbers follow the coordinate graph.
As the x value goes more negative, so does the numerator of the x value. It goes from -1 to the square root of -2 to the square root of -3. As for the denominator, all the points that aren't the easy points like (0, 1) have a denominator of 2. That's all you have to remember.
Okay, that's all well and good, but what's with the two different graphs? One has a whole lot of pis in it. Another good question! This second graph with the pis in it is showing you the equivalent radian measure of the degrees. So we have 90 degrees is the same as pi/2 radians and 60 degrees is pi/3 radians.
## Example 1
Let's take a look at what you might expect to find on tests and such. You might come across a problem that asks you something like this:
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# Multiplication by ‘t’ Property – Laplace Transform – Engineering Mathematics 3
Hello friends so today we are gonna learn a new theorem called as multiplication by T theorem of Laplace transform now what is the statement of multiplication by T Theorem so it is given like this if Laplace or function of T is some function of s then the Laplace of T raised to n into function of T is equal to minus 1 raise to n D by DS raised to n Phi of A that means nth derivative of Phi of S so your why it is called as multiplication by T so if you’ll observe then we have Laplace of P raise to N F of T means our function f of T is getting multiplied by T raise to n term or else A by P hence the name is given as multiplication by T theorem now we have to prove that this theorem is true for all natural numbers so now let’s start with the derivation of this theorem so since I want to prove that this theorem is true for all natural numbers n what I’ll do is Here I am going to use a method of mathematical induction because the method of mathematical induction is used to prove the theorem for all natural numbers or to prove any statement for all natural numbers so here and say that we will prove given state mein by method of mathematical induction now and what is the method of mathematical induction so it is a two-step process in which first step is called as base case and the second step is called as inductive step so in base case what we do that we prove that the given statement is true for first natural number so now the first natural number is 1 so we will prove that the given statement is true for n equal to 1 next in inductive step what we’ll do is we’ll assume that this statement is true for any natural number and we’ll prove that it is true for the next natural number so these are the two steps and after that we can infer that it is true for all the natural numbers so let’s start so in start with step number one which is called as base case where we prove that it is true for n equal to 1 now let’s prove for n equal to 1 so let’s start from here so it is given that let Phi of s is equal to Laplace of function of T now guys to prove that Laplace of function of TB use the definition of Laplace transform so what is the definition of Laplace transform failure I’d say it is integration from 0 to infinity e raised to minus s T into the function of T DT now as I want to find out or as I want to prove the theorem or the given statement for now first natural number what I’ll do is I take derivative on both sides and I’ll apply the rule of differentiation under the integral sign which is called as DUI s so here I say I’ll say this as equation number 1 and I’ll say differentiating both sides with respect to s and applying duis that is differentiation under integral sign so what is DUIs so in duis we get the differentiation of fire faces fire – of s and we applied differentiation under the integral sign so this is the integral sign so you and say zero to infinity as it is let us apply the differentiation so it is always a partial differentiation so it is dou by dou X of e raised to minus s T F of T DT so I write down the DT here now we have to find out the differentiation of this function with respect to X with respect to s partially so here F of T is constant which will come outside because it does not have s so we will get therefore Phi dash of s equal to integration 0 to infinity F of T outside next the derivative of e raised to minus s T with respect to s is e raised to minus s T into minus T so that is e raised to minus s T into minus T DT now we’ll rewrite this terms so it is integration 0 to infinity minus and outside a raised to minus s T into T f of T DT and now if I compare this term with the definition of Laplace transform then this integral is nothing but Laplace of T into F of T because here function of T is this term so we got the value of Phi dash of s so can I say that therefore this Laplace of P into function of T is equal to by sending this negative sign on the left hand side minus Phi dash of s and therefore I can say that Laplace of T into f of T is equal to minus off now Phi dash of S is nothing but derivative of Phi of s so it is minus D by D’s of Phi of s so guys if you’ll see carefully then I have derived or I have proved that the given statement is true for n equal to 1 where CR the power of T is 1 here we have minus 1 raised to 1 which is minus 1 and D raised to 1 D s raised to 1 so this was the original statement and if we’ll put n equal to 1 then we’ll get the exactly same result so here I’ll say that therefore the rule is true for and equal to 1 so in the base case we have proved that the given statement is true for first natural number that is n equal to 1 now let’s proceed to step number 2 that is inductive step so here and say that for step number two which is inductive step so an inductive step what we do is we consider that the given statement is true for any natural number and we prove that it is true for the next national number so here I say that now we assume that the rule is true for any natural number so let’s say that natural number is M so n is equal to M and we’ll prove that it is true for next natural number so the next natural number of M is M plus one so true for n equal to M plus one so let’s start with this so since n is equal to M so our statement so on statement Laplace of T raised to n will become T raised to n here we’ll get minus 1 raise to M D raise to M D s raise to M Phi of s so you will get Laplace of T raise to M f of T is equal to minus 1 raise to M D raise to M be s raise to M Phi of s so we have assumed that role is true for n equal to M hence I am substituting n equal to M everywhere now on left hand side we can apply the definition of Laplace transform so therefore on left hand side will get integration 0 to infinity e raise to minus s T into function of T so your function of T is T raise to M F of T DT which is equal to minus 1 raise to M D raise to M D s raise to M Phi of s now as we want to prove that this statement is true for n equal to M plus 1 we have to get M plus 1 everywhere and for that I will differentiate on both sides with respect to s so that here we can get M plus 1 but if you lobster the left hand side then we have integration so your we have to differentiate with respect to s as well as we have to apply the D you is that a differentiation under the integral sign so you will say differentiating both sides with respect to s and applying bu is that is differentiation and the integral sign so on the right hand side if you will see then minus 1 raise to M is a constant so I’ll keep it outside now as we want to differentiate this term so right now we have differentiation or I’ll say the M order differentiation so M the order differentiation when I differentiate once again with respect to s that will become M plus 1 to order so because we are taking the next order of derivative so we are taking to you to one more time hence that M is becoming M plus 1 so we go to value on right hand side on left hand side as we are applying the U is what will do is inside the integral sign will differentiate partially with respect to s so that will become dou by dou s of a raised to minus s T into T raise to M F of T DT now to differentiate this function with respect to s this P raised to M is constant because it does not have s as well as F of T is also constant so I take these two functions outside and the derivative of e raised to minus s T is e raised to minus s T into minus T because I am differentiating with respect to s so therefore integration 0 to infinity T raise to M F of T outside and I raise to minus s T into minus T DT is equal to right hand side as it is which is minus 1 raise to M D raised to M plus 1 upon D s raised to M plus 1 or M plus 1 derivative of Phi of s now you’re now here I can take minus sign outside so that will become minus in aggression 0 to infinity next arrays to minus s T as it is and T raised to M into T raised to 1 will become T raised to M plus 1 and f of T as it is DT is equal to minus 1 raise to M D raised to M plus 1 upon D s raised to n plus 1 into Phi of s so now on left hand side we have the definition of Laplace transform so erased 2 minus s T into this is my function of T and integration is with respect to 0 to infinity so I can say minus n as it is that it is Laplace of T raised to M plus 1 into f of T which is equal to minus 1 raise to M d raised to M plus 1 upon D s raised to M plus 1 into Phi of s so therefore we can say that Laplace of T raised to M plus 1 into f of T equal to now I’ll send this minus sign on other side so when it will go there that will become minus 1 raised to M plus 1 and here it is d raised to M plus 1 upon D s raise to n plus 1 or the M plus 1 derivative of Phi of s so there is now if you observe then we again got the same statement with n equal to M plus 1 so here I can say that therefore if this statement or the property is true for n equal to M then it is true for and equal to M plus one so you have seen here so we have completed the two steps of method of mathematical induction so from these two steps now will infer something so now I am going to step number three or the final step so here I say that since the property is true for n equal to one as we have seen in step number one we can say that by step number two we can say that it is true for n equal to M plus 1 where m is 1 that is 1 plus 1 that is 2 it means it is also true for n equal to 2 because in step number 2 we have proved that if the property is true for n equal to M then it is true for next natural number that is M plus 1 so since it is true for n equal to 1 from step number 2 we can say that it is true for n equal to 2 as well now I will repeat the concept once again for n equal to 2 so now I will say since the property is true for n equal to 2 then by step number 2 we can say it is true for the next natural number that is it is true for n equal to 2 plus 1 that is 3 and if I repeat this process again then you will find that the above property is true for all natural numbers so therefore you are and say that it is true for any value of n so here we accrued that multiplication by T theorem and we will use that result to solve the numerical based on multiplication by T of Laplace transform thank you |
jemh114 (1)
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MATHEMATICS
STATISTICS
14
14.1 Introduction
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives.
14.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
x1, x2,. . ., xn are observations with respective frequencies f1, f2, . . ., fn, then this
means observation x1 occurs f1 times, x2 occurs f2 times, and so on.
Now, the sum of the values of all the observations = f1x1 + f2x2 + . . . + fnxn, and
the number of observations = f1 + f2 + . . . + fn.
So, the mean x of the data is given by
f x + f 2 x2 + + f n xn
x = 1 1
f1 + f 2 + + f n
Recall that we can write this in short form by using the Greek letter Σ (capital
sigma) which means summation. That is,
2022-23
STATISTICS
261
n
fx
i i
x =
i =1
n
f
i
i =1
Σ fi xi
which, more briefly, is written as x =
, if it is understood that i varies from
Σ fi
1 to n.
Let us apply this formula to find the mean in the following example.
Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students.
Marks obtained 10
(x i )
Number of
students ( fi)
1
20
36
40
50
56
60
70
72
80
88
92 95
1
3
4
3
2
4
4
1
1
2
3
1
Solution: Recall that to find the mean marks, we require the product of each xi with
the corresponding frequency fi. So, let us put them in a column as shown in Table 14.1.
Table 14.1
.
Marks obtained (xi )
Number of students ( fi )
10
20
36
40
50
56
60
70
72
80
88
92
95
1
1
3
4
3
2
4
4
1
1
2
3
1
Total
Σfi = 30
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f ix i
10
20
108
160
150
112
240
280
72
80
176
276
95
Σfi xi = 1779
262
MATHEMATICS
x=
Now,
1779
Σ fi xi
=
= 59.3
30
Σ fi
Therefore, the mean marks obtained is 59.3.
In most of our real life situations, data is usually so large that to make a meaningful
study it needs to be condensed as grouped data. So, we need to convert given ungrouped
data into grouped data and devise some method to find its mean.
Let us convert the ungrouped data of Example 1 into grouped data by forming
class-intervals of width, say 15. Remember that, while allocating frequencies to each
class-interval, students falling in any upper class-limit would be considered in the next
class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped
frequency distribution table (see Table 14.2).
Table 14.2
Class interval
Number of students
10 - 25
25 - 40
40 - 55
55 - 70
70 - 85
85 - 100
2
3
7
6
6
6
Now, for each class-interval, we require a point which would serve as the
representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each
class can be chosen to represent the observations falling in the class. Recall that we
find the mid-point of a class (or its class mark) by finding the average of its upper and
lower limits. That is,
Class mark =
Upper class limit + Lower class limit
2
10 + 25
, i.e.,
2
17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 14.3. These class marks serve as our xi’s. Now, in general, for the ith
class interval, we have the frequency fi corresponding to the class mark xi. We can
now proceed to compute the mean in the same manner as in Example 1.
With reference to Table 14.2, for the class 10-25, the class mark is
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STATISTICS
263
Table 14.3
Class interval
Number of students ( fi )
Class mark (xi )
fi xi
10 - 25
2
17.5
35.0
25 - 40
3
32.5
97.5
40 - 55
7
47.5
332.5
55 - 70
6
62.5
375.0
70 - 85
6
77.5
465.0
85 - 100
6
92.5
555.0
Total
Σ fi = 30
Σ fi xi = 1860.0
The sum of the values in the last column gives us Σ fi xi. So, the mean x of the
given data is given by
Σfi xi 1860.0
x =
=
= 62
Σ fi
30
This new method of finding the mean is known as the Direct Method.
We observe that Tables 14.1 and 14.3 are using the same data and employing the
same formula for the calculation of the mean but the results obtained are different.
Can you think why this is so, and which one is more accurate? The difference in the
two values is because of the mid-point assumption in Table 14.3, 59.3 being the exact
mean, while 62 an approximate mean.
Sometimes when the numerical values of xi and fi are large, finding the product
of xi and fi becomes tedious and time consuming. So, for such situations, let us think of
a method of reducing these calculations.
We can do nothing with the fi’s, but we can change each xi to a smaller number
so that our calculations become easy. How do we do this? What about subtracting a
fixed number from each of these xi’s? Let us try this method.
The first step is to choose one among the xi’s as the assumed mean, and denote
it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xi
which lies in the centre of x1, x2, . . ., xn. So, we can choose a = 47.5 or a = 62.5. Let
us choose a = 47.5.
The next step is to find the difference di between a and each of the xi’s, that is,
the deviation of ‘a’ from each of the xi’s.
i.e.,
di = xi – a = xi – 47.5
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264
MATHEMATICS
The third step is to find the product of di with the corresponding fi, and take the sum
of all the fi di’s. The calculations are shown in Table 14.4.
Table 14.4
Class interval
10 - 25
25 - 40
40 - 55
55 - 70
70 - 85
85 - 100
Total
Number of
students ( fi )
Class mark
(xi )
di = xi – 47.5
fid i
2
3
7
6
6
6
Σfi = 30
17.5
32.5
47.5
62.5
77.5
92.5
–30
–15
0
15
30
45
–60
–45
0
90
180
270
Σfidi = 435
So, from Table 14.4, the mean of the deviations, d =
Σfi di
.
Σfi
Now, let us find the relation between d and x .
Since in obtaining di, we subtracted ‘a’ from each xi, so, in order to get the mean
x , we need to add ‘a’ to d . This can be explained mathematically as:
Mean of deviations,
d =
So,
d =
Σfi di
Σfi
Σfi ( xi − a )
Σfi
Σfi xi Σf i a
−
=
Σf i
Σfi
= x −a
Σfi
Σfi
= x−a
So,
x = a+ d
i.e.,
x = a+
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Σfi di
Σfi
STATISTICS
265
Substituting the values of a, Σfidi and Σfi from Table 14.4, we get
x = 47.5 +
435
= 47.5 + 14.5 = 62 .
30
Therefore, the mean of the marks obtained by the students is 62.
The method discussed above is called the Assumed Mean Method.
Activity 1 : From the Table 14.3 find the mean by taking each of xi (i.e., 17.5, 32.5,
and so on) as ‘a’. What do you observe? You will find that the mean determined in
each case is the same, i.e., 62. (Why ?)
So, we can say that the value of the mean obtained does not depend on the
choice of ‘a’.
Observe that in Table 14.4, the values in Column 4 are all multiples of 15. So, if
we divide the values in the entire Column 4 by 15, we would get smaller numbers to
multiply with fi. (Here, 15 is the class size of each class interval.)
So, let ui =
xi − a
, where a is the assumed mean and h is the class size.
h
Now, we calculate ui in this way and continue as before (i.e., find fi ui and
then Σ fi ui). Taking h = 15, let us form Table 14.5.
Table 14.5
Class interval
10 - 25
25 - 40
40 - 55
55 - 70
70 - 85
85 - 100
Total
Let
fi
xi
2
3
7
6
6
6
17.5
32.5
47.5
62.5
77.5
92.5
di = xi – a
–30
–15
0
15
30
45
Σfi = 30
ui =
xi – a
h
–2
–1
0
1
2
3
fi ui
–4
–3
0
6
12
18
Σfiui = 29
u =
Σfi ui
Σf i
Here, again let us find the relation between u and x .
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266
MATHEMATICS
We have,
Therefore,
ui =
u =
xi − a
h
Σfi
( xi − a )
1
h
=
Σf i
h
=
Σf
1 Σfi xi
−a i
h Σf i
Σfi
=
1
[ x − a]
h
Σfi xi − a Σfi
Σfi
hu = x − a
So,
x = a + hu
i.e.,
Σf u
x = a + h i i
Σfi
Now, substituting the values of a, h, Σfiui and Σfi from Table 14.5, we get
So,
29
x = 47.5 + 15 ×
30
= 47.5 + 14.5 = 62
So, the mean marks obtained by a student is 62.
The method discussed above is called the Step-deviation method.
We note that :
the step-deviation method will be convenient to apply if all the di’s have a
common factor.
The mean obtained by all the three methods is the same.
The assumed mean method and step-deviation method are just simplified
forms of the direct method.
The formula x = a + h u still holds if a and h are not as given above, but are
xi − a
.
h
Let us apply these methods in another example.
any non-zero numbers such that ui =
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267
Example 2 : The table below gives the percentage distribution of female teachers in
the primary schools of rural areas of various states and union territories (U.T.) of
India. Find the mean percentage of female teachers by all the three methods discussed
in this section.
Percentage of
15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85
female teachers
Number of
States/U.T.
6
11
7
4
4
2
1
Source : Seventh All India School Education Survey conducted by NCERT
Solution : Let us find the class marks, xi, of each class, and put them in a column
(see Table 14.6):
Table 14.6
xi
Percentage of female
Number of
teachers
States /U.T. ( f i )
15 - 25
6
20
25 - 35
11
30
35 - 45
7
40
45 - 55
4
50
55 - 65
4
60
65 - 75
2
70
75 - 85
1
80
Here we take a = 50, h = 10, then di = xi – 50 and ui =
We now find di and ui and put them in Table 14.7.
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xi − 50
.
10
268
MATHEMATICS
Table 14.7
Percentage of
female
teachers
15 - 25
25 - 35
35 - 45
45 - 55
55 - 65
65 - 75
75 - 85
Total
xi
di = xi – 50
20
30
40
50
60
70
80
–30
–20
–10
0
10
20
30
Number of
states/U.T.
( fi )
6
11
7
4
4
2
1
ui =
xi −50
10
fi xi
–3
–2
–1
0
1
2
3
120 –180 –18
330 –220 –22
280 –70 –7
200
0
0
240
40
4
140
40
4
80
30
3
35
fi di
fi ui
1390 –360 –36
From the table above, we obtain Σfi = 35, Σfixi = 1390,
Σfi di = – 360,
Σfiui = –36.
Σf i xi 1390
=
= 39.71
Σfi
35
Using the assumed mean method,
Using the direct method, x =
Σfi di
( −360)
= 39.71
= 50 +
Σfi
35
Using the step-deviation method,
x = a+
Σf u
– 36
x = a + i i × h = 50 +
× 10 = 39.71
35
Σf i
Therefore, the mean percentage of female teachers in the primary schools of
rural areas is 39.71.
Remark : The result obtained by all the three methods is the same. So the choice of
method to be used depends on the numerical values of xi and fi. If xi and fi are
sufficiently small, then the direct method is an appropriate choice. If xi and fi are
numerically large numbers, then we can go for the assumed mean method or
step-deviation method. If the class sizes are unequal, and xi are large numerically, we
can still apply the step-deviation method by taking h to be a suitable divisor of all the di’s.
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Example 3 : The distribution below shows the number of wickets taken by bowlers in
one-day cricket matches. Find the mean number of wickets by choosing a suitable
method. What does the mean signify?
Number of
wickets
20 - 60
Number of
bowlers
7
60 - 100 100 - 150 150 - 250 250 - 350 350 - 450
5
16
12
2
3
Solution : Here, the class size varies, and the xi,s are large. Let us still apply the stepdeviation method with a = 200 and h = 20. Then, we obtain the data as in Table 14.8.
Table 14.8
ui =
di
20
Number of
wickets
taken
Number of
bowlers
( fi )
xi
di = xi – 200
20 - 60
7
40
–160
–8
–56
60 - 100
5
80
–120
–6
–30
100 - 150
16
125
–75
–3.75
–60
150 - 250
12
200
0
0
0
250 - 350
2
300
100
5
10
350 - 450
3
400
200
10
30
Total
45
So, u =
u i fi
–106
−106
−106
⋅ Therefore, x = 200 + 20
= 200 – 47.11 = 152.89.
45
45
This tells us that, on an average, the number of wickets taken by these 45 bowlers
in one-day cricket is 152.89.
Now, let us see how well you can apply the concepts discussed in this section!
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Activity 2 :
Divide the students of your class into three groups and ask each group to do one of the
following activities.
1. Collect the marks obtained by all the students of your class in Mathematics in the
latest examination conducted by your school. Form a grouped frequency distribution
of the data obtained.
2. Collect the daily maximum temperatures recorded for a period of 30 days in your
city. Present this data as a grouped frequency table.
3. Measure the heights of all the students of your class (in cm) and form a grouped
frequency distribution table of this data.
After all the groups have collected the data and formed grouped frequency
distribution tables, the groups should find the mean in each case by the method which
they find appropriate.
EXERCISE 14.1
1. A survey was conducted by a group of students as a part of their environment awareness
programme, in which they collected the following data regarding the number of plants in
20 houses in a locality. Find the mean number of plants per house.
Number of plants
0-2
2-4
4-6
6-8
8 - 10
10 - 12
12 - 14
Number of houses
1
2
1
5
6
2
3
Which method did you use for finding the mean, and why?
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in )
Number of workers
500 - 520
520 -540
540 - 560
560 - 580
580 -600
12
14
8
6
10
Find the mean daily wages of the workers of the factory by using an appropriate method.
3. The following distribution shows the daily pocket allowance of children of a locality.
The mean pocket allowance is Rs 18. Find the missing frequency f.
Daily pocket
allowance (in )
Number of children
11 - 13
13 - 15
15 - 17
17 - 19
19 - 21
21 - 23
23 - 25
7
6
9
13
f
5
4
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4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per
minute were recorded and summarised as follows. Find the mean heartbeats per minute
for these women, choosing a suitable method.
Number of heartbeats 65 - 68
per minute
Number of women
2
68 - 71
71 - 74
74 - 77
77 - 80
80 - 83
83 - 86
4
3
8
7
4
2
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These
boxes contained varying number of mangoes. The following was the distribution of
mangoes according to the number of boxes.
Number of mangoes
50 - 52
53 - 55
56 - 58
59 - 61
62 - 64
15
110
135
115
25
Number of boxes
Find the mean number of mangoes kept in a packing box. Which method of finding
the mean did you choose?
6. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure
(in )
Number of
households
100 - 150
150 - 200
200 - 250
250 - 300
300 - 350
4
5
12
2
2
Find the mean daily expenditure on food by a suitable method.
7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data
was collected for 30 localities in a certain city and is presented below:
Concentration of SO2 (in ppm)
Frequency
0.00 - 0.04
4
0.04 - 0.08
9
0.08 - 0.12
9
0.12 - 0.16
2
0.16 - 0.20
4
0.20 - 0.24
2
Find the mean concentration of SO2 in the air.
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8. A class teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.
Number of
days
0-6
Number of
students
11
6 - 10 10 - 14 14 - 20 20 - 28 28 - 38 38 - 40
10
7
4
4
3
1
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
Literacy rate (in %)
45 - 55
55 - 65
65 - 75
75 - 85
85 - 95
3
10
11
8
3
Number of cities
14.3 Mode of Grouped Data
Recall from Class IX, a mode is that value among the observations which occurs most
often, that is, the value of the observation having the maximum frequency. Further, we
discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining
a mode of grouped data. It is possible that more than one value may have the same
maximum frequency. In such situations, the data is said to be multimodal. Though
grouped data can also be multimodal, we shall restrict ourselves to problems having a
single mode only.
Let us first recall how we found the mode for ungrouped data through the following
example.
Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows:
2
6
4
5
0
2
1
3
2
3
Find the mode of the data.
Solution : Let us form the frequency distribution table of the given data as follows:
Number of
wickets
0
1
2
3
4
5
6
Number of
matches
1
1
3
2
1
1
1
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Clearly, 2 is the number of wickets taken by the bowler in the maximum number
(i.e., 3) of matches. So, the mode of this data is 2.
In a grouped frequency distribution, it is not possible to determine the mode by
looking at the frequencies. Here, we can only locate a class with the maximum
frequency, called the modal class. The mode is a value inside the modal class, and is
given by the formula:
f1 − f0
Mode = l +
×h
2 f1 − f 0 − f 2
where l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f1 = frequency of the modal class,
f0 = frequency of the class preceding the modal class,
f2 = frequency of the class succeeding the modal class.
Let us consider the following examples to illustrate the use of this formula.
Example 5 : A survey conducted on 20 households in a locality by a group of students
resulted in the following frequency table for the number of family members in a
household:
Family size
1-3
3-5
5-7
7-9
9 - 11
Number of
families
7
8
2
2
1
Find the mode of this data.
Solution : Here the maximum class frequency is 8, and the class corresponding to this
frequency is 3 – 5. So, the modal class is 3 – 5.
Now
modal class = 3 – 5, lower limit (l ) of modal class = 3, class size (h) = 2
frequency ( f1 ) of the modal class = 8,
frequency ( f0 ) of class preceding the modal class = 7,
frequency ( f2 ) of class succeeding the modal class = 2.
Now, let us substitute these values in the formula :
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MATHEMATICS
f1 − f0
Mode = l +
×h
2 f1 − f 0 − f 2
8−7
2
= 3+
× 2 = 3 + = 3.286
7
2×8 − 7 − 2
Therefore, the mode of the data above is 3.286.
Example 6 : The marks distribution of 30 students in a mathematics examination are
given in Table 14.3 of Example 1. Find the mode of this data. Also compare and
interpret the mode and the mean.
Solution : Refer to Table 14.3 of Example 1. Since the maximum number of students
(i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore,
the lower limit ( l ) of the modal class = 40,
the class size ( h) = 15,
the frequency ( f1 ) of modal class = 7,
the frequency ( f0 ) of the class preceding the modal class = 3,
the frequency ( f2 ) of the class succeeding the modal class = 6.
Now, using the formula:
f1 − f0
Mode = l +
× h,
2 f1 − f 0 − f 2
we get
7−3
Mode = 40 +
× 15 = 52
14 − 6 − 3
So, the mode marks is 52.
Now, from Example 1, you know that the mean marks is 62.
So, the maximum number of students obtained 52 marks, while on an average a
student obtained 62 marks.
Remarks :
1. In Example 6, the mode is less than the mean. But for some other problems it may
be equal or more than the mean also.
2. It depends upon the demand of the situation whether we are interested in finding the
average marks obtained by the students or the average of the marks obtained by most
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275
of the students. In the first situation, the mean is required and in the second situation,
the mode is required.
Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations
assigned to the groups. Ask each group to find the mode of the data. They should also
compare this with the mean, and interpret the meaning of both.
Remark : The mode can also be calculated for grouped data with unequal class sizes.
However, we shall not be discussing it.
EXERCISE 14.2
1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years)
5 - 15
15 - 25
25 - 35
35 - 45
45 - 55
55 - 65
6
11
21
23
14
5
Number of patients
Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency.
2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components :
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100
100 - 120
10
35
52
61
38
29
Frequency
Determine the modal lifetimes of the components.
3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :
Expenditure (in )
Number of families
1000 - 1500
1500 - 2000
2000 - 2500
2500 - 3000
3000 - 3500
3500 - 4000
4000 - 4500
4500 - 5000
24
40
33
28
30
22
16
7
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4. The following distribution gives the state-wise teacher-student ratio in higher
secondary schools of India. Find the mode and mean of this data. Interpret the two
measures.
Number of students per teacher
Number of states / U .T.
15 - 20
3
20 - 25
8
25 - 30
9
30 - 35
10
35 - 40
3
40 - 45
0
45 - 50
0
50 - 55
2
5. The given distribution shows the number of runs scored by some top batsmen of the
world in one-day international cricket matches.
Runs scored
Number of batsmen
3000 - 4000
4
4000 - 5000
18
5000 - 6000
9
6000 - 7000
7
7000 - 8000
6
8000 - 9000
3
9000 - 10000
1
10000 - 11000
1
Find the mode of the data.
6. A student noted the number of cars passing through a spot on a road for 100
periods each of 3 minutes and summarised it in the table given below. Find the mode
of the data :
Number of cars
Frequency
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
7
14
13
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20
11
15
8
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277
14.4 Median of Grouped Data
As you have studied in Class IX, the median is a measure of central tendency which
gives the value of the middle-most observation in the data. Recall that for finding the
median of ungrouped data, we first arrange the data values of the observations in
n + 1
ascending order. Then, if n is odd, the median is the
th observation. And, if n
2
is even, then the median will be the average of the
n
n
th and the + 1 th observations.
2
2
Suppose, we have to find the median of the following data, which gives the
marks, out of 50, obtained by 100 students in a test :
Marks obtained
20
29
28
33
42
38
43
25
Number of students
6
28
24
15
2
4
1
20
First, we arrange the marks in ascending order and prepare a frequency table as
follows :
Table 14.9
Marks obtained
Number of students
(Frequency)
20
6
25
20
28
24
29
28
33
15
38
4
42
2
43
1
Total
100
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Here n = 100, which is even. The median will be the average of the
n th and the
2
n
+ 1 th observations, i.e., the 50th and 51st observations. To find these
2
observations, we proceed as follows:
Table 14.10
Marks obtained
Number of students
20
6
upto 25
6 + 20 = 26
upto 28
26 + 24 = 50
upto 29
50 + 28 = 78
upto 33
78 + 15 = 93
upto 38
93 + 4 = 97
upto 42
97 + 2 = 99
upto 43
99 + 1 = 100
Now we add another column depicting this information to the frequency table
above and name it as cumulative frequency column.
Table 14.11
Marks obtained
Number of students
Cumulative frequency
20
6
6
25
20
26
28
24
50
29
28
78
33
15
93
38
4
97
42
2
99
43
1
100
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279
From the table above, we see that:
50th observaton is 28
(Why?)
51st observation is 29
So,
Median =
28 + 29
= 28.5
2
Remark : The part of Table 14.11 consisting Column 1 and Column 3 is known as
Cumulative Frequency Table. The median marks 28.5 conveys the information that
about 50% students obtained marks less than 28.5 and another 50% students obtained
marks more than 28.5.
Now, let us see how to obtain the median of grouped data, through the following
situation.
Consider a grouped frequency distribution of marks obtained, out of 100, by 53
students, in a certain examination, as follows:
Table 14.12
Marks
Number of students
0 - 10
5
10 - 20
3
20 - 30
4
30 - 40
3
40 - 50
3
50 - 60
4
60 - 70
7
70 - 80
9
80 - 90
7
90 - 100
8
From the table above, try to answer the following questions:
How many students have scored marks less than 10? The answer is clearly 5.
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MATHEMATICS
How many students have scored less than 20 marks? Observe that the number
of students who have scored less than 20 include the number of students who have
scored marks from 0 - 10 as well as the number of students who have scored marks
from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, i.e., 8.
We say that the cumulative frequency of the class 10 -20 is 8.
Similarly, we can compute the cumulative frequencies of the other classes, i.e.,
the number of students with marks less than 30, less than 40, . . ., less than 100. We
give them in Table 14.13 given below:
Table 14.13
Marks obtained
Number of students
(Cumulative frequency)
Less than 10
5
Less than 20
5+3=8
Less than 30
8 + 4 = 12
Less than 40
12 + 3 = 15
Less than 50
15 + 3 = 18
Less than 60
18 + 4 = 22
Less than 70
22 + 7 = 29
Less than 80
29 + 9 = 38
Less than 90
38 + 7 = 45
Less than 100
45 + 8 = 53
The distribution given above is called the cumulative frequency distribution of
the less than type. Here 10, 20, 30, . . . 100, are the upper limits of the respective
class intervals.
We can similarly make the table for the number of students with scores, more
than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From
Table 14.12, we observe that all 53 students have scored marks more than or equal to
0. Since there are 5 students scoring marks in the interval 0 - 10, this means that there
are 53 – 5 = 48 students getting more than or equal to 10 marks. Continuing in the
same manner, we get the number of students scoring 20 or above as 48 – 3 = 45, 30 or
above as 45 – 4 = 41, and so on, as shown in Table 14.14.
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281
Table 14.14
Marks obtained
Number of students
(Cumulative frequency)
More than or equal to 0
More than or equal to 10
More than or equal to 20
More than or equal to 30
More than or equal to 40
More than or equal to 50
More than or equal to 60
More than or equal to 70
More than or equal to 80
More than or equal to 90
53
53 – 5 = 48
48 – 3 = 45
45 – 4 = 41
41 – 3 = 38
38 – 3 = 35
35 – 4 = 31
31 – 7 = 24
24 – 9 = 15
15 – 7 = 8
The table above is called a cumulative frequency distribution of the more
than type. Here 0, 10, 20, . . ., 90 give the lower limits of the respective class intervals.
Now, to find the median of grouped data, we can make use of any of these
cumulative frequency distributions.
Let us combine Tables 14.12 and 14.13 to get Table 14.15 given below:
Table 14.15
Marks
Number of students ( f )
Cumulative frequency (cf )
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70
70 - 80
80 - 90
90 - 100
5
3
4
3
3
4
7
9
7
8
5
8
12
15
18
22
29
38
45
53
Now in a grouped data, we may not be able to find the middle observation by
looking at the cumulative frequencies as the middle observation will be some value in
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MATHEMATICS
a class interval. It is, therefore, necessary to find the value inside a class that divides
the whole distribution into two halves. But which class should this be?
n
.
2
We now locate the class whose cumulative frequency is greater than (and nearest to)
n
n
⋅ This is called the median class. In the distribution above, n = 53. So,
= 26.5.
2
2
Now 60 – 70 is the class whose cumulative frequency 29 is greater than (and nearest
n
to)
, i.e., 26.5.
2
To find this class, we find the cumulative frequencies of all the classes and
Therefore, 60 – 70 is the median class.
After finding the median class, we use the following formula for calculating the
median.
n
2 − cf
Median = l +
× h,
f
where
l = lower limit of median class,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal).
n
= 26.5, l = 60, cf = 22, f = 7, h = 10
2
in the formula above, we get
Substituting the values
26.5 − 22
Median = 60 +
× 10
7
= 60 +
45
7
= 66.4
So, about half the students have scored marks less than 66.4, and the other half have
scored marks more than 66.4.
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283
Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school
was conducted and the following data was obtained:
Height (in cm)
Number of girls
Less than 140
4
Less than 145
11
Less than 150
29
Less than 155
40
Less than 160
46
Less than 165
51
Find the median height.
Solution : To calculate the median height, we need to find the class intervals and their
corresponding frequencies.
The given distribution being of the less than type, 140, 145, 150, . . ., 165 give the
upper limits of the corresponding class intervals. So, the classes should be below 140,
140 - 145, 145 - 150, . . ., 160 - 165. Observe that from the given distribution, we find
that there are 4 girls with height less than 140, i.e., the frequency of class interval
below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with
height less than 140. Therefore, the number of girls with height in the interval
140 - 145 is 11 – 4 = 7. Similarly, the frequency of 145 - 150 is 29 – 11 = 18, for
150 - 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with the
given cumulative frequencies becomes:
Table 14.16
Class intervals
Frequency
Cumulative frequency
Below 140
140 - 145
145 - 150
150 - 155
155 - 160
160 - 165
4
7
18
11
6
5
4
11
29
40
46
51
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Now
MATHEMATICS
n = 51. So,
n 51
= = 25.5 . This observation lies in the class 145 - 150. Then,
2 2
l (the lower limit) = 145,
cf (the cumulative frequency of the class preceding 145 - 150) = 11,
f (the frequency of the median class 145 - 150) = 18,
h (the class size) = 5.
n
2 − cf
× h , we have
Using the formula, Median = l +
f
25.5 − 11
Median = 145 +
×5
18
72.5
= 149.03.
18
So, the median height of the girls is 149.03 cm.
= 145 +
This means that the height of about 50% of the girls is less than this height, and
50% are taller than this height.
Example 8 : The median of the following data is 525. Find the values of x and y, if the
total frequency is 100.
Class interval
Frequency
0 - 100
100 - 200
200 - 300
300 - 400
400 - 500
500 - 600
600 - 700
700 - 800
800 - 900
900 - 1000
2
5
x
12
17
20
y
9
7
4
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285
Solution :
Class intervals
Frequency
Cumulative frequency
0 - 100
2
2
100 - 200
5
7
200 - 300
x
7+x
300 - 400
12
19 + x
400 - 500
17
36 + x
500 - 600
20
56 + x
600 - 700
y
56 + x + y
700 - 800
9
65 + x + y
800 - 900
7
72 + x + y
900 - 1000
4
76 + x + y
It is given that n = 100
So,
76 + x + y = 100,
i.e.,
x + y = 24
(1)
The median is 525, which lies in the class 500 – 600
So,
l = 500,
f = 20,
cf = 36 + x, h = 100
n
− cf
Median = l + 2
f
Using the formula :
h, we get
50 − 36 − x
525 = 500 +
× 100
20
i.e.,
525 – 500 = (14 – x) × 5
i.e.,
25 = 70 – 5x
i.e.,
5x = 70 – 25 = 45
So,
x= 9
Therefore, from (1), we get
i.e.,
9 + y = 24
y = 15
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MATHEMATICS
Now, that you have studied about all the three measures of central tendency, let
us discuss which measure would be best suited for a particular requirement.
The mean is the most frequently used measure of central tendency because it
takes into account all the observations, and lies between the extremes, i.e., the largest
and the smallest observations of the entire data. It also enables us to compare two or
more distributions. For example, by comparing the average (mean) results of students
of different schools of a particular examination, we can conclude which school has a
better performance.
However, extreme values in the data affect the mean. For example, the mean of
classes having frequencies more or less the same is a good representative of the data.
But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20,
21, 18, then the mean will certainly not reflect the way the data behaves. So, in such
cases, the mean is not a good representative of the data.
In problems where individual observations are not important, and we wish to find
out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical
productivity rate of workers, average wage in a country, etc. These are situations
where extreme values may be there. So, rather than the mean, we take the median as
a better measure of central tendency.
In situations which require establishing the most frequent value or most popular
item, the mode is the best choice, e.g., to find the most popular T.V. programme being
watched, the consumer item in greatest demand, the colour of the vehicle used by
most of the people, etc.
Remarks :
1. There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean
2. The median of grouped data with unequal class sizes can also be calculated. However,
we shall not discuss it here.
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287
EXERCISE 14.3
1. The following frequency distribution gives the monthly consumption of electricity of
68 consumers of a locality. Find the median, mean and mode of the data and compare
them.
Monthly consumption (in units)
Number of consumers
65 - 85
4
85 - 105
5
105 - 125
13
125 - 145
20
145 - 165
14
165 - 185
8
185 - 205
4
2. If the median of the distribution given below is 28.5, find the values of x and y.
Class interval
Frequency
0 - 10
5
10 - 20
x
20 - 30
20
30 - 40
15
40 - 50
y
50 - 60
5
Total
60
3. A life insurance agent found the following data for distribution of ages of 100 policy
holders. Calculate the median age, if policies are given only to persons having age 18
years onwards but less than 60 year.
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MATHEMATICS
Age (in years)
Number of policy holders
Below 20
2
Below 25
6
Below 30
24
Below 35
45
Below 40
78
Below 45
89
Below 50
92
Below 55
98
Below 60
100
4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and
the data obtained is represented in the following table :
Length (in mm)
Number of leaves
118 - 126
3
127 - 135
5
136 - 144
9
145 - 153
12
154 - 162
5
163 - 171
4
172 - 180
2
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median,
since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)
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5. The following table gives the distribution of the life time of 400 neon lamps :
Life time (in hours)
Number of lamps
1500 - 2000
14
2000 - 2500
56
2500 - 3000
60
3000 - 3500
86
3500 - 4000
74
4000 - 4500
62
4500 - 5000
48
Find the median life time of a lamp.
6. 100 surnames were randomly picked up from a local telephone directory and the
frequency distribution of the number of letters in the English alphabets in the surnames
was obtained as follows:
Number of letters
Number of surnames
1-4
4-7
7 - 10
10 - 13
13 - 16
16 - 19
6
30
40
16
4
4
Determine the median number of letters in the surnames. Find the mean number of
letters in the surnames? Also, find the modal size of the surnames.
7. The distribution below gives the weights of 30 students of a class. Find the median
weight of the students.
Weight (in kg)
Number of students
40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75
2
3
8
6
6
3
2
14.5 Graphical Representation of Cumulative Frequency Distribution
As we all know, pictures speak better than words. A graphical representation helps us
in understanding given data at a glance. In Class IX, we have represented the data
through bar graphs, histograms and frequency polygons. Let us now represent a
cumulative frequency distribution graphically.
For example, let us consider the cumulative frequency distribution given in
Table 14.13.
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MATHEMATICS
Recall that the values 10, 20, 30,
. . ., 100 are the upper limits of the
respective class intervals. To represent
the data in the table graphically, we mark
the upper limits of the class intervals on
the horizontal axis (x-axis) and their
corresponding cumulative frequencies
on the vertical axis ( y-axis), choosing a
convenient scale. The scale may not be
the same on both the axis. Let us now
plot the points corresponding to the
ordered pairs given by (upper limit,
corresponding cumulative frequency),
Fig. 14.1
i.e., (10, 5), (20, 8), (30, 12), (40, 15),
(50, 18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53) on a graph paper and join them
by a free hand smooth curve. The curve we get is called a cumulative frequency
curve, or an ogive (of the less than type). (See Fig. 14.1)
The term ‘ogive’ is pronounced as ‘ojeev’ and is derived from the word ogee.
An ogee is a shape consisting of a concave arc flowing into a convex arc, so
forming an S-shaped curve with vertical ends. In architecture, the ogee shape
is one of the characteristics of the 14th and 15th century Gothic styles.
Next, again we consider the cumulative frequency distribution given in
Table 14.14 and draw its ogive (of the more than type).
Recall that, here 0, 10, 20, . . ., 90
are the lower limits of the respective class
intervals 0 - 10, 10 - 20, . . ., 90 - 100. To
represent ‘the more than type’ graphically,
we plot the lower limits on the x-axis and
the corresponding cumulative frequencies
on the y-axis. Then we plot the points
(lower limit, corresponding cumulative
frequency), i.e., (0, 53), (10, 48), (20, 45),
(30, 41), (40, 38), (50, 35), (60, 31),
(70, 24), (80, 15), (90, 8), on a graph paper,
Fig. 14.2
and join them by a free hand smooth curve.
The curve we get is a cumulative frequency curve, or an ogive (of the more than
type). (See Fig. 14.2)
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STATISTICS
291
Remark : Note that both the ogives (in Fig. 14.1 and Fig. 14.2) correspond to the
same data, which is given in Table 14.12.
Now, are the ogives related to the median in any way? Is it possible to obtain the
median from these two cumulative frequency curves corresponding to the data in
Table 14.12? Let us see.
One obvious way is to locate
n 53
=
= 26.5 on the y-axis (see Fig.
2
2
14.3). From this point, draw a line parallel
to the x-axis cutting the curve at a point.
From this point, draw a perpendicular to
the x-axis. The point of intersection of
this perpendicular with the x-axis
determines the median of the data (see
Fig. 14.3).
Fig. 14.3
Another way of obtaining the
median is the following :
Draw both ogives (i.e., of the less
than type and of the more than type) on
the same axis. The two ogives will
intersect each other at a point. From this
point, if we draw a perpendicular on the
x-axis, the point at which it cuts the
x-axis gives us the median (see Fig. 14.4).
Fig. 14.4
Example 9 : The annual profits earned by 30 shops of a shopping complex in a
locality give rise to the following distribution :
Profit (Rs in lakhs)
Number of shops (frequency)
More than or equal to 5
More than or equal to 10
More than or equal to 15
More than or equal to 20
More than or equal to 25
More than or equal to 30
More than or equal to 35
30
28
16
14
10
7
3
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MATHEMATICS
Draw both ogives for the data above.
Hence obtain the median profit.
Solution : We first draw the coordinate
axes, with lower limits of the profit along
the horizontal axis, and the cumulative
frequency along the vertical axes. Then,
we plot the points (5, 30), (10, 28), (15, 16),
(20, 14), (25, 10), (30, 7) and (35, 3). We
join these points with a smooth curve to
get the ‘more than’ ogive, as shown in
Fig. 14.5.
Now, let us obtain the classes, their
frequencies and the cumulative frequency
from the table above.
Fig. 14.5
Table 14.17
Classes
5 - 10
10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40
No. of shops
2
12
2
4
3
4
3
Cumulative
frequency
2
14
16
20
23
27
30
Using these values, we plot the points
(10, 2), (15, 14), (20, 16), (25, 20), (30, 23),
(35, 27), (40, 30) on the same axes as in
Fig. 14.5 to get the ‘less than’ ogive, as
shown in Fig. 14.6.
The abcissa of their point of intersection is
nearly 17.5, which is the median. This can
also be verified by using the formula.
Hence, the median profit (in lakhs) is
17.5.
Remark : In the above examples, it may
be noted that the class intervals were
continuous. For drawing ogives, it should
be ensured that the class intervals are
continuous. (Also see constructions of
histograms in Class IX)
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STATISTICS
293
EXERCISE 14.4
1. The following distribution gives the daily income of 50 workers of a factory.
Daily income (in )
100 - 120
120 - 140
140 - 160
160 - 180
180 - 200
Number of workers
12
14
8
6
10
Convert the distribution above to a less than type cumulative frequency distribution,
and draw its ogive.
2. During the medical check-up of 35 students of a class, their weights were recorded as
follows:
Weight (in kg)
Number of students
Less than 38
0
Less than 40
3
Less than 42
5
Less than 44
9
Less than 46
14
Less than 48
28
Less than 50
32
Less than 52
35
Draw a less than type ogive for the given data. Hence obtain the median weight from
the graph and verify the result by using the formula.
3. The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield
(in kg/ha)
50 - 55
55 - 60
60 - 65
65 - 70
70 - 75
75 - 80
Number of farms
2
8
12
24
38
16
Change the distribution to a more than type distribution, and draw its ogive.
14.6 Summary
In this chapter, you have studied the following points:
1. The mean for grouped data can be found by :
(i) the direct method : x =
Σfi xi
Σfi
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MATHEMATICS
(ii) the assumed mean method : x = a +
Σ f i di
Σf i
Σf u
(iii) the step deviation method : x = a + i i × h ,
Σf i
with the assumption that the frequency of a class is centred at its mid-point, called its
class mark.
2. The mode for grouped data can be found by using the formula:
f1 − f 0
Mode = l +
×h
f
f
f
2
−
−
0
2
1
where symbols have their usual meanings.
3. The cumulative frequency of a class is the frequency obtained by adding the frequencies
of all the classes preceding the given class.
4. The median for grouped data is formed by using the formula:
n
2 − cf
Median = l +
× h,
f
where symbols have their usual meanings.
5. Representing a cumulative frequency distribution graphically as a cumulative frequency
curve, or an ogive of the less than type and of the more than type.
6. The median of grouped data can be obtained graphically as the x-coordinate of the point
of intersection of the two ogives for this data. |
# 3 Modular arithmetic
## 3.3 Operations in modular arithmetic
The Division Algorithm tells us that all the possible remainders on division by an integer n lie in the set
We denote this set by n. For each integer n ≥ 2 we have a set n, and it is on these sets that we perform modular arithmetic. The modular addition operations +n and modular multiplication operations ×n are defined as follows.
### Definitions
For any integer n ≥ 2,
For a and b in n, the operations +n and ×n are defined by:
a +n b is the remainder of a + b on division by n;
a ×n b is the remainder of a × b on division by n.
The integer n is called the modulus for this arithmetic.
Note: a +n b is read as a plus b (mod n), and may also be written a + b (mod n). Similarly, a ×n b is read as a times b (mod n) and may also be written as a × b (mod n).
For example, 7 = { 0,1,2,3,4,5,6} and we have
You have certainly met some modular arithmetic before, as the operations +12 and +24 are used in measuring time on 12-hour and 24-hour clocks, respectively.
### Exercise 41
Evaluate the following.
(a) 3 +5 7, 4 +17 5, 8 +16 12.
(b) 3 ×5 7, 4 ×17 5, 8 ×16 12.
### Solution
(a) 3 +5 7 = 0, 4 +17 5 = 9, 8 +16 12 = 4.
(b) 3 ×5 7 = 1, 4 ×17 5 = 3, 8 ×16 12 = 0.
In Sections 1 and 2 we listed some properties satisfied by the real and complex numbers. We now investigate whether the sets n satisfy similar properties.
We also investigate what equations we can solve in n; for example, can we solve the equations
These may look much simpler than the equations that we were trying to solve in , but they pose interesting questions. We shall see that the answers may depend on the modulus that we are using.
Before we discuss these questions further, we look at addition and multiplication tables, which provide a convenient way of studying addition and multiplication in n.
We consider addition first. Here are the addition tables for 4 and 7.
In order to evaluate 4 +7 2, say, we look in the row labelled 4 and the column labelled 2 in the second table to obtain the answer 6.
### Exercise 42
(a) Use the tables above to solve the following equations.
(i) x +4 3 = 2
(ii) x +7 5 = 2
(iii) x +4 2 = 0
(iv) x +7 5 = 0
(b) What patterns do you notice in the tables?
### Solution
(a)
(i) 3 +4 3 = 2, so x = 3.
(ii) 4 +7 5 = 2, so x = 4.
(iii) 2 +4 2 = 0, so x = 2.
(iv) 2 +7 5 = 0, so x = 2.
(b) You may have noticed that:
1. each element appears exactly once in each row and exactly once in each column;
2. there is a pattern of diagonal stripes running down from right to left.
### Exercise 43
(a) Construct the addition table for 6.
(b) Solve the equations x +6 1 = 5 and x +6 5 = 1.
### Solution
(a)
(b) x +6 1 = 5 has solution x = 4.
x +6 5 = 1 has solution x = 2.
For every integer n ≥ 2, the additive properties of n are the same as the additive properties of , as follows.
Property A1 follows from the Division Algorithm and the definition of n. The other properties can be deduced from the corresponding properties for integers.
### Exercise 44
By using the corresponding property for integers, prove property A5.
### Solution
By definition, a +n b and b +n a are the remainders of a + b and b + a, respectively, on division by n. But a + b = b + a, so a +n b = b+n a.
If a, b n and a +n b = 0, then we say that b is the additive inverse of a in n. For example, 4 and 5 belong to 9 and 4 +9 5 = 0, so 5 is the additive inverse of 4 in 9. Property A3 states that every element of n has an additive inverse in n.
Additive inverses are sometimes written in the form −na; that is, if a +n b = 0, then we write b = −na. For example, 5 = −94.
### Exercise 45
(a) Use the addition table for 7 (which appear above Exercise 42) to complete the following table of additive inverses in 7.
### Solution
(a)
(b)
The additive inverse of 0 is always 0, since 0 +n 0 = 0. For any integer r > 0 in n, nr n and r + (nr) = n, so r +n (nr) = 0.
The existence of additive inverses means that, as well as doing addition modulo n, we can also do subtraction. We define an b or, equivalently, ab (mod n), to be the remainder of ab on division by n.
(With this definition, an b is equal to a +n (−nb).)
For example, to find 2 −8 7, we have
Since 3 8, it follows that |
# Undetermined Coefficients
by Justin Skycak on
Undetermined coefficients can help us find a solution to a linear differential equation with constant coefficients when the right hand side is not equal to zero.
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Undetermined Coefficients. In Justin Math: Calculus. https://justinmath.com/undetermined-coefficients/
In the previous post, we learned how to solve differential equations of the form
\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = 0. \end{align*}
Now, we consider differential equations of the form
\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = f(x) \end{align*}
where the right hand side is no longer strictly $0$, but rather some function $f(x)$. The solution to such a differential equation is given by
\begin{align*} y(x) = y_0(x) + y_f(x) \end{align*}
where $y_0$ is the general solution to the “homogeneous” equation
\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = 0 \end{align*}
and $y_f$ is a particular solution that satisfies the “inhomogeneous” equation
\begin{align*} a_ny^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = f(x). \end{align*}
We already know how to use the characteristic polynomial to find $y_0$, and now we will learn how to use the method of undetermined coefficients to find $y_f$.
The method of undetermined coefficients involves guessing a solution $y_f(x)$ having the same form as $f(x)$, except possibly multiplied by some other coefficients. We then substitute this guess into the differential equation, and solve for the value of the coefficient that will make the guess correct.
## Case of Exponential Function
For example, to find a particular solution to the differential equation $y’^\prime+2y’+y=1-2e^{3x}$, we can guess that $y_f(x)=A+Be^{3x}$ for some values of $A$ and $B$. Substituting our guess into the equation, we can solve for the correct values of $A$ and $B$.
\begin{align*} y''+2y'+y &= 1-2e^{3x} \\ \left( A+Be^{3x} \right)'' + 2 \left( A+Be^{3x} \right)' + A+Be^{3x} &= 1-2e^{3x} \\ 9Be^{3x} + 6Be^{3x} + A+Be^{3x} &= 1-2e^{3x} \\ 16Be^{3x} + A &= 1-2e^{3x} \\ A = 1, & \hspace{.25cm} B = -\frac{1}{8} \end{align*}
Our particular solution is then given by $y_f(x)=1-\frac{1}{8}e^{3x}$. Then, using the characteristic polynomial method, we solve $y’^\prime+2y’+y=0$ to find $y_0(x)=(C_1+C_2)e^{-x}$. The full solution to the differential equation $y’^\prime+2y’+y=1-2e^{3x}$ is then given by
\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= (C_1+C_2x)e^{-x} + 1 - \frac{1}{8}e^{3x}. \end{align*}
## Case of Trig Functions with Same Angle
In cases where $f(x)$ contains $\sin \theta$ or $\cos \theta$, we include both $\sin \theta$ and $\cos \theta$ in our guess for $y_0$.
For example, to find a particular solution to the differential equation $y’’^\prime-3y’^\prime+2y’=-3\cos 2x$, we need to construct a guess that contains both $\sin 2x$ and $\cos 2x$. Our guess, then, is
\begin{align*} y_f(x) = A \sin 2x + B \cos 2x. \end{align*}
We substitute this guess into the differential equation and simplify.
\begin{align*} \begin{matrix} \left( A \sin 2x + B \cos 2x \right)''' - 3\left( A \sin 2x + B \cos 2x \right)'' \\ + 2\left( A \sin 2x + B \cos 2x \right)' \end{matrix} &= -3\cos 2x \\ \begin{matrix} \text{ } \\ \left( -8A \cos 2x + 8B \sin 2x \right) - 3\left( -4A \sin 2x -4B \cos 2x \right) \\ + 2\left( 2A \cos 2x - 2B \sin 2x \right) \\ \text{ } \end{matrix} &= -3\cos 2x \\ (12A+4B) \sin 2x + (-4A+12B) \cos 2x &= -3\cos 2x \end{align*}
Equating coefficients on the left and right sides of the equation yields a system of equations for $A$ and $B$.
\begin{align*} \begin{cases} 12A+4B &= 0 \\ -4A+12B &= -3 \end{cases} \end{align*}
Solving this system, we find $A=\frac{3}{40}$ and $B=-\frac{9}{40}$. The particular solution is then $y_f(x) = \frac{3}{40} \sin 2x - \frac{9}{40} \cos 2x$.
Using the characteristic polynomial to solve $y’’^\prime-3y’^\prime+2y’=0$ yields $y_0(x)=C_1+C_2 e^x + C_3 e^{2x}$, and the full solution of the differential equation $y’’^\prime-3y’^\prime+2y’=-3\cos 2x$ is then given by
\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= C_1 + C_2 e^x + C_3 e^{2x} + \frac{3}{40} \sin 2x - \frac{9}{40} \cos 2x. \end{align*}
## Case of Trig Functions with Different Angles
When we have multiple values of $\theta$, we end up with even more unknown coefficients in our guess.
For example, to find a particular solution to the differential equation $y’^\prime-2y=4\sin 3x - \cos 5x$, we need to construct a guess that contains both $\sin \theta$ and $\cos \theta$, for both $\theta = 3x$ and $\theta = 5x$. Our guess, then, is
\begin{align*} y_f(x) = A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x. \end{align*}
We substitute this guess into the differential equation and simplify.
\begin{align*} \begin{matrix} \left( A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x \right)'' \\ - 2\left( A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x \right) \end{matrix} &= 4\sin 3x - \cos 5x \\ \begin{matrix} \text{ } \\ \left( -9A\sin 3x -9 B \cos 3x - 25C \sin 5x - 25D \cos 5x \right) \\ - 2\left( A\sin 3x + B\cos 3x + C \sin 5x + D \cos 5x \right) \\ \text{ } \end{matrix} &= 4\sin 3x - \cos 5x \\ -11A \sin 3x - 11B \cos 3x - 27C \sin 5x - 27D \cos 5x &= 4\sin 3x - \cos 5x \end{align*}
Equating coefficients on the left and right sides of the equation yields $A=-\frac{4}{11}$, $B=0$, $C=0$, and $D=\frac{1}{27}$. The particular solution is then $y_f(x) = -\frac{4}{11} \sin 3x + \frac{1}{27} \cos 5x$.
Using the characteristic polynomial to solve $y’^\prime-2y=0$ yields $y_0(x) = C_1e^{\sqrt{2} x} + C_2e^{-\sqrt{2} x}$, and the full solution of the differential equation $y’^\prime-2y=4\sin 3x -\cos 5x$ is then given by
\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= C_1e^{\sqrt{2} x} + C_2e^{-\sqrt{2} x} -\frac{4}{11} \sin 3x + \frac{1}{27} \cos 5x . \end{align*}
## Case of Polynomial Functions
Lastly, the differential equation $y’^\prime+y’=x^3-2x+e^{2x}$ has a polynomial and an exponential term, so our guess for the particular solution needs to contain a polynomial and an exponential term.
The polynomial in the differential equation is of degree $3$, and the differential equation itself is of degree $2$, so our guess needs to contain a polynomial of degree $3+2=5$.
\begin{align*} y_f(x) = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F + Ge^{2x} \end{align*}
We substitute this guess into the differential equation and simplify.
\begin{align*} \begin{matrix} \left( Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F + Ge^{2x} \right)'' \\ + \left( Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F + Ge^{2x} \right)' \end{matrix} &= x^3-2x+e^{2x} \\ \begin{matrix} \text{ } \\ \left( 20Ax^3 + 12Bx^2 + 6Cx + 2D + 4Ge^{2x} \right) \\ + \left( 5Ax^4 + 4Bx^3 + 3Cx^2 + 2Dx + E + 2Ge^{2x} \right) \\ \text{ } \end{matrix} &= x^3-2x+e^{2x} \\ \begin{matrix} 5Ax^4+(20A+4B)x^3+(12B+3C)x^2 \\ + (6C+2D)x+2D+E + 6Ge^{2x} \end{matrix} &= x^3-2x+e^{2x} \end{align*}
Equating coefficients on the left and right sides of the equation yields $A=0$, $B=\frac{1}{4}$, $C=-1$, $D=2$, $E=-4$, and $G=\frac{1}{6}$. The coefficient $F$ can still be any number, so we leave it as-is. The particular solution is then
\begin{align*} y_f(x) = \frac{1}{4}x^4 -x^3 +2x^2 -4x + F +\frac{1}{6} e^{2x} . \end{align*}
Using the characteristic polynomial to solve $y’^\prime+y’=0$ yields $y_0(x) = C_1 + C_2 e^has {-x}$, and the full solution of the differential equation $y’^\prime+y’=x^3-2x+e^{2x}$ is then given by
\begin{align*} y(x) &= y_0(x) + y_f(x) \\ &= C_1 + C_2e^{-x} +\frac{1}{4}x^4 -x^3 +2x^2 -4x + F +\frac{1}{6} e^{2x}. \end{align*}
To eliminate redundancy, we can lump the $F$ constant into the $C_1$ constant, since $C_1+F$ is itself just another constant.
\begin{align*} y(x) = C_1 + C_2e^{-x} +\frac{1}{4}x^4 -x^3 +2x^2 -4x +\frac{1}{6} e^{2x} \end{align*}
## Exercises
Use the method of undetermined coefficients to solve the following differential equations. (You can view the solution by clicking on the problem.)
\begin{align*} 1) \hspace{.5cm} y''+y=4e^{5x} \end{align*}
Solution:
\begin{align*} y=C_1 \cos x + C_2 \sin x + \frac{2}{13}e^{5x} \end{align*}
\begin{align*} 2) \hspace{.5cm} y'+3y=\sin(2x) \end{align*}
Solution:
\begin{align*} y=C_1e^{-3x}+\frac{3\sin 2x - 2\cos 2x}{13} \end{align*}
\begin{align*} 3) \hspace{.5cm} y''-y'=\cos(\pi x) \end{align*}
Solution:
\begin{align*} y=C_1e^x + C_2 - \frac{\sin (\pi x)+ \pi \cos (\pi x)}{\pi + \pi^3} \end{align*}
\begin{align*} 4) \hspace{.5cm} y'''-2y'=x^2+1 \end{align*}
Solution:
\begin{align*} y=C_1e^{\sqrt{2}x} + C_2e^{-\sqrt{2}x} + C_3 - \frac{1}{6}x^3-x \end{align*}
\begin{align*} 5) \hspace{.5cm} 2y'-y=\sin(x)-\cos(2x) \end{align*}
Solution:
\begin{align*} y=C_1 e^{\frac{1}{2}x}-\frac{\sin x + 2 \cos x}{5} + \frac{\cos(2x)-4\sin(2x)}{17} \end{align*}
\begin{align*} 6) \hspace{.5cm} 2y'+y=e^x+3\sin(x) \end{align*}
Solution:
\begin{align*} y=C_1e^{-\frac{1}{2}x} + \frac{1}{3}e^{x} + \frac{3\sin x- 6\cos x}{5} \end{align*}
\begin{align*} 7) \hspace{.5cm} 4y''-9y=2x^4-3x^2+\cos(x+1) \end{align*}
Solution:
\begin{align*} y=C_1e^{\frac{3}{2}x} + C_2e^{-\frac{3}{2}x} - \frac{2}{9}x^4-\frac{23}{27}x^2-\frac{1}{13}\cos(x+1)-\frac{184}{243} \end{align*}
\begin{align*} 8) \hspace{.5cm} y'+y=\sin(2x+1)+\cos(5x)+1 \end{align*}
Solution:
\begin{align*} y=C_1e^{-x} + \frac{5\sin(5x)+\cos(5x)}{26} + \frac{\sin(2x+1)-2\cos(2x+1)}{5}+1 \end{align*}
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Undetermined Coefficients. In Justin Math: Calculus. https://justinmath.com/undetermined-coefficients/
Tags: |
## Thursday, December 5, 2013
### Thursday 12/5
Period 6: See Tuesdays post for homework and notes.
Period 7: We learned to multiply fractions, and started a worksheet on multiplying fractions
Period 2,3: We did inverse variation. There is a worksheet, ask me for it.
## Wednesday, December 4, 2013
### Wednesday 12/4
POW #4
A snail crawls up a 28 foot pole. Each morning it crawls up 4 feet, and each night it slipped down 3 feet. How long would it take for the snail to get to the top of the pole?
Get a printout for the worksheet on the POW page.
## Tuesday, December 3, 2013
### Tuesday 12/3
Period 2-5: We learned about direct variation ,and did a page finding direct variation equations, with some graphing on the back.
## Monday, December 2, 2013
### Monday 12/2
Homework for period 2-6: Workbook page 253 numbers 1-10. Here is a link. The user is CPMA, the password is math419.
Period 7: We rewrote mixed numbers as improper fractions, and improper fractions as mixed numbers.
## Thursday, November 21, 2013
### Thursday 11/21
Period 6: Check Tuesdays post for the worksheet and notes
Period 7: Finish the adding fractions worksheet. About half of you finished in class.
Period 2,3: Finish the parallel and perpendicular line worksheet. Be sure to have all the functions labeled with what they are.
Algebra(Ignore the Notes 5.2 - It was 4.7)
## Wednesday, November 20, 2013
### Wednseday 11/20
All classes: Do the Citizenshoip Rubric so I know what you think you deserve for your citizenship grade.
## Tuesday, November 19, 2013
### Tuesday 11/19
Periods 1-5: We learned about slope. Finish the 6 graphs on the backside. See below for ntoes and the slope worksheet. The functions to graph on the back.
f(x)= 2x - 3
f(x)= x
f(x)= 3/4x-2
f(x)=-4x+3
f(x)=1/2x
f(x)=-x - 3
## Monday, November 18, 2013
### Monday 11/18
No homework. Do the sub work.
## Friday, November 15, 2013
### Friday 11/15
We did Mangahigh today. Do the two challenges about functions I have on your homepage. Try to get the high score. There is no homework for the weekend.
Period 7: Finish simplifying fractions 2 worksheet.
## Thursday, November 14, 2013
### Thursday 11/14
All classes: No homework, there is a sub.
## Wednesday, November 13, 2013
### Wednseday 11/13
All classes: We turned in the problem of the week #3. No homework.
## Tuesday, November 12, 2013
### Tuesday 11/12
There was a sub. No homework
## Friday, November 8, 2013
### Friday - 11/8
See yesterdays post.
Period 7: Finish simplifying fractions worksheet, both sides.
## Thursday, November 7, 2013
### 11/7 Thursday
Period 6: Finish the function worksheet. Check Tuesdays post.
Period 7: Finish the equivalent fractions.
Period 2,3: Finish the graphs. The notes are below. The 10 equations are:
f(x)=2x+3
f(x)=-2x+5
f(x)=x+1
f(x)=-3x+2
f(x)=2x-7
f(x)=x
f(x)=-x+5
f(x)=x2
f(x)=-x2
f(x)=|x|
### Wednesday 11/6
POW.
While camping, Elijah and Tracy began to cook some rice. They needed to
measure 5 cups of water, but could not figure it out. They had one pot that
held 7 cups, and another that held 3 cups. Luckily Tracy is a genius
and figured it out.
How did she do it?
Check the POW page for the template.
## Tuesday, November 5, 2013
### Tuesday 11/5/13
Period 2-5: Finish 1-9 on the worksheet from today. If you were not here, see me when you get back.
### Monday 11/4
Period 2-6: We learned how to solve consecutive number problems. Do the worksheet from Thursday, except for the last two problems. They are extra credit.
Period 7: We learned about fractions. Finish placing the fractions on the number line for homework.
## Friday, November 1, 2013
### Thur-Fri 10/31 - 11/1
Periods 1-6: We took a test on Inequalities and Absolute Values. There was no homework.
Period 7: We used the worksheet from Monday, and learned how to find LCM. Use the work from Monday so you do not have to redo everything.
## Wednesday, October 30, 2013
### Wednesday 10/30
All classes: We did a notebook quiz on sections 2.2-3.5. If you missed a lot or were absent, you have to remind me later this week to take the test.
### Tuesday 10/29
Periods 2-5: We learned about absolute value equations, and we are doing #1-12 on the worksheet from class. You have to see me for the worksheet.
## Monday, October 28, 2013
### Monday 10/28
Period 2-3: Finish the challenge on Manga High.
Period 4-6: Check out last Thursdays post. We are making up some situation that can be described with a compound inequality.
Period 7: Finish the front side, Finding The GCF (Greatest Common Factor). Do the work on a separate sheet of paper.
## Thursday, October 24, 2013
### Thursday 10/24
Period 6: We took notes on conjunctions and disjunctions. You were supposed to do 1-12 and 19-22 on workbook page 235. See Tuesdays post for the links.
Period 7: You continued working on your factor and prime factorization pamphlet. If you did not turn it in today, they are due Friday first thing when you come in.
Period 2-3: We started a made-up compound inequality. Make up your own story that can be modeled with a compound inequality. Make it colorful. Due when you come back Monday.
Here was an example from class. Spell "Guard" correctly. Only inanimate violence allowed.
## Wednesday, October 23, 2013
### Wednesday 10/23
All Classes: No homework. We took a practice benchmark. I am going to a meeting on Friday where I will grade them, I think.
## Tuesday, October 22, 2013
### Tuesday 10/22
EDIT:
Period 2-5: We learned about conjunctions and disjunctions. Homework was 1-12 and 19-22 on Workbook page 235.
## Monday, October 21, 2013
### Monday 10/21
Periods 2-6: We learned about compound inequalities. HW was a matching activity. See me for the page.
Period 7: We learned about prime factorization last time, and today we started a project pamphlet where you will explain what Factors and Prime Factorization are, and give me examples of each.
## Friday, October 18, 2013
### Friday 10/18/13
Periods 2-6: We took notes, and solved these 10 equations.
Period 7: We learned about prime factorization, and did 2 pages from the worksheet. Many of you finished it in class.
## Thursday, October 17, 2013
### Thursday 10/17
Period 7: Finish shading the worksheet with the prime numbers. Prime numbers to 100are:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97
Period 2,3:
we did notes on why we flip the esign when multiplying or dividing by a negative number.
There were 10 inequalities to do in class.
1) -2(n+2) < -14
2) 2(x - 4) ≥ -6
3) 3x + 2 > -4x + 16
4) 8 - 4s > 16
5) 2(3x+7) > 28
6) 5m - 3 > -18
7) -4x - 2 ≤ 8
8) 5 - 3n ≥ -4
9) 4 - 2x ≤ 5 - x + 1
10) 4(x-2)-6x ≤ -9
### Wednesday 10/17
No homework, we did the problem of the week.
## Tuesday, October 15, 2013
### Tuesday 10/15
Homework for periods 2-5: Finish problems 1-8 from workbook practice 3-2. Below is the notes we took. We did double today.
## Monday, October 14, 2013
### Monday 10/14
Periods 2-6: finish up to problem 8 on the word problem page.
Period 7: we did factor houses, and you were supposed to finish for homework.
Ask me at school if you need either page.
## Friday, October 11, 2013
### Friday 10/11
Period 4-6: We finished the project. See these examples!
Period 7: We did manga high. Your goal is to get a medal in the Prime game. Many of you finished in class.
## Thursday, October 10, 2013
### Thursday 10/10
Period 6: we finished the 5 problems (check Tuesdays Post) and started a project. You explain one of the problems creatively to me. Period 7: We took a test, there was no homework. Period 2-3: We finished our project, there is no homework.
## Wednesday, October 9, 2013
### Wednesday 10/9
All classes, we did the citizenship rubric, and the problem of the week #2. Please see the links for the assignments
POW
Citizenship Rubric fill in
## Tuesday, October 8, 2013
### Tuesday 10/9
Period 2-5: We did 5 problems in class (which was also the homework). The project for the week is to take one of the problems and explain it back in pamphlet form. See below for the 3 we did together, and the 5 problems that were homework.
EDIT-I put this up Thursday, so if you are just getting this, you have until Tuesday. I wont be mad if its late, and it won't be marked down. It's my fault for not putting these up
## Monday, October 7, 2013
### Monday 10/7
Homework for periods 2-6: Finish the 8 problems that were on the board. See below.
Period 7: Page 209 (Practice 1-3) from the workbook, #1-20.
The 8 equations:
1) 7 - 2n = n - 14
2) 3d + 8 = 2d - 7
3) 6x = 2x - 4
4) 8z - 7 = 3z - 7 + 5z
5) -4x + 6 = -9x + 16
6) 4y - 12 = -5y + 51
7) 7x - 8 = 3x + 12
8) 5(x + 3) = 2x + 6
### Friday 10/4
Homework for periods 4-6: We did the word problems in class. Homework is the Manga high Algebra Meltdown challenge. You can access Manga High through the link at the top of this page.
Period 7: We did Distributive Property Worksheet. If you want extra credit, you can do ALL the problems from Thursdays Homework.
## Thursday, October 3, 2013
### Thursday 10/7/13
Homework for periods 6,7: do the 7 fraction problems, check Tuesdays post for the work and notes.
Period 2,3: We did the word problems on the back of Mondays homework, then went on to Manga high, your job is to get a medal in the Algebra Meltdown challenge
## Wednesday, October 2, 2013
### Wednesday 10/2
Today we took a Notebook Quiz, which goes over the warm-ups and notes we do in class. If you need to make it up, see me in the next few days. You need to remind me, because I will forget. You can redo the quiz at anytime.
## Tuesday, October 1, 2013
### Tuesday 10/1
Period 2-5: We messed with equations, and learned how to make fraction equations easier by multiplying them by some whole number to cancel out the fractions.
See below for the notes, and the 7 problems which were homework.
## Monday, September 30, 2013
### Monday 9/30
Homework for periods 2-6: Finish the first side of 2 step equations from page 34. Be sure to show both steps.
Period 7: Finish the multiplication page. Do only the negative answer problems
## Thursday, September 26, 2013
### Thursday 9/26
Period 6: no homework, we took a test.
Period 7: finish 2-1, identifying properties.
Period 2-3: it seemed confusing, so all I ask is you login to mangahigh, (use the link up in the tabs on this page) and you'll get credit
## Wednesday, September 25, 2013
### Wednesday 9/25
Today the problem of the week is due. There is no new assigned homework. Periods 6 and 7 have homework from Monday to turn in, and periods 1-5 have no homework.
## Tuesday, September 24, 2013
### Tuesday 9/24
Problem Of The Week is Due TOMORROW!!!
Periods 2-5: No homework today. We took a test, and I couldn't get the computers to log you in fast enough. We will do manga high next time I see you, and you can always get there by clicking the link at the top of this page.
## Monday, September 23, 2013
### Monday 9/23
Period 2-6: Homework was Practice 1-8, you can click the link to download it, user name is our school, in lowercase, password is "math" with my room number attached to it.
Period 7: Homework was Practice 1-6 you can click the link to download it, user name is our school, in lowercase, password is "math" with my room number attached to it.
## Friday, September 20, 2013
### Friday 9/20/13
Period 4-6: We went over like terms and distributing negatives. You can check yesterdays post for the notes.
Period 7: We went over how subtracting is just a negative sign, circling terms so we can see the negative sign, and the homework is the first two columns on the page 18 worksheet.
### Thursday 9/19
Period 6-7: You emailed me from class, which was the homework.
Period 2,3: You have to finish the distributive problems I gave you in class. Please only do #1-10.
See below for a pic of our notes.
## Wednesday, September 18, 2013
### Problem of the Week #1
Problem: In the sequence 16, 80, 48, 64, A, B, C, D , each term after the second is the mean (average) of the previous two terms. Wat is the value of D?
There will be a link to the POW template HERE .
Due Wednesday 9/25
## Tuesday, September 17, 2013
### Tuesday 9/17/13
Periods 2-5: you guys got on the net books, learned where my website was, and emailed me. Everyone in class got checked off for this. If you are absent, just email me and you will get credit next time you are in class.
## Monday, September 16, 2013
### Monday 9/16/13
Period 1-6: Today we went over the properties and did a worksheet on identifying properties. If you were absent, see me for a copy. It is due tomorrow, and we will keep track of these things in our planners from now on.
Period 7: We learned how to add integers. There are some problems that we did in class. For homework, do 1-20 on worksheet #12. If you need a copy, see me in class.
## Friday, September 13, 2013
### Friday 9/13/13
Homework for period 4-6: Finish the Rational Number identification page. Its attached below.
Period 7 - We did 14 absolute value questions. Check below for the notes and the problems.
## Wednesday, September 11, 2013
### Wednesday 9/11/13
No homework. We took notes. See below for a picture of them.
Period 6,7: Tomorrow we take the MDTP, don't worry, it doesnt count for points
Algebra Notes::
Pre-Algebra:
## Tuesday, September 10, 2013
### Tuesday 9/10/13
Periods 2-5: There is no homework. Today we did the MDTP for Geometry, which basically shows what you need to learn for the year. There is no score given for this. I will tell you your score later next week, but do not get discouraged, you do not know half of what is on the test. We will take this test again in June, and THEN you will be graded on it, but by that point you should know everything on the test.
## Monday, September 9, 2013
### Monday 9/9/13
Periods 1-6: No homework. We worked on calculating GPA.
Period 7: No homework. You kept the GPA worksheet. We will work on it again Wednesday, so do not lose it!
## Friday, September 6, 2013
### Friday 9/6/13
There is no real homework for this week. You were supposed to turn in your syllabus and tardy policy to me.
7th period: You also were supposed to turn in your brown packets for the school. Make sure either I or the front office gets these things. Have a good weekend
## Tuesday, September 3, 2013
### First Day of School
All Classes: We went over the syllabus and tardy policy. They need to be brought back by Friday.
7th period: Take-Home-Packets need to be returned to me by Friday
## Thursday, August 29, 2013
### Syllabus and Others
The new syllabus and other files for 2013/2014 have been posted. Check the links on the right.
