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# Prime Factors of 14443
Prime Factors of 14443 are 11, 13, and 101
#### How to find prime factors of a number
1. Prime Factorization of 14443 by Division Method 2. Prime Factorization of 14443 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
#### Steps to find Prime Factors of 14443 by Division Method
To find the primefactors of 14443 using the division method, follow these steps:
• Step 1. Start dividing 14443 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 14443, which is 11. Divide 14443 by 11 to obtain the quotient (1313).
14443 ÷ 11 = 1313
• Step 3. Repeat step 1 with the obtained quotient (1313).
1313 ÷ 13 = 101
101 ÷ 101 = 1
So, the prime factorization of 14443 is, 14443 = 11 x 13 x 101.
#### Steps to find Prime Factors of 14443 by Factor Tree Method
We can follow the same procedure using the factor tree of 14443 as shown below:
So, the prime factorization of 14443 is, 14443 = 11 x 13 x 101.
#### What does Prime factor in mathematics mean?
Prime numbers, in mathematics are all those whole numbers greater than 1 having exactly two divisors that is 1 and the number itself. When we express any number as the product of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 14443 are 11 x 13 x 101.
#### Properties of Prime Factors
• For any given number there is one and only one set of unique prime factors.
• 2 is the only even prime number. So, any given number can have only one even prime factor and that is 2.
• Two prime factors of a given number are always coprime to each other.
• 1 is neither a prime number nor a composite number and also 1 is the factor of every given number. So, 1 is the factor of 14443 but not a prime factor of 14443.
• Which is the smallest prime factor of 14443?
Smallest prime factor of 14443 is 11.
• Is 14443 a perfect square?
No 14443 is not a perfect square.
• What is the prime factorization of 14443?
Prime factorization of 14443 is 11 x 13 x 101.
• What is prime factorization of 14443 in exponential form?
Prime factorization of 14443 in exponential form is 11 x 13 x 101.
• Is 14443 a prime number?
false, 14443 is not a prime number.
• Which is the largest prime factors of 14443?
The largest prime factor of 14443 is 101.
• What is the product of all prime factors of 14443?
Prime factors of 14443 are 11 x 13 x 101. Therefore, their product is 14443.
• What is the sum of all odd prime factors of 14443?
Prime factors of 14443 are 11 , 13 , 101, out of which 11 , 13 , 101 are odd numbers. So, the sum of odd prime factors of 14443 is 11 + 13 + 101 = 125.
• What is the product of all odd prime factors of 14443?
Prime factors of 14443 are 11 , 13 , 101, out of which 11 , 13 , 101 are odd numbers. So, the product of odd prime factors of 14443 is 11 x 13 x 101 = 14443.
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## Quotient and Power Rules for Logarithms
### Learning Outcome
• Define and use the quotient and power rules for logarithms
For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to simplify division of like bases raised to powers by subtracting the exponents: $\frac{x^a}{x^b}={x}^{a-b}$. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.
### The Quotient Rule for Logarithms
The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.
${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$
We can show ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$.
Given positive real numbers M, N, and b, where $b>0$ we will show
${\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$.
Let $m={\mathrm{log}}_{b}M$ and $n={\mathrm{log}}_{b}N$. In exponential form, these equations are ${b}^{m}=M$ and ${b}^{n}=N$. It follows that
$\begin{array}{c}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\hfill & ={\mathrm{log}}_{b}\left(\frac{{b}^{m}}{{b}^{n}}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m-n}\right)\hfill & \text{Apply the quotient rule for exponents}.\hfill \\ \hfill & =m-n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}$
### Example
Expand the following expression using the quotient rule for logarithms.
$\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)$
In the previous example, it was helpful to first factor the numerator and denominator and divide common terms. This gave us a simpler expression to use to write an equivalent expression. It is important to remember to subtract the logarithm of the denominator from the logarithm of the numerator. Always check to see if you need to expand further with the product rule.
### Example
Expand ${\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)$.
## Analysis of the Solution
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for $x=-\frac{4}{3}$ and $x=2$. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that $x\gt0$, $x\gt1$, $x\gt-\frac{4}{3}$, and $x\lt2$. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.
In the following video, we show more examples of using the quotient rule for logarithms.
## Using the Power Rule for Logarithms
We have explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as ${x}^{2}$? One method is as follows:
$\begin{array}{c}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}$
Notice that we used the product rule for logarithms to simplify the example above. By doing so, we have derived the power rule for logarithms which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,
$\begin{array}{c}100={10}^{2}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}$
### The Power Rule for Logarithms
The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.
${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$
### Example
Expand ${\mathrm{log}}_{2}{x}^{5}$.
The power rule for logarithms is possible because we can use the product rule and combine like terms. In the next example, you will see that we can also rewrite an expression as a power in order to use the power rule.
### Example
Expand ${\mathrm{log}}_{3}\left(25\right)$ using the power rule for logs.
Now let us use the power rule in reverse.
### Example
Rewrite $4\mathrm{ln}\left(x\right)$ using the power rule for logs to a single logarithm with a leading coefficient of $1$.
## Summary
You can use the quotient rule of logarithms to write an equivalent difference of logarithms in the following way:
1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.
To use the power rule of logarithms to write an equivalent product of a factor and a logarithm, consider the following:
1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.
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Rd Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 8 Direction Cosines And Direction Ratios are provided here with simple step-by-step explanations. These solutions for Direction Cosines And Direction Ratios are extremely popular among Class 12 Science students for Math Direction Cosines And Direction Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2018 Book of Class 12 Science Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2018 Solutions. All Rd Sharma XII Vol 2 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.
Question 1:
If a line makes angles of 90°, 60° and 30° with the positive direction of x, y, and z-axis respectively, find its direction cosines.
Let the direction cosines of the line be l, m, n.
Now,
Question 2:
If a line has direction ratios 2, −1, −2, determine its direction cosines.
Question 3:
Find the direction cosines of the line passing through two points (−2, 4, −5) and (1, 2, 3).
Question 4:
Using direction ratios show that the points A (2, 3, −4), B (1, −2, 3) and C (3, 8, −11) are collinear.
Question 5:
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).
Question 6:
Find the angle between the vectors with direction ratios proportional to 1, −2, 1 and 4, 3, 2.
Question 7:
Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.
Question 8:
Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.
Question 9:
Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
Question 10:
Show that the line through points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and (1, 2, 5).
Question 11:
Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Thus, the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Question 12:
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).
Therefore, the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1) and (4, 3, -1).
Question 13:
Find the angle between the lines whose direction ratios are proportional to a, b, c and bc, ca, ab.
Question 14:
If the coordinates of the points A, B, C, D are (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2), then find the angle between AB and CD.
Question 15:
Find the direction cosines of the lines, connected by the relations: l + m +n = 0 and 2lm + 2lnmn = 0.
Question 16:
Find the angle between the lines whose direction cosines are given by the equations
(i) m + n = 0 and l2 + m2 − n2 = 0
(ii) 2l − m + 2n = 0 and mn + nl + lm = 0
(iii) l + 2m + 3n = 0 and 3lm − 4ln + mn = 0
(iv) 2l + 2m − n = 0, mn + ln + lm = 0
(iv) The given relations are
2l + 2m − n = 0 .....(1)
mn + ln + lm = 0 .....(2)
From (1), we have
n = 2l + 2m
Putting this value of n in (2), we get
When $l=-2m$, we have
$n=2×\left(-2m\right)+2m=-4m+2m=-2m$
When $l=-\frac{m}{2}$, we have
$n=2×\left(-\frac{m}{2}\right)+2m=-m+2m=m$
Thus, the direction ratios of two lines are proportional to
$-2m,m,-2m$ and $-\frac{m}{2},m,m$
Or $-2,1,-2$ and $-1,2,2$
So, vectors parallel to these lines are $\stackrel{\to }{a}=-2\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}$ and $\stackrel{\to }{b}=-\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}$.
Let $\theta$ be the angle between these lines, then $\theta$ is also the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$.
Thus, the angle between the two lines whose direction cosines are given by the given relations is $\frac{\mathrm{\pi }}{2}$.
Question 1:
Define direction cosines of a directed line.
Question 2:
What are the direction cosines of X-axis?
Question 3:
What are the direction cosines of Y-axis?
Question 4:
What are the direction cosines of Z-axis?
Question 5:
Write the distances of the point (7, −2, 3) from XY, YZ and XZ-planes.
Question 6:
Write the distance of the point (3, −5, 12) from X-axis?
Question 7:
Write the ratio in which YZ-plane divides the segment joining P (−2, 5, 9) and Q (3, −2, 4).
Question 8:
A line makes an angle of 60° with each of X-axis and Y-axis. Find the acute angle made by the line with Z-axis.
Question 9:
If a line makes angles α, β and γ with the coordinate axes, find the value of cos 2α + cos 2β + cos 2γ.
Question 10:
Write the ratio in which the line segment joining (a, b, c) and (−a, −c, −b) is divided by the xy-plane.
Question 11:
Write the inclination of a line with Z-axis, if its direction ratios are proportional to 0, 1, −1.
Question 12:
Write the angle between the lines whose direction ratios are proportional to 1, −2, 1 and 4, 3, 2.
Question 13:
Write the distance of the point P (x, y, z) from XOY plane.
Question 14:
Write the coordinates of the projection of point P (x, y, z) on XOZ-plane.
The projection of the point P (x, y, z) on XOZ-plane is (x, 0, z) as Y-coordinates of any point on XOZ-plane are equal to zero.
Question 15:
Write the coordinates of the projection of the point P (2, −3, 5) on Y-axis.
The coordinates of the projection of the point P ( 2, -3, 5) on the y-axis are ( 0, $-$3, 0) as both X and Z coordinates of each point on the y-axis are equal to zero.
Question 16:
Find the distance of the point (2, 3, 4) from the x-axis.
Question 17:
If a line has direction ratios proportional to 2, −1, −2, then what are its direction consines?
Question 18:
Write direction cosines of a line parallel to z-axis.
Question 19:
If a unit vector makes an angle and an acute angle θ with $\stackrel{^}{k}$, then find the value of θ.
Question 20:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question:
Write the distance of a point P(a, b, c) from x-axis.
We know that a general point (x, y, z) has distance $\sqrt{{y}^{2}+{z}^{2}}$ from the x-axis.
∴ Distance of a point P(a, b, c) from x-axis = $\sqrt{{b}^{2}+{c}^{2}}$
Question 21:
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Let the direction cosines of the line be l, m and n.
We know that
l2 + m2 + n2 = 1.
Let the line make angle θ with the positive direction of the z-axis.
cos2α+cos2β+cos2γ=1Here α=60 and β=45 and γ= θSo cos260+cos245+cos2θ=1cos2θ=11412=14cosθ=±12So θ= 60 degree or 120.and here it is given that we have to find the angle made by negative z axisSo cosθ=12θ=120 degree
Question 1:
For every point P (x, y, z) on the xy-plane,
(a) x = 0
(b) y = 0
(c) z = 0
(d) x = y = z = 0
(c) z = 0
The Z-coordinate of every point on the XY-plane is zero.
Question 2:
For every point P (x, y, z) on the x-axis (except the origin),
(a) x = 0, y = 0, z ≠ 0
(b) x = 0, z = 0, y ≠ 0
(c) y = 0, z = 0, x ≠ 0
(d) x = y = z = 0
Question 3:
A rectangular parallelopiped is formed by planes drawn through the points (5, 7, 9) and (2, 3, 7) parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is
(a) 2
(b) 3
(c) 4
(d) all of these
Question 4:
A parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7), parallel to the coordinate planes. The length of a diagonal of the parallelopiped is
(a) 7
(b) $\sqrt{38}$
(c) $\sqrt{155}$
(d) none of these
Question 5:
The xy-plane divides the line joining the points (−1, 3, 4) and (2, −5, 6)
(a) internally in the ratio 2 : 3
(b) externally in the ratio 2 : 3
(c) internally in the ratio 3 : 2
(d) externally in the ratio 3 : 2
Question 6:
If the x-coordinate of a point P on the join of Q (2, 2, 1) and R (5, 1, −2) is 4, then its z-coordinate is
(a) 2
(b) 1
(c) −1
(d) −2
Question 7:
The distance of the point P (a, b, c) from the x-axis is
(a) $\sqrt{{b}^{2}+{c}^{2}}$
(b) $\sqrt{{a}^{2}+{c}^{2}}$
(c) $\sqrt{{a}^{2}+{b}^{2}}$
(d) none of these
Question 8:
Ratio in which the xy-plane divides the join of (1, 2, 3) and (4, 2, 1) is
(a) 3 : 1 internally
(b) 3 : 1 externally
(c) 1 : 2 internally
(d) 2 : 1 externally
Question 9:
If P (3, 2, −4), Q (5, 4, −6) and R (9, 8, −10) are collinear, then R divides PQ in the ratio
(a) 3 : 2 internally
(b) 3 : 2 externally
(c) 2 : 1 internally
(d) 2 : 1 externally
Question 10:
A (3, 2, 0), B (5, 3, 2) and C (−9, 6, −3) are the vertices of a triangle ABC. If the bisector of ∠ABC meets BC at D, then coordinates of D are
(a) (19/8, 57/16, 17/16)
(b) (−19/8, 57/16, 17/16)
(c) (19/8, −57/16, 17/16)
(d) none of these
Disclaimer:This question is wrong, so the solution has not been provide.
Question 11:
If O is the origin, OP = 3 with direction ratios proportional to −1, 2, −2 then the coordinates of P are
(a) (−1, 2, −2)
(b) (1, 2, 2)
(c) (−1/9, 2/9, −2/9)
(d) (3, 6, −9)
Question 12:
The angle between the two diagonals of a cube is
(a) 30°
(b) 45°
(c) ${\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
(d) ${\mathrm{cos}}^{-1}\left(\frac{1}{3}\right)$
(d) ${\mathrm{cos}}^{-1}\left(\frac{1}{3}\right)$
Question 13:
If a line makes angles α, β, γ, δ with four diagonals of a cube, then cos2 α + cos2 β + cos2 γ + cos2 δ is equal to
(a) $\frac{1}{3}$
(b) $\frac{2}{3}$
(c) $\frac{4}{3}$
(d) $\frac{8}{3}$
View NCERT Solutions for all chapters of Class 12
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# 5 GMAT Statistics Practice Questions & Explanations
## Question 1:
The average weight of n students in a certain class was 24 kg. A new student named Jack joined the class and one existing student named Carol left the class. What is the new average?
I. Carol’s weight was 1 kg more than that of Jack
II. Carol’s weight was 24kg.
Let’s say the weight of Jack is J, weight of Carol is C and the sum of weights of all the students in the class is S
It is given that the average weight of n students in a certain class was 24 kg.
Since,
Average=Sum of the terms/Total number of terms
Thus, Average=S/n = 24
Or, S = 24n…..(i)
Since, Carol has left and Jack has joined the class, thus one person left and one person joined due to which the total number of students in the class will still be n.
Thus, the new average will be;
New Average = S - weight of Carol + weight of Jack/n..(ii)
1. Carol’s weight was 1 kg more than that of Jack
Or, C = J + 1
Using equation (ii), we get,
New Average = S - C + J/n
Or, New Average = S - ( J + 1 ) + J / n = S - J - 1 + J/n = S - 1 /n
Or, New Average = 24n - 1/n {as S = 24n}
=24- 1n
But since we don’t know the value of n, hence we cannot find a definite answer; NOT sufficient.
2. Carol’s weight was 24 kg
Using equation (ii), we get,
New Average = S - C + J/n = S - 24 + J/n = 24n - 24 + Jn {as S = 24n}
But since we don’t know the value of n and J, hence we cannot find a definite answer
The correct answer is E; both statements together are still not sufficient.
## Question 2:
The ratio of boys to girls in a class is 4 : 3. The average weight of all the students in the class is 45kg. If the average weight of girls is 7kg less than that of boys, what is the ratio of the average weight of boys to the average weight of girls?
1. 47 : 40
2. 48 : 41
3. 49 : 42
4. 50 : 43
5. 4 : 3
Ratio of Boys to Girls = 4:3, So let’s consider there are “4x” boys and “3x” girls in the class.
Let’s also consider the average weight of the boys = “y” kgs. Therefore, average weight of girls will be “y-7” kgs.
The average weight of Class = (Average weight of Boys * Number of Boys + (Average weight of Girls * Number of Girls)/(Total Number of Students)
=> 45 = y× 4x + (y-7)×3x/4x+3x
=> 45 = 7y - 21/7
y = 48, and also, y-7 = 41
Therefore, ratio of Average weight of Boys to Average weight of Girls = 48:41
## Question 3:
Each of the following linear equations defines y as a function of x for all integers x from 1 to 20. For which of the following equations is the standard deviation of the y – values corresponding to all the x – values, the greatest?
1. Y = -(x/5)
2. Y = -(x/5)
3. Y = x
4. Y = -2x + 7
5. Y = -4x - 9
We are given some linear equations which defines y as a function of x for all integers x from 1 to 20.
1 ≤ x ≤ 20
We have to find, in which case the standard deviation of y will be the greatest.
As y is a function of x, so we will have 20 different values of y for 20 values of integer x varying from 1 to 20
=> POINTS TO REMEMBER:-
• Whenever we add or subtract a constant in all the observations, the standard deviation of the data remains unchanged.
• Whenever we multiply or divide all the observations by a constant, the standard deviation will be respectively multiplied or divided by the absolute value of the constant.
(A) y = - x/5
In this option, we will get the values of ‘y’ by dividing the values of x by (-5). So, every observation in set x will be divided by (-5) to get the corresponding values in set y.
(Standard Deviation)Y = ((Standard Deviation of x))/5
(B) y = - x/2 + 2
In this option, we will first divide every value of x by (-2) and then we will add 2 in all the observations.
(Standard Deviation)Y = ((Standard Deviation of x))/2
(C) y = x
In this option, set y will have the same values as set x.
(Standard Deviation)Y = [(Standard Deviation)X]
(D) y = -2x + 7
In this option, we will first multiply every observation in x by (-2) and then we will add 7 in every observation.
(Standard Deviation)Y = 2 × [(Standard Deviation)X]
(E) y = - 4x – 9
In this option, we will first multiply every observation by (-4) in set x and then we will subtract 9 from all the observations.
(Standard Deviation)Y = 4 × [(Standard Deviation)X]
So, from all the options above, we will have the greatest standard deviation of ‘y’ in option E.
## Question 4:
An alumni meet is attended by 150 people. Are there any two people of the same age (rounded to the nearest year) who live in the same city?
(1) The attendees’ age ranges from 30 to 33 years, inclusive.
(2) The attendees live in 30 different cities.
Number of people attending the alumni meet = 150
We have to find – Are there any two people of the same age (rounded to the nearest year) who lives in the same city?
STATEMENT 1: The attendees’ age ranges from 30 to 33, inclusive.
From this statement, we are only getting the Information about the age but nothing is mentioned about the cities.
So, this statement is not alone sufficient to answer the question; NOT sufficient.
STATEMENT 2: The attendees’ live in 30 different cities.
In this statement we are mentioned about the cities but nothing is mentioned about the ages of the attendees.
So, this statement is also not alone sufficient to answer this question; NOT sufficient
Taking statement 1 and 2 together,
Now, the ages of attendees can be 30, 31, 32 and 33 only.
And, there are 30 different cities.
Now, If we assume there is one person each of age 30, 31, 32 & 33 in each city, so there will be 4 persons of different ages in each city.
So, total number of persons in 30 different cities will be = 30 × 4 = 120.
But, total number of attendees are 150.
So, there will be 30 persons left who will also be from one of the 30 cities with age either 30, 31, 32 or 33.
So, there will be at least two people of the same age living in the same city.
The correct answer is C; both statements together are sufficient.
## Question 5:
S is a set of n integers, where 0 < n < 11. If the arithmetic mean of set S is a positive integer b, which of the following could NOT be the median of set S?
(A) 0
(B) b
(C)(-b)
(D) n/5
(E) 5n/11
S is a set of n integers, where 0 < n < 11.
Arithmetic mean of Set S is a positive integer ‘b’.
We have to find which of the following could not be the median of Set S.
Now, We know that if the number of terms are odd, then there will be only one middle term and that will be the median, which will be an Integer value.
(A) And, If number of terms are even, then there will be two middle terms, and the median will be the average of those two middle terms, in the form of (a+b)/2, where a & b will be those two middle terms.
(F) So, the median can be either an integer or can be in the form of I/2, where I will be an integer.
So, option (A), (B) & (C) can be the median because all three are integers.
(A) Option (D), (n/5) can also be the median because 0
But, option (E), 5n/11 cannot be either an integer or in the form of I/2, as n is not a multiple of 11. Hence, this cannot be the median.
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# ACT Math : How to find a solution set
## Example Questions
← Previous 1
### Example Question #1 : How To Find A Solution Set
|2x – 25| – 3 = 7. There are two solutions to this problem. What is the sum of those solutions?
17
10
25
7.5
25
Explanation:
First, simplify the equation so the absolute value is all that remains on the left side of the equation:
|2x – 25| = 10
Now create two equalities, one for 10 and one for –10.
2x – 25 = 10 and 2x – 25 = –10
2x = 35 and 2x = 15
x = 17.5 and x = 7.5
The two solutions are 7.5 and 17.5. 17.5 + 7.5 = 25
### Example Question #1 : How To Find A Solution Set
Find the roots of the equation x+ 5x + 6.
1, 4
2, 3
2, –3
–2, –3
–2, 3
–2, –3
Explanation:
Factoring gives us (x + 2)(x + 3). This yields x = –2, –3.
### Example Question #1 : How To Find A Solution Set
When you divide a number by 3 and then add 2, the result is the same as when you multiply the same number by 2 then subtract 23. What is the number?
9
7
2
15
3
15
Explanation:
You set up the equation and you get: (x/3) + 2 = 2x – 23.
Add 23 to both sides: (x/3) + 25 = 2x
Multiply both sides by 3: x + 75 = 6x
Subtract x from both sides: 75 = 5x
Divide by 5 and get = 15
### Example Question #1 : How To Find A Solution Set
Find the sum of the solutions to the equation:
2x – 2x – 2 = 1 – x
Explanation:
First, we need to get everything on one side so that the equation equals zero.
2x- 2x -2 = 1-x
We need to add x to the left, and then subtract 1.
2x- 2x -2 +x - 1 = 0
2x- x - 3 = 0
Now we need to factor the binomial. In order to do this, we need to multiply the outer two coefficients, which will give us 2(-3) = -6. We need to find two numbers that will mutiply to give us -6. We also need these two numers to equal -1 when we add them, because -1 is the coefficient of the x term.
If we use +2 and -3, then these two numbers will multiply to give us -6 and add to give us -1. Now we can rewrite the equation as follows:
2x- x - 3 = 2x+ 2x - 3x - 3 = 0
2x+ 2x - 3x - 3 = 0
Now we can group the first two terms and the last two terms. We can then factor the first two terms and the last two terms.
2x(x+1) -3(x+1) = 0
(2x-3)(x+1) = 0
This means that either 2x - 3 = 0, or x + 1 = 0. So the values of x that solve the equation are 3/2 and -1.
The question asks us for the sum of the solutions, so we must add 3/2 and -1, which would give us 1/2.
### Example Question #3 : How To Find A Solution Set
If 3y = 2x – 7, then which of the following statements is correct?
not enough information given
x is greater
y is greater
they are equal
not enough information given
Explanation:
If we set one variable to the other we would get y = (2x – 7)/3 or x = (3y + 7)/2, but we aren't given any clues to what the values of x and y are and we can assume they could be any number. If x = 7/2, then y = 0. If y = -7/3, then x = 0. Let's try some other numbers. If y = –10, then x = –37/2. So for the first two examples, x is greater than y. In the last example, y is greater than x. We need more information to determine whether x or y is greater. The correct answer is not enough information given.
### Example Question #7 : How To Find A Solution Set
|10 2| – |1 – 9| = ?
16
2
8
0
0
Explanation:
When taking the absolute value we realize that both absolute value operations yield 8, which gives us a difference of 0.
### Example Question #2 : How To Find A Solution Set
When you multiply a number by 5 and then subtract 23, the result is the same as when you multiplied the same number by 3 then added 3. What is the number?
7
13
5
10
6
13
Explanation:
You set up the equation 5x – 23 = 3x + 3, then solve for x, giving you 13.
### Example Question #3 : Solution Sets
What is the product of the two values of that satisfy the following equation?
Explanation:
First, solve for the values of x by factoring.
or
Then, multiply the solutions to obtain the product.
### Example Question #2 : Solution Sets
Solve for y:
Explanation:
Collecting terms leaves
And dividing by yields
Solve for x.
|
# Permutations: Balls and boxes related problems
Problems related to balls and boxes can be classified mainly into 4 categories.
Balls – Distinct, Boxes – Distinct
Balls- Similar, Boxes – Distinct
Balls - Distinct, Boxes – Similar
Balls – Similar, Boxes - Similar
Case 1: Balls – Distinct, Boxes – Distinct
Condition 1: A box can take any Number of balls
We know that if a box can take any number of balls let us take a ball No.9 which can go into any of the four distinct boxes. Next ball can also go into any of these four boxes. Similarly, remaining balls also can go into any of these 4 boxes. So total number of ways are 4 x 4 x 4 x 4 = 256
Formula : ${(boxes)^{balls}}$
Condition 2: A box can take minimum one ball: (Assume there are 5 balls and 3 boxes)
This case is one of the most interesting case. This can be solved in many ways.
Method 1:
We first divide 5 balls into 3 different groups and then we allocate these groups into 3 boxes.
We divide 5 balls into 3 groups in two ways: 2, 2, 1 or 3, 3, 1
Now (m+n+p) objects may be divided into 3 groups containing m, n, p objects is given by $\displaystyle\frac{{(m + n + p)!}}{{m! \times n! \times p!}}$
But when any of m,n, or p are same we have to divide by that similar numbers. Here in 2, 2, 1 two 2's are same. So final answer should be divided by 2!.
$\displaystyle\frac{{(5)!}}{{2! \times 2! \times 1!}} \times \displaystyle\frac{1}{{2!}}$ = 15 ways
Similarly 5 balls may be divided into 3, 1, 1 are $\displaystyle\frac{{(5)!}}{{3! \times 1! \times 1!}} \times \displaystyle\frac{1}{{2!}}$ = 10 ways (Here 1, 1 are same)
Now each of these divisions are put into 3 boxes in 3! ways. So total ways are 3! x ( 15 + 10) = (3! x 25) = 150
Method 2:
Division of 5 balls into 3 groups of 2, 2, 1 can be done like this $^5{C_2}{ \times ^3}{C_2}{ \times ^1}{C_2} \times \displaystyle\frac{1}{{2!}}$ = 15
Division of 5 balls into 3 groups of 3, 1, 1 can be done like this $^5{C_3}{ \times ^2}{C_1}{ \times ^1}{C_1} \times \displaystyle\frac{1}{{2!}}$ = 10
Total divisios are 25 and total distributions are 25 x 3! = 150
Method 3:
Distribution of 5 distinct objects into 3 groups is equivalent to total onto mapping from a set of 5 to a set of 3 and which can be calculated by the formula :
${r^n} - {}^r{C_1}.{(r - 1)^n} + {}^r{C_2}.{(r - 2)^n} + {}^r{C_3}.{(r - 3)^n} - ......$
So ${3^5} - {}^3{C_1}.{(3 - 1)^5} + {}^3{C_2}.{(3 - 2)^5}$ = 150
Case 2: Balls – Similar, Boxes – Distinct
Suppose your dad has given you 100 rupees to buy ice-creasms. Let us say an ice cream costs you Rs.25. In an ice cream shop there are 5 varieties of ice cream available. Chololate, Vanilla, strawberry, Butter scotch, Pista.You can buy any varieties but you can buy maximum 4 ice creams as an ice cream costs you Rs.25 each.