## Thursday, May 30, 2013
### Alphabet Book Project
The final project for all classes. You have picked 3 words to define, give 2 examples of. Below you will find a link to the template. You need three of these. The word list is also below, in case you need it, but you should already know your words.
All work must be done by you. Do not cut and paste a definition from the internet. Use your own words. I know what you know. If you try defining things with heteroskedasticity or things like that, expect me to call you on it :)
ABC Template
Word List
## Friday, May 24, 2013
Pre-Algebra: Do as many Learning Upgrades as you can. The formula for scoring is as follows.
Basically the score is out of 50, and you get bonus points for getting golds or silvers.
The equation is:
{3(# of gold levels) + 2(# of silver levels) + (# of Bronze Levels)}*Levels Completed
this will be the score out of 50. You can get higher than 50 points on this assignment. That is extra credit.
### Algebra Dinner Menu Project Pages
Dinner menu project is due the day you come back after you come back from Memorial Day.
Dinner Menu Project - Rubric - Question Bank
## Thursday, May 9, 2013
### Algebra Graphing Calculator Worksheets
There are 3 assignments for the graphing calculator. One was an intro page, one was intro to using the calculator (8 pages), and the last was a series of images you had to replicate on your own graphing calculator. All three of these are due for the graphing calculator unit.
Graphing_Calculator_1
Graphing_Calculator_2
Graphing_Calculator_3
User name is CPMA, the other thing is math419
The files start with "Graphing Calculator" if you are searching for these yourself.
## Tuesday, May 7, 2013
### Week of 5/6
For all classes. Study for the CST. English and history. We will have no homework this week. Catch up on all the stuff you need to do from the last 12 weeks.
## Wednesday, May 1, 2013
### Week of 5/1/13
For the days this week: #cst
Monday: We reviewed. No homework, no nothing. We saved our brains for the CST
Tuesday, Wednesday, Thursday: Nothing. We took the CST, watched some cartoons, drew pictures. Anything but work.
Friday. Again, pictures, no homework. Save your brains, remember all that english and history for next week.
## Thursday, April 25, 2013
### Wednesday Thursday 4/25/13
Homework for periods 1,3,5,7: either do the review worksheet, (sorry I don't have an electronic copy) or do levels 50, 53 on learning upgrade.
Period 2,4:
Finish the box-and-whisker plots for the height of the teams. Bring the sets tomorrow so we can finish with weight.
## Tuesday, April 23, 2013
### 4/23/13
Homework for period 1,3,5: Do learning upgrade. level 42.
Homework for period 2,4: Do learning upgrade level 46 triangles
## Monday, April 22, 2013
### 4/22/13
Homework for periods 1,3,5,7: Do # 29, 30, 32, 33, 35 from page 393 in your workbook. Here is a link. CPMA is the user and math419 is the other thing.
Homework for period 2,4: Finish this worksheet.
## Friday, April 19, 2013
### 4/19/13
Homework for periods 1,3,5,7:
Homework for periods 2,4: Learning Upgrade levels 38, 39
## Thursday, April 18, 2013
### 4/18/13
Homework for periods 5,7: Do the problems from your workbook. Check yesterdays post for the link. It was problems 7, 9, 13, 16, 18, 20, 21, 22, 24, 26, 27
Period 4: Finish graphing the 10 functions from class using the slope and y-intercept.
## Wednesday, April 17, 2013
### 4/17/13
Homework for period 7: Finish the review of systems of equations.
Period 1,3: Simplify the fractions from class. Here is a link to the page. "CPMA" is the username, and use "math419" as the other thing..
Period 2: Graph the equations from class on your graph paper. They are listed below for your convenience.
### 4/16/13
Homework for periods 1,3,5: Do the review page of systems. Adding will solve all of them.
Period 2,4: Finish graphing the 7 functions from class. Here is a pic of the equations.
### Monday 4/15/13
Homework for periods 1,3,5,7: Do the practice quiz. We will do a real one next class period.
Period 2,4: Finish the backside of the slope worksheet, just the top 6 problems.
## Friday, April 12, 2013
### Friday 4/12/13
For periods 2,4: Finish the slope worksheet from class
Period 1,3,5,7: Finish the learning upgrade level 56. You can get a silver in it.
Algebra: There will be a quiz on radicals on Tuesday(or Wednesday). We will review on Monday.
## Thursday, April 11, 2013
### Wednesday 4/11/13
Homework for period 7: Finish 1-9 from the workbook, here is a link to the page.
Period 1,3: Graph the square root problems from the board. Be sure to list the domain and range.
Period 2: Finish graphing the equations from the board.
## Tuesday, April 9, 2013
### Tuesday 4/08/13
Homework for periods 1,3,5:
DO # 1-9 from 3-3, inequalities. Be sure to graph your answers on a number line. Here is a link to the page.
Period 2,4: Finish the function machine worksheets.
## Monday, April 8, 2013
### Monday 4/8/13
Homework for periods 1,3,5,7: Finish #1-20 from workbook page 373. Here is a link to the page
Period 2,4: Finish the SD reflection graphs.
## Thursday, March 28, 2013
### Spring Break
Hello all and happy Spring Break. Check your grades. Everything but the benchmark is in. Period 1,5 I didn't get your homework scores for the last two weeks. Missing does not mean zero. If you have stuff to turn in, take a picture of it and send it to me. That counts. Have fun. Mr. B
## Wednesday, March 20, 2013
### 3/20/13
Angry Birds Project!!!! : Period 1, no homework
Period 3: We started a project. This will be our work for the next week. See the page for it HERE
Period 7: Finish the Angry Birds Intro worksheet
Period 2: No homework, we took a test.
1) List two facts from the movie (Facts, tidbits, little pieces of knowledge)
2) Define "Terminal Velocity"
3) Write a three sentence summary of the movie
## Thursday, March 14, 2013
### 3/14/13
Period 5,7: Finish the worksheet found here. If there is no starting height, then C = 0
Period 4: Finish plotting the points for the triangles and find the perimeter and the area. Look a yesterdays post for the equations
### 3/13/13
Homework for period 7: Do the three word problems. Look at yesterdays post for them.
Period 1,3: Finish the worksheet (also three problems). If there is no starting height, assume C=0.
Period 2: Finish plotting the points for the right triangles, and graphing them. Find the perimeter and the area of each triangle.
Perimeter = side 1 + side 2 + side 3
Area = (Base x Height) / 2
## Tuesday, March 12, 2013
### 3/12/13
Homework for period 1,3,5: Solve the three problems from class using the equations
H = -16t2 + vt + c
Where v = initial velocity
c = starting height
t= time
H = H(t), the height we care about at our point of interest
Period 2,4: Finish the Pythagorean worksheet, both sides.
The three problems
1) If a rock is dropped off a bridge that is 30 feet above a
river, how long will it take for the rock to hit the water?
2) A tennis player hits a ball 1 foot above the ground with an initial velocity of 20 feet per second. How long will it take to hit the ground?
3) You drop a water balloon off a building 200 feet in the air. How long until it hits your friends head who is 6 feet tall?
## Friday, March 8, 2013
### Friday 3/8/13
Homework for period 1,3,4,5,7: Do the challenge on your Manga High page, get at least a bronze in it.
Period 2: Bring the drawing right triangles worksheet back.
Check your grades. Everything is current. I added to your grades based on the quadratic retake, or you redoing the scientific notation problems you missed. If I am missing something and did it, take a picture and email it to me so I can change your grade.
## Wednesday, March 6, 2013
### 3/6/13
Period 7: We took a test. No homework.
Period 1,3: Finish the graphs from class.
Here are the equations.(1-3 are notes)
4) x2 - 6x + 8 = y
5) x2 + 4x + 3 = y
6) x2 - 4 = y
7) x2 - 2x - 8 = y
8) x2 - 8x + 12 = y
9) x2 + 2x - 3 = y
10) x2 + 2x - 8 = y
11) x2 - 2x - 3 = y
12) x2 + 6x + 5 = y
Period 2: We finished the estimating square roots and I stamped you for it in class.
## Tuesday, March 5, 2013
### 3/5/13
Period 1,3,5: No homework, we took a test.
Period 2: You took the CST Writing test, no homework.
Period 4: You finished the estimating sqaure roots worksheet and got stamped for it.
## Monday, March 4, 2013
### 3/4/13
Homework for period 1,3,5,7: Finish the practice test
Period 2,4: Bring the square root paper back so we can do the back.
Advanced Algebra: Test next class period on solving quadratics, and some graphing function problems
## Friday, March 1, 2013
### 3/1/13
Homework for periods 1,3,5,7: Finish level 49 of Learning Upgrade.
Period 2,4: No homework, we took a test.
## Wednesday, February 27, 2013
### 2/27/13
Homework for period 7: Bring the white page back home
Period 2: Finish the scientific notation review. We have a test on Friday
Period 1,3: Finish graphing the 18 equations and labeling the axis of symmetry and vertex. Here is some graph paper
Here are the 18 equations by group
"A"
y = x2
y = 2x2
y = 3 x2
y = -2x2
y = -3x2
y = -4x2
"B"
y = x2 + 2x
y = x2 + 4x
y = x2 + x
y = x2 - 2x
y = x2 - 4x
y = x2 - 6x
"C"
y = x2 + 1
y = x2 + 3
y = x2 + 5
y = 2x2 - 3
y = 2x2 - 5
y = 2x2 - 7
## Tuesday, February 26, 2013
### 2/26/13
I was absent today. For all classes, finish BOTH sides of the worksheet and bring them back tomorrow for stamps. Also, do not forget to bring the homework from Monday.
If you need to make up the notebook quiz, any day is a good day to do it.
## Monday, February 25, 2013
### Homework for Algebra from Friday 2/22 and Monday 2/25
Homework for Friday:
Graph these 8 functions. Print out the six graphs from the right to do this.
1) x2 = y
2) -x2 = y
3) x2 - 9 = y
4) x2 + 4 = y
5) -x2 + 7 = y
6) -x2 + 1 = y
7) 2x2 + 1 = y
8) 3x2 - 10 = y
Homework for Monday:
1) f(x) = x2
2) f(x) = 2x2
3) f(x) = 1/4x2
4) f(x) = -x2
5) f(x) = -3x2 + 10
6) f(x) = -2x2 + 4
7) f(x) = -x2 + 9
8) f(x) = x2 - 9
### Pre-Algebra Notes 7.9 Scientific Notation
Scientific Notation
## Wednesday, February 20, 2013
### 2/20/13
Homework for Period 7: None, you turned in the work from last week, and I gave you time to finish it.
Period 1: Finish the 5 problems from the board. The first 3 work out real nice. The other 2, simplify the radical
Period 2: Bring back the simplifying fractions worksheet
Period 3: Get a bronze medal on the new MangaHigh site.
## Friday, February 8, 2013
### 2/8/13
For all classes:
No homework today. We finished the classwork. Check the Sub Lessons above to see what you are doing next week. Even if the sub messes something up, you are responsible for the three worksheets (And quiz Algebra) next week. Bring all three completed assignments back the following Tuesday the 19th.
## Wednesday, February 6, 2013
### 2/6/13
Homework for period 1: Finish page 333.
Period 2: Finish the workbook page to #18
Period 3: Finish page 333 (And the 4 classwork problems)
Period 7: Prepare to turn in the 10 problems with 19 answers. Work on page 333.
Sorry there are no notes posted, or the problems. Will post them tomorrow when I am back at school.
## Tuesday, February 5, 2013
### 2/5/13
Homework for period 2,4: Finish the ten problems shown below.
Period 1,3,5: If you did not finish the classwork, do it. Otherwise, do the 84 problems from page 333 in your workbook.
## Monday, February 4, 2013
### 2/4/13
Homework for period 2,4: Do the 12 problems from the overhead
Period 1,3,5,7: Finish the 5 problems from the overhead. Homework is the practice worksheet page 333. If you need a copy click the link below
Practice Page 333
## Friday, February 1, 2013
### 2/1/13
No homework, I was not there today.
## Thursday, January 31, 2013
### 1/31/13
No homework for periods 1,3,5,7. We finished the benchmark. Period 4 has the two graphs to make on the back of your classwork
## Tuesday, January 29, 2013
### 1/29/13
Period 1,2,3,4,5: No homework. We took benchmark. If you are in Adv. Algebra and did not do the 10 problems, check yesterdays post and do those problems. There are video links too if you forgot how to do them.
Bonus: If someone in Advanced Algebra wants to make a facebook group and invite people (and ME!), do it.
## Monday, January 28, 2013
### 1/28/13
Homework for periods 1,3,5,7: Factor the ten problems from class. Check down below for the pic of it and the notes. Links to videos I made about factoring polynomials:
Factoring Polynomials 1
Factoring Polynomials 2
Period 2,4: No homework.
All classes: Benchmark tomorrow and Wednesday |
# Word problem involving multiplication or division with mixed numbers Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Word problem involving multiplication or division with mixed numbers. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Brenda bought some packages of gum at a store and ate threeeighths of a package each week. How many packages would she have eaten after $\mathbf {7\frac{1}{2}}$ weeks?
### Explanation
Step 1:
Packages used in a week = $\frac{3}{8}$
Number of weeks = $7\frac{1}{2}$ = $\frac{15}{2}$
Step 2:
Number of packages of gum used in $7\frac{1}{2}$ weeks = $7\frac{1}{2} \times \frac{3}{8}$
= $\frac{15}{2} \times \frac{3}{8}$ = $\frac{45}{16}$ = $2\frac{13}{16}$
Q 2 - A food joint used $\mathbf {4\frac{1}{6}}$ pounds of potatoes during a lunch hour. If they used two-ninth as much meat, how many pounds of meat did they use?
### Explanation
Step 1:
Amount of potatoes used = $4\frac{1}{6}$ = $\frac{25}{6}$ pounds
Step 2:
Amount of meat used = $\frac{2}{9}$ of $4\frac{1}{6}$ pounds
= $\frac{2}{9} \times 4\frac{1}{6}$ pounds
= $\frac{2}{9} \times \frac{25}{6}$ = $\frac{25}{27}$ pounds
Q 3 - Ronnie stacked $\mathbf {4\frac{1}{3}}$ pieces of wood on top of one another. If each piece was two-thirds of a foot tall, how tall was his pile?
### Explanation
Step 1:
Number of pieces of wood stacked = $4\frac{1}{3}$ = $\frac{13}{3}$
Length of each wood piece = $\frac{2}{3}$ feet
Step 2:
Length of pieces of wood stacked = $4\frac{1}{3} \times \frac{2}{3}$
= $\frac{13}{3} \times \frac{2}{3}$ = $\frac{26}{9}$ = $2\frac{8}{9}$ feet
Q 4 - Every day an office uses $\mathbf {2\frac{1}{4}}$ of a box of paper. How many boxes would they have used after 7 days?
### Explanation
Step 1:
Number of boxes of paper used per day = $2\frac{1}{4}$ = $\frac{9}{4}$
Number of days = 7
Step 2:
Number of boxes used in 7 days = $7 \times 2\frac{1}{4}$
= $\frac{7}{1} \times \frac{9}{4}$ = $\frac{63}{4}$ = $15\frac{3}{4}$
Q 5 - A worker can clean $\mathbf {2\frac{1}{3}}$ windows in an hour. How many windows can he clean in three-eighths of an hour?
### Explanation
Step 1:
Number of windows cleaned in 1 hour = $2\frac{1}{3}$ = $\frac{7}{3}$
Number of hours = $\frac{3}{8}$
Step 2:
Number of windows cleaned in $\frac{3}{8}$ hour = $\frac{3}{8} \times 2\frac{1}{3}$
= $\frac{3}{8} \times \frac{7}{3}$ = $\frac{7}{8}$ window
Q 6 - A bag of walnuts was $\mathbf {5\frac{1}{2}}$ pounds. How many one-sixth of a pound servings are there in each bag?
### Explanation
Step 1:
Weight of a bag of walnuts = $5\frac{1}{2}$ = $\frac{11}{2}$ pounds
Weight of one serving = $\frac{1}{6}$ pound
Step 2:
Number of servings per bag = $5\frac{1}{2} \div \frac{1}{6}$
= $\frac{11}{2} \div \frac{1}{6}$ = $\frac{11}{2} \times \frac{6}{1}$ = $33$
Q 7 - A cook used $\mathbf {1\frac{1}{3}}$ of a bag of potatoes for a meal. If 6 people ate the meal, how much did each person eat?
### Explanation
Step 1:
Number of bags of potatoes used = $1\frac{1}{3}$ = $\frac{4}{3}$
Number of persons who ate = $6$
Step 2:
Amount of potatoes eaten by each person = $1\frac{1}{3} \div 6$
= $\frac{4}{3} \div \frac{6}{1}$= $\frac{4}{3} \times \frac{1}{6}$ = $\frac{2}{9}$ bag
Q 8 - A store had $\mathbf {4\frac{2}{3}}$ cartons of candles. How many days would it take to sell the candles if each day they sold three-eighths of a carton of candles?
### Explanation
Step 1:
Number of cartons of candles = $4\frac{2}{3}$ = $\frac{14}{3}$
Number of cartons sold per day = $\frac{3}{8}$
Step 2:
Number of days it would take to sell all candles = $4\frac{2}{3} \div \frac{3}{8}$
= $\frac{14}{3} \div \frac{3}{8}$ = $\frac{14}{3} \times \frac{8}{3}$ = $1\frac{12}{9}$
= $12\frac{4}{9}$ days
Q 9 - Jamie used $\mathbf {4\frac{1}{3}}$ cups of sugar to make lemonade. If he were to pour the lemonade into 6 glasses how much sugar would be there in each glass?
### Explanation
Step 1:
Number of cups of sugar used = $4\frac{1}{3}$ = $133$ cups
Number of glasses = $6$
Step 2:
Number of cups of sugar used per glass = $4\frac{1}{3} \div 6$
= $\frac{13}{3} \div \frac{6}{1}$ = $\frac{13}{3} \times \frac{1}{6}$
= $\frac{13}{18}$ cup
Q 10 - At the end of the day, a food joint had $\mathbf {3\frac{2}{3}}$ pounds of leftover food. If 4 employees wanted to split it, how much would each employee get?
### Explanation
Step 1:
Amount of leftover food = $3\frac{2}{3}$ = $\frac{11}{3}$ pounds
Number of employees = $4$
Step 2:
Amount of food per employee = $3\frac{2}{3} \div 4$
= $\frac{11}{3} \div \frac{4}{1}$ = $\frac{11}{3} \times \frac{1}{4}$
= $\frac{11}{12}$ pound
word_problem_involving_multiplication_or_division_with_mixed_numbers.htm |
Home » Math Theory » Algebra » Sets And Venn Diagrams
# Sets And Venn Diagrams
## Introduction
The branch of mathematical logic where we learn sets and their properties is known as set theory. The concept of set theory was initiated by German mathematician Georg Cantor (1845-1918) when he was working on “Problems on Trigonometric Series”. Set theory and its application are regarded as one of the fundamental concepts of mathematics. What are sets and how are they useful? Let us find out.
## Definition
A set is a well-defined collection of objects. Here, by collection, we mean an aggregate of objects or things while aggregate itself means a class of things. Let a be an element of a set A. It is written as a belongs to A, which in symbolic form is written as a ∈ A. If a is any other element but does not belong to the set A, it is written as a does not belong to A, which in symbolical form is written as a ∉ A.
Let us understand it through an example.
Suppose wish to create a collection of the vowels of the English alphabet. We know that the vowels are a, e, i, o and u. Hence, the collection of vowels in the English alphabet will be a, e, i, o and u.
### Some Standard Sets
In mathematics, we commonly refer to numbers as whole numbers, natural numbers etc. These numbers are standard sets of numbers based on their specific properties. Some of these standard sets are –
Natural Numbers – The set of natural numbers consists of numbers starting from 1, 2 , 3, ………. and so on. It is represented by N.
Whole Numbers – The set of whole numbers consists of numbers starting from 0, 1, 2 , 3, ……… and so on. It is represented by W.
Integers – The set of integers consists of numbers such as …….-3, -2, -1, 0, 1, 2 , 3……. and so on. It is represented by Z.
Rational Numbers – The set of rational numbers consists of numbers that can be represented in the form of $\frac{p}{q}$ where q ≠ 0. It is represented by Q.
Real Numbers – The set of integers consists of all real numbers such as …….-3, -2, -1, 0, 1, 2 , 3……. and so on. It is represented by R.
## Description of a Set
A set is often described in the following two forms –
1. Roster form
2. Set builder form
Let us discuss these one by one.
### Roster Form
In this form, a set is described by listing elements, separated by a comma and written within braces { }.
Let us understand it through an example.
Recall the set we described above for the vowels of the English alphabet. We obtained the elements of the set as a, e, i, o and u. let us name this set as A. This set in roster form will be represented as –
A = { a, e, i, o and u }
Similarly, the set of even natural numbers can be represented as –
A = { 2, 4, 6, 8, …… }
### Set Builder Form
In the set builder form a set is described by a characteristic property P ( x ) of its elements x. In such a case the set is described by { x : P ( x ) holds } or { x | P ( x ) holds }, which is read as “ the set of all x such that P ( x ) holds. The symbol “ | “ is read as “ such that “.
Let us understand it through an example.
Recall the set A = { a, e, i, o and u } which consisted of vowels of the English alphabet. This set in the set builder form will be written as –
A = { x : x is a vowel in the English alphabet }
Similarly, the set of even natural numbers can be represented as –
A = { x is a natural number and x = 2n for n N }
## Typesof Sets
Let us now learn about different types of sets depending upon the elements they contain.
Empty set – A set is said to be an empty or null or void set if it has no element. Such a set is denoted by ∅. In roster form, ∅ is denoted as { }.
For example, if we wish to create a set consisting of all natural numbers less than 1. We know that there does not exist any such natural number which is less than 1. Hence, this set will be represented as A = { x : x ∈ N and x < 1 } = ∅ or { }
Finite Set – A set is called a finite set if it is either void or it has a fixed number of elements. For example, the set of the first five natural numbers which will be given by A = { 1, 2, 3, 4, 5 } is a finite set as it has 5 fixed elements.
Infinite Set – A set is called an infinite set if it does not have a fixed number of elements. For example, the set of natural numbers which will be given by A = { 1, 2, 3, ……….. } is infinite as it has an infinite number of elements.
Equal Sets – Two finite sets are said to be equal if they have the same elements. For example, the set A = { 1, 2 , 3, 4 } and the set B = { 1, 2 , 3, 4 } are equal.
Subsets – Let A and B be two sets. If every element of A is an element of B, then A is called the subset of B. If A is a subset of B, we write it as A ⊆ B. Thus, if A ⊆ B, iff
a A ⇒ a ∈ B
For example,
{ 1 } ⊂ { 1, 2, 3 } but { 1, 4 } ⊄ { 1 , 2, 3 }
Universal Set – In any discussion in set theory, there always happens to be a set that contains all sets under consideration, i.e. it is a super set of each of the given sets. Such a set is called the universal set and is denoted by U. thus a set that contains all sets in a given context is called the universal set.
## Venn Diagrams
Swiss mathematician, Euler, was the first to come up with a pictorial representation of sets. Later on British mathematician John Venn ( 1834 – 1883) brought this idea to practice. Named after these two mathematicians, we have Venn-Euler diagrams or simply Venn diagrams. Venn diagrams are the pictorial representation of the sets. In Venn diagrams, the universal set, U is represented by points within a rectangle. The subsets of the universal set are represented by points in circles within a rectangle.
Let us understand it through an example.
### Subsets as Intersecting Circles
Let the universal set be the first 10 natural numbers.
Hence, let U = { 1, 2, 3, 4, 5, 6, , 8, 9, 10 }
Suppose we have two sets A and B where
A = { 1, 2, 3, 4, 5 } and B = { 1 , 3 , 5 , 7 , 9}
We can see that the two sets A and B have some elements in common. Therefore, they will be represented as intersecting circles. The common elements shall be placed in the common region of the two circles. Also, the elements that belong to the universal set but are not a part of any subset shall be listed separately.
### Subsets as Disjoint Circles
Again, let the universal set U = { 1, 2, 3, 4, 5, 6, , 8, 9, 10 }
Suppose we have two sets A and B where
A = { 1, 2, 3, 4, 5 } and B = { 6, 7, 8, 9, 10}
We can see that the two sets A and B have no element in common. Therefore, they will be represented as disjoint circles. The Venn diagram for this data will be represented as –
### Subsets as one circle within the other
Again, let the universal set U = { 1, 2, 3, 4, 5, 6, , 8, 9, 10 }
Suppose we have two sets A and B where
A = { 1, 2, 3, 4, 5 } and B = { 3, 4 }
We can see that all the elements of the set B are contained in the set A as well. This means that B ⊂ A. Since all the elements of the set are present in the set A, therefore, the set B will be shown inside the A with the common elements represented inside B. The other elements of A that are not present in B will be represented in the region that is not common to A and B. The Venn diagram for this data will be represented as –
## Operations on Sets
Let us now discuss various operations on sets using Venn diagrams –
### Union of Sets
Let A and B be two sets. The union of A and B is the set of all those elements which belong either to A or to b or to both A and B. The notation A B, read as A union B, is used to denote the union of two sets A and B.
Thus A U B = { x : x ∈ A or x ∈ B }
Let us now learn how to represent the union of two sets through a Venn diagram. The shaded portion in yellow shown in the below Venn diagram represents the union of two sets A and B.
Let us understand it through an example.
Suppose we have two sets A = { 1, 2, 3, } and B = { 1, 3, 5, 7 }
What will be A U B ? Let us find out.
A U B will be given by the union of all elements belonging either to A or to B. hence,
A U B = { 1 , 2, 3, 5, 7 }
### Intersection of Sets
Let A and B be two sets. The intersection of A and B is the set of all those elements which belong to both A and B. The notation A ∩ B, read as A intersection B, is used to denote the intersection of two sets A and B.
Thus A ∩ B = { x : x ∈ A and x ∈ B }
Let us now learn how to represent the intersection of two sets through a Venn diagram. The shaded portion in yellow shown in the below Venn diagram represents the intersection of two sets A and B.
Let us understand it through an example.
Suppose we have two sets A = { 1, 2, 3, } and B = { 1, 3, 5, 7 }
What will be A ∩ B? Let us find out.
A ∩ B will be given by finding the common elements that belong both to A and B. Hence,
A ∩ B = { 3 }
### Complement of a Set
Before we learn about the representation of complement of a set through the Venn diagram, let us learn what we mean by a complement of a set? Let U be the universal set and let A be a set such that A ⊂ U. then the complement of a with respect to U is denoted by A ‘ or A c or U – A and is defined as the set of all those elements of U which are not in A.
Let us understand it through an example.
Suppose we have universal set U = { 1, 2, 3, 4, 5, 6, , 8, 9, 10 }
Let A be a subset of U such that A = { 1, 3, 5, 7, 9 }. Now the complement of A will be given by A ‘ = { 2, 4, 6, 8, 10 } which means all those elements that are present in the universals et but are not present in A.
Now, let us learn how to represent the complement of a set using Venn diagram. The shaded portion in yellow shown in the below Venn diagram represents the complement of a set A.
## Laws of Algebra of Sets
Let us now learn about some fundamental laws of the algebra of sets.
1. For any set A, the intersection of a set A with itself results in the same set, i.e. A U A = A
2. For any set A, the union of a set A with empty set results in the same set, i.e. A U ∅ = A
3. For any set A, the intersection of a set A with the universal set results in the same set, i.e. A ∩ U = A
4. For any two sets A and B, the union of two sets A and B satisfies the commutative property, i.e. A U B = B U A
5. For any two sets A and B, the intersection of two sets A and B satisfies the commutative property, i.e. A ∩ B = B ∩ A
6. For any three sets A, B and C, the union of three sets A, B and C satisfies the associative property, i.e. ( A U B ) U C = ( A U ( B U C )
7. For any three sets A, B and C, the intersection of three sets A, B and C satisfies the associative property, i.e. ( A ∩ B ) ∩ C = ( A ∩ ( B ∩ C )
8. For any three sets A, B and C, the union and intersection of three sets A, B and C satisfies the distributive property, i.e. ( A U ( B ∩ C ) = ( A U B ) ∩ ( A ∪ C )
## Key Facts and Summary
1. A set is a well-defined collection of objects.
2. Let a be an element of a set A. It is written as a belongs to A, which in symbolic form is written as a ∈ A.
3. In roster form, a set is described by listing elements, separated by a comma and written within braces { }.
4. In set builder form a set is described by a characteristic property P ( x ) of its elements x.
5. A set is said to be an empty or null or void set if it has no element.
6. A set is called a finite set if it is either void or it has fixed number of elements.
7. A set is called an infinite set if it does not have a fixed number of elements.
8. The union of A and B is the set of all those elements which belong either to A or to b or to both A and B. The notation A U B, read as A union B, is used to denote the union of two sets A and B.
9. The intersection of A and B is the set of all those elements which belong to both A and B. The notation A ∩ B, read as A intersection B, is used to denote the intersection of two sets A and B.
10. Let U be the universal set and let A be a set such that A ⊂ U. then the complement of a with respect to U is denoted by A ‘ or A c or U – A and is defined as the set of all those elements of U which are not in A. |
# How do you simplify (√2+√5) /( √2-√5)?
Mar 16, 2018
The simplified expression is $- \frac{7 + 2 \sqrt{10}}{3}$.
#### Explanation:
$\textcolor{w h i t e}{=} \frac{\sqrt{2} + \sqrt{5}}{\sqrt{2} - \sqrt{5}}$
$= \frac{\left(\sqrt{2} + \sqrt{5}\right)}{\left(\sqrt{2} - \sqrt{5}\right)} \textcolor{red}{\cdot \frac{\left(\sqrt{2} + \sqrt{5}\right)}{\left(\sqrt{2} + \sqrt{5}\right)}}$
$= \frac{\left(\sqrt{2} + \sqrt{5}\right) \left(\sqrt{2} + \sqrt{5}\right)}{\left(\sqrt{2} - \sqrt{5}\right) \left(\sqrt{2} + \sqrt{5}\right)}$
$= \frac{\left(\sqrt{2} + \sqrt{5}\right) \left(\sqrt{2} + \sqrt{5}\right)}{{\sqrt{2}}^{2} + \sqrt{2} \sqrt{5} - \sqrt{2} \sqrt{5} - {\sqrt{5}}^{2}}$
$= \frac{\left(\sqrt{2} + \sqrt{5}\right) \left(\sqrt{2} + \sqrt{5}\right)}{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{+ \sqrt{10} - \sqrt{10}}}} - 5}$
$= \frac{\left(\sqrt{2} + \sqrt{5}\right) \left(\sqrt{2} + \sqrt{5}\right)}{2 - 5}$
$= \frac{\left(\sqrt{2} + \sqrt{5}\right) \left(\sqrt{2} + \sqrt{5}\right)}{- 3}$
$= \frac{{\sqrt{2}}^{2} + \sqrt{2} \sqrt{5} + \sqrt{2} \sqrt{5} + {\sqrt{5}}^{2}}{- 3}$
$= \frac{2 + \sqrt{10} + \sqrt{10} + 5}{- 3}$
$= \frac{2 + 2 \sqrt{10} + 5}{- 3}$
$= \frac{7 + 2 \sqrt{10}}{- 3}$
$= - \frac{7 + 2 \sqrt{10}}{3}$
This is the answer. You can verify using a calculator: |
0
# I am a number less than 100. I have 8 factors including myself and 1. If you reverse my digits I'll only have 4 factors. I'm a neighbor of a Fibonaccci number. What am I?
Updated: 9/16/2023
Wiki User
14y ago
The Fibonacci numbers less than 100 are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, and 89. The neighbors of each of these Fibonacci numbers, including the Fibonacci numbers that are neighbors of other Fibonacci numbers, are 0, 1, 2, 3, 4, 6, 7, 9, 12, 14, 20, 22, 33, 35, 54, 56, 88, and 90. We can eliminate the prime numbers because they have only two factors, which removes 2, 3, and 7. We can also remove 0 and 1. We can also remove 14 and 35 because when their digits are reversed, they are prime numbers with only two factors. The remaining numbers are 4, 6, 9, 12, 20, 22, 33, 54, 56, 88, and 90. Now, we check how many factors each of these numbers has:
The factors of 4 are 1, 2, and 4.
The factors of 6 are 1, 2, 3, and 6.
The factors of 9 are 1, 3, and 9.
The factors of 12 are 1, 2, 3, 4, 6, and 12.
The factors of 20 are 1, 2, 4, 5, 10, and 20.
The factors of 22 are 1, 2, 11, and 22.
The factors of 33 are 1, 3, 11, and 33.
The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.
The factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56.
The factors of 88 are 1, 2, 4, 8, 11, 22, 44, and 88.
The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Three of these numbers 54, 56, and 88, have eight factors. If we reverse the digits of 88, we have the same number which has eight factors instead of four, so we can eliminate it. That leaves us with 54 and 56.
The factors of 45 are 1, 3, 5, 9, 15, and 45.
The factors of 65 are 1, 5, 13, and 65.
The only one with four factors is 65.
Wiki User
14y ago
Earn +20 pts
Q: I am a number less than 100. I have 8 factors including myself and 1. If you reverse my digits I'll only have 4 factors. I'm a neighbor of a Fibonaccci number. What am I?
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# Service for Solving Linear Programming Problems
and other interesting typical problems
Ðóññêèé
## Example ¹4. Solving a Linear Programming Problem Using a Graphical Method. Function has a Minimum Value on the Line Segment
This solution was made using the calculator presented on the site.
Problem:
Find the minimum value of the function
F = 2 x1 + 2 x2
subject to the constraints:
x1 + x2 ≥ 6 3 x1 - x2 ≥ 3 x1 - x2 ≤ 2 x2 ≤ 6 x1 ≤ 5
x1 ≥ 0 x2 ≥ 0
Solution:
Points whose coordinates satisfy all the inequalities of the constraint system are called a region of feasible solutions.
It is necessary to solve each inequality of the constraint system to find the region of feasible solutions to this problem. (see step 1 - step 5)
The last two steps are necessary to get the answer.
(see step 6 - step 7)
This is a standard solution plan. If the region of feasible solutions is a point or an empty set then the solution will be shorter.
See the plan for solving this problem in pictures
By the condition of the problem: x1 ≥ 0 x2 ≥ 0.
Now we have the region of feasible solutions shown in the picture.
Step ¹1
Let's solve 1 inequality of the system of constraints.
x1 + x2 ≥ 6
We need to plot a straight line: x1 + x2 = 6
Let x1 =0 => x2 = 6
Let x2 =0 => x1 = 6
Two points were found: (0, 6) and (6 ,0)
Now we can plot the straight line (1) through the found two points.
Let's go back to the inequality.
x1 + x2 ≥ 6
We need to transform the inequality so that only x2 is on the left side.
x2 ≥ - x1 + 6
The inequality sign is ≥
Therefore, we must consider points above the straight line (1).
Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.
Step ¹2
Let's solve 2 inequality of the system of constraints.
3 x1 - x2 ≥ 3
We need to plot a straight line: 3 x1 - x2 = 3
Let x1 =0 => - x2 = 3 => x2 = -3
Let x2 =0 => 3 x1 = 3 => x1 = 1
Two points were found: (0, -3) and (1 ,0)
Now we can plot the straight line (2) through the found two points.
Let's go back to the inequality.
3 x1 - x2 ≥ 3
We need to transform the inequality so that only x2 is on the left side.
- x2 ≥ - 3 x1 + 3
x2 ≤ 3 x1 - 3
The inequality sign is ≤
Therefore, we must consider points below the straight line (2).
Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.
Step ¹3
Let's solve 3 inequality of the system of constraints.
x1 - x2 ≤ 2
We need to plot a straight line: x1 - x2 = 2
Let x1 =0 => - x2 = 2 => x2 = -2
Let x2 =0 => x1 = 2
Two points were found: (0, -2) and (2 ,0)
Now we can plot the straight line (3) through the found two points.
Let's go back to the inequality.
x1 - x2 ≤ 2
We need to transform the inequality so that only x2 is on the left side.
- x2 ≤ - x1 + 2
x2 ≥ x1 - 2
The inequality sign is ≥
Therefore, we must consider points above the straight line (3).
Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.
Step ¹4
Let's solve 4 inequality of the system of constraints.
x2 ≤ 6
We need to plot a straight line: x2 = 6
Now we can plot the straight line (4) through point (0,6) parallel to the OX1 axis
The inequality sign is ≤
Therefore, we must consider points below the straight line (4).
Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.
Step ¹5
Let's solve 5 inequality of the system of constraints.
x1 ≤ 5
We need to plot a straight line: x1 = 5
Now we can plot the straight line (5) through point (5,0) parallel to the OX2 axis
The inequality sign is ≤
Therefore, we must consider points to the left of the straight line (5).
Let's combine this result with the previous picture.
Now we have the region of feasible solutions shown in the picture.
Step ¹6
We need to plot the vector C = (2, 2), whose coordinates are the coefficients of the function F.
Step ¹7
We will move a "red" straight line perpendicular to vector C from the lower left corner to the upper right corner.
The function F has a minimum value at the point where the "red" straight line crosses the region of feasible solutions for the first time.
The function F has a maximum value at the point where the "red" straight line crosses the region of feasible solutions for the last time.
There is an assumption that the function F has a minimum value at points A and B at the same time (see picture). Let's check it.
Let's find the coordinates of point A.
Point A is on the straight line (1) and on the straight line (2) at the same time.
x1 + x2 = 6 => x1 = 9/4 3 x1 - x2 = 3 x2 = 15/4
Let's calculate the value of the function F at point A (9/4,15/4).
F (A) = 2 * 9/4 + 2 * 15/4 = 12
Let's find the coordinates of point B.
Point B is on the straight line (1) and on the straight line (3) at the same time.
x1 + x2 = 6 => x1 = 4 x1 - x2 = 2 x2 = 2
Let's calculate the value of the function F at point B (4,2).
F (B) = 2 * 4 + 2 * 2 = 12
F(A) = F(B).
Then we can conclude that the function F has a minimum value at any point on the line segment AB.
Result:
## F min = 12
Comment: changing the parameter t we can get the coordinates of any point of the line segment AB. |
$\newcommand{\dollar}{\} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## Section5.1Antiderivatives from Formulas
###### Motivating Questions
• What is an antiderivative of a function?
• If we know the equation of a function, can we find the equation of an antiderivative?
• What information do we need in order to find a single antiderivative of a function?
Up to this point, we have been focused on finding the derivative of a function. We will now examine how to find the function whose derivative is the function we are given. Our goal is the following: given a function $f(x)\text{,}$ can we find a function $F(x)$ such that $\displaystyle \frac{dF}{dx}=f(x)\text{?}$
###### The Antiderivative of a Function
An antiderivative of a function $f(x)$ is any function $F(x)$ such that
\begin{equation*} \frac{dF}{dx}=f(x). \end{equation*}
That is, $f(x)$ is the derivative of $F(x)\text{.}$
###### Example5.1
Consider the function $f(x)=2x\text{,}$ can we think of a possible antiderivative?
Solution
A possible antiderivative of this is $F(x)=x^2\text{,}$ since
\begin{equation*} F'(x)=2x=f(x). \end{equation*}
However, the antiderivative is not unique, and there are infinitely many antiderivatives of $f \text{.}$ Another possible antiderivative could be $F(x)=x^2+4$ since again $F'(x)=2x=f(x)\text{.}$ Note that any constant can be added to $x^2$ since the derivative of a constant is 0! Thus the most general antiderivative is
\begin{equation*} F(x)=x^2+C \end{equation*}
for a general constant C.
###### The Integral of a Function
If $F(x)$ is any antiderivative of a function $f(x)\text{,}$ then the indefinite integral of $f(x)$ is
\begin{equation*} \int f(x)dx=F(x)+C, \end{equation*}
where $C$ is called the constant of integration.
The general problem of finding an antiderivative is difficult. In part, this is due to the fact that we are trying to undo the process of differentiating, and the undoing is much more difficult than the doing. For example, while it is evident that an antiderivative of $f(x) = 2x$ is $F(x) = x^2$ and that an antiderivative of $g(x) = 3x^2$ is $G(x) = x^3\text{,}$ combinations of $f$ and $g$ can be far more complicated. This warrants the following question: in general, how do we find an antiderivative of a function given by a formula? We will begin to answer that question in this chapter.
### SubsectionBasic Antiderivatives
What is involved in trying to find an antiderivative for each function? From our experience with derivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of familiar functions are more complicated. Therefore, it stands to reason that antidifferentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text.
###### Constant Rule
For any constant $k\text{:}$
\begin{equation*} \int k dx=kx+C \end{equation*}
###### Example5.2
When evaluating the integral, the variable of integration is indicated by the $dx\text{,}$ or $dy\text{,}$ or $dt$ at the end of the integral.
\begin{equation*} \int 4dt=4t+C \end{equation*}
\begin{equation*} \int \pi dy=\pi y+C \end{equation*}
\begin{equation*} \int \frac{1}{4}dz=\frac{1}{4}z+C \end{equation*}
###### Power Rule
For any $n\neq -1\text{:}$
\begin{equation*} \int x^n dx=\frac{x^{n+1}}{n+1}+C \end{equation*}
###### Example5.3
Evaluate each of the following integrals:
\begin{equation*} \int x^2dx=\frac{x^3}{3}+C \end{equation*}
\begin{equation*} \int y^{3.4} dy=\frac{y^{4.4}}{4.4}+C \end{equation*}
If the function is in the form $x^n$ but is equivalent to this form, you may have to rewrite the function as we did when taking the derivative:
\begin{equation*} \int \frac{1}{z^2}dz=\int z^{-2}dz=\frac{z^{-1}}{-1}+C=\frac{-1}{z}+C \end{equation*}
\begin{equation*} \int \sqrt{x}dx=\int x^{1/2}dx=\frac{z^{3/2}}{3/2}+C=\frac{2}{3}z^{3/2}+C \end{equation*}
###### Natural Log and Exponential Rules
\begin{equation*} \int \frac{1}{x} dx=\ln|x|+C \end{equation*}
It is important to include the absolute values in the natural log since the domain of the natural log is $x\gt0\text{.}$
Exponential Rule
\begin{equation*} \int e^{ax}dx=\frac{e^{ax}}{a}+C \end{equation*}
###### Example5.4
1. \begin{equation*} \int e^{5x}dx=\frac{e^{5x}}{5}+C \end{equation*}
2. \begin{equation*} \int e^{y/3} dy=\frac{e^{y/3}}{1/3}+C=3e^{y/3}+C \end{equation*}
Before proceeding any further it will be useful to state two properties.
###### Properties of Antiderivatives: Sums and Constant Multiples
Let $f$ and $g$ be functions that have an antiderivative and $c$ a constant. Then
\begin{equation*} \int\left(f(x) \pm g(x)\right) dx= \int f(x)dx \pm \int g(x)dx \end{equation*}
In words, the antiderivative of a sum or difference is the sum or difference of the antiderivatives. Or put another way, you can integrate term by term.
\begin{equation*} \int cf(x) dx = c\int f(x)dx \end{equation*}
In words, a multiplied constant stays.
We can now put all these properties together, along with the rules we have stated and evaluate some more difficult examples.
###### Example5.5
We will start by showing every step
\begin{align*} \int (4x^2+5x+6)dx\mathstrut \amp =\int 4x^2dx+\int 5xdx+\int 6dx \\ \mathstrut \amp =4\int x^2dx+5\int x dx+\int 6dx \\ \mathstrut \amp =4\frac{x^3}{3}+5\frac{x^2}{2}+6x+C \end{align*}
In general, we do not need to show the steps of splitting the integral into separate integrals, rather we will simply evaluate term by term as we did when finding the derivative. Similarly, we will not show the step of factoring out a constant, rather the multiplied constant will simply stay along. Unlike taking the derivative, we cannot often simplify the constant in one step.
###### Example5.6
Evaluate each of the following integrals using the rules we have developed.
1. \begin{equation*} \int (5x^3+6x-3)dx \end{equation*}
2. \begin{equation*} \int \left(4x^{2.3}+5\sqrt{x}+\frac{3}{x^2}\right)dx \end{equation*}
3. \begin{equation*} \int \left(\frac{4}{x}+4e^{8x}+6\right)dx \end{equation*}
Solution
1. \begin{equation*} \int (5x^3+6x-3)dx=\frac{5}{4}x^4+3x^2-3x+C \end{equation*}
2. First rewrite the terms to apply the power rule
\begin{equation*} \int \left(4x^{2.3}+5\sqrt{x}+\frac{3}{x^2}\right)dx=\int \left(4x^{2.3}+5x^{1/2}+3x^{-2}\right)dx \end{equation*}
Then apply the rules of integration
\begin{equation*} \int \left(4x^{2.3}+5x^{1/2}+3x^{-2}\right)dx=4\frac{x^{3.3}}{3.3}+5\frac{x^{3/2}}{3/2}-3x^{-1}+C \end{equation*}
3. \begin{equation*} \int \left(\frac{4}{x}+4e^{8x}+6\right)dx=4\ln|x|+\frac{e^{8x}}{2}+6x+C \end{equation*}
### SubsectionApplications of Integration
We will now examine how we can find a specific function $f(x)$ given its derivative $f'(x)\text{.}$ Suppose you know $f'(x)=x^2+4\text{.}$ To find $f(x) \text{,}$ we can integrate:
\begin{equation*} f(x)=\int f'(x) dx=\int (x^2+4)dx=\frac{x^3}{3}+4x+C. \end{equation*}
The issue of course is the $+C\text{.}$ We want to find a single function $f(x)\text{;}$ however, we have found many possible functions, one for each possible value of the C. In order to find a specific function, we need more information. In particular, we need to know the value of the function at a point. If this is given, this is called an initial condition.
###### Example5.7
Find the function $f(x)$ given that $f'(x)=x^2+4$ and $f(0)=3\text{.}$
First we integrate
\begin{equation*} f(x)=\int f'(x)dx=\int (x^2+4) dx=\frac{x^3}{3}+4x+C. \end{equation*}
Then we use the initial condition, $f(0)=3\text{,}$ to find the value of $C \text{:}$
\begin{equation*} f(0)=\frac{0^3}{3}+4(0)+C=3 \end{equation*}
We use this to solve for $C \text{,}$ for which we get $C=3\text{.}$ Thus we have our solution
\begin{equation*} f(x)=\frac{x^3}{3}+4x+3. \end{equation*}
From this example, note that the initial condition does not always have to be at $x=0\text{.}$ Furthermore, note that to find the value of $C$ we must solve an equation.
###### Example5.8
Find the function $f(x)$ given that $f'(x)=x^2+4$ and $f(3)=3\text{.}$
Solution
The first step of evaluating the integral would remain the same as in the previous example. The difference here is in finding the value of $C\text{.}$
First we integrate
\begin{equation*} f(x)=\int f'(x)dx=\int (x^2+4) dx=\frac{x^3}{3}+4x+C. \end{equation*}
Then we use the initial condition, $f(3) = 3 \text{,}$ to find the value of $C \text{:}$
\begin{equation*} f(3)=\frac{3^3}{3}+4(3)+C=3. \end{equation*}
Solving, we get $C=-21\text{,}$ so we now have:
\begin{equation*} f(x)=\frac{x^3}{3}+4x-21. \end{equation*}
###### Example5.9
Find the function $g(t)$ given that $g'(x)=6e^{3t}+\frac{2}{t}$ and $g(1)=5\text{.}$
Solution
First we integrate
\begin{equation*} g(t)=\int g'(t)dt=\int (6e^{3t}+\frac{2}{t}) dt=\frac{6e^{3t}}{3}+2\ln|t|+C, \end{equation*}
then we use the initial condition: $g(1)=5\text{,}$ to find the value of C:
\begin{equation*} g(1)=2e^{3(1)}+2\ln|1|+C=5 \end{equation*}
to solve for $C=5-2e^3$ and thus we have our solution
\begin{equation*} g(t)=2e^{3(t)}+2\ln|t|+5-2e^3 \end{equation*}
Now let's examine some word problems in which we must find the function given the rate of change (derivative).
###### Example5.10
The marginal cost of producing the next unit, after $x$ units of a product have already been produced, is given by
\begin{equation*} C'(x)=x^3-6x, \text{ (in dollars per unit). } \end{equation*}
Find the total cost function, $C(x)\text{,}$ given that the fixed costs are $\ 1200\text{.}$
\begin{equation*} C(x)=\frac{x^4}{4}-3x+1200 \end{equation*}
Solution
We are given the marginal cost, $C'(x) \text{,}$ so to find the cost function we must first integrate.
\begin{equation*} C(x)=\int\left(x^3-6x\right)dx=\frac{x^4}{4}+\frac{6x^2}{2}+C \end{equation*}
Then to find the value of $C$ we will use the fact that the fixed costs are $\ 1200\text{.}$ Since this is fixed, this is the same as saying $C(0)=1200\text{,}$ then we plug in $x=0$ to find the value of $C\text{.}$
\begin{equation*} C(0)=\frac{0^4}{4}+\frac{6(0)^2}{2}+C=1200 \implies C=1200 \end{equation*}
Thus we find the total cost function
\begin{equation*} C(x)=\frac{x^4}{4}-3x+1200. \end{equation*}
###### Example5.11
A company finds that the rate at which the quantity of their item that consumers demand changes with respect to the price and is given by the marginal demand function
\begin{equation*} D'(p)=-\frac{4000}{p^2}, \end{equation*}
where $p$ is the price per item, in dollars. Find the demand function if $1015$ items are demanded by consumers when the price is $\8$ per item.
\begin{equation*} D(p)=\frac{4000}{p}+515 \end{equation*}
Solution
We are given the marginal demand function, $D'(p) \text{,}$ so to find the demand function we must first integrate.
\begin{equation*} D(p)=\int\left(\frac{-4000}{p^2}\right)dp=\int\left(-4000p^{-2}\right)dp=4000p^{-1}+C=\frac{4000}{p}+C. \end{equation*}
Then to find the value of $C$ we will use the fact that $1015$ items are demanded by consumers when the price is $p=8\text{,}$ thus $D(8)=1015\text{.}$
\begin{equation*} D(8)=\frac{4000}{8}+C=1015 \implies C=1015-500=515 \end{equation*}
Thus the demand function is
\begin{equation*} D(p)=\frac{4000}{p}+515. \end{equation*}
###### Example5.12
The rate of change in Cheryl's pulse (in beats per minute per minute) $t$ minutes after she stops exercising is given by
\begin{equation*} R'(t)=-46.964e^{-0.796t}. \end{equation*}
Find $R(t)$ if Cheryl's pulse is $78$ beats per minute 2 minutes after she stopped exercising.
\begin{equation*} R(t)=59e^{-0.796t}+66 \end{equation*}
Solution
We are given the rate of change, $R'(t)\text{,}$ so to find $R(t)$ we must first integrate.
\begin{equation*} R(t)=\int\left(-46.964e^{-0.796t}\right)dt=-46.964\frac{e^{-0.796t}}{-0.796}+C=59e^{-0.796t}+C. \end{equation*}
Then to find the value of $C$ we will use the fact that Cheryl's pulse is $78$ beats per minute 2 minutes after she stopped exercising; that is, $R(2)=78\text{.}$
\begin{equation*} R(2)=59e^{-0.796(2)}+C=78 \implies C=66 \end{equation*}
So we find that
\begin{equation*} R(t)=59e^{-0.796t}+66 \end{equation*}
### SubsectionSummary
• An antiderivative of a function $f(x)$ is any function $F(x)$ such that $\frac{dF}{dx} = f(x) \text{.}$ In other words, it is a function whose derivative is $f(x) \text{.}$ If $F(x)$ is any antiderivative of a function $f(x) \text{,}$ then the indefinite integral of $f(x)$ is
\begin{equation*} \int f(x)dx = F(x) + C, \end{equation*}
where $C$ is called the constant of integration. For a function $f(x) \text{,}$ there can be many antiderivatives of $f(x) \text{.}$
• As with derivatives, there are antiderivative rules which tell us how to integrate familiar functions, such as constant functions, power functions, logs, and exponents. There are also rules on how to integrate sums and constant multiples of functions. Integrating products, quotients, and compositions of functions are a little more involved, and we will explore these in later sections. |
# Standard Deviation and Coefficient of Variation
Define standard deviation.
A Standard deviation is defined as the positive square root of the mean of the square of deviation from the arithmetic mean. It is also known as root mean square deviation. It is denoted by the Greek letter 'σ' (sigma).
What is coefficient of standard deviation? Also, write its formula.
The coefficient of standard deviation is the ratio of standard deviation (σ) to the mean (X̅).
Its formula is:
Coefficient of standard deviation = $\frac{σ}{\overline{X}}$
Define coefficient of variation. Also, write its formula.
If the coefficient of standard deviation is multiplied by 100, it is known as coefficient of variation. It is denoted by C.V.
Its formula is:
CV = $\frac{\text{standard deviation}}{\text{mean}}$ × 100%
It is purely a number and independent of unit.
What is the meaning a low coefficient of variation?
Lower the coefficient of variation greater will be the consistency or uniformity or more homogenous or more stable.
Write the difference between standard deviation and mean deviation.
Standard deviation (σ) is the positive square root of mean of the square of deviation from the arithmetic mean. It is also known as root mean square deviation.
Mean deviation (MD) is the average of the absolute values of the deviation of each item from mean, median or mode. It is also known as average deviation.
Write the difference between coefficient of standard deviation and the coefficient of variation.
Coefficient of standard deviation is the ratio of standard deviation (σ) to the mean (X̅) whereas coefficient of variation is the multiplication of the coefficient of standard deviation by 100.
Coefficient of standard deviation = $\frac{σ}{\overline{X}}$
Coefficient of variation = $\frac{σ}{\overline{X}}$ × 100%
In a continuous series, ∑𝑓(𝑚 - X̅)2 = 484, N = 24 and X̅ = 25 find the standard deviation and its coefficient.
Here,
∑𝑓(𝑚 - X̅)2 = 484
N = 24
X̅ = 25
Standard deviation (σ) = ?
Coefficient of σ = ?
Now, we have,
σ = $\sqrt{\frac{∑𝑓(𝑚 - \overline{X})^2}{N}}$ = $\sqrt{\frac{484}{24}}$ = 4.49
Coefficient of σ = $\frac{σ}{\overline{X}}$ = $\frac{4.49}{25}$ = 0.179
In a continuous series ∑fd = 0, ∑fd2 = 848, N = 100, assumed mean (A) = 12 then find the standard deviation and its coefficient.
Here,
∑fd = 0
∑fd2 = 848
N = 100
Assumed mean (A) = X̅ = 12
Standard deviation (σ) = ?
Coefficient of σ = ?
Now, we have,
σ = $\sqrt{\frac{\sum fd^2}{N} - (\frac{\sum fd}{N})^2}$ = $\sqrt{\frac{848}{100} - 0}$ = 2.91
Coefficient of σ = $\frac{σ}{\overline{X}}$ = $\frac{2.91}{12}$ = 0.24
If ∑fd'2 = 125, ∑fd' = -4, N = 20, C = 10, find SD (σ).
Here,
∑fd'2 = 125
∑fd' = -4
N = 20
C = 10
Standard deviation (σ) = ?
Now, we have, σ = $\sqrt{\frac{\sum fd'^2}{N} - (\frac{\sum fd'}{N})^2}$ × C or, σ = $\sqrt{\frac{125}{20} - \frac{16}{400}}$ × 10
∴ σ = 24.92 |
How do you implicitly differentiate y^2+(y-x)^2-y/x^2-3y?
Nov 26, 2017
First expand the brackets to give:
${y}^{2} + {\left(y - x\right)}^{2} - \frac{y}{x} ^ 2 - 3 y$
$= {y}^{2} + {y}^{2} - 2 x y + {x}^{2} - \frac{y}{x} ^ 2 - 3 y$
$= 2 {y}^{2} - 2 x y + {x}^{2} - y \times {x}^{-} 2 - 3 y$
By implicit differentiation (presumably with respect to x here):
$\frac{d}{\mathrm{dx}} {y}^{a} = a {y}^{a - 1} \times \frac{\mathrm{dy}}{\mathrm{dx}}$.
Also using the product rule:
If $f \left(x\right) = u v$, then $f ' \left(x\right) = u ' v + u v '$
So the original expression differentiates to:
$4 {y}^{1} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y \times 1\right) + 2 {x}^{1} - \left(- 2 {x}^{-} 3 y + {x}^{-} 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 3 {y}^{0} \frac{\mathrm{dy}}{\mathrm{dx}}$
$= 4 y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + 2 x + \frac{2 y}{x} ^ 3 - \frac{1}{x} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}}$ |
Overview: Surds
# Overview: Surds | Quantitative Techniques for CLAT PDF Download
Table of contents Surds Rules for Surds Types of Surds You Need to Focus Conclusion
## Surds
Surd is a Latin word that means mute or deaf. In the past, Arabian mathematicians referred to rational & irrational numbers as audible & inaudible, respectively. Because surds are formed up of irrational numbers, they were given the Arabic name asamm (deaf, stupid), which was ultimately translated into Latin as surds.
Surds are the square roots of non-trivial numbers. It is impossible to convey accurately in a fraction. In another way, Surd is an irrational root of the whole number. Consider the number √2 ≈1.414213. If we keep it as a surd √2, it will be more correct.
It’s important to keep in mind that an irrational number cannot be stated as a fraction.
Surds are square root values that cannot be easily converted to whole numbers or integers in mathematics.
## Rules for Surds
Division of Surds
You can divide surds with various numbers inside the root by combining them into single root & dividing the numbers within the root as long as the indexes of roots are the same.
√a /√b = √a/ b
Example
√14 / √2 = √14/2= √7
### Multiplication of Surds
You can multiply surds with various numbers inside the root by combining them into single root & multiplying the numbers within the root as long as the indexes of roots are the same. Factors can also be used to divide a root into many roots.
√a ×√b= √a ×b
Example: √3 ×√5= √3 ×5= √15
### Multiplication of Square Root by Itself
You should get the original value if you multiply the square root of an integer by itself.
√a ×√a= √a ×a= √a²=a
Example: √5 ×√5= √5 ×5= √5²=5
### Multiplication of Surds containing Brackets
Each term in the 1st bracket should be multiplied by each term in the 2nd bracket in order to multiply brackets containing surds. You can then mix phrases that are similar.
(2+ √3)+(5+√3)=2×5+2√3+5 √3+√3²
=10+7 √3+√3=1√3+7√3
### Multiplication of Number by a Surd
The order of the elements does not matter when multiplying a number by a surd, & the result must be the number followed by surd.
x × √y=√yx=x√y
Example: 3 ×√5=√5×3=3√5
### Addition or Subtraction of Surds
The number within the roots should be the same to add or subtract surds. Outside the root, you add and subtract numbers.
a√ x+b√x=(a+b)√x
a √x-b√x=(a-b)√x
Examples: 5 √3+3√3=(5+3)√3=8√3
5 √3-3√3=(5-3)√3=2√3
## Types of Surds You Need to Focus
Simple Surds
• A Surd which has a single term is considered as a simple surd. Such as √2, √5, …
Pure Surds
• Surds that are completely irrational are considered as Pure Surds. Such as √3
Similar Surds
• The surds which have the same common Surds factor are termed as similar surds.
Binomial Surds
• A Surd which consists of 2 other surds are termed as Binomial Surds.
Compound Surds
• An expression which is the subtraction or addition of 2 or more Surds are termed as Compound Surds.
Mixed Surds
• Surds which are not complete irrational & Irrational numbers that can be written as a product of a rational and an irrational number.
## Conclusion
Surds are expressions which result in an irrational number having infinite decimals when they have a Square Root, Cube Root, or other root. They’ve been left in their original state to better portray them.
The index of roots should be the same to multiply & divide surds with various numbers inside the root.
The number within the roots have to be the same while adding or subtracting surds.
It may be important to simplify surds before subtracting or adding them.
If a square number is a factor in the number within the root of a surd, it can be simplified.
There are following types of surds which are given here;
• Simple Surds
• Pure Surds
• Similar Surds
• Binomial Surds
• Compound Surds
• Mixed Surds.
The document Overview: Surds | Quantitative Techniques for CLAT is a part of the CLAT Course Quantitative Techniques for CLAT.
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## SectionFSFour Subsets
There are four natural subsets associated with a matrix. We have met three already: the null space, the column space and the row space. In this section we will introduce a fourth, the left null space. The objective of this section is to describe one procedure that will allow us to find linearly independent sets that span each of these four sets of column vectors. Along the way, we will make a connection with the inverse of a matrix, so Theorem FS will tie together most all of this chapter (and the entire course so far).
### SubsectionLNSLeft Null Space
###### DefinitionLNS.Left Null Space.
Suppose $A$ is an $m\times n$ matrix. Then the left null space is defined as $\lns{A}=\nsp{\transpose{A}}\subseteq\complex{m}\text{.}$
The left null space will not feature prominently in the sequel, but we can explain its name and connect it to row operations. Suppose $\vect{y}\in\lns{A}\text{.}$ Then by Definition LNS, $\transpose{A}\vect{y}=\zerovector\text{.}$ We can then write
\begin{align*} \transpose{\zerovector} &=\transpose{\left(\transpose{A}\vect{y}\right)}&& \knowl{./knowl/definition-LNS.html}{\text{Definition LNS}}\\ &=\transpose{\vect{y}}\transpose{\left(\transpose{A}\right)}&& \knowl{./knowl/theorem-MMT.html}{\text{Theorem MMT}}\\ &=\transpose{\vect{y}}A&& \knowl{./knowl/theorem-TT.html}{\text{Theorem TT}}\text{.} \end{align*}
The product $\transpose{\vect{y}}A$ can be viewed as the components of $\vect{y}$ acting as the scalars in a linear combination of the rows of $A\text{.}$ And the result is a row vector, $\transpose{\zerovector}$ that is totally zeros. When we apply a sequence of row operations to a matrix, each row of the resulting matrix is some linear combination of the rows. These observations tell us that the vectors in the left null space are scalars that record a sequence of row operations that result in a row of zeros in the row-reduced version of the matrix. We will see this idea more explicitly in the course of proving Theorem FS.
We will find the left null space of
\begin{equation*} A= \begin{bmatrix} 1 & -3 & 1 \\ -2 & 1 & 1 \\ 1 & 5 & 1 \\ 9 & -4 & 0 \end{bmatrix}\text{.} \end{equation*}
We transpose $A$ and row-reduce,
\begin{equation*} \transpose{A}= \begin{bmatrix} 1 & -2 & 1 & 9 \\ -3 & 1 & 5 & -4 \\ 1 & 1 & 1 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 2 \\ 0 & \leading{1} & 0 & -3 \\ 0 & 0 & \leading{1} & 1 \end{bmatrix}\text{.} \end{equation*}
Applying Definition LNS and Theorem BNS we have
\begin{equation*} \lns{A}=\nsp{\transpose{A}}= \spn{\set{ \colvector{-2\\3\\-1\\1} }}\text{.} \end{equation*}
If you row-reduce $A$ you will discover one zero row in the reduced row-echelon form. This zero row is created by a sequence of row operations, which in total amounts to a linear combination, with scalars $a_1=-2\text{,}$ $a_2=3\text{,}$ $a_3=-1$ and $a_4=1\text{,}$ on the rows of $A$ and which results in the zero vector (check this!). So the components of the vector describing the left null space of $A$ provide a relation of linear dependence on the rows of $A\text{.}$
Similar to (right) null spaces, a left null space can be computed in Sage with the matrix method .left_kernel(). For a matrix $A\text{,}$ elements of the (right) null space are vectors $\vect{x}$ such that $A\vect{x}=\zerovector\text{.}$ If we interpret a vector placed to the left of a matrix as a 1-row matrix, then the product $\vect{x}A$ can be interpreted, according to our definition of matrix multiplication (Definition MM), as the entries of the vector $\vect{x}$ providing the scalars for a linear combination of the rows of the matrix $A\text{.}$ So you can view each vector in the left null space naturally as the entries of a matrix with a single row, $Y\text{,}$ with the property that $YA=\zeromatrix\text{.}$
Given Sage's preference for row vectors, the simpler name .kernel() is a synonym for .left_kernel(). Given your text's preference for column vectors, we will continue to rely on the .right_kernel(). Left kernels in Sage have the same options as right kernels and produce vector spaces as output in the same manner. Once created, a vector space all by itself (with no history) is neither left or right. Here is a quick repeat of Example LNS.
### SubsectionCCSComputing Column Spaces
We have three ways to build the column space of a matrix. First, we can use just the definition, Definition CSM, and express the column space as a span of the columns of the matrix. A second approach gives us the column space as the span of some of the columns of the matrix, and additionally, this set is linearly independent (Theorem BCS). Finally, we can transpose the matrix, row-reduce the transpose, kick out zero rows, and write the remaining rows as column vectors. Theorem CSRST and Theorem BRS tell us that the resulting vectors are linearly independent and their span is the column space of the original matrix.
We will now demonstrate a fourth method by way of a rather complicated example. Study this example carefully, but realize that its main purpose is to motivate a theorem that simplifies much of the apparent complexity. So other than an instructive exercise or two, the procedure we are about to describe will not be a usual approach to computing a column space.
Let us find the column space of the matrix $A$ below with a new approach.
\begin{equation*} A= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 \\ -16 & -1 & -4 & -10 & -13 \\ -6 & 1 & -3 & -6 & -6 \\ 0 & 2 & -2 & -3 & -2 \\ 3 & 0 & 1 & 2 & 3 \\ -1 & -1 & 1 & 1 & 0 \end{bmatrix}\text{.} \end{equation*}
By Theorem CSCS we know that the column vector $\vect{b}$ is in the column space of $A$ if and only if the linear system $\linearsystem{A}{\vect{b}}$ is consistent. So let us try to solve this system in full generality, using a vector of variables for the vector of constants. In other words, which vectors $\vect{b}$ lead to consistent systems? Begin by forming the augmented matrix $\augmented{A}{\vect{b}}$ with a general version of $\vect{b}$
\begin{equation*} \augmented{A}{\vect{b}}= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 & b_1 \\ -16 & -1 & -4 & -10 & -13 & b_2 \\ -6 & 1 & -3 & -6 & -6 & b_3 \\ 0 & 2 & -2 & -3 & -2 & b_4 \\ 3 & 0 & 1 & 2 & 3 & b_ 5\\ -1 & -1 & 1 & 1 & 0 & b_ 6 \end{bmatrix}\text{.} \end{equation*}
To identify solutions we will bring this matrix to reduced row-echelon form. Despite the presence of variables in the last column, there is nothing to stop us from doing this, except numerical computational routines cannot be used, and even some of the symbolic algebra routines do some unexpected maneuvers with this computation. So do it by hand. Yes, it is a bit of work. But worth it. We'll still be here when you get back. Notice along the way that the row operations are exactly the same ones you would do if you were just row-reducing the coefficient matrix alone, say in connection with a homogeneous system of equations. The column with the $b_i$ acts as a sort of bookkeeping device. There are many different possibilities for the result, depending on what order you choose to perform the row operations, but shortly we will all be on the same page. If you want to match our work right now, use row 5 to remove any occurrence of $b_1$ from the other entries of the last column, and use row 6 to remove any occurrence of $b_2$ from the last columns. We have
\begin{align*} \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 2 & b_3 - b_4 + 2 b_5 - b_6\\ 0 & \leading{1} & 0 & 0 & -3 & -2 b_3 + 3 b_4 - 3 b_5 + 3 b_6\\ 0 & 0 & \leading{1} & 0 & 1 & b_3 + b_4 + 3 b_5 + 3 b_6\\ 0 & 0 & 0 & \leading{1} & -2 & -2 b_3 + b_4 - 4 b_5\\ 0 & 0 & 0 & 0 & 0 & b_1 + 3 b_3 - b_4 + 3 b_5 + b_6\\ 0 & 0 & 0 & 0 & 0 & b_2 - 2 b_3 + b_4 + b_5 - b_6 \end{bmatrix}\text{.} \end{align*}
Our goal is to identify those vectors $\vect{b}$ which make $\linearsystem{A}{\vect{b}}$ consistent. By Theorem RCLS we know that the consistent systems are precisely those without a pivot column in the last column. Are the expressions in the last column of rows 5 and 6 equal to zero, or are they leading 1's? The answer is: maybe. It depends on $\vect{b}\text{.}$ With a nonzero value for either of these expressions, we would scale the row and produce a leading 1. So we get a consistent system, and $\vect{b}$ is in the column space, if and only if these two expressions are both simultaneously zero. In other words, members of the column space of $A$ are exactly those vectors $\vect{b}$ that satisfy
\begin{align*} b_1 + 3 b_3 - b_4 + 3 b_5 + b_6 & = 0\\ b_2 - 2 b_3 + b_4 + b_5 - b_6 & = 0\text{.} \end{align*}
Hmmm. Looks suspiciously like a homogeneous system of two equations with six variables. If you have been playing along (and we hope you have) then you may have a slightly different system, but you should have just two equations. Form the coefficient matrix and row-reduce (notice that the system above has a coefficient matrix that is already in reduced row-echelon form). We should all be together now with the same matrix
\begin{equation*} L= \begin{bmatrix} \leading{1} & 0 & 3 & -1 & 3 & 1 \\ 0 & \leading{1} & -2 & 1 & 1 & -1 \end{bmatrix}\text{.} \end{equation*}
So, $\csp{A}=\nsp{L}$ and we can apply Theorem BNS to obtain a linearly independent set to use in a span construction
\begin{equation*} \csp{A}=\nsp{L}=\spn{\set{ \colvector{-3\\2\\1\\0\\0\\0},\, \colvector{1\\-1\\0\\1\\0\\0},\, \colvector{-3\\-1\\0\\0\\1\\0},\, \colvector{-1\\1\\0\\0\\0\\1} }}\text{.} \end{equation*}
Whew! As a postscript to this central example, you may wish to convince yourself that the four vectors above really are elements of the column space. Do they create consistent systems with $A$ as coefficient matrix? Can you recognize the constant vector in your description of these solution sets?
OK, that was so much fun, let us do it again. But simpler this time. And we will all get the same results all the way through. Doing row operations by hand with variables can be a bit error prone, so let us see if we can improve the process some. Rather than row-reduce a column vector $\vect{b}$ full of variables, let us write $\vect{b}=I_6\vect{b}$ and we will row-reduce the matrix $I_6$ and when we finish row-reducing, then we will compute the matrix-vector product. You should first convince yourself that we can operate like this (this is the subject of a future homework exercise). Rather than augmenting $A$ with $\vect{b}\text{,}$ we will instead augment it with $I_6$ (does this feel familiar?).
\begin{equation*} M= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\ -16 & -1 & -4 & -10 & -13 & 0 & 1 & 0 & 0 & 0 & 0 \\ -6 & 1 & -3 & -6 & -6 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 2 & -2 & -3 & -2 & 0 & 0 & 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 2 & 3 & 0 & 0 & 0 & 0 & 1 & 0 \\ -1 & -1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*}
We want to row-reduce the left-hand side of this matrix, but we will apply the same row operations to the right-hand side as well. And once we get the left-hand side in reduced row-echelon form, we will continue on to put leading 1's in the final two rows, as well as making pivot columns that contain these two additional leading 1's. It is these additional row operations that will ensure that we all get to the same place, since the reduced row-echelon form is unique (Theorem RREFU).
\begin{equation*} N= \begin{bmatrix} 1 & 0 & 0 & 0 & 2 & 0 & 0 & 1 & -1 & 2 & -1 \\ 0 & 1 & 0 & 0 & -3 & 0 & 0 & -2 & 3 & -3 & 3 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 3 & 3 \\ 0 & 0 & 0 & 1 & -2 & 0 & 0 & -2 & 1 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 3 & -1 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 1 & -1 \end{bmatrix} \end{equation*}
We are after the final six columns of this matrix, which we will multiply by $\vect{b}\text{.}$
\begin{equation*} J= \begin{bmatrix} 0 & 0 & 1 & -1 & 2 & -1 \\ 0 & 0 & -2 & 3 & -3 & 3 \\ 0 & 0 & 1 & 1 & 3 & 3 \\ 0 & 0 & -2 & 1 & -4 & 0 \\ 1 & 0 & 3 & -1 & 3 & 1 \\ 0 & 1 & -2 & 1 & 1 & -1 \end{bmatrix} \end{equation*}
so
\begin{equation*} J\vect{b}= \begin{bmatrix} 0 & 0 & 1 & -1 & 2 & -1 \\ 0 & 0 & -2 & 3 & -3 & 3 \\ 0 & 0 & 1 & 1 & 3 & 3 \\ 0 & 0 & -2 & 1 & -4 & 0 \\ 1 & 0 & 3 & -1 & 3 & 1 \\ 0 & 1 & -2 & 1 & 1 & -1 \end{bmatrix} \colvector{b_1\\b_2\\b_3\\b_4\\b_5\\b_6} = \colvector{ b_3 - b_4 + 2 b_5 - b_6\\ -2 b_3 + 3 b_4 - 3 b_5 + 3 b_6\\ b_3 + b_4 + 3 b_5 + 3 b_6\\ -2 b_3 + b_4 - 4 b_5\\ b_1 + 3 b_3 - b_4 + 3 b_5 + b_6\\ b_2 - 2 b_3 + b_4 + b_5 - b_6\\ } \end{equation*}
So by applying the same row operations that row-reduce $A$ to the identity matrix (which we could do computationally once $I_6$ is placed alongside of $A$), we can then arrive at the result of row-reducing a column of symbols where the vector of constants usually resides. Since the row-reduced version of $A$ has two zero rows, for a consistent system we require that
\begin{align*} b_1 + 3 b_3 - b_4 + 3 b_5 + b_6 & = 0\\ b_2 - 2 b_3 + b_4 + b_5 - b_6 & = 0 \end{align*}
Now we are exactly back where we were on the first go-round. Notice that we obtain the matrix $L$ as simply the last two rows and last six columns of $N\text{.}$
This example motivates the remainder of this section, so it is worth careful study. You might attempt to mimic the second approach with the coefficient matrices of Archetype I and Archetype J. We will see shortly that the matrix $L$ contains more information about $A$ than just the column space.
Sage can very nearly reproduce the reduced row-echelon form we obtained from the augmented matrix with variables present in the final column. The first line below is a bit of advanced Sage. It creates a number system R mixing the rational numbers with the variables b1 through b6. It is not important to know the details of this construction right now. B is the reduced row-echelon form of A, as computed by Sage, where we have displayed the last column separately so it will all fit. You will notice that B is different than what we used in Example CSANS, where all the differences are in the final column.
However, we can perform row operations on the final two rows of B to bring Sage's result in line with what we used above for the final two entries of the last column, which are the most critical. Notice that since the final two rows are almost all zeros, any sequence of row operations on just these two rows will preserve the zeros (and we need only display the final column to keep track of our progress).
Notice that the last two entries of the final column now have just a single b1 and a single b2. We could continue to perform more row operations by hand, using the last two rows to progressively eliminate b1 and b2 from the other four expressions of the last column. Since the two last rows have zeros in their first five entries, only the entries in the final column would change. You will see that much of this section is about how to automate these final calculations.
### SubsectionEEFExtended Echelon Form
The final matrix that we row-reduced in Example CSANS should look familiar in most respects to the procedure we used to compute the inverse of a nonsingular matrix, Theorem CINM. We will now generalize that procedure to matrices that are not necessarily nonsingular, or even square. First a definition.