So number of ice creams you are going to buy is
Chololate +Vanilla + Strawberry + Butter scotch + Pista = 4
This is a problem of finding intezer solutions to the above equation where sum of all the numbers in the places of 5 different ice creams is equal to 4.
We can apply the same logic to the balls and boxes where balls are similar (as ice creams of same variety are similar) but the sum of the balls in all the boxes together must equal to 4.
Condition 1: A box can take any Number of balls
Now A + B + C + D = 4 where any of these numbers can be Zero.
Number of intezer solutions of these equation where 0 is allowed is = $\left( {n + k - 1} \right){C_{k - 1}}$ (here k= number of boxes, n= number of balls)
$\Rightarrow \left( {4 + 4 - 1} \right){C_{4 - 1}} \Rightarrow 7{C_3}$
Condition 2: A box can take minimum 1 Ball
In this case we should take only natural numbers as 0 is not allowed.
Number of intezer solutions where 0 is not allowed is = $\left( {n - 1} \right){C_{k - 1}}$ = $\Rightarrow \left( {4 - 1} \right){C_{4 - 1}} \Rightarrow 3{C_3} = 1$
Case 3: Balls – Distinct, Boxes – Similar
This is one of the typical probelm where we attempt to solve it by using a special recurrence table called Stirling numbers of second kind.
Condition 1: A box can take any number of balls
Here column N represents number of balls and K represents number of boxes.
Now 4 ball and 4 boxes can be arranged in 1 + 7 + 6 + 1 = 15 ways.
Condition 2: A box can take minimum one balls
Now we should take the intersection point of n = 4 and k= 4 so answer = 1
Case 4: Balls – Similar, Boxes – Similar
We use another recurrence table to solve this question
Condition 1: A box can take any number of balls
Add all the numbers in the 4th row and 4th column so answer = 5
Condition 2: A box can take minimum one balls
Here take the intersection of 4th row and 4th column = 1
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## LetsPlayMaths.com
WELCOME TO THE WORLD OF MATHEMATICS
# Class 1 Multiplication
Basic Concept of Multiplication
Order in Multiplication
Multiplying by 2
Multiplying by 3
Multiplying by 4
Multiplying by 5
Single Digit Multiplication
Multiplying 2-Digit Number by Single Digit Number
Multiplication Test
Multiplication Worksheet
## Basic Concept of Multiplication
When we add same number multiple times, then it is known as repeated addition. Multiplication is nothing but the repeated addition.
Symbol of multiplication is '×'
We have 3 balls in 3 groups.
So total number of balls = 3 + 3 + 3 = 9
We have 2 suns present in each group and we have 4 groups as shown in the above figure.
So total number of suns = 2 + 2 + 2 + 2 = 8
## Order in Multiplication
If we change the order of the numbers we are multiplying, the result remains unchanged.
2 groups of 3 = 2 × 3 = 6
3 groups of 2 = 3 × 2 = 6
So, 2 × 3 = 3 × 2 = 6
Let us take one more example
4 groups of 2 = 4 × 2 = 8
2 groups of 4 = 2 × 4 = 8
So, 4 × 2 = 2 × 4 = 8
## Multiplying by 2
1 group of 2 smileys = 2 2 groups of 2 smileys = 4
3 groups of 2 smileys = 6 4 groups of 2 smileys = 8
5 groups of 2 smileys = 10 6 groups of 2 smileys = 12
## Single Digit Multiplication
We can perform multiplication in two ways, they are shown below.
## Multiplying 2-Digit Number by Single Digit Number
1. Write both the numbers according to the place value
2. Bigger number should occupy the upper row as shown below
3. Multiply the digits present in ones place first
4. Multiply the digits present in tens place
Example 1. Multiply 13 by 2.
Solution. Write both the numbers in tabular format as shown below.
Multiply the digits present in ones place (3 × 2 = 6). Write the result in ones place as shown below.
Now, multiply the digits present in tens place (1 × 2 = 2). Write the result in tens place as shown below.
So, the result is 26.
## Class-1 Multiplication Worksheet
Multiplication Test - 1
Multiplication Test - 2
Multiplication Test - 3
## Class-1 Multiplication Worksheet
Multiplication Worksheet - 1
Multiplication Worksheet - 2
Multiplication Worksheet - 3
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# What’s 5 divided by 3 – Is it even possible?
5 divided by 3 equals 1 with a remainder of 2.
In simpler terms, the quotient of 5 divided by 3 is approximately equal to 1.6666666666666667.
If you want to express the result as a decimal, you can write 1.6666666666666667. If you want to express the result as a mixed number, you can write 1 2/3.
The more detailed response.
Division is a math operation that lets us find out how many times one number, called the dividend, can be divided by another number, called the divisor. For example, if you have 5 pieces of candy and you want to know how many times you can divide those 5 pieces among 3 of your friends, you can use division to find the answer.
How many sets of 3 can we create using 5 pieces? To clarify, we are attempting to determine the number of times 3 can be accommodated within 5. The divisor in this case is 3, while the dividend is 5. Now, let’s examine the equation 5 divided by 3 more closely.
The equation 3 ÷ 5 represents the division of 3 by 5. So, the symbol ÷, which is called the division symbol, is used in this equation to solve it.
Now, let’s utilize a few illustrations to aid in comprehending this equation more effectively.
Cars.
Let’s say we have 5 toy cars and we want to divide them among 3 friends using the equation ÷. How many times can we set up this equation to divide the toy cars?
By utilizing 5 toy vehicles, we can form a single set of 3 toy cars and retain 2 toy cars as surplus. In the event that we distribute the 5 toy cars among the 3 companions, each friend would receive 1 toy car. In order to resolve this mathematical expression, it is necessary to ascertain the number of sets of 3 toy cars that can be formed using 5 toy cars.
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After dividing the toy cars among the three friends, we are left with a remainder. The remainder is 2 when we divide 3 by 5. Therefore, the answer to the equation is 1.
Let’s divide the 10 cookies we have among the 3 friends and see how many times we can do it. Let’s try another example. We can set up the equation like this.
We can make 3 groups of cookies using 10 cookies, meaning that we would not have any cookies left over. If we divide the 10 cookies among 3 friends, each friend would get 3 cookies. We need to solve this equation to find out.
After dividing the 10 cookies among 3 friends, there are no cookies left because the equation 3 ÷ 10 gives an answer of 0 with a remainder of 3.
What is a Remainder in Division?
There will be 2 cookies left over, and each person will get 1 cookie. This means that if you have 5 cookies and you want to divide them equally among 3 people, you cannot divide them evenly because you will have a remainder of 2. For example, if there is one more cookie left when dividing by another number, the remainder is the amount that is left over.
Here’s another example:Output: Here’s one more instance:
Each person will get 3 cookies and there will be 4 cookies left over. In this case, you will be able to divide the cookies evenly because 12 can be divided by 4. If you want to divide them equally among 4 people and you have 12 cookies.
The remainder, in general, refers to the quantity that remains when division cannot be evenly done. In the case of dividing cookies among individuals, the remainder represents the number of cookies left over.
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We hope this helps! Division can be a little tricky at first, but with a bit of practice and a little bit of humor, you’ll soon be able to find out how many times one number can be divided by another number. Just remember that division is all about finding out how many times one number can be divided by another.
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Parametric Curves
by
In the following investgation, we are going to examine the following parametric equations.
In the above equations, a and b can be any value and h is any real number. For these equations, we are going to investigate what happens when a=b, a<b, and a>b. When we use the program X-Functions, we obtain the following graphs when we change a and b and let t run from 0 to 2*pi.
First, we are going to examine equations x=acos(t) and y=bsin(t).
By observing the above figures, you notice that we have circles and ellipses. Why is this? Let's look at the equations for these figues. The red circle is the graph of x=cos(t) and y=sin(t). The green circle is the graph of the equations x=2cos(t) and y=2sin(t). By observing these two graphs and equations, one notices that if a=b then we get a perfect circle with the length of the radius being the same as the value for a and b. The longer fushia ellipse and the dark blue ellipse demonstrate what happens when a<b. The equation for the fushia ellipse is x=1/2cos(t) and y=2sin(t) and the equation for the blue ellipse is x=1/2cos(t) and y=sin(t). The light blue ellipse and the smaller fushia ellipse demonstrate what happens when a>b. The equations of these ellipses are (x=2cos(t) , y=1/2sin(t)) and (x=cos(t), y=1/2sin(t)) respectively. Thus, if a>b then the ellipse is elongated along the x axis and if a<b then the ellipse is elongated along the y axis. From our observations, one notices that a and b determine the size of the shape of the figure. It will be either an ellipse or a circle.
Now, lets examine our second two equations.
First, lets examine what happens when we change a and b and fix h.
In the above graphs, I let h=-2 and I changed the values for a and b. I used the following values for a and b {-2,-1,-1/2,1/2,1,2}. We observe that some of the ellipses are postioned in a positve direction and some are postioned in a negative direction. The equation for the line is (x=2cos(t)-2sin(t), y=2sin(t)-2cos(t)). From observing this line, it seems that if a and b are positive and h is negative, then the figure is going to be postioned in the second and fourth quadrants. Let's see if this is true. The equations for the blue ellipses are (x=-1/2cos(t)-2sin(t), y=-2sin(t)-2cos(t)) and (x=1/2cos(t)-2sin(t), y=2sin(t)-2cos(t)). So which one is negative and which one is positive? The positive one is the one where a,b, and h are all negative. The same is true for the other ellipses as well. The fushia ellipses have the equations (x=-2cos(t)-2sin(t), y=(-1/2sin(t)-2cos(t)) and
(x=2cos(t)-2sin(t), y=(1/2sin(t)-2cos(t)). The equations for the red and green ellipses are
(x=cos(t)-2sin(t), y=sin(t)-2cos(t)) and (x=-cos(t)-2sin(t), y=-sin(t)-2cos(t)). In these graphs a and b determine the postion of the ellipse. If a<b, then they are postioned more toward the
y axis. If a>b, then they are closer to the x axis. If a=b, then the ellipses are in the center of the first and third quadrant or in the center of the second and forth quadrant.
Now let's see what happens when a=1/2, b=1, and we change h in the above equations. We get the following graphs.
By observing these graphs, it appears that if we keep a and b the same and change h then the size and postion of the ellipse change when h changes. The h value for the green ellipse is -2, for the blue ellipse 2, for the fushia ellipse 1, and for the red ellipse -1. As you can see, when we changed h from 1 to 2 the ellipse "grew" and when we changed h for 1 to -1 it rotated 90 degrees.
Now let's see what happens when we let a=1, b=1/2, and h={-3,-2,-1,1,2,3}. We get the following graphs.
The ellipses have the following h values:
Ellipse Color h Value
smaller fushia 1
red 2
blue 3
light blue -1
green -2
large fushia -3
By comparing these ellipses to the ellipses in the above problem, we notice that the postioning of the ellipse depends on if a>b or a<b. If a>b as in the above ellipses, then they are postioned closer to the x axis. If a<b as in the ellipses prior to these, then they are postioned closer to the
y axis. We also notice that the ellipses who have the same h value in these two graphs seem to be the same size. Let's make a chart to show the comparisons. The ellipses that are side by side are the ones that appear to be the same size.
In the chart, graph 1 is the graph where a=1/2 and b=1 and graph 2 is the graph where a=1 and b=1/2. As you notice, the ellipses that are the same size have the same h value.
Now, lets see what happens when we let a=b=(+ or -)h. In our first set of ellipses, we noticed that if a=b=-h then we had an equation of a line. Let's see if this holds true for all instances.
This holds true when a=b and h=(+ or -)a. The above lines have the following values for a, b, and h.
As you notice, if a=b=h then all of the lines lie on the same path. The path that bisects the first and third quadrant (the path y=x). If a=b=-h, then all of the lines lie on the path that bisects the second and fourth quadrant (the path y=-x).
What happens when a=-b and h is changing?
From observing these figures, one notices that they are all circles. Thus, if a=-b then we will get a circle no matter what h is. We have the following values for a, b, and h in the above circles.
These are only a few of the investigations that one can perform. There are many other investigations that one can perform on these equations.
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# Question: What Is The Result Of Divided 15 And 33 By 3 Respectively?
## What is the result of repeated division?
The result of division is to separate a group of objects into several equal smaller groups.
The starting group is called the dividend.
The number of groups that are separated out is called the divisor.
The number of objects in each smaller group is called the quotient..
## How do we use division in everyday life?
Division is something that you use daily. If you are in a Rock group of any kind or a band for instance. You might be the lead singer, the guitarist or even the drummer. Each one of you has their own place, and when you play a song, that song is divided into different parts where each one gets a piece to do.
## What is repeated division called?
the division which was already divide and than agin divide by any number it is called repeated division. rosariomividaa3 and 6 more users found this answer helpful.
## What is the result when you divide?
What is being divided is called the dividend, which is divided by the divisor, and the result is called the quotient.
## What are the possible remainder on dividing a number by 3?
To find the remainder of a number divided by 3, add the digits of the number and divide it by 3. So if the digits added together equal 8 then the number has a remainder of 2 since 8 divided by 3 has a remainder of 2.
## What is the remainder of the following division problem?
Answer: The remainder of the following division problem is 1. If one polynomial divided by another polynomial, then the remaining polynomial is known as remainder. The result of division is called the quotient.
## What is the remainder when 22005 is divided by 7?
The remainder when 8 is divided by 7 is 1.
## What is the answer to division called?
The number we divide by is the divisor, and the answer to a division problem is the quotient.
## What is the division fact?
Division facts are division number sentences related to times tables knowledge. For example, 50 ÷ 5 = 10, 25 ÷ 5 = 5, and 10 ÷ 5 = 2 are all division facts of the five times table.
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IMAT 2018 Q56 [Equation of a Line]
Which one of the following is an equation of the line that passes through (4, 3) and is perpendicular to the line y = 2x + 4?
A. 2x – y = 5
B. 2y + x = 11
C. 2y – x = 2
D. 2x + y = 11
E. 2y + x = 10
The basic equation of a line is y=kx+n.
Knowing that the line passes through (4,3) which is (x_1,y_1) we can put the information in the basic equation:
y_1=kx_1 +n \to 3= 4k +n \to n= 3-4k
Knowing that the if the line is perpendicular to the line y = 2x + 4. Perpendicular lines are lines that intersect at right angles . If you multiply the slopes of two perpendicular lines in the plane, you get −1 . That is, the slopes of perpendicular lines are opposite reciprocals .
The slopes are the k in the basic equation. So in this case, the slope needed is: k = \frac{-1}{2} = -\frac{1}{2}
So, using that information in the equation for n: n= 3-4k \to n= 3 - 4 \cdot (-\frac{1}{2})= 3+2 = 5
Integrating all of the data in the original form:
y = -\frac{1}{2} \cdot x +5 \to \text{multiplying ut by 2} \to 2y = -x +10 \to 2y + x = 10 .
So the final correct answer is (E)
5 Likes
But how is the slope of one line equall to 2. Clarify please!
1 Like
y = mx+ c ; y = 2x + 4. Therefore m (gradient) of the given line is 2
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# Popular 3rd Grade Math Word Problems
3rd grade word math problems are like stories that help us understand how math works in everyday life. Math 3rd grade word problems make us think and figure out solutions, which is essential for solving all kinds of problems. Reading and solving these math word problems 3rd grade teaches us how to use math to solve real situations. This helps us think smart and make good decisions.
Math word problems for 3rd grade also help us practice reading carefully and explaining our ideas clearly. They show us that math is not just about numbers but about thinking, solving, and understanding the world around us.
There are many 3rd grade math word problems in math. When teaching kids about word problems, it’s crucial to begin with simple ones. This is because word problems can quickly become confusing for kids.
By starting with fundamental problems and maybe 3rd grade math word problems games, we help kids learn how word problems work before they become more complicated. This way, they can build their skills and feel more confident as they learn.
## The Most Popular Math Word Problems 3rd Grade
Here are some 3rd grade math word problems you can use to teach your kids more about the real-life representation of math:
Problem:
Sarah has 8 apples. She gives 3 apples to her friend. How many apples does she have left?
Solution: Sarah has 5 apples left.
Problem:
There are 4 red pencils and 6 blue pencils. How many pencils are there in total?
Solution: There are 10 pencils in total.
Problem:
Liam has 15 marbles. He gives 7 marbles to his sister. How many marbles does Liam have now?
Solution: Liam has 8 marbles left.
Problem:
There are 9 birds in one tree and 5 birds in another. How many birds are there in both trees?
Solution: There are 14 birds in both trees.
Problem:
Emma had 20 candies. She ate 12 candies. How many candies does Emma have now?
Solution: Emma has 8 candies left.
Problem:
There are 3 boxes, each containing 4 crayons. How many crayons are there in total?
Solution: There are 12 crayons in total.
Problem:
Jason has 18 stickers. He gives 9 stickers to his friend. How many stickers does Jason have left?
Solution: Jason has 9 stickers left.
Problem:
There are 7 cats and 8 dogs at the park. How many animals are there in total?
Solution: There are 15 animals in total.
Problem:
Mia has 24 cookies. She shares 6 cookies with her brother. How many cookies does Mia have after sharing?
Solution: Mia has 18 cookies left.
Problem:
There are 5 cars in the parking lot and 3 cars on the street. How many cars are there in total?
Solution: There are 8 cars in total.
Problem:
Emily has 27 stickers. She gives 15 stickers to her classmates. How many stickers does Emily have now?
Solution: Emily has 12 stickers left.
Problem:
There are 10 balloons in one room and 6 balloons in another. How many balloons are there in both rooms?
Solution: There are 16 balloons in both rooms.
Problem:
Liam has 35 baseball cards. He trades 17 cards with his friend. How many cards does Liam have now?
Solution: Liam has 18 cards left.
Problem:
There are 8 girls and 12 boys on the playground. How many children are there in total?
Solution: There are 20 children in total.
Problem:
Sarah has 42 stickers. She gives 28 stickers to her cousin. How many stickers does Sarah have left?
Solution: Sarah has 14 stickers left.
Problem:
There are 9 ducks in one pond and 11 in another. How many ducks are there in both ponds?
Solution: There are 20 ducks in both ponds.
Problem:
There are 15 pencils in a box. If 8 pencils are taken out, how many pencils are left in the box?
Solution: There are 7 pencils left in the box.
Problem:
Ethan has 30 toy cars. He gives 10 toy cars to his friend. How many toy cars does Ethan have now?
Solution: Ethan has 20 toy cars left.
Problem:
There are 6 apples on the first tree and 9 apples on the second tree. How many apples are there in both trees?
Solution: There are 15 apples in both trees.
Problem:
Emma has 50 stickers. She gives 20 stickers to her sister. How many stickers does Emma have left?
Solution: Emma has 30 stickers left.
Problem:
There are 24 students in a classroom. If 15 students are present today, how many students are absent?
Solution: There are 9 students absent.
Problem:
A bakery had 36 cupcakes. They sold 19 cupcakes. How many cupcakes are left?
Solution: There are 17 cupcakes left.
Problem:
Amy has 56 stickers. She wants to share them equally among her 4 friends. How many stickers will each friend get?
Solution: Each friend will get 14 stickers.
Problem:
In a park, there are 7 benches and each bench has 5 seats. How many seats are there in total?
Solution: There are 35 seats in total.
Problem:
A box contains 48 crayons. If 30 crayons are taken out, how many crayons are left in the box?
Solution: There are 18 crayons left in the box.
Problem:
A farmer collected 63 eggs from the chickens. If 9 eggs are cracked, how many good eggs does the farmer have?
Solution: The farmer has 54 good eggs.
Problem:
There are 22 books on one shelf and 15 books on another shelf. How many books are there on both shelves?
Solution: There are 37 books on both shelves.
Problem:
Sarah has 40 stickers. She wants to give an equal number of stickers to her 5 friends. How many stickers will each friend receive?
Solution: Each friend will receive 8 stickers.
Problem:
There are 18 apples in a basket. If 6 apples are taken out, how many apples are left in the basket?
Solution: There are 12 apples left in the basket.
Problem:
Daniel has 75 marbles. He wants to share them equally among his 5 friends. How many marbles will each friend get?
Solution: Each friend will get 15 marbles.
Problem:
The garden has 16 red roses and 12 yellow roses. How many roses are there in total?
Solution: There are 28 roses in total.
Problem:
There are 25 students in a class. If 13 students are girls, how many students are boys?
Solution: There are 12 boys in the class.
Problem:
Ava saved \$18 from her allowance, and her brother saved \$14. How much money did they save together?
Solution: They saved \$32 together.
## Conclusion
Making math relatable in real life may just be the ultimate trick to cracking the complexities in math, which is why you must start early with these word problems that are listed above.
Brighterly.com is where awesome tutors use fun videos and games to teach these kinds of problems. If you’re a parent, teacher, or someone who wants to make math awesome for 3rd graders, register now for Brighterly.com
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# How to Use Percentages Instead of Odds in Gambling
Do you get confused when you see things like odds and fractions and percentages listed about a gambling game? Did you score well in school when you had to learn fractions? If you did, do you remember much about them now?
Most people know what they use on a regular basis and tend to forget the things that they don’t use often. Fractions fall into the category for most people because they don’t have to use them in everyday life.
Unless you’re a gambler, the chances are high that you don’t have any idea what odds really are. Odds are usually not used by most people on an everyday basis.
All of this makes it difficult for many gamblers to understand what they’re being told or shown when it comes to odds and fractions. But most people have a pretty good grasp on what percentages mean. And even if you aren’t great with percentages, you will be when you finish this page.
It’s easy to change odds and fractions to percentages if you know how. Anyone with a calculator can change them in seconds, and you probably carry around a calculator all the time, because most mobile phones have one built in.
Below you’re going to learn what odds are, how to convert odds to fractions, how to convert fractions to odds, and how all of this applies to gambling. In the final sections, I show you examples from specific games where you can use percentages.
## What Are Odds?
Odds are a way of expressing how likely something is to happen. When you flip a coin, the odds are 50/50. This means that you have an equal chance for the coin to land on heads or tails. Anytime the numbers on both sides of the slash line are equal, and it means the chance of each thing happening is the same.
If you’re looking at two possibilities and one thing happens twice as often as the other, you can write the odds as 1/2, which shows that the odds of the first thing happening are actually 1 out of 3, and the odds of the second thing happening are 2 out of 3.
I know this can be a little confusing, and this is why it’s important to understand how to make odds into fractions and percentages.
Odds are simply fractions that aren’t called fractions. A slash line, like the one between the two 50’s in 50/50 is also used to denote a fraction. So 50/50 is simply a fraction of 50 over 50. When you divide 50 by 50, you get one. Odds, or a fraction, that ends up as one when you divide it means that each possibility has the same chance of happening.
The best thing about fractions is you can easily convert them to percentages. Learn how to do this in the next section.
## Converting Fractions to Percentages
Now that you know that odds are simply fractions, all you need to learn is how to convert a fraction to a percentage. This is really easy, and you can use the calculator on your phone or computer to make the calculations quickly.
To change a fraction to a percentage, simply follow two rules:
1. Divide the top number of the fraction by the bottom number of the fraction.
2. Move the decimal point two places to the right and put a percentage sign, %, on the end.
Here’s an example:
If you have the fraction 7/28, you divide 7 by 28. This gives you a decimal of .25. Now move the decimal two places to the right. So you change .25 to 25. This makes it a whole number, so in this example, you simply drop the decimal. This means that 7/28 is the same as 25%.
If you have a 7 out of 28 chance of something happening, you have a 25% chance of it happening.
I mentioned that when the decimal is at the right end of the number, it means the number is a whole number so you can drop the decimal. You can write 25% as 25.0% or 25.00%. They all mean the same thing. The reason this is important is because some percentages have both whole numbers and decimals.
Here’s an example:
5/8 when divided gives you the decimal .625. When you move the decimal two places to the right, you have 62.5%. This means the same thing as 62 and ½.
The next thing you need to know about converting fractions to percentages involves rounding off your results. This is important because you don’t need to use long strings of decimals at the end of your percentages in most gambling application.
All you need are the first one or two numbers to the right of the decimal when you use them for gambling. Rounding is easy; you just follow a couple simple rules.
1. Find the number you want to round off to. For gambling purposes, this is the first or second number to the right of the decimal point.
2. Now look at the number to the immediate right of the number you want to round to. If the number to the immediate right is five or higher, round up. If the number is four or lower don’t round up.
3. After rounding, drop all of the numbers to the right of the rounded number.
Here’s an example:
If you want to round 42.336% to the second number to the right of the decimal, you look at the number to the immediate right of the number you want to round. In this case, the number to the immediate right is a six. This is five or more, so you round up. This makes the number after rounding 42.34%.
If you want to round 42.336 to the first number to the right of the decimal, you look at the number to the immediate right of the first number. In this case, the number to the immediate right is three, so you don’t round up. This makes the rounded number 42.3%.
Now there’s only one more thing you need to know about percentages. It can be a little bit confusing when you have to deal with percentages less than 1%. Look at a series of percentages and see if you know what they mean.
• 50%
• 5%
• .5%
• .05%
Don’t worry if you don’t know the exact difference between the four numbers, or if you only know what a couple of them mean for sure. It’s not hard to understand what different percentages mean once you get the hang of it.
The first number on the list, 50%, means that something happens half the time, 5 out of 10, or 50 out of 100 chances. It can also be 500 out of 1,000 times or 24 out of 48 times. Most people have an understanding of what 50% means.
The second number on the list is 5%. This means that something happens 5 out of 100 times or 50 out of 1,000 times, or 1 out of 10 times. 5% of a dollar is a nickel, is another way of looking at it.
The next two numbers on the list are where many people start struggling. This makes sense because most people don’t have to work with numbers smaller than 1% in their daily lives.
.5% is the same thing as half a percent. This is less than 1%, so this happens less than 1 out of 100 times. In this case, .5% happens 5 out of 1,000 times.
The final number on the list, .05%, is even smaller than .5%. This happens only 5 out of 10,000 times.
It might make it easier to understand if you look at how often each number on the list happens, one after the other.
• 50% 5 out of 10
• 5% 5 out of 100
• .5% 5 out of 1,000
• .05% 5 out of 10,000
As you can see, as the percentage gets smaller, the chances of it happening go down. When it moves down like the example above, the chances on the right go up by adding another zero to the end.
Don’t panic if this is still a bit confusing. You don’t need to use percentages smaller than 1% often, and if you do need to use them in gambling, you can use a calculator.
## Gambling Applications
Now that you know about odds and fractions and how to change them into percentages, you need to learn how this helps you when you gamble. I’ve put together some specific examples below using real gambling games and situations.
The main percentage you need to understand when you gamble is the house edge. The house edge is the percentage of every bet that the casino keeps as profit. The house edge is based on all the bets made on a game or machine. This means that it’s a long term percentage.
You play one hand at a time or take one spin at a time, but over the years of your life, you might play a million hands or take a million spins. Walk into a casino on a Saturday evening and look at the hundreds or thousands of people gambling.
Many casinos take over a million in bets on a busy day. Some take millions in action every day. Each game and machine has a house edge, and if you know how much action a game has on a given day and you know the house edge you can determine the theoretical profit for the day for that game.