###### DefinitionEEF.Extended Echelon Form.
Suppose $A$ is an $m\times n$ matrix. Extend $A$ on its right side with the addition of an $m\times m$ identity matrix to form an $m\times (n + m)$ matrix $M\text{.}$ Use row operations to bring $M$ to reduced row-echelon form and call the result $N\text{.}$ $N$ is the extended reduced row-echelon form of $A\text{,}$ and we will standardize on names for five submatrices ($B\text{,}$ $C\text{,}$ $J\text{,}$ $K\text{,}$ $L$) of $N\text{.}$
Let $B$ denote the $m\times n$ matrix formed from the first $n$ columns of $N$ and let $J$ denote the $m\times m$ matrix formed from the last $m$ columns of $N\text{.}$ Suppose that $B$ has $r$ nonzero rows. Further partition $N$ by letting $C$ denote the $r\times n$ matrix formed from all of the nonzero rows of $B\text{.}$ Let $K$ be the $r\times m$ matrix formed from the first $r$ rows of $J\text{,}$ while $L$ will be the $(m-r)\times m$ matrix formed from the bottom $m-r$ rows of $J\text{.}$ Pictorially,
\begin{equation*} M=[A\vert I_m] \rref N=[B\vert J] = \left[\begin{array}{c|c}C&K\\\hline\zeromatrix&L\end{array}\right]\text{.} \end{equation*}
We illustrate Definition EEF with the matrix $A\text{.}$
\begin{equation*} A= \begin{bmatrix} 1 & -1 & -2 & 7 & 1 & 6 \\ -6 & 2 & -4 & -18 & -3 & -26 \\ 4 & -1 & 4 & 10 & 2 & 17 \\ 3 & -1 & 2 & 9 & 1 & 12 \end{bmatrix} \end{equation*}
Augmenting with the $4\times 4$ identity matrix,
\begin{equation*} M= \begin{bmatrix} 1 & -1 & -2 & 7 & 1 & 6 & 1 & 0 & 0 & 0 \\ -6 & 2 & -4 & -18 & -3 & -26 & 0 & 1 & 0 & 0 \\ 4 & -1 & 4 & 10 & 2 & 17 & 0 & 0 & 1 & 0 \\ 3 & -1 & 2 & 9 & 1 & 12 & 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*}
and row-reducing, we obtain
\begin{equation*} N= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 & 0 & 1 & 1 & 1\\ 0 & \leading{1} & 4 & -6 & 0 & -1 & 0 & 2 & 3 & 0 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 & 0 & -1 & 0 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 & \leading{1} & 2 & 2 & 1 \end{bmatrix}\text{.} \end{equation*}
So we then obtain
\begin{align*} B&= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 \\ 0 & \leading{1} & 4 & -6 & 0 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ C&= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 \\ 0 & \leading{1} & 4 & -6 & 0 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 \end{bmatrix}\\ J&= \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 3 & 0 \\ 0 & -1 & 0 & -2 \\ \leading{1} & 2 & 2 & 1 \end{bmatrix}\\ K&= \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 3 & 0 \\ 0 & -1 & 0 & -2 \end{bmatrix}\\ L&= \begin{bmatrix} \leading{1} & 2 & 2 & 1 \end{bmatrix}\text{.} \end{align*}
You can observe (or verify) the properties of the following theorem with this example.
$J$ is the result of applying a sequence of row operations to $I_m\text{,}$ and therefore $J$ and $I_m$ are row-equivalent. $\homosystem{I_m}$ has only the zero solution, since $I_m$ is nonsingular (Theorem NMRRI). Thus, $\homosystem{J}$ also has only the zero solution (Theorem REMES, Definition ESYS) and $J$ is therefore nonsingular (Definition NM).
To prove the second part of this conclusion, first convince yourself that row operations and the matrix-vector product are associative operations. By this we mean the following. Suppose that $F$ is an $m\times n$ matrix that is row-equivalent to the matrix $G\text{.}$ Apply to the column vector $F\vect{w}$ the same sequence of row operations that converts $F$ to $G\text{.}$ Then the result is $G\vect{w}\text{.}$ So we can do row operations on the matrix, then do a matrix-vector product, or do a matrix-vector product and then do row operations on a column vector, and the result will be the same either way. Since matrix multiplication is defined by a collection of matrix-vector products (Definition MM), the matrix product $FH$ will become $GH$ if we apply the same sequence of row operations to $FH$ that convert $F$ to $G\text{.}$ (This argument can be made more rigorous using elementary matrices from the upcoming Subsection DM.EM and the associative property of matrix multiplication established in Theorem MMA.) Now apply these observations to $A\text{.}$
Write $AI_n=I_mA$ and apply the row operations that convert $M$ to $N\text{.}$ $A$ is converted to $B\text{,}$ while $I_m$ is converted to $J\text{,}$ so we have $BI_n=JA\text{.}$ Simplifying the left side gives the desired conclusion.
For the third conclusion, we now establish the two equivalences
\begin{align*} A\vect{x}&=\vect{y} & &\iff & JA\vect{x}&=J\vect{y} & &\iff & B\vect{x}&=J\vect{y} \end{align*}
The forward direction of the first equivalence is accomplished by multiplying both sides of the matrix equality by $J\text{,}$ while the backward direction is accomplished by multiplying by the inverse of $J$ (which we know exists by Theorem NI since $J$ is nonsingular). The second equivalence is obtained simply by the substitutions given by $JA=B\text{.}$
The first $r$ rows of $N$ are in reduced row-echelon form, since any contiguous collection of rows taken from a matrix in reduced row-echelon form will form a matrix that is again in reduced row-echelon form (Exercise RREF.T12). Since the matrix $C$ is formed by removing the last $n$ entries of each these rows, the remainder is still in reduced row-echelon form. By its construction, $C$ has no zero rows. $C$ has $r$ rows and each contains a leading 1, so there are $r$ pivot columns in $C\text{.}$
The final $m-r$ rows of $N$ are in reduced row-echelon form, since any contiguous collection of rows taken from a matrix in reduced row-echelon form will form a matrix that is again in reduced row-echelon form. Since the matrix $L$ is formed by removing the first $n$ entries of each these rows, and these entries are all zero (they form the zero rows of $B$), the remainder is still in reduced row-echelon form. $L$ is the final $m-r$ rows of the nonsingular matrix $J\text{,}$ so none of these rows can be totally zero, or $J$ would not row-reduce to the identity matrix. $L$ has $m-r$ rows and each contains a leading 1, so there are $m-r$ pivot columns in $L\text{.}$
Notice that in the case where $A$ is a nonsingular matrix we know that the reduced row-echelon form of $A$ is the identity matrix (Theorem NMRRI), so $B=I_n\text{.}$ Then the second conclusion above says $JA=B=I_n\text{,}$ so $J$ is the inverse of $A\text{.}$ Thus this theorem generalizes Theorem CINM, though the result is a “left-inverse” of $A$ rather than a “right-inverse.”
The third conclusion of Theorem PEEF is the most telling. It says that $\vect{x}$ is a solution to the linear system $\linearsystem{A}{\vect{y}}$ if and only if $\vect{x}$ is a solution to the linear system $\linearsystem{B}{J\vect{y}}\text{.}$ Or said differently, if we row-reduce the augmented matrix $\augmented{A}{\vect{y}}$ we will get the augmented matrix $\augmented{B}{J\vect{y}}\text{.}$ The matrix $J$ tracks the cumulative effect of the row operations that converts $A$ to reduced row-echelon form, here effectively applying them to the vector of constants in a system of equations having $A$ as a coefficient matrix. When $A$ row-reduces to a matrix with zero rows, then $J\vect{y}$ should also have zero entries in the same rows if the system is to be consistent.
### SubsectionFSFour Subsets
With all the preliminaries in place we can state our main result for this section. In essence this result will allow us to say that we can find linearly independent sets to use in span constructions for all four subsets (null space, column space, row space, left null space) by analyzing only the extended echelon form of the matrix, and specifically, just the two submatrices $C$ and $L\text{,}$ which will be ripe for analysis since they are already in reduced row-echelon form (Theorem PEEF).
First, $\nsp{A}=\nsp{B}$ since $B$ is row-equivalent to $A$ (Theorem REMES). The zero rows of $B$ represent equations that are always true in the homogeneous system $\homosystem{B}\text{,}$ so the removal of these equations will not change the solution set. Thus, in turn, $\nsp{B}=\nsp{C}\text{.}$
Second, $\rsp{A}=\rsp{B}$ since $B$ is row-equivalent to $A$ (Theorem REMRS). The zero rows of $B$ contribute nothing to the span that is the row space of $B\text{,}$ so the removal of these rows will not change the row space. Thus, in turn, $\rsp{B}=\rsp{C}\text{.}$
Third, we prove the set equality $\csp{A}=\nsp{L}$ with Definition SE. Begin by showing that $\csp{A}\subseteq\nsp{L}\text{.}$ Choose $\vect{y}\in\csp{A}\subseteq\complex{m}\text{.}$ Then there exists a vector $\vect{x}\in\complex{n}$ such that $A\vect{x}=\vect{y}$ (Theorem CSCS). Then for $1\leq k\leq m-r\text{,}$
\begin{align*} \vectorentry{L\vect{y}}{k} &=\vectorentry{J\vect{y}}{r+k}&& L\text{ a submatrix of }J\\ &=\vectorentry{B\vect{x}}{r+k}&& \knowl{./knowl/theorem-PEEF.html}{\text{Theorem PEEF}}\\ &=\vectorentry{\zeromatrix\vect{x}}{k}&& \text{Zero matrix a submatrix of } B\\ &=\vectorentry{\zerovector}{k}&& \knowl{./knowl/theorem-MMZM.html}{\text{Theorem MMZM}}\text{.} \end{align*}
So, for all $1\leq k\leq m-r\text{,}$ $\vectorentry{L\vect{y}}{k}=\vectorentry{\zerovector}{k}\text{.}$ So by Definition CVE we have $L\vect{y}=\zerovector$ and thus $\vect{y}\in\nsp{L}\text{.}$
Now, show that $\nsp{L}\subseteq\csp{A}\text{.}$ Choose $\vect{y}\in\nsp{L}\subseteq\complex{m}\text{.}$ Form the vector $K\vect{y}\in\complex{r}\text{.}$ The linear system $\linearsystem{C}{K\vect{y}}$ is consistent since $C$ is in reduced row-echelon form and has no zero rows (Theorem PEEF). Let $\vect{x}\in\complex{n}$ denote a solution to $\linearsystem{C}{K\vect{y}}\text{.}$
Then for $1\leq j\leq r\text{,}$
\begin{align*} \vectorentry{B\vect{x}}{j} &=\vectorentry{C\vect{x}}{j}&& C\text{ a submatrix of }B\\ &=\vectorentry{K\vect{y}}{j}&& \vect{x}\text{ a solution to }\linearsystem{C}{K\vect{y}}\\ &=\vectorentry{J\vect{y}}{j}&& K\text{ a submatrix of }J\text{.} \end{align*}
And for $r+1\leq k\leq m\text{,}$
\begin{align*} \vectorentry{B\vect{x}}{k} &=\vectorentry{\zeromatrix\vect{x}}{k-r}&& \text{Zero matrix a submatrix of }B\\ &=\vectorentry{\zerovector}{k-r}&& \knowl{./knowl/theorem-MMZM.html}{\text{Theorem MMZM}}\\ &=\vectorentry{L\vect{y}}{k-r}&& \vect{y}\in\nsp{L}\\ &=\vectorentry{J\vect{y}}{k}&& L\text{ a submatrix of }J\text{.} \end{align*}
So for all $1\leq i\leq m\text{,}$ $\vectorentry{B\vect{x}}{i}=\vectorentry{J\vect{y}}{i}$ and by Definition CVE we have $B\vect{x}=J\vect{y}\text{.}$ From Theorem PEEF we know then that $A\vect{x}=\vect{y}\text{,}$ and therefore $\vect{y}\in\csp{A}$ (Theorem CSCS). By Definition SE we now have $\csp{A}=\nsp{L}\text{.}$
Fourth, we prove the set equality $\lns{A}=\rsp{L}$ with Definition SE. Begin by showing that $\rsp{L}\subseteq\lns{A}\text{.}$ Choose $\vect{y}\in\rsp{L}\subseteq\complex{m}\text{.}$ Then there exists a vector $\vect{w}\in\complex{m-r}$ such that $\vect{y}=\transpose{L}\vect{w}$ (Definition RSM, Theorem CSCS). Then for $1\leq i\leq n\text{,}$
\begin{align*} \vectorentry{\transpose{A}\vect{y}}{i} &=\sum_{k=1}^{m}\matrixentry{\transpose{A}}{ik}\vectorentry{\vect{y}}{k}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{m}\matrixentry{\transpose{A}}{ik}\vectorentry{\transpose{L}\vect{w}}{k}&& \text{Definition of }\vect{w}\\ &=\sum_{k=1}^{m}\matrixentry{\transpose{A}}{ik}\sum_{\ell=1}^{m-r}\matrixentry{\transpose{L}}{k\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{m}\sum_{\ell=1}^{m-r}\matrixentry{\transpose{A}}{ik}\matrixentry{\transpose{L}}{k\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/property-DCN.html}{\text{Property DCN}}\\ &=\sum_{\ell=1}^{m-r}\sum_{k=1}^{m}\matrixentry{\transpose{A}}{ik}\matrixentry{\transpose{L}}{k\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/property-CACN.html}{\text{Property CACN}}\\ &=\sum_{\ell=1}^{m-r}\left(\sum_{k=1}^{m}\matrixentry{\transpose{A}}{ik}\matrixentry{\transpose{L}}{k\ell}\right)\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/property-DCN.html}{\text{Property DCN}}\\ &=\sum_{\ell=1}^{m-r}\left(\sum_{k=1}^{m}\matrixentry{\transpose{A}}{ik}\matrixentry{\transpose{J}}{k,r+\ell}\right)\vectorentry{\vect{w}}{\ell}&& L\text{ a submatrix of }J\\ &=\sum_{\ell=1}^{m-r}\matrixentry{\transpose{A}\transpose{J}}{i,r+\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{\ell=1}^{m-r}\matrixentry{\transpose{\left(JA\right)}}{i,r+\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/theorem-MMT.html}{\text{Theorem MMT}}\\ &=\sum_{\ell=1}^{m-r}\matrixentry{\transpose{B}}{i,r+\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/theorem-PEEF.html}{\text{Theorem PEEF}}\\ &=\sum_{\ell=1}^{m-r}0\vectorentry{\vect{w}}{\ell}&& \text{Zero rows in } B\\ &=0&& \knowl{./knowl/property-ZCN.html}{\text{Property ZCN}}\\ &=\vectorentry{\zerovector}{i}&& \knowl{./knowl/definition-ZCV.html}{\text{Definition ZCV}}\text{.} \end{align*}
Since $\vectorentry{\transpose{A}\vect{y}}{i}=\vectorentry{\zerovector}{i}$ for $1\leq i\leq n\text{,}$ Definition CVE implies that $\transpose{A}\vect{y}=\zerovector\text{.}$ This means that $\vect{y}\in\nsp{\transpose{A}}\text{.}$
Now, show that $\lns{A}\subseteq\rsp{L}\text{.}$ Choose $\vect{y}\in\lns{A}\subseteq\complex{m}\text{.}$ The matrix $J$ is nonsingular (Theorem PEEF), so $\transpose{J}$ is also nonsingular (Theorem MIT) and therefore the linear system $\linearsystem{\transpose{J}}{\vect{y}}$ has a unique solution. Denote this solution as $\vect{x}\in\complex{m}\text{.}$ We will need to work with two “halves” of $\vect{x}\text{,}$ which we will denote as $\vect{z}$ and $\vect{w}$ with formal definitions given by
\begin{align*} \vectorentry{z}{j}&=\vectorentry{x}{i} & &1\leq j\leq r,&&& \vectorentry{w}{k}&=\vectorentry{x}{r+k} & &1\leq k\leq m-r\text{.} \end{align*}
Now, for $1\leq j\leq r\text{,}$
\begin{align*} \vectorentry{\transpose{C}\vect{z}}{j} &=\sum_{k=1}^{r}\matrixentry{\transpose{C}}{jk}\vectorentry{\vect{z}}{k}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{C}}{jk}\vectorentry{\vect{z}}{k}+ \sum_{\ell=1}^{m-r}\matrixentry{\zeromatrix}{j\ell}\vectorentry{\vect{w}}{\ell}&& \knowl{./knowl/definition-ZM.html}{\text{Definition ZM}}\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{B}}{jk}\vectorentry{\vect{z}}{k}+ \sum_{\ell=1}^{m-r}\matrixentry{\transpose{B}}{j,r+\ell}\vectorentry{\vect{w}}{\ell}&& C,\,\zeromatrix\text{ submatrices of }B\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{B}}{jk}\vectorentry{\vect{x}}{k}+ \sum_{\ell=1}^{m-r}\matrixentry{\transpose{B}}{j,r+\ell}\vectorentry{\vect{x}}{r+\ell}&& \text{Definitions of } \vect{z}\, \vect{w}\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{B}}{jk}\vectorentry{\vect{x}}{k}+ \sum_{k=r+1}^{m}\matrixentry{\transpose{B}}{jk}\vectorentry{\vect{x}}{k}&& \text{Re-index second sum}\\ &=\sum_{k=1}^{m}\matrixentry{\transpose{B}}{jk}\vectorentry{\vect{x}}{k}&& \text{Combine sums}\\ &=\sum_{k=1}^{m}\matrixentry{\transpose{\left(JA\right)}}{jk}\vectorentry{\vect{x}}{k}&& \knowl{./knowl/theorem-PEEF.html}{\text{Theorem PEEF}}\\ &=\sum_{k=1}^{m}\matrixentry{\transpose{A}\transpose{J}}{jk}\vectorentry{\vect{x}}{k}&& \knowl{./knowl/theorem-MMT.html}{\text{Theorem MMT}}\\ &=\sum_{k=1}^{m}\sum_{\ell=1}^{m}\matrixentry{\transpose{A}}{j\ell}\matrixentry{\transpose{J}}{\ell k}\vectorentry{\vect{x}}{k}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{\ell=1}^{m}\sum_{k=1}^{m}\matrixentry{\transpose{A}}{j\ell}\matrixentry{\transpose{J}}{\ell k}\vectorentry{\vect{x}}{k}&& \knowl{./knowl/property-CACN.html}{\text{Property CACN}}\\ &=\sum_{\ell=1}^{m}\matrixentry{\transpose{A}}{j\ell}\left(\sum_{k=1}^{m}\matrixentry{\transpose{J}}{\ell k}\vectorentry{\vect{x}}{k}\right)&& \knowl{./knowl/property-DCN.html}{\text{Property DCN}}\\ &=\sum_{\ell=1}^{m}\matrixentry{\transpose{A}}{j\ell}\vectorentry{\transpose{J}\vect{x}}{\ell}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{\ell=1}^{m}\matrixentry{\transpose{A}}{j\ell}\vectorentry{\vect{y}}{\ell}&&\text{Definition of }\vect{x}\\ &=\vectorentry{\transpose{A}\vect{y}}{j}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\vectorentry{\zerovector}{j}&& \vect{y}\in\lns{A}\text{.} \end{align*}
So, by Definition CVE, $\transpose{C}\vect{z}=\zerovector$ and the vector $\vect{z}$ gives us a linear combination of the columns of $\transpose{C}$ that equals the zero vector. In other words, $\vect{z}$ gives a relation of linear dependence on the the rows of $C\text{.}$ However, the rows of $C$ are a linearly independent set by Theorem BRS. According to Definition LICV we must conclude that the entries of $\vect{z}$ are all zero, i.e. $\vect{z}=\zerovector\text{.}$
Now, for $1\leq i\leq m\text{,}$ we have
\begin{align*} \vectorentry{\vect{y}}{i} &=\vectorentry{\transpose{J}\vect{x}}{i}&& \text{Definition of }\vect{x}\\ &=\sum_{k=1}^{m}\matrixentry{\transpose{J}}{ik}\vectorentry{\vect{x}}{k}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{J}}{ik}\vectorentry{\vect{x}}{k}+ \sum_{k=r+1}^{m}\matrixentry{\transpose{J}}{ik}\vectorentry{\vect{x}}{k}&& \text{Break apart sum}\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{J}}{ik}\vectorentry{\vect{z}}{k}+ \sum_{k=r+1}^{m}\matrixentry{\transpose{J}}{ik}\vectorentry{\vect{w}}{k-r}&& \text{Definitions of }\vect{z},\,\vect{w}\\ &=\sum_{k=1}^{r}\matrixentry{\transpose{J}}{ik}0+ \sum_{\ell=1}^{m-r}\matrixentry{\transpose{J}}{i,r+\ell}\vectorentry{\vect{w}}{\ell}&& \vect{z}=\zerovector,\text{ re-index}\\ &=0+\sum_{\ell=1}^{m-r}\matrixentry{\transpose{L}}{i,\ell}\vectorentry{\vect{w}}{\ell}&& L\text{ a submatrix of }J\\ &=\vectorentry{\transpose{L}\vect{w}}{i}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\text{.} \end{align*}
So by Definition CVE, $\vect{y}=\transpose{L}\vect{w}\text{.}$ The existence of $\vect{w}$ implies that $\vect{y}\in\rsp{L}\text{,}$ and therefore $\lns{A}\subseteq\rsp{L}\text{.}$ So by Definition SE we have $\lns{A}=\rsp{L}\text{.}$
The first two conclusions of this theorem are nearly trivial. But they set up a pattern of results for $C$ that is reflected in the latter two conclusions about $L\text{.}$ In total, they tell us that we can compute all four subsets just by finding null spaces and row spaces. This theorem does not tell us exactly how to compute these subsets, but instead simply expresses them as null spaces and row spaces of matrices in reduced row-echelon form without any zero rows ($C$ and $L$). A linearly independent set that spans the null space of a matrix in reduced row-echelon form can be found easily with Theorem BNS. It is an even easier matter to find a linearly independent set that spans the row space of a matrix in reduced row-echelon form with Theorem BRS, especially when there are no zero rows present. So an application of Theorem FS is typically followed by two applications each of Theorem BNS and Theorem BRS.
The situation when $r=m$ deserves comment, since now the matrix $L$ has no rows. What is $\csp{A}$ when we try to apply Theorem FS and encounter $\nsp{L}\text{?}$ One interpretation of this situation is that $L$ is the coefficient matrix of a homogeneous system that has no equations. How hard is it to find a solution vector to this system? Some thought will convince you that any proposed vector will qualify as a solution, since it makes all of the equations true. So every possible vector is in the null space of $L$ and therefore $\csp{A}=\nsp{L}=\complex{m}\text{.}$ OK, perhaps this sounds like some twisted argument from Alice in Wonderland. Let us try another argument that might solidly convince you of this logic.
If $r=m\text{,}$ when we row-reduce the augmented matrix of $\linearsystem{A}{\vect{b}}$ the result will have no zero rows, and the first $n$ columns will all be pivot columns, leaving none for the final column, so by Theorem RCLS the system will be consistent. By Theorem CSCS, $\vect{b}\in\csp{A}\text{.}$ Since $\vect{b}$ was arbitrary, every possible vector is in the column space of $A\text{,}$ so we again have $\csp{A}=\complex{m}\text{.}$ The situation when a matrix has $r=m$ is known by the term full rank, and in the case of a square matrix coincides with nonsingularity (see Exercise FS.M50).
The properties of the matrix $L$ described by this theorem can be explained informally as follows. A column vector $\vect{y}\in\complex{m}$ is in the column space of $A$ if the linear system $\linearsystem{A}{\vect{y}}$ is consistent (Theorem CSCS). By Theorem RCLS, the reduced row-echelon form of the augmented matrix $\augmented{A}{\vect{y}}$ of a consistent system will have zeros in the bottom $m-r$ locations of the last column. By Theorem PEEF this final column is the vector $J\vect{y}$ and so should then have zeros in the final $m-r$ locations. But since $L$ comprises the final $m-r$ rows of $J\text{,}$ this condition is expressed by saying $\vect{y}\in\nsp{L}\text{.}$
Additionally, the rows of $J$ are the scalars in linear combinations of the rows of $A$ that create the rows of $B\text{.}$ That is, the rows of $J$ record the net effect of the sequence of row operations that takes $A$ to its reduced row-echelon form, $B\text{.}$ This can be seen in the equation $JA=B$ (Theorem PEEF). As such, the rows of $L$ are scalars for linear combinations of the rows of $A$ that yield zero rows. But such linear combinations are precisely the elements of the left null space. So any element of the row space of $L$ is also an element of the left null space of $A\text{.}$
We will now illustrate Theorem FS with a few examples.
In Example SEEF we found the five relevant submatrices of the extended echelon form for the matrix
\begin{equation*} A= \begin{bmatrix} 1 & -1 & -2 & 7 & 1 & 6 \\ -6 & 2 & -4 & -18 & -3 & -26 \\ 4 & -1 & 4 & 10 & 2 & 17 \\ 3 & -1 & 2 & 9 & 1 & 12 \end{bmatrix}\text{.} \end{equation*}
To apply Theorem FS we only need $C$ and $L\text{,}$
\begin{align*} C&= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 \\ 0 & \leading{1} & 4 & -6 & 0 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 2 & 2 & 1 \end{bmatrix}\text{.} \end{align*}
Then we use Theorem FS to obtain
\begin{align*} \nsp{A}&=\nsp{C}= \spn{\set{ \colvector{-2\\-4\\1\\0\\0\\0},\, \colvector{-1\\6\\0\\1\\0\\0},\, \colvector{-3\\1\\0\\0\\-2\\1} }}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}}\\ \rsp{A}&=\rsp{C}= \spn{\set{ \colvector{1\\0\\2\\1\\0\\3 },\, \colvector{0\\1\\4\\-6\\0\\-1},\, \colvector{0\\0\\0\\0\\1\\2} }}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}}\\ \csp{A}&=\nsp{L}= \spn{\set{ \colvector{-2\\1\\0\\0},\, \colvector{-2\\0\\1\\0},\, \colvector{-1\\0\\0\\1} }}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}}\\ \lns{A}&=\rsp{L}= \spn{\set{ \colvector{1\\2\\2\\1} }}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}}\text{.} \end{align*}
Boom!
Now let us return to the matrix $A$ that we used to motivate this section in Example CSANS
\begin{equation*} A= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 \\ -16 & -1 & -4 & -10 & -13 \\ -6 & 1 & -3 & -6 & -6 \\ 0 & 2 & -2 & -3 & -2 \\ 3 & 0 & 1 & 2 & 3 \\ -1 & -1 & 1 & 1 & 0 \end{bmatrix}\text{.} \end{equation*}
We form the matrix $M$ by adjoining the $6\times 6$ identity matrix $I_6\text{.}$
\begin{equation*} M= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\ -16 & -1 & -4 & -10 & -13 & 0 & 1 & 0 & 0 & 0 & 0 \\ -6 & 1 & -3 & -6 & -6 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 2 & -2 & -3 & -2 & 0 & 0 & 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 2 & 3 & 0 & 0 & 0 & 0 & 1 & 0 \\ -1 & -1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*}
and row-reduce to obtain $N$
\begin{equation*} N= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 2 & 0 & 0 & 1 & -1 & 2 & -1 \\ 0 & \leading{1} & 0 & 0 & -3 & 0 & 0 & -2 & 3 & -3 & 3 \\ 0 & 0 & \leading{1} & 0 & 1 & 0 & 0 & 1 & 1 & 3 & 3 \\ 0 & 0 & 0 & \leading{1} & -2 & 0 & 0 & -2 & 1 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 0 & 3 & -1 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & \leading{1} & -2 & 1 & 1 & -1 \end{bmatrix} \end{equation*}
To find the four subsets for $A\text{,}$ we only need identify the $4\times 5$ matrix $C$ and the $2\times 6$ matrix $L\text{.}$
\begin{align*} C&= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 2\\ 0 & \leading{1} & 0 & 0 & -3\\ 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & \leading{1} & -2 \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 0 & 3 & -1 & 3 & 1 \\ 0 & \leading{1} & -2 & 1 & 1 & -1 \end{bmatrix} \end{align*}
Then we apply Theorem FS
\begin{align*} \nsp{A}&=\nsp{C}= \spn{\set{ \colvector{-2\\3\\-1\\2\\1} }}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}}\\ \rsp{A}&=\rsp{C}= \spn{\set{ \colvector{1 \\ 0 \\ 0 \\ 0 \\ 2},\, \colvector{0 \\ 1 \\ 0 \\ 0 \\ -3},\, \colvector{0 \\ 0 \\ 1 \\ 0 \\ 1},\, \colvector{0 \\ 0 \\ 0 \\ 1 \\ -2} }}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}}\\ \csp{A}&=\nsp{L}= \spn{\set{ \colvector{-3\\2\\1\\0\\0\\0},\, \colvector{1\\-1\\0\\1\\0\\0},\, \colvector{-3\\-1\\0\\0\\1\\0},\, \colvector{-1\\1\\0\\0\\0\\1} }}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}}\\ \lns{A}&=\rsp{L}= \spn{\set{ \colvector{1 \\ 0 \\ 3 \\ -1 \\ 3 \\ 1},\, \colvector{0 \\ 1 \\ -2 \\ 1 \\ 1 \\ -1} }}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}}\text{.} \end{align*}
The next example is just a bit different since the matrix has more rows than columns, and a trivial null space.
Archetype G and Archetype H are both systems of $m=5$ equations in $n=2$ variables. They have identical coefficient matrices, which we will denote here as the matrix $G\text{.}$
\begin{equation*} G=\begin{bmatrix} 2 & 3\\ -1 & 4 \\ 3 & 10\\ 3 & -1\\ 6 & 9 \end{bmatrix} \end{equation*}
Adjoin the $5\times 5$ identity matrix, $I_5\text{,}$ to form
\begin{equation*} M= \begin{bmatrix} 2 & 3 & 1&0&0&0&0\\ -1 & 4 & 0&1&0&0&0\\ 3 & 10 & 0&0&1&0&0\\ 3 & -1 & 0&0&0&1&0\\ 6 & 9 & 0&0&0&0&1 \end{bmatrix}\text{.} \end{equation*}
This row-reduces to
\begin{equation*} N= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & \frac{3}{11} & \frac{1}{33}\\ 0 & \leading{1} & 0 & 0 & 0 & -\frac{2}{11} & \frac{1}{11}\\ 0 & 0 & \leading{1} & 0 & 0 & 0 & -\frac{1}{3}\\ 0 & 0 & 0 & \leading{1} & 0 & 1 & -\frac{1}{3}\\ 0 & 0 & 0 & 0 & \leading{1} & 1 & -1 \end{bmatrix}\text{.} \end{equation*}
The first $n=2$ columns contain $r=2$ leading 1's, so we obtain $C$ as the $2\times 2$ identity matrix and extract $L$ from the final $m-r=3$ rows in the final $m=5$ columns.
\begin{align*} C&= \begin{bmatrix} \leading{1} & 0\\ 0 & \leading{1} \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -\frac{1}{3}\\ 0 & \leading{1} & 0 & 1 & -\frac{1}{3}\\ 0 & 0 & \leading{1} & 1 & -1 \end{bmatrix} \end{align*}
Then we apply Theorem FS
\begin{align*} \nsp{G}=\nsp{C}&= \spn{\emptyset} =\set{\zerovector}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}}\\ \rsp{G}=\rsp{C}&= \spn{\set{ \colvector{1 \\ 0},\, \colvector{0 \\ 1} }} =\complex{2}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}}\\ \csp{G}=\nsp{L}&= \spn{\set{ \colvector{0\\-1\\-1\\1\\0},\, \colvector{\frac{1}{3}\\\frac{1}{3}\\1\\0\\1} }}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}}\\ &=\spn{\set{ \colvector{0\\-1\\-1\\1\\0},\, \colvector{1\\1\\3\\0\\3} }}\\ \lns{G}=\rsp{L}&= \spn{\set{ \colvector{1 \\ 0 \\ 0 \\ 0 \\ -\frac{1}{3}},\, \colvector{0 \\ 1 \\ 0 \\ 1 \\ -\frac{1}{3}},\, \colvector{0 \\ 0 \\ 1 \\ 1 \\ -1} }}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}}\\ &=\spn{\set{ \colvector{3 \\ 0 \\ 0 \\ 0 \\ -1},\, \colvector{0 \\ 3 \\ 0 \\ 3\ \\ -1},\, \colvector{0 \\ 0 \\ 1 \\ 1 \\ -1} }}\text{.} \end{align*}
As mentioned earlier, Archetype G is consistent, while Archetype H is inconsistent. See if you can write the two different vectors of constants from these two archetypes as linear combinations of the two vectors that form the spanning set for $\csp{G}\text{.}$ How about the two columns of $G\text{?}$ Can you write each individually as a linear combination of the two vectors that form the spanning set for $\csp{G}\text{?}$ They must be in the column space of $G$ also. Are your answers unique? Do you notice anything about the scalars that appear in the linear combinations you are forming?
Sage will compute the extended echelon form, an improvement over what we did “by hand” in Sage MISLE. And the output can be requested as a “subdivided” matrix so that we can easily work with the pieces $C$ and $L\text{.}$ Pieces of subdivided matrices can be extracted with indices entirely similar to how we index the actual entries of a matrix. We will redo Example FS2 as an illustration of Theorem FS and Theorem PEEF.
Notice how we can employ the uniqueness of reduced row-echelon form (Theorem RREFU) to verify that C and L are in reduced row-echelon form. Realize too, that subdivided output is an optional behavior of the .extended_echelon_form() method and must be requested with subdivide=True.
Additionally, it is the uniquely determined matrix J that provides us with a standard version of the final column of the row-reduced augmented matrix using variables in the vector of constants, as first shown back in Example CSANS and verified here with variables defined as in Sage RRSM. (The vector method .column() converts a vector into a 1-column matrix, used here to format the matrix-vector product nicely.)
Example COV and Example CSROI each describes the column space of the coefficient matrix from Archetype I as the span of a set of $r=3$ linearly independent vectors. It is no accident that these two different sets both have the same size. If we (you?) were to calculate the column space of this matrix using the null space of the matrix $L$ from Theorem FS then we would again find a set of 3 linearly independent vectors that span the range. More on this later.
So we have three different methods to obtain a description of the column space of a matrix as the span of a linearly independent set. Theorem BCS is sometimes useful since the vectors it specifies are equal to actual columns of the matrix. Theorem BRS and Theorem CSRST combine to create vectors with lots of zeros, and strategically placed 1's near the top of the vector. Theorem FS and the matrix $L$ from the extended echelon form gives us a third method, which tends to create vectors with lots of zeros, and strategically placed 1's near the bottom of the vector. If we do not care about linear independence we can also appeal to Definition CSM and simply express the column space as the span of all the columns of the matrix, giving us a fourth description.
With Theorem CSRST and Definition RSM, we can compute column spaces with theorems about row spaces, and we can compute row spaces with theorems about column spaces, but in each case we must transpose the matrix first. At this point you may be overwhelmed by all the possibilities for computing column and row spaces. Figure CSRST is meant to help. For both the column space and row space, it suggests four techniques. One is to appeal to the definition, another yields a span of a linearly independent set, and a third uses Theorem FS. A fourth suggests transposing the matrix and the dashed line implies that then the companion set of techniques can be applied. This can lead to a bit of silliness, since if you were to follow the dashed lines twice you would transpose the matrix twice, and by Theorem TT would accomplish nothing productive.
Although we have many ways to describe a column space, notice that one tempting strategy will usually fail. It is not possible to simply row-reduce a matrix directly and then use the columns of the row-reduced matrix as a set whose span equals the column space. In other words, row operations do not preserve column spaces (however row operations do preserve row spaces, Theorem REMRS). See Exercise CRS.M21.
###### 1.
Find a nontrivial element of the left null space of $A\text{.}$
\begin{equation*} A= \begin{bmatrix} 2 & 1 & -3 & 4\\ -1 & -1 & 2 & -1\\ 0 & -1 & 1 & 2 \end{bmatrix} \end{equation*}
###### 2.
Find the matrices $C$ and $L$ in the extended echelon form of $A\text{.}$
\begin{equation*} A= \begin{bmatrix} -9 & 5 & -3 \\ 2 & -1 & 1 \\ -5 & 3 & -1 \end{bmatrix} \end{equation*}
### ExercisesFSExercises
###### C20.
Example FSAG concludes with several questions. Perform the analysis suggested by these questions.
###### C25.
Given the matrix $A$ below, use the extended echelon form of $A$ to answer each part of this problem. In each part, find a linearly independent set of vectors, $S\text{,}$ so that the span of $S\text{,}$ $\spn{S}\text{,}$ equals the specified set of vectors.
1. The row space of $A\text{,}$ $\rsp{A}\text{.}$
2. The column space of $A\text{,}$ $\csp{A}\text{.}$
3. The null space of $A\text{,}$ $\nsp{A}\text{.}$
4. The left null space of $A\text{,}$ $\lns{A}\text{.}$
\begin{equation*} A= \begin{bmatrix} -5 & 3 & -1 \\ -1 & 1 & 1 \\ -8 & 5 & -1 \\ 3 & -2 & 0 \end{bmatrix} \end{equation*}
Solution
Add a $4\times 4$ identity matrix to the right of $A$ to form the matrix $M$ and then row-reduce to the matrix $N\text{,}$
\begin{equation*} M= \begin{bmatrix} -5 & 3 & -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & 1 & 0 & 1 & 0 & 0 \\ -8 & 5 & -1 & 0 & 0 & 1 & 0 \\ 3 & -2 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 2 & 0 & 0 & -2 & -5 \\ 0 & \leading{1} & 3 & 0 & 0 & -3 & -8 \\ 0 & 0 & 0 & \leading{1} & 0 & -1 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 1 & 3 \end{bmatrix} =N \end{equation*}
To apply Theorem FS in each of these four parts, we need the two matrices,
\begin{align*} C&= \begin{bmatrix} \leading{1} & 0 & 2 \\ 0 & \leading{1} & 3 \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 0 & -1 & -1 \\ 0 & \leading{1} & 1 & 3 \end{bmatrix} \end{align*}
(a)
\begin{align*} \rsp{A} &=\rsp{C}&& \knowl{./knowl/theorem-FS.html}{\text{Theorem FS}}\\ &=\spn{\set{\colvector{1\\0\\2},\,\colvector{0\\1\\3}}}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}} \end{align*}
(b)
\begin{align*} \csp{A} &=\nsp{L}&& \knowl{./knowl/theorem-FS.html}{\text{Theorem FS}}\\ &=\spn{\set{\colvector{1\\-1\\1\\0},\,\colvector{1\\-3\\0\\1}}}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}} \end{align*}
(c)
\begin{align*} \nsp{A} &=\nsp{C}&& \knowl{./knowl/theorem-FS.html}{\text{Theorem FS}}\\ &=\spn{\set{\colvector{-2\\-3\\1}}}&& \knowl{./knowl/theorem-BNS.html}{\text{Theorem BNS}} \end{align*}
(d)
\begin{align*} \lns{A} &=\rsp{L}&& \knowl{./knowl/theorem-FS.html}{\text{Theorem FS}}\\ &=\spn{\set{\colvector{1\\0\\-1\\-1},\,\colvector{0\\1\\1\\3}}}&& \knowl{./knowl/theorem-BRS.html}{\text{Theorem BRS}} \end{align*}
###### C26.
For the matrix $D$ below use the extended echelon form to find:
1. A linearly independent set whose span is the column space of $D\text{.}$
2. A linearly independent set whose span is the left null space of $D\text{.}$
\begin{align*} D&= \begin{bmatrix} -7 & -11 & -19 & -15\\ 6 & 10 & 18 & 14 \\ 3 & 5 & 9 & 7 \\ -1 & -2 & -4 & -3 \end{bmatrix} \end{align*}
Solution
For both parts, we need the extended echelon form of the matrix.
\begin{align*} \begin{bmatrix} -7 & -11 & -19 & -15 & 1 & 0 & 0 & 0\\ 6 & 10 & 18 & 14 & 0 & 1 & 0 & 0 \\ 3 & 5 & 9 & 7 & 0 & 0 & 1 & 0 \\ -1 & -2 & -4 & -3 & 0 & 0 & 0 & 1 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & -2 & -1 & 0 & 0 & 2 & 5 \\ 0 & \leading{1} & 3 & 2 & 0 & 0 & -1 & -3 \\ 0 & 0 & 0 & 0 & \leading{1} & 0 & 3 & 2 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & -2 & 0 \end{bmatrix} \end{align*}
From this matrix we extract the last two rows, in the last four columns to form the matrix $L\text{,}$
\begin{align*} L = \begin{bmatrix} \leading{1} & 0 & 3 & 2 \\ 0 & \leading{1} & -2 & 0 \end{bmatrix} \end{align*}
By Theorem FS and Theorem BNS we have
\begin{gather*} \csp{D}=\nsp{L}=\spn{\set{ \colvector{-3\\2\\1\\0},\, \colvector{-2\\0\\0\\1} }} \end{gather*}
By Theorem FS and Theorem BRS we have
\begin{gather*} \lns{D}=\rsp{L}=\spn{\set{ \colvector{1\\0\\3\\2},\, \colvector{0\\1\\-2\\0} }} \end{gather*}
###### C41.
The following archetypes are systems of equations. For each system, write the vector of constants as a linear combination of the vectors in the span construction for the column space provided by Theorem FS and Theorem BNS (these vectors are listed for each of these archetypes).
Archetype A, Archetype B, Archetype C, Archetype D, Archetype E, Archetype F, Archetype G, Archetype H, Archetype I, Archetype J
###### C43.
The following archetypes are either matrices or systems of equations with coefficient matrices. For each matrix, compute the extended echelon form $N$ and identify the matrices $C$ and $L\text{.}$ Using Theorem FS, Theorem BNS and Theorem BRS express the null space, the row space, the column space and left null space of each coefficient matrix as a span of a linearly independent set.
Archetype A, Archetype B, Archetype C, Archetype D/Archetype E, Archetype F, Archetype G/Archetype H, Archetype I, Archetype J, Archetype K, Archetype L
###### C60.
For the matrix $B$ below, find sets of vectors whose span equals the column space of $B$ ($\csp{B}$) and which individually meet the following extra requirements.
1. The set illustrates the definition of the column space.
2. The set is linearly independent and the members of the set are columns of $B\text{.}$
3. The set is linearly independent with a “nice pattern of zeros and ones” at the top of each vector.
4. The set is linearly independent with a “nice pattern of zeros and ones” at the bottom of each vector.
\begin{equation*} B= \begin{bmatrix} 2 & 3 & 1 & 1\\ 1 & 1 & 0 & 1\\ -1 & 2 & 3 & -4 \end{bmatrix} \end{equation*}
Solution
The definition of the column space is the span of the set of columns (Definition CSM). So the desired set is just the four columns of $B\text{,}$
\begin{equation*} S=\set{ \colvector{2\\1\\-1},\, \colvector{3\\1\\2},\, \colvector{1\\0\\3},\, \colvector{1\\1\\-4} }\text{.} \end{equation*}
Theorem BCS suggests row-reducing the matrix and using the columns of $B$ with the same indices as the pivot columns.
\begin{equation*} B\rref \begin{bmatrix} \leading{1} & 0 & -1 & 2\\ 0 & \leading{1} & 1 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
So the pivot columns are numbered by elements of $D=\set{1,\,2}\text{,}$ so the requested set is
\begin{equation*} S=\set{ \colvector{2\\1\\-1},\, \colvector{3\\1\\2}}\text{.} \end{equation*}
We can find this set by row-reducing the transpose of $B\text{,}$ deleting the zero rows, and using the nonzero rows as column vectors in the set. This is an application of Theorem CSRST followed by Theorem BRS.
\begin{equation*} \transpose{B}\rref \begin{bmatrix} \leading{1} & 0 & 3\\ 0 & \leading{1} & -7\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
So the requested set is
\begin{equation*} S=\set{ \colvector{1\\0\\3},\, \colvector{0\\1\\-7} } \end{equation*}
With the column space expressed as a null space, the vectors obtained via Theorem BNS will be of the desired shape. So we first proceed with Theorem FS and create the extended echelon form
\begin{equation*} \augmented{B}{I_3}= \begin{bmatrix} 2 & 3 & 1 & 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 1 & 0 & 1 & 0\\ -1 & 2 & 3 & -4 & 0 & 0 & 1 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & -1 & 2 & 0 & \frac{2}{3} & \frac{-1}{3}\\ 0 & \leading{1} & 1 & -1 & 0 & \frac{1}{3} & \frac{1}{3}\\ 0 & 0 & 0 & 0 & \leading{1} & \frac{-7}{3} & \frac{-1}{3} \end{bmatrix}\text{.} \end{equation*}
So, employing Theorem FS, we have $\csp{B}=\nsp{L}\text{,}$ where
\begin{equation*} L= \begin{bmatrix} \leading{1} & \frac{-7}{3} & \frac{-1}{3} \end{bmatrix} \end{equation*}
We can find the desired set of vectors from Theorem BNS as
\begin{equation*} S=\set{ \colvector{\frac{7}{3}\\1\\0},\, \colvector{\frac{1}{3}\\0\\1} }\text{.} \end{equation*}
###### C61.
Let $A$ be the matrix below, and find the indicated sets with the requested properties.
1. A linearly independent set $S$ so that $\csp{A}=\spn{S}$ and $S$ is composed of columns of $A\text{.}$
2. A linearly independent set $S$ so that $\csp{A}=\spn{S}$ and the vectors in $S$ have a nice pattern of zeros and ones at the top of the vectors.
3. A linearly independent set $S$ so that $\csp{A}=\spn{S}$ and the vectors in $S$ have a nice pattern of zeros and ones at the bottom of the vectors.
4. A linearly independent set $S$ so that $\rsp{A}=\spn{S}\text{.}$
\begin{equation*} A= \begin{bmatrix} 2 & -1 & 5 & -3\\ -5 & 3 & -12 & 7\\ 1 & 1 & 4 & -3 \end{bmatrix} \end{equation*}
Solution
First find a matrix $B$ that is row-equivalent to $A$ and in reduced row-echelon form.
\begin{equation*} B= \begin{bmatrix} \leading{1} & 0 & 3 & -2\\ 0 & \leading{1} & 1 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
By Theorem BCS we can choose the columns of $A$ that have the same indices as the pivot columns ($D=\set{1,2}$) as the elements of $S$ and obtain the desired properties. So
\begin{equation*} S=\set{\colvector{2\\-5\\1},\,\colvector{-1\\3\\1}} \end{equation*}
We can write the column space of $A$ as the row space of the transpose (Theorem CSRST). So we row-reduce the transpose of $A$ to obtain the row-equivalent matrix $C$ in reduced row-echelon form.
\begin{equation*} C= \begin{bmatrix} 1 & 0 & 8\\ 0 & 1 & 3\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
The nonzero rows (written as columns) will be a linearly independent set that spans the row space of $\transpose{A}\text{,}$ by Theorem BRS, and the zeros and ones will be at the top of the vectors,
\begin{equation*} S=\set{\colvector{1\\0\\8},\,\colvector{0\\1\\3}} \end{equation*}
In preparation for Theorem FS, augment $A$ with the $3\times 3$ identity matrix $I_3$ and row-reduce to obtain the extended echelon form.
\begin{equation*} \begin{bmatrix} 1 & 0 & 3 & -2 & 0 & -\frac{1}{8} & \frac{3}{8}\\ 0 & 1 & 1 & -1 & 0 & \frac{1}{8} & \frac{5}{8}\\ 0 & 0 & 0 & 0 & 1 & \frac{3}{8} & -\frac{1}{8} \end{bmatrix} \end{equation*}
Then since the first four columns of row 3 are all zeros, we extract
\begin{equation*} L= \begin{bmatrix} \leading{1} & \frac{3}{8} & -\frac{1}{8} \end{bmatrix} \end{equation*}
Theorem FS says that $\csp{A}=\nsp{L}\text{.}$ We can then use Theorem BNS to construct the desired set $S\text{,}$ based on the free variables with indices in $F=\set{2,3}$ for the homogeneous system $\homosystem{L}\text{,}$ so
\begin{equation*} S=\set{\colvector{-\frac{3}{8}\\1\\0},\,\colvector{\frac{1}{8}\\0\\1}} \end{equation*}
Notice that the zeros and ones are at the bottom of the vectors.
This is a straightforward application of Theorem BRS. Use the row-reduced matrix $B$ from part (a), grab the nonzero rows, and write them as column vectors,
\begin{equation*} S=\set{\colvector{1\\0\\3\\-2},\,\colvector{0\\1\\1\\-1}} \end{equation*}
###### M50.
Suppose that $A$ is a nonsingular matrix. Extend the four conclusions of Theorem FS in this special case and discuss connections with previous results (such as Theorem NME4).
###### M51.
Suppose that $A$ is a singular matrix. Extend the four conclusions of Theorem FS in this special case and discuss connections with previous results (such as Theorem NME4). |
# Convert zepto to deca
Learn how to convert 1 zepto to deca step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(zepto\right)={\color{rgb(20,165,174)} x}\left(deca\right)$$
Define the prefix value(s)
$$The \text{ } value \text{ } of \text{ } zepto \text{ } is \text{ } 10^{-21}$$
$$The \text{ } value \text{ } of \text{ } deca \text{ } is \text{ } 10.0$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(zepto\right)={\color{rgb(20,165,174)} x}\left(deca\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 10^{-21}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 10.0}}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 10^{-21}} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 10.0}$$
$$\text{Conversion Equation}$$
$$10^{-21} = {\color{rgb(20,165,174)} x} \times 10.0$$
Switch sides
$${\color{rgb(20,165,174)} x} \times 10.0 = 10^{-21}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{1.0}{10.0}\right)$$
$${\color{rgb(20,165,174)} x} \times 10.0 \times \dfrac{1.0}{10.0} = 10^{-21} \times \dfrac{1.0}{10.0}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{10.0}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{10.0}}} = 10^{-21} \times \dfrac{1.0}{10.0}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{10^{-21}}{10.0}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 0.1 \times 10^{-21}$$
$$\text{Conversion Equation}$$
$$1.0\left(zepto\right) = {\color{rgb(20,165,174)} 0.1 \times 10^{-21}}\left(deca\right)$$ |
# A Quick & Easy Fraction Plan
For some reason, fractions seem to have a reputation for being a lot harder to learn (and teach) than they really are. They can actually be fun and simple to master, as long as you have a simple plan for teaching them. Here’s a strategy you could use:
1. First, remember that the right time to start teaching fractions is right after the children have a good understanding of adding, subtracting, multiplying and dividing. Then begin with very basic, simple fractions for each operation.
1. We suggest using colored pencils for fractions, with one color consistently used for the top number, or numerator, and a different one for the bottom one, or denominator. To train your class to recognize the numerator or denominator by your designated color, write some fractions on the board, and have the kids come up and change either the numerator or denominator to the right color.
1. Ask each child to fold in half a piece of paper. Then fold this folded sheet of paper in half again. Use the squares to talk about the blank___ to mean one out of four and the blank___ to mean one out of two. Have them color to show first one fourth, and then one half, and then…
1. Gradually, you’ll begin working toward teaching the kids how each of the four operations is done with fractions. For instance, with addition and subtraction, you add or subtract the top number but leave the bottom number the same. With multiplication, you multiply the two top numbers and then the two bottom numbers. Then with division, you flip upside down the second fraction and multiply.
1. One of the best ways to help elementary age students remember fraction concepts is to teach it in terms of money, so either keep lots of change and dollar bills on hand, or perhaps cut out some fake coins and bills. You’ll find that many kids have an almost immediate recognition of how much a “half” dollar is. Also, as soon as you tell them that a “quarter” is another word for a “fourth,” it will be easy for them to grasp that in terms of money, a quarter is one-fourth of a dollar.
With these few simple tips, not only will fractions not be a chore to learn, but also you’ll find that most children actually enjoy the lessons!
For a refresher game on adding, subtracting, and low integer multiplication and division, we have Hex-a-Race!:
http://www.math-lessons.ca/activities/HexaRace.html
After the refresher, we have for equivalent fractions these ones:
http://www.math-lessons.ca/activities/cards.html
http://www.math-lessons.ca/activities/FractionsBoard5.html
http://www.math-lessons.ca/activities/FractionsCards.html
1. Jason Barnhard says:
this is very useful information to use, as a future elementary teacher. you ca never learn too many tips and tricks to get kids interested in their learning. thanks for the shared ideas.
2. Rob says:
Thank you Jason. Kudos to your becoming an elementary teacher!
3. Brian says:
That’s a great idea! I was definitely a victim of fraction fright, it would have been nice to have a simple way to remember them like this.
4. Henriette Gougeon says:
Merci pour le partage des conseils utiles.
5. Rob says:
Bienvenue, Henriette! :) |
# Holt CA Course 1 10-8 Volume of Prisms MG1.3 Know and use the formulas for the volume of triangular prisms and cylinders (area of base × height); compare.
## Presentation on theme: "Holt CA Course 1 10-8 Volume of Prisms MG1.3 Know and use the formulas for the volume of triangular prisms and cylinders (area of base × height); compare."— Presentation transcript:
Holt CA Course 1 10-8 Volume of Prisms MG1.3 Know and use the formulas for the volume of triangular prisms and cylinders (area of base × height); compare these formulas and explain the similarity between them and the formula for the volume of a rectangular solid. Also covered: AF3.1, AF3.2 California Standards
Holt CA Course 1 10-8 Volume of Prisms Vocabulary Volume Volume of Prisms Formula
Holt CA Course 1 10-8 Volume of Prisms Volume is the number of cubic units needed to fill a space.
Holt CA Course 1 10-8 Volume of Prisms You need 10, or 5 · 2, centimeter cubes to cover the bottom of this rectangular prism. You need 3 layers of 10 cubes each to fill the prism. It takes 30, or 5 · 2 · 3, cubes.
Holt CA Course 1 10-8 Volume of Prisms Volume is expressed in cubic units, so the volume of the prism is 5 cm · 2 cm · 3 cm = 30 cubic centimeters, or 30 cm 3.
Holt CA Course 1 10-8 Volume of Prisms The volume of a rectangular prism is the area of its base times its height. This formula can be used to find the volume of any prism.
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 1: Finding the Volume of a Rectangular Prism Find the volume of the rectangular prism. B = 26 · 11 The base is a rectangle. Multiply.B = 286 13 in. 26 in. 11 in. Step 1: Find the area of the base.
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 1 Continued: Find the volume of the rectangular prism. V = Bh Write the formula. V = 286 13Substitute for B and h. Multiply.V = 3,718 in 3 13 in. 26 in. 11 in. Step 2: Find the volume. The volume of the prism is 3,718 in 3.
Holt CA Course 1 10-8 Volume of Prisms Student Practice 1: Find the volume of the rectangular prism. B = 29 · 12 The base is a rectangle. Multiply.B = 348 Step 1: Find the area of the base. 16 in. 29 in. 12 in.
Holt CA Course 1 10-8 Volume of Prisms Student Practice 1 Continued: Find the volume of the rectangular prism. V = Bh Write the formula. V = 348 16Substitute for B and h. Multiply.V = 5,568 in 3 Step 2: Find the volume. The volume of the prism is 5,568 in 3. 16 in. 29 in. 12 in.
Holt CA Course 1 10-8 Volume of Prisms You can also use the formula V= Bh to find the volume of a triangular prism. For triangular prisms, B represents the area of a triangle, rather than a rectangle.
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 2: Finding the Volume of a Triangular Prism Find the volume of each triangular prism. A. Step 1: Find the area of the base. The base is a triangle. B = ( 3.9 1.3) 1 2 __ Multiply.B = 2.535
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 2A: Finding the Volume of a Triangular Prism Continued Find the volume of each triangular prism. A. Step 2: Find the volume. V = BhWrite the formula. V = 2.535 4 Substitute for B and h. Multiply.V = 10.14 m 3 The volume of the prism is 10.14 m 3.
Holt CA Course 1 10-8 Volume of Prisms The height of a prism is the distance between its two bases. Caution!
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 2B: Finding the Volume of a Triangular Prism Find the volume of each triangular prism. B. Step 1: Find the area of the base. The base is a triangle. B = ( 6.5 7) 1 2 __ Multiply.B = 22.75
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 2B: Finding the Volume of a Triangular Prism Continued Find the volume of each triangular prism. B. Step 2: Find the volume. V = BhWrite the formula. V = 22.75 6 Substitute for B and h. Multiply.V = 136.5 ft 3 The volume of the prism is 136.5 ft 3.
Holt CA Course 1 10-8 Volume of Prisms Student Practice 2A: Find the volume of each triangular prism. A. Step 1: Find the area of the base. The base is a triangle. B = ( 4.2 1.6) 1 2 __ Multiply.B = 3.36 1.6 m 7 m 4.2 m
Holt CA Course 1 10-8 Volume of Prisms Student Practice 2A Continued: Find the volume of each triangular prism. A. Step 2: Find the volume. V = BhWrite the formula. V = 3.36 7 Substitute for B and h. Multiply.V = 23.52 m 3 The volume of the prism is 23.52 m 3. 1.6 m 7 m 4.2 m
Holt CA Course 1 10-8 Volume of Prisms Student Practice 2B: Find the volume of each triangular prism. B. Step 1: Find the area of the base. The base is a triangle. B = ( 4.5 9) 1 2 __ Multiply.B = 20.25 4.5 ft 5 ft 9 ft
Holt CA Course 1 10-8 Volume of Prisms Student Practice 2B Continued: Find the volume of each triangular prism. B. Step 2: Find the volume. V = BhWrite the formula. V = 20.25 5 Substitute for B and h. Multiply.V = 101.25 ft 3 The volume of the prism is 101.25 ft 3. 4.5 ft 5 ft 9 ft
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 3: Application An artist wants to make glass paper-weights with the dimensions shown. He estimates that he will need less than 20 cubic centimeters of glass for each paperweight. Is his estimate reasonable? Explain. Step 1: Find the area of the base. The base is a triangle. B = 6 5.2 = 15.6 1 2 __ 5.2 cm 6 cm 3 cm
Holt CA Course 1 10-8 Volume of Prisms Teacher Example 3 Continued An artist wants to make glass paper-weights with the dimensions shown. He estimates that he will need less than 20 cubic centimeters of glass for each paperweight. Is his estimate reasonable? Explain. Step 2: Find the volume. 5.2 cm 6 cm 3 cm V = BhWrite the formula. V = 15.6 3 = 46.8 cm 3 Substitute for B and h. No; each paperweight will require about 47 cm 3 of glass.
Holt CA Course 1 10-8 Volume of Prisms Student Practice 3: An architect wants to make a model building with the dimensions shown. He estimates that he will need more than 60 cubic centimeters of paper for each building. Is his estimate reasonable? Explain. Step 1: Find the area of the base. The base is a triangle. B = 6 5.5 = 16.5 1 2 __ 5.5 cm 6 cm 5 cm
Holt CA Course 1 10-8 Volume of Prisms Step 2: Find the volume. V = BhWrite the formula. V = 16.5 5 = 82.5 cm 3 Substitute for B and h. No; each building will require about 83 cm 3 of paper. 5.5 cm 6 cm 5 cm Student Practice 3 Continued: An architect wants to make a model building with the dimensions shown. He estimates that he will need more than 60 cubic centimeters of paper for each building. Is his estimate reasonable? Explain.
Holt CA Course 1 10-8 Volume of Prisms 10.8 Warm-Up Find the volume of each figure. 1. triangular prism with a height of 12 cm and a triangular base with base length 7.3 cm and height 3.5 cm 2. Find the volume of the figure shown.
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# Unit 6Big Ideas
## Representing Situations of the Form px+q=r and p(x+q)=r
In this unit, your student will be representing situations with diagrams and equations. There are two main categories of situations with associated diagrams and equations.
Here is an example of the first type: A standard deck of playing cards has four suits. In each suit, there are 3 face cards and other cards. There are 52 total cards in the deck. A diagram we might use to represent this situation is:
and its associated equation could be . There are 4 groups of cards, each group contains cards, and there are 52 cards in all.
Here is an example of the second type: A chef makes 52 pints of spaghetti sauce. She reserves 3 pints to take home to her family, and divides the remaining sauce equally into 4 containers. A diagram we might use to represent this situation is:
and its associated equation could be . From the 52 pints of sauce, 3 were set aside, and each of 4 containers holds pints of sauce.
1. Draw a diagram to represent the equation
2. Draw a diagram to represent the equation
3. Decide which story goes with which equation-diagram pair:
• Three friends went cherry picking and each picked the same amount of cherries, in pounds. Before they left the cherry farm, someone gave them an additional 6 pounds of cherries. Altogether, they had 39 pounds of cherries.
• One of the friends made three cherry tarts. She put the same number of cherries in each tart, and then added 6 more cherries to each tart. Altogether, the three tarts contained 39 cherries.
Solution:
Diagram A represents and the story about cherry picking. Diagram B represents and the story about making cherry tarts.
## Solving Equations of the Form px+q=r and p(x+q)=r and Problems That Lead to Those Equations
Your student is studying efficient methods to solve equations and working to understand why these methods work. Sometimes to solve an equation, we can just think of a number that would make the equation true. For example, the solution to is 2, because we know that . For more complicated equations that may include decimals, fractions, and negative numbers, the solution may not be so obvious.
An important method for solving equations is doing the same thing to each side. For example, let's show how we might solve by doing the same thing to each side.
Another helpful tool for solving equations is to apply the distributive property. In the example above, instead of multiplying each side by , you could apply the distributive property to and replace it with . Your solution would look like this:
Elena picks a number, adds 45 to it, and then multiplies by . The result is 29. Elena says that you can find her number by solving the equation .
Find Elena’s number. Describe the steps you used.
Solution:
Elena’s number was 13. There are many different ways to solve her equation. Here is one example:
## Inequalities
This week your student will be working with inequalities (expressions with or instead of ). We use inequalities to describe a range of numbers. For example, in many places you need to be at least 16 years old to be allowed to drive. We can represent this situation with the inequality . We can show all the solutions to this inequality on the number line.
Noah already has $10.50, and he earns$3 each time he runs an errand for his neighbor. Noah wants to know how many errands he needs to run to have at least $30, so he writes this inequality: We can test this inequality for different values of . For example, 4 errands is not enough for Noah to reach his goal, because , and$22.50 is less than $30. 1. Will Noah reach his goal if he runs: 1. 8 errands? 2. 9 errands? 2. What value of makes the equation true? 3. What does this tell you about all the solutions to the inequality ? 4. What does this mean for Noah’s situation? Solutions 1. Yes, if Noah runs 8 errands, he will have , or$34.50.
2. Yes, since 9 is more than 8, and 8 errands was enough, so 9 will also be enough.
1. The equation is true when . We can rewrite the equation as , or . Then we can rewrite this as , or .
2. This means that when then Noah’s inequality is true.
3. Noah can’t really run 6.5 errands, but he could run 7 or more errands, and then he would have more than \$30.
## Writing Equivalent Expressions
This week your student will be working with equivalent expressions (expressions that are always equal, for any value of the variable). For example, and are equivalent expressions. We can see that these expressions are equal when we try different values for .
when is 5
when is -1
We can also use properties of operations to see why these expressions have to be equivalent—they are each equivalent to the expression . |
# Logarithm properties
Learn about the main properties of logarithms and learn how to use them to solve various types of algebraic operations.
Like many mathematical subjects, logarithms have properties that allow us to perform certain algebraic conversions as well as rewrite entire expressions. The purpose of logarithm properties is to simplify the resolution of difficult problems.
Before you can understand the logarithm properties you must be familiar with the concept. If you are unfamiliar with the concept of logarithm, click here and read the article where we explain you in detail about it.
## Basic logarithm properties
### Equivalence
The property states that:
if the logarithm of a number a in base b
equals the logarithm of another number c in the same base b
so a equals c
### Logarithm of 1(one)
The property states that:
othe logarithm of 1 on any base is always equal to 0(zero).
Let’s go to the demo:
very simple!
it is the same as
Which is true, because by the mathematical rule, any number raised to 0(zero) will always be equal to 1.
### Logarithm of a in the base a
The property states that:
the logarithm of a number a on a base of the same value a is always equal to 1.
Let’s go to the demo:
very simple!
it is the same as
Which is obviously true!
Based on the basic logarithm properties, find the value of x in the equations below:
## Operative logarithm properties
### Logarithm of Product
The property states that:
the product logarithm of two numbers a and c in base b is always equal to the logarithm of number a in base b plus the logarithm of number c in base b.
Let’s go to the demo:
### Logarithm of Quotient
The property states that:
the product logarithm of two numbers a and c in base b is always equal to the logarithm of number a in base b minus the logarithm of number c in base b
Let’s go to the demo:
### Logarithm of Power
The property states that:
the logarithm of a number a raised to exponent c in base b is always equal to exponent c multiplied by the logarithm of number a in base b.
Let’s go to the demo:
Based on the operative logarithm properties, find the value of x in the equations below:
### Conversion to logarithm
The property states that:
a number b raised to the logarithm of anumber a in base b is equal to a.
The property also states that the logarithm of ba in base b is equal to a.
Let’s go to the demo:
This is one of the most useful and interesting properties because it teaches us how to convert any number to a logarithmic representation. This allows us to solve many difficult expressions involving logarithms.
### Change of base
The property states that:
the logarithm of a number a in base b is equal to the logarithm of number a in a third base c over the logarithm of number b in that same third base c.
Let’s go to the demo:
This is also a very important logarithm property because it teaches us how to convert a logarithm from one base to another logarithm from any other base. This allows us to simplify and solve many difficult equations.
Based on the advanced logarithm properties, find the value of x in the equations below: |
# 12.5: Division of Polynomials
Difficulty Level: At Grade Created by: CK-12
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Practice Division of Polynomials
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What if you had a polynomial like \begin{align*}2x^2 + 5x - 3\end{align*} and you wanted to divide it by a monomial like \begin{align*}x\end{align*} or a binomial like \begin{align*}x + 1\end{align*}? How would you do so? After completing this Concept, you'll be able to divide polynomials like this one by monomials and binomials.
### Try This
To check your answers to long division problems involving polynomials, try the solver at http://calc101.com/webMathematica/long-divide.jsp. It shows the long division steps so you can tell where you may have made a mistake.
### Guidance
A rational expression is formed by taking the quotient of two polynomials.
Some examples of rational expressions are
\begin{align*}\frac{2x}{x^2-1} \qquad \frac{4x^2-3x+4}{2x} \qquad \frac{9x^2+4x-5}{x^2+5x-1} \qquad \frac{2x^3}{2x+3}\end{align*}
Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator. In special cases we can simplify a rational expression by dividing the numerator by the denominator.
Divide a Polynomial by a Monomial
We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator.
#### Example A
Divide.
a) \begin{align*}\frac{8x^2-4x+16}{2}\end{align*}
b) \begin{align*}\frac{3x^2+6x-1}{x}\end{align*}
c) \begin{align*}\frac{-3x^2-18x+6}{9x}\end{align*}
Solution
a) \begin{align*}\frac{8x^2-4x+16}{2}=\frac{8x^2}{2}-\frac{4x}{2}+\frac{16}{2}=4x^2-2x+8\end{align*}
b) \begin{align*}\frac{3x^3+6x-1}{x} = \frac{3x^3}{x}+\frac{6x}{x}-\frac{1}{x}=3x^2+6-\frac{1}{x}\end{align*}
c) \begin{align*}\frac{-3x^2-18x+6}{9x}=-\frac{3x^2}{9x}-\frac{18x}{9x}+\frac{6}{9x}=-\frac{x}{3}-2+\frac{2}{3x}\end{align*}
A common error is to cancel the denominator with just one term in the numerator.
Consider the quotient \begin{align*}\frac{3x+4}{4}\end{align*}.
Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.
The correct way to simplify is:
\begin{align*}\frac{3x+4}{4}=\frac{3x}{4}+\frac{4}{4}=\frac{3x}{4}+1\end{align*}
A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just \begin{align*}3x\end{align*}. This is incorrect, because the entire numerator needs to be divided by 4.
#### Example B
Divide \begin{align*}\frac{5x^3-10x^2+x-25}{-5x^2}\end{align*}.
Solution
\begin{align*}\frac{5x^3-10x^2+x-25}{-5x^2}=\frac{5x^3}{-5x^2}-\frac{10x^2}{-5x^2}+\frac{x}{-5x^2}-\frac{25}{-5x^2}\end{align*}
The negative sign in the denominator changes all the signs of the fractions:
\begin{align*}-\frac{5x^3}{5x^2}+\frac{10x^2}{5x^2}-\frac{x}{5x^2}+\frac{25}{5x^2}=-x+2-\frac{1}{5x}+\frac{5}{x^2}\end{align*}
Divide a Polynomial by a Binomial
We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.
#### Example C
Divide \begin{align*}\frac{x^2+4x+5}{x+3}\end{align*}.
Solution
When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor.
To start the division we rewrite the problem in the following form:
\begin{align*}& {x+3 \overline{ ) x^2+4x+5 }}\end{align*}
We start by dividing the first term in the dividend by the first term in the divisor: \begin{align*}\frac{x^2}{x}=x\end{align*}.
We place the answer on the line above the \begin{align*}x\end{align*} term:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\end{align*}
Next, we multiply the \begin{align*}x\end{align*} term in the answer by the divisor, \begin{align*}x + 3\end{align*}, and place the result under the dividend, matching like terms. \begin{align*}x\end{align*} times \begin{align*}(x + 3)\end{align*} is \begin{align*}x^2+3x\end{align*}, so we put that under the divisor:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \ \ x^2 + 3x\end{align*}
Now we subtract \begin{align*}x^2+3x\end{align*} from \begin{align*}x^2+4x+5\end{align*}. It is useful to change the signs of the terms of \begin{align*}x^2+3x\end{align*} to \begin{align*}-x^2-3x\end{align*} and add like terms vertically:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x}\\ & \qquad \qquad \quad \ x\end{align*}
Now, we bring down the 5, the next term in the dividend.
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\end{align*}
And now we go through that procedure once more. First we divide the first term of \begin{align*}x + 5\end{align*} by the first term of the divisor. \begin{align*}x\end{align*} divided by \begin{align*}x\end{align*} is 1, so we place this answer on the line above the constant term of the dividend:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\end{align*}
Multiply 1 by the divisor, \begin{align*}x + 3\end{align*}, and write the answer below \begin{align*}x + 5\end{align*}, matching like terms.
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\\ & \qquad \qquad \quad \ x + 3\end{align*}
Subtract \begin{align*}x + 3\end{align*} from \begin{align*}x + 5\end{align*} by changing the signs of \begin{align*}x + 3\end{align*} to \begin{align*}-x -3\end{align*} and adding like terms:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\\ & \qquad \qquad \ \ \underline{-x - 3}\\ & \qquad \qquad \qquad \quad 2\end{align*}
Since there are no more terms from the dividend to bring down, we are done. The quotient is \begin{align*}x + 1\end{align*} and the remainder is 2.
Remember that for a division with a remainder the answer is \begin{align*}\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\end{align*}. So the answer to this division problem is \begin{align*}\frac{x^2+4x+5}{x+3}=x+1+\frac{2}{x+3}\end{align*}.
Check
To check the answer to a long division problem we use the fact that
\begin{align*}(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}\end{align*}
For the problem above, here’s how we apply that fact to check our solution:
\begin{align*}(x+3)(x+1)+2 & = x^2+4x+3+2\\ & = x^2+4x+5\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
• A rational expression is formed by taking the quotient of two polynomials.
### Guided Practice
Divide \begin{align*}\frac{x^2+8x+17}{x+4}\end{align*}.
Solution
When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor.
To start the division we rewrite the problem in the following form:
\begin{align*}& {x+4 \overline{ ) x^2+8x+17 }}\end{align*}
We start by dividing the first term in the dividend by the first term in the divisor: \begin{align*}\frac{x^2}{x}=x\end{align*}.
We place the answer on the line above the \begin{align*}x\end{align*} term:
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\end{align*}
Next, we multiply the \begin{align*}x\end{align*} term in the answer by the divisor, \begin{align*}x + 4\end{align*}, and place the result under the dividend, matching like terms. \begin{align*}x\end{align*} times \begin{align*}(x + 4)\end{align*} is \begin{align*}x^2+4x\end{align*}, so we put that under the divisor:
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17\;}}\\ & \qquad \ \ x^2 + 4x\end{align*}
Now we subtract \begin{align*}x^2+4x\end{align*} from \begin{align*}x^2+8x+17\end{align*}. It is useful to change the signs of the terms of \begin{align*}x^2+4x\end{align*} to \begin{align*}-x^2-4x\end{align*} and add like terms vertically:
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x}\\ & \qquad \qquad \quad \ 4x\end{align*}
Now, we bring down the 17, the next term in the dividend.
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ 4x + 17\end{align*}
And now we go through that procedure once more. First we divide the first term of \begin{align*}4x + 17\end{align*} by the first term of the divisor. \begin{align*}4x\end{align*} divided by \begin{align*}x\end{align*} is 4, so we place this answer on the line above the constant term of the dividend:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 17\end{align*}
Multiply 4 by the divisor, \begin{align*}x + 4\end{align*}, and write the answer below \begin{align*}4x + 16\end{align*}, matching like terms.
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ 4x + 17\\ & \qquad \qquad \quad \ 4x + 16\end{align*}
Subtract \begin{align*}4x + 16\end{align*} from \begin{align*}4x + 17\end{align*} by changing the signs of \begin{align*}4x + 16\end{align*} to \begin{align*}-4x -16\end{align*} and adding like terms:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 17\\ & \qquad \qquad \ \ \underline{-4x - 16}\\ & \qquad \qquad \qquad \quad 1\end{align*}
Since there are no more terms from the dividend to bring down, we are done. The quotient is \begin{align*}x + 4\end{align*} and the remainder is 1.
Remember that for a division with a remainder the answer is \begin{align*}\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\end{align*}. So the answer to this division problem is \begin{align*}\frac{x^2+8x+17}{x+4}=x+4+\frac{1}{x+4}\end{align*}.
Check
To check the answer to a long division problem we use the fact that
\begin{align*}(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}\end{align*}
For the problem above, here’s how we apply that fact to check our solution:
\begin{align*}(x+4)(x+4)+1 & = x^2+8x+16+1\\ & = x^2+8x+17\end{align*}
### Practice
Divide the following polynomials:
1. \begin{align*}\frac{2x+4}{2}\end{align*}
2. \begin{align*}\frac{x-4}{x}\end{align*}
3. \begin{align*}\frac{5x-35}{5x}\end{align*}
4. \begin{align*}\frac{x^2+2x-5}{x}\end{align*}
5. \begin{align*}\frac{4x^2+12x-36}{-4x}\end{align*}
6. \begin{align*}\frac{2x^2+10x+7}{2x^2}\end{align*}
7. \begin{align*}\frac{x^3-x}{-2x^2}\end{align*}
8. \begin{align*}\frac{5x^4-9}{3x}\end{align*}
9. \begin{align*}\frac{x^3-12x^2+3x-4}{12x^2}\end{align*}
10. \begin{align*}\frac{3-6x+x^3}{-9x^3}\end{align*}
11. \begin{align*}\frac{x^2+3x+6}{x+1}\end{align*}
12. \begin{align*}\frac{x^2-9x+6}{x-1}\end{align*}
13. \begin{align*}\frac{x^2+5x+4}{x+4}\end{align*}
14. \begin{align*}\frac{x^2-10x+25}{x-5}\end{align*}
15. \begin{align*}\frac{x^2-20x+12}{x-3}\end{align*}
16. \begin{align*}\frac{3x^2-x+5}{x-2}\end{align*}
17. \begin{align*}\frac{9x^2+2x-8}{x+4}\end{align*}
18. \begin{align*}\frac{3x^2-4}{3x+1}\end{align*}
19. \begin{align*}\frac{5x^2+2x-9}{2x-1}\end{align*}
20. \begin{align*}\frac{x^2-6x-12}{5x^4}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Denominator The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Dividend In a division problem, the dividend is the number or expression that is being divided.
divisor In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Polynomial long division Polynomial long division is the standard method of long division, applied to the division of polynomials.
Rational Expression A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Root Theorem The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$.
Remainder Theorem The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
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How do you work out the gradient of a laser pipe?. In this article we will let you know details of your question. Also we will share with most asked related question by peoples end of this article. Let's check it out!
How do you work out the gradient of a laser pipe?
To calculate the gradient, divide the vertical fall by the horizontal length of the pipe run. So in our 1 in 40 example, the calculation would be (1/40) giving a gradient of 0.025.
Here are some related question people asked in various search engines.
How do you calculate drainage gradient?
To calculate the gradient, divide the vertical fall by the horizontal length of the pipe run. So in our 1 in 40 example, the calculation would be (1/40) giving a gradient of 0.025.
How do you calculate the slope of a pipeline?
Multiply the length of the pipe (x) by the inches needed to slope the line (y) This equals height difference (z) between the beginning and end of the pipe (x) ⋅ (y)=(z) Example: If your pipe is 10 feet, and it needs to slope ½ an inch per foot, the equation would be 10 ⋅ ½ = 5 inches.
1.19°1 : 482.08%
2.86°1 : 205%
4.76°1 : 128.3%
7.13°1 : 812.5%
What does it mean if the gradient is 1 50?
A 1:0.5 slope means that for every 1 metre along the ground, the slope height increases by 0.5 metres.
What is the gradient for waste pipe?
The most important bit of obvious advice ever: soil and waste pipes need to be on a downhill gradient! The “fall” or “drop” should be between 1/40 (1cm down for every 40cm across) and 1/110. Too steep (1/10) then the water runs quicker than the solids so doesn’t wash them away (ugh!).
How do you find the equation of a line with a gradient?
The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis. The equation of a straight line with gradient m and intercept c on the y-axis is y = mx + c.
What does a 30% grade mean? It means that if you travel a distance up the incline, the ratio of vertical to horizontal distance (times 100) would give you the grade. We usually represent the steepness of a slope with an angle, but this essentially does the same thing.
What is a 1 48 slope?
For the cross slope of an accessible route the maximum slope allowed is 1:48. The distance from the bottom edge of the level to the surface should be no more than ½ inch (. 5:24 = 1:48). The cross slope of an accessible route is the slope that is perpendicular to the direction of pedestrian travel.
What is a 1 in 60 gradient?
A gradient of 1:60 means that there will be 1 unit of fall for every 60 units of patio width. The patio is to be 4.2m wide, so if that distance (the run) is divided by 60, the result is the 1 unit of fall.
Slope
213.49
315.24
416.99
How steep is a 25 percent grade?
For example, a 25 percent slope is simply a ratio of 25:100. The 25 percent slope below shows that the slope rises . 25 inches for every inch of horizontal distance. The slope rises 2.5 centimeters or every 10 centimeters of horizontal distance, and it rises 1.25 inches for every 5 inches of horizontal distance.
How many degrees is a 1 in 40 fall?
The answers are 1 in 40 ratio and 1.4321 degrees.
What does a 1 in 80 fall mean?
The distance can be between sections of pipe or between manholes. … For example, calculate the fall in a 50 metre section of foul water pipe work if the gradient is to be 1 in 80. A gradient of 1 in 80 is converted to a number instead of a ratio – 1 / 80 = 0.0125. Fall = 0.0125 x 50. Fall = 0.625 metres or 625mm.
What is the minimum gradient for a 100mm sewage pipe?
Be laid at a minimum gradient of 1:60 for 100 ∅ and 1:100 for 150 ∅ pipes.
What is the gradient of the line?
The gradient of a line is how steep the straight line is. In the general equation of straight line, y = mx + c , the gradient is denoted by the letter m . Imagine walking up a set of stairs. Each step has the same height and you can only take one step forward each time you move.
How steep is a 20% hill?
It doesn’t matter exactly what it means, 20% is steeper than 10%. In surveying 20% is interpreted as 20% of a right angle (i.e. a brick wall) and so would be 18 degrees.
What is a 1 to 12 slope?
1:12 slope ratio (ADA Recommended) means that for every inch of rise, you will need one foot of ramp. As an example, a 12 inch rise would require a 12 foot ramp to achieve a 1:12 ratio. 2:12 slope ratio means that for every two inches of rise, you would need one foot of ramp.
How do you find the angle of a slope?
Angle of slope represents the angle that’s formed between the run (remember it’s an idealized flat surface that ignores elevation change) and your climb’s angular deviation from that idealized flat surface. To calculate this, you divide the rise divided by the run, and then obtain the inverse tangent of the result.
What is a 50% slope?
Slope is a measure of change in elevation. … A rise of 100 feet over a run of 100 feet yields a 100 percent slope. A 50-foot rise over a 100-foot run yields a 50 percent slope. Another way to express slope is as a slope angle, or degree of slope.
What is a 60% slope?
A 60 percent slope corresponds to a slope angle of 31°.
How do I calculate my plumbing grade?
Divide the pipe’s vertical fall by the length of the pipe, then multiply the result by 100 to find the percentage. The fall and length need to be in the same units (feet or inches) for this to work. For example, if the pipe fell by one foot and was 50 feet long, you divide 1 by 50 to get 0.02.
How do you convert grades to degrees?
To convert a gradian measurement to a degree measurement, multiply the angle by the conversion ratio. The angle in degrees is equal to the gradians multiplied by 0.9.
How do you calculate elevation grade?
How to Find Grade of an Elevation. Grade can be found by measuring the horizontal length of an elevation, the run, and the vertical height of the elevation, the rise. Grade is expressed as rise/run, so if the rise is 25 and the run is 80 the grade is 25/80.
What is the slope of a 22.5 degree angle?
A 6 in 12 slope = 26.5651 degrees, and 4.97056 in 12 slope = 22.5 degrees.
What degree is a 2 1 slope?
Angle (°)Percent (%)Ratio (H:V)24.245.026.149.026.650.02:130.057.7 |
# Limits at a point
• Jun 10th 2010, 12:12 AM
Glitch
Limits at a point
Hey guys,
I'm trying to understand how to determine the left and right side of this limit:
limit as x -> 2
|x-2|/(x-2)
Do I have to graph it, or is there an algebraic way to do this? Cheers.
• Jun 10th 2010, 12:27 AM
Prove It
Quote:
Originally Posted by Glitch
Hey guys,
I'm trying to understand how to determine the left and right side of this limit:
limit as x -> 2
|x-2|/(x-2)
Do I have to graph it, or is there an algebraic way to do this? Cheers.
Recall that
$\displaystyle |x - 2| = \begin{cases}\phantom{-(}x-2\phantom{)}\textrm{ if }x-2\geq 0\\-(x - 2)\textrm{ if }x-2<0\end{cases}$
$\displaystyle |x - 2| = \begin{cases}x-2\textrm{ if }x\geq 2\\2-x\textrm{ if }x<2\end{cases}$
When you are are approaching $\displaystyle x = 2$ from the left, $\displaystyle x < 2$.
So $\displaystyle \lim_{x \to 2^{-}}\frac{|x-2|}{x-2} = \lim_{x\to 2^{-}}\frac{2-x}{x-2}$
$\displaystyle = \lim_{x\to 2^{-}}\frac{-(x-2)}{x-2}$
$\displaystyle = \lim_{x\to 2^{-}}(-1)$
$\displaystyle = -1$.
When you are approaching $\displaystyle x=2$ from the right, $\displaystyle x>2$.
So $\displaystyle \lim_{x \to 2^{+}}\frac{|x-2|}{x-2} = \lim_{x\to 2^{+}}\frac{x-2}{x-2}$
$\displaystyle =\lim_{x \to 2^{+}}(1)$
$\displaystyle = 1$.
Clearly, since the left hand limit is not the same as the right hand limit, the limit does not exist.
• Jun 10th 2010, 12:30 AM
Glitch
Ahh, I see. I was thinking that that'd be 0/0 (when you sub in 2) which is undefined. Thanks. |
# HSPT Math : Problem Solving
## Example Questions
1 2 53 54 55 56 57 58 59 61 Next →
### Example Question #601 : Problem Solving
A certain cube has a side length of 25 m. How many square tiles, each with an area of 5 m2, are needed to fully cover the surface of the cube?
500
750
100
200
1000
750
Explanation:
A cube with a side length of 25m has a surface area of:
25m * 25m * 6 = 3,750 m2
(The surface area of a cube is equal to the area of one face of the cube multiplied by 6 sides. In other words, if the side of a cube is s, then the surface area of the cube is 6s2.)
Each square tile has an area of 5 m2.
Therefore, the total number of square tiles needed to fully cover the surface of the cube is:
3,750m2/5m= 750
Note: the volume of a cube with side length s is equal to s3. Therefore, if asked how many mini-cubes with side length n are needed to fill the original cube, the answer would be:
s3/n3
### Example Question #1 : How To Find The Surface Area Of A Cube
A company wants to build a cubical room around a cone so that the cone's height and diameter are 3 inch less than the dimensions of the cube. If the volume of the cone is 486π ft3, what is the surface area of the cube?
486 in2
69,984 in2
726 in2
73,926 in2
513.375 in2
73,926 in2
Explanation:
To begin, we need to solve for the dimensions of the cone.
The basic form for the volume of a cone is: V = (1/3)πr2h
Using our data, we know that h = 2r because the height of the cone matches its diameter (based on the prompt).
486π = (1/3)πr* 2r = (2/3)πr3
Multiply both sides by (3/2π): 729 = r3
Take the cube root of both sides: r = 9
Note that this is in feet. The answers are in square inches. Therefore, convert your units to inches: 9 * 12 = 108, then add 3 inches to this: 108 + 3 = 111 inches.
The surface area of the cube is defined by: A = 6 * s2, or for our data, A = 6 * 1112 = 73,926 in2
### Example Question #601 : Problem Solving
Angie is painting a 2 foot cube for a play she is in. She needs of paint for every square foot she paints. How much paint does she need?
It is impossible to convert between metric units and feet.
Explanation:
First we must calculate the surface area of the cube. We know that there are six surfaces and each surface has the same area:
Now we will determine the amount of paint needed
### Example Question #1 : How To Find The Surface Area Of A Sphere
A spherical orange fits snugly inside a small cubical box such that each of the six walls of the box just barely touches the surface of the orange. If the volume of the box is 64 cubic inches, what is the surface area of the orange in square inches?
256π
128π
64π
32π
16π
16π
Explanation:
The volume of a cube is found by V = s3. Since V = 64, s = 4. The side of the cube is the same as the diameter of the sphere. Since d = 4, r = 2. The surface area of a sphere is found by SA = 4π(r2) = 4π(22) = 16π.
### Example Question #1 : How To Find The Area Of A Rectangle
Steve's bedroom measures 20' by 18' by 8' high. He wants to paint the ceiling and all four walls using a paint that gets 360 square feet of coverage per gallon. A one-gallon can of the paint Steve wants costs $36.00; a one-quart can costs$13.00. What is the least amount of money that Steve can expect to spend on paint in order to paint his room?
Explanation:
Two of the walls have area ; two have area ; the ceiling has area
Therefore, the total area Steve wants to cover is
Divide 968 by 360 to get the number of gallons Steve needs to paint his bedroom:
Since , Steve needs to purchase either two gallon cans and three quart cans, or three gallon cans.
The first option will cost him ; the second option will cost him . The latter is the more economical option.
### Example Question #604 : Problem Solving
Give the surface area of the above box in square centimeters.
Explanation:
100 centimeters make one meter, so convert each of the dimensions of the box by multiplying by 100.
centimeters
centimeters
Use the surface area formula, substituting :
square centimeters
### Example Question #601 : Problem Solving
Note: Figure NOT drawn to scale.
Refer to the above diagram, which shows a square. Give the ratio of the area of the yellow region to that of the white region.
The correct answer is not given among the other choices.
Explanation:
The area of the entire square is the square of the length of a side, or
.
The area of the right triangle is half the product of its legs, or
.
The area of the yellow region is therefore the difference of the two, or
.
The ratio of the area of the yellow region to that of the white region is
; that is, 55 to 9.
### Example Question #121 : Plane Geometry
The above depicts a rectangular swimming pool for an apartment. The pool is five feet deep everywhere.
An apartment manager wants to paint the four sides and the bottom of the swimming pool. One one-gallon can of the paint he wants to use covers square feet. How many cans of the paint will the manager need to buy?
Explanation:
The bottom of the swimming pool has area
square feet.
There are two sides whose area is
square feet,
and two sides whose area is
square feet.
square feet.
One one-gallon can of paint covers 350 square feet, so divide:
Seven full gallons and part of another are required, so eight is the correct answer.
### Example Question #601 : Problem Solving
The surface area of the above cylinder is 80% of what number?
Explanation:
The surface area of the cylinder can be calculated by setting and in the formula
This is 80% of the number
### Example Question #608 : Problem Solving
What is the surface area of a cube with a side edge of 7? |
Licchavi Lyceum
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# Area of Triangle Calculator
The area of a triangle is a fundamental concept in geometry, used in various fields such as construction, architecture, and mathematics. Area of Triangle Calculator is a tool to calculate the its area. Understanding how to calculate the area of a triangle is essential for solving geometric problems and real-world applications.
Triangle Area Calculator
## Triangle Area Calculator
Understanding the Formula: The formula to calculate the area of a triangle is straightforward and relies on the length of the base and the height of the triangle. The general formula is: Area = (base * height) / 2
Types of Triangles: Triangular shapes come in various forms, each with its own characteristics. Let’s discuss a few common types of triangles:
• Equilateral Triangle: In an equilateral triangle, all sides and angles are equal. To find the area, you can use the formula: Area = (√3 * side^2) / 4
• Right Triangle: A right triangle has one angle measuring 90 degrees. The area can be calculated using different methods, such as: Area = (base * height) / 2 or Area = (a * b) / 2 (using the lengths of the two sides)
• Scalene Triangle: A scalene triangle has no equal sides or angles. The formula for finding the area of a scalene triangle is: Area = (base * height) / 2
Step-by-Step Calculation: To calculate the area of a triangle, follow these steps:
Step 1: Identify the base and height of the triangle.
Step 2: Ensure that the base and height are measured in the same unit (e.g., centimeters or inches).
Step 3: Apply the formula: Area = (base * height) / 2.
Step 4: Perform the multiplication and division operations to find the area.
Step 5: Round the result to an appropriate decimal place, depending on the required precision.
Real-World Applications: The concept of finding the area of a triangle extends beyond theoretical geometry. It finds practical applications in numerous fields. Some examples include:
• Construction: Architects and engineers use the area of triangles to determine materials needed for structures, such as roofs and flooring.
• Landscaping: Calculating the area of irregularly shaped garden beds or fields helps in estimating soil or turf requirements.
• Surveying: In surveying land or properties, knowledge of triangle area calculations is vital for accurate measurement.
Understanding how to calculate the area of a triangle is a valuable skill in various disciplines. Whether you’re solving math problems, designing structures, or working on real-world applications, the ability to find the area of a triangle opens doors to limitless possibilities. Remember the simple formula, explore the different types of triangles, and apply the step-by-step calculation process to unleash the power of triangles and their areas. |
# Worksheet on Percentage of a Number with Solutions | Finding Percentage Worksheets
Scholars who want to learn about the Percentage of the number can refer to this worksheet. This worksheet helps you with a detailed explanation of the percentage of a number by which you can answer any kind of question on it. And this Worksheet on Percentage of a Number will encourage you with certain tips and tricks to tackle the issues utilizing some alternate way, and also tells you how to find the percentage of the number. You can avail Percentage Worksheets of ours to clear your doubts on the concept. Let’s see below about percentage in detail by taking enough examples.
## What is the Percentage?
A Percentage refers to per hundred which is used to share the amount or any other thing in terms of hundred. The percentage can be represented with the symbol %. It is also used to define a portion of a fraction of a whole. This percentage can be applied for decimals, fractions.
## How to find the Percentage of a Number?
To find the percentage of the number, we will multiply the given fraction or a decimal by 100 and then we will add the percent symbol. For example, if X is the number to calculate the percentage that represents Y, then we will divide Y by X, and then we will multiply the result by 100. After that, we will add the percent symbol %. Below, we can see some examples on how to find the percentage of the numbers with solutions.
1. Solve the given below:
(i) 15 % of 150
(ii) 20 % of \$ 500
(iii) 6% of 20 km
(iv) 12.5 % of 2000 kg
Solution:
(i) 15 % of 150
Given 15% of 150, which means
15/100 ×150
on solving we will get
= 22.5.
(ii) 20 % of \$ 500
Given 20% of 500, which means
20/100 ×500
on solving we will get
= \$100.
(iii) 6% of 20 km
Given 6% of 20, which means
6/100 ×20
on solving we will get
= 1.2 km
(iv) 12.5% of 2000 kg
Given 12.5% of 2000, which means
12.5/100 ×2000
on solving we will get
= 25 kg.
2. How many liters is 12% of 90 liters?
Solution:
Given 12% of 90l, which means
12/100 × 90
on solving we will get
= 10.8 liters.
3. What will be 35% of 500 grams?
Solution:
Given 35% of 500, which means
35/100 ×500
on solving we will get
= 175 grams.
4. The bus was occupied by 60% and if the total seats on a bus are 70, how many seats were available?
Solution:
As the bus was occupied by 60%,
and total seats on a bus are 70, so
60/100 × 70
on solving we will get 42.
So 42 seats are occupied.
To find available seats, we will subtract the total seats with occupied seats.
On solving we will get 28 seats.
The number of seats available is 28.
5. Mike earns \$5280 per month and spends 45% of it. What will be the savings Mike for every month?
Solution:
Mike earns \$5280 per month and spends 45%, which means
\$5280 × 45/100 on solving we will get
\$2,376.
So Mike spends \$2,376 every month.
To find the savings of Mike, we will subtract his earning by savings
On solving we will get
\$2,904.
6. 75% of students cleared in their exams out of 1500 students. Find the number of students who have cleared the exams?
Solution:
The total number of students is 1500 and in them 75% of students cleared in the exams,
which means 1500 × 75/100
on solving we will get 1,125 students cleared the exams.
7. As a year-end sale the shopkeeper offers a discount of 25% on every item. What will be the discount on every item worth \$1500?
Solution:
The discount offers by the shopkeeper on each item is 25%,
and the price of goods sold with discount is \$1500.
So the discount offers are
\$1500 × 25/100
on solving \$375.
8. Mr. Jack bought a bicycle for a discount of 20% and the original price of the bicycle is \$6,000. What will be the price of the bicycle after the discount?
Solution:
The original price of the bicycle is \$6,000
and the discount that Jack got is 20%
which means 20% of \$6,000
20/100 × 6000
on solving we will get
\$1,200
So, the price of the bicycle after the discount is
\$6,000 – \$1,200= \$ \$4,800. |
Home > Geometry prompts >
### Reflections and coordinates inquiry
If point (a,b) is reflected in the line y = x, then the image will be at (b,a).
Michael Joseph, a teacher of mathematics at Haverstock School (Camden, London, UK), devised the prompt after his year 9 class struggled with a challenge task he gave them (below).
Michael explains the origins of the prompt: "When we were using y = x and y = -x as lines of reflection, the students' strategy of 'counting squares across then down on the other side of the mirror line' worked. The challenge task raised a lot of questions. I decided to go back to the beginning and encourage the students to generalise using y = x. Supported by GeoGebra software, the class came up with very interesting observations and questions. (See the picture below from a student’s exercise book.)"
"Translating the students' statements into mathematical language, this is what they came up with: If coordinate (a,b) is reflected on the line y
= x ± kthen the coordinate of the image is (b ± k, a ± kbut this is true only if the gradient is 1.The class then went onto explore what happens if the point is reflected in the line y = mx + c where m 1.
"For me, the prompt, which was inspired by the students' exploration, is a classic example of having less to it and more in it. It can lead to
• Plotting coordinates.
• Exploring the relationship between the coordinates of the object and image under reflection.
• Drawing straight line graphs.
• Changing the subject of an equation.
• Finding the inverse function.
• Finding the mid-point between two coordinates and working out the equation of a line that reflects an object onto its image.
• Exploring the relationship between the coordinates of the object and image under other transformations, such as rotation."
Notes
Matrices and transformations
Resources
Prompt sheet
Exploring the prompt
These are the questions and observations of a year 9 mixed attainment class. The teacher had structured the next part of the inquiry to explore the case of y = x by reflecting quadrilaterals in the line of reflection. Students created their own examples and generalised. The student who asked whether the point (b,a) goes back to (a,b) when reflected in the line y = x ended the first lesson of the inquiry by verifying that that is indeed the case. The second lesson addressed the question, "Could we use different lines of reflection?". Under the direction of the the teacher, the lesson started with an inquiry into the case of y = -x. The students' generalisations are shown in the table below.
Object Image after reflection in the line y = x Image after reflection in the line y = -x A (-3,5) A' (5,-3) A" (-5,3) B (2,5) B' (5,2) B" (-5,-2) C (1,2) C' (2,1) C" (-2,-1) D (0,4) D' (4,0) D" (-4,0) (a,b) (b,a) (-b,-a)
The second lesson continued with individual inquiries into different lines with students selecting which lines of reflection they would use. The teacher collated the results at the end of the lesson and, during a class discussion, the class co-constructed the generalisation.
Equation of line (a,b) under reflection in the line ... y = x + 1 (b - 1, a + 1) y = x + 2 (b - 2, a + 2) y = x + 3 (b - 3, a + 3) y = x + 4 (b - 4, a + 4) y = x + 5 (b - 5, a + 5) y = x + n (b - n, a + n)
In the last few seconds of the lesson, a student made this conjecture: If you swap the x and the y in the equation of the line of reflections (x = y + n), then (a,b) becomes (b + n,a - n).
At the start of the third lesson, the class considered the conjecture. The teacher showed that the equation (x = y + n) can be rearranged to give y = x - n. Students chose a value of n and verified that the conjecture is true for each particular case. The teacher moved the inquiry on to look at the change in the coordinates under rotation, as suggested by one student's question at the start of the inquiry. The centre of rotation was fixed at (0,0). The generalisations are in the table below.
Object Rotation 90o clockwise Rotation 180o clockwise Rotation 270o clockwise A (2,3) A' (3,-2) A" (-2,-3) A''' (-3,2) B (6,4) B' (4,-6) B" (-6,-4) B''' (-4,6) C (7,1) C' (1,-7) C" (-7,-1) C''' (-1,7) (a,b) (b,-a) (-a,-b) (-b,a)
The following questions are some that the class did not have time to explore:
• What happens if the centre of rotation is not (0,0)?
• How do the coordinates change under enlargement?
• Could we prove any of the generalisations? |
# How do solve the following linear system?: 2x-y=4 , 7x+3y=27 ?
Jun 19, 2018
See a solution process below:
#### Explanation:
Step 1) Solve the first equation for $y$:
$2 x - y = 4$
$2 x - \textcolor{b l u e}{4} - y + \textcolor{red}{y} = 4 - \textcolor{b l u e}{4} + \textcolor{red}{y}$
$2 x - 4 - 0 = 0 + y$
$2 x - 4 = y$
$y = 2 x - 4$
Step 2) Substitute $\left(2 x - 4\right)$ for $y$ in the second equation and solve for $x$:
$7 x + 3 y = 27$ becomes:
$7 x + 3 \left(2 x - 4\right) = 27$
$7 x + \left(3 \times 2 x\right) - \left(3 \times 4\right) = 27$
$7 x + 6 x - 12 = 27$
$\left(7 + 6\right) x - 12 = 27$
$13 x - 12 = 27$
$13 x - 12 + \textcolor{red}{12} = 27 + \textcolor{red}{12}$
$13 x - 0 = 39$
$13 x = 39$
$\frac{13 x}{\textcolor{red}{13}} = \frac{39}{\textcolor{red}{13}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}} x}{\cancel{\textcolor{red}{13}}} = 3$
$x = 3$
Step 3) Substitute $3$ for $x$ in the solution to the first equation at the end of Step 1 and calculate $y$:
$y = 2 x - 4$ becomes:
$y = \left(2 \times 3\right) - 4$
$y = 6 - 4$
$y = 2$
The Solution Is:
$x = 3$ and $y = 2$
Or
$\left(3 , 2\right)$ |
Question 1: Express the following complex numbers in the form $\displaystyle a + ib$
$\displaystyle \text{i) } (1+i)(1+2i) \hspace{1.0cm} \text{ii) } \frac{3+2i}{-2+i} \hspace{1.0cm} \text{iii) } \frac{1}{(2+i)^2} \hspace{1.0cm} \text{iv) } \frac{1-i}{1+i}$
$\displaystyle \text{v) } \frac{(2+i)^3}{2+3i} \hspace{1.0cm} \text{vi) } \frac{(1+i)(1+\sqrt{3} i)}{1-i} \hspace{1.0cm} \text{vii) } \frac{2+3i}{4+5i} \hspace{1.0cm} \text{viii) } \frac{(1-i)^3}{1-i^3}$
$\displaystyle \text{ix) } (1+2i)^{-3} \hspace{1.0cm} \text{x) } \frac{3-4i}{(4+2i)(1+i)}$
$\displaystyle \text{xi) } \Big( \frac{1}{1-4i} - \frac{2}{1+i} \Big) \Big( \frac{3-4i}{5+i} \Big) \hspace{1.0cm} \text{xii) } \frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$
$\displaystyle \text{i) } (1+i)(1+2i)$
$\displaystyle = 1 + 2i + i + 2i^2 = 1 + 3i - 2 = - 1 + 3i$
$\displaystyle \text{ii) } \frac{3+2i}{-2+i}$
$\displaystyle = \frac{3+2i}{-2+i} \times \frac{-2-i}{-2-i} = \frac{-6-3i-4i-2i^2}{4-i^2} = \frac{-6-7i+2}{4+1} = \frac{-4}{5} - i \big( \frac{7}{5} \big)$
$\displaystyle \text{iii) } \frac{1}{(2+i)^2}$
$\displaystyle = \frac{1}{4+i^2 + 4i} = \frac{1}{3+4i} \times \frac{3-4i}{3-4i} = \frac{3-4i}{9-16i^2} = \frac{3}{25} - i \big( \frac{4}{5} \big)$
$\displaystyle \text{iv) } \frac{1-i}{1+i}$
$\displaystyle = \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{1+i^2-2i}{1-i^2} = \frac{-2i}{2} = -i$
$\displaystyle \text{v) } \frac{(2+i)^3}{2+3i}$
$\displaystyle = \frac{( 4 + i^2 + 4 i )(2+i)}{2+3i} = \frac{8 + 2i^2 + 8i + 4i + i^3 + 4i^2}{2 + 3i} = \frac{2+11i}{2+3i} \times \frac{2-3i}{2-3i}$
$\displaystyle = \frac{4 - 6i + 22 i - 33i^2}{4 - 9i^2} = \frac{37+ 16i}{13} = \frac{37}{13} + i \big( \frac{16}{13} \big)$
$\displaystyle \text{vi) } \frac{(1+i)(1+\sqrt{3} i)}{1-i}$
$\displaystyle = \frac{(1+i)(1+\sqrt{3} i)}{1-i} \times \frac{1+i}{1-i} = \frac{(1+i^2+2i)(1+\sqrt{3}i}{1-i^2}$
$\displaystyle = \frac{2i ( 1 + \sqrt{3} i)}{2} = i(1 + \sqrt{3}i) = \sqrt{3} i^2 + i = - \sqrt{3}+ i$
$\displaystyle \text{vii) } \frac{2+3i}{4+5i}$
$\displaystyle = \frac{2+3i}{4+5i} \times \frac{4-5i}{4-5i} = \frac{8 - 10 i + 12 i - 15 i^2}{16 - 25 i^2} = \frac{23 + 2 i}{41} = \frac{23}{41} + i \big( \frac{2}{41} \big)$
$\displaystyle \text{viii) } \frac{(1-i)^3}{1-i^3}$
$\displaystyle = \frac{(1-i)^2 (1-i)}{1-i^3} = \frac{(1+i^2 - 2i)(1-i)}{1+i} = \frac{-2i(1-i)}{1+i} = \frac{-2(i - i^2)}{1+i} = \frac{-2(1+i)}{1+i} = - 2$
$\displaystyle \text{ix) } (1+2i)^{-3}$
$\displaystyle = \frac{1}{(1+2i)^2(1+2i)} = \frac{1}{(1+4i^2+4i)(1+2i)} = \frac{1}{(-3+4i)(1+2i)} = \frac{1}{-3+4i-6i+8^2}$
$\displaystyle = \frac{1}{-11-2i} = \frac{1}{-11-2i} \times \frac{11-2i}{11-2i} = \frac{-11+2i}{121+4} = \frac{-11}{125} + i \big( \frac{2}{125} \big)$
$\displaystyle \text{x) } \frac{3-4i}{(4+2i)(1+i)}$
$\displaystyle = \frac{3-4i}{4 - 2i + 4i - 2i^2} = \frac{3-4i}{6+2i} \times \frac{6-2i}{6-2i} = \frac{18 - 24i - 6i + 8i^2}{36 + 4}$
$\displaystyle = \frac{10-30i}{40} = \frac{1-3i}{4} = \frac{1}{4} + i \big( \frac{-3}{4} \big)$
$\displaystyle \text{xi) } \Big( \frac{1}{1-4i} - \frac{2}{1+i} \Big) \Big( \frac{3-4i}{5+i} \Big)$
$\displaystyle = \Big( \frac{1+i-2+8i}{1-4i+i-4i^2} \Big) \Big( \frac{3-4i}{5+i} \Big) = \Big( \frac{-1+9i}{5-3i} \Big) \Big( \frac{3-4i}{5+i} \Big) = \frac{-3+27i+4i-36i^2}{25-15i+5i-3i^2}$
$\displaystyle = \frac{33+ 31i}{28-10i} \times \frac{28+10i}{28+10i} = \frac{924+330i+868i-310}{784+100} = \frac{614+1198i}{884} = \frac{614}{884} + i \big( \frac{1198}{884} \big)$
$\displaystyle \text{xii) } \frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$
$\displaystyle = \frac{5+\sqrt{2} i}{1 - \sqrt{2} i} \times \frac{1 + \sqrt{2} i}{1 + \sqrt{2} i} = \frac{5 + \sqrt{2} i + 5 \sqrt{2} i + 2 i^2}{1+2}$
$\displaystyle = \frac{3 + 6 \sqrt{2} i}{3} = 1 + 2\sqrt{2} i = 1 + i ( 2\sqrt{2})$
$\displaystyle \\$
Question 2: Find the real values of $\displaystyle x \text{ and } y$, if
$\displaystyle \text{i) } (x+iy)(2-3i) = 4+i \hspace{1.0cm} \text{ii) } (3x-2iy)(2+i)^2 = 10(1+i)$
$\displaystyle \text{iii) } \frac{(1+i)x- 2i}{3+i} - \frac{(2-3i)y+i}{3-i} = i \hspace{1.0cm} \text{iv) } (1+i)(x+iy)= 2- 5i$
$\displaystyle \text{i) } (x+iy)(2-3i) = 4+i$
$\displaystyle \Rightarrow 2x + i 2y - 3x i - 3y i^2 = 4 + i$
$\displaystyle \Rightarrow ( 2x + 3y) + i ( -3x + 2y) = 4 + i$
Therefore comparing, we get
$\displaystyle 2x + 3y = 4$ … … … … … i)
$\displaystyle -3x + 2y = 1$ … … … … … ii)
Solving i) and ii), multiplying i) by $\displaystyle 3$ and ii) by $\displaystyle 2$ and adding we get
$\displaystyle { \hspace{1.0cm} 6x + 9y = 12} \\ \underline { (+) - 6x + 4 y = 2 } \\ { \hspace{2.0cm}13 y = 14}$
$\displaystyle \Rightarrow y = \frac{14}{13}$
Now substituting in ii) we get
$\displaystyle 2x + 3 \big( \frac{14}{13} \big) = 4 \Rightarrow 2x = \frac{10}{13} \Rightarrow x = \frac{10}{26}$
Hence $\displaystyle x = \frac{10}{26}$ and $\displaystyle y = \frac{14}{13}$
$\displaystyle \text{ii) } (3x-2iy)(2+i)^2 = 10(1+i)$
$\displaystyle \Rightarrow ( 3x - 2iy) ( 4 + i^2 + 4i) = 10 + 10 i$
$\displaystyle \Rightarrow ( 3x - i 2y) ( 3 + 4i) = 10 + 10i$
$\displaystyle \Rightarrow 9x - i 6 y + i 12 x + 8 y = 10 + 10 i$
$\displaystyle \Rightarrow ( 9x + 8 y) + i ( 12 x - 6y) = 10 + 10 i$
Therefore comparing, we get
$\displaystyle 9x + 8y = 10$ … … … … … i)
$\displaystyle 12 x - 6y = 10$ … … … … … ii)
Solving i) and ii), multiplying i) by $\displaystyle 6$ and ii) by $\displaystyle 8$ and adding we get
$\displaystyle { \hspace{1.0cm} 54x + 48 y = 60 } \\ \underline { (+) 96x - 48 y = 80 } \\ { \hspace{2.0cm} 150x = 140}$
$\displaystyle \Rightarrow x = \frac{14}{15}$
Now substituting in i) we get
$\displaystyle 9 \big( \frac{14}{15} \big) + 8y = 10 \Rightarrow 8y = \frac{8}{5} \Rightarrow y = \frac{1}{5}$
Hence $\displaystyle x = \frac{14}{15}$ and $\displaystyle y = \frac{1}{5}$
$\displaystyle \text{iii) } \frac{(1+i)x- 2i}{3+i} - \frac{(2-3i)y+i}{3-i} = i$
$\displaystyle \Rightarrow \frac{[ (1+i) x - 2 i ] (3-i)+[(2-3i)y + 1](3+i)}{9+1} = i$
$\displaystyle \Rightarrow 3x + i 3x - 6i - i x - i^2 x + 2 i^2+ 6y - i 9y + 3 i + i 2y - i^2 3y + i^2 = 10i$
$\displaystyle \Rightarrow 4x + i ( 2x) - 6i - 2 + 9y + i ( -7y) + 3i - 1 = 10i$
$\displaystyle \Rightarrow ( 4x + 9y - 3) + i ( 2x - 7y - 3) = 10i$
$\displaystyle \Rightarrow 4x + 9y - 3 = 0 \Rightarrow 4x + 9y = 3$ … … … … … i)
$\displaystyle \Rightarrow 2x - 7y - 3 = 10 \Rightarrow 2x - 7y = 13$ … … … … … ii)
Solving i) and ii), multiplying i) by $\displaystyle 7$ and ii) by $\displaystyle 9$ and adding we get
$\displaystyle 46 x = 117 + 21 \Rightarrow x = \frac{138}{46} = 3$
Substituting in i) we get
$\displaystyle 4 ( 3) + 9y = 3 \Rightarrow y = -1$
Hence $\displaystyle x = 3$ and $\displaystyle y = -1$
$\displaystyle \text{iv) } (1+i)(x+iy)= 2- 5i$
$\displaystyle x + ix + iy + i^2 y = 2 - 5i$
$\displaystyle (x-y) + i( x+y) = 2 - 5i$
Comparing we get
$\displaystyle x-y = 2$ … … … … … i)
$\displaystyle x+y = -5$ … … … … … ii)
Adding i) and ii) we get
$\displaystyle 2x = - 3 \Rightarrow x = \frac{-3}{2}$
Substituting in i) we get
$\displaystyle y = x - 2 = \frac{-3}{2} - 2 = \frac{-7}{2}$
Hence $\displaystyle x = \frac{-3}{2}$ and $\displaystyle y = \frac{-7}{2}$
$\displaystyle \\$
Question 3: Find the conjugates of the following complex numbers:
$\displaystyle \text{i) } 4-5i \hspace{1.0cm} \text{ii) } \frac{1}{3+5i} \hspace{1.0cm} \text{iii) } \frac{1}{1+i}$
$\displaystyle \text{iv) } \frac{(3-i)^2}{2+i} \hspace{1.0cm} \text{v) } \frac{(1+i)(2+i)}{3+i} \hspace{1.0cm} \text{vi) } \frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$
Note: $\displaystyle \text{If } z = x + iy$, then conjugate of $\displaystyle z$ is $\displaystyle \overline{z} = x - iy$
$\displaystyle \text{i) } \text{If } z = 4-5i \Rightarrow \overline{z} = 4 + 5i$
$\displaystyle \text{ii) } \text{If } z = \frac{1}{3+5i} = \frac{1}{3+5i} \times \frac{3 - 5i}{3 - 5i} = \frac{3 - 5i}{34} = \frac{3}{34} -i \frac{5}{34}$
Therefore $\displaystyle \overline{z} = \frac{3}{34} + i \frac{5}{34}$
$\displaystyle \text{iii) } \text{If } z = \frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i} = \frac{1-i}{2} = \frac{1}{2} - i \big( \frac{1}{2} \big)$
Therefore $\displaystyle \overline{z} = \frac{1}{2} + i \big( \frac{1}{2} \big)$
$\displaystyle \text{vi) } \text{If } z = \frac{(3-i)^2}{2+i} = \frac{9+i^2-6i}{2+i} = \frac{8-6i}{2+i} = \frac{8-6i}{2+i} \times \frac{2-i}{2-i}$
$\displaystyle = \frac{16-12i-8i-6}{4+1} = \frac{10-20i}{5} = 2-4i$
Therefore $\displaystyle \overline{z} = 2+4i$
$\displaystyle \text{v) } \text{If } z = \frac{(1+i)(2+i)}{3+i} = \frac{2+2i+i+i^2}{3+i} = \frac{1+3i}{3+i} = \frac{1+3i}{3+i} \times \frac{3-i}{3-i}$
$\displaystyle = \frac{3+9i - i-3i^2}{9+1} = \frac{6+8i}{10} = \frac{3}{4} + i \big( \frac{4}{5} \big)$
Therefore $\displaystyle \overline{z} = \frac{3}{4} - i \big( \frac{4}{5} \big)$
$\displaystyle \text{vi) } \text{If } z = \frac{(3-2i)(2+3i)}{(1+2i)(2-i)} = \frac{6-4i+9i-6i^2}{2+4i-i-2i^2} = \frac{12+5i}{4+3i} \times \frac{4-3i}{4-3i}$
$\displaystyle = \frac{48+20i - 36i - 15i^2}{16+9} = \frac{63}{25} - i \big( \frac{16}{25} \big)$
Therefore $\displaystyle \overline{z} = \frac{63}{25} + i \big( \frac{16}{25} \big)$
$\displaystyle \\$
Question 4: Find the multiplicative inverse of the following complex numbers:
$\displaystyle \text{i) } 1-i \hspace{1.0cm} \text{ii) } (1+i\sqrt{3})^2 \hspace{1.0cm} \text{iii) } 4-3i \hspace{1.0cm} \text{iv) } \sqrt{5}+3i$
$\displaystyle \text{If } z = x + iy$ is the complex number, then multiplication inverse of $\displaystyle z$ is $\displaystyle z^{-1}$ or $\displaystyle \frac{1}{z}$
$\displaystyle \text{i) } z = 1-i$
$\displaystyle z^{-1} = \frac{1}{1-i} \times \frac{1+i}{1+i} = \frac{1+i}{1-i^2} = \frac{1}{2} + \frac{1}{2} i$
$\displaystyle \text{ii) } z = (1+i\sqrt{3})^2 = 1+3i^2+i (2\sqrt{3})= -2 +i (2\sqrt{3})$
$\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{-2+i (2\sqrt{3})} \times \frac{-2-i (2\sqrt{3}) }{-2-i (2\sqrt{3}) } = \frac{-2-i (2\sqrt{3}) }{4 + 12} = \frac{-1}{8} -i \frac{\sqrt{3}}{6}$
$\displaystyle \text{iii) } z = 4-3i$
$\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{4-3i} \times \frac{4+3i}{4+3i} = \frac{4+3i}{16+9} = \frac{4}{25} + i \big( \frac{3}{25} \big)$
$\displaystyle \text{iv) } z = \sqrt{5}+3i$
$\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{\sqrt{5}+ 3i} \times \frac{\sqrt{5}- 3i}{\sqrt{5}-3i} = \frac{\sqrt{5}-3i}{5+9} = \frac{\sqrt{5}}{14} - \frac{3}{14} i$
$\displaystyle \\$
$\displaystyle \text{Question 5: If } z_1 = 2 - i, z_2 = 1+i , \text{ find } \Big| \frac{z_1+z_2+1}{z_1-z_2+1} \Big|$
$\displaystyle \text{If } z = x + iy$, then $\displaystyle |z| = \sqrt{x^2+y^2}$
$\displaystyle z_1 = 2 - i z_2 = 1 + i$
$\displaystyle \Rightarrow z_1 + z_2 + 1 = 2 - i + 1 + i + 1 = 4$
$\displaystyle \Rightarrow z_1 - z_2 + i = 2 - i - 1 - i + 1 = 1 - i$
$\displaystyle \Rightarrow \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} = \frac{4}{1-i} \times \frac{1+i}{1+i} = \frac{4+4i}{2} = 2+2i$
$\displaystyle \Rightarrow \Big| \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} \Big| = \sqrt{2^2+2^2} = \sqrt{8} =2\sqrt{2}$
$\displaystyle \\$
Question 6: $\displaystyle \text{If } z_1 = 2-i, z_2 = -2+i$, find $\displaystyle \text{i) } Re \Big( \frac{z_1z_2}{z_1} \Big) \hspace{1.0cm} \text{ii) } Im \Big( \frac{1}{z_1z_2} \Big)$
$\displaystyle z_1 = 2-i, \hspace{1.0cm} z_2 = -2+i \hspace{1.0cm} \overline{z_1} = 2 + i$
$\displaystyle \text{i) } z_1z_2 = ( 2 - i) ( -2 + i) = -4 + 2i + 2i - i^2 = - 3 + 4i$
$\displaystyle \therefore \frac{z_1z_2}{\overline{z_1}} = \frac{-3+4i}{2+i} \times \frac{2-i}{2-i} = \frac{-6+8i+3i-4i^2}{4+1} = \frac{-2+5i}{5} = \frac{-2}{5} + i$
$\displaystyle \therefore Re \big( \frac{z_1z_2}{\overline{z_1}} \big) = \frac{-2}{5}$
$\displaystyle \text{ii) } z_1 = 2 - i \hspace{1.0cm} \overline{z_1} = 2+i$
$\displaystyle \therefore z_1 \overline{z_1} = (2-i)(2+i) = 4 + 1 = 5$
$\displaystyle \therefore \frac{1}{z_1 \overline{z_1}} = \frac{1}{5}$
$\displaystyle \therefore Im \big( \frac{1}{z_1 \overline{z_1}} \big) = 0$
$\displaystyle \\$
$\displaystyle \text{Question 7: Find the modulus of } \frac{1+i}{1-i} - \frac{1-i}{1+i}$
$\displaystyle \frac{1+i}{1-i} - \frac{1-i}{1+i} = \frac{(1+i^2+2i)-(1+i^2-2i)}{1-i^2} = \frac{2i+2i}{2} = 2i$
$\displaystyle |2i| = \sqrt{2^2} = 2$
$\displaystyle \\$
$\displaystyle \text{Question 8: If } x+iy = \frac{a+ib}{a-ib} , \text{ prove that } x^2 + y^2 = 1$
$\displaystyle x+iy = \frac{a+ib}{a-ib}$
$\displaystyle \overline{x+iy} = \Big( \overline{\frac{a+ib}{a-ib}} \Big) = \frac{\overline{a+ib}}{\overline{a-ib}}$
$\displaystyle x-iy = \frac{a-ib}{a+ib}$
$\displaystyle \therefore ( x +iy)(x-iy) = \frac{a+ib}{a-ib} \times \frac{a-ib}{a+ib}$
$\displaystyle \Rightarrow x^2 + y^2 = \frac{a^2 +b^2}{a^2 +b^2} = 1$
Hence proved.
Therefore the general solution is given by $\displaystyle \theta = 2 n \pi \pm \frac{\pi}{2} , n \in Z$
$\displaystyle \\$
Question 9: Find the least positive integral value of $\displaystyle n$ for which $\displaystyle \Big( \frac{1+i}{1-i} \Big)^n$ is real.
For $\displaystyle n = 1$
$\displaystyle \Big( \frac{1+i}{1-i} \Big)^1 = \frac{1+i^2+2i}{2} = i \text{ which is not real. }$
For $\displaystyle n = 2$
$\displaystyle \Big( \frac{1+i}{1-i} \Big)^2 = \frac{1+i^2+2i}{1+i^2-2i} = \frac{2i}{-2i} = -1 \text{ which is real. }$
Therefore the least positive integral value of $\displaystyle n = 2$
$\displaystyle \\$
$\displaystyle \text{Question 10: Find the real values of } \theta \text{ or which the complex number } \\ \\ \frac{1+ i \cos \theta}{1 - 2 i \cos \theta} \text{ is purely real. }$
$\displaystyle \text{Let } z = \frac{1+ i \cos \theta}{1 - 2 i \cos \theta}$
$\displaystyle = \frac{1+ i \cos \theta}{1 - 2 i \cos \theta} \times \frac{1 + 2 i \cos \theta}{1 + 2 i \cos \theta}$
$\displaystyle = \frac{1+i \cos \theta + 2 i \cos \theta + 2 i \cos \theta + 2 i^2 \cos^2 \theta}{1+4 \cos^2 \theta}$
$\displaystyle = \frac{(1-2 \cos^2 \theta) + (i ( 3 \cos \theta)}{1+4 \cos^2 \theta}$
For $\displaystyle z$ to be purely Real, $\displaystyle Im (z) = 0$
$\displaystyle \therefore \frac{3 \cos \theta}{1+4 \cos^2 \theta} = 0$
$\displaystyle \Rightarrow \cos \theta = 0$
$\displaystyle \Rightarrow \theta = \frac{\pi}{2}$
$\displaystyle \\$
$\displaystyle \text{Question 11: Find the smallest positive integer value of n for which } \frac{(1+i)^n}{(1-i)^{n-2}} \\ \\ \text{ is a real number. }$
$\displaystyle \frac{(1+i)^n}{(1-i)^{n-2}}$
$\displaystyle = \frac{ (1+i)^n }{ (1-i)^{n-2} } \times (1-i)^2$
$\displaystyle = \Big( \frac{1+i}{1-i} \times \frac{1+i}{1+i} \Big)^m \times ( 1+i^2 - 2i)$
$\displaystyle = \Big( \frac{1+i^2+ 2i}{1-i^2} \Big)^m \times ( 1-1 - 2i)$
$\displaystyle = \Big( \frac{1-1+ 2i}{1+1} \Big)^m \times ( - 2i)$
$\displaystyle = -2i (i^m)$
$\displaystyle = - 2 i^{m+1}$
For this to be real , the smallest positive value of $\displaystyle n$ will be $\displaystyle 1$.
Thus $\displaystyle i^{1+1} = i^2 = - 1$ , which is real.
$\displaystyle \\$
$\displaystyle \text{Question 12: If } \Big( \frac{1+i}{1-i} \Big)^3 - \Big( \frac{1-i}{1+i} \Big)^3 = x+iy , \text{ find } ( x, y)$
$\displaystyle \Big( \frac{1+i}{1-i} \Big)^3 - \Big( \frac{1+i}{1-i} \Big)^3 = x+iy$
$\displaystyle \Rightarrow \Big( \frac{1+i}{1-i} \times \frac{1+i}{1+i} \Big)^3 - \Big( \frac{1+i}{1-i} \times \frac{1-i}{1-i} \Big)^3 = x+iy$
$\displaystyle \Rightarrow \Big( \frac{2i}{2} \Big)^3 - \Big( \frac{-2i}{2} \Big)^3 = x+iy$
$\displaystyle \Rightarrow i^3 - ( -i)^3 = x + i y$
$\displaystyle \Rightarrow -i - i = x + iy$
$\displaystyle \Rightarrow -2i = x + iy$
$\displaystyle \Rightarrow x = 0$ and $\displaystyle y = - 2$
$\displaystyle \\$
$\displaystyle \text{Question 13: If } \frac{(1+i)^2}{2-i} = x+iy , \text{ find } x+y$
$\displaystyle \frac{(1+i)^2}{2-i} = x + i y$
$\displaystyle \Rightarrow \frac{2i}{2-i} \times \frac{2+i}{2+i} = x+iy$
$\displaystyle \Rightarrow \frac{4i + 2i^2}{4+1} = x+iy$
$\displaystyle \Rightarrow \frac{-2+4i}{5} = x+iy$
$\displaystyle \Rightarrow x = \frac{-2}{5} , \ \ \ y = \frac{4}{5}$
$\displaystyle \therefore x + y = \frac{-2}{5} + \frac{4}{5} = \frac{2}{5}$
$\displaystyle \\$
$\displaystyle \text{Question 14: If } \Big( \frac{1-i}{1+ i} \Big)^{100} = a+ib , \text{ find } ( a, b)$
$\displaystyle \Big( \frac{1-i}{1+ i} \Big)^{100} = a+ib$
$\displaystyle \Rightarrow \Big( \frac{1-i}{1+ i} \times \frac{1-i}{1-i} \Big)^{100} = a+ib$
$\displaystyle \Rightarrow \Big( \frac{-2i}{2} \Big)^{100} = a+ib$
$\displaystyle \Rightarrow ( -i )^{100} = a+ib$
$\displaystyle \Rightarrow 1 = a + ib$
$\displaystyle \therefore a = 1 , b = 0$
$\displaystyle \therefore (a, b) = ( 1, 0)$
$\displaystyle \\$
$\displaystyle \text{Question 15: If } a = \cos \theta + i \sin \theta , \text{ find the value of } \frac{1+a}{1-a}$
$\displaystyle a = \cos \theta + i \sin \theta$
$\displaystyle \frac{1+a}{1-a} = \frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) }$
$\displaystyle = \frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) } \times \frac{(1-\cos \theta) + i (\sin \theta)}{(1-\cos \theta) + i (\sin \theta)}$
$\displaystyle = \frac{(1+i \sin \theta)^2 - \cos^2 \theta}{(1-\cos \theta)^2 + \sin^2 \theta}$
$\displaystyle = \frac{1 - \sin^2 \theta + 2 i \sin \theta- \cos^2 \theta}{1+ \cos^2 \theta - 2 \cos \theta + \sin^2 \theta}$
$\displaystyle = \frac{2 i \sin \theta }{2 ( 1 - \cos \theta)}$
$\displaystyle = \frac{i \sin \theta}{1 - \cos \theta}$
$\displaystyle = i \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}$
$\displaystyle = i \cot^2 \frac{\theta}{2}$
$\displaystyle \\$
Question 16:
$\displaystyle \text{i) } 2x^3+2x^2-7x+72 , \text{ when } x = \frac{3-5i}{2}$
$\displaystyle \text{ii) } x^4-4x^3+4x^2+8x+44 , \text{ when } x= 3 + 2i$
$\displaystyle \text{iii) } x^4+4x^3+6x^2+4x+9 , \text{ when } x = - 1 +i \sqrt{2}$
$\displaystyle \text{iv) } x^6 + x^4 + x^2+1 , \text{ when } x = \frac{1+i}{\sqrt{2}}$
$\displaystyle \text{v) } 2x^4 + 5x^3 + 7x^2 - x +41 , \text{ when } x = - 2 - \sqrt{3}i$
$\displaystyle \text{i) } x = \frac{3-5i}{2}$
$\displaystyle \Rightarrow 2x = 3 - 5i$
$\displaystyle \Rightarrow ( 2x - 3)^2 = ( -5 i)^2$
$\displaystyle \Rightarrow 4x^2 + 9 - 12 x = - 25$
$\displaystyle \Rightarrow 4x^2 - 12 x + 34 = 0$
$\displaystyle \Rightarrow 2x^2- 6x + 17 = 0$
$\displaystyle \Rightarrow 2x^3 + 2 x^2 - 7x + 72 = 0$
Now $\displaystyle 2x^3+2x^2-7x+72$
$\displaystyle = x ( 2x^2 - 6x + 17) + 6x^2 - 17x + 2x^2 - 7x + 72$
$\displaystyle = x(0) + 8x^2 - 24x + 72$
$\displaystyle = 4 ( 2x^2 - 6x +17) + 4$
$\displaystyle = 4(0) + 4 = 4$
$\displaystyle \text{ii) } x = 3 + 2i$
$\displaystyle \Rightarrow (x-3)^2 = (2i)^2$
$\displaystyle \Rightarrow x^2 + 9 - 6x = - 4$
$\displaystyle \Rightarrow x^2 - 6x + 13 = 0$
Now, $\displaystyle x^4-4x^3+4x^2+8x+44$
$\displaystyle = x^2( x^2 - 6x + 13) + 6x^2 - 13 x^2 - 4x^3 + 4x^2 + 8x + 144$
$\displaystyle = x^2 (0) + 2x^3 - 9x^2 + 8x + 44$
$\displaystyle = 2x ( x^2 - 6x + 13) + 12x^2 - 26 x - 9x^2 +8x+ 44$
$\displaystyle = 2x (0) + 3x^2 - 18x +44$
$\displaystyle = 3 ( x^2 - 6x + 13) + 5$
$\displaystyle = 3 ( 0) + 5 = 5$
$\displaystyle \text{iii) } x = - 1 + i \sqrt{2}$
$\displaystyle \Rightarrow ( x + 1) ^2 = ( i \sqrt{2})^2$
$\displaystyle \Rightarrow x^2 + 1 + 2 x = - 2$
$\displaystyle \Rightarrow x^2 + 2x + 3 = 0$
Now $\displaystyle x^4+4x^3+6x^2+4x+9$
$\displaystyle = x^2 ( x^2 + 2x + 3) + 2x^3 + 3x^2 + 4x + 9$
$\displaystyle = x (0) + 2x ( x^2 + 2x + 3) - ( x^2 - 2x + 9)$
$\displaystyle = 2x (0) - ( x^2 + 2x + 3) + 3 + 9$
$\displaystyle = 12$
$\displaystyle \text{iv) } x= \frac{1+i}{\sqrt{2}}$
$\displaystyle (\sqrt{2} x)^2 = ( 1 + i)^2$
$\displaystyle \Rightarrow 2x^2 = 2i$
$\displaystyle \Rightarrow x^2 = i$
$\displaystyle \Rightarrow x^4 = - 1$
$\displaystyle \Rightarrow x^4 + 1 = 0$
Now, $\displaystyle x^6 + x^4 + x^2+1$
$\displaystyle = x^2 ( x^4 + 1) + ( x^4 +1 )$
$\displaystyle = x^2 ( 0) + (0) = 0$
$\displaystyle \text{v) } x = - 2 - \sqrt{3} i$
$\displaystyle x^2 = ( -2 - \sqrt{3} i)^2 = 1 + 4 \sqrt{3} i$
$\displaystyle x^3 = ( 1+4\sqrt{3} i)( - 2 - \sqrt{3} i) = 10 - 9 \sqrt{3} i$
$\displaystyle x^4 = ( 10 - 9 \sqrt{3} i) ( - 2 - \sqrt{3} i) = - 20 + 18 \sqrt{3} i - 10 \sqrt{3} i + 27 i^2 = -47 + 8 \sqrt{3} i$
Now, $\displaystyle 2x^4 + 5x^3 + 7x^2 - x +41$
$\displaystyle = 2 ( -47 + 8\sqrt{3} i) + 5 ( 10 - 9 \sqrt{3} i) + 7 ( 1 + 4 \sqrt{3} i) - ( -2 - \sqrt{3} i) + 41$
$\displaystyle = - 94 + 16 \sqrt{3} i + 50 - 45 \sqrt{3} i + 7 + 28 \sqrt{3} i + 2 + \sqrt{3} i + 41$
$\displaystyle = ( - 94 + 50 + 7 + 2 + 41) + ( 16 \sqrt{3} i - 45 \sqrt{3} i + 28 \sqrt{3} i + \sqrt{3} i)$
$\displaystyle = 6+0 =6$
$\displaystyle \\$
$\displaystyle \text{Question 17: For a positive integer, find the value of } (1-i)^n \Big( 1 - \frac{1}{i} \Big)^n$
$\displaystyle (1-i)^n \big( 1 - \frac{1}{i} \big)^n = \Big[ (1-i) \big( 1 - \frac{1}{i} \big) \Big]^n = \Big[ \frac{(1-i)(i-1)}{i} \Big]^n = \Big[ \frac{i+1-1+i}{i} \Big]^n = 2^n$
$\displaystyle \\$
Question 18: $\displaystyle \text{If } (1 +i) z = (1-i) \overline{z}$ , then show that $\displaystyle z = -i \overline{z}$
$\displaystyle (1+i) z = ( 1- i) \overline{z}$
$\displaystyle \Rightarrow z = \Big( \frac{1-i}{1+i} \Big) \overline{z}$
$\displaystyle \Rightarrow z = \Big( \frac{1-i}{1+i} \times \frac{1-i}{1-i} \Big) \overline{z}$
$\displaystyle \Rightarrow z = \Big( \frac{-2i}{2} \Big) \overline{z}$
$\displaystyle \Rightarrow z = -i \overline{z}$
Hence proved.
$\displaystyle \\$
Question 19: Solve the system of equations $\displaystyle Re(z^2) = 0 , |z|= 2$.
$\displaystyle Re(z^2) = 0 , |z| = 2$
Let $\displaystyle z = x + iy$
$\displaystyle z^2 = ( x + iy)^2 = ( x^2 - y^2)+ i ( 2xy)$
$\displaystyle Re(z^2) = 0 \Rightarrow x^2 - y^2 = 0$ … … … … … i)
$\displaystyle |z| = 2$
$\displaystyle \Rightarrow \sqrt{x^2+y^2} = 2$
$\displaystyle \Rightarrow x^2 + y^2 = 4$ … … … … … ii)
Adding i) and ii) we get
$\displaystyle 2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$
$\displaystyle \therefore y = \pm \sqrt{2}$
Hence $\displaystyle x + iy = \pm \sqrt{2} \pm i \sqrt{2}$
For positive sign for $\displaystyle x : z = \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( 1 \pm i)$
For negative sign for $\displaystyle x : z = - \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( - 1 \pm i )$
$\displaystyle \\$
$\displaystyle \text{Question 20: If } \frac{z-1}{z+1} \text{ is purely imaginary number } (z \neq -1) , \text{ find the value of } |z|$
Let $\displaystyle z = x + iy$
$\displaystyle \frac{z-1}{z+1} = \frac{x + iy-1}{x + iy+1} = \frac{(x-1) + iy}{(x +1) + iy}$
$\displaystyle = \frac{(x-1) + iy}{(x +1) + iy} \times \frac{(x+1) - iy}{(x +1) - iy}$
$\displaystyle = \frac{[(x-1)+iy][(x+1)-iy]}{(x+1)^2 + y^2}$
$\displaystyle = \frac{x^2 -1 + i ( yx+y) - i ( xy - y) + y^2}{(x+1)^2 + y^2}$
$\displaystyle = \frac{(x^2 + y^2 -1) + i ( yx + y - xy + y)}{(x+1)^2 + y^2}$
$\displaystyle = \frac{(x^2 + y^2 -1) }{(x+1)^2 + y^2} + i \frac{ 2y}{(x+1)^2 + y^2}$
If it is purely imaginary, $\displaystyle Re(z) = 0$
$\displaystyle \Rightarrow x^2 + y^2 - 1 = 0$
$\displaystyle \Rightarrow x^2 + y^2 = 1$
$\displaystyle \Rightarrow \sqrt{x^2 + y^2} = 1$
$\displaystyle \Rightarrow |z| = 1$
$\displaystyle \\$
Question 21: $\displaystyle \text{If } z_1$ is a complex number other than $\displaystyle - 1$ such that $\displaystyle |z_1|= 1$ and $\displaystyle z_2 = \frac{z_1-1}{z_1+1} \text{ then show that the real parts of } z_2 \text{ is zero. }$
Let $\displaystyle z_1 = x_1 + i y_2$ and $\displaystyle z_2 = x_ 2 + i y_2$
$\displaystyle |z| = 1 \Rightarrow {x_1}^2 + {y_1}^2 = 1$ … … … … … i)
$\displaystyle z_2 = \frac{z_1 - 1}{z_1 + 1}$
$\displaystyle \Rightarrow z_2 = \frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}$
$\displaystyle = \frac{(x_1-1)+iy_1}{(x_1+1)+iy_1} \times \frac{(x_1+1)- iy_1}{(x_1+1)-iy_1}$
$\displaystyle = \frac{ ({x_1}^2-1) + i ( y_1x_1+y_1) - i ( y_1x_1-y_1)+{y_1}^2}{(x_1+1)^2+{y_1}^2}$
$\displaystyle = \frac{( {x_1}^2+{y_1}^2-1)+i ( 2y_1)}{(x_1+1)^2+{y_1}^2}$
$\displaystyle =0 + i \frac{2y_1}{(x_1+1)^2+{y_1}^2}$
Since there is no real part, $\displaystyle z_2$ is purely an imaginary numbers
$\displaystyle \Rightarrow x_2 = 0$
$\displaystyle \\$
Question 22: $\displaystyle \text{If } |z+1|= z + 2 (1+i)$ , find $\displaystyle z$
Let $\displaystyle z = x + i y$
$\displaystyle |z+1| = z + 2 ( 1 + i)$
$\displaystyle \Rightarrow |x+1+iy| = ( x+iy) + 2 ( 1+i)$
$\displaystyle \Rightarrow \sqrt{(x+1)^2 + y^2} = (x+2) + i ( y + 2)$
Comparing, $\displaystyle y + 2 = 0 \Rightarrow y = - 2$
$\displaystyle \therefore \sqrt{ (x+1)^2+y^2 } = (x+2)$
$\displaystyle \Rightarrow (x+1)^2 + 4 = ( x+2)^2$
$\displaystyle \Rightarrow x^2 + 1 + 2x + 4 = ( x+2)^2$
$\displaystyle \Rightarrow 5-4 = 2x$
$\displaystyle \Rightarrow x = \frac{1}{2}$
$\displaystyle \therefore z = \frac{1}{2} + i (-2)$
$\displaystyle \\$
Question 23: Solve the equation $\displaystyle |z| = z+1+2i$
Let $\displaystyle z = x + iy$
$\displaystyle |z| = z + 1 + 2i$
$\displaystyle |x+iy| = x + iy + 1 + 2i$
$\displaystyle \sqrt{x^2 + y^2} = (x+1) + i ( y+2)$
Comparing, $\displaystyle y+2 = 0 \Rightarrow y = -2$
and $\displaystyle \sqrt{x^2 + y^2} = (x+1)$
$\displaystyle \Rightarrow x^2 + y^2 = x^2 + 1 + 2x$
$\displaystyle \Rightarrow 2x = y^2 - 1 = ( -2)^2 - 1 = 3$
$\displaystyle \Rightarrow x = \frac{3}{2}$
$\displaystyle \therefore z = \frac{3}{2} - 2i$
$\displaystyle \\$
Question 24: What is the smallest positive integer $\displaystyle n$ for which $\displaystyle (1+i)^{2n}= ( 1- i)^{2n} \ ?$
$\displaystyle (1+i)^{2n}= ( 1- i)^{2n}$
$\displaystyle \Rightarrow [(1+i)^{2}]^n= [( 1- i)^{2}]^n$
$\displaystyle \Rightarrow (2i)^n = ( -2i)^n$
$\displaystyle \Rightarrow (2i)^n = ( -1)^n (2i)^n$
$\displaystyle \Rightarrow ( -1 )^n = 1$
$\displaystyle \therefore n$ is a multiple of $\displaystyle 2$.
Hence the smallest positive number $\displaystyle n$ for which $\displaystyle (1+i)^{2n} = (1-i)^{2n}$ is $\displaystyle 2$
$\displaystyle \\$
Question 25: $\displaystyle \text{If } z_1, z_2, z_3$, are complex numbers such that, $\displaystyle |z_1|= |z_2|=|z_3|= \Big| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \Big| = 1 , \text{ then find the value of } |z_1+z_2+z_3|$.
$\displaystyle |z_1+z_2+z_3| = \Big| z_1 \frac{\overline{z_1}}{z_1} + z_2 \frac{\overline{z_2}}{z_2} + z_3 \frac{\overline{z_3}}{z_3} \Big|$
$\displaystyle = \Big| \frac{| z_1|^2}{\overline{z_1}} + \frac{| z_2|^2}{\overline{z_2}} + \frac{| z_3|^2}{\overline{z_3}} \Big|$
$\displaystyle = \Big| \frac{1}{\overline{z_1}} + \frac{1}{\overline{z_2}} + \frac{1}{\overline{z_3}} \Big|$
$\displaystyle = 1$
$\displaystyle \therefore |z_1+z_2+z_3| = 1$
$\displaystyle \\$
Question 26: Find the number of solutions of $\displaystyle z^2 + |z|^2=0$
Let $\displaystyle z = x + iy \Rightarrow |z| = \sqrt{x^2 + y^2}$
Now $\displaystyle z^2 + |z|^2 = 0$
$\displaystyle \Rightarrow (x + iy)^2 + (\sqrt{x^2 + y^2})^2 = 0$
$\displaystyle \Rightarrow (x^2 - y^2) + 2i xy + (x^2 + y^2) = 0$
$\displaystyle \Rightarrow 2x^2 + 2ixy = 0$
$\displaystyle \Rightarrow 2x ( x+iy) = 0$
Therefore $\displaystyle x = 0 \text{ or } x + iy = 0 \Rightarrow z = 0$
For $\displaystyle x = 0, z = iy$
Thus there are infinite many solution of the form $\displaystyle z = 0 + iy , y \in R$ |
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Radical Numbers are used to represent Square root of a number. For Example: 3 is the square root of number 9. We will represent it as √9 = 3. Here '√' is the radical symbol and number under radical symbol is called as radicand. We can perform arithmetic operations like addition, subtraction, multiplication etc. Let us see how to add and subtract Radicals:
Addition of radicals: For addition of radicals we have to ensure that number under radical should be equal; if it is not equal then we will make it equal. Let us take example to understand it.
Assume that we want to add √3 and 4√3. Here numbers under radical sign are equal so we can add them like simple arithmetic numbers:
√3 + 4√3 = 5√3,
Now we will see subtraction of radicals. Say we want to subtract 3√2 from 6√2. Here radicands of both numbers are equal so we can perform subtraction as:
6√2 - 3√2 = 3√2.
If radicands are different in both numbers then make them same.
Now let us take another example to better understand this concept. Assume that we have to subtract √12 from √27. Here numbers under radical are not equal so we cannot add them. We have to make radicands equal. So we can write √12 as √2 * 2 * 3 and we can write √27 as √3 * 3 * 3. Now radicals are:
3√3 - 2√3,
Here radicands of both numbers are equal, so we can add them as:
3√3 - 2√3 = 1√3.
Just like addition, subtraction follows same property that radicands of both numbers should be same. |
Surface area of a cube is the number of square units that will exactly cover the surface of a cube. The total surface area of a cube is the sum of the areas of its surfaces. Cube is the special case of the cuboid whose faces are square of the same size and whose length, width and height are equal. The surface area of a cube is six times the surface area of one side of the cube. It unit of measurement is 'square units'.
## Surface Area of a Cube Formula
The surface area of a cube is six times the surface area of one side of the cube. Find the surface area of one side of a cube and multiply by six. It unit of measurement is m2, cm2, inches2 etc.
Formulas:
Surface Area of a Cube = 6S2
Lateral Surface Area of a Cube = 4S2
Where 's' is the length of any edge of the cube.
## How to Find Surface Area of a Cube
The surface area is the area on the outside of a three-dimensional shape. A cube has all edges the same length, means each of the cube's six faces is a square. The total surface area is therefore six times the area of one face. Similarly, the lateral surface area of a cube is the sum of the area of the lateral faces of a cube.
Steps for Finding the Surface Area of a Cube:
1. Measure the length of any one side of the cube, because all dimensions are equal.
2. Find the surface area of one side of a cube and multiply the result by 6. (Surface Area = 6 (Side)2)
3. Write answer in the proper square unit of measurement.
Example 1:
Find the surface area of a lunch box whose shape is of a cube of edge 13 cm.
Solution:
Edge(side) of cubic lunch box = 13
Surface area of a cube = 6 side2 .
=> Surface area of a cube = 6 (132)
= 6 * 13 * 13
= 1014
Hence surface area of a lunch box is 1014 sq. cm.
Example 2:
Find the cost of painting a wooden box which is in the shape of a cube of side 10 cm at the rate of 25 cents per square cm.(Write answer in dollars)
Solution:
Step 1:
Given
Side of a cubic wooden box = 10
Step 2:
To find the cost of painting, firstly we have to find the surface area of the wooden box.
Surface area of a cube = 6 side2 .
=> Surface area of the wooden box = 6(10)2
= 6 * 10 * 10
= 600
So the surface area of the wooden box is 600 sq. cm.
Step 3:
Since cost to paint per sq. unit is 25 cents.
=> The cost of painting a wooden box = 600 * 25
= 15000 cents.
Step 4:
1 dollar = 100 cents
or 1 cent = 0.01 dollar
=> 15000 * 0.01 = 150 dollars
Hence the cost of painting a wooden box is 150 dollars. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 8 Go to the latest version.
Chapter 1: Using Algebra
Difficulty Level: Basic Created by: CK-12
Introduction
Here you will get a preview of a variety of algebra topics. First, you will learn how to interpret data and represent data with bar graphs and histograms. Next, you will learn all about algebraic expressions, and how variables can help you to represent real-world problems with expressions. You will also be introduced to the idea of solving an equation, and the difference between an expression and an equation. You will see formulas as examples of equations that can be used to solve problems. Finally, you will learn how to make a problem solving plan.
Summary
You started by thinking about different kinds of data, and learning how to represent this data in either a bar graph or a histogram. You also learned how to interpret data that has been represented in a graph.
Next, you learned about expressions. You learned to evaluate expressions by replacing a variable with a number and then using the order of operations to find your answer. You also learned how to write expressions to represent different situations.
You then focused on equations, and learned that the difference between an expression and an equation is an equals sign. You learned to solve basic equations mentally by asking yourself what number the variable would have to be in order to make the equation true. You learned that you can write equations to represent real-world situations. Formulas are a good example of equations, and you used formulas to calculate perimeter, area, and distance/rate/time.
Finally, you started to think about problem solving. In math, you will often be presented with a problem that you don't know how to solve right away. It's important to have some problem solving strategies ready so that you don't just give up when presented with a difficult problem. Problem solving with a plan also helps you to communicate to other people what you are trying to do, the strategies you have tried, and why.
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How do you convert $0.3\overline{4}$ to a fraction?
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Hint: We first describe the concept of representation of non-terminating recurring decimals. We describe the process of converting them from decimal to fractions. Following those steps, we convert $0.3\overline{4}$ to a fraction. To understand the process better we also take an example of another complicated form and change that decimal into fraction.
Complete step-by-step solution:
The given decimal number is a representation of non-terminating recurring decimals. These types of decimal numbers are rational numbers. They can be expressed in the form of $\dfrac{p}{q}$.
The expansion of the given decimal is $0.3\overline{4}=0.343434........$
The process of converting into fraction form is below mentioned.
Step: 1
We have to find the numerator part of the fraction where we take the difference between the whole number without decimal point and the non-recurring part of the number without decimal point.
Step: 2
We have to find the denominator part of the fraction where we take the digits of 9 and 0. The number of digits of 9 is equal to the number of recurring digits in the given number after decimal and the number of digits of 0 is equal to the number of non-recurring digits in the given number after decimal. The 9s come first and the zeroes come after that.
Now we find the fraction form of $0.3\overline{4}$ applying the rules.
$0.3\overline{4}=\dfrac{34-3}{90}=\dfrac{31}{90}$.
The fractional value of $0.3\overline{4}$ is $\dfrac{31}{90}$.
Note: To understand the process better we take another example of $2.45\overline{74}$.
The fractional form of the decimal form will be $2.45\overline{74}=\dfrac{24574-245}{9900}=\dfrac{24329}{9900}$.
There are two recurring and two non-recurring digits in that number after decimal. That’s why we used two 9s and two 0s in the denominator. |
# Is 0.999… (repeating) equal to 1?
by Jakub Marian
Tip: See my list of the Most Common Mistakes in English. It will teach you how to avoid mistakes with commas, prepositions, irregular verbs, and much more.
The short answer is yes. $0.\bar 9$ (zero point 9 repeating) is exactly 1. However, a lot of people find this result counter-intuitive (people often feel that $0.\bar 9$ should be slightly less than 1), but this feeling stems from a misunderstanding of what $0.\bar 9$ means.
To satiate those looking for a quick and dirty argument, we can do the following simple calculation:
\begin{align*} x &= 0.999..{}.\\ 10x &= 9.999..{}.\\ 9x &= 10x - x = 9.999..{}.-0.999..{}. = 9\\ x &= 1 \end{align*}
This is not, however, a mathematical proof, because not all of the steps are obvious without further looking at the definition of the decimal notation. This will require a bit of preparation.
## The limit
You are probably familiar with the meaning of the decimal notation when the fractional part contains only a finite number of digits. For example:
$$32.58 = 3⋅10+2⋅1+5⋅÷{1}{10}+8⋅÷{1}{100}$$
To understand the meaning of the fractional part when the digits go on indefinitely, you first need to understand, at least purely intuitively, the notion of limit.
When you have a sequence of numbers, say $a_1 = 1$, $a_2 = ÷{1}{2}$, $a_3 = ÷{1}{3}$, …, $a_n = ÷{1}{n}$, you can compute what we call the limit of $a_n$ (at $∞$). The name sounds terrifying, but the concept is in fact really simple.
As you can see (or quickly check using a calculator), $÷{1}{n}$ gets closer and closer to 0 as $n$ increases, for example, $÷{1}{100} = 0.01$, $÷{1}{1000} = 0.001$, and so on. If a sequence gets arbitrarily close to a number, we say that the limit of the sequence is that number, and we write:
$$\lim_{n→∞} a_n = \text{the number to which a_n get arbitrarily close}$$
For example, in the case of $a_n = ÷{1}{n}$, we would write
$$\lim_{n→∞} a_n = 0$$ since $÷{1}{n}$ gets arbitrarily close to $0$ for large $n$. Now we are ready to define the decimal notation for never-ending sequences of digits.
## Real numbers and the decimal notation
It may come as a great surprise to those who don’t know any higher mathematics, but the real numbers are not defined to be “numbers like $0.53$, $3.1415$…, $86.51$”. In fact, we define them in three steps:
1) Define the integers (whole numbers), i.e. numbers like $1$, $5$, $-8$.
2) Define fractions of integers, i.e. numbers like $÷{5}{8}$, $÷{314}{100}$.
3) Add abstract numbers that fill in “gaps” between fractions.
The mathematical definition is somewhat more technical, but it follows the process outlined above. There is absolutely no mention of the decimal point anywhere in the process; we just have a bunch of numbers, such as $1$, $÷{3}{7}$, or $π$, and then, after we have defined the numbers, we define the decimal point.
This is simple if the decimal part is finite (we already did this above), but what if it is infinite? We use the limit! For example, to get the value of $x = 0.999$…, we define a sequence of the form $x_1 = 0.9$, $x_2 = 0.99$, $x_3 = 0.999$, and so on. Obviously, as $n$ gets larger, $x_n$ gets closer to the real value of $0.999$…, so we define:
$$0.\bar 9 = \lim_{n→∞} x_n$$
The only question that remains to be answered is: What is the limit of $x_n$? As $n$ gets larger, $x_n$ gets arbitrarily close to $1$, so we conclude that, by definition:
$$0.\bar 9 = \lim_{n→∞} x_n = 1$$
The phrase “by definition” is really important here. Symbols like “1.234” are not real numbers themselves; these are just just a way to label real numbers, just like we label humans with words like “Peter” or “Laura”it wouldn’t really make sense to say that a person is a series of letters, would it… And just like “Peter” may be also called “Pete”, some numbers can be written in two different ways in our notation: $0.999$… is just a different way of writing $1$.
By the way, I have written several educational ebooks. If you get a copy, you can learn new things and support this website at the same time—why don’t you check them out? |
New Zealand
Level 6 - NCEA Level 1
# Proportional Relationships as Equations
Lesson
We've already learnt about proportional relationships, where two quantities vary in such a way that one is a constant multiple of the other. In other words, they always vary by the same constant.
Proportional relationships can also be written as linear equations.
Remember!
Proportional relationships can be written generally in the form:
$y=kx$y=kx
where $k$k is the constant of proportionality
Notice how this general form of proportional relationships looks very similar to the $y=mx+b$y=mx+b form we use to solve equations. However, one of the features of proportional relationships is that, when graphed, they pass through the origin $\left(0,0\right)$(0,0), which is why there is no $b$b term.
So, we can solve proportional relationships like equations to make judgments about the relationship between two variables.
Let's look through some examples now.
#### Examples
##### Question 1
Frank serves $2$2 cups of coffee every $4$4 minutes.
Using $y$y for the number of cups of coffee and $x$x for the amount of minutes that have passed, write an equation that represents this proportional relationship.
##### Question 2
The amount of white and red paint needed to make 'flamingo pink' is shown in the graph.
a) Let $x$x represent the amount of white paint and y represent the amount of red paint needed. What is the equation of this line?
b) What does the equation of the line tell you?
##### Question 3
James wants to buy cereal, and sees that a $500$500 gram box is priced at $\$6.05$$6.05. a) What is the unit price of the cereal per 100100 grams? Give your answer to the nearest cent. b) Form an equation relating yy (the cost of the cereal) to xx (the weight of the cereal box in grams). c) James sees that a 750750 gram box of the same cereal is priced at \8.18$$8.18. Opting for the larger box, what is the saving per $100$100 grams? Give your answer to the nearest cent.