The actual percentage varies a little bit every day, but in the long run, it averages out to the house edge percentage.
Here’s an example:
The casino has a slot machine that has a house edge of 4%, and players make bets totaling \$20,000 on a Saturday on the machine. The expected profit on the machine for the day is \$800. You simply multiply the 4% house edge times the total amount of the wagers for the day. You change the percentage to a decimal by moving the decimal two places to the right.
Remember that a whole number percentage, like 4%, has a decimal to the right of the number. So 4% is the same thing as 4.0%. When you move the decimal, 4% becomes .04. .04 X \$20,000 = \$800.
You can use the house edge percentage to determine your expected loss on any game if you know the house edge.
Here’s an example:
You play Jacks or Better video poker on a 9/6 machine and are able to play close to perfect strategy. Because you make a few mistakes, the house edge is .5%. You play 200 hands per hour, and you bet \$5 per hand.
Your expected loss is determined by multiplying the house edge of .05%, or .005, times 200 hands per hour times \$5 per hand.
.005 X 200 X \$5 = \$5
Some hours you’re going to lose more than \$5, and some hours you’re going to win more than you lose. But if you play long enough using the same numbers, you’re going to average a \$5 an hour loss rate.
I always find it easier to learn and apply new skills by putting them into action. These examples can help you learn how to use percentages in real world gambling situations.
### Roulette
Roulette is an easy place to start because the fractions are easy to see and understand. When you bet on one number, the fractional chance of hitting the number is either 1/37 or 1/38. You use 37 on the bottom for a single zero wheel and 38 for a double zero wheel.
Here’s a list of fractions and the correlating percentages for popular roulette wagers.
Roulette Bet Fraction Percentage Single Zero Percentage Double Zero
One number 1/37 or 1/38 2.7% 2.6%
Two numbers 2/37 or 2/38 5.4% 5.3%
Three numbers 3/37 or 3/38 8.1% 7.9%
Four numbers 4/37 or 4/38 10.8% 10.5%
Six numbers 6/37 or 6/38 16.2% 15.8%
12 numbers 12/37 or 12/38 32.4% 31.6%
18 numbers 18/37 or 18/38 48.6% 47.4%
You can compare these numbers to the payout for winning a bet. The nice thing about roulette is that every bet has the same house edge, so you don’t have to try to figure out which bet is best. The house edge is the same on every wager on a single zero wheel, and the house edge is the same on every bet but one on a double zero wheel. The only bet you should never make playing roulette is the basket wager on a double zero wheel.
The house edge on all single zero roulette wagers is 2.7%, and the house edge on all double zero roulette bets other than the basket is 5.26%. The basket bet on a double zero wheel has a house edge of 7.89%.
### Slots
When you play slot machines you don’t have to deal with fractions, and the odds are already converted to percentages. The two most important percentages dealing with slot machines are pay back percentage and the house edge.
The pay back percentage is 100 minus the house edge. This means that the house edge is 100 minus the pay back percentage. In other words, the house edge and pay back percentage, when added together, equals 100%.
Here’s an example:
• A slot machine with a house edge of 6% has pay back percentage of 94%.
• A slot machine with a pay back percentage of 97% has a house edge of 3%.
Slot machines don’t usually have pay back percentages listed anywhere, and it’s even hard to find the information online. But it’s worth looking for it because you want to play on machines with the highest pay back percentage because these machines have the lowest house edge.
### Blackjack
Blackjack offers many opportunities to use percentages. One of the ways you can use percentages in blackjack is when the dealer offers you insurance when they have an ace as their up card.
When the dealer has an ace face up, they ask each player if they want insurance. If you take the insurance wager and the dealer has a blackjack, you get paid 2 to 1 on your bet. You’ve already made a bet at the start of the hand, so when you take the insurance wager, and the dealer has a blackjack, you basically break even of the hand.
For the dealer to have a blackjack, they have to have a face down card worth 10 points. This means that they have to have a 10, jack, queen, or king. What are the odds that they have one of these cards?
You know that the deck of cards has 13 ranks, from two to ace. Four of the 13 ranks complete a blackjack, and nine of the 13 ranks don’t. Here’s how to convert these numbers to percentages.
Four of the ranks complete a blackjack, so the fraction is 4/13. This gives a percentage of 30.77%.
Nine of the ranks don’t complete a blackjack, so the fraction is 9/13. This is a percentage of 69.23%.
The insurance bet pays 2 to 1. This means that for the payoff to be fair, the percentage chance of the dealer not completing a blackjack to the chance of them completing it needs to be 66.7% to 33.3%.
When you make the insurance wager you actually want the dealer to have a blackjack, so for the bet to be fair, you need the chance of them getting a blackjack to be 33.3%. As you can see, the chance is only 30.77%, so you should never take the insurance bet.
### Poker
Poker is where I use percentages the most. You play with a deck of 52 cards, so you can figure out the percentage chance of all kinds of things while you play. If you want to know your chance of hitting one of your outs on the turn or river, you can easily figure it out based on your number of outs and the number of unseen cards.
Here’s an example:
You’re playing Texas holdem and flop four to the nut flush. The turn is a blank, and your opponent shove their stack all in. You need to figure out the chance of hitting your flush on the river to determine if you want to call or fold.
You know that there are nine cards that complete your flush. You’ve seen the two hole cards in your hand and four board cards. This means there are 46 cards you haven’t seen, and nine of them complete your flush. Divide the nine outs by the 46 unseen cards, 9/46, and you know the chance of completing your flush on the river is 19.57%.
To make things easier, you can round this up to 20%. This alone is valuable information to have, but it’s even more valuable when you compare the size of the pot after your opponent’s bet to the amount you have to call.
If your opponent bets \$100 into a \$600 pot, the pot size is \$700, and you have to call \$100 to see the river. If you divide your call of \$100 by the pot size of \$700, 100/700, you get 14.3%.
If your opponent bets \$100 into a \$300 pot, the pot size is \$400, and you have to call \$100 to see the river. If you divide \$100 by \$400, 100/400, you get 25%.
Your chance of hitting your flush is roughly 20%, so you need the percentage of your call against the pot size to be lower than 20% for the call to be profitable. In the first example, the number is 14.3%, and in the second example, it’s 25%.
It’s profitable to call when the number 14.3% and you should fold when the number is 25%. This is called pot odds, and most players are taught to use ratios or fractions. Now you know how to use percentages to determine pot odds.
This is only one of the many ways you can use percentages at the poker table. Start looking for other ways to use them. Figure out the percentage chance of getting an ace as your first card. You know there are four aces and 52 cards in the deck, so you simply divide four by 52, 4/52.
Every time you play, look for opportunities to use percentages. Eventually, you’re going to find that you use percentages automatically to help you improve your results.
## Conclusion
I find it easier to use percentages than odds and fractions, even though I know how to use odds and fractions. Many gamblers I’ve talked to over the years struggle with odds, but when I show them how to make them into percentages, things start to become clear.
Now that you know how to use percentages when you gamble, you can make better betting decisions. When you make better betting decisions, it gives you the chance to win more often, and it makes your bankroll last longer.
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Michael Stevens
Michael Stevens has been researching and writing topics involving the gambling industry for well over a decade now and is considered an expert on all things casino and sports betting. Michael has been writing for GamblingSites.org since early 2016.
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# Find the domain and range of these functions.
• the function that assigns to each pair of positive integers the first integer of the pair.
• the function that assigns to each positive integer the largest decimal digit.
• the function that assigns to a bit string the number of ones minus the number of zeros in that string.
• the function that assigns to each positive integer the largest integer that does not exceed the square root of the integer.
• the function that assigns to a bit string the longest string of ones in that string.
This question aims to find the domain and range of the given functions.
A function is a relationship between a set of inputs and a set of permitted outputs. In a function, each input is related to precisely one output.
A domain takes a set of possible values for the components of a function. Suppose $f(x)$ is a function, the set of $x$ values in $f(x)$ is called domain of $f(x)$. In other words, we can define domain as the whole set of possible values for independent variables.
A range of the function is a set of values that the function can take. It is a set of values that the function returns after we enter an $x$ value.
• We have the function that assigns to each pair of positive integers, the first integer of the pair.
The positive integer is a natural number, and the only non-positive natural number is zero. This implies that $N-\{0\}$ refers to a set of positive integers under consideration. So its domain will be:
Domain $=\{(x,y)|x=1,2,3,\cdots\,\,\text{and}\,\, y=1,2,3,\cdots\}$
$=\{(x,y)|x\in N-\{0\}\wedge x\in N-\{0\}\}$
$=(N-\{0\})\times (N-\{0\})$
And range will be a positive first integer of the domain, that is:
Range $=\{1,2,3,\cdots\}=N-\{0\}$
• We have a function that assigns to each positive integer its largest decimal digit.
In this case, a domain will be a set of all positive integers:
Domain $=\{1,2,3,\cdots\}=N-\{0\}$
And the range will be a set of all the digits from $1$ to $9$, that is:
Range $=\{1,2,3,4,5,6,7,8,9\}$
• We have a function that assigns to a bit string the number of ones minus the number of zeros in the string.
The domain of such a function will be a set of all bit rings:
Domain $=\{\lambda,0,1,00,01,11,10,010,011,\cdots\}$
And according to the statement, the range may take on positive and negative values and a zero, since it will be a set of all differences between the number of ones and number of zeros in a string. Therefore:
Range $=\{\cdots,-2,-1,0,1,2,3,\cdots\}$
• We have the function that assigns to each positive integer the largest integer not exceeding the square root of the integer.
Here, the domain will be a set of all positive integers:
Domain $=\{1,2,3,\cdots\}=N-\{0\}$
The range is defined as the set of the largest integer that does not exceed the square root of a positive integer. We can see that the set contains all positive integers, so:
Range $=\{1,2,3,\cdots\}=N-\{0\}$
• Lastly, we have the function that assigns to a bit string the longest string of ones in the string.
The domain of such a function will be a set of all bit rings:
Domain $=\{\lambda,0,1,00,01,11,10,010,011,\cdots\}$
The range will be a set of all longest strings of ones in any string. As a result, the range only contains strings that contain the digit $1$:
Range $=\{\lambda,1,11,111,1111,11111,\cdots\}$
## Example
Find the domain and range of the function $f(x)=-x^2-4x+3$.
Since $f(x)$ has neither undefined points nor domain constraints, therefore:
Domain: $(-\infty,\infty)$
And $f(x)=-x^2-4x+3=-(x+2)^2+7$
Since, $-(x+2)^2\leq 0$ for all real $x$.
$\implies -(x+2)^2+7\leq 7$
Hence, the range is: $(-\infty,7]$
Graph of $f(x)$
Images/mathematical drawings are created with GeoGebra.
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# Package 3-2: Inside Seven Squares
Week 4 - Package 2 - Year 5 and 6 Mathematics - Inside Seven Squares
## Things you need
### Back up
• Pen or texta
• Grid paper such as that in your child’s mathematics book.
## Before you start
This is a reasoning activity adapted from nrich.maths and is expected to take some time. Try to encourage your child to solve the problem without stepping in to help at the first hurdle.
Your child will need to be able to draw lines accurately with a ruler using the dotty grid paper or ordinary grid paper as a guide. Drawing lines accurately is an important mathematics skill but is not necessary for the reasoning part of the activity. Your child will also need to know the basic properties of squares and right-angled triangles so you may want to do some research before you get started. In addition, your child will need to understand that area is the amount of space inside the boundary of a two-dimensional shape and can be measured in square units.
## What your child needs to know and do
### The problem
Seven squares are set inside each other. The centre point of each side of the outer square is joined to make a smaller square inside. This is repeated each time a square is drawn until there are 7 squares.
The centre square has an area of 1 square unit.
What is the total area of the four outside right-angled triangles (the ones in red)?
## What to do next
A useful way to get your child thinking about this problem is to make the beginning of the design by folding a square piece of paper. Using concrete materials is an important mathematical skill at all levels of learning. Remind your child that this is a tool for learning.
An A4 piece of paper can be made to represent a square as follows:
The following folding activity will help your child consider where the problem came from. It is really helpful to ask what they notice at each step of the process.
1. Once you have your square piece of paper fold it along both diagonals to find the centre.
1. Fold each right angle carefully into the centre.
1. Open out and use the folds to draw the next square.
1. Fold in again and then fold the next four right angles into the centre of the new square. At this point the paper is going to start looking like a Chatterbox. Making a Chatterbox would be a lovely fun activity to do after the maths! It is a great way to practise accurate folding.
1. Open out again and draw the next square.
1. Your child can continue this activity until it becomes too hard to fold the paper. I wonder whether 7 times really is the limit for folding a piece of paper? The folds will become less and less accurate as the paper gets thicker. (You might even like to view an old episode of Myth Busters about this!)
Next you will need the dotty paper or squared paper from your child’s maths book, a pencil and ruler.
It would be wonderful if you could do the activity alongside your child or if they could work with a sibling to share ideas. Make sure you give your child a chance to think about and discuss where on the paper they will start, what size square to start with, and whether to start with the largest or the smallest square. There may be a few trial runs.
Once all of the squares have been drawn it is important to remember that the problem asks you to find the combined area of the four outer triangles (the red ones in the diagram). There will be a number of ways to explain this. All strategies are good strategies and it is worth brainstorming as many ways as possible.
### Activity too hard?
If the drawing activity is too difficult there is still a lot of learning in the folding activity. Ask your child how much bigger the area of the original square is than the square created by the folds. Ask them how they know.
### Activity too easy?
Ask your child to determine how many times bigger the area of the outside square than each of the successive squares.
Ask if they can predict the area of the square outside the largest one they have drawn. Can they create a table that shows the areas of each square so they can predict the areas of both smaller and larger squares?
Can they add perimeter to their table or length of each side to discover the relationship between the perimeters of the squares and their areas?
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HOTS Questions: Pair of Linear Equations in Two Variables
# Class 10 Maths Chapter 3 HOTS Questions - Pair of Linear Equations in Two Variables
Q1. If two of the roots of f(x) = x3 – 5x2 – 16x + 80 are equal in magnitude but opposite in sign, then find all of its zeroes.
Hint: Let α and β are the two zeroes which are equal in magnitude but opposite in sign.
∴ α + β = 0
(Let the third zero is γ)
Sum of zeroes of
f(x)= α + β + γ =
∴ γ = 5 [α + β = 0]
Product of zeroes
⇒ – α2 = –16 [α + β = 0 ⇒ β = – α]
⇒ α2 = 16 ⇒ α = ± 4
α = ± 4 ⇒ β = ∓ 4
[∴ β = –α]
Thus, the zeroes are : [± 4, ∓ 4, and 5]
Q2.
Hint:
Adding (1) and (2), we get x = 1/3 and y = 1/2
Q3. Solve :
Hint: Put x + 2y = p and 2x – y = q
We have ...(1)
...(2)
Solving (1) and (2), we get p = 4 and q = 3
∴ x + 2y = 4 and 2x – y = 3
Solving these equations, we get x = 2 and y = 1
Q4. Solve : x + y = 18 ; y + z = 12 ; z + x = 16.
Hint: Adding the three equations, we get
⇒ x + y + z = 23
Now, (x + y + z = 23) – (x + y = 18)
⇒ z = 5 (x + y + z = 23) – (y + z = 12)
⇒ x = 11 (x + y + z = 23) – (z + x = 16)
⇒ y = 7
Thus, x = 11, y = 7 and z = 5
Q5. Solve :
Hint: Inverting the equations:
Adding (1), (2) and (3), we get
Now, subtracting (1), (2) and (3) turn by turn from (4), we get x = 2, y = 4 and z = 6
Q6. Solve:
Hint:
From
⇒ 11(x + y – 3) = 2(3x + y) ⇒ 5x + 9y = 33 ...(1)
From
⇒ 11(x + 2y – 4) = 3(3x + y)
⇒ 2x + 9y = 44 ...(2)
Solving (1) and (2), we get x = 3 and y = 2
The document Class 10 Maths Chapter 3 HOTS Questions - Pair of Linear Equations in Two Variables is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## FAQs on Class 10 Maths Chapter 3 HOTS Questions - Pair of Linear Equations in Two Variables
1. What is a pair of linear equations in two variables?
Ans. A pair of linear equations in two variables is a set of two equations that involve two variables, typically represented as x and y, and the power of both variables is 1. The general form of a linear equation is ax + by = c, where a, b, and c are constants.
2. How can we solve a pair of linear equations in two variables?
Ans. There are several methods to solve a pair of linear equations in two variables. Some common methods include the substitution method, elimination method, and graphical method. These methods involve manipulating the equations to eliminate one variable and solve for the other.
3. Can a pair of linear equations in two variables have multiple solutions?
Ans. Yes, a pair of linear equations in two variables can have multiple solutions. If the two equations represent the same line, they have infinitely many solutions. If the two equations represent parallel lines, they have no solution. Otherwise, they will intersect at a single point, which is the solution.
4. How can we determine if a given point is a solution to a pair of linear equations in two variables?
Ans. To determine if a given point is a solution to a pair of linear equations in two variables, substitute the values of the variables into both equations and check if the equations hold true. If both equations are satisfied, the point is a solution to the pair of equations.
5. What real-life applications involve the use of a pair of linear equations in two variables?
Ans. A pair of linear equations in two variables is commonly used to model real-life situations. Some examples include solving problems related to cost and revenue, distance and time, speed and time, mixture problems, and optimization problems. These equations help in finding the unknown variables and making predictions or decisions based on the given conditions.
## Mathematics (Maths) Class 10
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## College Algebra (10th Edition)
(a) Slope $= -3$ and y-intercept $=4$ (b) See the image. (c) Average rate of change $= -3$ (d) The linear function $h(x)=-3x+4$ is decreasing.
Step-1: Compare the given equation with the point-slope form of the linear equation, that is, $h(x) = mx+b$, where $m$ is the slope of the linear function and $b$ is its y-intercept. By comparing $$h(x)=-3x+4$$ to $$h(x) = mx+b$$ we that understand that the slope of the given function is $-3$ and its y-intercept is $4$. Step-2: Let us put values of $x$ from $-2$ to $2$ into the function to obtain corresponding $y$ values. Using this we can plot a graph. Thus, For $x=-2$, $h(x)=10$ For $x=-1$, $h(x)=7$ For $x=0$, $h(x)=4$ For $x=1$, $h(x)=1$ For $x=2$, $h(x) = -2$ This data obtains the graph shown. Step-3: The average rate of change is defined as follows: $$\frac{\Delta y}{\Delta x}=\frac{h(x_2)-h(x_1)}{x_2-x_1}$$ Let us calculate average rate of change between $x_2=2$ and $x_1=-2$, $$\frac{\Delta y}{\Delta x}=\frac{-2-10}{2-(-2)}=-3$$ Step-4: Since slope, $m=-3<0$, this linear function is decreasing.
|
# Number System: Factors and Co-primes
Number system is a very important chapter and you can find good number of questions in various competitive exams. Important formulas are the following.
A number can be written in its prime factorization format. For example 100 = 22 x 52
## Formula 1: The number of factors of a number
(p+1).(q+1).(r+1)... where N = ap x bq x cr ...
Example: Find the number of factors of 100.
Ans: We know that 100 = 22 x 52
So number of factors of 100 = (2 +1 ).(2 +1) = 9.
Infact the factors are 1, 2, 4, 5, 10, 20, 25, 50, 100
## Formula 2: The sum of factors of a number
$\frac{{\mathrm{a}}^{\mathrm{p}+1}-1}{\mathrm{a}-1}×\frac{{\mathrm{b}}^{\mathrm{q}+1}-1}{\mathrm{b}-1}×\frac{{\mathrm{c}}^{\mathrm{r}+1}-1}{\mathrm{c}-1}...$ where $N={a}^{p}×{b}^{q}×{c}^{r}×...$
Example: Find the sum of the factors of 72
Ans: 72 can be written as $\left({2}^{3}×{3}^{2}\right).$
Sum of all the factors of 72 = $\left(\frac{{2}^{3+1}-1}{2-1}×\frac{{3}^{2+1}-1}{3-1}\right)$= 15 x 13= 195
## Formula 3: The number of ways of writing a number as a product of two number
$\frac{1}{2}×\left[\left(\mathrm{p}+1\right).\left(\mathrm{q}+1\right).\left(\mathrm{r}+1\right)...\right]$ (if the number is not a perfect square)
If the number is a perfect square then two conditions arise:
1. The number of ways of writing a number as a product of two distinct numbers =$\frac{1}{2}×\left[\left(\mathrm{p}+1\right).\left(\mathrm{q}+1\right).\left(\mathrm{r}+1\right)...-1\right]$
2. The number of ways of writing a number as a product of two numbers and those numbers need not be distinct=$\frac{1}{2}×\left[\left(\mathrm{p}+1\right).\left(\mathrm{q}+1\right).\left(\mathrm{r}+1\right)...+1\right]$
Example: Find the number of ways of writing 140 as a product of two factors
Ans: The prime factorization of 140 = ${2}^{2}×5×7$
So number of ways of writing 140 as a product of two factors = $\frac{1}{2}×\left[\left(\mathrm{p}+1\right).\left(\mathrm{q}+1\right).\left(\mathrm{r}+1\right)...\right]$ = $\frac{1}{2}×\left[\left(2+1\right).\left(1+1\right).\left(1+1\right).\right]=6$
Example: Find the number of ways of writing 144 as a product of two factors subjected to the following conditions a. Both factors should be different b. Both factors need not be different.
Ans: The prime factorization of 144 = ${2}^{4}×{3}^{2}$
a. If both factors are different, then total ways = $\frac{1}{2}×\left[\left(\mathrm{p}+1\right).\left(\mathrm{q}+1\right).\left(\mathrm{r}+1\right)...-1\right]$ = $\frac{1}{2}×\left[\left(4+1\right).\left(2+1\right)-1\right]$ = 7
If both factors need not be different, then total ways = $\frac{1}{2}×\left[\left(\mathrm{p}+1\right).\left(\mathrm{q}+1\right).\left(\mathrm{r}+1\right)...+1\right]$ = $\frac{1}{2}×\left[\left(4+1\right).\left(2+1\right)+1\right]$ = 8
## Formula 4: The number of co-primes of a number
N=$\varphi \left(N\right)={a}^{p}.{b}^{q}.{c}^{r}...$ can be written as $N×\left(1-\frac{1}{a}\right)×\left(1-\frac{1}{b}\right)×\left(1-\frac{1}{c}\right)...$
Example: Find the number of co-primes to 144 which are less than that of it
Ans: The prime factorization of 144 = ${2}^{4}×{3}^{2}$
The number of co-primes which are less than that of 144 = $144×\left(1-\frac{1}{2}\right)×\left(1-\frac{1}{3}\right)$ = 48
## Formula 5: The sum of co-primes of a number
N= $\varphi \left(N\right)×\frac{N}{2}$
Example: Find the sum of all the co-primes of 144
Ans: The sum of co-primes of the 144 = $\varphi \left(N\right)×\frac{N}{2}$ = 48 x $\frac{144}{2}$ = 3456
## Formula 6: The number of ways of writing a number N as a product of two co-prime numbers
= ${2}^{n-1}$ where n=the number of prime factors of a number.
Example: Find the number of ways of writing 60 as a product of two co - primes
Ans: The prime factorization of 60 = ${2}^{2}×3×5$
The number of ways of writing 60 as a product of two co - primes = ${2}^{3-1}$ = 4
## Formula 7: Product of all the factors
${N}^{\left(\frac{\text{Number of factors}}{2}\right)}$ = ${N}^{\left(\frac{\left(p+1\right).\left(q+1\right).\left(r+1\right)....}{2}\right)}$
Example: Find the product of all the factors of 50
Ans: Prime factorization of 50 = $2×{5}^{2}$
Product of all the factors of 50 = ${50}^{\frac{\left(1+1\right).\left(2+1\right)}{2}}={50}^{3}$
Level - 2
1. P is the product of all the factors of 15552. If P = ${12}^{N}×M$, where M is not a multiple of 12, then find the value of M. [M and N are positive Integers]
Ans: 15552 = ${2}^{6}×{3}^{5}$
Product of all the factors of 15552 = ${\left({2}^{6}×{3}^{5}\right)}^{\frac{\left(6+1\right).\left(5+1\right)}{2}}={2}^{126}×{3}^{105}$
$⇒$${\left({2}^{2}\right)}^{63}×{3}^{63}×{3}^{42}$ = ${12}^{63}×{3}^{42}$
So M = ${3}^{42}$
2. Let M be the set of all the distinct factors of the number N=${6}^{5}×{5}^{2}×10$,Which are perfect squares. Find the product of the elements contained in the set M.
N = ${6}^{5}×{5}^{2}×10$ = ${2}^{6}×{3}^{5}×{5}^{3}$
Even powers of 2 available: ${2}^{0},{2}^{2},{2}^{4},{2}^{6}$
Even powers of 3 available: ${3}^{0},{3}^{2},{3}^{4}$
Even powers of 5 available: ${5}^{0},{5}^{2}$
Therefore number of factors of the number N that are perfect squares = 4 x 3 x 2 = 24
Product of the elements contained in M
${2}^{\left(2+4+6\right)×6}×{3}^{\left(2+4\right)×8}×{5}^{2×12}={2}^{72}×{3}^{48}×{5}^{24}$
3. In a hostel there are 1000 students in 1000 rooms. One day the hostel warden asked the student living in room 1 to close all the doors of the 1000 rooms. Then he asked the person living in room 2 to go to the rooms which are multiples of his room number 2 and open them. After he ordered the 3rd student to reverse the condition of the doors which are multiples of his room number 3. If He ordered all the 1000 students like the same, Finally how many doors of those 1000 rooms are in open condition?
We understand that a door is in open or in close condition depends on how many people visited the room.
If a door is visited by odd number of persons it is in close condition, and is visited by even number of persons it is in open condition.
The number of people who visit a certain door is the number of factors of that number. Let us say room no: 24 is visited by 1, 2, 3, 4, 6, 8, 12, 24 which are all factors of 24. Since the number of factors are even this door is in open condition.
we know that the factors of a number N=${a}^{p}.{b}^{q}.{c}^{r}...$ are (p+1).(q+1).(r+1)...
From the above formula the product is even if any of p, q, r... are odd, but the product is odd when all of p, q, r are even numbers.
If p, q, r ... are all even numbers then N=${a}^{p}.{b}^{q}.{c}^{r}...$ is a perfect square.
So for all the perfect squares below 1000 the doors are in closed condition.
There are 31 perfect squares below 1000 so total doors which are in open condition are (1000-31)= 969
4. What is the product of all factors of the number N = ${6}^{4}×{10}^{2}$ which are divisible by 5?Sol: N = ${6}^{4}×{10}^{2}$ = ${2}^{6}×{3}^{4}×{5}^{2}$
Total product of the factors = ${\left({2}^{6}×{3}^{4}×{5}^{2}\right)}^{\frac{\left(6+1\right).\left(4+1\right).\left(2+1\right)}{2}}$ = ${\left({2}^{6}×{3}^{4}×{5}^{2}\right)}^{\frac{105}{2}}$
So total product of the factors N which are not multiples of 5 =
${\left({2}^{6}×{3}^{4}\right)}^{\frac{\left(6+1\right).\left(4+1\right)}{2}}$ = ${\left({2}^{6}×{3}^{4}\right)}^{\frac{35}{2}}$
So, total product of the factors of N which are multiples of 5 = $\frac{{\left({2}^{6}×{3}^{4}×{5}^{2}\right)}^{\frac{105}{2}}}{{\left({2}^{6}×{3}^{4}\right)}^{\frac{35}{2}}}$ = ${2}^{210}×{3}^{140}×{5}^{105}$
5. Let N = ${2}^{3}×{3}^{17}×{5}^{6}×{7}^{4}$ and M = ${2}^{12}×{3}^{5}×{5}^{4}×{7}^{8}$. P is total number of even factors of N such that they are not factors of M. Q is the total number of even factors of M such that they are not factors of N. Then 2P -Q = ?
Sol:
N = ${2}^{3}×{3}^{17}×{5}^{6}×{7}^{4}$
M = ${2}^{12}×{3}^{5}×{5}^{4}×{7}^{8}$
Calculation of P:
Number of powers of 2 available are 3. i.e., ${2}^{1}$ to ${2}^{3}$(all these powers are even)
Since P is the total number of even factors of N such that they are not factors of M, the number of powers of 3 available are 17 - 5 = 12
For combination with each of these 12 powers of 3( i.e., ${3}^{6}$ to ${3}^{17}$ , number of powers of 5 are 7 ( i.e., ${5}^{0}$ to ${5}^{6}$) and number of powers of 7 are 5 ( i.e.,${7}^{0}$ to ${7}^{4}$) available respectively.
Therefore there are 3 x 12 x 7 x 5 = 1260 factors
Now consider powers of 5 that are in N but not in M: ${5}^{5},{5}^{6}$
So, for these two powers of 5, the number of powers of 2, 3, and 7 that are available are 3, 6, and 5 respectively. (Powers of 3 are only 6 (${3}^{0}$ to ${3}^{5}$), not 18 as we have already used ${3}^{6}$ to ${3}^{17}$ in caclulating P)
Therefore, there are 2 x 3 x 6 x 5 = 180 factors
P = 1260 + 180 = 1440
Calculation of Q:
Consider powers of 2 that are in M but not in N: ${2}^{4},{2}^{5},{2}^{6}....{.2}^{12}$
So for these 9 powers of 2, number of even factors of M such that they are not factors of N = 9 x 6 x 5 x 9 = 2430
Consider powers of 7 that are in M but not in N: ${7}^{5},{7}^{6},{7}^{7}$ and ${7}^{8}$
So for these 4 powers of 7, number of powers of 2, 3, and 5 that are available are 3, 6 and 5.
(Powers of 3 are only 6 (${3}^{0}$ to ${3}^{5}$), not 18 as we have already used ${3}^{6}$ to ${3}^{17}$ in calculating Q and (Powers of 5 are only 5 (${5}^{0}$ to ${5}^{4}$), not 7 as we have already used ${5}^{5}$ to ${5}^{6}$ in calculating Q))
So number of such factors = 4 x 3 x 6 x 5 = 360
Q = 2430 + 360 = 2790
Therefore, 2P - Q = 2880 - 2790 = 90
Alternative Method:
Many a student asked me to explain this question elaborately. I think if we use sets this question is easily done.
N = ${2}^{3}×{3}^{17}×{5}^{6}×{7}^{4}$
M = ${2}^{12}×{3}^{5}×{5}^{4}×{7}^{8}$
So by the above diagram, we can see that P is the number of even factors of N which are not factors of M and Q is number of even factors of M which are not factors of N.
Now to calculate common even factors of N and M, we take HCF of N and M which is ${2}^{3}×{3}^{5}×{5}^{4}×{7}^{4}$ ($\because$ take minimum powers of both numbers).
Even factors of above number = 3 × (5 + 1) × (4 + 1) × (4 + 1) = 450 ( $\because$ We have to include atleast one multiple of 2 to make it even number.
Even factors of P = 3 × 18 × 7 × 5 = 1890
Even factors of Q = 12 × 6 × 5 × 9 = 3240
So 2P - Q = 90
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Math Expressions Grade 2 Unit 3 Lesson 4 Answer Key Draw, Estimate, and Measure
Math Expressions Common Core Grade 2 Unit 3 Lesson 4 Answer Key Draw, Estimate, and Measure
Math Expressions Grade 2 Unit 3 Lesson 4 Homework
Estimate and measure each side. Then find the distance around the triangle.
Answer Key Math Expressions Grade 2 Unit 3 Lesson 4 Question 1.
a. Complete the table.
Explanation:
A triangle is a three-sided polygon, which has three vertices.
The three sides are connected with each other end to end at a point, which forms the angles.
b. Find the distance around the triangle.
_________ cm + _________ cm + _________ cm = _________ cm
2cm + 2cm + 2cm = 6cm
Explanation:
A triangle is a three-sided polygon, which has three vertices.
The three sides are connected with each other end to end at a point, which forms the equilateral triangle.
Grade 2 Unit 3 Lesson 4 Answer Key Math Expressions Question 2.
a. Complete the table.
Explanation:
The given triangle is isosceles triangle.
A triangle in which two sides are equal is called an isosceles triangle.
If two sides of a triangle are equal, then the angles opposite to those sides are also equal.
b. Find the distance around the triangle.
_________ cm + _________ cm + _________ cm = _________ cm
_____2____ cm + ____1_____ cm + ____2_____ cm = ____5_____ cm
Explanation:
If the two sides are of equal length, then it is isosceles triangle.
Lesson 4 Answer Key Math Expressions Grade 2 Unit 3 Question 3.
a. Complete the table.
Explanation:
The given triangle is isosceles triangle.
A triangle in which two sides are equal is called an isosceles triangle.
If two sides of a triangle are equal, then the angles opposite to those sides are also equal.
b. Find the distance around the triangle.
_________ cm + _________ cm + _________ cm = _________ cm
__1.5_ cm + __0.5__ cm + _1.5_ cm = __3.5_ cm
Explanation:
If two sides of a triangle are equal, then the given triangle is called isosceles triangle.
Math Expressions Grade 2 Unit 3 Lesson 4 Remembering
Find the total or partner.
Math Expressions Grade 2 Unit 3 Lesson 4 Answer Key Question 1.
Explanation:
In the Math mountain, the top part of the mountain to called Total or Sum and it consists of two parts called Partners or Addends
The addition of two whole numbers results in the total amount or sum of those values combined.
Make a drawing for each number. Write <, >, or =.
Question 2.
131 122
Explanation:
The symbol used to represent greater than is “>”.
The given value 131 is larger than the other value 122.
So, we use greater than symbol.
Question 3.
27 35
Explanation:
The symbol used to represent less than is “<”.
The given value 27 is less than the other value35
So, we use less than symbol.
Question 4.
List or draw objects that show rectangles.
Explanation:
A rectangle is a closed two-dimensional figure with four sides and four corners.
The length of the opposite sides is equal and parallel to each other.
Question 5.
Stretch Your Thinking Draw and label two different triangles. Each shape should have a distance around it of 12 cm.
Explanation:
The three sides are connected with each other end to end at a point, which forms the equilateral triangle.
Triangle ABC
Distance from
A to B = 4 cm
B to C = 4 cm
C to A = 4 cm
AB + BC + CA = 4 + 4 + 4 = 12 cm
If two sides of a triangle are equal, then the angles opposite to those sides are also equal.
Triangle LMN
Distance from
L to M = 4 cm
M to N = 5 cm
N to L = 3 cm
LM + MN + NL = 12 cm
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# Factorising quadratic expressions, and solving quadratic equations
## What do we mean by factorising?
Factorising - or factoring, to non-British English speakers - is the process of breaking an expression into its factors.
You may have seen how to factorise a linear expression by taking out a common factor:
2x + 4 = 2(x + 2) as 2 is a common factor of both 2x and 4.
This process is the opposite of expanding the bracket - and the same is true for quadratics.
## What does "quadratic" mean?
A quadratic is an algebraic expression where the highest power of the unknown variable is 2 - in other words, anything in the form ax2 + bx + c. Generally, we write them in descending order; the x2 term first, then the x term, then the constant c.
## Factorising as the inverse of expanding
When you worked on expanding a double bracket (x + a)(x + b), you may have noticed that:
• The numbers in the brackets always add up to the number of "xs," or the "coefficient of x."
• The numbers in the brackets always multiply to make the +? on the end, or "the constant term."
If we expand (x + a)(x + b) we get x2 + ax + bx + ab, or x2 + (a+b)x + ab - which supports the observations above...
So, to factorise a quadratic we need to find two numbers that add up to the x coefficient, and multiply to make the constant on the end. For example:
"Factorise x2 + 7x + 12"
We need two numbers that:
• Sum to 7
• Have a product of 12
From this we can (hopefully) see that the numbers we need are 3 and 4.
Top tip: Look for factors of the constant first, as there are a finite number of possibilities! In this case, the only possible integer pairs are: 1 & 12, 2 & 6, 3 & 4, -3 & -4, -2 & -6, -1 & -12. Then find the pair that sum to the number you need.
## So how do I solve quadratic equations?
The "solutions," or roots, of a quadratic are the values of x where the graph would cross - or touch - the x axis. In other words, they are the x values that make the whole equation equal zero.
Because we are multiplying our two linear factors (the pair of brackets) together to make our quadratic, we can make the whole expression zero by making either bracket zero; anything multiplied by zero is still zero!
So to solve x2 + 7x + 12 = 0:
1. Factorise: (x + 3)(x + 4)
2. What x value makes the first bracket zero? -3 + 3 = 0, so x = -3.
3. What x value makes the second bracket zero? -4 + 4 = 0, so x = -4.
This means our roots are x = -3 and x = -4. If you plotted the graph of y = x2 + 7x + 12, it would cross the x-axis at these points.
If you have a quadratic that won't factorise, you need to learn the Quadratic Formula - but it's not too hard!
## Take the factorising quiz:
view quiz statistics
## 2 comments
Babbler87 5 years ago from Ho Chi Minh City (Saigon), Viet Nam
Funniest video on quadratics!
Spirit Whisperer 5 years ago from Isle of Man
Brilliant1 This is a great treatment of solving quadratics and the video adds an element of fun that would definitely make it stick in any student's long term memory. I will be visiting regularly as result of reading this hub. Thank you.
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# SKETCHING A TRIANGLE AND DETERMINE THE MEASURE OF INDICATED SIDE
Sketch each triangle. Determine the measure of the indicated side.
Example 1 :
In triangle ABC, ∠A = 57°, ∠B = 73°, and AB = 24 cm. Find the length of AC.
Solution :
AB = c = 24 cm, BC = a, AC = b, <A = 57 and <B = 73
In triangle ABC,
<A + <B + <C = 180
57 + 73 + <C = 180
<C = 180 - 130 = 50
Using sin formula,
a/sin A = b/sin B = c/sin C
a/sin 57 = b/sin 73 = 24/sin 50
We have to solve for b.
Equating 1 and 3, we get
b/sin 73 = 24/sin 50
b/0.9563 = 24/0.7660
b = 31.33(0.9563)
b = 29.96
b = 30 cm (approximately)
Hence the indicated side is 30 cm.
Example 2 :
In triangle ABC, ∠B = 38°, ∠C = 56°, and BC = 63 cm. Find the length of AB.
Solution :
AB = c, BC = a = 63, AC = b, <C = 56 and <B = 38
In triangle ABC,
<A + <B + <C = 180
<A + 38 + 56 = 180
<A = 180 - 94 = 86
Using sin formula,
a/sin A = b/sin B = c/sin C
63/sin 86 = b/sin 38 = c/sin 56
We have to solve for c.
Equating 1 and 3, we get
63/sin 86 = c/sin 56
63/0.9975 = c/0.8290
63.15 = c/0.8290
c = 63.15(0.8290)
c = 52.35
c = 52.4 cm (approximately)
Hence the indicated side is 52.4 cm.
Example 3 :
In triangle ABC, ∠A = 50°, ∠B = 50°, and AC = 27 m. Find the length of AB.
Solution :
AB = c, BC = a = 27 m, AC = b = 27 m, <A = 50 and <B = 50
In triangle ABC,
<A + <B + <C = 180
50 + 50 + <C = 180
<C = 180 - 100 = 80
Using sin formula,
a/sin A = b/sin B = c/sin C
27/sin 50 = 27/sin 50 = c/sin 80
We have to solve for c.
Equating 1 and 3, we get
27/sin 50 = c/sin 80
27/0.7660 = c/0.9848
35.24 = c/0.9848
c = 35.24(0.9848)
c = 34.70 m
Hence the indicated side is 34.7 m.
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# 2-3-11 PowerPoint PPT Presentation
2-3-11. Please have hw out to correct. Equations with Two Variables. Lesson 8-2 p.391. Equations with Two Variables. In the other chapters, we learned how to solve equations like this: 5x + 3 = 2x +9 In this type of equation, there was only one kind of variable—”x”.
2-3-11
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### 2-3-11
• Please have hw out to correct.
## Equations with Two Variables
Lesson 8-2 p.391
### Equations with Two Variables
• In the other chapters, we learned how to solve equations like this:
• 5x + 3 = 2x +9
• In this type of equation, there was only one kind of variable—”x”.
• Now we will learn how to solve variables like: y = 2x + 3
y = 2x + 3
### Equations with Two Variables
• y = 2x + 3
• What do you notice about this equation? Yes there are two kinds of variables—an x and a y.
• We will find in this chapter that the solution to this type of equation is an ordered pair and if we graph the ordered pairs of the equation, we get a straight line when the points are connected.
### Equations with Two Variables
• We will also find that an equation like y = 2x + 3 can have many solutions, not just one, but it is the graph of the solutions that will be our answer.
• Let’s start with one way to solve this type of problem. . .a t-table or t-chart
### Equations with Two Variables
• y = 2x + 3
• One strategy is to make a table of values or a t-table.
• It looks like this:
XY
### Equations with Two Variables
• y = 2x + 3
• We begin by choosing any value we want for x. This may seem odd to you, but the reason will become apparent later.
• I like to choose one positive number, one negative number and the number zero.
### Equations with Two Variables
• y = 2x + 3 Let’s choose 1, 0 and -2
xy
1Place the x values in
0the chart. This reminds
-2us which numbers to
substitute for x.
### Equations with Two Variables
• y = 2x + 3Then we substitute each
value one at a time and
x ysolve for “y”
1 52(1) + 3 = 5
0
-2
### Equations with Two Variables
• y = 2x + 3Then we substitute each
value one at a time and
x ysolve for “y”
1 52(1) + 3 = 5
0 32(0) + 3 = 3
-2
### Equations with Two Variables
• y = 2x + 3Then we substitute each
value one at a time and
x ysolve for “y”
1 52(1) + 3 = 5
0 32(0) + 3 = 3
-2 -12(-2) + 3 = -1
### Equations with Two Variables
• The information in the t-table is a series of ordered pairs that when graphed on the coordinate plane, will result in a straight line like this
### Equations with Two Variables
x y
1 5
0 3
-2 -1
First, plot point (1,5)
### Equations with Two Variables
x y
1 5
0 3
-2 -1
First, plot point (1,5)
### Equations with Two Variables
x y
1 5
0 3
-2 -1
First, plot point (1,5)
Then plot point (0,3)
### Equations with Two Variables
x y
1 5
0 3
-2 -1
First, plot point (1,5)
Then plot point (0,3)
### Equations with Two Variables
x y
1 5
0 3
-2 -1
First, plot point (1,5)
Then plot point (0,3)
Then plot point (-2,-1)
### Equations with Two Variables
x y
1 5
0 3
-2 -1
First, plot point (1,5)
Then plot point (0,3)
Then plot point (-2,-1)
Finally draw a line that connects and goes through the points.
### Equations with Two Variables
This is the graph of the equation:
y = 2x + 3
We will find that each equation has its own unique graph.
### Try This
• Make a t-table for the equation
y = 3x -2 using the following values
for x
xy
3
0
-1
### Try This
• Make a t-table for the equation
y = 3x -2 using the following values
for x
xy
373(3) – 2 = 7
0
-1
### Try This
• Make a t-table for the equation
y = 3x -2 using the following values
for x
xy
373(3) – 2 = 7
0 -23(0) – 2 = -2
-1
### Try This
• Make a t-table for the equation
y = 3x -2 using the following values
for x
xy
373(3) – 2 = 7
0 -23(0) – 2 = -2
-1 -53(-1) – 2 = -5
xy
37
0 -2
-1 -5
Now graph the
Ordered pairs
xy
37
0 -2
-1 -5
### One more Thing
• Sometimes, an equation will be given as well as a sample ordered pair, and you will be asked “Is this a solution to the equation?”
• For example, is (4,3) a solution to this equation: y = -2x + 2
• Substitute the ordered pair in the solution: 3 = -2(4) + 2
In this case 3 = -8 + 2 or
3 = -6 is not true, so no it is not a solution.
### Try This
• Is (3,0) a solution to y = 2x – 6
• 0 = 6 – 6
• 0=0
• Is (-2,5) a solution to y = -3x + 1
• 5 = 7
### Try This
• Is (3,0) a solution to y = 2x – 6 yes
• Is (-2,5) a solution to y = -3x + 1 no
### 2-3-11 Agenda
PA#13:
Pp.394-395
#12-18 even, 20-30 even
|
# Trigonometry Basics
Definition of Trigonometry: Trigonometry considers the properties of angles and certain ratios associated with angles, and applies the knowledge of these properties to the solution of triangles and various other algebraic and geometric problems. Incidentally trigonometry considers also certain time-saving aids in computation such as logarithms, which are generally employed in the solution of triangles. Briefly stated, Trigonometry is the science of angular magnitudes and the art of applying the principles of this science to the solution of problems. The word Trigonometry comes from two Greek words, trigonon = triangle, and metron = measure. The method was originated in the second century B.C. by Hipparchus and other early Greek astronomers in their attempts to solve certain spherical triangles. The term trigonometry was not used until the close of the sixteenth century. Before we get into the basic definitions of Trigonometric Functions, let us look at the basic definition of a function. Definition of Function: When two variables are so related that the value of the one depends upon the value of the other, the one is said to be a function of the other. EXAMPLES: The area of a square is a function of its side. The volume of a sphere is a function of its radius. The velocity of a falling body is a function of the time elapsed since it began to fall. The output of a factory is a function of the number of men employed. In the expression y depends upon x for its value, hence y is a function of x.
Definition of Reciprocal: If the product of two quantities equals unity, each is said to be the reciprocal of the other. For example, if xy = 1, x is the reciprocal of y, and y is the reciprocalof x. 1/2 is the reciprocal of 2, and 2 is the reciprocal of 1/2, for 1/2X2=1. In general, a/b and b/a are reciprocals since a/bxb/a=1 . From xy = 1 it follows that x = 1/y, and y = 1/x, that is, The reciprocal of any quantity is unity divided by that quantity.
Six Trigonometric Functions of an Acute Angle: Let A be any acute angle, B any point on either side of the angle, and ABC the right triangle formed by drawing a perpendicular from B to the other side of the angle. Denote AC, the side adjacent to the angle A, by b (for base), BC, the side opposite the angle A, by a (for altitude), and the hypotenuse AB by h. The three sides of the right triangle form six different ratios, namely, and their reciprocals
Since these ratios depend upon the angle for their values, they are the functions of the angle according to the general definition of a function that we discussed at the beginning of our lesson. Each of these functions has received a special name. The six functions just defined are variously known as the trigonometric, circular, or goniometric functions: trigonometric, because they form the basis of the science of trigonometry; circular, because of their relations to the arc of a circle; goniometric, because of their use in determining angles, from gonia, a Greek word meaning angle. The terms sine of angle A, cosine of angle A, etc., are abbreviated to sin A, cos A, tan A, cosec A , sec A , and cot A. The definitions of the first six trigonometric functions must be thoroughly memorized. The first three are especially important and should be memorized.The remaining three functions may be remembered most readily by the aid of the reciprocal relations, reciprocal relations, Sin A.Cosec A = 1 Cos A.Sec A = 1 Tan A.Cot A=1 It should be noticed that while a, b, and h are lines, the ratio of any two of them is an abstract number; that is, the trigonometric functions are abstract numbers. Also, the expressions sin A cos A, tan A etc., are single symbols which cannot be separated, sin has no meaning except as it is associated with some angle. EXAMPLE: The sides of a right triangle are 3, 4, 5. Find all the trigonometric functions of the angle A opposite the side 4.
Solution: The hypotenuse of the triangle equals 5. Hence, applying the definitions, we have
Basic Identities
Phythagorean Identities Symmetry Properties
Graphs of the Six Trigonometric Functions
Advanced Questions on Trignometry Part - 1
Ques: 1 Prove that First Solution: We will show that Indeed, by the addition and subtraction formulas, we obtain
Second Solution: Note that by the addition and subtraction formula, we have
Hence
and so
that is, Ques: 2 In triangle ABC, show that
, as desired.
Solution: By the extended law of sines, we have
Applying the double-angle formulas and sum-to-product formulas in the above relation gives
by noting that
because
Note: By symmetry, we have analogous formulas and
Ques: 3 Let numbers x. Solution: We need to show that
for
Prove that
for all real
for all real numbers x. Indeed, the left-hand side is equal to
Ques: 4 A circle of radius 1 is randomly placed in a 15Ã-36 rectangleABCD so that the circle lies completely within the rectangle. Compute the probability that the circle will not touch diagonal AC. Note : In order for the circle to lie completely within the rectangle, the center of the circle must lie in a rectangle that is (15-2)Ã-(36-2), or 13Ã-34. The requested probability is equal to the probability that the distance from the circle's center to the diagonalAC is greater than 1, which equals the probability that the distance from a randomly selected point in the 13 Ã- 34 rectangle to each side of triangles ABC and CDA is greater than 1. Let |AB| = 36 and |BC| = 15 (and so |AC| = 39). Drawthree segments that are 1 unit away from each side of triangle ABC and whose endpoints are on the sides. Let E,F, and Gbe the three points of intersection nearest to A,B, and C, respectively, of the three segments. Because the corresponding sides of triangle ABC and EFG are parallel, the two triangles are similar to each other. The desired probability is equal to
Because E is equidistant from sides AB and AC, E lies on the bisector of Similarly, F and G lie on the bisectors of and respectively. Hence lines AE, BF and CG meet I , the in center of triangle ABC.
First Solution: Let Then
and be the feet of the perpendiculars from E and F to segment AB, respectively. It is not difficult to see that Set and By either the double-angle formulas or
Then the half-angle formulas, or and we obtain Second Solution: Set It follows that Hence
or and and
Consequently, Because E lies on the angle bisector of Consequently,
has the same slope as that is, the slope of line AE is and the rest of the solution proceeds like that of the first solution.
Third Solution: Because the corresponding sides of triangles ABC and EFG are parallel, it follows that I is also the incenter of triangle EFG and that the triangles are homothetic (with I as the center). If r is the inradius of triangle ABC, then r âÖ' 1 is the inradius of triangle EFG; that is, the ratio of the similarity between triangles EFG and ABC is Note that Hence the desired probability is
Solving the last equation gives r = 6, and so Ques: 5 Prove that Solution: The equality is equivalent to for all where k is in Z.
or
That is,
which is evident. where k is in Z, such that
Note: More generally, if are real numbers different from then the relation
holds. Ques: 6 Let a, b, c, d be numbers in the interval such that
Prove that Solution: Rewrite the two given equalities as
By squaring the last two equalities and adding them, we obtain and the conclusion follows from the addition formulas. Ques: 7 Express Solution: By the sum-to-product formulas, we have as a monomial.
By the double-angle formulas, we have
Thus
by the sum-to-product formulas. Note: In exactly the sameway, we can showthat if a, b, and c are real numbers with a + b + c = 0, then
In this question, we have Ques: 8 Prove that Solution: We have Thus
and so for all
as desired. Ques: 9 Prove that for all real numbers with and Solution: Expanding both sides, the desired inequality becomes
By the arithmetic-geometric means inequality, we obtain
By the double-angle formulas, we have
and so
and
Combining the last three inequalities gives the the desired result. Ques: 10 In triangle ABC, Prove that
Solution: Without loss of generality, we assume that law of sines and the triangle inequality imply that It follows that gives that that is, as desired. Ques: 11 Let ABC be a triangle. Prove that
We need to prove that so and the inequality
The
Solution: By the addition and subtraction formulas, we have
Because Thus
and so
as desired.
Note: An equivalent form of this equation is:
Ques: 12 Let ABC be a triangle. Prove that
Solution: By the arithmetic-geometric means inequality, we have
from which follows the desired equation. Ques: 13 Let ABC be an acute-angled triangle. Prove that
Solution: Note that because of the condition
all the above expressions are well defined.
The proof of the identity in part (1) is similar to that of Question 11. By the arithmetic-geometric means inequality,
By (1), we have
from which follows equation (2) Note: Indeed, the identity in (1) holds for all angles where k and m are in Z. Ques: 14 Let ABC be a triangle. Prove that with and
Conversely, prove that if are real numbers with then there exists a triangle ABC such that and Solution: If ABC is a right triangle, then without loss of generality, assume that Then and and so implying the desired result. If by then is well defined. Multiplying both sides of the desired identity reduces the desired result to Question 13(1). to
The second claim is true because is a bijective function from the interval Ques: 15 Let ABC be a triangle. Prove that
Conversely, prove that if x, y, z are positive real numbers such that
then there is a triangle ABC such that and Solution: Solving the second given equation as a quadratic in x gives
We make the trigonometric substitution
and
where
Then
Set Because
and
Because implying that Then angles of a triangle. and or where A, B and C are the
If ABC is a triangle, all the above steps can be reversed to obtain the first given identity Ques: 16 Let ABC be a triangle. Prove that
Solution: By Question 2, we have
The arithmetic-geometric means inequality yields
Combining the last two equalities gives part (a). Part (b) then follows from (a) and Question 15. Part (c ) then follows from part (b) by noting that Finally, by (c ) and by the arithmetic- geometric means inequality, we have
implying (d). Again by Question 2, we have
and analogous formulas for means inequality.
and
. Then part (e) follows routinely from the arithmetic-geometric
Note: We present another approach to part (a). Note that It suffices to show that have
are all positive. Let By the arithmetic-geometric means inequality, we
By Question 15, we have
Thus,
Consequently,
establishing (a).
Ques: 17 In triangle ABC, show that
Conversely, if x, y, z are positive real numbers such that acute triangle ABC such that Solution: Parts (c.) and (d) follow immediately from (b) because Thus we show only (a) and (b). (a) Applying the sum-to-product formulas and the fact that
show that there is an
we find that
establishing (a).
(b) By the sum-to-product formulas, we have
because Note that It suffices to show that
or formula
which is evident by the sum-to-product
From the given equality, we have and thus we may set where Because is an increasing function of z, there is at most one nonnegative value c such that the given equality holds. We know that one solution to this equality is where Because we know that Because and we have Therefore, we must have implying that as desired. Thus,
Nevertheless, we present a cool proof of part (d). Consider the system of equations
Using the addition and subtraction formulas, one can easily see that nontrivial solution. Hence the determinant of the system is 0; that is,
is a
as desired. Ques: 18 In triangle ABC, show that
Solution: By the extended law of sines,
establishing (a). By the same token, we have
which is (b). Note that: By the extended law of sines, we obtain
from which (c ) follows. By the law of cosines,
Hence, by the half-angle formulas, we have
where 2s = a + b + c is the perimeter of triangle ABC. It follows that
and the analogous formulas for
and
Hence
by Heron's formula. It follows that
from which (d) follows. Now we prove (e). By the extended law of sines, we have a
Likewise,
and
By (a) and (b), we have
It suffices to show that
which is Question 17(a). Ques: 19 Let s be the semiperimeter of triangle ABC. Prove that
Solution: It is well known that
or
By Question 18 (b) and (d), part (a) follows from by the double-angle formulas. We conclude part (b) from (a) and Question 16 (d). Ques: 20 In triangle ABC, show that
Solution: By the sum-to-product and the double-angle formulas, we have
and
It suffices to show that
or,
which follows from the sum-to-product formulas, and hence (a) is established. Recalling Question 18 (c.), we have
Euler's formula states that triangle ABC. Because we have
where O and I are the circumcenter and incenter of or from which (b) follows.