### Outcomes
#### NA6-1
Apply direct and inverse relationships with linear proportions
#### 91026
Apply numeric reasoning in solving problems |
Elementary arithmetic
Elementary arithmetic is a branch of mathematics involving basic numerical operations, namely addition, subtraction, multiplication, and division. Due to its low level of abstraction, broad range of application, and position as the foundation of all mathematics, elementary arithmetic is commonly the first branch of mathematics taught to school children.
Digits
Digits are used to represent the value of numbers in a numeral system. The most commonly used digits are the Arabic numerals (0 to 9 or base 10). The Hindu-Arabic numeral system is the most commonly used numeral system, being a positional notation system used to represent numbers using these digits.
Successor function and size
In elementary arithmetic, the successor of a natural number (including zero) is the result of adding a value of one to that number. The predecessor of a natural number (excluding zero) is the result obtained by subtracting a value of one from that number. For example, the successor of zero is one and the predecessor of eleven is ten (${\displaystyle 0+1=1}$ and ${\displaystyle 11-1=10}$). Every natural number has a successor, and all natural numbers (except zero) have a predecessor.
If one number is greater than (${\displaystyle >}$) another number, then the latter is less than (${\displaystyle <}$) the former. For example, three is less than eight (${\displaystyle 3<8}$), and eight is greater than three (${\displaystyle 8>3}$).
Counting
Counting involves assigning a natural number to each object in a set, starting with one for the first object and increasing by one for each subsequent object. The number of objects in the set is the count, which is equal to the highest natural number assigned to an object in the set. This count is also known as the cardinality of the set.
Counting can also be the process of tallying using tally marks, drawing a mark for each object in a set.
In more advanced mathematics, the process of counting can be thought of as constructing a one-to-one correspondence (or bijection), between the elements of a finite set and the set ${\displaystyle \{1,2,3,...,n\}}$, where ${\displaystyle n}$ is a natural number, and the size of the set is ${\displaystyle n}$.
Informally, two sets have the same cardinality if both of the sets' elements can be matched with one-to-one correspondence. (Example: 4 apples and 4 bananas have the same cardinality as each apple can be matched to each banana without any left-over.)
Addition is a mathematical operation that combines two or more numbers, called addends or summands, to produce a final number, called the sum. The addition of two numbers is expressed using the plus sign "+" and is performed according to the following rules:
• The sum of two numbers is equal to the number obtained by adding their individual values.
• The order in which the addends are added does not affect the sum. This is known as the commutative property of addition.
• The sum of two numbers is unique, meaning that there is only one correct answer for the sum of any given pair of numbers.
• Addition has an inverse operation, called subtraction, which can be used to find the difference between two numbers.
Addition is used in a variety of contexts, including comparing quantities, joining quantities, and measuring. When the sum of a pair of digits results in a two-digit number, the "tens" digit is referred to as the "carry digit" in the addition algorithm. In elementary arithmetic, students typically learn to add whole numbers, and may also learn about topics such as negative numbers and fractions.
Example
Adding the numbers 653 and 274, starting with the ones column (bolded in the figure below), the sum of three and four is seven.
Hundreds Tens Ones 6 5 3 + 2 7 4 7
Moving on to the tens column (bolded below), the sum of 5 tens (50) and 7 tens (70) is 12 tens (120). The tens-digit from 120 (2, in this case) is written under the tens column, while the hundreds-digit (1) is written above the hundreds column as a carry digit.
Hundreds Tens Ones 1 6 5 3 + 2 7 4 2 7
In the hundreds column, the sum of 6 hundreds (600) and 2 hundreds (200) is 8 hundreds (800), but a carry digit of 1 hundred (100) is present and must also be added to the total, resulting in 9 hundreds (900).
Hundreds Tens Ones 1 6 5 3 + 2 7 4 9 2 7
The result:
${\displaystyle 653+274=927}$
Subtraction
Subtraction is used to evaluate the difference between two numbers, where the minuend is the number being subtracted from, and the subtrahend is the number being subtracted. It is represented using the minus sign (-).
Subtraction is not commutative, meaning the order of the numbers in the operation can change the result. 3 - 5 is not the same as 5 - 3. In elementary arithmetic, the minuend is always larger than the subtrahend to produce a positive result.
Subtraction is also used to separate, combine (e.g. find the cardinal of a subset of the set we are interested in), and find quantities in other contexts. For example, "Tom has 8 apples. He gives away 3 apples. How many is he left with?" represents separation, while "Tom has 8 apples. Three of the apples are green, and the rest are red. How many are red?" represents combination. In some cases, subtraction can also be used to find the total number of objects in a group, as in "Tom had some apples. Jane gave him 3 more apples, so now he has 8 apples. How many did he start with?"
There are several methods to accomplish subtraction. The traditional mathematics method teaches elementary school students to subtract using methods suitable for hand calculation. Reform mathematics is distinguished generally by the lack of preference for any specific technique, replaced by guiding 2nd-grade students to invent their own methods of computation, such as using properties of negative numbers in the case of TERC.
American schools currently teach a method of subtraction using borrowing[citation needed]. However, a method of borrowing had been known and published in prior textbooks. Crutches are the invention of William A. Brownell, who used them in a study in November 1937. In the method of borrowing, a subtraction problem such as 86 - 39 can be solved by borrowing a 10 from the tens place to add to the ones place in order to facilitate the subtraction. For example, subtracting 9 from 6 involves borrowing a 10 from the tens place, making the problem into 70 + 16 - 39. This is indicated by crossing out the 8, writing a 7 above it, and writing a 1 above the 6. These markings are called "crutches."
The Austrian method, also known as the additions method, is taught in certain European countries and employed by some American people from previous generations. In contrast to making use of crutches, there is no borrowing in this method. There are also crutches that vary according to country. The method of addition involves augmenting the subtrahend, rather than reducing the minuend, as in the borrowing method. This transforms the problem into (80 + 16) - (39 + 10). A small 1 is marked below the subtrahend digit as a reminder.
Example
Subtracting the numbers 792 and 308, starting with the ones-column, 2 is smaller than 8, borrowing 10 from 90, making 90 become 80. Adding this 10 to 2, changes the problem to 12 - 8, which is 4.
Hundreds Tens Ones 8 12 7 9 2 − 3 0 8 4
Taking 10 from 90, it is now 80. The difference between 80 and 0, is 80.
Hundreds Tens Ones 8 12 7 9 2 − 3 0 8 8 4
The difference between 700 and 300 is 400.
Hundreds Tens Ones 8 12 7 9 2 − 3 0 8 4 8 4
The result:
${\displaystyle 792-308=484}$
Multiplication
Multiplication is a mathematical operation of repeated addition. When two numbers are multiplied, the resulting value is a product. The numbers being multiplied are called multiplicands and multipliers and are altogether known as factors. For example, if there are five bags, each containing three apples, and the apples from all five bags are placed into an empty bag, the empty bag will contain 15 apples. This can be expressed as "five times three equals fifteen" "five times three is fifteen" or "fifteen is the product of five and three".
Multiplication is represented using the multiplication sign (×), the asterisk (*), parentheses (), or a dot (⋅). Therefore, the statement "five times three equals fifteen" can be written as "${\displaystyle 5\times 3=15}$", "${\displaystyle 5\ast 3=15}$", "${\displaystyle (5)(3)=15}$", or "${\displaystyle 5\cdot 3=15}$". The asterisk notation is most commonly used in computer programming languages. In algebra, the multiplication symbol may be omitted; for example, ${\displaystyle xy}$ represents ${\displaystyle x\times y}$.
The order in which two numbers are multiplied does not affect the result. This is known as the commutative property of multiplication. The grouping of three or more numbers in parentheses also does not affect the result. This is known as the associative property of multiplication.
In the multiplication algorithm, the "tens" digit of the product of a pair of digits is referred to as the "carry digit". To multiply a pair of digits using a table, one must locate the intersection of the row of the first digit and the column of the second digit, which will contain the product of the two digits. Most pairs of digits result in two-digit numbers.
Example of multiplication for a single-digit factor
Multiplying 729 and 3, starting on the ones-column, the product of 9 and 3 is 27. 7 is written under the ones column and 2 is written above the tens column as a carry digit.
Hundreds Tens Ones 2 7 2 9 × 3 7
The product of 2 and 3 is 6, and the carry digit adds 2 to 6, so 8 is written under the tens column.
Hundreds Tens Ones 7 2 9 × 3 8 7
The product of 7 and 3 is 21, and since this is the last digit, 2 will not be written as a carry digit, but instead beside 1.
Hundreds Tens Ones 7 2 9 × 3 2 1 8 7
The result:
${\displaystyle 3\times 729=2187}$
Example of multiplication for multiple-digit factors
Multiplying 789 and 345, starting with the ones-column, the product of 789 and 5 is 3945.
7 8 9 × 3 4 5 3 9 4 5
4 is in the tens digit. The multiplier is 40, not 4. The product of 789 and 40 is 31560.
7 8 9 × 3 4 5 3 9 4 5 3 1 5 6 0
3 is in the hundreds-digit. The multiplier is 300. The product of 789 and 300 is 236700.
7 8 9 × 3 4 5 3 9 4 5 3 1 5 6 0 2 3 6 7 0 0
7 8 9 × 3 4 5 3 9 4 5 3 1 5 6 0 + 2 3 6 7 0 0 2 7 2 2 0 5
The result: ${\displaystyle 789\times 345=272205}$
Division
Division is an arithmetic operation that is the inverse of multiplication.
Specifically, given a number a and a non-zero number b, if another number c times b equals a, that is ${\displaystyle c\times b=a}$, then a divided by b equals c.
That is: ${\displaystyle {\frac {a}{b}}=c}$. For instance, ${\displaystyle {\frac {6}{3}}=2}$.
The number a is called the dividend, b the divisor, and c the quotient. Division by zero is considered impossible at an elementary arithmetic level, and is generally disregarded.
Division can be shown by placing the dividend over the divisor with a horizontal line, also called a vinculum, between them. For example, a divided by b is written as:
${\displaystyle {\frac {a}{b}}}$
This can be read verbally as "a divided by b" or "a over b".
Another way to express division all on one line is to write the dividend, then a slash, then the divisor, as follows:
${\displaystyle a/b}$
This is the usual way to specify division in most computer programming languages.
A handwritten or typographical variation uses a solidus (fraction slash) but elevates the dividend and lowers the divisor:
ab
Any of these forms can be used to display a fraction. A common fraction is a division expression where both dividend and divisor are numbers (although typically called the numerator and denominator), and there is no implication that the division needs to be evaluated further.
A more basic way to show division is to use the obelus (or division sign) in this manner:
${\displaystyle a\div b.}$
In some non-English-speaking cultures, "a divided by b" is written a : b. However, in English usage, the colon is restricted to the concept of ratios ("a is to b").
Two numbers can be divided on paper using the method of long division. An abbreviated version of long division, short division, can be used for smaller divisors as well.
A less systematic method involves the concept of chunking, involving subtracting more multiples from the partial remainder at each stage.
To divide by a fraction, one can simply multiply by the reciprocal (reversing the position of the top and bottom parts) of that fraction. For example:
${\displaystyle \textstyle {5\div {1 \over 2}=5\times {2 \over 1}=5\times 2=10}}$
${\displaystyle \textstyle {{2 \over 3}\div {2 \over 5}={2 \over 3}\times {5 \over 2}={10 \over 6}={5 \over 3}}}$
Example
Dividing 272 and 8, starting with the hundreds-digit, 2 is not divisible by 8, adding 20 to 7, to get 27. The largest number that the divisor 8 can be multiplied by without exceeding 27 is 3, so the digit 3 is written under the tens-column to start constructing the quotient. Subtracting 24 (product of 3 and 8) from 27 gives 3 as remainder.
2 7 2 ÷ 8 3
8 is bigger than remainder 3. Going to the ones-digit to continue the division, where the number is 2. Adding 30 and 2 to get 32, which is divisible by 8, and the quotient of 32 and 8 is 4. 4 is written under the ones-column.
2 7 2 ÷ 8 3 4
The result:
${\displaystyle 272\div 8=34}$
Educational standards
Elementary arithmetic is typically taught at the primary or secondary school levels and is governed by local educational standards. In the United States and Canada, there has been debate about the content and methods used to teach elementary arithmetic.[citation needed] One issue has been the use of calculators versus manual computation, with some arguing that calculator use should be limited to promote mental arithmetic skills. Another debate has centered on the distinction between traditional and reform mathematics, with traditional methods often focusing more on basic computation skills and reform methods placing a greater emphasis on higher-level mathematical concepts such as algebra, statistics, and problem-solving.
In the United States, the 1989 National Council of Teachers of Mathematics (NCTM) standards led to a shift in elementary school curricula that de-emphasized or omitted certain topics traditionally considered to be part of elementary arithmetic, in favor of a greater focus on college-level concepts such as algebra and statistics. This shift has been controversial, with some arguing that it has resulted in a lack of emphasis on basic computation skills that are important for success in later math classes. |
# Volume of Prism
Contributed by:
The formula for the volume of a prism is V=Bh, where B is the base area and h is the height. The base of the prism is a rectangle.
1. Volume: Prisms and Cylinders
Lesson 10-7 p.538
2. The volume of a three-dimensional figure
is the amount that fills the figure. The
volume is given in cubic units.
3. Consider this prism.
If we cut it into cubic units, it would
look like this:
2 in
3 in
4 in
4. Consider this prism.
If we cut it into cubic units, it would
look like this:
2 in
3 in
4 in
5. Consider this prism.
The common formula for finding
Volume of a rectangular prism is
V = lwh
where l is the length, w is the width
and h is the height.
2 in
3 in
4 in
6. Consider this prism.
In this example,
V = lwh = 4 (3) (2) = 24 cu. in.
2 in
3 in
4 in
7. Another way to look at this problem is take the area of the
base (meaning bottom of the figure) and multiply it by the
height. Solving in this way lets us apply the method to
other shapes as well.
Area of the base = l x w
therefore, area of the base x height
is the volume.
2 in
3 in
4 in
8. Let’s try this method with a triangular prism
6 ft Volume = area of the base x height
The base is the triangle. A = bh
2
9 ft
A = (8) (6)
8 ft
2
A = 24 sq. ft.
Volume = area of the base x height
= 24 x 9
= 216 cu. Ft.
9. Try This
Find the volume:
4 in
5 in. 4 ft
6 in. 6 ft
3 ft
10. Try This
Find the volume:
4 in
5 in. 4 ft
6 in. 6 ft
3 ft
120 cu. in.
11. Try This
Find the volume:
4 in
5 in. 4 ft
6 in. 6 ft
3 ft
120 cu. in. 36 cu. ft.
12. This works for finding the volume of cylinders
also.
3m Volume = area of the base x height.
Area of base = r 2
5m = 3.14 x 32
= 28.26 sq. m
Volume = area of base x height.
= 28.26 x 5
= 141.3 cu. meters
13. Try This
Find the volume of the cylinder:
11 ft
5 ft.
14. Try This
Find the volume of the cylinder:
11 ft
5 ft. 1,899.7 cu. ft.
15. P.540 #1-16
16. Bellwork #
HW, red pen, book
on desk.
17. Workbook pp.86 & 87
Workbook p.89
18. Workbook p.89 |
## Thomas' Calculus 13th Edition
For a product, we have at least two factors (a product of one term makes no sense), so we prove that the formula is true for any $n\geq 2$ $1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=2$. We are given by the problem text that $f(x_{1}\cdot x_{2})=f(x_{1})+f(x_{2})$ $\qquad$ Yes, it is true when $n=2.$ $2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k$, $f(x_{1}\cdot x_{2}\cdots\cdot x_{k})=f(x_{1})+f(x_{2})+\ldots+f(x_{k})$ (we call this the step hypothesis) For $n=k+1$, the LHS equals $f(x_{1}\cdot x_{2}\cdots\cdot x_{k}\cdot x_{k+1})$, which we rewrite as $LHS= f[(x_{1}\cdot x_{2}\cdots\cdot x_{k})\cdot(x_{k+1})]$ $f$ of a product of TWO factors, for which we use the given formula $=f(x_{1}\cdot x_{2}\cdots\cdot x_{k})+f(x_{k+1})$ the first term is, by the hypothesis, equal to $f(x_{1})+f(x_{2})+\ldots+f(x_{k})$ $=f(x_{1})+f(x_{2})+\ldots+f(x_{k})+f(x_{k+1})=RHS$ for $n=k+1$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n\geq 2.$ |
### Theory:
In the previous topics, we have learnt how to find the area of planar regions. And, also familiar with the topic "Congruence of triangles". Let us recall them.
Consider $$2$$ figures $$PQRS$$ and $$ABCD$$, whose areas are $$(4 \times 4) \ cm^2$$ and $$(8 \times 2) \ cm^2$$. We know the statement that "Two figures are said to be congruent if they have the same shape and the same size". That is, we can also say that "Two figures are congruent if their areas are equal". But, the converse of this statement is not true. That is, "Two figures having equal areas need not be congruent".
We can see that the areas of the figures $$PQRS$$ and $$ABCD$$ are equal, but they are not congruent because the $$2$$ figures have equal area, which doesn't mean they have equal sides.
Now, consider the below planar region.
We can see that the planar region $$X$$ is made up of $$2$$ planar regions $$A$$ and $$B$$. Therefore, the area of the planar region $$X$$ can be written as:
Area of figure $$X =$$ Area of figure $$A +$$ Area of figure $$B$$
Important!
We can denote the area of figure $$A$$ as $$ar(A)$$ and area of figure $$B$$ as $$ar(B)$$, and so on.
Now, we arrive at $$2$$ properties associated with the planar region. They are:
Important!
1. If $$A$$ and $$B$$ are two congruent figure, then $$ar(A) = ar(B)$$.
2. If a planar region $$X$$ is formed by $$2$$ non-overlapping planar regions $$A$$ and $$B$$, then $$ar(T) = ar(A) + ar(B)$$.
In this chapter, let us discuss the areas of certain geometrical figures under the condition when they lie on the same base and between the same parallels. |
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# Find the four numbers in G.P, whose sum is $85$ and the product is $4096$.A.$64,16,4,1$B.$64,16,40,1$C.$64,19,4,1$D.$67,16,4,1$
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Hint: Using the concept of G.P as let the first term be a and general ratio be r, then nth term can be given as ${{\text{T}}_{\text{n}}}{\text{ = a}}{{\text{r}}^{{\text{n - 1}}}}$. So from the given concept form all the terms and then solve it using the condition given in the question.
Let the four terms of a G.P be ,
${\text{a,ar,a}}{{\text{r}}^{\text{2}}}{\text{,a}}{{\text{r}}^{\text{3}}}$
Now, as per the given that sum of all the terms is $85$
$\Rightarrow {\text{a + ar + a}}{{\text{r}}^{\text{2}}}{\text{ + a}}{{\text{r}}^{\text{3}}} = 85 \\ \Rightarrow {\text{a(1 + r + }}{{\text{r}}^{\text{2}}}{\text{ + }}{{\text{r}}^{\text{3}}}{\text{) = 85}} \\$
And the product of the four terms is $4096$
$\Rightarrow {\text{a(ar)(a}}{{\text{r}}^{\text{2}}}{\text{)(a}}{{\text{r}}^{\text{3}}}) = 4096 \\ \Rightarrow {\text{(}}{{\text{a}}^4}{{\text{r}}^6}{\text{) = 4096}} \\$
Calculating the factors of
$\Rightarrow {\text{4096 = }}{{\text{a}}^{\text{4}}}{{\text{r}}^{\text{6}}}{\text{ = }}{{\text{2}}^{{\text{12}}}} \\ \Rightarrow {{\text{a}}^{\text{4}}}{{\text{r}}^{\text{6}}}{\text{ = 1}}{\text{.}}{{\text{4}}^{\text{6}}} \\ \Rightarrow {\text{a = 1,r = 4}} \\$
And so from this we can conclude that the terms are,
${\text{a = 1,r = 4}} \\ {\text{1,4,16,64}} \\$
Hence, option (a) is the correct answer.
Note: In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
Properties :
If all the terms of G.P are multiplied or divided by the same non-zero constant then the sequence remains in G.P with the same common ratio.
The reciprocals of the terms of a given G.P. form a G.P |
1 / 21
# Hawkes Learning Systems: College Algebra - PowerPoint PPT Presentation
Hawkes Learning Systems: College Algebra. Section 4.6: Inverses of Functions. Objectives. Inverses of relations. Inverse functions and the horizontal line test. Finding inverse function formulas. Inverses of Relations. Let R be a relation. The inverse of R , denoted , is the set
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### Hawkes Learning Systems:College Algebra
Section 4.6: Inverses of Functions
• Inverses of relations.
• Inverse functions and the horizontal line test.
• Finding inverse function formulas.
Let R be a relation. The inverse of R, denoted , is the set
In other words, the inverse of a relation is the set of ordered pairs of that relation with the first and second coordinates of each exchanged.
Consider the relation
The inverse of Ris
Determine the inverse of the relation. Then graph the relation and its inverse and determine the domain and range of both.
In the graph to the left, R is in blue and its inverse is in red. R consists of three ordered pairs and its inverse is simply these ordered pairs with the coordinates exchanged. Note: the domain of the relation is the range of its inverse and vice versa.
Determine the inverse of the relation. Then graph the relation and its inverse and determine the domain and range of both.
In this problem, R is described by the given equation in xand y. The inverse relation is the set of ordered pairs in R with the coordinates exchanged, so we can describe the inverse relation by just exchanging x and y in the equation, as shown at left.
Note:
A relation and its inverse are mirror images of one another (reflections) with respect to the line
Even if a relation is a function, its inverse is not necessarily a function.
Verify these two facts against the previous examples.
In practice, we will only be concerned with whether or not the inverse of a function f, denoted , is itself a function. Note that has already been defined: stands for the inverse of f, where we are making use of the fact that a function is also a relation.
Caution!
does not stand for when fis a function!
The Horizontal Line Test
Let f be a function. We say that the graph of f passes the horizontal line test if every horizontal line in the plane intersects the graph no more than once.
One-to-One Functions
A function f is one-to-one if for everypair of distinct elements and in the domain of f, we have . This means that every element of the range of f is paired with exactly one element of the domain of f.
Note: If a function is one-to-one, it will pass the horizontal line test.
In Example 1you have with and , then Ris one- to-one, so its inverse must be a function. But, if you notice in Example 2, the graph of is a parabola and obviously fails the horizontal line test. Thus, Ris not one-to-one so its inverse is not a function.
Tip!
The inverse of a function fis also a function if and
only if fis one-to-one.
Does have an inverse function?
No.
We can see by graphing this function that it does not pass the horizontal line test, as it is an open “V” shape. By this, we know that f is not one-to-one and can conclude that it does not have an inverse function.
Does have an inverse function?
Yes.
We know that the standard cube shape passes the horizontal line test, so g has an inverse function. We can also convince ourselves of this fact algebraically:
Consider Example 3 again:
We stated in the previous slide that because fis not one-to-one, it does not have an inverse function. However, if we restrict the domain of fexplicitly by specifying that the domain is the interval , the new function, with its restricted domain, is one-to-one and has an inverse function.
Let’s think about this graphically: what shape does the graph f have now that we restricted the domain? Notice that it is a diagonal line beginning at the point
or, simply, the right half of the graph.
This process is called restriction of domain.
To Find a Formula for
Let fbe a one-to-one function, and assume that f is defined by a formula. To find a formula for , perform the following steps:
1. Replace in the definition of f with the variable y. The result is an equation in x and y that is solved for y.
2. Interchange x and y in the equation.
3. Solve the new equation for y.
4. Replace the y in the resulting equation with .
For example, to find the inverse function formula for the function
1. Replace with y.
2. Interchange x and y.
3. Solve for y.
4. Replace y with
If you noticed, finding the inverse function formula for with the defined algorithm was a relatively long process for how simple the function is. Notice that ffollows a sequence of actions: first it multipliesxby 5, then it adds1. To obtain the inverse of fwe could “undo” this process by negating these actions in the reverse order. So, we would first subtract 1and then divide by 5:
Find the inverse of the following function.
We can always find the inverse function formula by using the algorithm we defined. However, this function is simple enough to easily undo the actions of f in reverse order. The application of the algorithm would be:
As you might notice, for this particular function, “undoing” the actions of f in reverse order is much simpler than applying the algorithm.
Find the inverse of the following function.
We will apply the algorithm to find the inverse.
Substitute y for
Interchange x and y.
Substitute
for y.
Solve for y.
The graph of a relation and its inverse are mirror images of one another with respect to the line
This is still true of functions and their inverses.
Consider the function , its inverse
and the graph of both:
Note: the key characteristic of the inverse of a function is that it “undoes” the function. This means that if a function and its inverse are composed together, in either order, the resulting function has no effect on any allowable input; specifically:
For reference, observe the graph on the previous slide.
Consider the function for and its inverse (you should verify that this is the inverse of f ).
Below are both of the compositions of fand : |
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# There are 3450 employees in an organisation out of which 42% got promoted , how many such employees are there who got promoted ?A.1449B.1518C.1587D.1656
Last updated date: 20th Jun 2024
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Hint: It is given that the total number of employees in that organisation is 3450 and 42 % of them were promoted. From this to find the number of employees who got promoted we need to find the value of 42% of the total number of employees. The value of x% of a number N can be given by the formula $\dfrac{x}{{100}}*N$.Using this formula we can find the value 42% of 3450.
Step 1 :
We are given that there are 3450 employees in an organisation
Let N be the number of employees in the organisation
Therefore , N = 3450
Step 2:
We are given that 42% of the total number got promoted .
In order to find the number of employees we need to find the value of 42% of the total number of employees
That is , 42% of N
The value of x% of a number N can be given by the formula $\dfrac{x}{{100}}*N$
Therefore now we can find the value of 42% of N by using the above formula
Number of employees promoted = 42% of N
$= \dfrac{{42}}{{100}}*N \\ = \dfrac{{42}}{{100}}*3450 \\ = 0.42*3450 \\ = 1449 \\$
Therefore , the number of employees promoted is 1449
The correct option is a.
Note:1. We can also calculate the number of employees not promoted by subtracting this value from the total number of employees.
2.That decimal fraction which has 100 as its denominator is known as Percentage. The numerator of such a fraction is known as Rate Per Cent.
3.The percent rate is calculated by dividing the new value by the original value and multiplying by 100%. The percentage value or new value is calculated by multiplying the original value by the percent rate and dividing by 100%. |
3 Motion Along a Straight Line
3.2 Instantaneous Velocity and Speed
Learning Objectives
By the end of this section, you will be able to:
• Explain the difference between average velocity and instantaneous velocity.
• Describe the difference between velocity and speed.
• Calculate the instantaneous velocity given the mathematical equation for the velocity.
• Calculate the speed given the instantaneous velocity.
We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.
Instantaneous Velocity
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is
. To find the instantaneous velocity at any position, we let
and
. After inserting these expressions into the equation for the average velocity and taking the limit as
, we find the expression for the instantaneous velocity:
Instantaneous Velocity
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point
is the rate of change of the position function, which is the slope of the position function
at
. (Figure) shows how the average velocity
between two times approaches the instantaneous velocity at
The instantaneous velocity is shown at time
, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times,
, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.
Example
Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of (Figure), find the velocity-versus-time graph.
Strategy
The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.
Solution
[hidden-answer a=”508307″]Time interval 0 s to 0.5 s:
Time interval 0.5 s to 1.0 s:
Time interval 1.0 s to 2.0 s:
The graph of these values of velocity versus time is shown in (Figure).
Significance
During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in (Figure). In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has a negative velocity.
Speed
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:
Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:
If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.
Speeds of Various Objects*Escape velocity is the velocity at which an object must be launched so that it overcomes Earth’s gravity and is not pulled back toward Earth.
Speed m/s mi/h
Continental drift
Brisk walk 1.7 3.9
Cyclist 4.4 10
Sprint runner 12.2 27
Rural speed limit 24.6 56
Official land speed record 341.1 763
Speed of sound at sea level 343 768
Space shuttle on reentry 7800 17,500
Escape velocity of Earth* 11,200 25,000
Orbital speed of Earth around the Sun 29,783 66,623
Speed of light in a vacuum 299,792,458 670,616,629
Calculating Instantaneous Velocity
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment, let’s use polynomials
, because they are easily differentiated using the power rule of calculus:
The following example illustrates the use of (Figure).
Example
Instantaneous Velocity Versus Average Velocity
The position of a particle is given by
.
1. Using (Figure) and (Figure), find the instantaneous velocity at
s.
2. Calculate the average velocity between 1.0 s and 3.0 s.
Strategy(Figure) gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in t. Therefore, we can use (Figure), the power rule from calculus, to find the solution. We use (Figure) to calculate the average velocity of the particle.
Solution
1.
.Substituting t = 2.0 s into this equation gives
.
2. To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of x(1.0 s) and x(3.0 s):
Then the average velocity is
Significance
In the limit that the time interval used to calculate
goes to zero, the value obtained for
converges to the value of v.
Example
Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is
.
1. What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
2. What is the speed of the particle at these times?
Strategy
The instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use (Figure) and (Figure) to solve for instantaneous velocity.
Significance
The velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually (Figure). In (a), the graph shows the particle moving in the positive direction until t = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle’s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of x(t) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.
The position of an object as a function of time is
. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at t = 1.0 s?
(a) Taking the derivative of x(t) gives v(t) = −6t m/s. (b) No, because time can never be negative. (c) The velocity is v(1.0 s) = −6 m/s and the speed is
.
Summary
• Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle’s motion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the position function, which gives us the functional form of instantaneous velocity v(t).
• Instantaneous velocity is a vector and can be negative.
• Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.
• Average speed is total distance traveled divided by elapsed time.
• The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.
Conceptual Questions
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.
Does the speedometer of a car measure speed or velocity?
If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?
Average speed. They are the same if the car doesn’t reverse direction.
How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
Problems
A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Given the following velocity-versus-time graph, sketch the position-versus-time graph.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
A particle moves along the x-axis according to
. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b.
; (c)
Unreasonable results. A particle moves along the x-axis according to
. At what time is the velocity of the particle equal to zero? Is this reasonable?
Glossary
instantaneous velocity
the velocity at a specific instant or time point
instantaneous speed
the absolute value of the instantaneous velocity
average speed
the total distance traveled divided by elapsed time |
# A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Given
Length of a piece of land = 0.7km
The breadth of a piece of land = 0.5km
Find out
We have to determine the length of the wire needed to fence the piece of land given
Solution
The length of the fence required is given by the perimeter of the rectangle
We know that
Perimeter of rectangle = 2 (Length + Breadth)
Perimeter of the plot = 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 km
Each side is to be fenced with 4 rows = 4 × 2.4
= 9.6 km
∴ Total length of the required wire is 9.6 km |
# A helicopter rises from rest on the ground vertically upwards with a constant acceleration g.
Question:
A helicopter rises from rest on the ground vertically upwards with a constant acceleration $g$. A food packet is dropped from the helicopter when it is at a height $h$. The time taken by the packet to reach the ground is close to [ $g$ is the accelertion due to gravity] :
1. (1) $t=\frac{2}{3} \sqrt{\left(\frac{h}{g}\right)}$
2. (2) $t=1.8 \sqrt{\frac{h}{g}}$
3. (3) $t=3.4 \sqrt{\left(\frac{h}{g}\right)}$
4. (4) $t=\sqrt{\frac{2 h}{3 g}}$
Correct Option: , 3
Solution:
(3) For upward motion of helicopter,
$v^{2}=u^{2}+2 g h \Rightarrow v^{2}=0+2 g h \Rightarrow v=\sqrt{2 g h}$
Now, packet will start moving under gravity.
Let ' $t$ ' be the time taken by the food packet to reach the ground.
$s=u t+\frac{1}{2} a t^{2}$
$\Rightarrow-h=\sqrt{2 g h} t-\frac{1}{2} g t^{2} \Rightarrow \frac{1}{2} g t^{2}-\sqrt{2 g h} t-h=0$
or, $t=\frac{\sqrt{2 g h} \pm \sqrt{2 g h+4 \times \frac{g}{2} \times h}}{2 \times \frac{g}{2}}$
or, $t=\sqrt{\frac{2 g h}{g}}(1+\sqrt{2}) \Rightarrow t=\sqrt{\frac{2 h}{g}}(1+\sqrt{2})$
or, $t=3.4 \sqrt{\frac{h}{g}}$
Pranav
May 26, 2024, 6:35 a.m.
If the helicopter leaves the packet then the packet will also move upwards for some time then it will start comming down Then how are you calculating just the time from the position where it was dropped to ground
Rohit
May 17, 2024, 6:35 a.m.
Hi Sarah Sir 👋
sarah
May 21, 2024, 6:35 a.m.
hi beta
April 5, 2023, 6:35 a.m.
How h is negative
Ayesha
Sept. 30, 2023, 6:35 a.m.
Becoz we have taken downward direction as - ve
Manasvi Bhushan
Sept. 14, 2022, 9:29 a.m.
sir koi dus ra approach baato na ye to sab soch lete hai |
# Worksheet on Multiplication on Fraction
This worksheet will provide a variety of sums on multiplication of fraction for practice. It will help in clear understanding of the topic multiplication on fraction. It will also develop confidence and better in-depth knowledge on the topic.
The worksheet is based on multiplication of fractions with whole numbers, multiplication of mixed fraction with another mixed fraction. It also contains multiplication using properties of multiplicative inverse and other properties of multiplication. This worksheet also contains story sums based on multiplication of fractions with another fraction or whole number or mixed fraction. The aim of this worksheet is to provide students a complete practice on every type of multiplication sums based on fractions.
Here are few problems on multiplication on fraction for practice:
1. Find:
(i) 6/7 × 70
(ii) 6/25 × 1500
(iii) 37/12 × 480
Solution:
(i) 60
(ii) 360
(iii) 1480
2. Find the product:
(i) 110 × 2/5
(ii) 65 × 5/13
(iii) 95 × 3/5
Solution:
(i) 44
(ii) 25
(iii) 57
3. (i) 7 of 12 6/7
(ii) 1 4/3 of 5 2/14
(iii) 25 1/2 of 2
Solution:
i) 90
ii) 12
iii) 51
4. (i) Find the multiplicative inverse of 65
(ii) Find the multiplicative inverse of 43/3
(iii) Find the multiplicative inverse of 7/11
Solution:
(i) 1/65
(ii) 3/43
(iii) 11/7
5. Fill in the blanks and state the property wherever applicable:
(i) 3/5 × 7 1/8 × 8/9 = …………………..
(ii) 56 × 9/8 = …………………..
(iii) 17/3 × 0 = …………………..
(iv) 31/37 × 37/31 = …………………..
(v) 11/13 × 1 = …………………..
(vi) 18/17 × 5/17 = ………………….. × 18/17
Solution:
(i)19/5
(ii) 63
(iii) 0; Property : Anything multiplied by 0 is 0
(iv) 1; Property: A fraction multiplied by its inverse is always 1
(v) 11/13 ; Property : Any fraction multiplied by 1 is the fraction itself
(vi ) 5/17 ; Property: If two or more fractions are multiplied then the order of multiplication does not effect the result
6. There 255 marbles in a bag consisting red and white marbles. Out of which 4/5 are red marbles. Find the number of white marbles.
Solution:
51 white marbles
7. A rope is cut into 15 pieces. The length of each piece is 3/5 cm . Find the total length of the rope.
Solution:
9 cm
8. There are 1500 students in a school. Out of which 4/5 are boys. Find the number of boys and girls in the school.
Solution:
Boys = 1200
Girls = 300
9. Rohan’s mother gave him 50 pens. He gave 3/5 pens to his sister. What is the fraction of pens given to his sister?
Solution:
30
10. In a birthday party 20 children were invited. Each child got 3/2 of a chocolate bar. How many chocolate bars were required all together?
Solution:
30 chocolate bars
11. There are 50 toffees in a packet. Ram gave 4/5 of the toffees to his sister. How many toffees did Ram gave to his sister?
Solution:
40 toffees
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This topic i on divisibility rule of 4 and will provide examples on the same. Divisibility Rule of 4: If the last two digits of the number is divisible by 4 then the whole number is divisible by 4 For eg: 524 The last two digit of the number is 24 which is divisible by 4.
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# Irregular Polygons
Home » Math Vocabulary » Irregular Polygons
## Irregular Polygons Introduction
We know that a polygon is a two-dimensional enclosed figure made by joining three or more straight lines. A regular polygon is a type of a polygon that has equal sides and all interior angles of equal measure. If any one of the conditions is not met, it becomes an irregular polygon. Check out the irregular polygon shapes based on three conditions.
Note that non-polygon shapes are the shapes that do not fulfill the criterion of a polygon. For example: Circle
## What Are Irregular Polygons?
Irregular polygons are polygons that do not have equal sides and equal angles.
Common examples of irregular polygons are a scalene triangle, kite, rectangle.
The image below shows the difference between a regular hexagon and an irregular hexagon.
## Irregular Polygons: Definition
A polygon is said to be an irregular polygon or non-regular polygon if all the sides are not equal in length and and all the interior angles may not be of equal measure.
## How to Classify Irregular Polygons
We can classify irregular polygons based on the number of sides. A three-sided polygon is a triangle, a four-sided polygon is a quadrilateral, a five-sided polygon is a pentagon, and so on. Here are a few examples showing the names of irregular polygons and the number of sides:
## Properties of Irregular Polygons
Let’s see some properties of irregular polygons.
1. First, a regular polygon has equal sides and equal angles. If the polygon fails to meet one of these two conditions, it becomes an irregular polygon.
2. Second, there are irregular polygons having equal sides but unequal angles and vice versa.
3. In the figure shown below, the shape is a rhombus. Is a rhombus a polygon? Yes, because it is a closed figure made of four line segments. A rhombus is an irregular polygon with equal sides and unequal angles. Its opposite angles are equal but all angles do not have the same measure.
In the above figure, shape B is a rectangle. Now, is a rectangle a polygon? Yes, it is, as it is a closed figure made of line segments. However, it is an irregular polygon with equal angles (90 degrees each) and unequal sides.
## Common Types of Irregular Polygons
We will discuss some common irregular polygon examples along with their important properties.
### Scalene Triangle
A scalene triangle has three unequal sides. So, it is an irregular polygon.
In the figure, the scalene triangle PQR is an irregular polygon as PQ, QR, and PR are of different lengths.
### Isosceles Triangle
An isosceles triangle has two equal sides $(\text{AC} = \text{AB})$ and two equal angles (angle C $=$ angle B). Since the third side and the third angle are not equal to the other two, it is an irregular polygon.
### Rectangle
A rectangle ABCD has congruent angles (all angles are right angles), but it does not have equal sides. Since only the opposite sides are equal, a rectangle is an irregular polygon.
### Right Triangle
In the right triangle ABC, if the angle B is right angle, the angles A and B will be acute angles (the sum of all angles of a triangle is 180 degrees). Therefore, the three angles cannot be congruent. So, it is an irregular polygon. The following image shows an example of one such right triangle.
### Irregular Pentagon
An irregular pentagon is a polygon with five unequal sides. Since it does not have equal sides, it is an irregular polygon.
### Irregular Hexagon
An irregular hexagon (or a non regular hexagon) has six unequal sides. So, a non-regular hexagon is an irregular polygon.
## Formulas Associated with Irregular Polygons
There are three formulas related to irregular polygons: area formula, perimeter formula, and angle-sum formula.
### Area of Irregular Polygons
Since irregular polygons come in different shapes and sizes, we do not have a standard formula for finding the area of an irregular polygon. So, we can divide it into regular polygons for which the area can be calculated and then we can add them to get the area of an irregular polygon.
In the following figure, the irregular polygon can be divided into two triangles PQM and RSN, and a rectangle PQRS.
So, the area of the given irregular polygon $=$ Area of $\angle$PQM $+$ Area of $\angle$RSN $+$ Area of rectangle PQRS
We can find the area of a triangle, provided that the base and height is known, by the formula:
Area of a triangle $= \frac{1}{2}$ base $\times$ height.
### Perimeter of Irregular Polygons
To find the perimeter of an irregular polygon, we have to add up the length of all its sides.
Consider the irregular polygon given below. The side lengths are a, b, c, d, e, f, g, h.
Perimeter $= a + b + c + d + e + f + g + h$
### Interior Angles and Exterior Angles
When you extend a side of an irregular polygon, the angle formed between the extended side and its adjacent side is the exterior angle. On the other hand, interior angles are angles between a polygon’s two adjacent sides.
#### Sum of Interior Angles of Irregular Polygons
The interior angle-sum formula for an irregular polygon is the same as the sum of interior angle in a regular polygon.
Sum of interior angles in a n-sided polygon $= (n − 2) 180^{\circ}$, where n is the number of sides of the irregular polygon.
Example: What is the sum of the interior angles in a pentagon?
Solution: A pentagon has 5 sides.
So, $n = 5$
The sum of interior angles of a regular polygon $= (5 − 2) \times 180^{\circ}$
$= (5 − 2) \times 180^{\circ}$
$= 3 \times 180$
$= 540^{\circ}$
Therefore, the sum of interior angles of a pentagon is $540^{\circ}$.
#### Sum of Exterior Angles of Irregular Polygons
For any polygon, regular or non-regular, the sum of its exterior angles is 360 degrees.
## Regular Polygons vs. Irregular Polygons
Let’s now compare and contrast regular and irregular polygons.
1. All sides of a regular polygon are congruent.
Irregular polygons generally don’t have equal sides.
Rhombus is an example of an irregular polygon; it has equal sides, but unequal angles.
1. Regular polygons have equal interior/exterior angles.
In irregular polygons, the measure of all interior/ exterior angles is not equal.
A rectangle is an example of an irregular polygon where all the angles are equal but all sides are not of equal length.
1. Regular polygons have equal sides and equal angles. Conversely, irregular polygons can either have equal sides or equal angles or none.
## Solved Examples
1. Find the perimeter of the below figure:
Solution:
The perimeter of the irregular pentagon $ABCDE = AB + BC + CD + DE + AE$
$= 12 + 10 + 6 + 4 + 7$
$= 39$ inches
2. Find the area of the below right triangle ABC.
Solution:
Area of the right triangle ABC $= \frac{1}{2} \times$ base $\times$ height
$= \frac{1}{2} \times$ BC $\times$ AB
$= \frac{1}{2} \times 6 \times 3$
$= 9$ $\text{cm}^{2}$.
3. How many exterior angles are there in an irregular octagon?
Solution: An irregular octagon has eight sides and eight interior angles. Therefore, it has eight exterior angles because the number of interior angles of a polygon is equal to its exterior angles.
4. Find the sum of the interior angles of a polygon with eleven sides.
Solution: The sum of interior angles of an n-gon is $(n − 2) \times 180^{\circ}$.
Here, $n = 11$.
So, interior angle sum $= 9 \times 180^{\circ} = 1620^{\circ}$ degrees.
5. If the sum of the interior angles of an irregular polygon with “n” sides is 1800 degrees, what is n?
Solution: Given the sum of the interior angles of an n-gon $= 1800^{\circ}$
$(n − 2) × 180^{\circ} = 1800^{\circ}$
$n − 2 = 10$
So, $n = 12$.
Therefore, the polygon has 12 sides.
## Practice Problems
1
### The sum of interior angles of an irregular octagon is:
$540^{\circ}$
$720^{\circ}$
$900^{\circ}$
$1080^{\circ}$
CorrectIncorrect
Correct answer is: $1080^{\circ}$
Sum of interior angles of an irregular polygon $= (n - 2) \times 180^{\circ}$
An octagon has 8 sides. So, $n = 8$
$(n - 2) \times 180^{\circ} = (8 - 2) \times 180^{\circ} = 6 \times 180^{\circ} = 1080^{\circ}$
2
### The sum of exterior angles of an irregular triangle is:
$180^{\circ}$
$360^{\circ}$
$540^{\circ}$
$720^{\circ}$
CorrectIncorrect
Correct answer is: $360^{\circ}$
The sum of exterior angles of all polygons is $360^{\circ}$.
3
### The rhombus is irregular since it has
all sides unequal
all angles equal
only opposite sides equal
all sides equal
CorrectIncorrect
Correct answer is: all sides equal
All sides of a rhombus are equal. Its opposite angles are equal.
4
### A rectangle is irregular because:
all its sides are equal
all its angles are equal
all its sides are not equal
all its angles are not equal
CorrectIncorrect
Correct answer is: all its sides are not equal
A rectangle is an irregular polygon. All angles of a rectangle are right angles, and thus congruent. However, all sides of a rectangle are not equal; only opposite sides are congruent.
5
### Which of the following triangles is not an irregular polygon?
Scalene triangle
Isosceles triangle
Equilateral triangle
Right triangle
CorrectIncorrect
Equilateral triangle has all sides congruent and all angles measure $60^{\circ}$. So, it is a regular polygon.
The number of triangles required to form a polygon is equal to the number of sides of a polygon minus two. If the polygon has “n” sides, then the number of triangles in a polygon is $(n – 2)$. For examples, to form a rectangle (which has 4 sides), we need $4 – 2 = 2$ triangles. |
Evaluating Limits Mathematically
So here's an example of the kind of problem i'm talking about: limits. How exactly do I go about evaluating something like that. I can see and kind of figure it out and the answers make sense, but how do I do this mathematically? Thanks
-
If you look at the fractional exponent $\frac {3}{3-x}$ you can see where you're going. You're approaching 3 from the right side, so the numbers you're going through are larger than 3. For example, 4, 5, 6, etc. So if you were to plug in these values you would get a negative fraction, and $e^{-number}$ =$\frac{1}{e^{number}}$, so you can see that as you get closer to 3 (from the right direction), the $e^{\frac{3}{3-x}}$ gets smaller, i.e. approaches zero.
-
so you are trying to figure out $\lim_{x\rightarrow 3^+}\exp(3/(3-x))$ and $\lim_{x\rightarrow 3^-}\exp(3/(3-x))$.
$exp(z)$ is a continuous function so when taking limits, they will pass inside. So $\lim_{x\rightarrow 3^+}\exp(3/(3-x))=\exp(\lim_{x\rightarrow 3^+} 3/(3-x))$ and similarly from the other side. Now, $\lim_{x\rightarrow 3^+}3/(3-x)=\infty$ and $\lim_{x\rightarrow 3^-}3/(3-x)=-\infty$, so it amounts to figuring out $\exp(+\infty)$ and $\exp(-\infty)$ one of which is infinity and the other 0, respectively.
-
Mathematically: $\lim_{x \to a} f(x) = L$ means: $$\forall \epsilon > 0 \exists \delta > 0$$ so that $$|x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$$
- |
Post from Marilyn’s Blog: Where’s the Math?
by admin, August 04th, 2015
All Blog Posts
Lydia’s idea emerged when I was teaching multiplication in her third grade class. To engage the students with connecting real-world situations to multiplication, and also with estimating and calculating accurate answers in several different ways, I presented a word problem:
There are 7 tricycles.
How many wheels are there altogether?
The students wrote a multiplication equation, estimated, and then figured out the answer. We discussed their different strategies for computing and agreed that 7 x 3 = 21.
I then gave a related assignment. This time I presented the students with a multiplication equation―8 x 4 = ? I asked them to estimate the product, write a word problem that matched the equation, and figure out the answer in at least two different ways.
As I circulated, Lydia was eager to explain to me how she had arrived at the estimate of 32. She had noticed that one less than each of the factors in 8 x 4 resulted in 7 x 3, the factors in the tricycle problem we had solved. So she figured that the answer to 8 x 4 should follow the same pattern, and she changed 21 to 32. Here’s what she wrote.
I was astonished. And speechless. (I think it’s hard to teach and think at the same time, especially when we experience unexpected responses from students.)
Lydia happily continued with the assignment. She wrote an appropriate word problem―There were 8 dancers. Each dancer held 4 canes. How many canes all together? When she solved the problem (she did this in five different ways), she was pleased that her estimate was the exact answer.
Meanwhile, I was wondering about what Lydia had done and why it had worked. I grabbed a piece of paper and tried her method on other problems, increasing each factor and the digits in the product by 1 . It didn’t work the first few times, but then it worked again!
I returned to Lydia and asked her to try her method on 6 x 2, a problem I knew wasn’t too hard for her to figure. Also, the factors were each 1 less than in 7 x 3, so it related back to the tricycle problem. Lydia wrote 6 x 2 = 12 and, underneath, increased each digit by 1, producing the incorrect result that 7 x 3 = 23.
“Hey,” she said, “it wouldn’t work. I don’t get it. It worked over here.” She referred again to how she had estimated 8 x 4 from 7 x 3.
While I found Lydia’s conjecture surprising and intriguing, I discounted it as just a lucky occurrence . . . until I investigated further.
My challenge was to make sense of the mathematics from a pattern that was orderly but seemed based solely on luck.
That night, I phoned a math friend (Nicholas Branca, who was a mathematical marvel) and told him what had happened. I gave him my conclusion that Lydia’s reasoning had no mathematical grounding, even though it worked at times. Nicholas’s response was, “If a student does something in an orderly way, following a clear and predictable pattern, then it’s likely that there’s some mathematical grounding. I’ll get back to you.” (Let’s hear it for the Common Core Mathematical Practice Standard 7: Look for and make use of structure.)
I spent a good deal of time that night trying Lydia’s method on problems with different pairs of factors, looking for patterns relating to when her method did and didn’t work, and trying to figure out why. Nicholas and I talked several times.
I love investigations like these, where the math isn’t particularly difficult or complex so the problem is accessible, but there’s something mathematically wonderful, and often surprising and delightful, to discover.
I’m now convinced that Lydia’s reasoning indeed has mathematical grounding, that it’s possible to predict when her method will and won’t work, and to be able to explain why. Interested in finding out more? Try these investigations.
Suggestions for Investigating
The investigations below helped me make sense of the mathematics underlying Lydia’s method. I’ve used the first two suggestions with third, fourth, and fifth graders.
Having choices about what to investigate and how is effective for motivating and encouraging learning.
More recently, I put together the full list of investigations for a professional learning session. Teachers were free to investigate some or all of them, and in any order. Then, after everyone had some time to think by themselves, I organized them into areas in the room numbered for the seven suggestions. They had the choice to continue working on their own or to join a group that was focusing on the same question. It was OK for them to switch to another group as they worked. I think that having choices about what to investigate and how is valuable, both for professional learning sessions and in classrooms with students. I think this would be a good set of investigations and structure for a middle school or high school class.
1. Does Lydia’s method work for other pairs of equations? Why?
2. What can you discover about pairs of equations for which it does and doesn’t work?
3. How might using rectangular arrays inform the problem?
4. What about adding 2 to each digit (or any other number) instead of adding 1?
5. How can you use algebra to explain what’s happening?
6. What happens with negative factors?
7. What about factors that aren’t whole numbers; e.g., 2.6 or 3.05?
As always, I’m interested in any feedback and/or questions. |
# Subtraction of Matrices
A mathematical system of subtracting one matrix from another matrix is called subtraction of matrices.
Subtraction of matrices is a mathematical operation, used to subtract one matrix from another matrix in mathematics.
Subtraction of matrices is performed by subtracting the elements of a matrix from the elements of another matrix. Therefore, the number of elements in one matrix must be equal to the number of elements in another matrix to perform the subtraction but it is not possible to perform subtraction with matrices of different order.
## Example
1.
### Subtraction of Matrices of same order
Example: 1
$A$ and $B$ are two matrices. $A$ is a matrix of order $3×3$ and $B$ is also a matrix of order $3×3$.
and
Assume, matrix $B$ is subtracted from matrix $A$ and it is written in mathematics by placing a minus sign between them.
Both $A$ and $B$ are $3×3$ matrices. Every matrix is formed by the elements which are arranged in it in rows and columns. Subtraction of matrices means subtraction of elements.
Nine elements are there in each matrix. All nine elements are in nine different positions in every matrix. So, subtract every element of matrix $B$ from every element in matrix $A$ in same position to complete the process of subtraction with matrices.
Step: 1
Subtract elements in First row First column of Matrices $A$ and $B$
$–7$ is an element in first row, first column in matrix $A$. $4$ is an element in first row, first column in matrix $B$. Subtract $4$ from $–7$ and place the operation in first row, first column.
Step: 2
Subtract elements in First row second column of Matrices $A$ and $B$
$3$ is an element in first row, second column in matrix $A$. $–9$ is an element in first row, second column in matrix $B$. Subtract $–9$ from $3$ and place the operation in first row, second column.
Step: 3
Subtract elements in First row third column of Matrices $A$ and $B$
$–2$ is an element in first row, third column in matrix $A$. $–3$ is an element in first row, third column in matrix $B$. Subtract $–3$ from $–2$ and place the operation in first row, third column.
Step: 4
Subtract elements in Second row First column of Matrices $A$ and $B$
$1$ is an element in second row, first column in matrix $A$. $–5$ is an element in second row, first column in matrix $B$. Subtract $–5$ from $1$ and place the operation in second row, first column.
Step: 5
Subtract elements in Second row second column of Matrices $A$ and $B$
$6$ is an element in second row, second column in matrix $A$. $7$ is an element in second row, second column in matrix $B$. Subtract $7$ from $6$ and place the operation in second row second column.
Step: 6
Subtract elements in Second row third column of Matrices $A$ and $B$
$9$ is an element second row third column in matrix $A$. $8$ is an element in second row second column in matrix $B$. Subtract $8$ from $9$ and place the operation in second row third column.
Step: 7
Subtract elements in Third row first column of Matrices $A$ and $B$
$2$ is an element of matrix $A$ in third row and first column. $1$ is an element of matrix $B$ in third row first column. Subtract $1$ from $2$ and place the operation in third row first column.
Step: 8
Subtract elements in Third row second column of Matrices $A$ and $B$
$5$ is an element of matrix $A$ in third row second column. $9$ is an element of matrix $B$ in third row second column. Subtract $9$ from $5$ and place the operation in third row second column.
Step: 9
Subtract elements in Third row third column of Matrices $A$ and $B$
$3$ is an element of matrix $A$ in third row third column. $8$ is an element of matrix $B$ in third row third column. Subtract $8$ from $3$ and place the operation in third row third column.
The matrix $B$ is subtracted from matrix $A$ in $9$ steps and it is time to simplify the matrix.
It is a matrix formed by the subtraction of matrix $B$ from matrix $A$.
Example: 2
Use same process to perform subtraction of matrix $A$ from matrix $B$ for practicing the subtraction of matrices.
The nine steps of subtraction is done in only one step.
2.
### Subtraction of Matrices of different order
Example: 1
and
$C$ is a matrix of order $1×3$ and $D$ is another matrix of order $2×2$. The order of matrices $C$ and $D$ are different. In other words, the matrix $C$ contains $3$ elements in $1$ row $3$ columns and the matrix $D$ contains $4$ elements in $2$ rows $2$ columns. $4$ elements cannot be subtracted from $3$ elements and vice-versa.
Firstly, perform subtraction of matrix $D$ from matrix $C$.
1. The element $0$ is an element in first row first column of matrix $D$ and it subtracted the element $2$ belongs to first row first column of matrix $C$. The operation of the elements in first row first column of both matrices is placed in same position in matrix.
2. The element $3$ in first row second column of matrix $D$ subtracted the element $1$ in first row second column of matrix $C$. Place the operation in same position in matrix.
3. There is no position called first row third column in matrix $D$ but matrix $C$ has. However, matrix $D$ cannot subtract matrix $C$ in this position. Therefore, subtraction of matrices is failed.
Example: 2
and
$E$ is a matrix of order $4×1$ and $F$ is a matrix of order $2×2$. Matrix $E$ is having $4$ elements and Matrix $F$ is also having $4$ elements. Four elements of one matrix can subtract four elements of another matrix.
Now, perform subtraction with matrices $E$ and $F$.
1. The element $7$ is an element belongs to first row first column in matrix $E$ and $5$ is an element in first row first column in matrix $F$. Subtract the element $5$ from element $7$ and place the operation in same position of the matrix.
2. There is no position, called first row second column in matrix $E$ but the matrix $F$ is having $0$ as element in first row second column. Due to lack of first row second column, the matrix $F$ cannot subtract the matrix $E$. Therefore, subtraction of matrix $F$ from matrix $E$ is not possible.
In first example, matrix $C$ is having three elements and matrix $D$ is having four elements. Three elements cannot subtract four elements but in the second example, subtraction is also failed even though both matrix $E$ and $F$ are having four elements.
Subtraction does not depend on the number of elements completely but it depends on the order of the matrices. Subtraction of two matrices is possible if they are in same order.
Therefore, first check the order of the matrices. If order of matrices is same, then perform the subtraction. Otherwise, it is not possible to perform the subtraction of matrices in mathematics. |
# The indefinite integral, or which call the "antiderivative", of f'(x)=6x-8x^3
is f(x) = 3x^2 - 2x^4 + C
where C is any constant.
Plug in x = 2 to that to see what C is.
3 = 3*4 - 2*16 + C
Determine the antiderivative of the following function given the initial condition: f'(x)=6x-8x^3 given that
f(2)=3
f'(x)=6x-8x^3
f(x)= 6x^2/2 - 8x^4/4 + C
Put in x=2 and solve for C
Explain Bot answered
8 months ago
## To determine the antiderivative of the function f'(x) = 6x - 8x^3, we can apply the power rule of integration.
The power rule states that the integral of x^n with respect to x is (x^(n+1))/(n+1) + C, where C is the constant of integration.
Using the power rule, we can integrate each term of f'(x) separately:
∫(6x - 8x^3) dx = (6 ∫ x dx) - (8 ∫ x^3 dx)
For the first term, the integral of x with respect to x is (x^2)/2.
For the second term, the integral of x^3 with respect to x is (x^4)/4.
Therefore, the antiderivative of f'(x) is:
f(x) = (6x^2)/2 - (8x^4)/4 + C
= 3x^2 - 2x^4 + C
Now, we need to determine the constant of integration, C, using the initial condition f(2) = 3.
Plugging in x = 2 into our expression for f(x), we get:
f(2) = 3(2)^2 - 2(2)^4 + C
= 3(4) - 2(16) + C
= 12 - 32 + C
= -20 + C
Since f(2) = 3, we can set -20 + C = 3 and solve for C:
-20 + C = 3
C = 3 + 20
C = 23
Therefore, the antiderivative of f'(x) = 6x - 8x^3, given that f(2) = 3, is f(x) = 3x^2 - 2x^4 + 23.
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# Multiplication Chart Nz
Discovering multiplication after counting, addition, and subtraction is good. Children find out arithmetic through a all-natural progression. This advancement of learning arithmetic is usually the adhering to: counting, addition, subtraction, multiplication, and lastly section. This document contributes to the question why discover arithmetic in this sequence? Moreover, why find out multiplication after counting, addition, and subtraction before division?
## The following details respond to these inquiries:
1. Kids learn counting first by associating visible objects using their fingers. A tangible case in point: Just how many apples are there any inside the basket? Far more abstract case in point is how outdated are you currently?
2. From counting numbers, the next rational phase is addition then subtraction. Addition and subtraction tables can be quite helpful educating aids for children as they are graphic equipment generating the cross over from counting simpler.
3. Which ought to be learned up coming, multiplication or section? Multiplication is shorthand for addition. At this stage, children use a firm knowledge of addition. Therefore, multiplication will be the up coming plausible form of arithmetic to discover.
## Assess basic principles of multiplication. Also, evaluate the essentials utilizing a multiplication table.
Allow us to assessment a multiplication illustration. By using a Multiplication Table, grow a number of times about three and obtain a solution 12: 4 by 3 = 12. The intersection of row three and column several of the Multiplication Table is twelve; a dozen is the solution. For the kids starting to discover multiplication, this really is straightforward. They are able to use addition to solve the trouble therefore affirming that multiplication is shorthand for addition. Illustration: 4 x 3 = 4 4 4 = 12. It is an outstanding overview of the Multiplication Table. The additional benefit, the Multiplication Table is aesthetic and demonstrates straight back to discovering addition.
## Where by can we get started understanding multiplication using the Multiplication Table?
1. Initially, get acquainted with the table.
2. Get started with multiplying by 1. Start at row number 1. Go on to line number one. The intersection of row a single and line the first is the answer: one.
3. Replicate these techniques for multiplying by 1. Multiply row one particular by columns one by way of a dozen. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Replicate these techniques for multiplying by two. Flourish row two by columns one particular by way of 5. The solutions are 2, 4, 6, 8, and 10 respectively.
5. Allow us to jump forward. Recurring these actions for multiplying by several. Grow row five by columns 1 via 12. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now let us boost the quantity of trouble. Replicate these actions for multiplying by 3. Flourish row three by columns 1 by means of 12. The replies are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are at ease with multiplication to date, use a examination. Remedy these multiplication issues in your mind then examine your responses on the Multiplication Table: multiply six and two, flourish 9 and 3, flourish 1 and 11, grow several and 4, and increase 7 as well as 2. The issue responses are 12, 27, 11, 16, and 14 respectively.
When you obtained 4 from 5 various difficulties proper, make your very own multiplication exams. Determine the responses in your head, and look them using the Multiplication Table. |
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# How do you evaluate the integral $\int {\dfrac{{x - 1}}{{x + 1}}dx}$?
Last updated date: 15th Jun 2024
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Hint: Start by splitting the integral so that it comes down to form which we can integrate easily. If there are any constants then get them out of the integral. Then mention all the standard integrands and use them to solve the integral.
Complete step by step solution:
First we will start off by splitting the integrand function.
$= \int {\dfrac{{x - 1}}{{x + 1}}dx} \\ = \int {\dfrac{{x + 1 - 2}}{{x + 1}}dx} \\ = \int {\left( {1 - \dfrac{2}{{x + 1}}} \right)dx} \\$
Now further we will divide the integral over the signs.
$\int {1dx - \int {\dfrac{2}{{x + 1}}dx} }$
Now we will use the standard integrals such as $\int {1dx} = x + c$ and $\int {\dfrac{1}{x}dx = \ln x}$.
Now we substitute these integrals in our expression.
Also, the constant here which is $2$ will get out of the integral.
$= \int {1dx - \int {\dfrac{2}{{x + 1}}dx} } \\ = \int {1dx - 2\int {\dfrac{1}{{x + 1}}dx} } \\ = x - 2(\ln |x + 1|) + c \\$
Hence, the value of the integral $\int {\dfrac{{x - 1}}{{x + 1}}dx}$ will be $x - 2(\ln |x + 1|) + c$.
Additional Information: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.
Note: While substituting the terms make sure you are taking into account the signs of the terms as well. While using any standard integral check if your integral satisfies all the required conditions. While splitting the interval make sure you split terms along with their respective signs. |
Math Worksheets
# Angles of Triangles Worksheets
The worksheets on this page require grade school students to solve problems related to the angles of triangles, including calculating interior angles, calculating exterior angles and calculating the lengths of line segments created from triangle angle bisectors. Click the images below to open each PDF worksheet with answer keys and get your triangle geometry practice started!
## Calculating Interior Angles of Triangles
The sum of the angles of a triangle postulate can be used to find a missing interior angle of a triangle given the other two angles.
### Triangle Sum Theorem
Given a triangle ABC, the sum of the measurements of the three interior angles will always be 180°:
∠A + ∠B + ∠C = 180°
If you know two of the three angles of a triangle, you can use this postulate to calculate the missing angle's measurement. For example, if ∠A and ∠B are known, ∠C can be calculated using the following equation...
### Formula for Calculating a Missing Interior Angle of a Triangle
Given a triangle ABC, where ∠A and ∠B are known, ∠C is calculated as follows:
∠C = 180° − (∠A + ∠B)
A common problem is calculating a missing angle in a right triangle. In these triangle problems, we know that one of the angles must be exactly 90° therefore only one other angle must be known to calculate the third missing angle:
### Formula for Calculating a Missing Interior Angle of a Right Triangle
Given a right triangle ABC, where ∠A is the right angle and where ∠B is known, ∠C is calculated as follows:
∠C = 180° − (90° + ∠B)
Or, more succinctly...
∠C = 90° − ∠B
Another common problem involves finding the two base angles for an isosceles triangle given the triangles vertex or apex angle. Because the legs of an isosceles triangle are equal, the corresponding adjacent angles at the triangles base are also equal, so calculating the measure of the leg angles involves substracting the vertex angle from 180° and dividing the result among the two remaining angles:
### Formula for Calculating Base Angles of an Isosceles Triangle
Given an isosceles triangle ABC, where ∠A is the apex angle, the base angles ∠B and ∠C are calculated as follows:
∠B = ∠C =
180° − ∠A
2
## Calculating Exterior Angles of Triangles
The exterior angle of a triangle is calculated using the properties of supplementary angles. By extending a line from one of the triangle's sides, the exterior angle can be calculated from the adjacent interior angle by subtracting that angle from 180°. Conversely, the interior angle can be calculated by from the exterior angle using procedure. The triangle worksheets on this page provide practice calculating interior angles given exterior angles, or calculating exterior angles given interior angles.
## Triangle Angle Bisector Theorem
An angle bisector of a triangle is a line that divdes one of the triangle's interior angles in half. The resulting line segment splits to opposite side of the triangle into two line segments. The Triangle Angle Bisector theorem states that the angle bisector divides the opposite side into two line segments that are proportionate to their adjacent sides. The triangle angle bisector worksheets on this page allow students to practice determining the length of these bisected sides for different types of triangles, starting with equilateral and isosceles triangles (where the bisected segments are equal), and progressing more generally to obtuse and acute triangles where the proportionate relationship is used to calculate two different lengths. |
# Class 11 RD Sharma Solutions – Chapter 17 Combinations- Exercise 17.2 | Set 2
Last Updated : 30 Apr, 2021
### Question 12. In an examination, a student to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make a choice.
Solution:
Total number of questions = 5
Total number of questions to be answered = 4
As 2 questions are compulsory to answer, student can choose only 2 questions (4−2) out of the remaining 3 questions (5−2).
So, number of ways = 3C2
=
=
= 3
Therefore, the number of ways answering the questions is 3.
### Question 13. A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions?
Solution:
Total number of questions = 12
Total number of questions to be answered = 7
Total number of questions in each set = 6
Now a student can’t attempt 5 questions from either group but has to answer 7 questions in total.
So, number of ways = (6C5 × 6C2) + (6C4 × 6C3) + (6C3 × 6C4) + (6C2 × 6C5)
=
=
= (6×15) + (15×20) + (20×15) + (15×6)
= 90 + 300 + 300 + 90
= 780
Therefore, the number of ways of answering 7 questions is 780.
### Question 14. There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.
Solution:
Total number of points = 10
Number of collinear points = 4
Now we know, number of lines formed will be the difference between total number of lines formed by all 10 points and number of lines formed by collinear points added with 1.
Here, we add 1 because only 1 line can be formed by the given four collinear points.
So, number of ways = 10C2 – 4C2 + 1
=
=
= 45 – 6 + 1
= 40
Therefore, the total number of ways of drawing different lines is 40.
### Question 15. Find the number of diagonals of
(i) a hexagon
Solution:
A hexagon has 6 angular points. By joining any two angular points we get a line which is either a side or a diagonal.
So number of lines formed = 6C2
=
=
= 3×5
= 15
We know number of sides of hexagon is 6.
So, number of diagonals = 15 – 6 = 9
Therefore, the total number of diagonals of a hexagon is 9.
(ii) a polygon of 16 sides
Solution:
A polygon of 16 sides has 16 angular points. By joining any two angular points we get a line which is either a side or a diagonal.
So number of lines formed = 16C2
=
=
= 8×15
= 120
Number of sides of given polygon = 16
So, number of diagonals = 120 – 16 = 104
Therefore, the total number of diagonals of a hexagon is 104.
### Question 16. How many triangles can be obtained by joining 12 points, five of which are collinear?
Solution:
We know that 3 points are required to draw a triangle.
Since 5 out of 12 points are collinear, so number of triangles that can be formed would be the difference between the number of triangles formed by all 12 points and the number of triangles formed by collinear points.
So, number of triangle= 12C3 – 5C3
=
=
= (2×11×10) – (5×2)
= 220 – 10
= 210
Therefore, the total number of triangles that can be formed is 210.
### Question 17. In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one women has to be necessarily selected?
Solution:
Total number of men = 6
Total number of women = 4
Total number of persons to be selected = 5.
Now it is given that we must choose at least one woman. It means we can choose from 1 to all 4 women in our committee at a time. Number of men will change according to it.
So, number of ways = (4C1 × 6C4) + (4C2 × 6C3) + (4C3 × 6C3) + (4C4 × 6C1)
=
=
= 60 + 120 + 60 + 6
= 246
Therefore, the number of ways of selection is 246.
### Question 18. In a village, there are 87 families of which 52 families have at most 2 children. In a rural development program, 20 families are to be helped chosen for assistance, of which 18 families must have at most 2 children. In how many ways can the choice be made?
Solution:
Its given that 52 families have at most 2 children out of 87. Therefore the remaining 35 families have exactly 2 children.
Now to choose any 20 families of which 18 families must have at most 2 children can be done in 3 ways. Either we can choose all the 20 families out of those 52 who have at most children, or 19 out of 52 and remaining 1 out of 35, or 18 families from 52 and remaining 2 out of 35.
So, total number of ways = (52C18 × 35C2) + (52C19 × 35C1) + (52C20 × 35C0)
### Question 19. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has
(i) no girls
(ii) at least one boy and girl
(iii) at least 3 girls
Solution:
Number of girls = 4
Number of boys = 7
Number of members to be selected = 5
(i) no girls
As it is given that the team cannot have a girl, we have to choose 5 members out of 7 boys.
So, number of ways = 7C5
=
=
= 21
Therefore, the number of ways of selection such that team has no girls is 21.
(ii) at least one boy and girl
To select a team which consists of at least one boy and girl, we can choose from 1 to all the 4 girls at a time. Number of boys will change according to it.
So, number of ways = (7C1 × 4C4) + (7C2 × 4C3) + (7C3 × 4C3) + (7C4 × 4C1)
=
=
= 7+84+210+140
= 441
Therefore, the number of ways of selection such that team has at least one boy and girl is 441.
(iii) at least 3 girls
To select a team which consists of at least 3 girls, we can choose either 3 or 4 girls at a time. Number of boys will change according to it.
So, number of ways = (7C2 × 4C3) + (7C1 × 4C4)
=
=
= 84 + 7
= 91
Therefore, the number of ways of selection such that team has at least 3 girls is 91.
### Question 20. A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Solution:
Total number of men = 2
Total number of women = 3
So total number of persons = 2+3 = 5
Number of persons to be selected = 3
It is given that a committee of 3 persons has to be formed. So we have to choose 3 persons from 5 persons.
So, number of ways = 5C3
=
=
= 10
Also we have to find the number of committees consisting of 1 man and 2 women. So we have to choose 1 man out of 2 men and 2 women out of 3 women.
So, number of ways = 2C1 × 3C2
=
= 6
### Question 21. Find the number of
(i) diagonals formed in a decagon.
Solution:
A decagon has 10 sides. By joining any two angular points we get a line which is either a side or a diagonal.
So number of lines formed = 10C2
=
=
= 45
Number of sides = 10
So, number of diagonals = 45−10 = 35.
Therefore, number of diagonals formed in a decagon is 35.
(ii) triangles formed in a decagon.
Solution:
A decagon has 10 sides. By joining any 3 angular points we get a triangle.
So number of lines formed = 10C3
=
=
=
= 120
Therefore, number of triangles formed in a decagon is 120.
### Question 22. Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king?
Solution:
Out of a deck of 52 cards, we have to choose 5 cards combinations where at least one of the 5 cards has to be a king.
We know there are 4 kings in a deck of 52 cards. So, we can choose from 1 to all the four kings at a time and selection of remaining cards will change accordingly.
So, number of ways = (4C1 × 48C4) + (4C2 × 48C3) + (4C3 × 48C2) + (4C4 × 48C1)
=
=
= 778320 + 103776 + 4512 + 48
= 886656
Therefore, number of ways of selecting 5 card combinations if at least one of the 5 cards has to be a king is 886656.
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# UNIT 4 APPLICATIONS OF PROBABILITY Lesson 1: Events. Instruction. Guided Practice Example 1
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1 Guided Practice Example 1 Bobbi tosses a coin 3 times. What is the probability that she gets exactly 2 heads? Write your answer as a fraction, as a decimal, and as a percent. Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} There are 8 outcomes in the sample space. 2. Identify the outcomes in the event and count the outcomes. There are 3 outcomes in the event exactly 2 heads: HHT, HTH, and THH. 3. Apply the formula for the probability of an event. Divide the number of outcomes for the event Bobbi is hoping for (3) by the total number of possible outcomes (8). number of outcomes inthe sample space P( exactly2heads ) = 3 8 = = 37.5% Example 2 Donte is playing a card game with a standard -card deck. He s hoping for a club or a face card on his first draw. What is the probability that he draws a club or a face card on his first draw? The sample space is the set of all cards in the deck, so there are outcomes. U4-37
2 2. Identify the outcomes in the event and count the outcomes. You can use a table to show the sample space. Then identify and count the cards that are either a club or a face card or both a club and a face card. Suit J Q K A Spade Club Diamond Heart The event club or face card has 22 outcomes. 3. Apply the formula for the probability of an event. number of outcomes inthe sample space P( club or face card ) = = Apply the Addition Rule to verify your answer. Let A be the event club. There are 13 clubs, so A has 13 outcomes. Let B be the event face card. There are 12 face cards, so B has 12 outcomes. The event A and B is the event club and face card, which has 3 outcomes: jack of clubs, queen of clubs, and king of clubs. Apply the Addition Rule. P(A or B) = P(A) + P(B) P(A and B) P(club or face card) = P(club) + P(face card) P(club and face card) P(club or face card) = = = The Addition Rule answer checks out with the probability found in step 3, so the probability of the event is approximately U4-38 CCSS IP Math II Teacher Resource Walch Education
3 Example 3 Corrine is playing a board game. To find the number of spaces to move, she rolls a pair of dice. On her next roll she wants doubles or a sum of 10. What is the probability that she rolls doubles or a sum of 10 on her next roll? Interpret your answer in terms of a uniform probability model. (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) There are outcomes in the sample space. 2. Apply the Addition Rule to find the probability that she rolls doubles or a sum of 10 on her next roll. Let A be the event doubles. Event A has 6 outcomes: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6). Let B be the event sum of 10. Event B has 3 outcomes: (4, 6), (5, 5), and (6, 4). The event A and B is the event doubles and sum of 10, which has 1 outcome: (5, 5). Apply the Addition Rule. P(A or B) = P(A) + P(B) P(A and B) P(doubles or sum of 10) = P(doubles) + P(sum of 10) P(doubles and sum of 10) P(doubles or sum of 10) = = = 9. % U4-39
4 3. Interpret the answer in terms of a uniform probability model. The probabilities used in the application of the Addition 6 Rule,, 3, and 1, are found by using the formula number of outcomes inthe sample space. This formula is a uniform probability model, which requires that all outcomes in the sample space be equally likely. It is reasonable to assume that the dice Corrine rolls are fair, so that all outcomes in the sample space are equally likely. Therefore, the answer is valid and can serve as a reasonable predictor. You can predict that the relative frequency of getting doubles or a sum of 10 will be about 22% for a large number of dice rolls. That is, you can predict that Corrine will get doubles or a sum of 10 about 22% of the time if she rolls the dice a large number of times. Example 4 Students at Rolling Hills High School receive an achievement award for either performing community service or making the honor roll. The school has 500 students and 180 of them received the award. There were 125 students who performed community service and 75 students who made the honor roll. What is the probability that a randomly chosen student at Rolling Hills High School performed community service and made the honor roll? 1. Define the sample space and state its number of outcomes. The sample space is the set of all students at the school; it has 500 outcomes. U4-40 CCSS IP Math II Teacher Resource Walch Education
5 2. Define events that are associated with the numbers in the problem and state their probabilities. Let A be the event performed community service. Then P( A) = Let B be the event made the honor roll. Then P( B) = The event A or B is the event performed community service or made the honor roll, and can also be written A B. P( A B) = because 180 students received the award for either community service or making the honor roll. 3. Write the Addition Rule and solve it for P(A and B), which is the probability of the event performed community service and made the honor roll. P(A and B) can also be written P( A B). P( A B) = P( A) + P( B) P( A B) = + P( A B) Substitute the known probabilities = P( A B) Simplify. 20 = P( A B) 500 Subtract 200 from both sides = ( A B ) Multiply both sides by 1. The probability that a randomly chosen student at Rolling Hills High School has performed community service and is on the honor roll 20 1 is =, or 4% U4-41
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### Exam III Review Problems
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Algebra: A Combined Approach (4th Edition)
Chapter 6 - Section 6.3 - Exercise Set: 12 (Answer) $36r^2 - 5r - 24 = (4r + 3)(9r - 8)$
Chapter 6 - Section 6.3 - Exercise Set: 12 (Solution) Factorize : $36r^2 - 5r - 24$ Factors of $36r^2$ : $36r\cdot r$ or $18r\cdot 2r$ or $12r\cdot 3r$ or $9r\cdot 4r$ or $6r\cdot 6r$ Factors of -24 : $1\cdot -24$ or $2\cdot -12$ or $3\cdot -8$ or $4\cdot -6$ or $6\cdot -4$ or $8\cdot -3$ or $12\cdot -2$ or $24\cdot -1$ Try combination of those factors, ignoring those pairs with GCF that can be further factorized $(36r + 1)(r - 24) = 36r^2 - 863r - 24$ (Incorrect middle term) $(18r + 2)(2r - 12)$ (Not to be considered as it can be factorized as 4(9r + 1)(r – 6)) $(12r + 3)(3r - 8)$ (Not to be considered as it can be factorized as 3(4r + 1)(3r – 8)) $(4r + 3)(9r - 8) = 36r^2 - 5r - 24$ (Correct middle term) Thus, $36r^2 - 5r - 24 = (4r + 3)(9r - 8)$ |
# Spot Factor Pairs
In this worksheet, students must identify the factor pairs for the given number.
Key stage: KS 2
Curriculum topic: Number: Multiplication and Division
Curriculum subtopic: Recognise/Use Pairs in Mental Calculation
Difficulty level:
### QUESTION 1 of 10
Two whole numbers that multiply together to give a number are a factor pair of that given number.
2 and 6 are a factor pair of 12 because 2 × 6 = 12.
But 12 has other factor pairs too.
3 and 4 are a factor pair of 12 because 3 × 4 = 12.
1 and 12 are the simplest factor pair because 1 × 12 = 12.
This means that:
12 =
1 × 12 and 12 × 1
2 × 6 and 6 × 2
3 × 4 and 4 × 3
Example
Which of these are a factor pair of 64?
3 and 18 or 4 and 16.
3 × 18 = 54
4 × 16 = 64
So 4 and 16 are a factor pair of 64.
Want to understand this further and learn how this links to other topics in maths?
Why not watch this video?
Tick all the factor pairs of:
8
1 and 8
2 and 6
4 and 4
2 and 4
3 and 5
6 and 2
Tick all the factor pairs of:
10
1 and 9
2 and 5
4 and 6
2 and 4
1 and 10
6 and 4
Tick all the factor pairs of:
18
2 and 9
2 and 12
3 and 6
2 and 8
1 and 18
6 and 12
Tick all the factor pairs of:
20
2 and 9
2 and 10
4 and 5
2 and 8
1 and 19
20 and 1
Tick all the factor pairs of:
25
2 and 5
2 and 10
5 and 5
2 and 23
1 and 25
20 and 1
Tick all the factor pairs of:
31
2 and 15
3 and 11
1 and 30
11 and 21
1 and 31
3 and 1
Tick all the factor pairs of:
56
2 and 28
4 and 14
5 and 6
2 and 23
1 and 56
8 and 7
Tick all the factor pairs of:
63
1 and 63
2 and 32
3 and 21
6 and 3
60 and 3
9 and 7
Tick all the factor pairs of:
45
1 and 44
1 and 45
3 and 15
5 and 9
9 and 3
4 and 5
Tick all the factor pairs of:
108
1 and 108
9 and 12
2 and 54
3 and 36
4 and 27
6 and 18
• Question 1
Tick all the factor pairs of:
8
1 and 8
2 and 4
EDDIE SAYS
8 = 1 × 8 = 2 × 4
• Question 2
Tick all the factor pairs of:
10
2 and 5
1 and 10
EDDIE SAYS
10 = 1 × 10 = 2 × 5
• Question 3
Tick all the factor pairs of:
18
2 and 9
3 and 6
1 and 18
EDDIE SAYS
18 = 1 × 18 = 2 × 9 = 3 × 6
• Question 4
Tick all the factor pairs of:
20
2 and 10
4 and 5
20 and 1
EDDIE SAYS
20 = 1 × 20 = 2 × 10 = 4 × 5
• Question 5
Tick all the factor pairs of:
25
5 and 5
1 and 25
EDDIE SAYS
25 = 1 × 25 = 5 × 5
• Question 6
Tick all the factor pairs of:
31
1 and 31
EDDIE SAYS
31 = 1 × 31
• Question 7
Tick all the factor pairs of:
56
2 and 28
4 and 14
1 and 56
8 and 7
EDDIE SAYS
56 = 1 × 56 = 2 × 28 = 4 × 14 = 8 × 7
• Question 8
Tick all the factor pairs of:
63
1 and 63
3 and 21
9 and 7
EDDIE SAYS
63 = 3 × 21 = 7 × 9
• Question 9
Tick all the factor pairs of:
45
1 and 45
3 and 15
5 and 9
EDDIE SAYS
45 = 1 × 45 = 3 × 15 = 5 × 9
• Question 10
Tick all the factor pairs of:
108
1 and 108
9 and 12
2 and 54
3 and 36
4 and 27
6 and 18
EDDIE SAYS
108 = 1 × 108 = 2 × 54 = 3 × 36 = 4 × 27 = 6 × 18 = 9 × 12
---- OR ----
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# Finding the right most non zero digit of a factorial
Suppose,we are asked to find the last non zero digit of 100!, How do we find it?
We know that the number of zeros at the end of 100! is 24. Our task is to find the leading digit from the right.
Before that, we will see some interesting observations.
100! = 1 × 2 × 3 × 4 × 5 × ........× 99 × 100
Let us write the above format like below
= (5 × 10 × 15 ×......× 95 × 100) × (1 × 2 × 3 × 4 ) × (6 × 7 × 8 × 9) × (11 × 12 × 13 × 14) × (16 × 17 × 18 × 19 ) × ............ × (91 × 92 × 93 × 94) × (96 × 97× 98 × 99)
= (1 × 5) × (2 × 5) × (3 × 5) ×......× (20 × 5) × (1 × 2 × 3 × 4 ) × (6 × 7 × 8 × 9) × (11 × 12 × 13 × 14) × (16 × 17 × 18 × 19 ) × ............ × (91 × 92 × 93 × 94) × (96 × 97× 98 × 99)
By taking 5's common and multiplying the remaining terms,
520 × 20! × (1 × 2 × 3 × 4) × (6 × 7 × 8 × 9) × (11 × 12 × 13 × 14) × (16 × 17 × 18 × 19 ) × ............ × (91 × 92 × 93 × 94) × (96 × 97 × 98 × 99)
You can see the black terms gives units digits of 2 if we divide it by 2
Proof:
Let us consider (10a + 1)(10a + 2)(10a + 3)(10a +4)
This expression is a generalized format of the terms (1 × 2 × 3 × 4 )
For a = 0 we get the first term in the bracket, for a = 1 we get third term in the bracket
By pulling out one 2 and multiplying the remaining terms gives us,
2 × (10a + 1)(5a +1) (10a + 3) (10 a + 4)
2 × (50a2 + 15a + 1)(100a2 +70 a + 12)
2 × (5000a4 + 3500a3 + 600a2 + 1500a3 + 1050a2 + 180a + 100a2 + 70a + 12)
Each term contains zero except the last term. So after pulling one 2 out, unit digit of the remaining expression is 2.
Similarly, (10a + 6)(10a + 7)(10a + 8)(10a + 9)
By pulling out one 2 and multiplying the remaining terms gives us,
2 × (5a + 3)(10a + 7) (10 a + 8) (10 a + 9)
2 × (50a2 + 65a + 21)(100a2 +170 a + 72)
2 × (5000a4 + 8500a3 + 3600a2 + 6500a3 + 11050a2 + 4680a + 2100a2 + 3570a + 1512)
Each term contains zero except the last term. So after pulling one 2 out, unit digit of the remaining expression is 2.
Therefore, unit digit of 100! = ${5}^{20}$ x 20! x (2 x 2) x (2 x 2) x ......20 times = ${5}^{20}×20!×{2}^{20}×{2}^{20}={10}^{20}×20!×{2}^{20}$
So unit digit of 100! = Unit digit of $20!×{2}^{20}$
From the above discussion leading non zero digit of 100 ! = $\left[\frac{100}{5}\right]!×{2}^{\left(100/5\right)}=20!×{2}^{20}$
Applying the above logic, the last non zero digit of 20! = $\left[\frac{20}{5}\right]!×{2}^{\left(20/5\right)}=4!×{2}^{4}$
So Last non zero digit of 100! = $\left(20!×{2}^{20}\right)=\left(4!×{2}^{4}\right)×{2}^{20}$
= 24 x 16 x 6 = 4
2. How do we proceed If the given number is not divisible by 5 exactly?? For example we have to find the leading non zero digit of 28!
In this case we consider the leading non zero digit upto 25 and manually we multiply the units digits of the remaining 3 terms with the result.
25! = ${5}^{5}×5!×{2}^{5}×{2}^{5}⇒{10}^{5}×5!×{2}^{5}$
So the unit digit of 25! = 120 x 32 = 4 (We omit the zero)
Now unit digit of 28! = 4 x 26 x 27 x 28 = 4
The generalized formula to find the non zero digit of N = $\left[\frac{N}{5}\right]!×{2}^{\left[N/5\right]}×{\left[\frac{N}{5}\right]}_{\mathrm{Re}m}!$
Here [ ] denotes greatest integer function. [3.6] = 3 and rem means remainder when N is divided by 5.
3. Find the last non zero digit of 2000!
We know that Last non zero digit of 2000! = $\left[\frac{2000}{5}\right]!×{2}^{\left(2000/5\right)}=\left(400!×{2}^{400}\right)$
Again Last non zero digit of 400! = $\left[\frac{400}{5}\right]!×{2}^{\left(400/5\right)}=\left(80!×{2}^{80}\right)$
Again last non zero digit of 80! = $\left[\frac{80}{5}\right]!×{2}^{\left(80/5\right)}=\left(16!×{2}^{16}\right)$
Now 16! = $\left[\frac{16}{5}\right]!×{2}^{\left[16/5\right]}×{\left[\frac{16}{5}\right]}_{\mathrm{R}\mathrm{e}m}!$ = $3!×{2}^{3}×1!=8$
Now combining the whole discussion last non zero digit of 2000! = ${2}^{400}×{2}^{80}×{2}^{16}×8$ = 6 x 6 x 6 x 8 = 8 |
# How to Calculate the Geometric Average Return. Overview and Example
In this article, we're going to talk about how to calculate the geometric average rate of return which is also known as the compound annual rate of return or the compounded annual growth rate. So when we think about this compounded rate of return what we're going to do to calculate it is, we're going to take the return of each period, so let's say we've got years 1, years 2, and 3 here and we've got a rate of return of 25% in year 1, a rate of return of negative 40% in year 2 and a rate of return a 30% in year 3. So we're going to take those rates of return and we're going to compound them over time.
## The Formula of Calculating Geometric Average Return
So our formula is going to be as follows we're going to say 1.25 and we'll just take one point two five forget the percent for a moment we'll convert back to that later, (1.25 ✕ .6 1.3). Now here's how I come up with these numbers if we had \$100 at the beginning of the year and we're at the beginning of year 1 and we're thinking "How does that increase while there's a 25% rate of return." So we would multiply that \$100 by 1.25 that's how we would get what we have at the end of the period, and then the next period we're saying hey we have to multiply by (1 - 0.4) because we're losing actually 40%. So what are we going to keep if we're losing 0.4 proportion or 40%? that means we're keeping 60% or 0.6 of our investment. So we multiply our initial investment would be (1.25 .6 1.3) as we go 30% at the end of the period.
Now all of this we're going to take we're going to raise that to power and that power is going to be 1 divided by the number of periods, so we'll say 1 divided by N and in this case "n" is going to be 3 because we have three years here. Then we're going to take this whole thing and then we're going to subtract 1 from this whatever results in we get here. So if you do the math there that's going to give you, and this is going to be rounded here but it's going to give you a negative 0.0008. Let's say to make it easier to interpret we are going to round this number and this will be negative 0.01, which is the same as saying negative 1%. |
# Question: What Does 2nd Derivative Tell You?
## When can you not use the second derivative test?
If f'(x) doesn’t exist then f”(x) will also not exist, so the second derivative test is impossible to carry out..
## Why does the second derivative test fail?
If f ′(c) = 0 and f ″(c) < 0, then f has a local maximum at c. Else, the test fails (if f ′(c) doesn't exist, or f ″(c) = 0, or f ″(c) doesn't exist). Note: Even though it is often easier to use than the first derivative test, the second derivative test can fail at some points, as noted above.
## Why does the second derivative determine concavity?
If the second derivative of a function f(x) is defined on an interval (a,b) and f ”(x) > 0 on this interval, then the derivative of the derivative is positive. Thus the derivative is increasing! In other words, the graph of f is concave up. Similarly, if f ”(x) < 0 on (a,b), then the graph is concave down.
## How do you do the second derivative test?
Set f ‘ (x) = 0, and solve for x. Plug your solution(s) from step 2 into f ‘ ‘ (x) and use the rules set forth in the second derivative test to determine if there is a maximum or minimum point at these values. Plug the same values (from step 2) back into f(x) to find the actual value of the relative maxima or minima.
## How do you interpret the second derivative graph?
This is read aloud as “the second derivative of f. If f″(x) is positive on an interval, the graph of y = f(x) is concave up on that interval. We can say that f is increasing (or decreasing) at an increasing rate. If f″(x) is negative on an interval, the graph of y = f(x) is concave down on that interval.
## What is the difference between first and second derivative?
The first derivatives are used to find critical points while the second derivative is used to find possible points of inflection. By itself, a first derivative equal to 0 at a point does not tell you whether that point is actually an extrema.
## What does it mean when the second derivative test fails?
If f (x0) = 0, the test fails and one has to investigate further, by taking more derivatives, or getting more information about the graph. Besides being a maximum or minimum, such a point could also be a horizontal point of inflection.
## What is the second derivative rule?
If the second derivative is positive over an interval, indicating that the change of the slope of the tangent line is increasing, the graph is concave up over that interval. … CONCAVITY TEST: If f ”(x) < 0 over an interval, then the graph of f is concave upward over this interval.
## What happens when the second derivative is 0?
The second derivative is zero (f (x) = 0): When the second derivative is zero, it corresponds to a possible inflection point. If the second derivative changes sign around the zero (from positive to negative, or negative to positive), then the point is an inflection point.
## What does a positive second derivative mean?
As stated above, if the second derivative is positive, it implies that the derivative, or slope is increasing, while if it is negative, implies that the slope is decreasing. As a graphical example, consider the graph, y=(x)(x−2)(x−3) which looks like this.
## What does the first derivative tell you?
The first derivative of a function is an expression which tells us the slope of a tangent line to the curve at any instant. Because of this definition, the first derivative of a function tells us much about the function. If is positive, then must be increasing. If is negative, then must be decreasing.
## What is the second derivative test used for?
The second derivative may be used to determine local extrema of a function under certain conditions. If a function has a critical point for which f′(x) = 0 and the second derivative is positive at this point, then f has a local minimum here. |
## Simple Problems On Geometric Progression
Below are four problems on geometric series with simple calculations
Question 1
Find the n-th term of the series -1/2, 1/4, -1/8,...(when n is an even number)
a) 1/2n b)2n c) -1/4n d)1/4n
Answer : a) 1/2n
Solution :
The given series -1/2, 1/4, -1/8,... is a geometric series where a = -1/2
and r = [1/4 / (-1/2)] = -1/2
Then, the nth term of the G.P = tn = a(r(n-1))
tn = (-1/2)x(-1/2)(n-1)
tn = (-1/2)n = (-1)n x (1/2)n
tn = 1/2n (since (-1)n = 1 when n is an even number)
Hence, the answer is 1/2n.
Question 2
The total number of terms in the series 1, 3, 9,....,19683 is:
a) 11 b) 10 c) 21 c) 17
Solution:
The given geometric series is 1, 3, 9,....,19683 where a = 1, r = 3 and last term = 19683
Let us find the total number of terms
i.e, we have to find the position of the last term(19683).
The nth term in G.P represented by tn = ar(n-1)
Now, tn = 1 x 3(n-1) = 19683
1 x 3(n-1) = 19683
3(n-1) = 39
n-1 = 9
n = 10.
Therefore, the nth term of the series is 19683.
Hence, there are 10 terms in the given G.P
Question 3
The sum of the first 15 terms of the series 45 + -9 + 9/5 + -9/25 + .... is equal to:
(here C is a constant)
a) 75C/2 b) 45C/2 c) 45C/4 d) 54C
Solution:
The given G.P series is 45, (-9), 9/5, (-9/25), ....
Here, a = 45, r = -1/5 and n = 15
The sum of the n terms in G.P series when r < 1 = S(n) = a(1 - rn) / (1 - r)
Now, S(n) = 45[1 - (-1/5)15 ] /(1-(-1/5))
= 45 x [1 + 1/515] / (6/5)
= 45 x 5/6 x [1 + 1/515]
= 75/2 x [1 + 1/515]
= 75C/2 where C = 1 + 1/515 is a constant.
Hence the answer is option a.
Question 4
If there are 20 terms in the geometric series 2, 8, 32, 128,.... then the sum of all the terms is equal to:
a) 3(420 - 1)/2 b) 2(420 - 1)/3 c) 2(1-420) d) 2(1+420)
Answer: b) 2(420 - 1)/3
Solution:
The Given G.P series is 2, 8, 32, 128,....
Then we have a = 2, r = 8/2 = 4 and n = 20
The sum of the n terms in G.P series when r > 1 = S(n) = a(rn - 1) / (r-1)
Now, S(n) = 2(420 - 1) / (4-1)
= 2(420 - 1)/3
Hence, the answer is option b.
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Solving Linear Inequalities - Systems of Equations and Inequalities - High School Algebra I Unlocked (2016)
## High School Algebra I Unlocked (2016)
### Lesson 5.3. Solving Linear Inequalities
Now that you’re a pro at graphing and finding solutions for linear equations and simultaneous equations, let’s turn our focus to graphing and solving linear inequalities. The approach for working with linear inequalities is similar to that for working with linear equations, with one slight difference: Since an inequality provides a range of values, your solution will also be a range of values, rather than a single value, graphed on the coordinate plane.
Graphing a System of Inequalities
1.Solve each inequality for y.
2.Graph each inequality independently from one another.
•If the inequality contains a ≥ or ≤, the boundary line will be solid.
•If the inequality contains a > or <, the boundary line will be dashed.
3.Select a point not on the line and see if the inequality holds true.
•If the inequality holds true, shade the area on the side of the boundary line that contains the test point.
•If the inequality is false, shade the area on the side of the boundary line that does not contain the test point.
If boundary lines enclose the solution region, and the shaded region is contained, the solution is bounded.
If boundary lines do not enclose the solution region, and the shaded region extends forever, the solution is unbounded.
line does not go
through the origin,
use the point (0, 0)
much easier!
Let’s see how we’d use this process to solve a system of linear inequalities.
EXAMPLE
y ≥ 2x + 10
yx + 8
Solve the system of inequalities graphically and state whether the solution region is bounded or unbounded.
Since the question requires us to solve the system of inequalities, follow the process for graphing linear inequalities outlined on the previous page. Here, both equations are already in slope-intercept form, and the signs in each equation are greater than or equal to, so you can graph each with a solid boundary line, as shown on the graph below.
Choose a point not on the first boundary line to determine where on the graph the inequality holds true. Let’s start by plugging in the point (0, 0) into the first equation, y ≥ 2x + 10.
y ≥ 2x + 10
0 ≥ 2(0) + 10
0 ≥ 10
Since 0 is not greater than 10, the point (0, 0) is not a solution to the first inequality. So you will shade the side of the boundary line that does not contain the point (0, 0).
Now choose a point not on the second boundary line to determine where on the graph the inequality holds true. Again, the point (0, 0) is not on the line, so plug in the point (0, 0) into the second equation, yx + 8.
yx + 8
0 ≥ 0 + 8
0 ≥ 8
Like before, the inequality is not true; 0 is not greater than 8. Therefore, the point (0, 0) is not a solution to the first inequality, and you will shade the side of the boundary line that does not contain the origin.
Congratulations! You’ve finished graphing your first system of linear inequalities. The solution to this system, called the solution region, is the area where the two solutions overlap. Additionally, since the graph extends beyond the shown boundaries, the solution is referred to as unbounded, which means it continues in a direction forever; in this case, the solution to the system is unbounded in the upper half-plane.
Planes and Corner Points
When discussing the solution to a system of linear inequalities, you may be asked to refer to the shaded area according to planes. This means that when you graph an inequality in the coordinate plane, the graph is divided into half-planes, which can then be further split into either left and right half-planes by a vertical boundary line or upper and lower half-planes by a non-vertical boundary line.
You may also be asked to find the corner point(s) in a system of inequalities, which is the point(s) in the solution region where two boundary lines intersect.
Let’s try a question that involves corner points.
Here is how you may see a systems of equations problem involving parallel lines on the ACT.
Side AB of parallelogram ABCD is shown in the figure below. If the coordinates of A are (7, 6) and those of B are (5, 1), then CD could lie on which of the following lines?
A. y = x + 9
B. y = x + 5
C. y = x − 4
D. y = −x + 4
E. y = −x − 9
EXAMPLE
2y + 2 + 6x < 4
4 − 2y ≤ 12x + 2
Solve the system of inequalities graphically. Then state whether the solution region is bounded or unbounded, and find the coordinates of each corner point.
As in the previous example, follow the process for graphing linear inequalities. Before we can graph the lines, however, we need to rewrite them in slope-intercept form, or in the form y = mx + b.
2y + 2 + 6x < 4 4 − 2y ≤ 12x + 2 2y < −6x + 4 − 2 −2y ≤ 12x + 2 − 4 2y < −6x + 2 −2y ≤ 12x − 2 y < −3x + 1 y ≥ −6x + 1
Great! Now that the equations are in slope-intercept form, graph each line. Note that the first equation has a less than sign and will be graphed with a dashed boundary line, while the second equation has a greater than or equal to sign and will be graphed with a solid boundary line.
Once you’ve graphed the lines, choose a point not on the first boundary line to determine where on the graph the inequality holds true. Since the point (0, 0) is not on the first boundary line, let’s plug the point (0, 0) into the equation, y < −3x + 1.
y < −3x + 1
0 < −3(0) + 1
0 < 1
Since 0 is less than 1, the point (0, 0) is in the solution set of the first inequality. Therefore, you will shade the side of the boundary line that contains the point (0, 0).
Next, select a point not on the second boundary line to determine where on the graph the inequality holds true. As with the first equation, the point (0, 0) is not on the second boundary line, so you can plug the point (0, 0) into the second equation, y ≥ −6x + 1.
y ≥ −6x + 1
0 −6(0) + 1
0 1
This question requires you to find the equation of line CD, which is a side of the parallelogram ABCD. But how can you find the equation of line CD if it isn’t even shown? Well, because line CD does not include either point A or point B, line CD must be parallel to line AB. Since parallel lines have identical slopes, start by finding the slope of line AB using the coordinates of point A, (7, 6), and point B, (5, 1).
Thus, the slope of line AB is and, since parallel lines have identical slopes, you know that line CD must also have a slope of . The only equation that has a slope of is (A), y = x + 9. The correct answer is (A).
In this case, 0 is not greater than 1, so the point (0, 0) is not a solution to the second inequality, and we need to shade the side of the boundary line that does not contain the point (0, 0).
Now that we’ve found the solution region, you need to find the corner point, or the point at which the two boundary lines intersect. Find the corner point in a system of inequalities using the same methods we used to find the point of intersection in a system of equations. Since the equations are already in y = mx + b form, and y is isolated in the equations y < −3x + 1 and y ≥ −6x + 1, use the algebraic method to find x:
−3x + 1 −6x + 1
−3x + 6x ≥ 1 − 1
3x ≥ 0
x ≥ 0
Now that we know x ≥ 0, we can solve for y by plugging x = 0 into either of the equations:
y < −3x + 1 y ≥ −6x + 1 y < −3(0) + 1 y ≥ −6(0) + 1 y < 1 y ≥ 1
Here we find that y < 1 and y ≥ 1, indicating that y = 1. Thus, the corner point in this system is (0, 1). You’ll notice that this is the exact point where the boundary lines intersect on your graph! If the boundary lines are straightforward, and the corner point can be easily determined by looking at the graph, you do not need to use the algebraic method to find the point of intersection. Instead, you can determine the corner point directly from the graph.
Now we’re done. The system is unbounded in the lower plane, the corner point is (0, 1), and the solution region is the area in which the shaded regions overlap.
EXAMPLE
2yx ≤ 2
7x < 14y − 28
Solve the system of inequalities graphically. Then state whether the solution region is bounded or unbounded, and find the coordinates of each corner point.
Again, follow the process for graphing linear inequalities. The first step is to rewrite the inequalities in slope-intercept form.
2y − x ≤ 2 7x < 14y − 28 2y ≤ x + 2 14y > 7x + 28 y ≤ x + 1 y > x + 2
Now that the equations are in slope-intercept form, graph each line. Here, the first equation has a less than or equal to sign and requires a solid boundary line, while the second equation has a less than sign and requires a dashed boundary line.
After graphing the boundary lines, select a point that is not on the first line to determine which side of the boundary line to shade. Here, the origin is not on the line, so you can plug in the point (0, 0) into the equation, y ≤ 1/2x + 1:
y ≤ 1/2x + 1
0 ≤ 1/2(0) + 1
0 ≤ 1
The statement 0 ≤ 1 is true, so (0, 0) is in the solution set of the first inequality, and you will shade the side of the boundary line that contains the origin.
Next, select a point not on the second boundary line to determine which side of the boundary line to shade. Again, the origin is not on the second line, so plug in the point (0, 0) into the second equation, y > 1/2x + 2:
y > 1/2x + 2
0 > 1/2(0) + 2
0 > 2
Since the statement 0 > 2 is false, the origin is not in the solution set of the second inequality, and you need to shade the side of the boundary line that does not contain the origin.
This system of inequalities does not look the other systems we’ve looked at, as there is no overlapping region. But don’t worry: The lack of an overlapping region is the solution. Whenever you graph a system of inequalities and find that there is no overlapping region, the system has no solution region and no corner point. And that is exactly what happened with this system—there is no solution.
So can you have a system of inequalities that consists of more than two inequalities? Absolutely! Luckily, the process for tackling these questions is identical to the process we’ve been using to tackle systems of inequalities with two inequalities. Look at the following example.
EXAMPLE
y + x ≥ 8
y ≥ 0
2 − x ≤ 0
Solve the system of inequalities graphically. Then state whether the solution region is bounded or unbounded, and find the coordinates of each corner point.
Follow the process for graphing linear inequalities. The first step, once again, is to rewrite the inequalities in slope-intercept form.
y + x ≥ 8 y ≥ 0 2 − x ≤ 0 y ≥ −x + 8 y ≥ 0 x ≥ 2
Unlike the previous questions we’ve done, we have three inequalities here, two of them containing only one variable. That’s okay, though. Recall that single-variable linear equations are either horizontal or vertical lines; the same is true for single-variable linear inequalities. Graph solid boundary lines, as all the inequalities contain greater than or equal to signs.
As we’ve done in previous questions, select a point that is not on the first boundary line to determine which side of the line to shade. Here, the origin is not on the line, so plug the point (0, 0) into the equation, y ≥x + 8.
y ≥x + 8
0 −0 + 8
0 8
The statement 0 8 is false, so the origin is not in the solution set of the first inequality, and we’ll shade the side of the boundary line that does not contain the origin.
Next, select a point not on the second boundary line to determine which side of the line to shade. Here, the origin is on the second line, so you must plug in a different point; let’s plug the point (1, 1) into the second equation, y ≥ 0:
y ≥ 0
1 0
Since the statement 1 0 is true, the origin is in the solution set of the second inequality, and you need to shade the side of the line that contains the point (1, 1).
Now, select a point not on the third boundary line to determine which side of the line to shade. Here, the origin is not on the second line, so you can plug in (0, 0) into the second equation, x ≥ 2.
x ≥ 2
0 2
The statement 0 2 is false, so the origin is not in the solution set of the third inequality, and you need to shade the side of the boundary line that does not contain the origin.
Fantastic! You successfully found the solution region of this system of inequalities and, because this system extends forever, it is considered an unbounded system. Now, you just need to find the corner points in this system. By looking at the graph, we see that the lines intersect at points (2, 6) and (8, 0).
Thus, this is an unbounded system with corner points at (2, 6) and (8, 0).
We’ve covered tons of information in this chapter, so before moving on, make sure you understand the concepts discussed here. You can assess your skills by completing the practice questions on the next page, as well as by completing the Reflect activity at the end of the chapter.
DRILL
CHAPTER 5 PRACTICE QUESTIONS
Directions: Complete the following problems as specified by each question. For extra practice, try using an alternative method to solve the problem or check your work.
1. y + 6x = 5 − 4x
y = 2x + 3
Solve the system of linear equations by graphing.
2. What is the slope of a line perpendicular to the line x − 4y = 10 + 2xy ?
3. 4y + 12 − 2x = 2y + 20
12 − 3x = 3y
Solve the system of equations by substitution.
4. 3y + 6x −120 = −90 + 5x
8 + 2x = 4y − 16
Solve the system of equations algebraically, graph the system, and find the value of + 2x.
5. Which of the following lines is the graph of a line parallel to the line defined by the equation 14x + 4y − 4 = 0 ?
A)
B)
C)
D)
6. y ≤ 4x − 7
2y ≥ −2x +
Solve the system of inequalities graphically and state whether the solution region is bounded or unbounded.
7. −8y > 16x + 4
y > −x + 2
Solve the system of inequalities graphically and state whether the solution region is bounded or unbounded.
8. y + 2x ≤ −2
3y ≥ 9 − 3x
−18x ≥ −36
Solve the system of inequalities graphically and state whether the solution region is bounded or unbounded. Then find the coordinates of each corner point.
SOLUTIONS TO CHAPTER 5 PRACTICE QUESTIONS
1. (1, −5)
To solve the system of equations graphically, start by graphing each equation in the coordinate plane. First, though, you need to rewrite each equation in slope-intercept form. The first equation, y + 6x = 5 − 4x, becomes y = 5 − 4x − 6x and y = − 10x + 5; thus, the first equation has a slope of −10 and a y-intercept at (0, 5). Repeat this process with the second equation to find that −y = 2x + 3 becomes y = −2x − 3; thus, the second equation has a slope of −2 and a y-intercept at (0, −3). Graph each equation as follows.
Now, to find the solution to the system of equations, you need to identify the point of intersection. Therefore, refer to the graph and find the point where the two lines meet.
Here, the two lines intersect at (1, −5).
2. 3
This question requires you to find the slope of a line perpendicular to the line x − 4y = 10 + 2xy. First, you need to identify the slope of the given line, so rewrite the equation in y = mx + b form:
x − 4y = 10 + 2xy
x − 2x − 10 = 4yy
x − 10 = 3y
3y = −x − 10
Therefore, the slope of the original line is −1/3. Now you need to find the slope of a line perpendicular to the original line; recall that perpendicular lines have slopes that are negative reciprocals of one another. Therefore, a line perpendicular to the original line will have a slope of −(−3) = 3.
3. (0, 4)
This question asks you to solve the system of equations by substitution. To use the substitution method, start by isolating either the x or y value in either equation. You can isolate y in the first equation, 4y + 12 − 2x = 2y + 20, as follows:
4y + 12 − 2x = 2y + 20
4y − 2y = 2x + 20 − 12
2y = 2x + 8
y = x + 4
Now, substitute the value y = x + 4 into the second equation, 12 − 3x = 3y, to solve for x as follows:
12 − 3x = 3y
12 − 3x = 3(x + 4)
12 − 3x = 3x + 12
12 − 12 = 3x + 3x
0 = 6x
x = 0
Next, substitute x = 0 into either of the original equations. When you substitute x = 0 into the second equation, 12 − 3x = 3y, you will find that
12 − 3(0) = 3y
12 − 0 = 3y
3y = 12
y = 4
Accordingly, the lines intersect at (0, 4), which is the solution to this system of equations.
4. 13.8
In order to solve this system of equations algebraically, you first need to isolate either the x- or y-variable in each equation. Start by isolating the y-variable, putting the equations in slope-intercept form so that you can graph them easily.
3y + 6x − 120 = −90 + 5x 8 + 2x = 4y − 16 3y = −6x + 5x − 90 + 120 4y = 2x + 8 + 16 3y = −x + 30 4y = 2x + 24 y = −1/3x + 10 y = 1/2x + 6
Next, graph the two equations in the coordinate plane, noting that the first equation has a slope of −1/3 and a y-intercept at (0, 10), and the second equation has a slope of 1/2 and a y-intercept at (0, 6). The graph will appear as follows:
In addition to graphing the equations, isolating the y-variable allows you to set the two equations equal to one another and solve for x:
−1/3x + 10 = 1/2x + 6
1/2x + 1/3x = 10 − 6
3/6x + 2/6x = 4
5/6x = 4
5x = 24
x = 24/5 or 4.8
Next, plug x = 4.8 into either equation to solve for y. If you plug x = 4.8 into the second equation, you will find that
y = 1/2x + 6
y = 1/2(4.8) + 6
y = 2.4 + 6
y = 8.4
Therefore, the solution to this system of equations is (4.8, 8.4). However, the question asks you to find the value of y/2 + 2x, so plug in the x- and y-values to find that
y/2 + 2x =
8.4/2 + 2(4.8) =
4.2 + 9.6 = 13.8
Accordingly, the value of y/2 + 2x is 13.8.
5. B
For this question, you need to identify the graph of a line parallel to the line defined by the equation 14x + 4y − 4 = 0. The first step is to identify the slope of the line. When an equation is written in standard form, ax2 + by + c = 0, the slope of a line is equal to −a/b. In the given equation, a = 14 and b = 4, so the slope is −a/b = −14/4 = −7/2. Since the question asks you to find a line parallel to the given equation, you need to identify a graph that also has a slope of −7/2. You can quickly eliminate (A) and (D) because they depict equations that have a positive slope. Now you must determine which of the remaining graphs has a slope of −7/2. The equation in (B) passes through points (3, 0) and (1, 7) and has a slope of ; keep (B). Conversely, (C) passes through points (2, 0) and (0, 2) and has a slope of = −1; eliminate (C). Thus, the correct answer is (B).
6. Unbounded
Follow the process for graphing linear inequalities. First, graph the two inequalities in the coordinate plane, as you would a linear equation. While the first equation, y ≤ 4x − 7, is in slope-intercept form, you need to rewrite the second equation, 2y ≥ −2x + 1/2, in slope-intercept form as follows: 2y ≥ −2x + 1/2 and y ≥x + 1.
Since the sign in the first equation is less than or equal to, and the sign in the second equation is greater than or equal to, you will graph each with a solid boundary line:
Next, pick a point not on the first boundary line to determine where on the graph the inequality holds true. Let’s plug the point (0, 0) into the first equation, y ≤ 4x − 7 to find that 0 ≤ 4(0) − 7, 0 ≤ 0 − 7, and 0 ≤ −7. Since this is a false statement, you will shade the side of the boundary line that does not contain the point (0, 0).
Now pick a point not on the second boundary line to determine where on the graph the inequality holds true. Again, the point (0, 0) is not on the line, so plug the point (0, 0) into the second equation, y ≥x + 1 to find that 0 −0 + 1 and 0 1. Since this is a false statement, you will shade the side of the boundary line that does not contain the point (0, 0).
As shown in the graph, the solution for the system of inequalities extends forever along the right half-plane; thus, the solution is unbounded.
7. No solution
In order to solve this system of inequalities, you must first rewrite the inequalities in slope-intercept form.
Next, graph each line, using dashed lines due to the less than and greater than signs used in each equation.
Now select a point that is not on the first line to determine which side of the boundary line to shade. Here, the origin is not on the line, so you can plug the point (0, 0) into the equation, y < −2x − 1/2 to find that 0 < −2(0) − 1/2, 0 < 0 − 1/2, and 0 < −1/2. Since the statement 0 < −1/2 is false, (0, 0) is in the solution set of the first inequality, and you will shade the side of the boundary line that does not contain the origin.
Next, select a point not on the second boundary line to determine which side of the boundary line to shade. Again, the origin is not on the second line, so plug the point (0, 0) into the second equation, y > −2x + 4, to find that 0 > −2(0) + 4, 0 > 0 + 4, and 0 > 4. Since the statement 0 > 4 is false, the origin is not in the solution set of the second inequality, and you need to shade the side of the boundary line that does not contain the origin.
Here, there is no overlapping region, so the system has no solution region and no corner point.
8. Unbounded; Corner point: (−7/3, 16/3)
Follow the process for graphing linear inequalities. The first step is to rewrite the inequalities in slope-intercept form.
Next, graph each inequality with solid boundary lines, as all the inequalities contain greater than or equal to or less than or equal to signs.
Next, select a point that is not on the first boundary line, y ≤ −4x − 4, to determine which side of the line to shade. Since the origin is not on the line, plug the point (0, 0) into the equation, y ≤ −4x − 4 to find that 0 ≤ −4(0) − 4, 0 ≤ 0 − 4, and 0 ≤ −4. Since 0 ≤ −4 is a false statement, the origin is not in the solution set of the first inequality, and you will shade the side of the boundary line that does not contain the origin.
Next, select a point not on the second boundary line to determine which side of the line to shade. Here, the origin is not on the second line, so plug the point (0, 0) into the second equation, y ≥x + 3, to find that 0 −(0) + 3 and 0 3. Since the statement 0 3 is false, the origin is not in the solution set of the second inequality, and you need to shade the side of the line that does not contain the origin.
Now, select a point not on the third boundary line to determine which side of the line to shade. Here, the origin is not on the second line, so you can plug in (0, 0) into the second equation, x ≤ 2 to find that 0 ≤ 2. The statement 0 ≤ 2 is true, so the origin is in the solution set of the third inequality, and you need to shade the side of the boundary line that contains the origin.
In this scenario, the solution region is an unbounded system, outlined as follows.
Finally, you need to find the corner points in this system. Unfortunately, due to the scale of our graph, you can’t find the corner points just by looking at the graph. Instead, you need to find the intersection points of the first two inequalities, y ≤ −4x − 4 and y ≥x + 3. When you convert these inequalities into equations and set them equal to one another, you will find that −x + 3 = −4x − 4, 3x = −7, and x = −7/3. Now, substitute x = −7/3 into the second equation to find y: y = −x + 3, y = −(−7/3) + 3, y = 7/3 + 9/3, and y = 16/3. Thus, the corner point is (−7/3, 16/3).
REFLECT
Congratulations on completing Chapter 5!
Here’s what we just covered.
•Define and categorize systems of linear equations
1 2 3 4 5
•Solve systems of equations questions with methods such as graphing, substitution, addition or subtraction, and algebra
1 2 3 4 5
•Calculate the slopes of parallel and perpendicular lines
1 2 3 4 5
•Graph and solve linear inequalities
1 2 3 4 5
If you rated any of these topics lower than you’d like, consider reviewing the corresponding lesson before moving on, especially if you found yourself unable to correctly answer one of the related end-of-chapter questions.
CHAPTER 5 KEY POINTS
A system of equations is a set of equations that share variables and work together in the coordinate plane.
The solution to a system of equations will be the point(s) of intersection. A system of equations can have an infinite number of solutions, one solution, or no solution.
When a system of equations has no solution, the equations are referred to as inconsistent. Conversely, if a system of equations has either an infinite number of solutions or exactly one solution, the equations are referred to as consistent.
A system of equations with exactly one solution is known as an independent system, as each equation provides unique information. Conversely, a system of equations with an infinite number of solutions is known as a dependent system, because the equations provide identical information.
Systems of equations can be solved in multiple ways: graphing, substitution, adding or subtracting equations, or by using algebra.
If you use the substitution method and cannot find the value of the second variable, and instead find that both sides of the equations are equivalent numerical values, the system of equations has an infinite number of solutions.
Parallel lines have equal slopes. If the original line has a slope of 4, all lines parallel to the original line will also have slopes of 4.
The slope of a perpendicular line is equal to the negative reciprocal of the original line. If the original line has a slope of 4, a line perpendicular to the original line will have a slope of −1/4.
Use the following process for graphing a system of linear inequalities:
○ Solve each inequality for y.
○ Graph each inequality independently of one another.
• If the inequality contains a or ≤, the boundary line will be solid.
• If the inequality contains a > or <, the boundary line will be dashed.
○ Select a point not on the line to see if the inequality holds true.
• If the inequality holds true, shade the area on the side of the boundary line that contains the test point.
• If the inequality is false, shade the area on the side of the boundary line that does not contain the test point.
The solution region of a system of inequalities is the area where the solutions overlap.
When you graph an inequality in the coordinate plane, the graph is divided into half-planes.
○ A vertical line divides the coordinate plane into left and right half-planes.
○ A non-vertical line divides the coordinate plane into upper and lower half-planes.
A corner point in a system of inequalities is the point in the solution region where two boundary lines intersect.
|
This activity is a game with two dice as an intro to probability. The activity was taken from Math from Three to Seven (again)
`Supplies/Preparation: Two dice and prepare a game board by drawing an 8x15 table (or print out the one in the download). Label the bottom row with numbers from 1 through 15.`
```What to do: Each person gets two chips (or beads) and puts each over a number (put one of yours on the number 1). Then we roll two dice. They add the numbers on the dice. If the sum match where one of their chips number, he/she gets to move that chip up one cell. Repeat rolling the dice, moving chips if appropriate.
The winner is the one who gets to the top row. ```
After playing a couple of times the full game, you can ask why the chip in cell 1 never moved. If they don’t know, that’s ok, you can leave them to think about it and figure it out later.
We will draw the addition table and investigate the possible numbers for the sum in a later activity.
### How it went
It went well. We each had three chips. Nia got to place the first chip, then Bel, then me. We had chips on 1, 3, 5, 7, 9, 10, 11, 13, and 15. Nia managed to give the correct sum most of the times by counting on her fingers. After a few rolls, Bel said “it looks like there is a pattern” (she noticed that we chose all the odd numbers – although she called them even and I corrected her) then commented on the numbers in the middle appearing more often.
When I lamented that my chip on 1 didn’t move up. Bel asked what the lowest number on the dice were then told me that 1+1 is 2 so it would never happen. After a few turns, she also said that 13, 14, and 15 would not happen either. When I asked Nia what that meant, she wasn’t interested. BUT, on the second game, she did not put any chip on 1, 13, 14, or 15 so she must have heard our conversation.
At the end of the second game, Bel won again (with 7) and Nia started to complain that Bel always chose 7. I told her that she got the first pick and could have chosen 7. She wanted to play a third time. I didn’t want to continue but said they could keep playing with the dice. I was hoping Nia would practice her counting.
They made up their own game: They wanted to see what the second number to win would be. Bel wanted to rule out 7, 8, and 9, and see which of the other numbers would win. They lost interest really fast because I had taken the game board away (and in retrospect should have let them play a little more but I didn’t want it to get lost in the papers around the house).
### A riddle
There are two men. One of them is wearing a red shirt, and the other is wearing a blue shirt. The two men are named Andrew and Bob, but we do not know which is Andrew and which is Bob.
The guy in the blue shirt says, “I am Andrew.”
The guy in the red shirt says, “I am Bob.”
If we know that at least one of them lied, then what color shirt is Andrew wearing?
For more puzzles, go to https://brilliant.org/wiki/truth-tellers-and-liars/ |
# How do you integrate int x^3/(x^2+2)^3 by integration by parts method?
Jan 23, 2017
Integration by parts is: $\int u \mathrm{dv} = u v - \int v \mathrm{du}$
choose $u$ so that $\mathrm{dv}$ can be integrated by variable substitution.
#### Explanation:
let $u = {x}^{2} \mathmr{and} \mathrm{dv} = \frac{x}{{x}^{2} + 2} ^ 3 \mathrm{dx}$, because this will allow:
$v = \int \frac{x}{{x}^{2} + 2} ^ 3 \mathrm{dx}$
to be integrated by letting $t = {x}^{2} + 2$, then $\mathrm{dt} = 2 x \mathrm{dx}$ or $\frac{\mathrm{dt}}{2} = x \mathrm{dx}$
This makes the integral become:
$v = \frac{1}{2} \int {t}^{-} 3 \mathrm{dt}$
$v = - \frac{1}{4} {t}^{-} 2$
Reversing the substitution:
v=-1/(4(x^2+2)^2
and $\mathrm{du} = 2 x \mathrm{dx}$
$\int {x}^{3} / {\left({x}^{2} + 2\right)}^{3} \mathrm{dx} = \left({x}^{2}\right) \left(- \frac{1}{4 {\left({x}^{2} + 2\right)}^{2}}\right) - \int - \frac{1}{{x}^{2} + 2} ^ 2 \left(2 x\right) \mathrm{dx}$
$\int {x}^{3} / {\left({x}^{2} + 2\right)}^{3} \mathrm{dx} = - {x}^{2} / \left(4 {\left({x}^{2} + 2\right)}^{2}\right) + \frac{1}{4} \int \frac{2 x}{{x}^{2} + 2} ^ 2 \mathrm{dx}$
The last integral is the same sort of variable substitution:
$\int {x}^{3} / {\left({x}^{2} + 2\right)}^{3} \mathrm{dx} = - {x}^{2} / \left(4 {\left({x}^{2} + 2\right)}^{2}\right) - \frac{1}{4 \left({x}^{2} + 2\right)} + C$ |
What is the sum of the roots of all the quadr...
What is the sum of the roots of all the quadratic equations that can be formed such that both the roots of the quadratic equation are common with the roots of equation (x – a) (x – b) (x – c) = 0?
(2014)
• a)
3(a + b + c)
• b)
2(a + b + c)
• c)
(a + b + c)
• d)
4(a + b + c)
What is the sum of the roots of all the quadratic equations that can b...
The equations formed by the roots of the equation
(x – a) (x – b)(x – c) can be as follows:
(i) (x – a)(x – b) ⇒ Roots are a, b
(ii) (x – b)(x – c) ⇒ Roots are b, c
(iii) (x – c)(x – a) ⇒ Roots are c, a
(iv) (x – a)2 ⇒ Roots are a, a
(v) (x – b)2 ⇒ Roots are b, b
(vi) (x – c)2 ⇒ Roots are c, c
Adding all these roots, we get 4(a + b + c).
What is the sum of the roots of all the quadratic equations that can b...
The quadratic equation (x - a)(x - b) = 0, where a and b are the roots, can be expanded as x^2 - (a + b)x + ab = 0.
For the quadratic equation to have roots in common with (x - a)(x - b) = 0, the discriminant of (x - a)(x - b) = 0, which is (a + b)^2 - 4ab, must be zero.
Setting the discriminant to zero, we have (a + b)^2 - 4ab = 0.
Expanding this equation, we get a^2 + 2ab + b^2 - 4ab = 0.
Simplifying, we have a^2 - 2ab + b^2 = 0.
This equation can be factored as (a - b)^2 = 0.
Taking the square root of both sides, we have a - b = 0.
Therefore, the sum of the roots of all the quadratic equations that can be formed such that both the roots of the quadratic equation are common with the roots of (x - a)(x - b) = 0 is a + b.
So, the sum of the roots is a + b.
Related Test
Test: Algebra - 130 Ques | 50 Mins
What is the sum of the roots of all the quadratic equations that can b...
The equations formed by the roots of the equation
(x – a) (x – b)(x – c) can be as follows:
(i) (x – a)(x – b) ⇒ Roots are a, b
(ii) (x – b)(x – c) ⇒ Roots are b, c
(iii) (x – c)(x – a) ⇒ Roots are c, a
(iv) (x – a)2 ⇒ Roots are a, a
(v) (x – b)2 ⇒ Roots are b, b
(vi) (x – c)2 ⇒ Roots are c, c
Adding all these roots, we get 4(a + b + c).
### Learn this topic in detail
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What is the sum of the roots of all the quadratic equations that can be formed such that both the roots of the quadratic equation are common with the roots of equation (x – a) (x – b) (x – c) = 0?(2014)a)3(a + b + c)b)2(a + b + c)c)(a + b + c)d)4(a + b + c)Correct answer is option 'D'. Can you explain this answer?
Question Description |
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# wikiHow to Find the Domain and Range of a Function
Every function contains two types of variables: independent variables and dependent variables, whose values literally “depend” on the independent variables. For example, in the function y = f(x) = 2x + y, x is independent and y is dependent (in other words, y is a function of x). The valid values for a given independent variable x are collectively called the “domain.” The valid values for a given dependent variable y are collectively called the “range.”[1]
### Part 1 Finding the Domain of a Function
1. 1
Determine the type of function you’re working with. The domain of the function is all of the x-values (horizontal axis) that will give you a valid y-value output. The function equation may be quadratic, a fraction, or contain roots. To calculate the domain of the function, you must first evaluate the terms within the equation.
• A quadratic function has the form ax2 + bx + c:[2] f(x) = 2x2 + 3x + 4
• Examples of functions with fractions include: f(x) = (1/x), f(x) = (x + 1)/(x - 1), etc.
• Functions with a root include: f(x) = √x, f(x) = √(x2 + 1), f(x) = √-x, etc.
2. 2
Write the domain with proper notation. Writing the domain of a function involves the use of both brackets [,] and parentheses (,). You use a bracket when the number is included in the domain and use a parenthesis when the domain does not include the number. The letter U indicates a union that connects parts of a domain that may be separated by a gap.[3]
• For example, a domain of [-2, 10) U (10, 2] includes -2 and 2, but does not include number 10.
• Always use parentheses if you are a using the infinity symbol, ∞.
3. 3
Draw a graph of the quadratic equation. Quadratic equations make a parabolic graph that either points up or down. Given that the parabola will continue infinitely outward on the x-axis, the domain of most quadratic function is all real numbers. Stated another way, a quadratic equation encompasses all of the x-values on the number line, making its domain R (the symbol for all real numbers).[4]
• To get an idea of the function choose any x-value and plug it into the function. Solving the function with this x-value will output a y-value. These x- and y-values are a coordinate (x, y) of the graph of the function.
• Plot this coordinate and repeat the process with another x-value.
• Plotting a few values in this fashion should give you a general idea of shape of the quadratic function.
4. 4
Set the denominator equal to zero, if it’s a fraction. When working with a fraction, you can never divide by zero. By setting the denominator equal to zero and solving for x, you can calculate the values that will be excluded in the function.[5]
• For example: Identify the domain of the function f(x) = (x + 1)/(x - 1).
• The denominator of this function is (x - 1).
• Set it equal to zero and solve for x: x – 1 = 0, x = 1.
• Write the domain: The domain of this function cannot include 1, but includes all real numbers except 1; therefore, the domain is (-∞, 1) U (1, ∞).
• (-∞, 1) U (1, ∞) can be read as the set of all real numbers excluding 1.The infinity symbol, ∞, represents all real numbers. In this case, all real numbers greater than 1 and less than one are included in the domain.
5. 5
Set the terms inside the radical to be greater than or equal to zero, if there’s a root function. You cannot take the square root of a negative number; therefore, any x-value that leads to a negative number must be excluded from the domain of that function.[6]
• For example: Identify the domain of the function f(x) = √(x + 3).
• The terms within the radical are (x + 3).
• Set them greater than or equal to zero: (x + 3) ≥ 0.
• Solve for x: x ≥ -3.
• The domain of this function includes all real numbers greater than or equal to -3; therefore, the domain is [-3, ∞).
### Part 2 Finding the Range of a Quadratic Function
1. 1
Confirm that you have a quadratic function. A quadratic function has the form ax2 + bx + c: f(x) = 2x2 + 3x + 4. The shape of a quadratic function on a graph is parabola pointing up or down. There are different methods to calculating the range of a function depending on the type you are working with.[7]
• The easiest way to identify the range of other functions, such as root and fraction functions, is to draw the graph of the function using a graphing calculator.
2. 2
Find the x-value of the vertex of the function. The vertex of a quadratic function is the tip of the parabola. Remember, a quadratic equation is of the form ax2 + bx + c. To find the x-coordinate use the equation x = -b/2a. This equation is a derivative of the basic quadratic function which represents the equation with a zero slope (at the vertex of the graph, the slope of the function is zero).[8]
• For example, find the range of 3x2 + 6x -2.
• Calculate x-coordinate of vertex: x = -b/2a = -6/(2*3) = -1
3. 3
Calculate the y-value of the vertex of the function. Plug the x-coordinate into the function to calculate the corresponding y-value of the vertex. This y-value denotes the edge of your range for the function.
• Calculate y-coordinate: y = 3x2 + 6x – 2 = 3(-1)2 + 6(-1) -2 = -5.
• The vertex of this function is (-1, -5).
4. 4
Determine the direction of the parabola by plugging in at least one more x-value. Choose any other x-value and plug it into the function to calculate the corresponding y-value. If the y-value is above the vertex, the parabola continues to +∞. If the y-value is below the vertex, the parabola continues to -∞.
• Use the x-value -2: y = 3x2 + 6x – 2 = y = 3(-2)2 + 6(-2) – 2 = 12 -12 -2 = -2.
• This yields the coordinate (-2, -2).
• This coordinate tells you that the parabola continues above the vertex (-1, -5); therefore, the range encompasses all y-values above -5.
• The range of this function is [-5, ∞)
5. 5
Write the range with proper notation. Like the domain, the range is written with the same notation. Use a bracket when the number is included in the domain and use a parenthesis when the domain does not include the number. The letter U indicates a union that connects parts of a domain that may be separated by a gap.[9]
• For example, a range of [-2, 10) U (10, 2] includes -2 and 2, but does not include number 10.
• Always use parentheses if you are a using the infinity symbol, ∞.
### Part 3 Finding the Range of a Function Graphically
1. 1
Graph the function. Oftentimes, it is easiest to determine the range of a function by simply graphing it. Many root functions have a range of (-∞, 0] or [0, +∞) because the vertex of the sideways parabola is on the horizontal, x-axis. In this case, the function encompasses all of the positive y-values if the parabola goes up, or all of the negative y-values if the parabola goes down. Fraction functions will have asymptotes that define the range.[10]
• Some root functions will start above or below the x-axis. In this case, the range is determined by the point the root function starts. If the parabola starts at y = -4 and goes up, then the range is [-4, +∞).
• The easiest way to graph a function is to use a graphing program or a graphing calculator.
• If you do not have a graphing calculator, you can draw a rough sketch of a graph by plugging x-values into the function and getting the corresponding y-values. Plot these coordinates on the graph to get an idea of the shape of the graph.
2. 2
Find the minimum of the function. Once you have graphed the function, you should be able to clearly see the lowest point of the graph. If there is no obvious minimum, know that some functions will continue on to -∞.
• A fraction function will include all points except those at the asymptote. They often have ranges such as (-∞, 6) U (6, ∞).
3. 3
Determine the maximum of the function. Again, after graphing, you should be able to identify the maximum point of the function. Some functions will continue on to +∞ and therefore, will not have a maximum.
4. 4
Write the range with proper notation. Like the domain, the range is written with the same notation. Use a bracket when the number is included in the domain and use a parenthesis when the domain does not include the number. The letter U indicates a union that connects parts of a domain that may be separated by a gap.[11]
• For example, a range of [-2, 10) U (10, 2] includes -2 and 2, but does not include number 10.
• Always use parentheses if you are a using the infinity symbol, ∞. |
1.
Introduction-
2.
Solution Approach
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Last Updated: Mar 27, 2024
# Find the Array which when sorted forms an AP and has Least Maximum
Riya
1 upvote
## Introduction-
This blog will discuss the problem to find the array which when sorted forms an AP and has least maximum. Before going into details of the problem, let's recall what an AP is:
"Arithmetic progression (AP) is a sequence of numbers in which difference of any two consecutive numbers is constant called the common difference."
In this problem, we will be given three positive integers' N', 'X,' and 'Y,' where X is less than Y. We have to find an array of positive integers containing both X and Y, which when sorted forms an AP and the length of the array should be 'N.' One more condition is that the maximum element of the array should be the least possible.
Let's understand the problem with an example:
Suppose N = 5, X = 20 and Y = 50
We have to find an array whose length is 5 containing 20 and 30 and which, when sorted forms an AP and has least maximum.
Consider the array: arr = {20, 30, 40, 50, 10}
Its sorted form will be: {10, 20, 30, 40, 50}
Here we can see that array "arr" has 5 elements containing both 20 and 50 and, when sorted, forms an AP with a common difference of 10.
Now that we understand the problem, let's discuss the approach to solve it in the next section.
## Solution Approach
This section will discuss the approach to find the array which when sorted forms an AP and has least maximum. In simple words, we have to find the 'N' elements of the required array, out of which two elements, ' X' and 'Y,' are already given.
So, we have to find the remaining 'N-2' elements of the array such that when the array is sorted, it forms an AP, and the maximum element of the array should be the least possible.
The simple logic behind this approach is if we have to make the maximum element of the array the least possible, then we have to insert as many elements as possible between X and Y. The maximum number of elements that we can insert between X and Y is 'N-2'. So, we have to start with 'num = N-2' and keep checking whether it is possible to insert the 'num' number of elements between X and Y. If it is not possible to insert, decrease 'num' by 1.
After finding the number of elements to be inserted between X and Y, calculate the common difference of the AP, which will be formed after sorting the required array. Then find the required elements of the array. To check whether it is possible to insert the 'num' number of elements between X and Y, check if we can get an integer 'common difference 'by inserting the 'num' number of elements between X and Y.
Also see, Euclid GCD Algorithm
## Algorithm -
Step 1. Create a function "findRequiredArray()" to find the array which when sorted forms an AP and has least maximum, which takes 'N,' 'X,' and 'Y' as inputs and return the vector of size 'N' containing both 'X' and 'Y' and when sorted forms an AP.
Step 2. Inside the function "findRequiredArray()," first initialize a variable 'num = N-2' to store the number of elements to be inserted between X and Y. Run a 'while loop' and check if we can get an integer 'common difference' by inserting the 'num' number of elements between X and Y, else decrease 'num' by 1. Break the while loop if the 'num' becomes zero as the minimum number of elements between X and Y can be zero.
Step 3. After finding the number of elements to be inserted between X and Y, calculate the common difference of the AP, which will be formed after sorting the required array. Then create a vector and store the elements between X and Y (both inclusive). Also, create a variable to keep the count of elements inserted in the vector.
Step 4. Check if the number of elements inserted in the vector is less than N, then insert the elements less than X and greater than 0, satisfying the common difference criteria. If still the number of elements in the vector is less than N, insert the elements greater than Y.
Step 5. Finally, return the vector. It will contain the elements which when sorted, will form an AP.
## Algorithm Complexity:
Time Complexity: O(N )
In the function "findRequiredArray()" to find the array which when sorted forms an AP and has least maximum, loops are run, each having a maximum iteration of N. So, the overall time complexity is O(N), where 'N' is the length of the required array.
Space Complexity: O(N)
In the function "findRequiredArray()" to find the array which when sorted forms an AP and has least maximum, we have created a vector to store the required array of integers of size N. So, the space complexity is O(N), where 'N' is the length of the required array.
## FAQs
1. What is AP (Arithmetic Progression)?
Arithmetic progression (AP) is a sequence of numbers in which the difference of any two consecutive numbers is constant called the common difference.
2. In the approach to solve the problem, why we have tried to maximize the number of elements between ‘X’ and ‘Y’?
In this problem, there is a condition to minimize the maximum possible element of the AP that we are generating and we have to include ‘X’ and ‘Y’ as well. So, if the number of elements between ‘X’ and ‘Y’ will be more, the common difference of the generated AP will be less. And we know that the number of elements of the AP is fixed to be ‘N’ in the problem.
Hence the largest element of the AP will be the least possible if we will take the least possible common difference of the AP.
## Key takeaways-
This article discussed the problem "Find the array which when sorted forms an AP and has least maximum," the solution approach to this problem, its C++ implementation, and its time and space complexity.
If you want to solve similar problems on data structures and algorithms for practice, you can visit Coding Ninjas Studio.
If you think that this blog helped you share it with your friends!. Refer to the DSA C++ course for more information.
Until then, All the best for your future endeavors, and Keep Coding.
Live masterclass |
## Lesson: Comparing and Ordering Integers and Decimals Developing the Concept
Now that students have created their own number lines and acquired an initial understanding of ordering and comparing decimals, they can expand their understanding by comparing and ordering integers.
Materials: your model number line and the number line students completed from the first lesson; large model of a number line from 0 to 5; blank number line with write-on lines, for each student (see below)
Preparation: Post your model number line from 0 to 5 where students can see it. Construct a model number line from 0 to 5. Prepare student number lines with write-on lines. Use Learning Tool 5 in the Learning Tools Folder.
• Say: Take out your number lines, because we are going to use them to answer some questions.
• Ask: Where would you find 4, or positive 4, on your number line? (Write 4 and +4 on the board.)
Your students should say that +4 is to the right of 0 on the number line. Explain that positive numbers can be written with or without a positive sign and are found to the right of 0. Have your students put positive signs in front of 1, 2, 3, 4, and 5 on their number lines while you put positive signs on your model number line from 0 to 5.
• Ask: Where would you find negative 4? (Write 4 on the board.)
Students should say that 4 is to the left of 0 on the number line. If they don't, tell them that negative numbers are found to the left of 0. Show students the model number line from 0 to 5. Attach it to your model number line from 0 to +5, overlapping the number lines so that only one zero is showing. Explain that negative numbers must always have a negative sign in front of them. Point out that zero is neither positive nor negative, so neither sign is used with zero.
• Say: These two number lines together make up the set of integers. Integers include zero, the positive numbers, and the negative numbers. (Point to zero, positive numbers, and negative numbers as you name them.)
• Ask: Where is the number +1.5 on the number line? Where is the number 1.5 on the number line?
Your students should use their number lines to point out where +1.5 is. They may be able to tell you where 1.5 is on your number line. If they don't, tell them that 1.5 is between 1 and 2. Then help them determine exactly where 1.5 is on your number line.
• Say: The numbers +1.5 and 1.5 are called opposites. This means that they are both the same distance from 0 on the number line.
Help students count the number of units from 0 for each number.
Ask several other questions in which students locate numbers on the number line and identify and then locate the opposites of those numbers.
• Ask: Which is greater, +1.5 or 1.5?
Students should say that positive numbers are always greater than negative numbers; therefore +1.5 is greater than 1.5. Remind them that numbers to the right are always greater than the numbers to their left. Write +1.5 > 1.5 on the board.
• Ask: Which is greater, 2.1 or 1.7?
Students may incorrectly say that 2.1 is greater than 1.7. If they do, have them locate 2.1 and 1.7 on the number line. Then remind them that numbers to the right are always greater than the numbers to their left.
Write 1.7 > 2.1 on the board.
Ask several other questions in which students locate two numbers on the number line and then identify which of the numbers is the greater or lesser of the two.
• Say: Now that you have a good idea of how you can find and compare integers on a number line, each of you will extend your number line to include negative numbers, using the number line I will pass out to you.
Ask students to write neatly and to carefully place the numbers on the number line.
Have the students join the two number lines so that they have one number line that looks like the number line from 5 to +5 that you have modeled.
Wrap-Up and Assessment Hints
Create several true-or-false statements and have students use their number lines to tell you whether a statement is true or false. If the statement is false, have students rewrite it to make a true statement.
+2.3 > 2.3 (True)
1.8 > +0.9 (False; +0.9 > 1.8 or 1.8 < +0.9)
Zero is an integer. (True)
The number 8 is to the left of 150 on a number line. (True)
Negative 3 can be written as 3 or 3. (False; negative 3 can only be written as 3; positive 3 can be written as 3 or +3.)
The numbers 2.4 and +2.4 are opposites of each other. (True)
All positive numbers are to the left of zero on a number line. (False; all positive numbers are to the right of zero on a number line; all negative numbers are to the left of zero on a number line.) |
### Teddy Town
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Four Triangles Puzzle
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Three Squares
What is the greatest number of squares you can make by overlapping three squares?
# Rod Ratios
##### Age 7 to 11Challenge Level
We had just a few solutions sent in for this task.
Monty from the British School in Brussels in Belgium said:
I decided to use the grid mode and count the squares in order to work out the ratios. Here are some of my answers:
Gianna, also from the International School of Brussels in Belgium sent in this good piece of work:
Well done also to Isis (South Island School), Rohaan (Long Bay Primary) and Samantha (The Hamlin School) for completing this task! Samantha explained how she manged to find a pair with the same ratio as the yellow and pink rods:
The pink rod had 4 units and the yellow one had 5 units. This meant that they had a ratio of 4:5. I then needed to find a like ratio. This can be done by multiplying both numbers in the ratio by the same number. In this case the number had to be 2 because the longest rod is 10 units. So the 8 (brown coloured) and the 10 (orange coloured) rods would have the same ratio as the 4 (pink coloured) and the 5 (yellow coloured).
She also found pairs with the same ratio as the combined yellow+orange to blue:
The ratio of yellow (5) + orange (10) to blue (9) is equal to the ratio 15 to 9. This ratio can be simplified to a 5:3 ratio by dividing both 15 and 9 by 3. Therefore, the yellow and green rods form an equal ratio to that of the yellow + orange to the blue. Another equal ratio would be 10 to 6. Therefore, the dark green and orange rods also form an equal ratio.
Finally she noted that the rods with the same ratio as 9:6 have already been given!
The ratio of 9:6 can be simplified to the ratio 3:2. The possible combinations with rods for that ratios are shown at the start of the question!
Thank you to all those who sent in their work, thoughts and ideas. |
AP 10th Class Maths Important Questions Chapter 5 Quadratic Equations
These AP 10th Class Maths Chapter Wise Important Questions Chapter 5 Quadratic Equations will help students prepare well for the exams.
AP State Syllabus 10th Class Maths 5th Lesson Important Questions and Answers Quadratic Equations
Question 1.
If b2 – 4ac ≥ 0, then write the roots of a quadratic equation ax2 + bx + c = 0
Solution:
When b2 – 4ac ≥ 0 then the roots of given quadratic equation ax2 + bx + c = 0 are
$$\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$$ and $$\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$$
Question 2.
Find the Quadratic polynomial with zeroes -2 and $$\frac{1}{3}$$
Solution:
Let α = -2; β = $$\frac{1}{3}$$
= -2 + $$\frac{1}{3}$$ = $$\frac{-6+1}{3}=\frac{-5}{3}$$
α.β = -2. $$\frac{1}{3}$$ = $$\frac{-2}{3}$$
Quadratic polynomial is [x2 – (α + β )x + αβ ] = [x2 + $$\frac{5}{3}$$x – $$\frac{-2}{3}$$
the quadratic polynomial will be 3x2 + 5x – 2
Question 3.
Two angles are complementary and one angle is 18° more than tne other, then find angles.
Solution:
Let smaller angle be x°
bigger angle be y°
Since these two angles are comple-mentary
x + y = 90° ………… (1)
Since bigger angle is more than smaller angle
by 18°, y – x = 18° ………….(2)
By solving (1) & (2), we get x = 36°, y = 54°
Question 4.
Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0.
Solution:
ax2 + bx + c = 0 is b2 – 4ac
Now comparing the given quadratic equation 2x2 – 4x + 3 = 0 with stan¬dard form of quadratic equation.
We get a = 2, b = -4, c = 3 then its discriminant = b2 – 4ac
= (-4)2 – 4(2) (3)
= 16 – 24 = -8
∴ Discriminant = -8.
Question 1.
Find the roots of x + $$\frac { 6 }{ x }$$ = 7, x ≠ 0
Solution:
x + $$\frac { 6 }{ x }$$ = 7 ⇒ $$\frac{x^{2}+6}{x}$$ = 7
⇒ x2 – 7x + 6 =0
⇒ (x – 6) (x – 1) = 0
⇒ x = 6 or 1
Roots = 6, 1
Question 2.
Length of a rectangle is 2 units greater than its breadth. If the area of the rect-angle is 120 sq. units then find its length.
Solution:
Let breadth of the rectangle = x
Length = x + 2 ,
Area = 120 sq. units
x(x+2) = 120
x2 + 2x- 120 = 0
(x + 12) (x- 10) = 0
x = – 12 or 10
∴ Breadth of the rectangle = x
= 10 units
∴ Length = x + 2 = 10 + 2 = 12 units
Question 3.
Find the zeroes of the quadratic poly-nomial x2 – x – 30 and verify the rela¬tion between the zeroes and its co¬efficients.
Solution:
Given quadratic polynomial = x2 – x – 30
⇒ x2 – x – 30 = 0 = 0
x2 – 6x + 5x – 30 = 0
⇒ x(x – 6) + 5(x – 6) = 0
⇒ (x – 6) (x + 5) = 0
⇒ x – 6 = 0
x = 6
x + 5 = 0 x = -5
∴ Zeroes are α = 6 and β = – 5
Sum of zeroes = α + β = $$\frac{-\mathrm{b}}{\mathrm{a}}$$
⇒ 6 – 5 = $$\frac{-(-1)}{1}$$
⇒ 1 = 1
Product of zeroes α + β = 6(-5) = $$\frac{\mathrm{c}}{\mathrm{a}}$$
= -30 = $$\frac{-30}{1}$$
Hence the relation was verified.
Question 4.
Find the roots of the quadratic equa¬tion (3x – 2)2 – 4(3x – 2) + 3 = 0.
Solution:
(3x – 2)2 – 4(3x – 2) + 3 = 0
9x2 + 4 – 12x – 12x + 8 + 3 = 0
9x2 – 24x +15 = 0
3x2 – 8x + 5 = 0
3x2 – 3x – 5x + 5 = 0
3x(x – 1) – 5 (x – 1) = 0
(x- 1) (3x – 5) = 0
x = 1 (or) x = $$\frac{5}{3}$$
Roots of quadratic equation are 1, $$\frac{5}{3}$$.
Question 5.
Two numbers differ by 4 and their product is 192. Find the numbers.
Solution:
Let the larger number be ‘x’
Since the difference of two numbers is 4,
Then the smaller number is (x – 4)
Their product = x(x – 4)
Given that product = 192
∴ x(x – 4) = 192
⇒ x2 – 4x – 192 = 0
⇒ x2 – 16x + 12x- 192 = 0
⇒ x(x – 16) + 12(x – 16) = 0
⇒ (x – 16)(x + 12) = 0
⇒ x = 16 or x = -12
If x = 16, then x – 4 = 16 – 4 = 12
then the numbers are 16 and 12.
If x = – 12, thenx-4 = – 12 – 4 = – 16
then the numbers are – 12 and – 16.
Question 1.
Find the roots of the equation 5x2 – 7x – 6 = 0 by the method of completing the square.
Solution:
Given that 5x2 – 7x – 6 = 0
5x2 – 7x = 6
x2 – $$\frac{7}{5}$$x = $$\frac{6}{5}$$
⇒ x2.2.$$\frac{7}{10}$$x = $$\frac{6}{5}$$
adding $$\frac{49}{100}$$ on the both sides
The roots of the quadratic equation are 2 and $$\frac{-3}{5}$$
Question 2.
Find the roots of the quadratic equation 3x2 + 11x + 10 = 0 by method of – completing the Square.
Solution:
Given : 3x2 + 11x + 10 = 0
Dividing both sides by 3.
Question 3.
Solve the Quadratic equation 9x2 – 9x + 2 = 0 by the method of com-pleting the square.
Solution:
Given : 9x2 – 9x + 2 = 0
⇒ x2 – x + $$\frac{2}{9}$$ = 0
⇒ x2 – x = –$$\frac{2}{9}$$ |
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During the sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and same number of pencils and crayons, how many of each would you need to buy?
Last updated date: 15th Jul 2024
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Hint- In order to find the same number of crayons and pencils, try to solve using L.C.M.
Number of colour pencils to be packed in a packet $= 24$
Number of crayons to be packed in a packet $= 32$
We have to find the L.C.M of $24$ and $32$.
$24 = 2 \times 2 \times 3 \\ 32 = 2 \times 2 \times 2 \times 2 \times 2 \\$
L.C.M of $24$ and $32$ ${\text{ = }}2 \times 2 \times 2 \times 2 \times 2 \times 3 = 96$
Capacity of $1$ packet of colour pencils $= 24$
So, for 96 pencils, number of packets needed $= \dfrac{{96}}{{24}} = 4$
$3$Now, capacity of $1$ packet of crayons $= 32$
SO, for 96 crayons, number of packets needed $= \dfrac{{96}}{{32}} = 3$
$\therefore$ In order to buy full packs of both and same number of pencils and crayons, we need to buy $4$ packets of colour pencils and $3$ packets of crayons.
Note- L.C.M stands for Lowest Common Multiple. For any two numbers a and b, L.C.M is the smallest positive integer that is divided by both a and b. Hence, whenever you see problems like these, L.C.M is the shortest way to find solutions. |
Maths - Combined Rotation and Translation
Prerequisites
How can we combine rotations and translations without using matrices?
If you are not familiar with this subject you may like to look at the following pages first:
Combined Rotation and Translation
In VRML and related standards there is the concept of a transform group. This is a node in a scenegraph which contains the following parameters:
center="0 0 0"
rotation="0 0 1 0"
scale="1 1 1"
scaleOrientation="0 0 1 0"
translation="0 0 0"
To simplify things at this stage assume that the node contains only:
center="0 0 0"
rotation="0 0 1 0"
translation="0 0 0"
Here we cover the rotation about a point (not necessarily the origin). This can be solved using matrix methods as shown here.
First as a comparison consider a rotation about the origin.
• Angle = theta
• translation = 0
Now imagine the same object being rotated by the same angle, but about point P.
• angle = theta
• translation = T = C - rotate(C)
Where:
• T = translation caused by the rotation.
• C = the point we are rotating about.
• R(C) = the rotate function applied to point C
So the angle of rotation is the same whatever point we rotate about, but if we rotate about a point other than the origin this will translate the centre of the object by C - R(C).
Rotating Sum of two points
What is the effect of applying the rotation function R() to the sum of two points?
What this diagram is trying to show (I realise that the diagram is a bit messy) is that if:
Point P1, in absolute coordinates, is rotated to point R(P1) in local coordinates.
Point P2, in absolute coordinates, is rotated to point R(P2) in local coordinates.
Then:
Point P1+P2, in absolute coordinates, is rotated to point R(P1+P2) in local coordinates.
So:
R(P1+P2) = R(P1) + R(P2)
Transforms
Let us take a transform with the following parameters:
• T = translation
• R() = rotation function
• C = centre of rotation
To calculate the total transform in terms of these parameters, we need to first shift the centre of rotation to the origin, then rotate, then shift back to the original centre.
So first shift C to origin: Pabsolute = Plocal - C Now rotate about origin: Pabsolute = R(Plocal - C) Now move back to original point: Pabsolute = R(Plocal - C) + C
Pabsolute = R(Plocal - C) + C + T
The terms in the rotation function can be spilt up as described in 'Rotating Sum of two points' above.
Pabsolute = R(Plocal) - R(C) + C + T
Combining Transform Nodes
What is the combined effect of putting one transform node under another:
It is possible to find the resulting transform in terms of a 4x4 matrix. But what is an equivalent Transform node that will have the same effect as the two nodes above:
For on level of transform we have seen that the absolute position is given by:
Pabsolute = R(Plocal - C) + C + T
To cascade more than one of these we can calculate the position in coordinates of layer 'n' from the position in the coordinates of the layer below:
Pn-1 = Rn(Pn - Cn) + Cn + Tn
where:
• Pn = position of point in coordinates of layer n
• Rn = rotation function of layer n in coordinates of layer n-1
• Pn-1 = position of point in coordinates of the layer below n
• Cn = centre of rotation of layer n in coordinates of layer n-1
• Tn = translation on layer n in coordinates of layer n-1
So Pabsolute = Ra(Pa - Ca) + Ca + Ta
and Pa = Rb(Pb - Cb) + Cb + Tb
where:
• Ra = rotation function of transform a in absolute coordinates
• Ca = centre of rotation of transform a in absolute coordinates
• Ta = translation of transform a in absolute coordinates
• Rb = rotation function of transform b in coordinates of transform a
• Cb = centre of rotation of transform b in coordinates of transform a
• Tb = translation of transform b in coordinates of transform a
So substituting the Pa from the second equation into the first gives:
Pabsolute =Ra(Rb(Pb - Cb) + Cb + Tb - Ca) + Ca + Ta
Expanding out gives:
Pabsolute =Ra Rb(Pb - Cb) +Ra (Cb + Tb - Ca) + Ca + Ta
We want to put this into this format:
Pabsolute = Rt(Pb - Ct) + Ct + Tt
where:
• Pabsolute = position of point in absolute coordinates
• Rt = equivalent total rotation
• Pb = position of point in coordinates of transform b
• Ct = centre of equivalent total rotation
• Tt = equivalent total translation
so to make this equivalent to the two separate rotations makes:
• Rt = equivalent total rotation = Ra Rb
• Ct = centre of equivalent total rotation = Cb
• Tt = equivalent total translation = Ra (Cb + Tb - Ca) + Ca + Ta - Cb
So we rotate around a point which is equal to the centre of rotation of the first level rotation, but we have to adjust the translation by a factor which depends on the absolute rotation.