Note: Relation (âÖ-) also has a geometric interpretation.
As shown in the Figure, let O be the circumcenter, and let be the feet of the perpendiculars from O to sides BC,CA,AB, respectively. (Thus are the midpoints of sides BC,CA,AB, respectively.) Because and triangle AOB is isosceles with we have Likewise, and
It suffices to show that
Note that and Hence Let s denote the semiperimeter of triangleABC.Applying Ptolemy's theorem to cyclic quadrilaterals yields
Adding the above gives
from which our desired result follows. Ques: 21 Let ABC be a triangle. Prove that
Solution: For part (a), if triangle ABC is non-acute, the left-hand side of the inequality is nonpositive, and so the inequality is clearly true. If ABC is acute, then are all positive. To establish (a) and (d), we need only note that the relation between (a) and (d) and Question 17 (d) is similar to that of Question 16 (a) and (b) and Question 15. (Please see the note after the solution of Question 16.) The two inequalities in parts (d) and (e) are equivalent because By (e) and by the arithmetic-geometric means inequality, we have
from which (b) follows. From that (c.) Part (f) follows from (e) and or by application of Cauchy-Schwarz inequality, we can show By (e) and by setting Finally, (g) follows from (b) and the identity we obtain
proved in Question 18(e). Ques: 22 Prove that
for all where k is in Z. Solution: From the triple-angle formulas, we have
for all
where k is in Z.
Ques: 23 [AMC12P 2002] Given that
find n. First Solution: Note that
Hence
It follows that
implying that n = 23. Second Solution: Note that
Hence
implying that n = 23. Ques: 24 [AIME 2003] Let and be points in the coordinate plane. Let ABCDEF be a and and the y The area of the hexagon can be
convex equilateral hexagon such that coordinates of its vertices are distinct elements of the set
written in the form where m and n are positive integers and n is not divisible by the square of any prime. Find Note: Without loss of generality, we assume that b > 0. (Otherwise, we can reflect the hexagon across the y axis.) Let the x coordinates of C,D,E, and F be c, d, e, and f , respectively. Note that the y coordinate of C is not 4, since if it were, the fact |AB| = |BC| would imply that A,B, and C are collinear or that c = 0, implying that ABCDEF is concave. Therefore, F = (f, 4). Since conclude that b = d. Since need to compute and and so Because the y coordinates of B,C, and D are 2, 6, and 10, respectively, and |BC| = |CD|, we Let a denote the side length of the hexagon. Then f < 0. We
Solution:
First Solution: Note that Note that
Apply the law of cosines
in triangle ABF to obtain We have three independent equations in three variables. Hence we can solve this system of equations. The quickest way is to note that
implying that Squaring both sides gives
or
Hence
and so
and
Therefore, and the answer to the problem is 51. Second Solution: Let denote the measure (in degrees) of the standard angle formed by the line AB and and the x axis. Then the standard angle formed by the line AF and the x axis is By considering the y coordinates of B and F, we have and
by the addition and subtraction formulas. Hence Thus, by considering the x coordinates of B and F, we have and It follows that Note: The vertices of the hexagon are
and Ques: 25 Show that one can use a composition of trigonometry buttons such as, and to replace the broken reciprocal button on a calculator. Solution: Because and
for
we have for any x >0,
as desired. It is not difficult to check that Ques: 26 Prove that in a triangle ABC,
will also do the trick.
Solution: From the law of sines and the sum-to-product formulas, we have
as desired. Ques: 27 Let a, b, c be real numbers, all different from âÖ'1 and 1, such that a +b+c = abc. Prove that
Solution: Let where for all integers k. The condition a + b + c = abc translates to tan(x + y + z) = 0, as indicated in notes after Question 13(1). From the double-angle formulas, it follows that
Hence
using a similar argument to the one in Question 13(1). This implies that
and the conclusion follows. Ques: 28 Prove that a triangle ABC is isosceles if and only if
Solution: By the extended law of sines, a = 2R sin A, b = 2R sin B, and c = 2R sin C. The desired identity is equivalent to
or
Because
The last equality simplifies to
which in turn is equivalent to
by Question 7. The conclusion now follows.
Ques: 29 Prove that is an irrational number. Solution: Assume, for the sake of contradiction, that Using the identity
is rational. Then so is
we obtain by strong induction that is rational for all integers for example, is not rational, yielding a contradiction.
But this is clearly false, because,
Note: For the reader not familiar with the idea of induction. We can reason in the following way. Under the assumption that both and are rational, relation (âÖ-) implies that is rational, by setting n = 2 in the relation (âÖ-). Similarly, by the assumption that both and are rational, relation (âÖ-) implies that is rational, by setting n = 4 in the relation (âÖ-). And so on.We conclude that is rational, for all positive integers n, under the assumption that is rational. Ques: 30 Prove that for all Solution: Multiplying the two sides of the inequality by we obtain the equivalent form
But this follows from Cauchy-Schwarz inequality because according to this inequality, the left-hand side is greater than or equal to Ques: 31 If Solution: Note that what are the possible values of ?
Because that
it follows that we conclude that
Similarly, because Combining the above results shows
But we have not shown that indeed, we consider
can obtain all values in the interval
To do this,
Let equation
and
Then If and
and
Consider the range of the sum then x and y are the roots of the quadratic
Thus,
By checking the boundary condition we obtain By checking similar boundary conditions, we conclude that the equation (âÖ-) for all is Because both the sine and cosine is is , and so the
has a pair of solutions x and y with for . Thus, the range of
functions are surjective functions from R to the interval [-1,1], the range of . Thus the range of reange of is Ques: 32 Let a, b, c be real numbers. Prove that
Solution: Let a = tan x, b = tan y, c = tan z with Then Multiplying by and on both sides of the desired inequality gives
Note that
and
Consequently, we obtain
as desired. Ques: 33 Prove that
Solution: If cos x = 0, the desired inequality reduces to assume that Dividing both sides of the desired inequality by
which is clearly true. We gives
Set t = tan x. Then
The above inequality reduces to
or
The last inequality is equivalent to
which is evident. Ques: 34 Prove that
Solution: We proceed by induction on n. The base case holds, because
For the inductive step, in order to prove that
it suffices to show that
for all real numbers inequality becomes we have
Let
For k = 1,2,.....n+1. The last Indeed by the addition and subtraction formulas,
as desired Ques: 35 [Russia 2003, by Nazar Agakhanov] Find all angles α for which the three element set is equal to the set . Solution: The answers are for all intergers k
Because S = T, the sums of the elements in S and T are equal to each other that is,
Applying the sum-to-product formulas to the first and the third summands on each side of the last equation gives or
if to check that It follows that
then , and so for all intergers k. It is then not difficult and both of S and T are not three-elements sets. , implying that ;that is, . The possible It not difficult to
answers are for all intergers k. Because check that all such angles satisfy the conditions of the problem. Ques: 36 Let be the sequence of polynomials such that
for all positive integers i. The polynomial called the nth Chebyshev polynomial. (a) Prove that (b) Prove that (c) Prove that (d) Determine all the roots of (e) Determine all the roots of ; and are odd and even functions,respectively; for real numbers x with x > 1; for all nonegative intergers n;
is
Solution: Parts (a) and (b) are simple facts that will be useful in establishing (e).We present them together. (a). We apply strong induction on n. Note that and are even and odd, respectively. Assume that and are odd and even, respectively. Then is odd, and so is odd. Thus is even, and so is even. This completes our induction. (b). We apply strong induction on n.For for x > 1 and induction hypothesis yields for x > 1. Assume that where k is some nonnegative integer. For n = k + 1, the
completing our induction.
|
# What is the perimeter and area of a right triangle with a hypotenuse of 26 and a leg of 24?
May 23, 2018
$\text{Perimeter} = 60$ and $\text{Area} = 120$
#### Explanation:
Assign the sides $b = 24 \mathmr{and} c = 26$
Use the Pythagorean theorem,
${c}^{2} = {a}^{2} + {b}^{2}$
,to find the length of side a:
$a = \sqrt{{26}^{2} - {24}^{2}}$
$a = 10$
The perimeter is:
$\text{Perimeter} = a + b + c$
$\text{Perimeter} = 10 + 24 + 26$
$\text{Perimeter} = 60$
The area is:
$\text{Area} = \frac{1}{2} b h$
If we select side b as the base, then side a is the height:
$\text{Area} = \frac{1}{2} \left(24\right) \left(10\right)$
$\text{Area} = 120$
|
# ICSE Solutions for Selina Concise Chapter 21 Solids Class 9 Maths
### Exercise 21(A)
1. The length breadth and height of a rectangular solid are in the ratio of 5 : 4 : 2 if the total surface area of 1216 cm². Find the length breadth and height of the solid
Let the length, breadth and height of rectangular solid are 5x, 4x, 2x.
Total surface area = 1216 cm2
2(5x. 4x + 4x.2x + 2x.5x) = 1216
⇒ 20x2 + 8x2 + 10x2 = 608
⇒ 38x2 = 608
⇒ x2 = (608/38) = 16
⇒ x = 4
Therefore, the length, breadth and height of rectangular solid are
5×4 = 20cm, 4×4 = 16cm, 2×4 = 8cm.
2. The volume of a cube is 729 cm3. Find its total surface area.
Let a be the one edge of a cube.
Volume = a3
729 = a3
⇒ 93 = a3
⇒ 9 = a
⇒ a = 9 cm
Total surface area 6a2 = 6×92 = 486 cm2 .
3. The dimensions of a Cinema Hall are 100 m, 60 m and 15 m. How many persons can sit in the hall, if each requires 150 m3 of air?
Volume of cinema hall = 100×60×15 = 90000 m3
⇒ 150m3 requires = 1 person
⇒ 90000m3 requires = (1/150)×90000 = 600 persons
Therefore, 600 persons can sit in the hall.
4. 75 persons can sleep in a room 25 m by 9.6 m. If each persons requires 16 m3 of air; find the height of the room.
Let h be height of the room.
1 person requires 16m3
75 person requires 75×16 m3 = 1200m3
Volume of room is 1200 m3
1200 = 25×9.6×h
⇒ h = 1200/(25×9.6)
⇒ h = 5m
5. The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube.
Volume of melted single cube = 33 + 43 + 53 cm3
= 27 + 64 + 125 cm3
= 216 cm3
Let a be the edge of the new cube.
Volume = 216 cm3
a3 = 216
⇒ a3 = 63
⇒ a = 6cm
Therefore, 6 cm is the edge of cube.
6. Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 12 cm. Find ‘x’.
Volume of melted single cube x3 + 83 + 103 cm3
= x3 + 512 + 1000 cm3
= x3 + 1512 cm3
Given that 12cm is edge of the single cube.
123 = x3 + 1512 cm3
⇒ x3 = 123 - 1512
⇒ x3 = 1728 - 1512
⇒ x3 = 216
⇒ x3 = 63
⇒ x = 6 cm
7. Three equal cubes are placed adjacently in a row. Find the ratio of the total surfaced area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.
Let the side of a cube be 'a' units.
Total surface area of one cube = 6a2
Total surface area of 3 cubes = 3×6a2 = 18a2
after joining 3 cubes in a row, length of Cuboid = 3a
Breadth and height of cuboid = a
Total surface area of cuboid = 2 [3a2 + a2 + 3a2] = 14a2
Ratio of total surface area of cuboid to the total surface area of 3 cubes = 14a2/18a2 = 7/9.
8. The cost of papering the four walls of a room at 75 paisa per square meter Rs. 240. The height of the room is 5 metres. Find the length and the breadth of the room, if they are in the ratio 5 : 3.
Let the length and breadth of the room is 5x and 3x respectively.
Given that the four walls of a room at 75 paise per square met Rs. 240.
Thus,
240 = Area ×0.75
⇒ Area = 240/0.75
⇒ Area = 24000/75
⇒ Area = 320 m
⇒ Area = 2×Height (Length + Breadth)
⇒ 320 = 2×5(5x+ 3x)
⇒ 320 = 10×8x
⇒ 32 = 8x
⇒ x = 4
Length = 5x
= 5(4)m
= 20 m
= 3(4)m
= 12 m .
9. The area of a playground is 3650 m2. Find the cost of covering it with gravel 1.2 cm deep, if the gravel costs Rs. 6.40 per cubic metre.
The area of the playground is 3650 m2 and the gravels are 1.2 cm deep. Therefore the total volume to be covered will be:
3650×0.012 =43.8 m3.
Since the cost of per cubic meter is Rs. 6.40, therefore the total cost will be:
43.8×Rs.6.40 = Rs.280.32
10. A square plate of side ‘x’ cm is 8 mm thick. If its volume is 2880 cm3; find the value of x.
We know that
1mm = 1/10 cm
8mm = 8/10cm
Volume = Base area ×Height
⇒ 2880 cm3 = x × x × 8/10
⇒ 2880 ×(10/8) = x2
⇒ x2 = 3600
⇒ x = 60 cm
11. The external dimensions of a closed wooden box are 27 cm, 19 cm and 11 cm. If the thickness of the wood in the box is 1.5 cm; find:
(i) Volume of the wood in the box;
(ii) The cost of the box, if wood costs Rs. 1.20 per cm3;
(iii) Number of 4 cm cubes that could be placed into the box.
External volume of the box = 27×19×11 cm3 = 5643 cm
Since, external dimensions are 27 cm, 19cm, 11cm; thickness of the wood is 1.5 cm.
∴ Internal dimensions
= (27 - 2×1.5)cm, (19 - 2×1.5)cm, (11-2×1.5)cm
= 24cm, 16cm, 8cm
Hence interval volume of box = (24×16×8)cm3 = 3072 cm3
(i) Volume of wood in the box = 5643 cm3 - 3072 cm3 = 2571 cm3
(ii) Cost of wood = Rs. 1.20×2571 = Rs 3085.2
(iii) Vol. of 4 cm cube = 43 = 64cm3
Number of 4 cm cubes that could be placed into the box
= 3072/64 = 48
12. A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. If it is open at the top. Determine the cost of iron-sheet, at the rate of Rs. 12.50 per metre, if the sheet is 2.5 m wide.
Area of sheet = Surface area of the tank
⇒ Length of the sheet × its width = Area of 4 walls of the tank + Area of its base
⇒ length of the sheet × 2.5m = 2(20+12)×8 m2 + 20×12 m2
⇒ length of the sheet = 300.8 m
Cost of the sheet = 300.8 × Rs. 12.50 = Rs 3760
13. A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.
Let exterior height is h cm. Then interior dimensions are 78-3=75, 19-3=16 and h-3 (subtract two thicknesses of wood). Interior volume = 75 x 16 x (h-3) which must = 15 cu dm
= 15000 cm^3
(1 dm = 10cm, 1 cu dm = 103 cm3).
15000 cm3 = 75×16×(h – 3)
⇒ h -3 = 15000/(75×16) = 12.5cm
⇒ h =15.5cm.
14. The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate:
(i) The side of the cube.
(ii) The total surface area of the cube.
(i) If the sides of the cube = a cm
The length of its diagonal = a√3 cm
And,
(a√3)2 = 1875
⇒ a = 25cm
(ii) Total surface area of the cube = 6a2
= 6(25)2 = 3750 cm2
15. A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm3 of iron in this tube. Find its thickness.
Given that the volume of the iron in the tube 192 cm2
Let the thickness of the tube = x cm
∴ Side of the external square = (5+2x)cm
∵ Ext. vol. of the tube - its internal vol. = volume of iron in the tube, we have,
(5+2x)(5+2x) × 8-5×5×8 = 192
⇒ (25+4x2 + 20x)×8 - 200 = 192
⇒ 200 + 32x2 + 160x - 200 = 192
⇒ 32x2 + 160x - 192 = 0
⇒ x2 + 5x - 6 = 0
⇒ x2 +6x- x- 6 = 0
⇒ x(x+6)-(x+6) =0
⇒ (x+6)(x-1) = 0
⇒ x- 1= 0
⇒ x = 1
therefore, thickness is 1 cm.
16. Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid as 648 m2; find the length of edge of each cube.
Also, find the ratio between the surface area of resulting cuboid and the surface area of a cube.
Let L be the length of the edge of each cube.
The length of the resulting cuboid = 4×L = 4L cm
Let width(b) = L cm and its height(h) = L cm
∴ The total surface area of the resulting cuboid
= 2(l× b + b× h + h× l)
⇒ 648 = 2(4l × l + l× l + l ×4l)
⇒ 4l2 + l2 + 4l2 = 324
⇒ 9l2 = 324
⇒ l2 = 36
⇒ l = 6 cm
Therefore , the length of each cube is 6 cm.
(Surface area of the resulting cuboid)/(Surface area of cube) = 648/6l2
(Surface area of the resulting cuboid)/(Surface area of cube)= 648/6(6)2
(Surface area of the resulting cuboid)/(Surface area of cube) = 648/216 = 3/1 = 3: 1
### Exercise 21(B)
1. The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres.
Assume that all angles in the figures are right angles.
The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, 3 cm, and 9 cm respectively. Hence, volume of solid
= 9×4×3 + 6×4×3
= 108+72
= 180 cm3
2. A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.
Area of cross section of the solid = (1/2)(1.5 +3)×(40) cm2
= (1/2)(4.5)×40 cm2
= 90 cm2
Volume of solid = Area of cross section × length
= 90 × 15cm3
= 1350 cm3
= 1350000 liters [since 1cm3 = 1000 lt.]
3. The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:
AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:
(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m2 (sq. metre).
(ii) The cost of paving the floor at the rate of Rs. 18 per m2.
The cross section of a tunnel is of the trapezium shaped ABCD in which AB = 7m, CD = 5m and AM = BN. The height is 2.4 m and its length is 40m.
(i) AM = BN = (7-5)/2 =2/2 = 1 m
AD2 = AM2 + DM2 [Using pythagoras theorem]
= 12 + (2.4)2
= 1 + 5.76
= 6.76
= (2.6)2
Perimeter of the cross - section of the tunnel = (7+2.6 + 2.6 + 5)m = 17.2m
Length = 40 m
∴ Internal surface area of the tunnel (except floor)
= (17.2×40 - 40×7) m2
= (688 - 280)m2
= 408 m2
Rate of painting = Rs 5 per m2
Hence, total cost of painting = Rs 5×408 = Rs 2040
(ii) Area of floor of tunnel l×b = 40×7 = 280 m2
Rate of cost of paving = Rs 18 per m2
Total cost = 280×18 = Rs 5040
4. Water is discharged from a pipe of cross-section area 3.2 cm2 at the speed of 5m/s. Calculate the volume of water discharged:
(i) In cm3 per sec.
(ii) In litres per minute.
(i) The rate of speed =5 m/s = 500 cm/s
Volume of water flowing per sec = 3.2×500 cm3 = 1600 cm3
(ii) Vol. of water flowing per min = 1600×60 cm3 = 96000 cm3
Since 1000 cm3 = 1lt
Therefore, Vol. of water flowing per min = 96000/1000 = 96 litres
5. A hose-pipe of cross-section area 2 cm2 delivers 1500 litres of water in 5 minutes. What is the speed of water in m/s through the pipe?
Vol. of water flowing in 1sec = (1500×1000)/5×60 = 5000 cm3
Vol. of water flowing = area of cross section × speed of water
= 5000 cm3/s = 2 cm2 × speed of water
⇒ speed of water = 5000/2 cm/s
⇒ speed of water = 2500 cm/s
⇒ speed of water = 25 m/s
6. The cross-section of a piece of metal 4 m in length is shown below. Calculate:
(i) The area of the cross-section;
(ii) The volume of the piece of metal in cubic centimetres.
If 1 cubic centimetre of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg.
(i) Area of total cross section = Area of rectangle abce + area of △def
= (12×10) + (1/2)(16-10)(12-7.5)
= 120 + (1/2)×6×4.5 cm2
= 120 + 13.5 cm2
= 133.5 cm2
(ii) The volume of the piece of metal in cubic centimeters = Area of total cross section × length
= 133.5 cm2 × 400 cm = 53400 cm3
1 cubic centimetre of the metal weighs 6.6g
53400 cm3 of the metal weighs 6.6× 53400 g = (6.6×53400)/1000 kg
= 352.440 kg
The weight of the piece of metal to the nearest Kg is 352 kg.
7. A rectangular water-tank measuring 80 cm 60 cm 60 cm is filled form a pipe of cross-sectional area 1.5 cm2, the water emerging at 3.2 m/s. How long does it take to fill the tank?
Vol. of rectangular tank = 80×60×60 cm3 = 288000 cm3
One liter = 1000 cm3
Vol. of water flowing in per sec =
1.5 cm2 ×3.2 m/s = 1.5 cm2 × (3.2 ×100)cm/s
= 480 cm3/s
Vol. of water flowing in 1 min = 480×60 = 28800 cm3
Hence,
28800 cm3 can be filled = 1 min
⇒ 28800 cm3 can be filled = [(1/28800)×288000] min = 10 min
8. A rectangular card-board sheet has length 32 cm and breadth 26 cm. Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed.
Length of sheet = 32 cm
Breadth of sheet = 26 cm
Side of each square = 3cm
∴ Inner length = 32 - 2×3 = 32 - 6 = 26 cm
Inner breadth = 26 - 2×3 = 26 -6 = 20 cm
By folding the sheet, the length of the container = 26 cm
Breadth of the container = 20 cm and height of the container = 3cm
∴ Vol. of the container = l ×b×h
= 26cm ×20cm×3cm = 1560 cm3
9. A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1.2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly.
Length of pool = 18 m
Breadth of pool = 8 m
Height of one side = 2m
Height of second side = 1.2 m
∴ Volume of pool = 18×8×(2+1.2)/2 m3
= (18×8×3.2)/2
= 230.4 m3
10. The following figure shows a closed victory-stand whose dimensions are given in cm.
Find the volume and the surface are of the victory stand.
Consider the box 1
Thus, the dimensions of box 1 are: 60cm, 40cm and 30 cm.
Therefore, the volume of box 1 = 60×40×30 = 72000 cm3
Surface area of box 1 = 2 (lb + bh + lh)
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 1 = 40 ×40 + 40×30 + 40×30+ 2(60×30)
= 1600 + 1200 +1200 +3600
= 7600 cm2
Consider the box 2
Thus, the dimensions of box 2 are : 40 cm , 30 cm and 30 cm.
Therefore, the volume of box 2 = 2(lb + bh + lh)
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 2 = 40×3 + 40×30 + 2(30×30)
= 1200 + 1200 + 1800
= 4200 cm2
Consider the box 3
Thus, the dimensions of box 2 are : 40 cm , 30 cm and 20 cm .
Therefore, the volume of box 3 = 40×30×20 = 24000 cm3
Surface area of box 3 = 2(lb + bh + lh)
Since the box is open at the bottom and from the given figure, we have
Surface area of box 3 = 40×30 + 40×20 + 2(30×20)
= 1200 + 800 + 1200
= 3200 cm2
Total volume of the box = volume of box 1 + volume of box 2 + volume of box 3
= 72000 + 36000+ 24000
= 132000 cm3
Similarly, total surface area of the box
= surface area of box 1 + surface area of box 2 + surface area of box 3
= 7600 +4200 + 3200
= 15000 cm3
### Exercise 21 (C)
1. Each face of a cube has perimeter equal to 32 cm. Find its surface area and its volume.
The perimeter of a cube formula is, Perimeter = 4a where, a = length.
Given that perimeter of the face of the cube is 32 cm
⇒ 4a = 32cm
⇒ a = 32/4
⇒ a = 8cm
We know that surface area of a cube with side 'a' = 6a2
Thus, Surface area = 6×82 = 6×64 = 384 cm2
We know that the volume of a cube with side 'a' = a3
Thus, volume = 83 = 512 cm3
2. A school auditorium is 40 m long, 30 m broad and 12 m high. If each student requires 1.2 m2 of the floor area; find the maximum number of students that can be accommodated in this auditorium. Also, find the volume of air available in the auditorium, for each student.
Given dimensions of the auditorium are: 40m × 30m ×12 m
The area of the floor = 40×30
Also given that each student requires 1.2 m2 of the floor area.
Thus, Maximum number of students = (40×30)/1.2 = 1000
Volume of the auditorium
= 40×30×12 m3
= Volume of air available for 1000 students
Therefore, Air available for each student = (40×30×12)/1000 m3 = 14.4 m3 .
3. The internal dimensions of a rectangular box are 12 cm x cm 9 cm. If the length of the longest rod that can be placed in this box is 17 cm; find x.
Length of longest rod = Length of the diagonal of the box
⇒ 172 = 122 + x2 + 92
⇒ x2 = 172 - 122 - 92
⇒ x2 = 289 - 144 - 81
⇒ x2 = 64
⇒ x = 8 cm
4. The internal length, breadth and height of a box are 30 cm, 24 cm, and 15 cm. Find the largest number of cubes which can be placed inside this box if the edge of each cube is
(i) 3 cm (ii) 4 cm (iii) 5 cm
(i) No. of cube which can be placed along length = 30/3 = 10
No. of cube along the breadth = 24/3 = 8
No. of cubes along the height = 15/3 = 5.
∴ The total no. of cubes placed = 10×8×5 = 400
(ii) Cubes along length = 30/4 = 7.5 = 7
Cubes along width = 24/4 = 6 and cubes along height = 15/4 = 3.75 = 3
∴ The total no. of cubes placed = 7×6×3 = 126
(iii) Cubes along length = 30/5 = 6
Cubes along width = 24/5 = 4.5 = 4 and cubes along height = 15/5 = 3
∴ The total no. of cubes placed = 6×4×3 = 72
5. A rectangular field is 112 m long and 62 m broad. A cubical tank of edge 6 m is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.
Vol. of the tank = vol. of earth spread
4×63 m3 = (112×62 -4×62) m2 × Rise in level
⇒ Rise in level = (4×63)/(112×62 - 4×62)
= 864/6800
= 0.127 m
= 12.7cm
6. When length of each side of a cube is increased by 3 cm, its volume is increased by 2457 cm3. Find its side. How much will its volume decrease, if length of each side of it is reduced by 20%?
Let a be the side of the cube.
Side of the new cube = a+ 3
Volume of the new cube = a3 + 2457
That is, (a+3)2 = a3 + 2457
⇒ a3 + 3×a×3(a+3) + 33 = a3 + 2457
⇒ 9a3 + 27a + 27 = 2457
⇒ 9a3 + 27a - 2430 = 0
⇒ a3 + 3a -270 = 0
⇒ a3 + 18a - 15a - 270 = 0
⇒ a(a+18) - 15(a+18) = 0
⇒ (a- 15)(a + 18) = 0
⇒ a -15 = 0 or a+18 = 0
⇒ a = 15 or a = -18
⇒ a = 15 cm [since side cannot be negative]
Volume of the cube whose side is 15 cm = 153 = 3375 cm3
Suppose the length of the given cube is reduced by 20%
Thus new side anew = a- ( 20/100)×a
= a[1-(1/5)]
= (4/5)×15
= 12 cm
Volume of the new cube whose side is 12 cm = 123 = 1728 cm3
Decrease in volume = 3375 - 1728 = 1647 cm3
7. A rectangular tank 30 cm × 20 cm × 12 cm contains water to a depth of 6 cm. A metal cube of side 10 cm is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in litres, that must be poured in the tank so that the metal cube is just submerged in the water.
The dimensions of rectangular tank = 30 cm× 20cm ×12cm
Side of the cube = 10cm
volume of the cube = 103 = 1000 cm3
The height of the water in the tank is 6 cm.
Volume of the cube till 6 cm = 10×10×6 = 600 cm3
Hence when the cube is placed in the tank,
then the volume of the water increase by 600 cm3 .
The surface area of the water level is 30 cm×20 cm = 600 cm2
Out of this area, let us subtract the surface area of the cube.
Thus, the surface area of the shaded part in the above figure is 500 cm2
The displaced water is spread out in 500 cm2 to a height of 'h' cm.
And hence the volume of the water displaced is equal to the volume of the part of the cube in water.
Thus, we have ,
500 × h = 600 cm3
⇒ h = 600/500 cm
⇒ h = 1.2 cm
Thus, now the level of the water in the tank is = 6+1.2 = 7.2 cm
Remaining height of the water level, so that the metal cube is just
submerged in the water = 10- 7.2 = 2.8 cm
Thus the volume of the water that must be poured in the tank so that the metal
cube is just submerged in the water = 2.8 ×500 = 1400 cm3
we know that 1000 cc = 1 litre
Thus, the required volume of water = 1400/1000 = 1.4 litres.
8. The dimensions of a solid metallic cuboid are 72 cm × 30 cm × 75 cm. It is melted and recast into identical solid metal cubes with each of edge 6 cm. Find the number of cubes formed.
Also, find the cost of polishing the surfaces of all the cubes formed at the rate Rs. 150 per sq. m.
The dimensions of a solid cuboid are: 72 cm, 30cm, 75cm
Volume of the cuboid = 72cm ×30 cm ×75 cm = 162000 cm3
Side of a cube = 6 cm
Volume of a cube = 63 = 216 cm3
The number of cubes = 162000/216 = 750
The surface area of a cube = 6a2 = 6×62 = 216 cm2
Total surface area of 750 cubes = 750 ×216 = 162000 cm2
Total surface area in square metres = 162000/10000
= 16.2 square metres
Rate of polishing the surface per square metre = Rs. 150
Total cost of polishing the surface = 150×16.2 = Rs. 2430
9. The dimensions of a car petrol tank are 50 cm × 32 cm × 24 cm, which is full of petrol. If car’s average consumption is 15 km per litre, find the maximum distance that can be covered by the car.
The dimensions of a car petrol tank are = 50cm × 32 cm ×24 cm
Volume of the tank = 38400 cm3
We know that 1000 cm3 = 1 litre
Thus volume of the tank = 38400/1000 = 38.4 litres
The average consumption of the car = 15 km/litres
Thus, the total distance that can be covered by the car = 38.4×15 = 576 km.
10. The dimensions of a rectangular box are in the ratio 4 : 2 : 3. The difference between cost of covering it with paper at Rs. 12 per m2 and with paper at the rate of 13.50 per m2 is Rs. 1,248. Find the dimensions of the box.
Given dimensions of a rectangular box are in the ratio 4:2:3
Therefore, the total surface area of the box
= 2[4x ×2x + 2x ×3x + 4x ×3x]
= 2[8x2 + 6x2 + 12x2 ] m2
Difference between cost of covering the box with paper at Rs. 12 per m2 and with paper at Rs. 13.50 per m2 = Rs. 1,248
⇒ 52x2 [13.5 - 12] = 1248
⇒ 52× x2 ×1.5 = 1248
⇒ 78× x2 = 1248
⇒ x2 = 1248/78
⇒ x2 = 16
⇒ x = 4 [Length, width and height cannot be negative]
Thus, the dimensions of the rectangular box are = 4×4 m, 2×4m, 3×4 m
Thus, the dimensions are 16m, 8m and 12 m .
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How Many Miles is 10 Acres of Land?
Some might think this is a weird thing to be wondering about, but you have come to just the right place to get your answers. Because we think this is a very valid and interesting question. Not only that, but we also have a few fun examples to help you understand the vast acreage and size you are thinking about.
An acre is a unit of area equivalent to 43,560 square feet. So, 10 acres of land is 435,600 square feet. But in some parts of the world, including the United States, miles are used to measure distance. So how many miles is 10 acres of land?
10 Acres of Land in Miles
Well, first of all, miles is a unit of distance, whereas acres is a unit of area. To calculate an area in miles, the unit we use is square miles (like square meters or square feet).
One square mile is equal to 640 acres. So 10 acres is equivalent to 0.015625 square miles or 1/64th of a square mile.
Visualizing 10 Acres of Land
Since the numbers are not easy to remember, you must be having a hard time estimating just how big of land we are talking about here. Understanding that, we have come up with a few fun examples to help you visualize the entire 10 acres of land.
1/15 of the Largest Office Building
The Pentagon is the largest office building in the world, standing over a mind-blowing 6,500,000 square feet, which is 150 acres of land. If we were to divide it into 15 equal sections, each room would cover an area of 10 acres.
8 American Football Fields
If you are an avid football fan like us, you would know just how big a football field is. Including the end zones, a football field is 57,600 square feet or 1.32 acres of land.
Now, imagine a football field being just a little over an acre of land. How many football fields would cover an area of 10 acres?
Yes, almost 8 American football fields lying down from end zone to end zone would cover an area of 10 acres. Let that sink in; that’s how big a 10 acres piece of land would be.
155 Double Tennis Courts
Tennis courts are used for both single and double play, but we will consider doubles courts since they are comparatively larger, covering 2808 square feet in area. Taking these measurements, try visualizing 155 double tennis courts side by side. That is how big of an area we are talking about in terms of 10 acres or 1/64 square miles.
1360 Shipping Containers
Shipping containers most often come in different sizes, but here, we will consider the most commonly used containers that are about 8 feet in width and 40 feet in length. Hence, each one of these containers takes up 0.007 acres of land. Now imagine 1360 of these shipping containers lined on a piece of land. That’s 10 acres for you!
Two to Three Walmart Supercenters
If you have ever been inside a Walmart supercenter, you must already have an idea about how huge and spacious they are. Each one takes up an average of 4 acres of land. So, on the piece of land that we are talking about here, one would be able to build about 2.5 such Walmart supercenters. Just imagine!
30 Pontoon Boats
When it comes to pontoon boats, the average size is 22 feet. Taking that into consideration, it is safe to assume that we can line up 30 electric pontoon boats end to end on 10 acres of land.
15,800 Potatoes
You must already have a pretty clear idea of the vastness of 10 acres of land by now, but we couldn’t resist throwing in this fun example for you to visualize.
An average sack of potatoes we get is 5 inches in length, so some quick math tells us that we can line up 15,840 potatoes on 10 acres of land. Some potato party that would be!
Wrap Up
Whether you were thinking about buying 10 acres of land or whether this was a past-bedtime random thought that crept in and wouldn’t let you sleep peacefully until you did some research, we hope this article provided you with some great information. Not to forget the fun examples that you can quote and show off at the next party or business meeting.
Warwick Braith
Warwick Braith is a thrill seeker at heart. He loves getting outdoors and testing his limits in the wild. As a blogger for YapQ, Warwick provides readers with insights and tips on how to get the most out of their outdoor experiences. Whether it's hiking, camping, or simply exploring nature, Warwick knows how to make the most of it.
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52
Requisite background: high school level programming and calculus. Explanation of backprop is included, skim it if you know it. This was originally written as the first half of a post on organizational scaling, but can be read standalone.
Backpropagation
If you’ve taken a calculus class, you’ve probably differentiated functions like . But if you want to do math on a computer (for e.g. machine learning) then you’ll need to differentiate functions like
function f(x): # Just wait, the variable names get worse… a = x^(1/2) # Step 1 b = 1–1/(a+1) # Step 2 c = ln(b) # Step 3 return c
This is a toy example, but imagine how gnarly this could get if contains loops or recursion! How can we differentiate , without having to write it all out in one line? More specifically, given the code to compute , how can we write code to compute ? The answer is backpropagation. We step through the function, differentiating it line-by-line, in reverse. At each line, we compute the derivative with respect to one of the internal variables. As we work back, we differentiate with respect to earlier and earlier variables until we reach x itself.
The key to backprop is the chain rule. Look the bottom left arrow: . That’s just the chain rule! A similar formula applies to each arrow in the bottom row.
Let’s work through the example. Keep an eye on the diagram above to see where we are as we go. First, we differentiate the last line:
return c: df_dc = 1
Then, we work backward:
Step 3: c = ln(b) -> df_db = df_dc * dc_db = df_dc ∗ (1/b)
Notice what happened here: we used the chain rule, and one of the two pieces to come out was — the result of the previous step. The other piece is the derivative of this particular line. Continuing:
Step 2: b = 1–1/(a+1) -> df_da = df_db * db_da = df_db ∗ (1/(a+1)^2) Step 1: a = x^(1/2) -> df_dx = df_da * da_dx = df_da ∗ (1/2*x^(−1/2))
Each step takes the result of the previous step and multiplies it by the derivative of the current line. Now, we can assemble it all together into one function:
function df_dx(x): a = x^(1/2) # Step 1 b = 1–1/(a+1) # Step 2 c = ln(b) # Step 3 df_dc = 1 df_db = df_dc * 1/b # df_dc times derivative of step 3 df_da = df_db * (1/(a+1)^2) # df_db times derivative of step 2 df_dx = df_da * (1/2*x^(-1/2)) # df_da times derivative of step 1 return df_dx
The name “backpropagation” derives from the df_d* terms, which “propagate backwards” through the original function. These terms give us the derivative with respect to each internal variable.¹ In the next section, we’ll relate this to intermediate prices in supply chains.
Price Theory
Suppose we have a bunch of oversimplified profit-maximizing companies, each of which produces one piece in the supply chain for tupperware. So, for instance:
1. Company A produces ethylene from natural gas
2. Company B produces plastic (polyethylene) from ethylene
3. Company C molds polyethylene into tupperware
We’ll give each company a weird made-up production function:
1. Company A can produce units of ethylene from units of natgas
2. Company B can produce units of polyethylene from units of ethylene
3. Company C can produce units of tupperware from units of polyethylene
You may notice that these weird made-up cost functions look suspiciously similar to steps 1, 2 and 3 in our function from the previous section. Indeed, from the previous section tells us how much tupperware can be made from a given amount of natgas: we compute how much ethylene can be made from the natgas (step 1, company A), then how much polyethylene can be made from the ethylene (step2, company B), then how much tupperware can be made from the polyethylene (step 3, company C).
Each company wants to maximize profit. If company 3 produces units of tupperware (at unit price ) from units of polyethylene (unit price ), then their profit is : value of the tupperware minus value of the polyethylene. In order to maximize that profit, we set the derivative to zero, then mutter something about KKT and pretend to remember what that means:
Company C: ddb[Pc∗c(b)—Pb∗b]=0→Pb=Pcdcdb=Pc∗1b
We’ve assumed competitive markets here: no single company is large enough to change prices significantly, so they all take prices as fixed when maximizing profit. Then, at a whole-industry-level, the above formula lets us compute the price of polyethylene in terms of the price of tupperware.
Compute prices just like derivatives.
Well, now we can work back up the supply chain. Maximize profit for company B, then company A:
Company B: dda[Pb∗b(a)—Pa∗a]=0→Pa=Pb∗(1(a+1)2) Company A: ddx[Pa∗a(x)—Px∗x]=0→Px=Pa∗(12x−12)
Notice that these formulas are exactly the same as the formulas we used to compute in the previous section. Just replace by , by , etc — the price of the intermediate good is just the derivative of the production function with respect to that good. (Actually, the price is proportional to the derivative, but it’s equal if we set the price of tupperware to 1 — i.e. price things in tupperware rather than dollars.)
So the math is the same, but how physically realistic is this picture? So far we've assumed that the market is at equilibrium, which makes it tricky to assign causality. The causality of backprop does apply directly if the final output (tupperware in our example) is perfectly elastic, meaning that it always has the exact same price. But more generally, once we look at out-of-equilibrium markets, the behavior isn't just distributed backpropagation - it's an unusual variant of gradient descent. It's using local information to move in a general "downhill" direction.
To sum it all up: we can think of profit-maximizing firms in a competitive market at equilibrium as a distributed backpropagation algorithm. Each firm “backpropagates” price information from their output prices to their input prices. This isn't always a perfect reflection of physical causality, but it does match the math for markets at equilibrium.
Footnotes
¹In ML we usually have multiple inputs, in which case we compute the gradient. Other than single-variable, I’ve also simplified the example by only reading each variable in one line, strictly sequentially — otherwise we sometimes need to update the derivatives in place. All of this also carries over to price theory, for supply chains with multiple inputs which can be used in multiple ways.
[UPDATE Dec 2019: removed a couple comments toward the end about a future post which I never got around to writing, and replaced them with a few sentences on out-of-equilibrium markets. Someday I will learn not to promise posts which I have not yet written.]
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There are two separate lenses through which I view the idea of competitive markets as backpropagation.
First, it's an example of the real meat of economics. Many people - including economists - think of economics as studying human markets and exchange. But the theory of economics is, to a large extent, general theory of distributed optimization. When we understand on a gut level that "price = derivative", and markets are just implementing backprop, it makes a lot more sense that things like markets would show up in other fields - e.g. AI or biology.
Second, competitive markets as backpropagation is a great example of the power of viewing the world in terms of DAGs (or circuits). I first got into the habit of thinking about computation in terms of DAGs/circuits when playing with generalizations of backpropagation, and the equivalence of markets and backprop was one of the first interesting results I stumbled on. Since then, I've relied more and more on DAGs (with symmetry) as my go-to model of general computation.
The most interesting dangling thread on this post is, of course, the extension to out-of-equilibrium markets. I had originally promised a post on that topic, but I no longer intend to ever write that post: I have generally decided to avoid publishing anything at all related to novel optimization algorithms. I doubt that simulating markets would be a game-changer for optimization in AI training, but it's still a good idea to declare optimization off-limits in general, so that absence of an obvious publication never becomes a hint that there's something interesting to be found.
A more subtle dangling thread is the generalization of the idea to more complex supply chains, in which one intermediate can be used in multiple ways. Although I've left the original footnote in the article, I no longer think I know how to handle non-tree-shaped cases. I am quite confident that there is a way to integrate this into the model, since it must hold for "price=derivative" to hold, and I'm sure the economists would have noticed by now if "price=derivative" failed in nontrivial supply chain models. That said, it's nontrivial to integrate into the model: if a program reads the same value in two places, then both places read the same value, whereas if two processes consume the same supply-chain intermediate, then they will typically consume different amounts which add up to the amount produced. Translating a program into a supply chain, or vice versa, thus presumably requires some nontrivial transformations in general.
People sometimes make grand claims like "system X is basically just Y" (10-sec off-the-cuff examples: "the human brain is just a predictive processing engine", "evolution is just an optimisation algorithm", "community interaction is just a stag hunt", "economies in disequilibrium is just gradient descent"...)
When I first saw you mention this in a comment on one my posts, I felt a similar suspicion as I usually do toward such claims.
So, to preserve high standards in an epistemic community, I think it's very valuable to actually write up the concrete, gears-like models behind claims like that -- as well as reviewing those derivations at a later time to see if they actually support the bold claim.
___
Separately, I think it's good to write-up things which seem like "someone must have thought of this" -- because academia might have a "simplicity tax", where people avoiding clearly explaining such things since it might make them look naïve.
___
For the record, I have not engaged a lot with this post personally, and it has not affected my own thinking in a substantial way.
• Not assigning prices out of the gate makes it more obvious that this could be used for things like energy/resource consumption directly. Lately this has been of interest to me.
• Do non-equilibrium markets as gradient descent imply that this method could be used to determine in what way the market is out of equilibrium? If the math sticks well, a gradient can contain a lot more information than the lower/higher price expectation I usually see.
• It seems like applying this same logic inside the firm should be possible, and so if we use price and apply recursively we could get a fairly detailed picture; I don't see any reason the gradient trick wouldn't apply to the labor market as well, for example.
The original piece continues where this post leaves off to discuss how this logic applies inside the firm. The main takeaway there is that most firms do not have competitive internal resource markets, so each part of the company usually optimizes for some imperfect metric. The better those metrics approximate profit in competitive markets, the closer the company comes to maximizing overall profit. This model is harder to quantify, but we can predict that e.g. deep production pipelines will be less efficient than broad pipelines.
I'm still writing the piece on non-equilibrium markets. The information we get on how the market is out of equilibrium is rather odd, and doesn't neatly map to any other algorithm I know. The closest analogue would be message-passing algorithms for updating a Bayes net when new data comes in, but that analogy is more aesthetic than formal.
The information we get on how the market is out of equilibrium is rather odd, and doesn't neatly map to any other algorithm I know
I don't want to put the cart before the horse or anything, but this increases my expectation that the information is valuable, rather than decreases it.
Neat! Is this connection well known? It sounds simple enough that someone probably thought of this before.
"Price = derivative" is certainly well-known. I haven't seen anyone else extend the connection to backprop before, but there's no way I'm first person to think of it.
When the goods are contingent and a market maker is used to construct the price formation process there is a line of work stemming from https://arxiv.org/abs/1003.0034 on this ( http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.296.6481&rep=rep1&type=pdf , https://papers.nips.cc/paper/4529-interpreting-prediction-markets-a-stochastic-approach.pdf )
Thanks, these are great!
Seconding Jacob.
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# Calculating the area of compound rectilinear shapes
Introduced in the Year 5 curriculum as: "Calculate and compare the area of rectangles (including squares), and including using standard units, square centimetres (cm2 ) and square metres (m2 ) and the area of irregular shapes"
The area of a shape is how many square cm are inside.
A compound shape is 2 smaller shapes joined together.
To work out the total area, split the shape into the smaller shapes, work out the area of each, then add them together.
Example 1:
Find the area of the two following shapes:
First, we need to split the shapes into smaller shapes.
To find the area of
this shape by multiplying 2 by 7.
2cm x 7cm = 14²
We must first find the correct
length of this shape as it is
not the full 8cm.
8cm - 2cm = 6cm
6cm x 4cm = 24²
The area of the two shapes is
and 2 x 7 = 14cm²
4cm x 6cm = 24cm²
Finally, we need to add the area of each shape together to find the total area:
24cm² + 14cm² = 38cm²
Example 2:
Area = 4 x 10 + 4 x 3
= 52cm²
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An effective method to solve ODE in terms of power series with regular singular point
• Jul 19th 2011, 04:33 AM
Fractalus
An effective method to solve ODE in terms of power series with regular singular point
The goal of this post is to see if what I did is correct for an exercice but also to show an effective method to use when you need to find power series of an ODE with regular singular point. The method is probably well known but I think that having a detailed example of it can help other students.
$\displaystyle \underline{QUESTION}$: Find around the origin, the solutions in power series the following equation.
$\displaystyle 4xy'' + 3y' + 3y = 0$
$\displaystyle \underline{SOLUTION}$:
$\displaystyle \underline{Step 1}$: Find the singular points
$\displaystyle P(x)= 4x, Q(x)=3, R(x)=3$
The only x such that $\displaystyle P(x)=0$ is $\displaystyle x=0$. Therefore, $\displaystyle x=0$ is the only singular point and all other real points are ordinary points.
$\displaystyle \underline{Step 2}$: Determine if the singular points are regular or irregular
$\displaystyle \lim_{x\to 0} xp(x)=\lim_{x\to 0} x\frac{Q(x)}{P(x)} = \lim_{x\to 0} x\frac{3}{4x}= \frac{3}{4}$ which exists.
$\displaystyle \lim_{x\to 0} x^2p(x)=\lim_{x\to 0} x^2\frac{R(x)}{P(x)} = \lim_{x\to 0} x^2\frac{3}{4x}= \lim_{x\to 0} x\frac{3}{4} = 0$ which exists.
Therefore, $\displaystyle x=0$ is a regular singular point.
$\displaystyle \underline{Step 3: Find \sum_{n=0}^\infty p_nx^n and \sum_{n=0}^\infty q_n x^n}$
Because $\displaystyle x = 0$ is a regular point, this means that $\displaystyle xQ(x)/P(x) = xp(x)$ and $\displaystyle x^2R(x)/P(x) = x^2q(x)$ have infinite limits when $\displaystyle x \to 0$and are analytics in $\displaystyle x = 0$. Therefore, they have a converging power series expansion of the form:
$\displaystyle xp(x) = \sum_{n=0}^\infty p_nx^n, x^2q(x)= \sum_{n=0}^\infty q_n x^n$
in a neighbourhoud \$\displaystyle \mid x \mid < \rho$ around the origin, where $\displaystyle \rho > 0$
In this example,
$\displaystyle xp(x) = 3/4$ and $\displaystyle x^2q(x) = \frac{3}{4}t$ which means that
$\displaystyle p_0 = \frac{3}{4}, p_1 = \cdots = p_n = \cdots = 0$ and
$\displaystyle q_1 = \frac{3}{4}, q_0 = q_2 = \cdots = q_n = \cdots = 0$
$\displaystyle \underline{Step 4}$: : Find the roots of $\displaystyle F(r)$.
$\displaystyle F(r) = r(r-1) + p_0r + q_0 = 0$
Here, we have $\displaystyle F(r) = r(r-1) + \frac{3}{4}r = r^2 - \frac{1}{4}r = r(r - \frac{1}{4}) = 0$. Therefore, $\displaystyle r_1 = 0$ and $\displaystyle r_2 = \frac{1}{4}$.
$\displaystyle \underline{Step 5}$: Find the $\displaystyle F(r+n)$ for the first root and the second root.
We first have to compute $\displaystyle F(r+n)$ for some n (let say five), and for $\displaystyle r = 0$. We are going to do the same thing for $\displaystyle r = \frac{1}{4}$.
For $\displaystyle \underline{r = 0}$:
$\displaystyle F(0+1) = F(1) = 1(1-1) + \frac{3}{4}(1) = \frac{3}{4}$
$\displaystyle F(0+2) = F(2) = 2(2-1) + \frac{3}{4}(2) = \frac{7}{2}$
$\displaystyle F(0+3) = F(3) = 3(3-1) + \frac{3}{4}(3) = \frac{33}{4}$
$\displaystyle F(0+4) = F(4) = 4(4-1) + \frac{3}{4}(4) = 15$
$\displaystyle F(0+5) = F(5) = 5(5-1) + \frac{3}{4}(5) = \frac{95}{4}$
For $\displaystyle \underline{r = \frac{1}{4}}$:
$\displaystyle F(\frac{1}{4}+1) = F(\frac{5}{4}) = \frac{5}{4}(\frac{5}{4}-1) + \frac{3}{4}(\frac{5}{4}) = \frac{5}{4}$
$\displaystyle F(\frac{1}{4}+2) = F(\frac{9}{4}) = \frac{9}{4}(\frac{9}{4}-1) + \frac{3}{4}(\frac{9}{4}) = \frac{9}{2}$
$\displaystyle F(\frac{1}{4}+3) = F(\frac{13}{4}) = \frac{13}{4}(\frac{13}{4}-1) + \frac{3}{4}(\frac{13}{4}) = \frac{39}{4}$
$\displaystyle F(\frac{1}{4}+4) = F(\frac{17}{4}) = \frac{17}{4}(\frac{17}{4}-1) + \frac{3}{4}(\frac{17}{4}) = 17$
$\displaystyle F(\frac{1}{4}+5) = F(\frac{21}{4}) = \frac{21}{4}(\frac{21}{4}-1) + \frac{3}{4}(\frac{21}{4}) = \frac{105}{4}$
$\displaystyle \underline{Step 6}$: Find the coefficients for the first root and the second root.
The recurrence relation is given by
$\displaystyle a_n = - \frac{\sum_{k=0}^{n-1} a_k[kp_{n-k}+q_{n-k}]}{F(r+n)}$ $\displaystyle n \geq 1$ and $\displaystyle a_0 = 1$
For $\displaystyle \underline{r = 0}$
$\displaystyle a_1 = - \frac{a_0(0\cdot p_1 + q_1)}{F(1)} = \frac{-3/4}{3/4} = -1$
$\displaystyle a_2 = - \frac{a_1q_1)}{F(2)} = \frac{-(-1)(3/4)}{7/2} = 3/14$
$\displaystyle a_3 = - \frac{a_2(q_1)}{F(3)} = \frac{-(3/14)(3/4)}{33/4} = -3/154$
$\displaystyle a_4 = - \frac{a_3q_1}{F(4)} = \frac{-(-3/154)(3/4)}{15} = 3/3080$
$\displaystyle a_5 = - \frac{a_4q_1}{F(5)} = \frac{(3/3080)(3/4)}{95/4} = -9/292800$
For $\displaystyle \underline{r = \frac{1}{4}}$
$\displaystyle a_1 = - \frac{a_0q_1}{F(1)} = \frac{(-3/4)}{5/4} = -3/5$
$\displaystyle a_2 = - \frac{a_1q_1}{F(2)} = \frac{-(-3/5)(3/4)}{9/2} = 1/10$
$\displaystyle a_3 = - \frac{a_2q_1}{F(3)} = \frac{-(1/10)(3/4)}{39/4} = -1/130$
$\displaystyle a_4 = - \frac{a_3q_1}{F(4)} = \frac{-(-1/130)(3/4)}{17} = 3/8840$
$\displaystyle a_5 = - \frac{a_4q_1}{F(5)} = \frac{-(3/8840)(3/4)}{105/4} = -9/928200$
$\displaystyle \underline{Step 7}$: Write the solutions
The solutions are given by:
$\displaystyle y_1(x) = 1 - x + \frac{3}{14}x^2 - \frac{3}{154}x^3 + \frac{3}{3080}x^4 -\frac{9}{292600}x^5 + \cdots$
and
$\displaystyle y_2(x) = \mid x \mid^{-1/4} \left[1 - \frac{3}{5}x + \frac{1}{10}x^2 - \frac{1}{130}x^3 + \frac{3}{8840}x^4 -\frac{9}{928200}x^5 + \cdots \right]$
And that's all! Did I made a mistake somewhere? Can I find a better solution than that?
Regards,
Fractalus
• Jul 19th 2011, 04:12 PM
Fractalus
Re: An effective method to solve ODE in terms of power series with regular singular p
Anyone knows if what I did is correct? This was the main point of my post. (Wink)
Regards,
Fractalus
• Jul 20th 2011, 01:41 PM
Fractalus
Need my method to be checked!
I would really appreciate if someone tells if my solution is correct or not and why. I gave a detailed solution to help people finding my errors.