Example 1
Imagine that we are implementing a trackerball control. To do this we are translating an object in a transform under another transform, the lowest level transform rotates everything by 45 degrees. We pull the object across the screen and we want it to move across the screen but how do we correct for the bottom transform.
We are setting the lowest level transform: but we want it to be equivalent to the following equivalent total translation = Ra (Tb) equivalent total translation = Tin
Therefore Tb = Ra'(Tin)
In other words we apply the inverse rotation to the input translation which is minus 45degrees.
Example 2
As example 1 but with translation and off-centre..
We are setting the lowest level transform: but we want it to be equivalent to the following equivalent total translation = Ra (Tb - Ca) + Ca + Ta equivalent total translation = Ta + Tin
Therefore Ra (Tb - Ca) + Ca + Ta = Ta + Tin
Ra (Tb - Ca) + Ca = Tin
Tb = Ra'(Tin - Ca) + Ca
Scaling, Translation and Rotation.
We now want to add scaling to the following parameters:
• T = translation
• R() = rotation function
• C = centre of rotation
We add the following two parameters:
• The S (scale) field specifies a non-uniform scale of the coordinate system. Scale values shall be greater than zero.
• The SR (scaleOrientation) specifies a rotation of the coordinate system before the scale (to specify scales in arbitrary orientations). The scaleOrientation applies only to the scale operation.
The position in terms of local position is:
Pabsolute = R(SR(S × SR'(Plocal-C))) + T + C
Combining Transform Nodes with Scale
What is the combined effect of putting one transform node under another:
For on level of transform we have seen that the absolute position is given by:
Pabsolute = R(SR(S × SR'(Plocal-C))) + T + C
To cascade more than one of these we can calculate the position in coordinates of layer 'n' from the position in the coordinates of the layer below:
Pn-1 = Rn(SRn(Sn × SRn'(Pn-Cn))) + Tn + Cn
where:
• Pn = position of point in coordinates of layer n
• Rn = rotation function of layer n in coordinates of layer n-1
• Pn-1 = position of point in coordinates of the layer below n
• Cn = centre of rotation of layer n in coordinates of layer n-1
• Tn = translation on layer n in coordinates of layer n-1
• Sn = scale of layer n in coordinates of layer n-1
• SRn = scale rotation function of layer n in coordinates of layer n-1
So Pabsolute = Ra(SRa(Sa × SRa'(Pa-Ca))) + Ta + Ca
and Pa = Rb(SRb(Sb × SRb'(Pb-Cb))) + Tb + Cb
where:
• Ra = rotation function of transform a in absolute coordinates
• Ca = centre of rotation of transform a in absolute coordinates
• Ta = translation of transform a in absolute coordinates
• Sa = scale of transform a in coordinates of layer n-1
• SRa = scale rotation function of transform a in coordinates of layer n-1
• Rb = rotation function of transform b in coordinates of transform a
• Cb = centre of rotation of transform b in coordinates of transform a
• Tb = translation of transform b in coordinates of transform a
• Sb = scale of transform b in coordinates of layer n-1
• SRb = scale rotation function of transform b in coordinates of layer n-1
So substituting the Pa from the second equation into the first gives:
Pabsolute =Ra(SRa(Sa × SRa'((Rb(SRb(Sb × SRb'(Pb-Cb))) + Tb + Cb)-Ca))) + Ta + Ca
Can anyone help me expand this out and find an equivilant single transform?? |
# 6.1 Exponential functions (Page 3/16)
Page 3 / 16
• Let $\text{\hspace{0.17em}}b=1.\text{\hspace{0.17em}}$ Then $\text{\hspace{0.17em}}f\left(x\right)={1}^{x}=1\text{\hspace{0.17em}}$ for any value of $\text{\hspace{0.17em}}x.$
To evaluate an exponential function with the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ we simply substitute $x\text{\hspace{0.17em}}$ with the given value, and calculate the resulting power. For example:
Let $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.\text{\hspace{0.17em}}$ What is $f\left(3\right)?$
To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:
Let $\text{\hspace{0.17em}}f\left(x\right)=30{\left(2\right)}^{x}.\text{\hspace{0.17em}}$ What is $\text{\hspace{0.17em}}f\left(3\right)?$
Note that if the order of operations were not followed, the result would be incorrect:
$f\left(3\right)=30{\left(2\right)}^{3}\ne {60}^{3}=216,000$
## Evaluating exponential functions
Let $\text{\hspace{0.17em}}f\left(x\right)=5{\left(3\right)}^{x+1}.\text{\hspace{0.17em}}$ Evaluate $\text{\hspace{0.17em}}f\left(2\right)\text{\hspace{0.17em}}$ without using a calculator.
Follow the order of operations. Be sure to pay attention to the parentheses.
Let $f\left(x\right)=8{\left(1.2\right)}^{x-5}.\text{\hspace{0.17em}}$ Evaluate $\text{\hspace{0.17em}}f\left(3\right)\text{\hspace{0.17em}}$ using a calculator. Round to four decimal places.
$5.5556$
## Defining exponential growth
Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.
## Exponential growth
A function that models exponential growth grows by a rate proportional to the amount present. For any real number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and any positive real numbers and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}b\ne 1,$ an exponential growth function has the form
where
• $a\text{\hspace{0.17em}}$ is the initial or starting value of the function.
• $b\text{\hspace{0.17em}}$ is the growth factor or growth multiplier per unit $\text{\hspace{0.17em}}x$ .
In more general terms, we have an exponential function , in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function $\text{\hspace{0.17em}}A\left(x\right)=100+50x.\text{\hspace{0.17em}}$ Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function $\text{\hspace{0.17em}}B\left(x\right)=100{\left(1+0.5\right)}^{x}.$
A few years of growth for these companies are illustrated in [link] .
Year, $x$ Stores, Company A Stores, Company B
$0$ $100+50\left(0\right)=100$ $100{\left(1+0.5\right)}^{0}=100$
$1$ $100+50\left(1\right)=150$ $100{\left(1+0.5\right)}^{1}=150$
$2$ $100+50\left(2\right)=200$ $100{\left(1+0.5\right)}^{2}=225$
$3$ $100+50\left(3\right)=250$ $100{\left(1+0.5\right)}^{3}=337.5$
$x$ $A\left(x\right)=100+50x$ $B\left(x\right)=100{\left(1+0.5\right)}^{x}$
The graphs comparing the number of stores for each company over a five-year period are shown in [link] . We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.
Notice that the domain for both functions is $\text{\hspace{0.17em}}\left[0,\infty \right),$ and the range for both functions is $\text{\hspace{0.17em}}\left[100,\infty \right).\text{\hspace{0.17em}}$ After year 1, Company B always has more stores than Company A.
Now we will turn our attention to the function representing the number of stores for Company B, $\text{\hspace{0.17em}}B\left(x\right)=100{\left(1+0.5\right)}^{x}.\text{\hspace{0.17em}}$ In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and $\text{\hspace{0.17em}}1+0.5=1.5\text{\hspace{0.17em}}$ represents the growth factor. Generalizing further, we can write this function as $\text{\hspace{0.17em}}B\left(x\right)=100{\left(1.5\right)}^{x},$ where 100 is the initial value, $\text{\hspace{0.17em}}1.5\text{\hspace{0.17em}}$ is called the base , and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is called the exponent .
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# 2.1: Solving Linear Equations in One Variable using Integers
Difficulty Level: At Grade Created by: CK-12
Objectives
The lesson objectives for Solving Linear Equations with Variables are:
• Solving linear equations with variables on one side of the equation
• Solving linear equations with variables on both sides of the equation
• Solving linear equations that apply the distributive property
## Solving Linear Equations with Variables on One Side of the Equation
Introduction
In this concept you will begin your study of mathematical equations by learning how to solve equations with only one variable. What this means is that the mathematical expression will have a letter and possibly some number on one side of the equal sign and numbers on the other side of the equal sign. It is your job to find a value for the letter that makes the expression true. Once you have mastered this, you can solve for two variables and also add more complex numbers.
It is interesting to note that the variables are known as literal coefficients. Literal means “letter”. So the literal coefficient is the unknown quantity in the mathematical expression represented by a letter (also called a variable). What do you think a numerical coefficient is?
Right! A numerical coefficient is a number in the mathematical expression. In the algebraic term 5abc\begin{align*}5abc\end{align*}, the numerical coefficient is 5.
Watch This
Guidance
Erin, Jillian, Stephanie and Jacob went to the movies. The total bill for the tickets and snacks came to $72.00. How much should each teen pay to split the bill evenly. There are four teens going to the movies (Erin, Jillian, Stephanie, and Jacob). The total bill was$72.00. Therefore our equation is 4x=72\begin{align*}4x = 72\end{align*}. We divide by 4 to find our answer.
xx=724=18\begin{align*}x &= \frac{72}{4}\\ x &= 18\end{align*}
Therefore each teen will have to pay 18.00 for their movie ticket and snack. Example A 5a+2=17\begin{align*}5a + 2 = 17\end{align*} The problem can be solved if we think about the problem in terms of a balance (or a seesaw). We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance. In order to solve the equation, we have to get the literal coefficient (or the a\begin{align*}a\end{align*}) all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side. Let’s first subtract 2 from both sides to get rid of the 2 on the left. Since 5 is multiplied by a\begin{align*}a\end{align*}, we can get a by itself (or isolate it) by dividing by 5. Remember that whatever we do to one side, we have to do to the other. If we simplify this expression, we get: Therefore a=3\begin{align*}a = 3\end{align*}. We can check our answer to see if we are correct. 5a+25(3)+215+217=17=17=17=17 Y\begin{align*}5a + 2 &= 17\\ 5({\color{red}3}) + 2 &= 17\\ 15 + 2 &= 17\\ 17 &= 17 \ \ Y\end{align*} Example B 7b7=42\begin{align*}7b - 7 = 42\end{align*} Again, we can solve the problem if we think about the problem in terms of a balance (or a seesaw). We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance. In order to solve the equation, we have to get the literal coefficient (or the b\begin{align*}b\end{align*}) all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side. Let’s first add 7 from both sides to get rid of the 7 on the left. Since 7 is multiplied by b\begin{align*}b\end{align*}, we can get a by itself (or isolate it) by dividing by 7. Remember that whatever we do to one side, we have to do to the other. If we simplify this expression, we get: Therefore b=7\begin{align*}b = 7\end{align*}. We can check our answer to see if we are correct. 7b77(7)749742=42=42=42=42 Y\begin{align*}7b - 7 &= 42\\ 7({\color{red}7}) - 7 &= 42\\ 49 - 7 &= 42\\ 42 &= 42 \ \ Y\end{align*} Example C This same method can be extended by using algebra tiles. If we let white tiles represent the variable, green tiles represent the positive numbers and white tiles to represent the negative numbers, we can solve the equations using an alternate method. The green algebra x\begin{align*}x-\end{align*}tiles represent variables (or the literal coefficients). Therefore there are 3 c\begin{align*}c\end{align*} blocks for the equation. The other green blocks represent the numbers or constants. There is a 2 on the left side of the equation so there are 2 green blocks. There is an 11 on the right side of the equation so there are 11 green blocks on the right side of the equation. To solve, add two negative tiles to the right and left hand sides. The same rule applies to this problem as to all of the previous problems. Whatever we do to one side we have to do to the other. This leaves us with the following: We can reorganize these to look like the following: Organizing the remaining algebra tiles allows us to realize the answer to be x=3\begin{align*}x = 3\end{align*} or for our example b=3\begin{align*}b = 3\end{align*}. Let’s do our check as with the previous two problems. 3c+23(3)+29+211=11=11=11=11 Y\begin{align*}3c + 2 &= 11\\ 3({\color{red}3}) + 2 &= 11\\ 9 + 2 &= 11\\ 11 &= 11 \ \ Y\end{align*} Vocabulary Constant A constant is also a numerical coefficient but does not contain a variable. For example in the equation 4x+72=0\begin{align*}4x + 72 = 0\end{align*}, the 72 is a constant. Equation An equation is a mathematical equation with expressions separated by an equal sign. Numerical Coefficient In mathematical equations, the numerical coefficients are the numbers associated with the variable. For example, with the expression 4x\begin{align*}4x\end{align*}, 4 is the numerical coefficient and x\begin{align*}x\end{align*} is the literal coefficient. Variable A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient. Guided Practice 1. Use a model to solve for the variable in the problem x5=12\begin{align*}x-5=12\end{align*}. 2. Use a different model than used in question (1) to solve for the variable in the problem 3y+9=12\begin{align*}3y+9=12\end{align*}. 3. Using one of the models from the lesson, solve for x\begin{align*}x\end{align*} in the equation 3x2x+16=3\begin{align*}3x-2x+16=-3\end{align*}. Answers 1. x5=12\begin{align*}x-5=12\end{align*} Therefore, x=17\begin{align*}x=17\end{align*}. Let's do a check to make sure. x5(17)517512=12=12=12=12 Y\begin{align*}x - 5 &= 12\\ ({\color{red}17}) - 5 &= 12\\ 17 - 5 &= 12\\ 12 &= 12 \ \ Y\end{align*} 2. 3y+9=12\begin{align*}3y+9=12\end{align*} First you have to subtract 9 from both sides of the equation in order to start to isolate the variable. Now, in order to get y\begin{align*}y\end{align*} all by itself, you have to divide both sides by 3. This will isolate the variable y\begin{align*}y\end{align*}. Therefore, y=1\begin{align*}y=1\end{align*}. Let's do a check to make sure. 3y+93(1)+93+912=12=12=12=12 Y\begin{align*}3y + 9 &= 12\\ 3({\color{red}1}) +9 &= 12\\ 3 + 9 &= 12\\ 12 &= 12 \ \ Y\end{align*} 3. 3x2x+16=3\begin{align*}3x-2x+16=-3\end{align*} You can use any method to solve this equation. Remember to isolate the x\begin{align*}x\end{align*} variable. You will notice here that there are two x\begin{align*}x\end{align*} values on the left. First let’s combine these terms. 3x2x+16x+16=3=3\begin{align*}3x-2x+16 &= -3\\ x+16 &= -3\end{align*} Now you can use any method to solve the equation. You now should just have to subtract 16 from both sides to isolate the x\begin{align*}x\end{align*} variable. x+1616x=316=19\begin{align*}x+16 {\color{red}-16} &= -3 {\color{red}-16}\\ x &= -19\end{align*} Let's do a check to make sure. 3x2x+16x+1619+163=3(original problem)=3(simplified problem)=3=3 Y\begin{align*}3x-2x+16 &=-3 \quad \text{(original problem)}\\ x+16 &= -3 \quad \text{(simplified problem)}\\ {\color{red}-19}+16 &= -3\\ -3 &= -3 \ \ Y\end{align*} Summary Solving equations with variables on one side can be done using models such as a balance (or a seesaw) or by using algebra tiles. Remember that when solving equations with variables on one side of the equation there is one main rule to follow: whatever you do to one side of the equal sign you must do the same to the other side of the equal sign. For example, if you add a number to the left side of an equal sign, you must add the same number to the right side of the equal sign. Problem Set Use the model of the balance (or seesaw) to solve for each of the following variables. 1. a+3=5\begin{align*}a+3=-5\end{align*} 2. 2b1=5\begin{align*}2b-1=5\end{align*} 3. 4c3=9\begin{align*}4c-3=9\end{align*} 4. 2d=3\begin{align*}2-d=3\end{align*} 5. 4=3e=2\begin{align*}4=3e=-2\end{align*} Use algebra tiles to solve for each of the following variables. 1. x+3=14\begin{align*}x+3=14\end{align*} 2. 2y7=5\begin{align*}2y-7=5\end{align*} 3. 3z+6=9\begin{align*}3z+6=9\end{align*} 4. 5+3x=3\begin{align*}5+3x=-3\end{align*} 5. 2x+2=4\begin{align*}2x+2=-4\end{align*} Use the models that you have learned to solve for the variables in the following problems. 1. \begin{align*}-4x+13=5\end{align*} 2. \begin{align*}3x-5=22\end{align*} 3. \begin{align*}11-2x=5\end{align*} 4. \begin{align*}2x-4=4\end{align*} 5. \begin{align*}5x+3=28\end{align*} For each of the following models, write a problem involving an equation with a variable on one side of the equation expressed by the model and then solve for the variable. Answers Using the balance (seesaw) model... 1. \begin{align*}a+3=-5\end{align*} Therefore \begin{align*}a = - 8\end{align*} \begin{align*}\text{Check} &\\ a+3 &=-5\\ ({\color{red}-8})+3 &= -5\\ -5 &= -5 \ \ Y\end{align*} 1. \begin{align*}4c-3=9\end{align*} Therefore \begin{align*}c = 3\end{align*} \begin{align*}\text{Check} &\\ 4c-3 &= 9\\ 4({\color{red}3})-3 &= 9\\ 12-3 &= 9\\ 9 &= 9 \ \ Y\end{align*} 1. \begin{align*}4-3e=-2\end{align*} Therefore \begin{align*}e = 2 \end{align*} \begin{align*}\text{Check} &\\ 4-3e &= -2\\ 4-3({\color{red}2}) &= -2\\ 4-6 &= -2\\ -2 &= -2 \ \ Y\end{align*} Using algebra tiles... 1. \begin{align*}x+3=14\end{align*} Therefore \begin{align*}x = 11\end{align*} \begin{align*}\text{Check} &\\ x+3 &= 14\\ {\color{red}11}+3 &= 14\\ 14 &= 14 \ \ Y\end{align*} 1. \begin{align*}3z+6=9\end{align*} Therefore \begin{align*}z = 1\end{align*} \begin{align*}\text{Check} &\\ 3z+6 &= 9\\ 3({\color{red}1})+6 &= 9\\ 3+6 &= 9\\ 9 &= 9 \ \ Y\end{align*} 1. \begin{align*}2x+2=-4\end{align*} Therefore \begin{align*}x = -3\end{align*} \begin{align*}\text{Check}&\\ 2x+2 &= -4\\ 2({\color{red}-3})+2 &= -4\\ -6 + 2 &= -4\\ -4 &= -4 \ \ Y\end{align*} Use a model of your choice... 1. \begin{align*}-4x+13=5\end{align*} Therefore \begin{align*}x = 2\end{align*} \begin{align*}\text{Check} &\\ -4x+13 &= 5\\ -4({\color{red}2})+13 &= 5\\ -8+13 &= 5\\ 5 &= 5 \ \ Y\end{align*} 1. \begin{align*}11-2x=5\end{align*} Therefore \begin{align*}x = 3\end{align*} \begin{align*}\text{Check} &\\ 11-2x &= 5\\ 11-2(\color{red}3) &= 5\\ 11-6 &= 5\\ 5 &= 5 \ \ Y\end{align*} 1. \begin{align*}5x+3=28\end{align*} Therefore \begin{align*}x = 5\end{align*} \begin{align*}\text{Check} &\\ 5x+3 &= 28\\ 5({\color{red}5})+3 &= 28\\ 25+3 &= 28\\ 28 &= 28 \ \ Y\end{align*} For each of the following models... 1. This is \begin{align*}\text{Check} &\\ x+3 &= 7\\ {\color{red}4}+3 &= 7\\ 7 &= 7 \ \ Y\end{align*} \begin{align*}x+3 &= 7\\ x+3 {\color{red}-3} &= 7 {\color{red}-3}\\ x &= 4\end{align*} 1. This is \begin{align*}3x+6 &= 3\\ 3x+6 {\color{red}-6} &= 3 {\color{red}-6}\\ 3x &= -3\\ \frac{3x}{3} &= \frac{-3}{3}\\ x &= -1\end{align*} \begin{align*}\text{Check} &\\ 3x+6 &= 3\\ 3({\color{red}-1})+6 &= 3\\ -3+6 &= 3\\ 3 &= 3 \ \ Y\end{align*} 1. This is \begin{align*}-3x+5 &= 8\\ -3x+5 {\color{red}-5} &= 8 {\color{red}-5}\\ -3x &= 3\\ \frac{-3x}{-3} &= \frac{3}{-3}\\ x &= -1\end{align*} \begin{align*}\text{Check} &\\ -3x+5 &= 8\\ -3({\color{red}-1})+5 &= 8\\ 3+5 &= 8\\ 8 &= 8 \ \ Y\end{align*} ## Solving Linear Equations with Variables on Both Sides of the Equation Introduction In this concept you will learn to solve equations where there are variables on both sides of the equal sign. You will use the same models to solve these equations but the number of steps increases in these problems because now you have to combine like terms for the variables. Combining like terms means that you are putting all of the variables that match on the same side of the equation. A like term is one in which the degrees match and the variables match. So, for example, \begin{align*}3x\end{align*} and \begin{align*}4x\end{align*} are like terms, \begin{align*}3x\end{align*} and \begin{align*}4z\end{align*} are not. Three apples and four apples are like terms, three apples and four oranges are not. Remember that your goal for solving any of these problems is to get the variables on one side and the constants on the other side. You do this by adding and subtracting terms from both sides of the equal sign. Then you isolate the variables by multiplying or dividing. Learning the following lesson in solving equations with variables will help you when you learn the rules for solving more complex problems involving the distributive property in the next lesson. Watch This Guidance Thomas has50 and Jack has $100. Thomas is saving$10 per week for his new bike. Jack is saving 5 a week for his new bike. How long will it be before the two boys have the same amount of money? If we let \begin{align*}w\end{align*} be the number of weeks, we can write the following equation. \begin{align*}\underbrace{ 10x+50 }_{\text{Thomas's money:} \ \10 \ \text{per week} + \50}= \underbrace{ 5x+100 }_{\text{Jack's money:} \ \5 \ \text{per week} + \100}\end{align*} We can solve the equation now by first combining like terms. \begin{align*}10x+50 &= 5x+100\\ 10x {\color{red}-5x}+50 &= 5x {\color{red}-5x}+100 && \text{-combining the} \ x \ \text{variables to left side of the equation}\\ 5x+50 {\color{red}-50} &= 100 {\color{red}-50} && \text{-combining the constants to left side of the equation}\\ 5x &= 50\end{align*} We can now solve for \begin{align*}x\end{align*} to find the number of weeks until the boys have the same amount of money. \begin{align*}5x &= 50\\ \frac{5x}{5} &= \frac{50}{5}\\ x &= 10\end{align*} Therefore in 10 weeks Jack and Thomas will each have the same amount of money! Example A \begin{align*}x+4=2x-6\end{align*} We will solve this problem using the balance (or seesaw) used in the previous lesson. You could first try to get the variables all on one side of the equation. You do this by subtracting \begin{align*}x\end{align*} from both sides of the equation. Next, isolate the \begin{align*}x\end{align*} variable by adding 6 to both sides. Therefore \begin{align*}x = 10\end{align*}. \begin{align*}\text{Check}&\\ x+4 &= 2x-6\\ ({\color{red}10})+4 &= 2({\color{red}10})-6\\ 14 &= 20-6\\ 14 &= 14 \ \ Y\end{align*} Example B \begin{align*}14-3y=4y\end{align*} We will solve this problem using algebra tiles used in the previous lesson. We first have to combine our variables \begin{align*}(x)\end{align*} tiles onto the same side of the equation. We do this by adding \begin{align*}3 x\end{align*} tiles to both sides of the equal sign. In this way the \begin{align*}-3y\end{align*} will be eliminated from the left hand side of the equation. By isolating the variable \begin{align*}(y)\end{align*} we are left with these algebra tiles. Rearranging we will get the following. {Note: Remember that rearranging is not necessary, it simply allows you to quickly see what the value for the variable is.} \begin{align*}\text{Check} &\\ 14 - 3y &= 4y\\ 14-3({\color{red}2}) &= 4({\color{red}2})\\ 14-6 &= 8\\ 8 &= 8 \ \ Y\end{align*} Therefore \begin{align*}y = 2\end{align*}. Example C We can use these same methods for any of the equations involving variables. Sometimes, however, numbers are so large that one method is more valuable than the other. Let’s look at the following problem. \begin{align*}53a-99=42a\end{align*} To solve this problem, we may need to have a large number of algebra tiles! It might be more efficient to use the balance method to solve this problem. \begin{align*}\text{Check} &\\ 53a-99 &= 42a\\ 53({\color{red}9}) &= 42({\color{red}9})\\ 477-99 &= 378\\ 378 &= 378 \ \ Y\end{align*} Therefore \begin{align*}a = 9\end{align*}. Vocabulary Degrees The degree is the exponent on the variable in a term. For example, in the term \begin{align*}4x\end{align*}, the exponent is 1 so the degree is 1. Like Terms Like terms refer to terms in which the degrees match and the variables match. For example \begin{align*}3x\end{align*} and \begin{align*}4x\end{align*} are like terms. Variable A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient. Guided Practice 1. Use algebra tiles to solve for the variable in the problem \begin{align*}6x+4=5x-5\end{align*}. 2. Use the balance (seesaw) method to solve for the variable in the problem \begin{align*}7r-4=3+8r\end{align*}. 3. Determine the most efficient method to solve for the variable in the problem \begin{align*}10b-22=29-7b\end{align*}. Explain your choice of method for solving this problem. Answers 1. \begin{align*}6x+4=5x-5\end{align*} Therefore \begin{align*}x = -9\end{align*}. \begin{align*}\text{Check}&\\ 6x+4 &= 5x-5\\ 6({\color{red}-9})+4 &= 5({\color{red}-9})-5\\ -54+4 &= -45-5\\ -50 &= -50 \ \ Y\end{align*} 2. \begin{align*}7r-4=3+8r\end{align*} You can begin by combining the \begin{align*}r\end{align*} terms. Subtract \begin{align*}8r\end{align*} from both sides of the equation. You next have to isolate the variable. To do this, add 4 to both sides of the equation. But there is still a negative sign with the \begin{align*}r\end{align*} term. You now have to divide both sides by -1 to finally isolate the variable. Therefore \begin{align*}r = -7\end{align*}. \begin{align*}\text{Check} &\\ 7r-4 &= 3+8r\\ 7({\color{red}-7})-4 &= 3+8({\color{red}-7})\\ -49-4 &= 3-56\\ -53 &= -53 \ \ Y\end{align*} 3. \begin{align*}10b-22=29-7b\end{align*} You could choose either method but there are larger numbers in this equation. With larger numbers, the use of algebra tiles is not an efficient manipulative. You should solve the problem using the balance (or seesaw) method. Work through the steps to see if you can follow them. Therefore \begin{align*}b = 17\end{align*}. \begin{align*}\text{Check} &\\ 10b-22 &= 29-7b\\ 10({\color{red}3})-22 &= 29-7({\color{red}3})\\ 30-22 &= 29-21\\ 8 &= 8 \ \ Y\end{align*} Summary The methods used for solving equations with variables on both sides of the equation are the same as the methods used to solve equations with variables on one side of the equation. What differs in this lesson is that there is the added step of combining like terms with the variables before isolating the variable to find the solution. You must remember in these problems, as with any math problems involving an equal sign, whatever function (addition, subtraction, multiplication, or division) you do to one side of the equal sign, you must do to the other side. This is a big rule to remember in order for equations to remain equal or to remain in balance. Problem Set Use the balance (seesaw) method to find the solution for the variable in each of the following problems. 1. \begin{align*}5p+3=-3p-5\end{align*} 2. \begin{align*}6b-13=2b+3\end{align*} 3. \begin{align*}2x-5=x+6\end{align*} 4. \begin{align*}3x-2x=-4x+4\end{align*} 5. \begin{align*}4t-5t+9=5t-9\end{align*} Use algebra tiles to find the solution for the variable in each of the following problems. 1. \begin{align*}6-2d=15-d\end{align*} 2. \begin{align*}8-s=s-6\end{align*} 3. \begin{align*}5x+5=2x-7\end{align*} 4. \begin{align*}3x-2x=-4x+4\end{align*} 5. \begin{align*}8+t=2t+2\end{align*} Use the methods that you have learned for solving equations with variables on both sides to solve for the variables in each of the following problems. Remember to choose an efficient method to solve for the variable. 1. \begin{align*}4p-7=21-3p\end{align*} 2. \begin{align*}75-6x=4x-15\end{align*} 3. \begin{align*}3t+7=15-t\end{align*} 4. \begin{align*}5+h=11-2x\end{align*} 5. \begin{align*}9-2e=3-e\end{align*} For each of the following models, write a problem to represent the model and then find the variable for the problem. Answers Using the balance (seesaw) method... 1. Therefore \begin{align*}p = -1\end{align*}. \begin{align*}\text{Check} &\\ 5p+3 &= -3p-5\\ 5({\color{red}-1})+3 &= -3({\color{red}-1})-5\\ -5+3 &= 3-5\\ -2 &= -2 \ \ Y\end{align*} 1. Therefore \begin{align*}x = 11\end{align*}. \begin{align*}\text{Check}&\\ 2x-5 &= x+6\\ 2({\color{red}11})-5 &= ({\color{red}11})+6\\ 22-5 &= 11+66\\ 17 &= 17 \ \ Y\end{align*} 1. Therefore \begin{align*}t = 3\end{align*}. \begin{align*}\text{Check} &\\ 4t-5t+9 &= 5t-9\\ 4({\color{red}3})-5({\color{red}3})+9 &= 5{\color{red}(3)}-9\\ 12-15+9 &= 15-9\\ 6 &= 6 \ \ Y\end{align*} Using algebra tiles... 1. Therefore \begin{align*}x = -9\end{align*} \begin{align*}\text{Check}&\\ 6-2d &= 15-d\\ 6-2({\color{red}-9}) &= 15-({\color{red}-9})\\ 6+18 &= 15+9\\ 24 &= 24 \ \ Y\end{align*} 1. Therefore \begin{align*}x = -4\end{align*}. \begin{align*}\text{Check} &\\ 5x+5 &= 2x-7\\ 5({\color{red}-4})+5 &= 2({\color{red}-4})-7\\ -20+5 &= -8-7\\ -15 &= -15 \ \ Y\end{align*} 1. Therefore \begin{align*}t = 6\end{align*}. \begin{align*}\text{Check} &\\ 8+t &= 2t+2\\ 8+ {\color{red}6} &= 2({\color{red}6})+2\\ 14 &= 12+2\\ 14 &= 14 \ \ Y\end{align*} Choose a method... 1. Therefore \begin{align*}p = 4\end{align*}. \begin{align*}\text{Check} &\\ 4p-7 &= 21-3p\\ 4({\color{red}4})-7 &= 21-3({\color{red}4})\\ 16-7 &= 21-12\\ 9 &= 9\end{align*} 1. Therefore \begin{align*}t = 2\end{align*}. \begin{align*}\text{Check} &\\ 3t+7 &= 15-t\\ 3({\color{red}2})+7 &= 15-({\color{red}2})\\ 6+7 &= 13\\ 13 &= 13 \ \ Y\end{align*} 1. Therefore \begin{align*}e = 6\end{align*}. \begin{align*}\text{Check} &\\ 9-2e &= 3-e\\ 9-2({\color{red}6}) &= 3-({\color{red}6})\\ 9-12 &= -3\\ -3 &= -3 \ \ Y\end{align*} For each of the following models... 1. \begin{align*}-3x+17 &= 5-x\\ -3x {\color{red}+x}+17 &= 5-x {\color{red}+x}\\ -2x+17 &= 5\\ -2x+17 {\color{red}-17} &= 5 {\color{red}-17}\\ -2x &= -12\\ \frac{-2x}{-2} &= \frac{-12}{-2}\\ x &= 6\end{align*} \begin{align*}\text{Check} &\\ -3x+17 &= 5-x\\ -3({\color{red}6})+17 &= 5-({\color{red}6})\\ -18+17 &= 5-6\\ -1 &= -1 \ \ Y\end{align*} 1. \begin{align*}\text{Check} &\\ 5x+2 &= 3x-6\\ 5 ({\color{red}-4})+2 &= 3({\color{red}-4})-6\\ -20+2 &= -12-6\\ -18 &= -18 \ \ Y\end{align*} \begin{align*}5x+2 &= 3x-6\\ 5x {\color{red}-3x}+2 &= 3x {\color{red}-3x}-6\\ 2x+2 &= -6\\ 2x+2 {\color{red}-2} &= -6{\color{red}-2}\\ 2x &= -8\\ \frac{2x}{2} &= \frac{-8}{2}\\ x &= -4\end{align*} 1. \begin{align*}7x+8 &= 4x-4\\ 7x {\color{red}-4x}+8 &= 4x {\color{red}-4x}-4\\ 3x+8 &= -4\\ 3x+8 {\color{red}-8} &= -4 {\color{red}-8}\\ 3x &= -12\\ \frac{3x}{3} &= \frac{-12}{3}\\ x &= -4\end{align*} \begin{align*}\text{Check} &\\ 7x+8 &= 4x-4\\ 7({\color{red}-4})+8 &= 4({\color{red}-4})-4\\ -28+8 &= -16-4\\ -20 &= -20 \ \ Y\end{align*} ## Solving Linear Equations that Apply to the Distributive Property Introduction In this concept you will learn to solve equations with variables in which the equations involve the distributive property. The distributive property is a mathematical way of grouping terms. The distributive property states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. Wow! Let’s see what this looks like in math terms. Say you had \begin{align*}{\color{red}3}({\color{blue}x + 5})\end{align*}. The distributive property states that the product of a number (\begin{align*}{\color{red}3}\end{align*}) and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}3})\end{align*} and the addends (\begin{align*}{\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5}\end{align*}). When solving equations using the distributive property, you simply have one more step to follow than you have experienced in the previous lessons. You still have the same goal for solving problems with variables. That is to get the variables on one side and the constants on the other side. You do this by adding and subtracting terms from both sides of the equal sign. Then you isolate the variables by multiplying or dividing. In this lesson, your first step will be to use the distributive property to remove any brackets. Watch This Guidance Morgan, Connor, and Jake are going on a class field trip with 12 of their class mates. Each student is required to have a survival kit with a flashlight, a first aid kit, and enough food rations for the trip. A flashlight costs10. A first aid kit costs $9. Each days food ration costs$7. If the class has only 500 for the survival kits, how many days can they go on their trip? If First let’s write down what we know: \begin{align*}\#\end{align*} Students going on the trip = 15 (Morgan, Connor, Jake + 12 others) Total money available =500
Each flashlight costs $10 Each first aid kit costs$9
Each days food ration costs 7 One survival kit contains 1 flashlight + 1 first aid kit + \begin{align*}x\end{align*} day’s rations Therefore cost of each survival kit: \begin{align*}\10 + \9 + \7x = \19 + \7x\end{align*} For the 15 students, the total cost of the survival kits would be: \begin{align*}15 (\19+\7x)=\285+\105x\end{align*} Since the class has1000 to spend, you can calculate how many days they can go on their trip.
\begin{align*}\1000 &= \285+\105x\\ \1000 {\color{red}- \285} &= \285 {\color{red}-\285}+\105x\\ \715 &= \105x\\ \frac{\715}{\105} &= \frac{\105x}{\105}\\ 6.81 &= x\end{align*}
Since the class does not have enough money to buy 7 days of food rations for each student, they will buy six days of food rations and the class will go on a class trip for six days.
Example A
\begin{align*}2(3x+5)=-2\end{align*}
We will solve this problem using the balance (or seesaw) used in the previous lesson.
Your first step is to remove the brackets. To do this multiply the 2 by the numbers inside the brackets. Therefore multiply 2 by \begin{align*}3x\end{align*} and 2 by 5.
Next, isolate the \begin{align*}x\end{align*} variable by subtracting 10 from both sides.
Simplifying you get:
Now divide by 6 to solve for the \begin{align*}x\end{align*} variable.
To finally solve for the variable, simplify each side.
Therefore \begin{align*}x = -2\end{align*}.
\begin{align*}\text{Check}&\\ 2(3x+5) &= -2\\ 2(3({\color{red}-2})+5) &= -2\\ 2(-6+5) &= -2\\ 2(-1) &= -2\\ -2 &= -2 \ \ Y\end{align*}
Example B
\begin{align*}3(4+3y)=-6\end{align*}
We will solve this problem using algebra tiles used in previous lessons.
Since all of our variables (\begin{align*}x\end{align*} tiles) are on the same side, you only need to subtract the 3 groups of 4 from both sides of the equal sign to isolate the variable.
Simplifying the right side of the equation and rearranging the tiles leave the following:
Therefore \begin{align*}y = -2\end{align*}.
\begin{align*}\text{Check} &\\ 3(4+3y) &= -6\\ 3(4+3({\color{red}-2})) &= -6\\ 3(4-6) &= -6\\ 3(-2) &= -6\\ -6 &= -6 \ \ Y\end{align*}
Example C
\begin{align*}6(4x+1)=-18\end{align*}
Therefore \begin{align*}x = -1\end{align*}.
\begin{align*}\text{Check} &\\ 6(4x+1) &= -18\\ 6(4({\color{red}-1})+1) &= -18\\ 6(-4+1) &= -18\\ 6(-3) &= -18\\ -18 &= -18 \ \ Y\end{align*}
Vocabulary
Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*}{\color{red}3}({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}({\color{red}3})\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}3})\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.
Guided Practice
1. Use balance (seesaw) method to solve for the variable in the problem \begin{align*}6(-3x+2)=-6\end{align*}.
2. Use the algebra tiles to solve for the variable in the problem \begin{align*}2(r-2)=-6\end{align*}.
3. Determine the most efficient method to solve for the variable in the problem \begin{align*}2(j+1)+3=3(j-1)\end{align*}. Explain your choice of method for solving this problem.
1. \begin{align*}6(-3x+2)=-6\end{align*}.
You can begin by using the distributive property on the left side of the equation.
Next subtract 12 from both sides of the equation to get the variable term alone.
Simplifying, you get:
You next have to isolate the variable. To do this, divide both sides of the equation by -18.
Finally, simplifying leaves you with:
Therefore \begin{align*}x = 1\end{align*}.
\begin{align*}\text{Check} &\\ 6(-3x+2) & =-6\\ 6(-3({\color{red}1})+2) &= -6\\ 6(-3+2) &= -6\\ 6(-1) &= -6\\ -6 &= -6 \ \ Y\end{align*}
2. \begin{align*}2(r-2)=-6\end{align*}
First, you need to add 4 to both sides to get the variables alone on the one side of the equal sign.
Simplifying leaves you with:
Therefore \begin{align*}r = -1\end{align*}.
\begin{align*}\text{Check} &\\ 2(r-2) &= -6\\ 2(({\color{red}-1})-2) &= -6\\ 2(-3) &= -6\\ -6 &= -6 \ \ Y\end{align*}
3. \begin{align*}2(j+1)+3=3(j-1)\end{align*}.
You could choose either method for this problem. Below you will see the balance method used to solve the problem. Work through the steps to see if you can follow them. Remember since there are brackets, you must start with the distributive property and remove the brackets.
Therefore \begin{align*}j = 5\end{align*}.
\begin{align*}\text{Check} &\\ 2(j+1) &= 3(j-1)\\ 2(({\color{red}5})+1) &= 3(({\color{red}5})-1)\\ 2(6) &= 3(4)\\ 12 &= 12 \ \ Y\end{align*}
Summary
The previous three lessons began your study of solving equations with variables. In this final lesson of the section, you were working with the distributive property. Equations with one variable that involve the distributive property are solved in the exact same way as any other equations with one variable. The only difference is that the very first step with distributive property problems is to remove the brackets.
To remove brackets, remember that you must multiply all the numbers inside the brackets by the number outside the brackets. After you remove brackets, you can then solve the equation by combining like terms, moving constants to one side of the equal sign, variables to the other side of the equal sign, then isolating the variable to find the solution.
Again, as with previous lessons, you must remember in these problems, as with any math problems involving an equal sign, whatever function (addition, subtraction, multiplication, or division) you do to one side of the equal sign, you must do to the other side. This is a big rule to remember in order for equations to remain equal or to remain in balance.
Problem Set
Use the balance (seesaw) method to find the solution for the variable in each of the following problems.
1. \begin{align*}5(4x+3)=75\end{align*}
2. \begin{align*}3(s-4)=15\end{align*}
3. \begin{align*}5(k-4)=10\end{align*}
4. \begin{align*}43=4(t+6)-1\end{align*}
5. \begin{align*}6(x+4)=3(5x+2)\end{align*}
Use algebra tiles to find the solution for the variable in each of the following problems.
1. \begin{align*}2(d-3)=4\end{align*}
2. \begin{align*}5+2(x+7)=20\end{align*}
3. \begin{align*}2(3x-4)=22\end{align*}
4. \begin{align*}2(3x+2)-x=-6\end{align*}
5. \begin{align*}2(x+4)-x=9\end{align*}
Use the methods that you have learned for solving equations with variables to solve for the variables in each of the following problems. Remember to choose an efficient method to solve for the variable.
1. \begin{align*}-6=-6(3x-8)\end{align*}
2. \begin{align*}-2(x-2)=11\end{align*}
3. \begin{align*}2+3(-2x+1)=20\end{align*}
4. \begin{align*}3(x+2)-x=12\end{align*}
5. \begin{align*}5(2-3x)=-8-6x\end{align*}
Using the balance (seesaw) method...
1. Therefore \begin{align*}x = 3\end{align*}. \begin{align*}\text{Check} &\\ 5(4x+3) &= 75\\ 5(4({\color{red}3})+3) &= 75\\ 5(12+3) &= 75\\ 5(15) &= 75\\ 75 &= 75 \ \ Y\end{align*}
1. Therefore \begin{align*}k = 6\end{align*}. \begin{align*}\text{Check} &\\ 5(k-4) &= 10\\ 5(({\color{red}6})-4) &= 10\\ 5(2) &= 10\\ 10 &= 10 \ \ Y\end{align*}
1. Therefore \begin{align*}x = 2\end{align*}. \begin{align*}\text{Check} &\\ 6(x+4) &= 3(5x+2)\\ 6(({\color{red}2})+4) &= 3(5({\color{red}5})+2)\\ 6(6) &= 3(10+2)\\ 36 &= 3(12)\\ 36 &= 36 \ \ Y\end{align*}
Using algebra tiles...
1. Therefore \begin{align*}x = 5\end{align*} \begin{align*}\text{Check} &\\ 2(d-3) &= 4\\ 2(({\color{red}5})-3) &= 4\\ 2(2) &= 4\\ 4 &= 4 \ \ Y\end{align*}
1. Therefore \begin{align*}x = 5\end{align*}. \begin{align*}\text{Check}&\\ 2(3x-4) &= 22\\ 2(3({\color{red}5})-4) &= 22\\ 2(15-4) &= 22\\ 2(11) &= 22\\ 22 &= 22 \ \ Y\end{align*}
1. Therefore \begin{align*}x = 1\end{align*}. \begin{align*}\text{Check} &\\ 2(x+4)-x &= 9\\ 2(({\color{red}1})+4)-({\color{red}1}) &= 9\\ 2(5)-1 &= 9\\ 10-1 &= 9\\ 9 &= 9 \ \ Y\end{align*}
Choose a method...
1. Therefore \begin{align*}x = 3\end{align*}. \begin{align*}& \text{Check} \\ -6 &= -6(3x-8)\\ -6 &= -6(3({\color{red}3})-8)\\ -6 &= -6(9-8)\\ -6 &= -6(1)\\ -6 &= -6 \ \ Y\end{align*}
1. Therefore \begin{align*}x = -3\end{align*}. \begin{align*}\text{Check} &\\ 2+3(-2x+1) &= 23\\ 2+3(-2(-3)+1) &= 23\\ 2+3(6+1) &= 23\\ 2+3(7) &= 23\\ 23 &= 23 \ \ Y\end{align*}
1. Therefore \begin{align*}x = 2\end{align*}. \begin{align*}\text{Check} &\\ 5(2-3x) &= -8-6x\\ 5(2-3({\color{red}2})) &= -8-6({\color{red}2})\\ 5(2-6) &= -8-12\\ 5(-4) &= -20\\ -20 &= -20 \ \ Y\end{align*}
## Summary
In this lesson, you have worked with linear equations in one variable. You have learned how to solve problems where variables were on one side of the equation, where variables were on both sides of the equation, and also when there are brackets involved in the equations. The key to solving these equations is to make sure that whatever you do to one side of the equal sign, you do the other. In other words, if you add a number to the left hand side of the equal sign, add it to the right hand side. In this way the equation stays in balance.
When solving equations, you want to get variables on one side of the equal sign by adding or subtracting constant terms and then solving for the variable by multiplying or dividing by whatever term is attached to your variable. When you have brackets, your very first step is to use the distributive property and remove these brackets.
As well, you used two main methods to solve these equations as you begin learning to solve linear equations. You learned to solve using a balance (or seesaw) where the middle of the seesaw represents the equal sign. The other method is using algebra tiles. Although algebra tiles are great for solving these types of equations, when numbers get large, they become very cumbersome and therefore not very efficient. It may be more helpful to use another method.
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CK.MAT.ENG.SE.1.Algebra-I---Honors.2.1 |
# How to solve the equation with a logarithm
The logarithmic equations are the equations supporting the unknown under the sign of a logarithm and/or in its basis. The simplest logarithmic equations are the equations of a type of logaX=b, or the equation which can be reduced to this look. Let's consider as different types of the equation can be reduced to this type and to solve.
## Instruction
1. Follows from definition of a logarithm that to solve the equation of logaX=b it is necessary to make equivalent transition of a^b=x if a> 0 and an is not equal 1, that is 7=logX on the basis 2, then x=2^5, x=32.
2. At the solution of the logarithmic equations often pass to not equivalent transition therefore check of the received roots, by substitution in this equation is necessary. For example, the equation of log(5+2x) on the basis 0.8=1, by not equivalent transition is given, log(5+2x) on the basis 0.8=log0.8 on the basis 0.8 turns out, it is possible to lower the sign of a logarithm, then the equation 5+2kh =0.8 turns out, solving this equation we receive also =-2.1. When checking x =-2.1 5+2kh> 0 that corresponds to properties of logarithmic function (the range of definition of logarithmic area is positive), therefore, x =-2.1 - an equation root.
3. If unknown is in the logarithm basis, then the similar equation is solved in the same ways. For example, the equation, log9 on the basis (x-2) =2 is given. Working also as well as in the previous examples, we receive (x-2) ^2=9, x^2-4x+4=9, x^2-4x-5=0, solving this equation of X1=-1, X2=5. As the basis of function has to be more than 0 and is not equal 1, there is only root X2=5.
4. Often at the solution of the logarithmic equations it is necessary to apply properties of logarithms: 1) logaXY=loda[X]+loda[Y] logbX/Y=loda[X] - loda[Y]2) logfX^2n=2nloga[X] (2n - even number) logfX^(2n+1)= (2n+1) of logaX (2n+1 - odd number) 3) logX about the basis of a^2n=(1/2n) log [a] X logX about the basis of a^(2n+1)= (1/2n+1) of logaX4) logaB=1/logbA, b is not equal to 15) logaB=logcB/logcA, c is not equal to 16) to a^logaX=X, X> 07) a^logbC=clogbAИспользуя these properties, you can reduce the logarithmic equation to simpler type, and further solve in already above-stated ways.
Author: «MirrorInfo» Dream Team |
# The rate of change of the angle sum S of a polygon with n sides is a constant 180. If S is 360 when n=4, find S when n=7.
embizze | Certified Educator
We are given that `(dS)/(dn)=180` and a point `(S,n)=(360,4)` .
`dS=180dn`
`intdS=int180dn`
`S=180n+C` Using (360,4) to solve for C we get:
`360=180(4)+C==>C=-360`
Thus `S=180n-360`
---------------------------------------------------------------
When n=7 we have S=180(7)-360=900
---------------------------------------------------------------
Without calculus: If a rate of change is constant the function is linear. The function will be of the form S=180n+k where k is the S intercept. Plugging in known values we get 360=180(4)+k so S=180n-360 as before.
jeew-m | Certified Educator
The rate of change is 180 per one change in n.
For n=7 we have n=7-4=3 changes in n.
Therefore angle sum will change 180*3 than n=4 when n=7.
So when n=7; S= 360+180*3 = 900
najm1947 | Student
As the rate of change for angle is constant (180) for the angle sum S, therefore the angle sum follows a linear equation that can be written as:
S = 180.n +C where C is a constant and A is the number of angles
for n = 4, S = 360 (given) and substituting this value in the equation we get:
360 = 180*4 + C
=> C = 360-720 = -360
The angle sum is thus given by
S = 180.n-360
for n = 7
S = 180*7-360 = 900
The angle sum is 900 when n = 7 |
1. ## Easy inequality
Let $\displaystyle x,y$ be real. Show that if $\displaystyle x^2 + y^2 > 2$, then $\displaystyle x^4 + y^4 > x^2 + y^2$
2. Originally Posted by EinStone
Let $\displaystyle x,y$ be real. Show that if $\displaystyle x^2 + y^2 > 2$, then $\displaystyle x^4 + y^4 > x^2 + y^2$
As $\displaystyle x^2 + y^2 > 2$ it means x and y are both greater than 1. so,
$\displaystyle x^4>x^2 ...............(i)$
$\displaystyle y^4>y^2................(ii)$
adding (i) & (ii), we have
$\displaystyle x^4 + y^4 > x^2 + y^2$
3. Originally Posted by slovakiamaths
As $\displaystyle \color{red}x^2 + y^2 > 2$ it means x and y are both greater than 1. so,
What makes you think that is true?
It is not true: $\displaystyle x=2~\&~y=0.5$
4. Originally Posted by EinStone
Let $\displaystyle x,y$ be real. Show that if $\displaystyle x^2 + y^2 > 2$, then $\displaystyle x^4 + y^4 > x^2 + y^2$
If $\displaystyle x = r \cos \theta$ and $\displaystyle y = r \sin \theta$, then you would like to establish that
$\displaystyle r^2 \left( \cos^4 \theta + \sin^4 \theta\right) > 1 \; \text{if}\; r^2 > 2$ or
$\displaystyle \cos^4 \theta + \sin^4 \theta \ge \frac{1}{2}.$
Since $\displaystyle \cos^2\theta = 1 - \sin^2 \theta$ then we wish to prove that
$\displaystyle \sin^4 \theta +(1 - \sin^2 \theta)^2 \ge \frac{1}{2},$
or
$\displaystyle 2 \sin^4 \theta - 2\sin^2 \theta + 1 \ge \frac{1}{2}$
Completing the square gives
$\displaystyle \left( \sin^2 \theta - \frac{1}{2}\right)^2 + \frac{1}{4} \ge \frac{1}{4}$
which obviously is true.
5. Assume, without loss of generality, that $\displaystyle x^2\ge 1 \ge y^2$.
$\displaystyle x^2+y^2 > 2 \Leftrightarrow x^2-1 > 1-y^2$,
where $\displaystyle x^2-1 \ge 0$ and $\displaystyle 1-y^2 \le 0$,
so $\displaystyle x^2(x^2-1)>y^2(1-y^2)$, and thus $\displaystyle x^4+y^4 > x^2+y^2$. |
# How many ways can you draw 5 cards from a standard deck and get 4 aces?
How many ways can you draw 5 cards from a standard deck and get 4 aces? a) There are (525)=2,598,960 ways of choosing 5 cards. There are (44)=1 way to select the 4 aces. So
## How many ways can you draw 5 cards from a standard deck and get 4 aces?
a) There are (525)=2,598,960 ways of choosing 5 cards. There are (44)=1 way to select the 4 aces. So there are (481)=48 ways to select the remaining card. Thus there are a total of 48 ways to select 5 cards such that 4 of them are aces, and the probability is: 482,598,960=154,145.
## What is the probability you draw 4 aces from a deck of 52 cards?
= 3.6938e-006. The first answer you provided (152×151×150×149×4!) is correct. If we draw four cards from 52 cards, then the total possible outcomes are C5244!.
What is the probability of drawing 4 aces from a deck of cards?
There are 52 cards, 4 of them aces. The probability the first card drawn is an ace is 4/52= 1/13. There are then 51 cards, 3 of them aces.
How many ways can you draw 4 aces?
Method One: There are C(52,5) = 2,598,960 possible hands. 48 of these contain 4 aces (4 aces and the king of spades, 4 aces and the queen of spades, 4 aces and the jack of spades, etc.).
### What are the odds of getting four Aces?
Chances of getting four Aces in 5-Card Draw: 1 in 54,145 hands on average. Chances of getting any four of a kind in Hold ‘Em: 1 in 4,165 hands on average.
### What is the probability of drawing a black card?
However, The probability of getting a black card or an ace [which we may denote as P(black or ace)] is not P(black) + P(ace) since the former is 28/52 (there are 26 black cards and 2 red aces) while the latter is 26/52 + 4/52.
What is the probability of drawing a red card in a standard deck of 52 cards?
6 out of
Explanation: In a deck of cards, there are 6 red-faced cards. So, the probability of drawing a red-faced card is 6 out of 52 cards.
What are the odds of getting dealt 4 Aces?
1 in 54,145
Chances of getting four Aces in 5-Card Draw: 1 in 54,145 hands on average. Chances of getting any four of a kind in Hold ‘Em: 1 in 4,165 hands on average.
## What is the probability of getting 4 of a kind?
0.02401%
Frequency of 5-card poker hands
Hand Distinct hands Probability
Four of a kind 156 0.02401%
Full house 156 0.1441%
Flush (excluding royal flush and straight flush) 1,277 0.1965%
Straight (excluding royal flush and straight flush) 10 0.3925%
## What are the odds of a royal flush on the flop?
There are 2598960 unique 5-card poker hands (C(n,r) = C(52, 5) = 2598960). 4 of those are royal flushes. So, the odds of one specific player flopping a royal flush would be 4-in-2598960, or 1-in-649740.
What are the odds of winning Queen of Hearts?
Assuming there are 500,000 tickets sold and you bought one of them, you have a one in 500,000 chance in being picked. That’s the easy part. |
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In this article for teachers, Bernard uses some problems to suggest that once a numerical pattern has been spotted from a practical starting point, going back to the practical can help explain. . . .
### Little Boxes
##### Age 7 to 11 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Two on Five
##### Age 5 to 11 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Making Cuboids
##### Age 7 to 11 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Triangle Shapes
##### Age 5 to 11 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### Sticks and Triangles
##### Age 7 to 11 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### Cutting Corners
##### Age 7 to 11 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Cuisenaire Rods
##### Age 7 to 11 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
### Shaping Up
##### Age 7 to 11 Challenge Level:
Are all the possible combinations of two shapes included in this set of 27 cards? How do you know?
### Two by One
##### Age 7 to 11 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Putting Two and Two Together
##### Age 7 to 11 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
### Cereal Packets
##### Age 7 to 11 Challenge Level:
How can you put five cereal packets together to make different shapes if you must put them face-to-face?
### Creating Cubes
##### Age 7 to 11 Challenge Level:
Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour.
### Dice Stairs
##### Age 7 to 11 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Egyptian Rope
##### Age 7 to 11 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
### Map Folding
##### Age 7 to 11 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### Sort Them Out (2)
##### Age 7 to 11 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
### It's a Fence!
##### Age 5 to 11 Challenge Level:
In this challenge, you will work in a group to investigate circular fences enclosing trees that are planted in square or triangular arrangements.
### The Numbers Give the Design
##### Age 7 to 11 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Fit These Shapes
##### Age 5 to 11 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Four Layers
##### Age 5 to 11 Challenge Level:
Can you create more models that follow these rules?
### Three Sets of Cubes, Two Surfaces
##### Age 7 to 11 Challenge Level:
How many models can you find which obey these rules?
### Sports Equipment
##### Age 7 to 11 Challenge Level:
If these balls are put on a line with each ball touching the one in front and the one behind, which arrangement makes the shortest line of balls?
### Four Colours
##### Age 5 to 11 Challenge Level:
Kate has eight multilink cubes. She has two red ones, two yellow, two green and two blue. She wants to fit them together to make a cube so that each colour shows on each face just once.
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Escher Tessellations
##### Age 7 to 11 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
##### Age 7 to 11 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
##### Age 7 to 11 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Fencing
##### Age 7 to 11 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### Polydron
##### Age 7 to 11 Challenge Level:
This activity investigates how you might make squares and pentominoes from Polydron.
### Cuboid-in-a-box
##### Age 7 to 11 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### Tri.'s
##### Age 7 to 11 Challenge Level:
How many triangles can you make on the 3 by 3 pegboard?
### Cunning Card Trick
##### Age 11 to 14 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Domino Sets
##### Age 7 to 11 Challenge Level:
How do you know if your set of dominoes is complete?
### Two Squared
##### Age 7 to 11 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Counter Ideas
##### Age 7 to 11 Challenge Level:
Here are some ideas to try in the classroom for using counters to investigate number patterns.
### Making Maths: Double-sided Magic Square
##### Age 7 to 14 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Making Maths: Make a Magic Circle
##### Age 7 to 11 Challenge Level:
Make a mobius band and investigate its properties.
### Jig Shapes
##### Age 5 to 11 Challenge Level:
Can you each work out what shape you have part of on your card? What will the rest of it look like?
### Making Maths: Walking Through a Playing Card?
##### Age 7 to 14 Challenge Level:
It might seem impossible but it is possible. How can you cut a playing card to make a hole big enough to walk through?
### Paper Curves
##### Age 7 to 11 Challenge Level:
This is a simple paper-folding activity that gives an intriguing result which you can then investigate further.
### Midpoint Triangle
##### Age 7 to 11 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Cut and Make
##### Age 7 to 11 Challenge Level:
Cut a square of paper into three pieces as shown. Now,can you use the 3 pieces to make a large triangle, a parallelogram and the square again?
### Making Maths: Birds from an Egg
##### Age 7 to 11 Challenge Level:
Can you make the birds from the egg tangram? |
# Two Step Equations
The two step equations worksheets on this page include problems with integers, decimals and fractions for a variety of math operations. These beginning algebra worksheets are appropriate practice for 6th grade, 7th grade and 8th grade students. Full answer keys are included on the second page of each PDF file.
Two Step Equations
These worksheets will help students practice their algebraic skills in solving two step equations including integers, fractions, and decimals. The first set of worksheets start with solving basic algebraic equations involving integers - all operations included from addition, subtraction, multiplication and division.The second set involves fraction problems and finally the last set will let you solve decimal problems from small numbers to bigger ones.All of these worksheets include answer keys with solutions! Go ahead and check out these worksheets or continue reading below to discover tips and more facts about 2 step equations.
What Are Two Step Equations?
With the meaning in its name – a two step equation is an algebraic equation that will take you two steps to solve for the unknown variable. You've solved the equation when you get the variable by itself, with no numbers in front of it, on one side of the equal sign.Once you've solved it, you can the check the equation by substituting your final answer to the variable and if both sides are equal then you've found the value of the variable that makes the equation true.Let us check how…
Example:
3x + 1 = 10
3x = 10 - 1
3x = 9
x = 9/3
x = 3
To check if the equation is true, substitute “x” with “3”.
3x + 1 = 10
3(3) + 1 = 10
9 + 1 = 10
10 = 10
Both sides are equal, making the equation “3x + 1 = 10” true.
Solving Two Step Equations
Positive and Negative Integers Equations
I’ve discussed about positive and negative integers in my post about one step equation, but let me refresh you again.When we hear integers, some of us will say “these are whole numbers.”but what exactly are integers and what includes whole numbers?Integers are whole numbers, but they also include negative numbers, including 0.It is a number with no decimal or fractional part.These are all of the whole numbers ( 2, 4, 5, 7…) and their opposites (-2, -4, -5, -7…) and don’t forget, 0 is also an integer.0 is neither a positive nor a negative integer. Anything above 0 is a positive integer, and anything below it is a negative integer.There you have it!Now that we already know about what exactly integers are, let us do some examples and know the step by step procedure to solve different kinds of two step equation problems.After we’re done with all these examples, solving two step equations will be a piece of cake for you!
Steps to Solving an Equation
Example 1:Integers
4 + 2x = 8
Step 1:Isolate “x” by subtracting “4” from both sides of the mathematical sentence and this will cancel out the 4 on the left side part of the equation. Remember this:When moving a positive number to the opposite part of the equation, positive should turn into negative; “4” becomes “-4”. This will make the equation as shown below.
4 + 2x - 4 = 8 - 4
2x = 4
Step 2:Divide both sides by 2 to cancel out “2” in “2x”
2x/2 = 4/2
x = 4/2
Step 3:Solve for the division question “4/2”
x = 2
4 + 2x = 8
4 + 2(2) = 8
4 + 4 = 8
8 = 8
Our answer “x = 2” is correct because both sides of the equation are equal.
Example 2: Integers
4 - 2x = 8
Step 1:Subtract “4” on both sides.
4 - 2x - 4 = 8 - 4
-2x = 4
Step 2: Divide both sides by “-2”
-2x/2 = 4/2
-x = 2
Step 3:Again, let me remind you about the rule that the “x” should never be negative. In the example given above, “x” is negative.To change this, put the “-” sign onto the other side, to the “2”. To conclude, the answer is “-2”.
x = -2
Let us check if our answer will make the equation true.Substitute the variable “x” with “-2”.
4 - 2x = 8
4 - 2(-2) = 8
4 + 4 = 8
8 = 8
Our answer “x = -2” is correct because both sides of the equation are equal.
Example 3:Integers
x/2 + 4 = 9
Step 1:Start by simplifying the equation by subtracting “4” on both sides.
x/2 + 4 -4 = 9 - 4
x/2 = 5
Step 2:Multiply both sides by “2”.This will cancel out “2” on the left side of the equation.
x/2 * 2 = 5 * 2
x = 5 * 2
Step 2:Solve for “5 * 2”
x = 10
To check, substitute “x” with our final answer “10”
x/2 + 4 = 9
10/2 + 4 = 9
5 + 4 = 9
9 = 9
Our answer “x = 10” is correct because both sides of the equation are equal.
Example 4:Integers
x + 3 - 2 = 10
4
Step 1: Add “2” on both sides to simplify the equation.
x + 3- 2 + 2 = 10 + 2
4
x + 3= 12
4
Step 2: Mutliply both sides by “4” to cancel out “4” on the left side of the equation.
x + 3 * 4 = 12 * 4
4
x + 3 = 48
Step 3:Subtract “3” from both sides.This will leave you with the final answer that is “x = 45”.
x + 3 - 3 = 48 - 3
x = 45
Let’s now check our final answer by substituting “x” with “45”
x + 3- 2 = 10
4
45 + 3- 2 = 10
4
48- 2 = 10
4
48= 10 + 2
4
12 = 12
We got it right again!Both sides of the equation are equal.
Two Step Equations With Decimals
Now that we have practiced solving problems on positive and negative integers, solving for decimal problems will be easier.Same rules apply and just solve the problem as you would solve any two step equation.
Let’s go try some examples at solving two step equations involving decimals this time.
Example 1:Decimals
3.1x + 2 = 5.5
Step 1:Subtract “2” from both sides of the equation.
3.1x + 2 - 2 = 5.5 - 2
3.1x = 3.5
Step 2:Divide both sides by “3.1”.
3.1x ÷ 3.1 = 3.5 ÷ 3.1
x = 3.5 ÷ 3.1
Step 3:Do the calculations to solve for x.
x = 1.13
3.1x + 2 = 5.5
3.1(1.13) + 2 = 5.5
3.1(1.13) + 2 - 2 = 5.5 - 2
3.1(1.13) = 3.5
3.5 = 3.5
Yes, our answer is correct because both sides of the equation are equal.
Example 2: Decimals
12.1x = x - 9.3
Step 1:Combine like terms.Move “x” to the left hand side of the equation.
12.1x - x = -9.3
11.1x = -9.3
Step 2:Divide both sides by “11.1” to isolate “x”.
11.1x ÷ 11.1 = -9.3 ÷ 11.1
x = -9.3 ÷ 11.1
Step 3:Solve for x.
x = -0.8378
Let’s check our answer.Substitute “x” with “-0.8378”
12.1x = x - 9.3
12.1 (-0.8378) = -0.8378 - 9.3
-10.1378 = -10.1378
Our answer “x = -0.8378” is correct because both sides of the equation are equal.
Two Step Equations With Fractions
The procedure for solving 2 step equations involving fractions is the same as integers and decimals.If you have mastered solving 2 step equations for integers and decimals, then solving for fractions will be easy too.The first goal is still the same: to isolate the variable on one side. Let me walk you through the steps by solving these example problems below…
Example 1: Fraction
x + 1/2 = 3/4
Step 1:Isolate “x” by subtracting “1/2” from both sides of the equation.This will cancel out 1/2 on the left-hand side of the equation and will leave “x” alone.
x + 1/2 - 1/2 = 3/4 - 1/2
x = 3/4 - 1/2
Step 2:Find the LCM (least common multiple) of the denominators “4” and “2”.In this case, the LCM is “4”.The smallest positive number that is both divisible by “4” and “2”.Adjust the fractions based on LCM.
x = 3/4 - 2/4
Step 3:Now that the denominators are the same, combine the fractions “3/4” and “2/4”.
The numerators “3” and “-2” will become “1” since “3 - 2” is “1”.
x = 1/4
Let’s now check our final answer by substituting “x” with “1/4”.
x + 1/2 = 3/4
1/4 + 1/2 = 3/4
1/4 + 2/4 = 3/4
3/4 = 3/4
Our answer “x = 1/4” is correct because both sides of the equation are equal.
Example 2:Fraction
2/3 x + 1/2 = 1/3
2/3 x + 1/2 - 1/2 = 1/3 - 1/2
2/3 x = 1/3 - 1/2
2/3 x = 2/6- 3/6
2/3 x = -1/6
36
2/3 x (3) = -1/6 (3)
2x = -3/6
x = -3/6 ÷ 2
x = -3/12
x = -1/4
Time to check if we got the answer right… Substitute with “-1/4” for “x”.
2/3 x + 1/2 = 1/3
2/3* -1/4 + 1/2 = 1/3
2/3* -1/4 + 1/2 - 1/2 = 1/3 - 1/2
2/3 * -1/4 = 2/6 - 3/6
2/3 * -1/4 = -1/6
1/3 * -1/2 = -1/6
-1/6 = -1/6
Yes, both sides are equal so our answer “-1/4” is correct.
Solving 2 step equations is extremely easy right? The key is practice! practice! practice!Just remember the steps and don’t forget about changing the signs when you move the numbers.If you are still feeling a little lost, that’s okay.You can go over our 2 step equations examples again above or check our pdf worksheets whenever you’re ready. All of our worksheets have answer keys and solutions so you can always validate your answers. Good luck! |
Class 8 – Relations
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Topic Sub Topic Online Practice Test
Relations
• Introduction to relations
• Types of relations
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Study Material
Introduction: In our daily life, we come across many relations such as father-son, brother-sister, teacher-student and many more. In mathematics too, we come across relation such as A is a subset of B, Line l is parallel to line m. Number m is less than number n In all these situations, we notice that a relation involves pairs of objects in certain order. This synopsis deals with the study of relations in mathematics. Ordered pair: In mathematics, an ordered pair is a combination of two objects. Two elements a and b listed in a specific order, form an ordered pair, denoted by (a, b). In an ordered pair (a, b), we call a as first component or first coordinate and b as second component or second coordinate. By changing the position of the components, the ordered pair is changed. Thus, (a, b) ¹ (b, a) Equality of ordered pairs: Two ordered pairs (a, b) and (c, d) are said to be equal if and only if a = c and b = d. i.e., (a, b) = (c, d) Example: Given Find a and b ? Solution: Cartesian product of two sets: For the given two sets A and B, the set containing all the ordered pairs where the first element is taken from A and second element is taken from B is called the Cartesian product of two sets A and B. Cartesian product of two sets is denoted by and read as A cross B. The set builder form Example: Let then,
Example: Let A= {1,2,3}, B = {a, b}
A × B = {(1,a),(1,b)(2,a),(2,b),(3,a),(3,b)}
B × A= {(a,1),(a,2),(a,3),(b,1),(b,2),(b,3)}
We notice from the above example that
i.e
further
= 3 × 2 =6
i) A × B≠ B ×A unless A = B or A = Φ or B = Φ
ii) If n(A)= m; n(B) = n then n(A × B) = mn
iii) n(A ×B) = n(B × A)
iv) A × Φ = Φ × A = Φ
v) If A × B = Φ then either
a) A = Φ b)B = Φ c)A = Φ and B = Φ
Representation of Cartesian product:
Cartesian product of sets can be represented in the following ways.
1. Arrow diagram 2. Tree diagram 3.Graphical representation
4. Roster form 5. Set builder form
Representation of A×B in Roster form: In this form of representation, the elements of A × B (ordered pairs) are listed and each separated from the other by means of a comma.
Example: Let A= { p, q, r } B = { 2, 4 } then find A×B in Roster form
In the above example A×B can also be written in set builder form as
Representation of A × B in arrow diagram: If A= {x, y, z} and B = {1, 3, 5} then find A × B. In order to find A × B, represent the elements of A and B as shown in the diagram below:
Now draw the arrows from each element of A to each element of B, as shown in the above figure.
Now represent all the elements related by arrows in ordered pairs in a set, which is the required Cartesian product A × B.
A × B= {(x, 1), (x, 3), (x, 5), (y, 1), (y, 3), (y, 5), (z, 1),(z, 3), (z, 5)}
Representation of A×B using a tree diagram:
Example: If A = {2, 4, 6}, B = {1, 3, 5} thenfind A × B using tree diagram.
To represent A × B using tree diagram, write all the elements of A vertically and then for each element of A, write all the elements of B as shown and arrows as shown in the figure below.
∴ A ×B = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}
Graphical Represention of A × B:
Example: Let A= {1, 3, 4} B = {2, 5, 6} then find A × B.
Considerthe elements of A along the X-axis and the elements of B along the
Y-axisand mark the points.
∴ A ×B = {(1,2), (1,5), (1,6), (3,2), (3,5), (3,6), (4,2), (4,5), (4,6)}
Proof: We have
Relation: Let A and B be two non-empty sets. Then a relation R from A to B is the set of all ordered pairs such that and a is associated with b according to a definite rule. If we say that is related to b and we write .
Clearly, every relation R from A to B is a subset of A × B. i.e., .
A“relation” is just a relationship between any two sets of information. Consider a relationship between all the people in a class and their heights. The pairing of names and heights is a relation. This pairing is done so that either if a person’s name is known we can give that person’s height, or else a height is given, the names of all the people who are that tall can be known. We also come across various relations in real life like ‘is mother of ’, is son of’,‘is teacher of ’. Let us now define the mathematical relation.
Definition: Let A and B be two non-empty sets and R is called a relation from the set A to B. (Any subset of is called a relation from A to B).
∴ A relation contains ordered pairs as elements. Hence “A relation is a set of ordered pairs”.
Example:
1.Let A= {1, 3, 4}, B = {1, 2}
Then,A × B = {(1,1), (1,2), (3,1), (3,2), (4,1), (4,2)}
Let R1= {(3, 1), (3, 2), (4, 1), (4, 2)}
We can observe that and also that forevery ordered pair (a, b) ∈ R1, a>b.
Hence R1 is the relation “is greater than” from A to B.
2. In the previous example, let Here also, and we also notice that for every ordered pair .
Hence, R2 is the relation “is less than or equal to” from A to B.
i) If then the number of possible relations from A to B is 2mn. [Since every relation from A to B is a subset of A × B, the number of relations is equal to number of subsets of A × B].
ii) A relation R is said to be defined on a set A if .
iii) If n(A) = the number of relations defined on A, i.e.,
Example: Let A = {1, 2, ………10} then there are relations on A.
Domain and Range of a Relation: Let A and B be two non-empty sets and R be a relation from A to B, we note that
i) The set of first co-ordinates of all ordered pairs in R is called the domain of R.
ii) The set of second co-ordinates of all ordered pairs in R is called the range of R.
Example: Let A = {1, 2, 4}, B = {a, b, c} and
R ={(1, a), (1, b), (2, a), (2, c), (4, c)} be a relation from A to B.
Then,domain of R = {1, 2, 4} and range of R = {a, b, c}
Representation of Relation:
i) Roster-method (or) List method: In this method we list all the ordered pairs that satisfy the rule or property given in the relation.
Example:
Let A = {1, 3, 5}, B = {3, 4, 6}
If R is a relation from A to B having property “is less than” then the roster form of R is
R = {(1, 3), (1, 4), (1, 6), (3, 4), (3, 6), (5, 6)}
ii) Set-builder method: In this method, a relation is described by using a representation and stating the property or properties, which the first and second co-ordinates of every ordered pair of the relation satisfy through the representation.
Example:
Let A = {1, 2, 4}
If R is a relation on A defined as “is greater than or equal to”, the set builder form of R is .
iii) Arrow diagram: In this method, a relation is described by drawing arrows between the elements which satisfy the property or properties given in the relation.
Example:
Let A = {2, 3, 4} B = {1, 3, 4}
Let R be a relation from A to B with property “is equal to”.
The arrow diagram of R is
Let A and B be two non-empty sets. The Cartesian product of A and B, denoted by A · B is the set of all ordered pairs (a, b), such that a A and b B.
i.e., A · B = {(a, b)/a A,b B}
1. A · B ≠ B · A, unless A = B
2. For any two sets A and B. n(A · B) = n(B · A)
3. If n(A) = p and n(B) = q, then n(A · B) = pq
If A = {1, 2, 3} and B = {3, 4, 5}, then find A · B.
Consider the elements of A on the X-axis and the elements of B on the Y-axis and mark the points |
Engage NY Eureka Math 3rd Grade Module 5 Lesson 1 Answer Key
Eureka Math Grade 3 Module 5 Lesson 1 Problem Set Answer Key
Question 1.
A beaker is considered full when the liquid reaches the fill line shown near the top. Estimate the amount of water in the beaker by shading the drawing as indicated. The first one is done for you.
Explanation :
The First Beaker contains the liquid of half of the beaker. whereas , the second beaker should be marked to 1 fourth that means where it takes four parts to make a whole. so, half of the first beaker gives the 1 fourth quantity of the second beaker and it is marked .
The Third beaker should be marked to 1 third that means One third is one part of three equal parts. the beaker quantity is divided into 3 equal parts and marked only one part that gives 1 third part as shown in above figure .
Question 2.
Juanita cut her string cheese into equal pieces as shown in the rectangles below. In the blanks below, name the fraction of the string cheese represented by the shaded part.
Explanation :
Figure 1 shows the cheese is divided into 3 parts and 1 part is shaded so it is 1 shaded part / total parts 3 means 1 third part
Figure 2 shows the cheese is divided into 6 parts and 1 part is shaded so it is 1 shaded part / total parts 6 means 1 Sixth part .
Figure 3 shows the cheese is divided into 4 parts and 1 part is shaded so it is 1 shaded part / total parts 4 means 1 Fourth part .
Question 3.
a. In the space below, draw a small rectangle. Estimate to split it into 2 equal parts. How many lines did you draw to make 2 equal parts? What is the name of each fractional unit?
b. Draw another small rectangle. Estimate to split it into 3 equal parts. How many lines did you draw to make 3 equal parts? What is the name of each fractional unit?
c. Draw another small rectangle. Estimate to split it into 4 equal parts. How many lines did you draw to make 4 equal parts? What is the name of each fractional unit?
Explanation :
Figure a – Divided into 2 equal parts and 1 line is used to divide the rectangle into 2 equal parts . Each is 1/2 fractional unit .
Figure b – Divided into 3 equal parts and 2 lines are used to divide the rectangle into 3 equal parts . Each is 1/3 fractional unit .
Figure c – Divided into 4 equal parts and 3 lines are used to divide the rectangle into 4 equal parts . Each is 1/4 fractional unit .