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2 arctan(x) = arctan($$\frac{2x}{1 - x^{2}}$$) = arcsin($$\frac{2x}{1 + x^{2}}$$) = arccos($$\frac{1 - x^{2}}{1 + x^{2}}$$)
We will learn how to prove the property of the inverse trigonometric function, 2 arctan(x) = arctan($$\frac{2x}{1 - x^{2}}$$) = arcsin($$\frac{2x}{1 + x^{2}}$$) = arccos($$\frac{1 - x^{2}}{1 + x^{2}}$$)
or, 2 tan$$^{-1}$$ x = tan$$^{-1}$$ ($$\frac{2x}{1 - x^{2}}$$) = sin$$^{-1}$$ ($$\frac{2x}{1 + x^{2}}$$) = cos$$^{-1}$$ ($$\frac{1 - x^{2}}{1 + x^{2}}$$)
Proof:
Let, tan$$^{-1}$$ x = θ
Therefore, tan θ = x
We know that,
tan 2θ = $$\frac{2 tan θ}{1 - tan^{2}θ}$$
tan 2θ = $$\frac{2x}{1 - x^{2}}$$
2θ = tan$$^{-1}$$($$\frac{2x}{1 - x^{2}}$$)
2 tan$$^{-1}$$ x = tan$$^{-1}$$($$\frac{2x}{1 - x^{2}}$$) …………………….. (i)
Again, sin 2θ = $$\frac{2 tan θ}{1 + tan^{2}θ}$$
sin 2θ = $$\frac{2x}{1 + x^{2}}$$
2θ = sin$$^{-1}$$($$\frac{2x}{1 + x^{2}}$$ )
2 tan$$^{-1}$$ x = sin$$^{-1}$$($$\frac{2x}{1 + x^{2}}$$) …………………….. (ii)
Now, cos 2θ = $$\frac{1 - tan^{2}θ}{1 + tan^{2}θ}$$
cos 2θ = $$\frac{1 - x^{2} }{1 + x^{2} }$$
2θ = cos$$^{-1}$$ ($$\frac{1 - x^{2} }{1 + x^{2} }$$)
2 tan$$^{-1}$$ x = cos ($$\frac{1 - x^{2} }{1 + x^{2} }$$) …………………….. (iii)
Therefore, from (i), (ii) and (iii) we get, 2 tan$$^{-1}$$ x = tan$$^{-1}$$ $$\frac{2x}{1 - x^{2}}$$ = sin$$^{-1}$$ $$\frac{2x}{1 + x^{2}}$$ = cos$$^{-1}$$ $$\frac{1 - x^{2}}{1 + x^{2}}$$ Proved.
Solved examples on property of inverse circular function 2 arctan(x) = arctan($$\frac{2x}{1 - x^{2}}$$) = arcsin($$\frac{2x}{1 + x^{2}}$$) = arccos($$\frac{1 - x^{2}}{1 + x^{2}}$$):
1. Find the value of the inverse function tan(2 tan$$^{-1}$$ $$\frac{1}{5}$$).
Solution:
tan (2 tan$$^{-1}$$ $$\frac{1}{5}$$)
= tan (tan$$^{-1}$$ $$\frac{2 × \frac{1}{5}}{1 - (\frac{1}{5})^{2}}$$), [Since, we know that, 2 tan$$^{-1}$$ x = tan$$^{-1}$$($$\frac{2x}{1 - x^{2}}$$)]
= tan (tan$$^{-1}$$ $$\frac{\frac{2}{5}}{1 - \frac{1}{25}}$$)
= tan (tan$$^{-1}$$ $$\frac{5}{12}$$)
= $$\frac{5}{12}$$
2. Prove that, 4 tan$$^{-1}$$ $$\frac{1}{5}$$ - tan$$^{-1}$$ $$\frac{1}{70}$$ + tan$$^{-1}$$ $$\frac{1}{99}$$ = $$\frac{π}{4}$$
Solution:
L. H. S. = 4 tan$$^{-1}$$ $$\frac{1}{5}$$ - tan$$^{-1}$$ $$\frac{1}{70}$$ + tan$$^{-1}$$ $$\frac{1}{99}$$
= 2(2 tan$$^{-1}$$ $$\frac{1}{5}$$) - tan$$^{-1}$$ $$\frac{1}{70}$$ + tan$$^{-1}$$ $$\frac{1}{99}$$
= 2(tan$$^{-1}$$ $$\frac{2 × \frac{1}{5}}{1 - (\frac{1}{5})^{2}}$$) - tan$$^{-1}$$ $$\frac{1}{70}$$ + tan$$^{-1}$$ $$\frac{1}{99}$$, [Since, 2 tan$$^{-1}$$ x = tan$$^{-1}$$($$\frac{2x}{1 - x^{2}}$$)]
= 2 (tan$$^{-1}$$ $$\frac{2\frac{1}{5}}{1 - (\frac{1}{25})}$$)- tan$$^{-1}$$ $$\frac{1}{70}$$ + tan$$^{-1}$$ $$\frac{1}{99}$$,
= 2 tan$$^{-1}$$ $$\frac{5}{12}$$ - (tan$$^{-1}$$ $$\frac{1}{70}$$ - tan$$^{-1}$$ $$\frac{1}{99}$$)
= tan$$^{-1}$$ ($$\frac{2 × \frac{5}{12}}{1 - (\frac{5}{12})^{2}}$$) - tan$$^{-1}$$ ($$\frac{\frac{1}{70} - \frac{1}{99}}{1 + \frac{1}{77} × \frac{1}{99}}$$)
= tan$$^{-1}$$ $$\frac{120}{199}$$ - tan$$^{-1}$$ $$\frac{29}{6931}$$
= tan$$^{-1}$$ $$\frac{120}{199}$$ - tan$$^{-1}$$ $$\frac{1}{239}$$
= tan$$^{-1}$$ ($$\frac{\frac{120}{199} - \frac{1}{239}}{1 + \frac{120}{119} × \frac{1}{239}}$$)
= tan$$^{-1}$$ 1
= tan$$^{-1}$$ (tan $$\frac{π}{4}$$)
= $$\frac{π}{4}$$ = R. H. S. Proved.
`
Inverse Trigonometric Functions
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How To Add Fractions With The Same Denominator. Find the lowest common denominator by multiplying each denominator by the other. To add fractions with the same denominators, the denominator remains the same and we add the numerators together.
Here’s the way to do it: To add fractions with the same denominators, the denominator remains the same and we add the numerators together. Multiply the two denominators together to get the denominator of the answer.
Again, We Notice That Both Fractions Have The Same Denominator.
Multiply the two denominators together to get the denominator of the answer. Add and subtract fractions with the same denominator; To add fractions with like or the same denominator, simply add the numerators then copy the common denominator.
Instead of dividing by 1/3, if we were to divide by 2/3. What is the total of 2/8 + 4/8? Add or subtract the numerators and keep the denominator the same.
For example, 153/24 +217/24 = 370/24. Here’s the way to do it: Multiplying fractions typically has four to five steps.
To Add Fractions There Are Three Simple Steps:
Add the top numbers (the numerators ), put that answer over the denominator. In the above image a/c and b/c are the fractions with same denominator c. Next you need to find the least common denominator (lcd) step 3:
Simplify The Fraction (If Possible)
{eq}\frac{3}{4} + \frac{3}{4} = \frac{3+3}{4} = \frac{6}{4} {/eq}. To add fractions with the same denominators, the denominator remains the same and we add the numerators together. Here, you want to start off by adding the 2 and the 4.
Categories: how to make
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# 11.1 Systems of linear equations: two variables (Page 2/20)
Page 2 / 20
Another type of system of linear equations is an inconsistent system , which is one in which the equations represent two parallel lines. The lines have the same slope and different y- intercepts. There are no points common to both lines; hence, there is no solution to the system.
## Types of linear systems
There are three types of systems of linear equations in two variables, and three types of solutions.
• An independent system has exactly one solution pair $\text{\hspace{0.17em}}\left(x,y\right).\text{\hspace{0.17em}}$ The point where the two lines intersect is the only solution.
• An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
• A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.
[link] compares graphical representations of each type of system.
Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.
1. Substitute the ordered pair into each equation in the system.
2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.
## Determining whether an ordered pair is a solution to a system of equations
Determine whether the ordered pair $\text{\hspace{0.17em}}\left(5,1\right)\text{\hspace{0.17em}}$ is a solution to the given system of equations.
$\begin{array}{l}\text{\hspace{0.17em}}x+3y=8\hfill \\ \text{\hspace{0.17em}}2x-9=y\hfill \end{array}$
Substitute the ordered pair $\text{\hspace{0.17em}}\left(5,1\right)\text{\hspace{0.17em}}$ into both equations.
The ordered pair $\text{\hspace{0.17em}}\left(5,1\right)\text{\hspace{0.17em}}$ satisfies both equations, so it is the solution to the system.
Determine whether the ordered pair $\text{\hspace{0.17em}}\left(8,5\right)\text{\hspace{0.17em}}$ is a solution to the following system.
$\begin{array}{c}5x-4y=20\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+1=3y\end{array}$
Not a solution.
## Solving systems of equations by graphing
There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.
## Solving a system of equations in two variables by graphing
Solve the following system of equations by graphing. Identify the type of system.
$\begin{array}{c}2x+y=-8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-y=-1\end{array}$
Solve the first equation for $\text{\hspace{0.17em}}y.$
$\begin{array}{c}2x+y=-8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=-2x-8\end{array}$
Solve the second equation for $\text{\hspace{0.17em}}y.$
$\begin{array}{c}x-y=-1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=x+1\end{array}$
Graph both equations on the same set of axes as in [link] .
The lines appear to intersect at the point $\text{\hspace{0.17em}}\left(-3,-2\right).\text{\hspace{0.17em}}$ We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
The solution to the system is the ordered pair $\text{\hspace{0.17em}}\left(-3,-2\right),$ so the system is independent.
Solve the following system of equations by graphing.
The solution to the system is the ordered pair $\text{\hspace{0.17em}}\left(-5,3\right).$
Can graphing be used if the system is inconsistent or dependent?
Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
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Factoring
# Factor Trinomials
### Learning Objectives
By the end of this section, you will be able to:
• Factor trinomials of the form
• Factor trinomials of the form using trial and error
• Factor trinomials of the form using the ‘ac’ method
• Factor using substitution
Before you get started, take this readiness quiz.
1. Find all the factors of 72.
If you missed this problem, review (Figure).
2. Find the product:
If you missed this problem, review (Figure).
3. Simplify:
If you missed this problem, review (Figure).
### Factor Trinomials of the Form
You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.
To figure out how we would factor a trinomial of the form such as and factor it to let’s start with two general binomials of the form and
Foil to find the product. Factor the GCF from the middle terms. Our trinomial is of the form
This tells us that to factor a trinomial of the form we need two factors and where the two numbers m and n multiply to c and add to b.
How to Factor a Trinomial of the form
Factor:
Factor:
Factor:
Let’s summarize the steps we used to find the factors.
Factor trinomials of the form
1. Write the factors as two binomials with first terms x.
2. Find two numbers m and n that
• multiply to
3. Use m and n as the last terms of the factors.
4. Check by multiplying the factors.
In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.
How do you get a positive product and a negative sum? We use two negative numbers.
Factor:
Again, with the positive last term, 28, and the negative middle term, we need two negative factors. Find two numbers that multiply 28 and add to
Factors of Sum of factors
Factor:
Factor:
Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.
How do you get a negative product and a positive sum? We use one positive and one negative number.
When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.
Factor:
Factors of Sum of factors
Factor:
Factor:
Sometimes you’ll need to factor trinomials of the form with two variables, such as The first term, is the product of the first terms of the binomial factors, The in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in How To Factor trinomials.
Factor:
We need r in the first term of each binomial and s in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
Factors of Sum of factors
Factor:
Factor:
Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.
Factor:
We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
Factors of Sum of factors
Note there are no factor pairs that give us as a sum. The trinomial is prime.
Factor:
prime
Factor:
prime
Let’s summarize the method we just developed to factor trinomials of the form
Strategy for Factoring Trinomials of the Form
When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.
### Factor Trinomials of the form ax2 + bx + c using Trial and Error
Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form
Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.
Factor completely:
Factors of Sum of factors
Factor completely:
Factor completely:
What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.
Let’s factor the trinomial
From our earlier work, we expect this will factor into two binomials.
We know the first terms of the binomial factors will multiply to give us The only factors of are We can place them in the binomials.
Check: Does
We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.
Which factors are correct? To decide that, we multiply the inner and outer terms.
Since the middle term of the trinomial is the factors in the first case will work. Let’s use FOIL to check.
Our result of the factoring is:
How to Factor a Trinomial Using Trial and Error
Factor completely using trial and error:
Factor completely using trial and error:
Factor completely using trial and error:
Factor trinomials of the form using trial and error.
1. Write the trinomial in descending order of degrees as needed.
2. Factor any GCF.
3. Find all the factor pairs of the first term.
4. Find all the factor pairs of the third term.
5. Test all the possible combinations of the factors until the correct product is found.
6. Check by multiplying.
Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.
Factor completely using trial and error:
The trinomial is already in descending order. Find the factors of the first term. Find the factors of the last term. Consider the signs. Since the last term, 5, is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.
Consider all the combinations of factors.
Possible factors Product
Factor completely using trial and error:
Factor completely using trial and error:
When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.
Factor completely using trial and error:
The trinomial is already in descending order. Find the factors of the first term. Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors.
Consider all the combinations of factors.
Factor completely using trial and error
Factor completely using trial and error:
Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.
Factor completely using trial and error:
Notice the greatest common factor, so factor it first. Factor the trinomial.
Consider all the combinations.
Factor completely using trial and error:
Factor completely using trial and error:
### Factor Trinomials of the Form using the “ac” Method
Another way to factor trinomials of the form is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!
How to Factor Trinomials using the “ac” Method
Factor using the ‘ac’ method:
Factor using the ‘ac’ method:
Factor using the ‘ac’ method:
The “ac” method is summarized here.
Factor trinomials of the form using the “ac” method.
1. Factor any GCF.
2. Find the product ac.
3. Find two numbers m and n that:
4. Split the middle term using m and n.
5. Factor by grouping.
6. Check by multiplying the factors.
Don’t forget to look for a common factor!
Factor using the ‘ac’ method:
Is there a greatest common factor? Yes. The GCF is 5. Factor it. The trinomial inside the parentheses has a leading coefficient that is not 1. Find the product Find two numbers that multiply to and add to b. Split the middle term. Factor the trinomial by grouping. Check by multiplying all three factors.
Factor using the ‘ac’ method:
Factor using the ‘ac’ method:
### Factor Using Substitution
Sometimes a trinomial does not appear to be in the form. However, we can often make a thoughtful substitution that will allow us to make it fit the form. This is called factoring by substitution. It is standard to use u for the substitution.
In the the middle term has a variable, x, and its square, is the variable part of the first term. Look for this relationship as you try to find a substitution.
Factor by substitution:
The variable part of the middle term is and its square, is the variable part of the first term. (We know If we let we can put our trinomial in the form we need to factor it.
Rewrite the trinomial to prepare for the substitution. Let and substitute. Factor the trinomial. Replace u with Check:
Factor by substitution:
Factor by substitution:
Sometimes the expression to be substituted is not a monomial.
Factor by substitution:
The binomial in the middle term, is squared in the first term. If we let and substitute, our trinomial will be in form.
Rewrite the trinomial to prepare for the substitution. Let and substitute. Factor the trinomial. Replace u with Simplify inside the parentheses.
This could also be factored by first multiplying out the and the and then combining like terms and then factoring. Most students prefer the substitution method.
Factor by substitution:
Factor by substitution:
Access this online resource for additional instruction and practice with factoring.
### Key Concepts
• How to factor trinomials of the form
1. Write the factors as two binomials with first terms x.
2. Find two numbers m and n that
3. Use m and n as the last terms of the factors.
4. Check by multiplying the factors.
• Strategy for Factoring Trinomials of the Form : When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.
• How to factor trinomials of the form using trial and error.
1. Write the trinomial in descending order of degrees as needed.
2. Factor any GCF.
3. Find all the factor pairs of the first term.
4. Find all the factor pairs of the third term.
5. Test all the possible combinations of the factors until the correct product is found.
6. Check by multiplying.
• How to factor trinomials of the form using the “ac” method.
1. Factor any GCF.
2. Find the product ac.
3. Find two numbers m and n that:
4. Split the middle term using m and n.
5. Factor by grouping.
6. Check by multiplying the factors.
#### Practice Makes Perfect
Factor Trinomials of the Form
In the following exercises, factor each trinomial of the form
In the following exercises, factor each trinomial of the form
Prime
Prime
Factor Trinomials of the Form Using Trial and Error
In the following exercises, factor completely using trial and error.
Factor Trinomials of the Form using the ‘ac’ Method
In the following exercises, factor using the ‘ac’ method.
Factor Using Substitution
In the following exercises, factor using substitution.
Mixed Practice
In the following exercises, factor each expression using any method.
#### Writing Exercises
Many trinomials of the form factor into the product of two binomials Explain how you find the values of m and n.
Tommy factored as Sara factored it as Ernesto factored it as Who is correct? Explain why the other two are wrong.
List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form
How is the “ac” method similar to the “undo FOIL” method? How is it different?
#### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
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# Direct Variation
Related Topics:
Math Worksheets
Videos, worksheets, solutions, and activities to help Algebra students learn about direct variation or direct proportion. What is direct variation and how to solve direct variation problems?
What is Direct Variation?
We often use the term direct variation to describe a form of dependence of one variable on another. An equation that makes a line and crosses the origin is a form of direct variation, where the magnitude of x increases or decreases directly as y increases or decreases. Direct variation and inverse variation are used often in science when modeling activity, such as speed or velocity.
Some real life examples of direct variation are:
The number of hours you work and the amount of your paycheck
The amount of weight on a spring and the distance the spring will stretch
The speed of a car and the distance traveled in a certain amount of time
How to solve direct variation problems?
In general, if two quantities vary directly, if one goes up the other goes up or down proportionally.
Graphically, we have a line that passed through the origin. The slope is k, which is called the constant of proportionality.
Examples:
1. Y varies directly with x. y = 54 when x = 9. Determine the direct variation equation and then determine y when x = 3.5.
2. Hooke's Law states that the displacement, d that a spring is stretched by a hanging object caries directly as the mass, m of the object. If the distance is 10 cm when the mass is 3 kg, what is the distance when the mass is 5 kg?
3. Y varies directly with with the square of x. y = 32 when x = 4. Determine the direct variation equation and then determine y when x = 6. What is the Direct Variation or Direct Proportionality Formula?
Ever heard of two things being directly proportional? Well, a good example is speed and distance. The bigger your speed, the farther you'll go over a given time period. So as one variable goes up, the other goes up too, and that's the idea of direct proportionality. But you can express direct proportionality using equations, and that's an important thing to do in algebra. See how to do that in the tutorial.
Algebra Word Problem: Variation Application
Solving a Variation problem
Example:
Art's wages are directly proportional to the number of hours he works per week. If Art works 36 hours in a week, he earns \$540. What are his wages if he works 40 hours in a week? Direct Variation Models
Examples:
1. The current standard for low-flow showerheads is 2.5 gallons per minute. Calculate how long it would take to fill a 30 gallon bathtub using such a showerhead to supply the water.
2. Amen is using a hose to fill his swimming pool for the first time. He starts the hose at 10 P.M. and leaves it running all night. At 6 A.M. he measures the depth and calculates that the pool is four-sevenths full. At what time will his new pool be full? Direct Variation Word Problems
Determine the constant of variation and find the other missing value using the given information.
Solve a direct variation equation y = kx, where k is the constant of variation.
A direct variation graph is a straight line graph that passes through the origin.
Example:
The distance required to stop a car varies directly as the square of its speed. If 200 feet are required to stop a car traveling 60 miles per hour, how many feet are required to stop a car traveling 100 miles per hour?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Unit 2 Review - Differentiation: Definition & Fundamental Properties
279Β resources
See Units
## Defining Average and Instantaneous Rates of Change at a Point π
(Photo courtesy of slideshare.net)
An important concept to grasp in this lesson is the difference between the average rate of change and the instantaneous rate of change. The average rate of change is simply the slope of the secant line between two points. The instantaneous rate of change is the slope of the tangent line at any given point (the derivative).
The rate of change formula (pictured below) is the slope of the secant line between two points. "f(b)" represents your y-value to your first point and "f(a)" represents your y-value to your second point, "b" and "a" are the corresponding x-values to those coordinates.
• The instantaneous rate of change formula (pictured below) is the slope of the tangent line at a given point. Many people are puzzled with f(x+h) but it simply means that whenever you see an x in your original equation, you insert "x+h." For example, if f(x) = xΒ² + 3x - 9 then f(x+h) = (x+h)Β² + 3(x+h) - 9. The second part of this formula is subtracting f(x) from f(x+h); however, a common error is that students forget to distribute the "-" sign to all the terms in the original function. The last part is to put everything over h and simplify the entire equation by combining like terms and factoring). If there are any h terms left you evaluate them as 0 and then simplify.
## Defining the Derivative of a Function and Using Derivative Notation π€
The function that gives the instantaneous rate of change is the derivative, and the derivative is the slope of the tangent line to the graph at a given point. The formula for the derivative is the same as the instantaneous rate of change formula.
The AP exam loves using the notation for the derivative, so don't be scared as they all mean the same thing; however in certain scenarios in calculus, we may use one notation over the other. Here are some of the ways we can express the derivative (pictured below). For now, only review the notations concerning the first derivative. The bottom line is that y' is synonymous with f'(x) or dy/dx or d/dx f(x)!
Β (Photo courtesy of education.fcps.org)
## Connecting Differentiability and Continuity: Determining when Derivatives DO or DO NOT EXIST β
If you need a complete refresher on continuity you can watch a replay of our stream on Continuity here!
Being differentiable means that a derivative exists. It is important to know that being differentiable is being continuous however being continuous does not mean you are differentiable.
On top of being continuous in order to be differentiable, the function must have NO corners, cusps, and no vertical tangent lines.
## Applying the Power Rule π¦ΈββοΈπ¦ΈββοΈ
Using the definition of the derivative for every single problem you encounter is a time-consuming and it is also open to careless errors and mistakes. However, one great mathematician decided to bless us with a fundamental rule known as the Power Rule, pictured below.
• Your "n" term is the power your x-term is being raised to.
For example, in xΒ² your "n" term would be 2. Additionally, you multiply that "n" value to the term's coefficient (in xΒ² your coefficient is 1) and then decrease your terms exponent by 1 (using the power rule the derivative of xΒ² is 2x).
• Remember: the derivative is the slope of the line tangent to any point on the graph and using the power rule (like the definition of the derivative) simply gives you the formula to the slope of the line tangent. If you would like to know what that point is then you would have to evaluate using your x-coordinate.
## Derivative Rules: Constant, Sum, Difference, and Constant MultipleΒ ββ
These first set of derivative rules are simple but absolutely crucial to your understanding of calculus.Β
• The sum rule states that the derivative of a sum of functions is the same as the sum of their derivatives.
• Just like the sum rule, the difference rule states the derivative of a difference in functions is the same as the difference in their derivatives.
While these two rules may seem confusing, they are actually straightforward. The sum and difference rules are essentially applications of the power rule to every term, as well as combining them (if possible). Here is an example of using both these formulas, pictured below:
(Photo courtesy of https://magoosh.com)
• The constant multiple rule states that the derivative of aΒ constantΒ times a function is equal to theΒ constantΒ times the derivative.
For example, the derivative of 5γ»xΒ², (5xΒ²) is equal to 5 times the derivative of 2x.
• The constant rule states that the derivative of any constant is 0 (for example, the derivative of 5 is 0).
## Derivatives of cos, sin, natural exponentialΒ functions, and the natural log π€
These rules must be committed to memory as they are used throughout the year in calculus.
• The derivative of sin (x) is cos (x).
• The derivative of cos (x) is -sin (x).
If you would like to find a derivative of a trig function with a constant (such as 5sin(x)), you would use the constant multiple rule to get 5cos(x).
• It is important to note that the derivative of these functions only work when you are only using "x" in the function. For example, the derivative of sin(3x) is not cos(3x), in order to get the correct derivative you would need to apply the chain rule. (Do not worry about the chain rule in this unit as it is covered in Unit 3 of calculus.)
The derivative of e^x is e^x. No matter how many times you take the derivative of this function, the derivative of e^x will remain as e^x.
• The derivative of e^x only works when it is raised to only the "x" power. For example, the derivative e^2x is not e^2x, in order to get the correct derivative you would need to apply the chain rule.
The derivative of ln(x) is pictured below. If you would want to find the derivative of ln(4x), you would need to apply the chain rule.
## The Product RuleΒ β
By now, we know how to add and subtract derivative functions, but what about multiplying them? With the product rule we can finally multiply derivatives together. Here is the rule (pictured below)
f(x) and g(x) represent two different functions that are being multiplied together. Here is an example of how to apply this rule (pictured below).
Let's call the first function f(x) and the second function g(x). For our first part, we will take the derivative of the first function and multiply it by the original second function. For the second part, we then multiply the original first function and multiply that by the derivative of the second function. Finally, we will add up both parts. If it is helpful to remember the derivative of first times second plus derivative of second times first, go for it!
## The Quotient Rule β
Let's now move on to the product rule's partner: the Quotient Rule! With the quotient rule, we can finally divide derivatives. Here is what the quotient rule looks like. pictured below:
(Photo courtesy of andymath.com)
The quotient rule states that f(x) is the top function (the dividend) and g(x) is the bottom function (divisor). The first part would be to multiply g(x) by the derivative of f(x) and the second part would be to multiply f(x) by the derivative of g(x). After that, you would subtract the two parts (don't forget to distribute the negative sign!). Lastly, you would divide everything by g(x)Β². If you need another visual, here is an example (pictured below).
(Photo courtesy of www.studygeek.org)
Note that (4x-2) is your top function, f(x) and (xΒ²+1) is your bottom function, g(x).
## Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions π
Here is a helpful chart with the derivatives of the rest of the trigonometric functions besides sine and cosine:
(Photo courtesy of slideplayer.com)
However, it is important to take note that AP Calculus mainly focuses on the derivatives of sin, cos, and tan.
## Final Note π
Make sure you get the basics down of unit 2 of AP Calculus AB for this unit sets the foundations of calculus, which is essentially the rest of the course. As a result, it is essential for you to understand these concepts!
Β
Browse Study Guides By Unit
πUnit 1 β Limits & Continuity
π€Unit 2 β Fundamentals of Differentiation
π€π½Unit 3 β Composite, Implicit, & Inverse Functions
πUnit 4 β Contextual Applications of Differentiation
β¨Unit 5 β Analytical Applications of Differentiation
π₯Unit 6 β Integration & Accumulation of Change
πUnit 7 β Differential Equations
πΆUnit 8 β Applications of Integration
π¦Unit 9 β Parametric Equations, Polar Coordinates, & Vector-Valued Functions (BC Only)
βΎUnit 10 β Infinite Sequences & Series (BC Only)
πStudy Tools
π€Exam Skills
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Congruent Shapes
What can you say about these shapes?
That's right!Β
They have the same shape and size.
We call these congruent shapes.
What Are Congruent Shapes?
Congruent shapes are exactly the same shape and size.
If you put two congruent shapes on top of each other, the top one would completely cover the second one.Β
Let's look at some triangles:
π They have the same number of sides and their sizes are equal.
It does not matter that their colors are different. We only look at their shapes and sizes.
Now, look at these rectangles.
Are they congruent?
At first, you would think that they're not, because they're laid out differently.Β
π But if you turn the second rectangle, you will see that it matches the first rectangle exactly.
π Tip: Congruent shapes match even if you flip, slide, or turn them.
How to Check for Congruence
Example 1
When shapes are very different from each other, it is easy to know that they are not congruent.
π Just by looking at these two shapes, we know they're not the same.Β
One is a circle and the other one is a square.
Example 2
But sometimes it is hard to tell just by looking.Β
Especially when the shapes almost look the same.
So what do we do?
π We count their sides.Β
The first shape has 6 sides. The second shape has 5 sides.
This means that the two shapes are not congruent.
Example 3
Sometimes figures have exactly the same shape, but they're not congruent.
Do you know why?
They aren't the same size!
π When shapes look the same, we need to check their sizes.