Question 4.
Each rectangle represents 1 sheet of paper.
a. Estimate to show how you would cut the paper into fractional units as indicated below.
b. What do you notice? How many lines do you think you would draw to make a rectangle with 20 equal parts?
a :
Explanation :
To Divided the given rectangle into 7 equal parts and 6 lines are used. Each part is 1/7 fractional unit .
To Divided the given rectangle into 9 equal parts and 8 lines are used. Each part is 1/9 fractional unit .
I notice that you draw one line less than the number of parts you want . you draw 19 lines to make a rectangle with 20 equal parts .
Question 5.
Rochelle has a strip of wood 12 inches long. She cuts it into pieces that are each 6 inches in length. What fraction of the wood is one piece? Use your strip from the lesson to help you. Draw a picture to show the piece of wood and how Rochelle cut it.
Explanation :
Length of the wooden strip = 12 inches .
Length of wooden strip after cutting = 6 inches .
Fraction of the 1 wooden piece after cutting =1 wooden piece length after cutting ÷ total length = 6 ÷ 12 = 1 ÷ 2
Eureka Math Grade 3 Module 5 Lesson 1 Exit Ticket Answer Key
Question 1.
Name the fraction that is shaded.
Explanation :
Fractional Shaded part = shaded part ÷ Total parts = 1 ÷ 4 = 1 Fourth
Question 2.
Estimate to partition the rectangle into thirds.
Explanation :
Given figure is divided into 3 equal parts and Fraction of each part is one third . 2 lines are used to divided the rectangle into 3 equal parts .
Question 3.
A plumber has 12 feet of pipe. He cuts it into pieces that are each 3 feet in length. What fraction of the pipe would one piece represent? (Use your strip from the lesson to help you.)
Explanation :
Length of the pipe = 12 feet
length of each strip = 3 feet
Number of strips = Length of the pipe ÷ length of each strip = 12 ÷ 3 = 4 strips
Fraction of the pipe strip = length of each strip ÷ Length of the pipe = 3 ÷ 12 = 1 ÷ 4 .
Question 1.
A beaker is considered full when the liquid reaches the fill line shown near the top. Estimate the amount of water in the beaker by shading the drawing as indicated. The first one is done for you.
Explanation :
The second Beaker is divided into 5 equal parts and one part is shaded that shows the 1 fifth part as shown in above figure .
The Third Beaker is divided into 6 equal parts and one part is shaded that shows the 1 sixth part as shown in above figure .
Question 2.
Danielle cut her candy bar into equal pieces as shown in the rectangles below. In the blanks below, name the fraction of candy bar represented by the shaded part.
Explanation :
Figure a – Divided into 3 equal parts and 3 lines are used to divide the rectangle into 3 equal parts .1 part is shaded so, Each is 1/3 fractional unit .
Figure b – Divided into 4 equal parts and 3 lines are used to divide the rectangle into 4 equal parts .1 part is shaded so, Each is 1/4 fractional unit .
Figure c – Divided into 7 equal parts and 6 lines are used to divide the rectangle into 7 equal parts . 1 part is shaded so, Each is 1/7 fractional unit .
Question 3.
Each circle represents 1 whole pie. Estimate to show how you would cut the pie into fractional units as indicated below.
Explanation :
Explanation :
Figure a shows the 1 pie is divided into 2 parts and 1 part is shaded so Fractional part is 1 shaded part / 2 total parts
so, Each part is 1 half part
Figure b shows the 1 pie is divided into 3 parts and 1 part is shaded so Fractional part is 1 shaded part / 3 total parts
so, Each part is 1 Thirds part
Figure c shows the 1 pie is divided into 6 parts and 1 part is shaded so Fractional part is 1 shaded part / 6 total parts
so, Each part is 1 sixths part
Question 4.
Each rectangle represents 1 sheet of paper. Estimate to draw lines to show how you would cut the paper into fractional units as indicated below.
Explanation :
Explanation :
Figure a shows The Rectangle is divided into 2 parts and 1 Fractional part is 1 half part.
Figure b shows The Rectangle is divided into 4 parts and 1 Fractional part is 1 Fourth part
Figure c shows The Rectangle is divided into 8 parts and 1 Fractional part is 1 Eights part
Question 5.
Each rectangle represents 1 sheet of paper. Estimate to draw lines to show how you would cut the paper into fractional units as indicated below.
Explanation :
Fig a – To Divided the given rectangle into 6 equal parts and 5 lines are used. Each part is 1/6 fractional unit .
Fig b – To Divided the given rectangle into 3 equal parts and 2 lines are used. Each part is 1/3 fractional unit .
Question 6.
Yuri has a rope 12 meters long. He cuts it into pieces that are each 2 meters long. What fraction of the rope is one piece? Draw a picture. (You might fold a strip of paper to help you model the problem.)
Explanation :
Explanation :
Length of the pipe = 12 meters
length of each strip = 2 meters
Number of strips = Length of the pipe ÷ length of each strip = 12 ÷ 2 = 6 strips
Fraction of the pipe strip = length of each strip ÷ Length of the pipe = 2 ÷ 12 = 1 ÷ 6 = 1 sixth Fraction . .
Question 7.
Dawn bought 12 grams of chocolate. She ate half of the chocolate. How many grams of chocolate did she eat? |
Question
1. # What Is The Probability Of Making An Even Number
Have you ever wondered what the probability of making an even number is? It might sound like a simple question, but it can actually be quite complex. After all, probabilities are calculated based on factors such as the randomness of the data set, the number of trials, and more. In this blog post, we will explore the basics of probability and answer the question: What is the probability of making an even number? We’ll look at examples from dice rolls to coin flips to help explain how these calculations are made. Ready to learn more? Read on!
## The definition of probability
The probability of making an even number is the likelihood or chance of something happening. Probability is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur. A probability of 0.5 indicates that there is an equal chance of the event occurring or not occurring.
There are two types of events: dependent and independent. Dependent events are those in which the outcome of one event affects the probability of another event occurring. For example, if you roll a dice and get a 4, then the probability of rolling a 4 on the second roll decreases. Independent events are those in which the outcome of one event does not affect the probability of another event occurring. For example, if you flip a coin and it comes up heads, that doesn’t affect the probability of it coming up heads on the next flip.
To calculate probability, you need to know two things:
the number of possible outcomes
and
the number of favorable outcomes.
The number of possible outcomes is simply how many different things could happen overall (e.g., with a die there are 6 possibilities). To find the number of favorable outcomes, you need to figure out how many ways there could be for your chosen event to happen (e.g., rolling a 6). Once you have these numbers, you can use them in this formula: Probability =
## The different types of probability
There are three different types of probability: Classical, Relative Frequency, and Subjective.
Classical Probability is based on the idea of equally likely outcomes. If you flip a coin, there is a 50% chance of it landing on heads and a 50% chance of it landing on tails. The probability of an event happening is calculated by taking the number of ways it can happen divided by the total number of possible outcomes.
Relative Frequency Probability is based on what has happened in the past. For example, if you flip a coin 10 times and it lands on heads 6 times, then the relative frequency probability of Heads is 6/10 or 60%. This method can be used when there are too many possible outcomes to calculate using classical probability, or when information about all possible outcomes is not available.
Subjective Probability is based on someone’s personal opinion about whether an event will happen or not. For example, if you ask someone if they think it will rain tomorrow, their answer would be based on their subjective probability of rain. This type of probability is often used in cases where there is no objective way to calculate the probability of an event occurring.
## How to calculate probability
If you’re working with a deck of cards, the probability of making an even number is pretty simple to calculate. Just take the total number of even cards in the deck (26) and divide it by the total number of cards in the deck (52). That gives you a probability of 0.5, or 50%.
But what if you’re not dealing with a standard deck of cards? What if, for example, you’re trying to calculate the probability of flipping a coin and it landing on heads? In that case, you would use a different method.
To calculate the probability of an event happening, you need to know two things:
1. How many ways can the event happen?
2. How many ways can the event NOT happen?
For our coin flip example, there is only one way for the event to happen (the coin can land on heads or tails), so we’ll focus on the second part of the equation. To find out how many ways the event CANNOT happen, we need to know how many outcomes there are in total. Since there are two possible outcomes when flipping a coin (heads or tails), that means there are two ways for the event NOT to happen. So our equation would look like this:
1 / (2 * 2) = 0.25
That gives us a probability of 25%.
## The probability of making an even number
The probability of making an even number is 2/3.
This means that if you make threeeven numbers, the probability of making an even number on the fourth try is 2/3. Conversely, if you make an odd number on the first try, the probability of making an even number on the second try is 1/3.
## Conclusion
In conclusion, we can see that the probability of making an even number when rolling a single die is 50%. We have explored how this works and discussed why it is important to understand basic probability. Knowing this information will be useful in many dice games, as well as understanding statistics or predicting outcomes in other situations. Whether you are playing your favorite game with friends or trying to figure out how likely it is that something will happen, having an understanding of probability can be helpful!
2. What is the probability of making an even number?
It’s a question we’ve all asked ourselves at some point: what are the chances of rolling an even number? The answer, of course, depends on the type of dice you’re using and the number of sides it has.
In the case of a regular six-sided die, the probability of rolling an even number is 50%. This is because on a six-sided die, there are three even numbers (2, 4 and 6) and three odd numbers (1, 3 and 5). As such, the chance of rolling any one of the even numbers is one-third, or 33.3%. This means that the probability of rolling an even number is 50%.
When it comes to other types of dice, the probability of rolling an even number can vary significantly. For example, the probability of rolling an even number with an eight-sided die is 63%. This is because, on an eight-sided die, there are four even numbers (2, 4, 6 and 8) and four odd numbers (1, 3, 5 and 7). This means that the probability of rolling an even number is four-eighths, or 50%.
The probability of rolling an even number can also be affected by the number of dice you’re using. For instance, if you’re rolling two six-sided dice, the probability of rolling an even number is 75%. This is because, when you roll two dice, the chance of either one or both of them rolling an even number is greater than when you roll just one die.
Finally, it’s important to remember that the probability of rolling an even number is always based on the type of dice you’re using and the number of sides it has. As such, it’s important to keep these things in mind when calculating the probability of rolling an even number.
So, to summarise: the probability of rolling an even number on a six-sided die is 50%, on an eight-sided die is 63%, and when you roll two dice the probability of rolling an even number is 75%. |
# How do you solve Log (x +3)(x-2) = Log (7x-11)?
Nov 17, 2015
This equation has one solution $x = 5$
#### Explanation:
First we have to find the domain of this equation. Since $\log$ functions are only defined for positive real numbers we have to solve inequalities:
$\left(x + 3\right) \left(x - 2\right) > 0$ and $7 x - 11 > 0$
From first inequality we get: $x < - 3 \vee x > 2$
From the second we get: $x > 1 \frac{4}{7}$
So we can write, that the domain is: D=(2;+oo)
Now we can solve the equation:
$\log \left(x + 3\right) \left(x - 2\right) = \log \left(7 x - 11\right)$
Since the base of logarythm is the same on both sides (none is written, so I suppose it is $10$), we can skip $\log$ signs and solve equation:
$\left(x + 3\right) \left(x - 2\right) = 7 x - 11$
${x}^{2} + x - 6 = 7 x - 11$
${x}^{2} - 6 x + 5 = 0$
$\Delta = {6}^{2} - 4 \cdot 1 \cdot 5$
$\Delta = 36 - 20 = 16$
$\sqrt{\Delta} = 4$
${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a}$
${x}_{1} = \frac{6 - 4}{2} = 1$
${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a}$
${x}_{2} = \frac{6 + 4}{2}$
${x}_{2} = 5$
Since ${x}_{1} \notin D$, the only solution of this equation is ${x}_{2} = 5$ |
How do you calculate the ‘number of digits in a number’ using a square root? October 27, 2021 October 27, 2021
The numbers in your smartphone’s numeric keypad might look different from the numbers you use in your pocketbook, but the square root of a number, commonly called a ‘numerical constant’, is just a constant that you use to calculate how many digits there are in a particular number.
If you’ve ever wondered how to calculate a number like 2 + 3 + 4 + 5 + 6, you’ve probably seen the squareroot of 2.
So what is a squareroot?
The square root is the smallest value of a value that is the same for all of its values, or the smallest number that is divisible by itself.
A constant, in other words, is one that has a fixed number of values that are the same as any other constant.
In computing a square of any value, you multiply its number by itself (not the number that you multiply the square of), then divide the result by itself, and add the result to the square.
So, if you’re interested in finding the square Root of 2, multiply 2 by 2 and then multiply the result that comes out by itself by 1, the square is 2.
But what is the squareRoot of 2?
It’s the smallest square of a square value.
How do we know?
To find the square value of 2 on the phone’s numeric keys, we multiply 2 and multiply the number of digits by itself 3 times.
This results in 2.5.
So we know that we have 2.
If we multiply 3 times the square, we get 2.6.
So 2.4 equals 2.8, which equals 2 squared.
Now we know how to find the exact square root.
This is how to determine how many squares we have in a square.
Now, a square has an area equal to the area of the whole number in which it is divided.
That’s why you can multiply 2 square roots by itself to get 3.4.
This works for a number of different things, but it’s the square that we use in the calculations above.
If, instead, you were to divide the number 2 by the square 4 times, you’d get 3/4.
But the area would be a lot smaller, so you’d have to divide by 4 again.
We use the square to find our number of squares in the square as well.
2.25/4 = 3.3 Now we can determine how to add the square and squareRoot together.
3.6/4 + 3.5 = 6.5 Now we have a number that we can use to divide a number by.
That number, 1.6, equals 1.8 or one-half the square-root.
But remember that the square roots of 2 and 3 and 6 and 9 are always 0.
So 1/4 of the square equals 0.
This gives us the square we need to multiply with.
So that’s the number we’re trying to find in the phone.
The squareRoot method is simple to understand, but there are some things to consider when it comes to figuring out how to use it.
What does the square mean?
The equation 2.45 divided by 4 gives you the number in the circle, or, to use its mathematical term, the circle root.
So if you have two numbers that are equal to each other, like 2.2 and 2.46, you can use the equation to find out how many times they must be divided by 3 to get the answer you want.
That means the answer will be 1/3 of the number, or one half of the area.
This isn’t always the case.
Say you have a line that is 3/8th the width of the phone and that has 3 sides.
The first side has a width of 1.4, so the area is 3.
So the answer to the number will be 0.25 of the width, or 3.8 of the length.
But if you divide by 3, you get 1.25, which is 0.9 of the height.
This means the area will be the same.
If the number has a diameter of 1 inch, the area that will be 3 inches will be 7.6 inches.
So your answer to this question will be 2.9 inches of diameter.
So how do you divide a square by a square?
The answer to that question is 0, or -2.
That is, 2 divided by -3 is 1.5 or -3 divided by 0 is 1 or -1 divided by 2 is 0 or -0 divided by 1 is -0.
This tells you how many of the sides of the line are equal.
If a side has width 1.2, then the area (width of the two sides of a line) is 3 times 1.1, or 1.65 of the side’s width.
The area of a circle is the area between its two sides.
So a circle has a circumference of |
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# RS Aggarwal Class 11 Chapter 15 Solutions (Trigonometric or Circular Functions)
RS Aggarwal Class 11 Maths Solutions for Chapter 15 will explain to you all the elementary functions and identities that help in solving trigonometric equations and problems. The chapter consists of 5 exercises that contain 88 questions in total. The important topics covered in the chapter are angles, trigonometric functions, sum and difference of two angles, trigonometric equations, measurement of an angle and trigonometric ratios of compound angles.
RS Aggarwal Class 11 Chapter 15 – Trigonometric and Circular Functions will help you prepare the concepts for CBSE exams along with entrance tests such as JEE, BITSAT and NEET. We ensure that you get maximum marks in these exams and hence we have designed the best RS Aggarwal Solutions for you. Refer to them anytime and solve your doubts in minutes. If you have more queries related to the chapter, just send us the question and we will provide you with an adequate solution immediately.
## Important Points for RS Aggarwal Class 11 Maths Solutions for Chapter 15
1. Relations between different sides and angles of a right-angled triangle are called trigonometric ratios or T-ratios.
2. The algebraic sum of two or more angles is generally called compound angles and the angles are known as the constituent angles.
3. Trigonometric functions are elementary functions, the argument of which is an angle.
4. Trigonometric functions describe the relationship between the sides and angles of a right triangle.
5. Applications of trigonometric functions are extremely diverse.
6. The measure of an angle is the amount of rotation from the initial side to the terminal side.
7. The trigonometric functions include the following functions:
• Sine
• Cosine
• Tangent
• Cotangent
• Secant
• Cosecant
For each of these functions, there is an Inverse Trigonometric function
Important Trigonometric Identities
1. Cos²x + sin²x = 1
2. Sec²x + tan²x = 1
3. Cosec²x + cot²x = 1
4. cos(2nπ + x) = cosx
5. sin(2nπ + x) = sinx
6. cos(-x) = cosx
7. sin(-x) = -sinx
8. Cos(x+y) = cos x cos y – sin x sin y
9. Cos(x-y) = cos x cos y + sin x sin y
10. Sin(x+y) = sin x cos y + cos x sin y
11. Sin(x-y) = sin x cos y – cos x sin y
12. Cos 2x = cos²x – sin²x = 2cos²x – 1 = 1 – 2sin²x = 1-tan²x/1-tan²x
13. Sin 2x = 2cosx . sinx = 2tanx/1+tan²x
## Exercise Discussion of RS Aggarwal Class 11 Maths Solutions for Chapter 15
Exercise 15A
This exercise 15A has 10 questions in which you have to find the value of all trigonometric functions, the value of the particular trigonometric function, the value of the trigonometric function at a given angle or prove the trigonometric equations.
Exercise 15B
This exercise 15B has 20 questions in which you have to find the value of trigonometric functions at a given angle by splitting the angle, prove the given trigonometry equations by using identities and find the value of the given trigonometric function using identities.
Exercise 15C
This exercise 15C has 27 questions in which you have to prove the given trigonometric equations by splitting angles and using properties, express the given trigonometric equations as a result, express the given trigonometric equations as an algebraic sum of sines or cosines and prove the given trigonometric equation using identities and formulas.
Exercise 15D
This exercise 15D has 21 questions in which you have to find the value of a trigonometric function when the value of another function is given, prove the given trigonometric equation and prove the given trigonometric equations using multiplication and division.
Exercise 15E
This exercise 15E has 10 questions in which you have to find the value of a trigonometric function when the value of another function is given using identities.
## Benefits of RS Aggarwal Class 11 Maths Solutions for Chapter 15
Our subject matter experts have designed all the RS Aggarwal Class 11 Maths Solutions for Chapter 15 in a manner that you can easily understand all the concepts of the chapter. If you face an issue in solving any question, you can contact our team at any time as we are available for you round the clock to solve your issues. Instasolv is going to be your best companion when it comes to preparing for CBSE Class 11 Maths exam.
More Chapters from Class 11 |
How to find roots of polynomials
Integer roots
If the coefficients of a polynomial are integers, it is natural to look for roots which are also integers. Any such root must divide the constant term. We can often guess" one or more roots by trying all possibilities.
Example:
r4-2r2-3r-2=0.
If there is an integer root, it must divide -2. This leaves only four possibilities: 1, -1, 2, and -2. By plugging in, we find that
(-1)4-2(-1)2-3(-1)-2=0,
(-2)4-2(-2)2-3(-2)-2=12.
Therefore -1 and 2 are roots, but 1 and -2 are not.
Reducing the degree
If a root has been found, you can reduce the degree of the polynomial to get a simpler problem for the remaining roots. For example, you know from above that
r4-2r2-3r-2=0
has the roots -1 and 2. This tells you that the polynomial must contain the factors r+1 and r-2. You can use the long division algorithm to find
(r4-2r2-3r-2)/(r+1)=r3-r2-r-2,
and
(r3-r2-r-2)/(r-2)=r2+r+1.
Therefore, the remaining roots must solve
r2+r+1=0.
In summary, we have found that the roots of
r4-2r2-3r-2=0
are -1, 2, and .
Multiple roots
It can happen that a polynomial contains the same linear factor more than once. For instance,
r3-r2-r+1=(r-1)2(r+1).
Since the factor r-1 appears twice, we call 1 a double root of the polynomial, while -1 is a simple root. Note that
also has a root at r=1. This happens in general: If a polynomial P(r) as a k-fold root at r=c, then
P(c)=P'(c)=...=P(k-1)(c)=0,
but
If roots are counted by multiplicity (i.e. a double root counts twice, a triple root three times etc.), then a polynomial of nth degree has n roots.
The nth root of a complex number
We want to find the roots of the equation
rn=w,
where w is a given number. w may be complex, but the following procedure is important even if w is real. The solution of the equation requires writing w in polar form
That is, if x and y are the real an imaginary parts of w, we want to find and is such a way that and . In other words, and are polar coordinates of the point (x,y) in the Cartesian plane.
Examples:
1. w=1. In this case x=1 and y=0. We can choose and .
2. w=-1. Now x=-1 and y=0. We have and .
3. w=1+i. We have and .
The point of using polar form is that it is very easy to take the nth root. We find
Since is an angle, it is not unique; it is determined only up to a multiple of . That is, in the polar form of w, we could have written
instead of . Although, we have
if n>1. To find all nth roots of w, you must consider the representations
where k takes the integer values , 1, ..., n-1 (if you go further, then k=n will give you the same as k=0). The nth roots of w are then
Example:
r3=-1.
We put -1 into polar form
For the third roots, we find
We could have found the same result from the factorization
r3+1=(r+1)(r2-r+1).
for the roots of the second factor.
WARNING
Beware of the difference between
(r+1)3=0
and
r3+1=0.
The first equation has a triple root at -1, the second has three different roots: -1 and . In general, the equation rn=w always has n DIFFERENT ROOTS. Be prepared for a whipping with the cat of nine tails if you should ever say that the equation rn=w has an n-fold root. |
# percentage formula for marks
posted in: Uncategorised | 0
Ste… In this article, we will discuss in detail what is a percentage, how to calculate the percent increase and decrease, how to estimate the percent of a number using the percentage formula, with examples. Calculate the percentage of men present in the hall. Question 4: There are 150 people present in a cinema hall. For example, the given number is 0.45. The formula for percentage increase is given by: Percent Decrease (% Decrease) = (Decrease in Value/Original Value) x 100, Decrease in Value = Original Value – New Value, Also check: Percentage Increase and Decrease. Percentage formula = (Value/Total value) ×100. After the criteria are defined, then we need to calculate the Total marks of students and the Percentage achieved by the student. To obtain the percentage, multiply the given number by 100. To find the percentage of the marks, divide the marks obtained in the examination with the maximum marks and multiply the result with 100. To determine the percentage, we have to divide the value by the total value and then multiply the resultant to 100. Lets calculate how much percentage of marks that you got in the Mathematics => (70/100) x 100 => 70% marks. M = A candidate scoring x % in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more than the minimum required passing marks. Percentage means a number or a ratio expressed in terms of fractions of 100. Solution: Given, the total number of apples = 60, Percentage of ripe apples = (10/60) × 100. Percentage increase and decrease formulas. Written by co-founder Kasper Langmann, Microsoft Office Specialist.. Markup in very simple terms is basically the difference between the selling price per unit of the product and the cost per unit associated in making that product. Percentage= 0.9633 x 100. Calculating Aggregate Percentage Aggregate percentage is the total percentage calculated from the sum of marks obtained in all the subjects divided by the sum of maximum possible marks of each subject (n), which is multiplied by 100. An online exam marks percentage calculator to calculate the percentage of mark scored in exams. Enter the formula shown below. After you set up the formula, you get: 36/of = 24/100 Replace of by y and cross multiply to get: 36/y = 24/100 y × 24 = 36 × 100 y × 24 = 3600 Divide 3600 by 24 to get y 3600/24 = 150, y = 150 Therefore, 24 % of 150 is 36 How to use the other formula for percentage on the right. Also, read: Difference Between Percentage And Percentile. How do you calculate aggregate percentage? Percent means per cent (hundreds), i.e., a ratio of the parts out of 100. 200 + 200 = 600Step 4:To find the result, let us substitute the marks of each subject in the given formula.From Step 2 and Step 3,Aggregate Percentage = x1 + x2 + x3 + .......+ xn / n x 100= 440 / 600 x 100= 73.3, How to clear eustachian tube blockage naturally, How to calculate Fixed Deposit(FD) interest, How To Enable IonCube Loader In WHM Panel. 1. The topics related to percentage have been introduced to the students in Class 7 and Class 8 syllabus. Calculate the percentage of shaded parts in the circle. The symbol of percent is %.We generally count percentage of marks obtained, return on investment etc. Both are different. marks obtained in all the subjects divided by the sum of maximum possible marks of each subject (n), which is multiplied by 100.FormulaAggregate Percentage = x1 + x2 + x3 + .......+ xn / n x 100Step 1: Let us calculate the aggregate percentage for the marks obtained in Maths, Physics and Chemistry.For Example:Maths = 130 / 200Physics = 140 / 200Chemistry = 170 / 200Step 2:Calculate the total marks obtained in all the subjects.Sum of marks obtained = 130 + 140 + 170 = To find the exact marks out of percentage, you have to know the total marks. Your email address will not be published. CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, Difference Between Percentage And Percentile, Important Questions Class 8 Maths Chapter 1 Rational Numbers, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths, The percentage represents the number out of 100, It can be written as ratios or proportions, It cannot be written as ratios or proportions, A percentage is obtained by multiplying the ratio of two numbers with 100, A percentile is a percentage of values that can be found below a specific value, It does not rely on the normal distribution, Decimal form – If the number is given in decimal format, it is easy to find the percentage. |
# Texas Go Math Grade 8 Module 13 Quiz Answer Key
Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 13 Quiz Answer Key.
## Texas Go Math Grade 8 Module 13 Quiz Answer Key
Texas Go Math Grade 8 Module 13 Ready to Go On? Answer Key
13.1 Properties of Dilations
Determine whether one figure is a dilation of the other. Justify your answer.
Question 1.
Triangle XYZ has angles measuring 54° and 29°. Triangle XYZ’ has angles measuring 29° and 92°.
We need to see if triangle X’Y’Z’ is dilation of triangle XYZ Triangles are similar if alt corresponding angles are equal.
Triangle XYZ has angLes measuring 54° and 29°. Let’s calculate the measure of third angle:
180° – (54° + 29°) = 180° – 83° (Calculate) ………………. (1)
= 97° (Simplify) ………………….. (2)
Triangle X’Y’Z’ has angles measuring 29° and 92°. The measure of third angle is
180° – (29° + 92°) = 180° – 121° (Calculate) …………….. (3)
= 69° (Simplify) ……………… (4)
Since this triangles have only one pair of congruent angles, the triangle X’Y’Z’ is not a dilation of XYZ.
Question 2.
Quadrilateral DEFG has sides measuring 16 m, 28m, 24m, and 20m. Quadrilateral D’E’F’G’ has sides measuring 20 m, 35 m, 30 m, and 25 m.
Quadrilateral DEFG has sides measuring 16 m, 28 m, 24 m, and 20 m.
Quadrilateral D’E’F’G’ has sides measuring 20 m, 35 m, 30 m, and 25 m
Lets find the scale factor of the corresponding sides
$$\frac{20}{16}=\frac{35}{28}=\frac{30}{24}=\frac{25}{20}=\frac{5}{4}$$
Since the scale factor is the same, the quadrilateral D’E’F’G’ is a dilation of quadrilateral DEFG.
13.2 Algebraic Representations of Dilations
Dilate each figure with the origin as the center of dilation.
Question 3.
(x, y) → (0.8x, 0.8y)
Dilatation is (x, y) → (0.8x, 0.8y)
(0, -5) → (0 ∙ 0.8, -5 ∙ 0.8) → (0, -4) …………… (1)
(5, 0) → (5 ∙ 0.8, 2 ∙ 0.8) → (4, 0) ……………. (2)
(0, 5) → (0 ∙ 0.8, 5 ∙ 0.8) → (0, 4) ……………. (3)
(-5, 0) → (- 5 ∙ 0.8, 0 ∙ 0.8) → ( 4, 0) …………… (4)
Coordinates of vertices of the dilated figure are (0, -4), (4, 0), (0, 4) and (-4, 0)
Question 4.
(x, y) → (2.5x, 2.5y)
Dilation equation (x, y) → (2.5x, 2.5y)
(2, -1) → (2 ∙ 2.5, -1 ∙ 2.5) → (5, -2.5) ………………. (1)
(2, 2) → (∙ 2.5, 2 ∙ 2.5) → (5, 5) ………… (2)
(1, 1) → (1 ∙ 2.5, 1 ∙ 2.5) → (2.5, 2.5) ……………. (3)
Coordinates of dilated vertices are (5, 2.5), (5, 5), and (2.5, 2.5).
13.3 Dilations and Measurement
Question 5.
A rectangle with length 8 cm and width 5 cm is dilated by a scale factor of 3. What are the perimeter and area of the image?
Given,
A rectangle with a length of 8 cm and a width of 5 cm is dilated by a scale factor of 3.
length = 8 cm
width = 5 cm
We know that,
Area of the rectangle = length × width
A = 8 × 5
A = 40 sq. cm
The scale factor is 3
Area of the rectangle with scale factor = 40 × 3 = 120 sq. cm
We know that,
Perimeter of the rectangle = 2L + 2W
P = (2 × 8 + 2 × 5)
P = 16 + 10
P = 26
The scale factor is 3
The perimeter of the rectangle with a scale factor of 3 is 26 × 3 = 78 cm.
Essential Question
Question 6.
How can you use dilations to solve real-world problems?
A dilations in real life are used for making models of buildings in architecture, projects, making maps.
Texas Go Math Grade 8 Module 13 Mixed Review Texas Test Prep Answer Key
Selected Response
Question 1.
Quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm. Which could be the side lengths of a dilation of HIJK?
(A) 24 cm, 50 cm, 28 cm, 60 cm
(B) 6 cm, 15 cm, 7 cm, 15 cm
(C) 18 cm, 39 cm, 21 cm, 45 cm
(D) 30 cm, 78 cm, 35 cm, 75 cm
Answer: (C) 18 cm, 39 cm, 21 cm, 45 cm
Explanation:
The quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm.
Scale factor = 1.5
12 × 1.5 = 18 cm
26 × 1.5 = 39 cm
14 × 1.5 = 21 cm
30 × 1.5 = 45 cm
Thus the side lengths of dilation of HIJK are 18 cm, 39 cm, 21 cm, 45 cm
So, the correct answer is option C.
Question 2.
A rectangle has vertices (6, 4), (2, 4), (6, -2), and (2, -2). What are the coordinates of the vertices of the image after a dilation with the origin as its center and a scale factor of 2.5?
(A) (9, 6), (3, 6), (9, -3), (3, -3)
(B) (3, 2), (1, 2), (3, -1), (1, -1)
(C) (12, 8), (4, 8), (12, -4), (4, -4)
(D) (15, 10), (5, 10), (15, -5), (5, -5)
(A) (9, 6), (3, 6), (9, -3), (3, -3)
Explanation:
Let’s find coordinates of vertices after a dilation, by multiplying each coordinate by 1.5, and we can easily find the correct answer.
Coordinates of vertices of the image after a dilation are:
(6, 4) → (6 ∙ 1.5, 4 ∙ 1.5) → (9, 6) …………… (1)
(2, 4) → (2 ∙ 1.5, 4 ∙ 1.5) → (3, 6) ………….. (2)
(6, -2) → (6. 1.5, -2 ∙ 1.5) → (9, -3) ……………….. (3)
(2, 2) → (2 ∙ 1.5, 2 ∙ 1.5) → (3, 3) …………… (4)
So, the correct answer is (A)
Question 3.
Which represents the dilation shown where the black figure is the preimage?
(A) (x, y) → (1 .5x, 1 .5y)
(B) (x, y) → (2.5x, 2.5y)
(C) (x, y) → (3x, 3y)
(D) (x, y) → (6x, 6y)
(B) (x, y) → (2.5x, 2.5y)
Explanation:
First, we can see that one shape is dilation of other. Lets look at the picture and see which sides can we find measures. It’s easiest to count units on sides parallel with axis. Count, and write the measures.
Use that measures to calculate scale factor
$$\frac{5}{2}$$ = 2.5 (horizontal sides) …………… (1)
$$\frac{10}{4}$$ = 2.5 (vertical sides) …………… (2)
Scale factor is 2.5
A dilation with scale factor 2.5 is (x, y) → (2.5x, 2.5y)
Question 4.
Solve -a + 7 = 2a – 8.
(A) a = – 3
(B) a = –$$\frac{1}{3}$$
(C) a = 5
(D) a = 15
(C) a = 5
Explanation:
Lets solve equation a + 7 = 2a – 8 and than we can choose correct answer.
Given equation
-a + 7 = 2a – 8 (Rewrite) ……………… (1)
-a + 7 – 7 = 2a – 8 – 7 (Take 7 from both sides) …………….. (2)
-a = 2a – 15 (Simplify) ………….. (3)
-a – 2a = -2a – 15 – 2a (Take 2a from both sides) …………….. (4)
-3a = -15 (Simplify) …………… (5)
a = $$\frac{-15}{-3}$$ (divide both sides by -3) …………….. (6)
a = 5 (Calculate) ………………… (7)
Question 5.
An equilateral triangle has a perimeter of 24 centimeters. If the triangle is dilated by a factor of 0.5, what is the length of each side of the new triangle?
(A) 4 centimeters
(B) 12 centimeters
(C) 16 centimeters
(D) 48 centimeters
Answer: (A) 4 centimeters
Explanation:
Given,
An equilateral triangle has a perimeter of 24 centimeters.
24/3 = 8
The triangle is dilated by a factor of 0.5 and the triangle is equilateral, the side lengths of the new triangle is
8 × 0.5 = 4
The correct answer is option A.
Question 6.
Which equation does not represent a line with an x-intercept of 3?
(A) y = -2x + 6
(B) y = –$$\frac{1}{3}$$x + 1
(C) y = $$\frac{2}{3}$$x – 2
(D) y = 3x – 1
(D) y = 3x – 1
Explanation:
Coordinates of the point where the Line intersects the x-axis are (x, 0). So, if an x-intercept is 3, then the coordinates (3, 0) satisfy the equation.
(A): y = 2x + 6 → y = 0, x = 3:
0 = -2 ∙ 3 + 6
0 = -6 + 6 (Calculate)
0 = 0 (Simplify)
= This equation represent a line with an x-intercept of 3
(B): y = –$$\frac{1}{3}$$x + 1
0 = –$$\frac{1}{3}$$ ∙ 3 + 1
0 = -1 + 1 (Calculate
0 = 0 (Simplify)
⇒ This equation represent a line with an x-intercept of 3
(C): y = $$\frac{2}{3}$$x – 2
0 = $$\frac{2}{3}$$ ∙ 3 – 2
0 = 2 – 2 (Calculate)
0 = 0 (Simplify)
= This equation represent a line with an x-intercept of 3
(D): y = 3x – 1 → y = 0, x = 3
0 = 3 ∙ 3 – 1
0 = 9 – 1 (Calculate)
0 ≠ 8 (Simplify)
= This equation does not represent a line with an x-intercept of 3
Gridded Response
Question 7.
A car is traveling at a constant speed. After 2.5 hours, the car has traveled 80 miles. If the car continues to travel at the same constant speed, how many hours will it take to travel 270 miles? |
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### Course: 5th grade foundations (Eureka Math/EngageNY) >Unit 6
Lesson 3: Topic C: Foundations
# Angle types review
Review the following types of angles: acute, right, obtuse, and straight. Identify and draw types of angles in some practice problems.
### Summary of angle sizes
Angle typeAngle size (in degrees)
AcuteBetween ${0}^{\circ }$ and ${90}^{\circ }$
RightExactly ${90}^{\circ }$
ObtuseBetween ${90}^{\circ }$ and ${180}^{\circ }$
StraightExactly ${180}^{\circ }$
ReflexBetween ${180}^{\circ }$ and ${360}^{\circ }$
CompleteExactly ${360}^{\circ }$
### Right angles
A right angle is a ${90}^{\circ }$ angle. A right angle is in the shape of a perfect corner, like the corner of a rectangular sheet of paper. Below is an example of a right angle.
### Straight angles
A straight angle is a ${180}^{\circ }$ angle. A straight angle looks like a straight line. Below is an example of a straight angle.
### Acute angles
An acute angle is an angle whose degree measure is less than ${90}^{\circ }$. Below is an example of an acute angle.
When we compare an acute angle to a $\text{right angle}$, we can see that an acute angle is less than ${90}^{\circ }$.
### Obtuse angles
An obtuse angle is an angle whose degree measure is greater than ${90}^{\circ }$ but less than ${180}^{\circ }$. Below is an example of an obtuse angle.
When we compare an obtuse angle to a $\text{right angle}$, we can see that an obtuse angle is greater than ${90}^{\circ }$.
When we compare an obtuse angle to a $\text{straight angle}$, we can see that an obtuse angle is less than ${180}^{\circ }$.
### Reflex angles
A reflex angle is an angle whose degree measure is greater than ${180}^{\circ }$ but less than ${360}^{\circ }$. Below is an example of a reflex angle.
When we compare a reflex angle to a $\text{straight angle}$, we can see that a reflex angle is greater than ${180}^{\circ }$.
### Complete angles
A complete angle is a ${360}^{\circ }$ angle. A complete angle has both rays pointing the exact same direction. That is because we've turned a full circle from the first ray. Below is an example of a complete angle.
### Practice set 1: Identify angles
Problem 1A
Is the following angle acute, right, obtuse, or straight?
Want to try more problems like this? Check out this exercise.
### Practice set 2: Angles in diagrams and images
Problem 2A
What type of angle is the angle highlighted in green below?
Want to try more problems like this? Check out this exercise.
### Practice set 3: Drawing angle types
Problem 3A
Drag the vertex of the angle to place the vertex at point $\text{A}$.
Drag another point on the angle to make one of the rays go through point $\text{B}$.
Make the other ray go through one of the unlabeled black points to create a straight angle.
The arc symbol near the vertex indicates the angle being measured.
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• is there any angle that could be greater than 180 degrees?
• no. After 180 degrees it starts bending backwards, and it would be a acute angle again. A circle equals 360 degrees though.
• the last one confused me a little bit
• Yeah it was very confusing
• How is this "Right". Saying that everything around me is ATLEAST Right, Obtuse or strait? How and why? And if you go on drawing right, acute and obtuse angles then they don't let you make one how ever you want on the dots?
• Is there an angle over 720 degrees?
• Yes, angles can have any angle measure. But since there's only 360 degrees in a circle, we usually don't write angles any bigger than 360. A 720 degree angle is the same as a 360 degree angle.
• who likes math cuz i do
• I'm with you. I like math.
• I HAVE A JOKE how do u keep warm in the winter go to a corner its all was 90 digress.
• u spelled degrees wrong
• hi am a new person for khan academy
• so am I |
What Is 18 Multiplied By 23?
18 * 23 = 414
What is the Answer to the following Calculation: 18 x 23
In this section, we'll address a single issue: What is the product of 18 multiplied by 23? (Or, alternatively, what is 23 times 18?) The following is the response: 23 times 18 is equal to 414.
Here are some alternative ways to show or express the fact that 18 x 23 = 414
Twenty Three times Eighteen equals Four Hundred And Fourteen
Here are some alternative ways to show or express the fact that 18 x 23 = 414
• Twenty Three times Eighteen equals Four Hundred And Fourteen
• 18 (23) = 414
Consider the sum of 18 multiplied by 23 to understand better what "23 times 18" means. Add the Twenty Three digits together to get the result, or you may write down the number 18 23 times.
When using a handheld calculator, you can double-check the solution to 414 by pressing 18 x 23 and then = to get the result.
The answer to this riddle is 18 x 23 is 414.
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In what ways is it useful to multiply numbers?
Finding the total number of things requires some multiplication. To do this, we'll consider the total number of items and the number of groups of similar size.
What do we name the numbers used in multiplication?
Typically, the "factors" are the input integers into a multiplication problem. The "multiplicand" refers to the number that is multiplied, whereas the "multiplier" refers to the number that is used to do the multiplication.
What is the inverse of multiplication?
Subtraction has an inverse operation for multiplication. Addition can be cancelled out by subtraction. Adding and subtracting are polar opposites in the mathematical world.
How many multiplication facts are there?
Since integers can be multiplied forever, there are an unlimited number of multiplication tables. Typically, students study 144 of these multiplication facts, ranging from 1 to 12. A times table or multiplication chart is a common way to display data using multiplication.
What are the 4 rules of multiplication?
1. Multiplying 0 by any number always results in 0.
2. Any number times one is always the same number.
3. When multiplying by 10, add a zero to the first digit of the original integer.
4. When multiplying by two numbers that have the same sign, the result is always a positive number.
How many properties of multiplication are there?
There are three properties of multiplication:
1. Commutative
2. Associative
3. Distributive |
In fourth grade math, students begin to explore more complex math concepts. Don’t get overwhelmed! The best way to support your child is to keep your attitude about math positive as you tackle fourth grade math together. Get started by learning more about what your child will be working on.
Over the year, your child will:
### 1. Multiply bigger numbers
Now that your child has a basic understanding of multiplication, it’s time to move on to working with bigger numbers. In fourth grade, students learn how to multiply four-digit numbers by one-digit numbers and multiply two two-digit numbers together.
Support your child to develop understanding by using different strategies when multiplying. For example, your child can use pictures, knowledge of place value, or the traditional way to solve problems. Give your child a multiplication problem and challenge her to solve it in two different ways.
### 2. Divide with remainders
Just like with multiplication, your fourth grader is ready to move onto more complex division problems. In fourth grade, your child will learn how to divide numbers of up to four digits. These problems will often involve remainders, which may be a new concept for your child to work with.
Again, these division problems can and should be solved using multiple strategies. Encourage your child to use pictures, rectangular arrays, their knowledge of place value, and/or multiplication. Give your child the freedom to experiment by using different strategies and comparing/contrasting how they work.
### 3. Find factors of numbers 1-100
Though your child is working on multiplying and dividing bigger numbers, she probably isn’t 100 percent fluent in multiplication and division facts.
A great way to practice these facts is by figuring out the factors of two-digit numbers. Have your child list all the factors of a given number between 1-100. Or have a “factor race,” where you both try to list the factors of a number as quickly as possible.
### 4. Solve real-world word problems
Your child has been working on developing fluency in addition, subtraction, multiplication and division, so she’s ready to start solving problems using all of them (though maybe not all at once!). Your fourth grader is ready to solve multi-step word problems involving real-life scenarios around distances, intervals of time, liquid volumes, masses of objects, and money.
Help your child solve word problems by giving real-world scenarios to think about. Come up with a multi-step problem but don’t tell your child what they should use to solve it. Talk through the answer together. You can even have your child create word problems for you and check your answers against each other’s.
### 5. Understand large numbers in various forms
In previous grades, your child has developed an understanding of place value. In fourth grade, it’s time to read and write multi-digit numbers in a variety of forms. Your child will learn to work with base-ten numerals, number names, and expanded form.
Support your child to write the same number using expanded form and number names. For example, give your child the number 39,420 to write in expanded form: 30,000 + 9,000 + 400 + 20. Challenge your child to do this with even bigger numbers!
### 6. Work with fractions
Fourth graders generally have a basic understanding of fractions, but now they’ll learn more about equivalence and multiplying fractions. In fourth grade, students will learn how to compare two fractions with different denominators or different numerators. They will also work on multiplying fractions by a whole number.
Cooking and baking are great places to find real-world examples of fractions for students to compare (e.g. is ¼ cup of rice bigger than ⅓ cup of broth?)
### 7. Compare decimals
Students in fourth grade are developing an understanding of both large numbers and numbers less than 1, like fractions and decimals. In fourth grade, students will compare two decimals to the hundredths place.
Give your child two different decimals (.24 and .48, for example) to compare. Have your child tell you not only which decimal is bigger, but how many tenths and hundredths are in each number.
There’s a lot of math learning that will happen over the course of fourth grade - make time to check in with your child regularly to hear all about it!
Written by Lily Jones, Lily loves all things learning. She has been a kindergarten & first grade teacher, instructional coach, curriculum developer, and teacher trainer. She loves to look at the world with curiosity and inspire people of all ages to love learning. She lives in California with her husband, two kids, and a little dog.
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# Solving Equations with Double-Angle Identities
## Solve sine, cosine, and tangent of angles multiplied or divided by 2.
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Solving Trig Equations using Double and Half Angle Formulas
Trig Riddle: I am an angle x such that \begin{align*}0\le x <2\pi\end{align*}. I satisfy the equation \begin{align*}\sin 2x - \sin x=0\end{align*}. What angle am I?
### Solve Trigonometric Equations
We can use the half and double angle formulas to solve trigonometric equations.
Let's solve the following trigonometric equations.
1. Solve \begin{align*}\tan 2x+\tan x=0\end{align*} when \begin{align*}0\le x <2\pi\end{align*}.
Change \begin{align*}\tan 2x\end{align*} and simplify.
\begin{align*}\tan 2x + \tan x &=0 \\ \frac{2\tan x}{1-\tan ^2 x}+\tan x &=0 \\ 2\tan x +\tan x(1-\tan ^2x) &=0 \quad \rightarrow \text{Multiply everything by} \ 1-\tan^2x \text{ to eliminate denominator.}\\ 2\tan x +\tan x -\tan ^3 x &=0 \\ 3\tan x - \tan ^3 x &=0 \\ \tan x(3-\tan^2 x) &=0\end{align*}
Set each factor equal to zero and solve.
\begin{align*}&\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3-\tan ^2x =0 \\ &\ \ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ -\tan ^2 x =-3 \\ &\tan x=0 \qquad \qquad \qquad and \qquad \qquad \ \tan^2 x=3 \\ & \qquad x=0 \ and \ \pi \ \qquad \qquad \qquad \quad \qquad \tan x =\pm \sqrt{3} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x =\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\end{align*}
1. Solve \begin{align*}2\cos \frac{x}{2}+1=0\end{align*} when \begin{align*}0\le x<2\pi\end{align*}.
In this case, you do not have to use the half-angle formula. Solve for \begin{align*}\frac{x}{2}\end{align*}.
\begin{align*}2\cos \frac{x}{2}+1 &=0 \\ 2\cos \frac{x}{2} &=-1 \\ \cos \frac{x}{2} &= -\frac{1}{2}\end{align*}
Now, let’s find \begin{align*}\cos a = -\frac{1}{2}\end{align*} and then solve for \begin{align*}x\end{align*} by dividing by 2.
\begin{align*}\frac{x}{2} &=\frac{2\pi}{3},\frac{4\pi}{3} \\ &=\frac{4\pi}{3}, \frac{8\pi}{3}\end{align*}
Now, the second solution is not in our range, so the only solution is \begin{align*}x=\frac{4\pi}{3}\end{align*}.
1. Solve \begin{align*}4\sin x \cos x = \sqrt{3}\end{align*} for \begin{align*}0\le x < 2\pi\end{align*}.
Pull a 2 out of the left-hand side and use the \begin{align*}\sin 2x\end{align*} formula.
\begin{align*}4\sin x \cos x &=\sqrt{3} \\ 2\cdot 2\sin x \cos x &=\sqrt{3} \\ 2 \cdot \sin 2x &=\sqrt{3} \\ \sin 2x &=\frac{\sqrt{3}}{2} \\ 2x &= \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3} \\ x &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the angle x, where \begin{align*}0\le x <2\pi\end{align*}, such that x satisfies the equation \begin{align*}\sin 2x - \sin x=0\end{align*}.
Use the double angle formula and simplify.
\begin{align*}\sin 2x - \sin x = 0\\ 2\sin x \cos x - \sin x = 0\\ \sin x(2\cos x - 1)= 0\\ \sin x = 0 \text OR \cos x = \frac{1}{2}\end{align*}
Under the constraint \begin{align*}0\le x <2\pi\end{align*}, \begin{align*}\sin x = 0\end{align*} when \begin{align*}x=0\end{align*} or when \begin{align*}x=\pi\end{align*}. Under this same constraint, \begin{align*}\cos x = \frac{1}{2}\end{align*} when \begin{align*}x=\frac{\pi}{3}\end{align*} or when \begin{align*}x=\frac{5\pi}{3}\end{align*}.
#### Example 2
Solve the following equation for \begin{align*}0\le x <2\pi\end{align*}.
\begin{align*}\sin \frac{x}{2}=-1\end{align*}
\begin{align*}\sin \frac{x}{2} &=-1 \\ \frac{x}{2} &=\frac{3\pi}{2} \\ x &=3\pi\end{align*}
From this we can see that there are no solutions within our interval.
#### Example 3
Solve the following equation for \begin{align*}0\le x <2\pi\end{align*}.
\begin{align*}\cos 2x-\cos x=0\end{align*}
\begin{align*}\cos 2x - \cos x &=0 \\ 2\cos ^2x-\cos x -1 &=0 \\ (2\cos x -1)(\cos x +1) &=0\end{align*}
Set each factor equal to zero and solve.
\begin{align*}2\cos x-1 &=0 \\ 2\cos x &=1 \qquad \qquad \qquad \qquad\cos x +1=0\\ \cos x &=\frac{1}{2} \qquad \qquad and \qquad \qquad \cos x =-1\\ x &=\frac{\pi}{3}, \frac{5\pi}{3} \qquad \qquad \qquad \qquad \quad \ x=\pi \\ \end{align*}
### Review
Solve the following equations for \begin{align*}0\le x < 2\pi\end{align*}.
1. \begin{align*}\cos x -\cos \frac{1}{2}x=0\end{align*}
2. \begin{align*}\sin 2x \cos x=\sin x\end{align*}
3. \begin{align*}\cos 3x - \cos ^3x=3\sin ^2x\cos x\end{align*}
4. \begin{align*}\tan 2x - \tan x =0\end{align*}
5. \begin{align*}\cos 2x -\cos x =0\end{align*}
6. \begin{align*}2\cos ^2\frac{x}{2}=1\end{align*}
7. \begin{align*}\tan \frac{x}{2}=4\end{align*}
8. \begin{align*}\cos \frac{x}{2}=1+\cos x\end{align*}
9. \begin{align*}\sin 2x +\sin x=0\end{align*}
10. \begin{align*}\cos ^2x-\cos 2x =0\end{align*}
11. \begin{align*}\frac{\cos 2x}{\cos ^2x}=1\end{align*}
12. \begin{align*}\cos 2x-1=\sin^2x\end{align*}
13. \begin{align*}\cos 2x =\cos x\end{align*}
14. \begin{align*}\sin 2x-\cos 2x =1\end{align*}
15. \begin{align*}\sin^2x-2=\cos 2x\end{align*}
16. \begin{align*}\cot x+\tan x=2\csc 2x\end{align*}
To see the Review answers, open this PDF file and look for section 14.17.
### Notes/Highlights Having trouble? Report an issue.
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# Difference between revisions of "2018 AMC 10A Problems/Problem 14"
## Problem
What is the greatest integer less than or equal to $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?$$
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
## Solution 1
We write $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.$$ Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is $\boxed{\textbf{(A) }80}$.
## Solution 2
Let's set this value equal to $x$. We can write $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.$$ Multiplying by $3^{96}+2^{96}$ on both sides, we get $$3^{100}+2^{100}=x(3^{96}+2^{96}).$$ Now let's take a look at the answer choices. We notice that $81$, choice $B$, can be written as $3^4$. Plugging this into our equation above, we get $$3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4\cdot 2^{96}.$$ The right side is larger than the left side because $$2^{100} \leq 2^{96}\cdot 3^4.$$ This means that our original value, $x$, must be less than $81$. The only answer that is less than $81$ is $80$ so our answer is $\boxed{A}$.
~Nivek
## Solution 3
$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}(\frac{3^{100}}{2^{96}})+2^{96}(2^{4})}{2^{96}(\frac{3}{2})^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{\frac{3^{100}}{2^{100}}*2^{4}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}$.
We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.
$\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}<\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}$
$\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81$
So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is $\boxed{(A) 80}$
## Solution 4
Let $x=3^{96}$ and $y=2^{96}$. Then our fraction can be written as $\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}$. Notice that $\frac{65x}{x+y}<\frac{65x}{x}=65$. So , $16+\frac{65x}{x+y}<16+65=81$. And our only answer choice less than 81 is $\boxed{(A) 80}$ (RegularHexagon)
## Solution 5
Let $x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$. Multiply both sides by $(3^{96}+2^{96})$, and expand. Rearranging the terms, we get $3^{96}(3^4-x)+2^{96}(2^4-x)=0$. The left side is decreasing, and it is negative when $x=81$. This means that the answer must be less than $81$; therefore the answer is $\boxed{(A)}$.
## Solution 6 (eyeball it)
A faster solution. Recognize that for exponents of this size $3^{n}$ will be enormously greater than $2^{n}$, so the terms involving $2$ will actually have very little effect on the quotient. Now we know the answer will be very close to $81$.
Notice that the terms being added on to the top and bottom are in the ratio $\frac{1}{16}$ with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: $\boxed{(A)}$.
## Solution 7
Notice how $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ can be rewritten as $\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}$. Note that $\frac{65(2^{96})}{3^{96}+2^{96}}<1$, so the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ is $80$ or $\boxed{\textbf{(A)}}$ ~blitzkrieg21
## Solution 8
For positive $a, b, c, d$, if $\frac{a}{b}<\frac{c}{d}$ then $\frac{c+a}{d+b}<\frac{c}{d}$. Let $a=2^{100}, b=2^{96}, c=3^{100}, d=3^{96}$. Then $\frac{c}{d}=3^4$. So answer is less than 81, which leaves only one choice, 80.
• Note that the algebra here is synonymous to the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down.
~ ccx09
## Solution 9
Try long division, and notice putting $3^4=81$ as the denominator is too big and putting $3^4-1=80$ is too small. So we know that the answer is between $80$ and $81$, yielding $80$ as our answer.
## Solution 10 (Using the answer choices)
### Solution 10.1
We can compare the given value to each of our answer choices. We already know that it is greater than $80$ because otherwise there would have been a smaller answer, so we move onto $81$. We get:
$\frac{3^{100}+2^{100}}{3^{96}+2^{96}} \text{ ? } 3^4$
Cross multiply to get:
$3^{100}+2^{100} \text{ ? }3^{100}+(2^{96})(3^4)$
Cancel out $3^{100}$ and divide by $2^{96}$ to get $2^{4} \text{ ? }3^4$. We know that $2^4 < 3^4$, which means the expression is less than $81$ so the answer is $\boxed{(A)}$.
### Solution 10.2
We know this will be between 16 and 81 because $\frac{3^{100}}{3^{96}} = 3^4 = 81$ and $\frac{2^{100}}{2^{96}} = 2^4 = 16$. $80=\boxed{(A)}$ is the only option choice in this range.
## Explanation for why 80 is indeed the floor
We need $3^{100}+2^{100} > 80 \cdot 3^{96} + 5 \cdot 2^{100}$. Since $3^{100} = 81\cdot 3^{96}$, this translates to $$3^{96} > 4\cdot 2^{100} = 64\cdot 2^{96}.$$ We now prove that $(3/2)^k > k$ for all positive integers $k$. Clearly, $(3/2)^2 = 2.25 > 2$. Assume $(3/2)^k > k$ where $k\ge 2$. Then $\left(\frac{3}{2}\right)^{k+1} > \frac{3k}{2} = k + \frac{k}{2}$. But since $k/2 \ge 1$, we have that $(3/2)^{k+1} > k+1$. By induction (and $k=1$ is trivial), the claim is proven.
Thus, $\left(\frac{3}{2}\right)^{96} > 96 > 64$. Writing this proof backwards and dividing both sides of the initial equation by $80$ yields $80 < \frac{3^{100}+2^{100}}{3^{96}+2^{96}} < 81$.
## Solution 11
We know that in this problem, $3^{96}+2^{96}$ times some number is equal to $3^{100}+2^{100}$. Multiplying answer $\boxed{\textbf{(B)}}$ or 81 to $3^{96}+2^{96}$ gives us $3^{100}+2^{96}\cdot3^4$. We know that $3^4\cdot2^{96}$ is greater than $2^{100}$, so that means $\boxed{\textbf{(B)}}$ or 81 is too big. That leaves us with only one solution: $80=\boxed{\textbf{(A) } 80}.$
~ Terribleteeth |
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# The Golden Ratio
Volkan UYGUN
In mathematics and the arts, two quantities are in the golden ratio if the ratio between the sum of those quantities and the larger one is the same as the ratio between the larger one and the smaller. The golden ratio is a mathematical constant approximately 1.6180339887. The golden ratio is also known as the most aesthetic ratio between the two sides of a rectangle. The golden ratio is often denoted by the Greek letter (phi). .
Also known as: Extreme and mean ratio, Medial section, Divine proportion, Divine section (Latin: sectio divina), Golden proportion, Golden cut, Mean of Phidias
## Construction of the Golden Section
Firstly, divide a square such that it makes two precisely equal rectangles.
Take the diagonal of the rectangle as the radius to contsruct a circle to touch the next side of the square. Then, extend the base of the square so that it touches the circle.
When we complete the shape to a rectangle, we will realize that the rectangle fits the optimum ratio of golden. The base lenght of the rectangle (C) divided by the base lenght of the square (A) equals the golden ratio. C / A =A / B = 1.6180339 = The Golden Ratio
Each time we substract a square from the golden rectangle, what we will get is a golden rectangle again.
## The Golden Spiral
After doing the substraction infinitely many times, if we draw a spiral starting from the square of the smallest rectangle, (Sidelenght of the square=Radius of the spiral) we will get a Golden Spiral. The Golden spiral determines the structure and the shape of many organic and inorganic assets.
## Relations with the Fibonacci Numbers
Fibonacci Numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, ... Fibonacci numbers have an interesting attribute. When we divide one of the fibonacci numbers to the previous one, we will get results that are so close to each other. Moreover, after the 13th number in the serie, the divison will be fixed at 1.618, namely the golden number.
Golden ratio= 1,618 233 / 144 = 1,618 377 / 233 = 1,618 610 / 377 = 1,618 987 / 610 = 1,618 1597 / 987 = 1,618 2584 / 1597 = 1,618
## Golden Ratio In Arts and Architecture
Leonardo Da Vinci
Many artists who lived after Phidias have used this proportion. Leonardo Da Vinci called it the "divine proportion" and featured it in many of his paintings, for example in the famous "Mona Lisa". Try drawing a rectangle around her face. You will realize that the measurements are in a golden proportion. You can further explore this by subdividing the rectangle formed by using her eyes as a horizontal divider.
## The Vitruvian Man
Leonardo did an entire exploration of the human body and the ratios of the lengths of various body parts. Vitruvian Man illustrates that the human body is proportioned according to the Golden Ratio.
The Parthenon
Phi was named for the Greek sculptor Phidias. The exterior dimensions of the Parthenon in Athens, built in about 440BC, form a perfect golden rectangle.
The baselenght of Egyptian pyramids divided by the height of them gives the golden ratio..
## Golden Ratio in Human Hand and Arm
Look at your own hand: You have ... 2 hands each of which has ... 5 fingers, each of which has ... 3 parts separated by ... 2 knuckles
The length of different parts in your arm also fits the golden ratio.
## Golden Ratio in the Human Face
The dividence of every long line to the short line equals the golden ratio. Lenght of the face / Wideness of the face Lenght between the lips and eyebrows / Lenght of the nose, Lenght of the face / Lenght between the jaw and eyebrows Lenght of the mouth / Wideness of the nose, Wideness of the nose / Distance between the holes of the nose, Length between the pupils / Length between teh eyebrows. All contain the golden ratio.
The Golden Spiral can be seen in the arrangement of seeds on flower heads.
## Golden Ratio In The Sea Shells
The shape of the inner and outer surfaces of the sea shells, and their curves fit the golden ratio..
## The ratio of the braches of a snowflake results in the golden ratio.
Disputed Sightings
Some specific proportions in the bodies of many animals (including humans) and parts of the shells of mollusks and cephalopods are often claimed to be in the golden ratio. There is actually a large variation in the real measures of these elements in a specific individual and the proportion in question is often significantly different from the golden ratio The proportions of different plant components (numbers of leaves to branches, diameters of geometrical figures inside flowers) are often claimed to show the golden ratio proportion in several species. In practice, there are significant variations between individuals, seasonal variations, and age variations in these species. While the golden ratio may be found in some proportions in some individuals at particular times in their life cycles, there is no consistent ratio in their proportions
Referances |
University Physics Volume 1
# 7.3Work-Energy Theorem
University Physics Volume 17.3 Work-Energy Theorem
## Learning Objectives
By the end of this section, you will be able to:
• Apply the work-energy theorem to find information about the motion of a particle, given the forces acting on it
• Use the work-energy theorem to find information about the forces acting on a particle, given information about its motion
We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested in the motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, or the net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion.
Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is the dot product of the net force and the displacement: $dWnet=F→net·dr→.dWnet=F→net·dr→.$ Newton’s second law tells us that $F→net=m(dv→/dt),F→net=m(dv→/dt),$ so $dWnet=m(dv→/dt)·dr→.dWnet=m(dv→/dt)·dr→.$ For the mathematical functions describing the motion of a physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is,
$dWnet=m(dv→dt)·dr→=mdv→·(dr→dt)=mv→·dv→,dWnet=m(dv→dt)·dr→=mdv→·(dr→dt)=mv→·dv→,$
where we substituted the velocity for the time derivative of the displacement and used the commutative property of the dot product [Equation 2.30]. Since derivatives and integrals of scalars are probably more familiar to you at this point, we express the dot product in terms of Cartesian coordinates before we integrate between any two points A and B on the particle’s trajectory. This gives us the net work done on the particle:
$Wnet,AB=∫AB(mvxdvx+mvydvy+mvzdvz)=12m|vx2+vy2+vz2|AB=|12mv2|AB=KB−KA.Wnet,AB=∫AB(mvxdvx+mvydvy+mvzdvz)=12m|vx2+vy2+vz2|AB=|12mv2|AB=KB−KA.$
7.8
In the middle step, we used the fact that the square of the velocity is the sum of the squares of its Cartesian components, and in the last step, we used the definition of the particle’s kinetic energy. This important result is called the work-energy theorem (Figure 7.11).
## Work-Energy Theorem
The net work done on a particle equals the change in the particle’s kinetic energy:
$Wnet=KB−KA.Wnet=KB−KA.$
7.9
Figure 7.11 Horse pulls are common events at state fairs. The work done by the horses pulling on the load results in a change in kinetic energy of the load, ultimately going faster. (credit: modification of work by “Jassen”/ Flickr)
According to this theorem, when an object slows down, its final kinetic energy is less than its initial kinetic energy, the change in its kinetic energy is negative, and so is the net work done on it. If an object speeds up, the net work done on it is positive. When calculating the net work, you must include all the forces that act on an object. If you leave out any forces that act on an object, or if you include any forces that don’t act on it, you will get a wrong result.
The importance of the work-energy theorem, and the further generalizations to which it leads, is that it makes some types of calculations much simpler to accomplish than they would be by trying to solve Newton’s second law. For example, in Newton’s Laws of Motion, we found the speed of an object sliding down a frictionless plane by solving Newton’s second law for the acceleration and using kinematic equations for constant acceleration, obtaining
$vf2=vi2+2g(sf−si)sinθ,vf2=vi2+2g(sf−si)sinθ,$
where s is the displacement down the plane.
We can also get this result from the work-energy theorem in Equation 7.1. Since only two forces are acting on the object-gravity and the normal force-and the normal force doesn't do any work, the net work is just the work done by gravity. The work dW is the dot product of the force of gravity or $F→=−mgj^F→=−mgj^$ and the displacement $dr→=dxi^+dyj^dr→=dxi^+dyj^$. After taking the dot product and integrating from an initial position $yiyi$ to a final position $yfyf$, one finds the net work as
$Wnet=Wgrav=−mg(yf−yi),Wnet=Wgrav=−mg(yf−yi),$
where y is positive up. The work-energy theorem says that this equals the change in kinetic energy:
$−mg(yf−yi)=12m(vf2−vi2).−mg(yf−yi)=12m(vf2−vi2).$
Using a right triangle, we can see that $(yf−yi)=(sf−si)sinθ,(yf−yi)=(sf−si)sinθ,$ so the result for the final speed is the same.
What is gained by using the work-energy theorem? The answer is that for a frictionless plane surface, not much. However, Newton’s second law is easy to solve only for this particular case, whereas the work-energy theorem gives the final speed for any shaped frictionless surface. For an arbitrary curved surface, the normal force is not constant, and Newton’s second law may be difficult or impossible to solve analytically. Constant or not, for motion along a surface, the normal force never does any work, because it’s perpendicular to the displacement. A calculation using the work-energy theorem avoids this difficulty and applies to more general situations.
## Problem-Solving Strategy
### Work-Energy Theorem
1. Draw a free-body diagram for each force on the object.
2. Determine whether or not each force does work over the displacement in the diagram. Be sure to keep any positive or negative signs in the work done.
3. Add up the total amount of work done by each force.
4. Set this total work equal to the change in kinetic energy and solve for any unknown parameter.
5. Check your answers. If the object is traveling at a constant speed or zero acceleration, the total work done should be zero and match the change in kinetic energy. If the total work is positive, the object must have sped up or increased kinetic energy. If the total work is negative, the object must have slowed down or decreased kinetic energy.
## Example 7.9
### Loop-the-Loop
The frictionless track for a toy car includes a loop-the-loop of radius R. How high, measured from the bottom of the loop, must the car be placed to start from rest on the approaching section of track and go all the way around the loop?
Figure 7.12 A frictionless track for a toy car has a loop-the-loop in it. How high must the car start so that it can go around the loop without falling off?
### Strategy
The free-body diagram at the final position of the object is drawn in Figure 7.12. The gravitational work is the only work done over the displacement that is not zero. Since the weight points in the same direction as the net vertical displacement, the total work done by the gravitational force is positive. From the work-energy theorem, the starting height determines the speed of the car at the top of the loop,
$–mg(y2−y1)=12mv22,–mg(y2−y1)=12mv22,$
where the notation is shown in the accompanying figure. At the top of the loop, the normal force and gravity are both down and the acceleration is centripetal, so
$atop=Fm=N+mgm=v22R.atop=Fm=N+mgm=v22R.$
The condition for maintaining contact with the track is that there must be some normal force, however slight; that is, $N>0N>0$. Substituting for $v22v22$ and N, we can find the condition for $y1y1$.
### Solution
Implement the steps in the strategy to arrive at the desired result:
$N=−mg+mv22R=−mgR+2mg(y1−2R)R>0 or y1>5R2.N=−mg+mv22R=−mgR+2mg(y1−2R)R>0 or y1>5R2.$
### Significance
On the surface of the loop, the normal component of gravity and the normal contact force must provide the centripetal acceleration of the car going around the loop. The tangential component of gravity slows down or speeds up the car. A child would find out how high to start the car by trial and error, but now that you know the work-energy theorem, you can predict the minimum height (as well as other more useful results) from physical principles. By using the work-energy theorem, you did not have to solve a differential equation to determine the height.
Suppose the radius of the loop-the-loop in Example 7.9 is 15 cm and the toy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop?
## Interactive
Watch this video to see a looping rollercoaster.
In situations where the motion of an object is known, but the values of one or more of the forces acting on it are not known, you may be able to use the work-energy theorem to get some information about the forces. Work depends on the force and the distance over which it acts, so the information is provided via their product.
## Example 7.10
### Determining a Stopping Force
A bullet has a mass of 40 grains (2.60 g) and a muzzle velocity of 1100 ft./s (335 m/s). It can penetrate eight 1-inch pine boards, each with thickness 0.75 inches. What is the average stopping force exerted by the wood, as shown in Figure 7.13?
Figure 7.13 The boards exert a force to stop the bullet. As a result, the boards do work and the bullet loses kinetic energy.
### Strategy
We can assume that under the general conditions stated, the bullet loses all its kinetic energy penetrating the boards, so the work-energy theorem says its initial kinetic energy is equal to the average stopping force times the distance penetrated. The change in the bullet’s kinetic energy and the net work done stopping it are both negative, so when you write out the work-energy theorem, with the net work equal to the average force times the stopping distance, that’s what you get. The total thickness of eight 1-inch pine boards that the bullet penetrates is $8×34in.=6in.=15.2cm.8×34in.=6in.=15.2cm.$
### Solution
Applying the work-energy theorem, we get
$Wnet=−FaveΔsstop=−Kinitial,Wnet=−FaveΔsstop=−Kinitial,$
so
$Fave=12mv2Δsstop=12(2.6×10−3kg)(335m/s)20.152m=960N.Fave=12mv2Δsstop=12(2.6×10−3kg)(335m/s)20.152m=960N.$
### Significance
We could have used Newton’s second law and kinematics in this example, but the work-energy theorem also supplies an answer to less simple situations. The penetration of a bullet, fired vertically upward into a block of wood, is discussed in one section of Asif Shakur’s recent article [“Bullet-Block Science Video Puzzle.” The Physics Teacher (January 2015) 53(1): 15-16]. If the bullet is fired dead center into the block, it loses all its kinetic energy and penetrates slightly farther than if fired off-center. The reason is that if the bullet hits off-center, it has a little kinetic energy after it stops penetrating, because the block rotates. The work-energy theorem implies that a smaller change in kinetic energy results in a smaller penetration. You will understand more of the physics in this interesting article after you finish reading Angular Momentum.
## Interactive
Learn more about work and energy in this PhET simulation called “the ramp.” Try changing the force pushing the box and the frictional force along the incline. The work and energy plots can be examined to note the total work done and change in kinetic energy of the box.
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# Chapter 7: Sampling Distributions
## Presentation on theme: "Chapter 7: Sampling Distributions"— Presentation transcript:
Chapter 7: Sampling Distributions
Section 7.2 Sample Proportions The Practice of Statistics, 4th edition – For AP* STARNES, YATES, MOORE
Chapter 7 Sampling Distributions
7.1 What is a Sampling Distribution? 7.2 Sample Proportions 7.3 Sample Means
Section 7.2 Sample Proportions
Learning Objectives After this section, you should be able to… FIND the mean and standard deviation of the sampling distribution of a sample proportion DETERMINE whether or not it is appropriate to use the Normal approximation to calculate probabilities involving the sample proportion CALCULATE probabilities involving the sample proportion EVALUATE a claim about a population proportion using the sampling distribution of the sample proportion
The Sampling Distribution of
Consider the approximate sampling distributions generated by a simulation in which SRSs of Reese’s Pieces are drawn from a population whose proportion of orange candies is either 0.45 or 0.15. What do you notice about the shape, center, and spread of each? Sample Proportions
The Sampling Distribution of
What did you notice about the shape, center, and spread of each sampling distribution? Sample Proportions
As sample size increases, the spread decreases.
The Sampling Distribution of In Chapter 6, we learned that the mean and standard deviation of a binomial random variable X are Sample Proportions As sample size increases, the spread decreases.
Sampling Distribution of a Sample Proportion
The Sampling Distribution of Sample Proportions As n increases, the sampling distribution becomes approximately Normal. Before you perform Normal calculations, check that the Normal condition is satisfied: np ≥ 10 and n(1 – p) ≥ 10. Sampling Distribution of a Sample Proportion
Using the Normal Approximation for
Sample Proportions A polling organization asks an SRS of 1500 first-year college students how far away their home is. Suppose that 35% of all first-year students actually attend college within 50 miles of home. What is the probability that the random sample of 1500 students will give a result within 2 percentage points of this true value? STATE: We want to find the probability that the sample proportion falls between 0.33 and 0.37 (within 2 percentage points, or 0.02, of 0.35). PLAN: We have an SRS of size n = 1500 drawn from a population in which the proportion p = 0.35 attend college within 50 miles of home. DO: Since np = 1500(0.35) = 525 and n(1 – p) = 1500(0.65)=975 are both greater than 10, we’ll standardize and then use Table A to find the desired probability. CONCLUDE: About 90% of all SRSs of size 1500 will give a result within 2 percentage points of the truth about the population.
Section 7.2 Sample Proportions
Summary In this section, we learned that… In practice, use this Normal approximation when both np ≥ 10 and n(1 - p) ≥ 10 (the Normal condition).
Looking Ahead… In the next Section…
We’ll learn how to describe and use the sampling distribution of sample means. We’ll learn about The sampling distribution of Sampling from a Normal population The central limit theorem In the next Section…
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#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 33 Maths Textbook Solution.
$\frac{3}{4}log3-log2$
Hint:Use indefinite integral formula then put the limits to solve this integral
Given:
$\int_{1}^{3} \frac{\log x}{(x+1)^{2}} d x=\int_{1}^{3} \log x \frac{1}{(x+1)^{2}} d x$
Integrating by parts then
\begin{aligned} &\int_{1}^{3} \log x \cdot \frac{1}{(x+1)^{2}} d x=\left[\log x \int \frac{1}{(x+1)^{2}} d x\right]_{1}^{3}-\int_{1}^{3}\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{(x+1)^{2}} d x\right\} d x \\ &=\left[\log x \int(x+1)^{-2} d x\right]_{1}^{3}-\int_{1}^{3}\left\{\frac{1}{x} \cdot \int(x+1)^{-2} d x\right\} d x \\ &=\left[\log x\left(\frac{(x+1)^{-2+1}}{-2+1}\right)\right]_{1}^{3}-\int_{1}^{3} \frac{1}{x}\left(\frac{(x+1)^{-2+1}}{-2+1}\right) d x \end{aligned}
\begin{aligned} &=\left[\log x\left(\frac{(x+1)^{-1}}{-1}\right)\right]_{1}^{3}-\int_{1}^{3} \frac{1}{x}\left(\frac{(x+1)^{-1}}{-1}\right) d x \\ &=-\left[\frac{\log x}{x+1}\right]_{1}^{3}+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\left[\frac{\log 3}{3+1}-\frac{\log 1}{1+1}\right]_{1}^{3}+\int_{1}^{3} \frac{1}{x(x+1)} d x \end{aligned}
\begin{aligned} &=-\left[\frac{\log 3}{4}-\frac{\log 1}{2}\right]+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\left[\frac{\log 3}{4}-0\right]+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\frac{1}{4} \log 3+\int_{1}^{3} \frac{1}{x(x+1)} d x \ldots . .(1) \\ &\therefore \int_{1}^{3} \frac{1}{x(x+1)} d x \end{aligned}
To solve this integral first we need to find its partial fraction then we will integrate and put the limits. So
\begin{aligned} &\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1} \\ &\Rightarrow 1=\frac{A}{x}(x+1)(x)+\frac{B}{x+1} x(x+1) \\ &\Rightarrow 1=A(x+1)+B x \\ &\Rightarrow 1=A x+A+B x \\ &\Rightarrow 1=(A+B) x+A \end{aligned}
Equating coefficient of x from both sides
$0=A+B\Rightarrow A=-B\Rightarrow B=-A$
Again Equating coefficient of constant term from both side, then
$1=A$
$\Rightarrow B=-A=-1$
$\therefore A=1,B=-1$
Thus
$\frac{1}{x\left (x+1 \right )}=\frac{1}{x}-\frac{1}{x+1}$
Now
$\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int_{1}^{3} \frac{1}{x} d x-\int_{1}^{3} \frac{1}{x+1} d x$
put $x+1=u$
$\Rightarrow dx=du$ in the 2nd integral, then when x=1
$\Rightarrow u=2$ and when x=3,u=4 then
\begin{aligned} &\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int_{1}^{3} \frac{1}{x} d x-\int_{2}^{4} \frac{1}{u} d u \\ &=[\log |x|]_{1}^{3}-[\log |u|]_{2}^{4} \\ &{\left[\because \int \frac{1}{x} d x=\log |x|\right]} \end{aligned}
\begin{aligned} &=[\log 3-\log 1]-[\log 4-\log 2] \\ &=[\log 3-0]-\left[\log \frac{4}{2}\right] \\ &=\log 3-\log 2 \ldots \ldots . .(2) \end{aligned}
putting the value of this integral eq(2) in eq(1) then
$\int_{1}^{3} \frac{\log x}{(x+1)^{2}} d x=-\frac{1}{4} \log 3+\log 3-\log 2$
\begin{aligned} &=\left(-\frac{1}{4}+1\right) \log 3-\log 2 \\ &=\left(\frac{-1+4}{4}\right) \log 3-\log 2 \\ &=\frac{3}{4} \log 3-\log 2 \end{aligned} |
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