The first circle is bigger than the second one.Β
So these circles not congruent.
Example 4
Now, look at these triangles.
Are they congruent?
Yes, they are!Β
These triangles have the same shape and size.
Watch and Learn
Great job! π
Now, you can move on to practice. πͺ
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# What are the 4 steps for solving algebra word problems?
Contents
The four steps for solving algebra word problems are: read and understand the problem, identify the unknown variables, create an equation or system of equations based on the information given, and solve for the unknown variables.
Solving algebra word problems can be a challenging task for many students. However, by following these four steps, students can simplify the process and increase their chances of success.
Step 1: Read and understand the problem
The first step to solving any algebra word problem is to read and understand the problem. It is important to read the problem carefully and identify any key information that may be useful in solving the problem. This step can be challenging, but as Albert Einstein said, “If you can’t explain it simply, you don’t understand it well enough.”
Step 2: Identify the unknown variables
The next step is to identify the unknown variables in the problem. This means determining what variables are being asked to solve for in the problem. For example, if the problem states, “If John has 12 apples and gives 3 to his friend, how many apples does he have left?” The unknown variable would be the number of apples John has left.
Step 3: Create an equation or system of equations based on the information given
Once the unknown variables have been identified, the next step is to create an equation or system of equations based on the information given in the problem. This can be done by using the information given in the problem to write an equation that represents the relationship between the variables. A helpful tip for creating equations is to use the “plug and chug” method, where you plug in known values and solve for the unknown variable.
Step 4: Solve for the unknown variables
The final step is to solve for the unknown variables using the equation or system of equations created in step 3. This can be done by using algebraic techniques such as simplifying equations, combining like terms, and isolating the unknown variable. It is important to check the solution by plugging it back into the original equation to ensure that it is correct.
IT\\\'S IMPORTANT: Top answer to: is a math interventionist a teacher?
• Algebra is derived from the Arabic word “al-jabr,” which means “reunion of broken parts.”
• The earliest known mathematical textbook on algebra was written in 820 AD by an Islamic mathematician named Al-Khwarizmi.
• Algebra is used in various fields such as economics, engineering, physics, and computer science.
Table: Common algebraic equations and their meanings
Equation Meaning
y = mx + b Slope-intercept form of a linear equation
a/b Ratio
(x + y)^2 Binomial expansion
x + y > 10 Inequality
F = ma Formula for force
a^2 + b^2 = c^2 Pythagorean theorem
## See the answer to “What are the 4 steps for solving algebra word problems?” in this video
In “4 Steps in Solving Problems”, the video provides a comprehensive guide on how to solve math word problems in four main steps, namely, understanding the problem, making a plan, carrying out the plan, and checking your answer. The tutorial emphasizes the importance of reading word problems carefully, looking for hints, defining the problem-solving strategy to be considered, and rigorously checking one’s answer to ensure it makes sense in the context of the problem.
There are other opinions on the Internet
To solve an algebraic word problem:
• Define a variable.
• Write an equation using the variable.
• Solve the equation.
• If the variable is not the answer to the word problem, use the variable to calculate the answer.
You can tackle any word problem by following these five steps: Read through the problem carefully, and figure out what it’s about. Represent unknown numbers with variables. Translate the rest of the problem into a mathematical expression. Solve the problem. Check your work.
To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer.
If a mathematical problem is expressed in words, then first try to express all the relevant information in the problem in terms of symbols representing unknowns, in the form of equations, inequalities, etc.
Once you have a symbolic representation of the problem – typically in the form of a set of simultaneous equations to be solved, try eliminating variables one at a time by rearranging terms, using basic arithmetic operations and substitutions.Eventually this process will probably lead to values (or at least constraints) for all of the unknowns.
Convert these values or constraints back into the context and language of the problem to express the solution.
For example, suppose Andrew is half the age of Brian, Brian is three times older than Charles and the sum of their ages is 44 years. How old is Charles?
Use A to represent Andrew’s age, B to represent Brian’s age and C to represent Charles’s age.
Then expressing the information in the problem symbolically we have:
A=12B
B=3C
A…
## Furthermore, people are interested
Also, What are the 4 steps in solving word problems?
As an answer to this: Beta Program
• Read through the problem and set up a word equation — that is, an equation that contains words as well as numbers.
• Plug in numbers in place of words wherever possible to set up a regular math equation.
• Use math to solve the equation.
IT\\\'S IMPORTANT: You requested: what is a trivial solution to a homogeneous system of equations?
In this manner, What are the 4 steps to solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division.
Also Know, How do you solve algebraic word problems easily? The answer is: You can tackle any word problem by following these five steps:
1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
Similarly one may ask, How do you solve algebra problems step by step? Answer will be: Now all we need to do is separate the 3 from the x the opposite of multiplication is division. So we need to divide both sides by 3.. So x is 12 divided by 3 which is 4. And this is the answer.
Secondly, How do you solve word problems in Algebra? Response to this: There are many different kinds of algebra word problems out there, but they can all be solved by following these four simple steps: Define and identify your variable. Write an equation that uses the variable. Solve the equation. If the variable is not the correct answer to the word problem, use that variable to figure out the answer.
What are the 5 steps of algebra problem solving?
While the 5 steps of Algebra problem solving are listed below, the following will help you learn how to first identify the problem. Identify the problem. Identify what you know. Make a plan. Carry out the plan. Verify that the answer makes sense. Back away from the calculator; use your brain first.
IT\\\'S IMPORTANT: Your request — what is the second most important thing in mathematics?
Thereof, How do you solve a word problem with only one variable? Response: Solve an equation for one variable. If you have only one unknown in your word problem,isolate the variable in your equation and find which number it is equal to. Use the normal rules of algebra to isolate the variable. Remember that you need to keep the equation balanced.
Keeping this in view, Is it hard to solve word problems?
Response: Solving word problems may seem difficult, but when you read through the problem and can figure out what the specific equation is, it’s no harder than a regular algebra problem. Here are some tips for getting a solid system of steps to follow when you are solving algebra problems: 1. Read through the entire problem before trying to solve it.
How do you solve algebra word problems? Solving Algebra word problems is useful in helping you to solve earthly problems. While the 5 steps of Algebra problem solving are listed below, the following will help you learn how to first identify the problem. Identify the problem. Identify what you know. Make a plan. Carry out the plan. Verify that the answer makes sense.
What are the 5 steps of algebra problem solving? The answer is: While the 5 steps of Algebra problem solving are listed below, the following will help you learn how to first identify the problem. Identify the problem. Identify what you know. Make a plan. Carry out the plan. Verify that the answer makes sense. Back away from the calculator; use your brain first.
Keeping this in view, How do you solve math problems step-by-step? As a response to this: To solve math problems step-by-step start by reading the problem carefully and understand what you are being asked to find. Next, identify the relevant information, define the variables, and plan a strategy for solving the problem. Math notebooks have been around for hundreds of years. You write down problems, solutions and notes to go back…
Is it hard to solve word problems?
Solving word problems may seem difficult, but when you read through the problem and can figure out what the specific equation is, it’s no harder than a regular algebra problem. Here are some tips for getting a solid system of steps to follow when you are solving algebra problems: 1. Read through the entire problem before trying to solve it.
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# Question Video: Finding the Limit of a Rational Function by Factoring a Sum or Difference of Two Cubes Mathematics • Higher Education
Find lim_(π₯ βΆ β2) ((π₯Β³ + 8)/(π₯ + 2)).
04:37
### Video Transcript
Find the limit as π₯ approaches negative two of π₯ cubed plus eight all divided by π₯ plus two.
In this question, weβre asked to evaluate the limit of the quotient of two polynomials. This is a rational function. And we recall we can attempt to evaluate the limit of any rational function by using direct substitution. However, if we substitute π₯ is equal to negative two into this function, we get negative two cubed plus eight all divided by negative two plus two, which if we evaluate gives us zero divided by zero, an indeterminate form. Therefore, we canβt evaluate this limit by using direct substitution alone.
Instead, weβre going to need to use some other method. One way of evaluating this limit is to notice that negative two is a root of the polynomial in our numerator and itβs a root of the polynomial in our denominator. And by using the remainder theorem, this tells us that π₯ plus two is a factor of both the polynomial in the numerator and the denominator. And of course we already knew this for our denominator. However, we can use this to factor our numerator. And thereβs a few different ways of doing this. One way of doing this would be to use algebraic division. We would divide π₯ cubed plus eight by π₯ plus two.
However, since we know when we divide a cubic polynomial by a linear polynomial we get a quadratic polynomial, we can write this as follows. π₯ cubed plus eight will be equal to π₯ plus two multiplied by some quadratic ππ₯ squared plus ππ₯ plus π. We can then distribute our parentheses on the right-hand side of our equation. We get ππ₯ cubed plus ππ₯ squared plus ππ₯ plus two ππ₯ squared plus two ππ₯ plus two π. We can then simplify this further by taking out the shared factors of π₯ on the right-hand side of our equation. We get ππ₯ cubed plus π plus two π multiplied by π₯ squared plus π plus two π multiplied by π₯ plus two π. And remember, this expression is equal to π₯ cubed plus eight. We can find the values of π, π, and π by equating coefficients on both sides of our equation.
Letβs start by equating the coefficients of π₯ cubed. On the left-hand side of our equation, the coefficient of π₯ cubed is one. On the right-hand side of our equation, the coefficient of π₯ cubed is π. Therefore, our value of π is one. And we can substitute this into the expression on the right-hand side of our equation. Doing this then gives us the following equation.
Now, letβs equate the constant terms on both sides of our equation. We see that π is equal to two π. Therefore, π must be equal to four. And once again we can substitute this value of π back into our equation. This then gives us the following equation. Finally, we can equate the coefficients of π₯ squared on both sides of the equation or the coefficients of π₯ on both sides of the equation to find the value of π. However, on the left-hand side of our equation, thereβs no π₯ squared or π₯-term. So the coefficients of these terms are zero. And therefore, for the coefficients on the right-hand side of our equation to match, these two coefficients have to be zero. π plus two is equal to zero, and four plus two π is equal to zero, which means π is negative two.
And now we can substitute the value of π equal to one, the value of π equal to negative two, and the value of π equal to four into our factored expression. This gives us that π₯ cubed plus eight is equal to π₯ plus two multiplied by π₯ squared minus two π₯ plus four. This is exactly the same answer we wouldβve got if weβd used algebraic division.
We can now use this to help us evaluate our limit. Weβll substitute this expression for the numerator inside of our limit. Doing this, we get the limit as π₯ approaches negative two of π₯ plus two multiplied by π₯ squared minus two π₯ plus four all divided by π₯ plus two. And now we can simplify this limit. We just need to recall weβre taking the limit as π₯ approaches negative two. This means weβre interested in what happens to the output of our function as our values of π₯ get closer and closer to negative two. The output of the function when π₯ is equal to negative two will not affect its limit.
Therefore, we can cancel the shared factor of π₯ plus two in the numerator and the denominator of this expression. Because when π₯ is not equal to negative two, π₯ plus two divided by π₯ plus two is just equal to one. Therefore, weβve rewritten our limit as the limit as π₯ approaches negative two of π₯ squared minus two π₯ plus four.
This is then the limit of a polynomial. And we can evaluate the limits of polynomials by using direct substitution. Substituting π₯ is equal to negative two into our quadratic gives us negative two all squared minus two times negative two plus four, which we can evaluate is equal to 12. Therefore, we were able to show the limit as π₯ approaches negative two of π₯ cubed plus eight all divided by π₯ plus two is equal to 12.
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### Proportions
```Constructed Response Assessment
October 17th
Day 1: Ratios
A ratio is a comparison of two quantities
using division.
Ratios can be written in three different
7
ways: 7 to 5, 7:5, and
5
Order matters when writing a ratio.
Find the ratio of boys to girls in Donnelly’s class.
16 cows, 8 sheep, and 6 pigs
cows to sheep
pigs to total animals
sheep to pigs
Always simplify your ratio to the lowest term.
The ratio of wings to beaks in the bird house
at the zoo was 2:1, because for every 2 wings
there was 1 beak.
For every vote candidate A received,
Make a table or model to represent one of the
above situations.
Beak
Wing
1
2
2
4
3
6
4
8
Remember, a ratio makes a
comparison.
The ratio of green aliens to total
aliens is 3 to 7.
****Make sure you write the ratio
just like they ask for it!****
The ratio of total aliens to purple
aliens is 7 to 4. Not 4 to 7
1. What is the ratio of
blue balloons to red
balloons?
2. What is the ratio of
total balloons to orange
balloons?
3. What is the ratio of
yellow balloons to total
balloons?
4. What is the ratio of
green balloons to purple
balloons?
Day 3: Equivalent Ratios
Ratios that make the same comparison
are equivalent ratios. To check
whether two ratios are equivalent, you
can write both in simplest form.
20 cars : 30 trucks
10 : 15
2 : 3 80 : 120
Check It Out! Example 1
Write the ratio 24 shirts to 9 jeans in
simplest form.
Write the ratio as
a fraction.
shirts = 24
jeans
9
=
24 ÷ 3
9÷3
= 8
3
Simplify.
The ratio of shirts to jeans is 8 , 8:3, or 8 to 3.
3
Lesson Quiz: Part I
Write each ratio in simplest form.
1
1. 22 tigers to 44 lions
2
30
2. 5 feet to 14 inches
7
Find a ratios that is equivalent to each given
ratio.
3. 4
15
30 45
4. 7
21
3 42
Determining Whether Two Ratios Are Equivalent
Simplify to tell whether the ratios are
equivalent.
A. 3 and 2
27
18
B. 12 and 27
15
36
Lesson Quiz: Part II
Simplify to tell whether the ratios are
equivalent.
5. 16 and 32 8 = 8; yes
10
20 5 5
6. 36 and 28
24
18
3 14 ; no
2 9
7. Kate poured 8 oz of juice from a 64 oz bottle.
Brian poured 16 oz of juice from a 128 oz bottle.
Are the ratios of poured juice to starting amount
of juice equivalent?
8 and 16 ; yes, both equal 1
64
8
128
A rate is a ratio that compares quantities that
are measured in different units.
This spaceship travels at a certain speed.
Speed is an example of a rate.
This spaceship can travel
100 miles in 5 seconds is a rate.
It can be written 100 miles
5 seconds
A rate is a ratio that compares quantities that
are measured in different units.
One key word that often identifies a rate is PER.
•Example: Miles per gallon, Points per free throw,
Dollars per pizza, Sticks of gum per pack
What other examples of rates can
Remember: A rate is a ratio that compares
two quantities measured in different units
(miles, inches, feet, hours, minutes, seconds).
The unit rate is the rate for one unit of a
given quantity. Unit rates have a
denominator of 1.
A unit rate compares a quantity to one unit of
another quantity. These are all examples of
unit rates.
2 eyes per alien
1 tail per body
1 foot per leg
3 windows per spaceship
3 riders per spaceship
Rate
150 heartbeats
2 minutes
Unit Rate (divide to get it):
150 ÷ 2 = 75
75 heartbeats to 1minute OR
75 heartbeats per minute
Amy can read 88 pages in 4 hours (rate).
What is the unit rate? (How many pages can
88 pages
4 hours
22 pages / hour
Try this by yourself!
Unit rates are rates in which the second
quantity is 1.
The ratio 90 can be simplified by dividing:
3
90 = 30
3
1
unit rate: 30 miles, or 30 mi/h
1 hour
Check It Out! Can you solve?
Penelope can type 90 words in 2 minutes. How
many words can she type in 1 minute?
90 words
2 minutes
Write a rate.
90 words ÷ 2 = 45 words
2 minutes ÷ 2 1 minute
Divide to find words
per minute.
Penelope can type 45 words in one minute.
Unit price is a unit rate used to compare
price per item.
Use division to find the unit prices of the two
products in question.
The unit rate that is smaller (costs less) is the
better value.
Juice is sold in two different sizes. A 48fluid ounce bottle costs \$2.07. A 32-fluid
ounce bottle costs \$1.64. Which is the
\$2.07
48 fl.oz.
\$1.64
32 fl.oz.
0.04312
5
0.05125
\$0.04 per
fl.oz.
\$0.05 per
fl.oz.
The 48 fl.oz. bottle is the better value.
Additional Example: Finding Unit Prices to Compare
Costs
Pens can be purchased in a 5-pack for \$1.95
or a 15-pack for \$6.20. Which pack has the
lower unit price?
price for package = \$1.95 = \$0.39 Divide the price
by the number
number of pens
5
of pens.
price for package = \$6.20 \$0.41
number of pens
15
The 5-pack for \$1.95 has the lower unit price.
Try this by yourself
John can buy a 24 oz bottle of ketchup for
\$2.19 or a 36 oz bottle for \$3.79. Which
bottle has the lower unit price?
price for bottle
= \$2.19 \$0.09
number of ounces
24
Divide the price
by the number
of ounces.
price for bottle
= \$3.79 \$0.11
number of ounces
36
The 24 oz jar for \$2.19 has the lower unit price.
Day 7: A proportion is an equation
stating that two ratios are equal.
x3
7 , 21
10 30
Yes, these two ratios DO form a proportion,
because the same relationship exists in both the
numerators and denominators.
x3
÷4
No, these ratios do NOT form a
proportion, because the ratios are not
equal.
8 , 2
9 3
÷3
A proportion is an equation stating that two
ratios are equal.
Example:
Example:
A piglet can gain 3 pounds in 36 hours. If
this rate continues, the pig will reach 18
pounds in _________ hours.
Jessica drives 130 miles every two hours.
If this rate continues, how long will it take
her to drive 1,000 miles?
Joe’s car goes 25 miles per gallon of
gasoline. How far can it go on 8
gallons of gasoline?
x8
Unit Rate
25 miles
1 gallon
=
8 gallons
x8
25 x 8 = 200. Joe’s car can go 200
miles on 8 gallons of gas.
```
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# CLASS-4DECIMALS - HUNDREDTHS
HUNDREDTHS -
Example.1)
Suppose, there are 100 small squares in a large square. One small square is
1
one-hundredth or -------- of the large square.
100
In above fraction, there are two zero in denominator, so decimal point or dot should move two steps left of numerator but there is only one numeric digit 1. So, now to justify two steps movement of decimal points we have to place one zero before given numerator and we have to place point towards two steps left.
1
So, ------- = 0.01
100
Example.2)
Let, if 8 small squares out of 100 squares are considered, then eight
8
small squares make eight hundredths or ------- of the large square.
100
In above fraction, there are two zero in denominator, so decimal point or dot should move two steps left of numerator but there is only one numeric digit 8. So, now to justify two steps movement of decimal points we have to place one zero before given numerator and we have to place point towards two steps left.
8
so, ------ = 0.08
100
Example.3)
Let, if 25 small squares out of 100 squares are considered, then twenty five
25
small squares make eight hundredths or ------- of the large square.
100
In above fraction, there are two zero in denominator, so decimal point or dot should move two steps left of numerator
25
so, -------- = 0.25
100
Example.4)
25 8425
84 ------ = --------
100 100
In above fraction, there are two zero in denominator, so decimal point or dot should move two steps left of numerator.
8425
-------- = 84.25
100
It is read as i) eighty four and twenty-five hundredths
ii) eighty four point two five
NOTE: The digits after the decimal point are read one by one. (like 84.25 is read as eighty four point two five and not as eighty four point twenty five)
In 84.25, the integral part (the part written before decimal point) is 84 and the decimal part (the part written after the decimal point) is 25.
There are some more examples are given below for your better understanding -
Tens Ones Dots Tenths Hundredths
eight point three four 8 . 3 4
seventeen point seven nine 1 7 . 7 9
point two seven 0 . 2 7
thirty nine point three five 3 9 . 3 5
seventy four point five six 7 4 . 5 6
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# Spectrum Math Grade 2 Chapter 6 Lesson 8 Answer Key Measuring Length in Inches
Go through the Spectrum Math Grade 2 Answer Key Chapter 6 Lesson 6.8 Measuring Length in Inches and get the proper assistance needed during your homework.
## Spectrum Math Grade 2 Chapter 6 Lesson 6.8 Measuring Length in Inches Answers Key
Perimeter is the length around an object.
The perimeter of this hexagon is 1 + 1 + 1 + 1 + 1 + 1 = 6 inches.
Use an inch ruler to measure length.
___________ inch
Step 1:Place the object beside the inch scale, starting from point 0 on it.
Step 2: Look at the endpoint and mark the value.
The estimated length of the given object will be 1 inch.
____________ inches
Step 1:Place the object beside the inch scale, starting from point 0 on it.
Step 2: Look at the endpoint and mark the value.
The estimated length of the given object will be 5 inches.
____________ inches
Step 1:Place the object beside the inch scale, starting from point 0 on it.
Step 2: Look at the endpoint and mark the value.
The estimated length of the given object will be 3 inches.
____________ inches
Step 1:Place the object beside the inch scale, starting from point 0 on it.
Step 2: Look at the endpoint and mark the value.
The estimated length of the given object will be 4 inches.
Measure the length of each side. Add the lengths of all sides to get the perimeter.
________ + _________ + __________ = ___________ inches
To measure the side length of the given figure, take an inch scale and place 0 point of the scale on one of the endpoint of a triangle side.
Look at the other endpoint and note down the value from the scale.
The above triangle has all equal sides, each measuring 2 inches.
To get the perimeter of the given figure, we need to add all the sides.
2+2+2 = 6 inches.
________ + ________ + ________ + ________ = ________ inches
To measure the side length of the given figure, take an inch scale and place 0 point of the scale on one of the endpoint of a square side.
Look at the other endpoint and note down the value from the scale.
The above figure has all equal sides, each measuring 2 inches.
To get the perimeter of the given figure, we need to add all the sides.
2+2+2+2 = 8 inches.
________ + ________ + ________ + ________ = ________ inches
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1. ## Circle problem
find a formula for the following function
the top half of a circle with center (a,b) and radius r
need some help with this one thanks...
2. Every point $\displaystyle (x,y)$ on the graph of a circle of radius $\displaystyle r$ satisfies the following relationship:
$\displaystyle x^2+y^2 = r^2$
So to express $\displaystyle y$ as a function of $\displaystyle x$ rewrite:
$\displaystyle y = \pm \sqrt{r^2-x^2}$
The top half of the circle is just $\displaystyle y = \sqrt{r^2-x^2}$
Now to shift this function to be centered at $\displaystyle (a,b)$ we have to shift the function up by $\displaystyle b$ and over by $\displaystyle a$.
$\displaystyle y = \sqrt{r^2-(x-a)^2}+b$
3. Originally Posted by mathlete
find a formula for the following function
the top half of a circle with center (a,b) and radius r
need some help with this one thanks...
The equation of a circle with center (a,b) is given by
$\displaystyle (x-a)^2+(y-b)^2=r^2$
to find the top half we need to isolate y and take the positive square root
$\displaystyle (y-b)^2=r^2-(x-a)^2$ taking the postive root
$\displaystyle y-b=\sqrt{r^2-(x-a)^2 }$
so we get...
$\displaystyle y=b+\sqrt{r^2-(x-a)^2 }$
4. The equation of a circle is:
$\displaystyle (x-h)^2 + (y-k)^2 = r^2$
Where (h,k) is the center of the circle.
The equation for the two halves of a circle are given by isolating y.
$\displaystyle y = \pm \sqrt {r^2 - (x-h)^2} + k$
We are only interested in the top half so we only take the positive square root:
$\displaystyle y = \sqrt {r^2 - (x-h)^2} + k$
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# Find the equation for the ellipse that satisfies the given conditions : Length of minor axis $16$, foci $(0, \pm 6).$
$\begin {array} {1 1} (A)\;\large\frac{x^2}{100}+\large\frac{y^2}{64}=1 & \quad (B)\;\large\frac{x^2}{64}+\large\frac{y^2}{100}=1 \\ (C)\;\large\frac{x^2}{100}-\large\frac{y^2}{64}=1 & \quad (D)\;\large\frac{x^2}{64}-\large\frac{y^2}{100}=1 \end {array}$
Toolbox:
• Length of the major axis is 2a
• Length of the minor axis is 2b
• Equation of an ellipse whose major axis is along x - axis is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
• Equation of an ellipse whose minor axis is along y - axis is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
• $c = \sqrt{a^2-b^2}$ whose $c$ is the foci.
Step 1 :
Given length of the minor axis = 16.
Foci = $(0, \pm 6)$
Since the foci is on the y - axis, the major axis is along y - axis.
Hence the equation of the ellipse should be of the form.
$\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
Step 2 :
Length of the minor axis = 16.
(i.e) $2b=16$
$\Rightarrow b = 8$
also $c = 6$
$a^2=b^2+c^2$
$= 8^2+6^2$
$= 64 + 36$
$= 100$
Hence the equation of the ellipse is
$\large\frac{x^2}{64}$$+\large\frac{y^2}{100}$$=1$
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# What is the make a 10 strategy?
## What is the make a 10 strategy?
In 1st grade, as students begin learning their basic addition facts, they apply that knowledge in a strategy known as “make a ten” to help make sense of facts that might otherwise be hard to memorize, such as 8 + 4 or 9 + 5. To use the strategy, students decompose one of the addends to make a ten from the other.
Why do we use 10 frames?
Ten frames are an amazing tool used in kindergarten and first grade to help your children understand counting, place value (e.g. where the digit in a number is), adding, subtracting, and more. Young learners who have a solid understanding of this number, can relate that understanding to all different areas of math.
### Why is the make a ten strategy helpful?
Why Is the Making 10 Strategy Great for Addition? The make-ten strategy is great for addition! It helps students understand place value and the relationships between numbers. Ten-frames help students develop a good “mind picture” for the make-ten strategy because our place-value system is based on making groups of ten.
What are the counting strategies?
Counting from one. The most basic strategy for solving addition and subtraction problems is to count from one. For example, to solve 6 + 3, students could count a set of 6 objects, then count a set of 3 objects, then join the two together and count to find that there are 9 in the combined group.
#### What is a ten frame used for?
Ten frames are an amazing tool used in kindergarten and first grade to help your children understand counting, place value (e.g. where the digit in a number is), adding, subtracting, and more.
How do you explain ten frames?
Ten-Frames are two-by-five rectangular frames into which objects like counters can be placed to show numbers less than or equal to ten. They’re a common teaching tool for LKS1 Maths students and are often used to develop children’s number sense within the context of ten.
## What are two ways to decompose a number?
When you are given a number with two digits, the number has a “ones” place piece and a “tens” place piece. To decompose this number, you will need to separate it into its separate pieces. Example: Decompose the number 82. The 8 is in the “tens” place, so this part of the number can be separated and written as 80.
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GreeneMath.com - Solving Quadratic Equations using the Square Root Property Test
Square Root Property Test
About Square Root Property:
We previously spoke about quadratic equations and learned how to solve a quadratic equation when it is factorable. Here we begin to develop techniques to solve any quadratic equation, whether it is factorable or not. Specifically we will focus on solving simple quadratic equations using the square root property.
Test Objectives:
•Demonstrate a general understanding of the square root property
•Demonstrate the ability to solve a quadratic equation of the form: x2 = k
•Demonstrate the ability to solve a quadratic equation of the form: (x + a)2 = k
Square Root Property Test:
#1:
Instructions: Solve each equation.
a) $$a^2 = 36$$
b) $$n^2 = 44$$
Watch the Step by Step Video Solution
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View the Written Solution
#2:
Instructions: Solve each equation.
a) $$5r^2 + 7 = 132$$
b) $$-4 - 6v^2 = -388$$
Watch the Step by Step Video Solution
|
View the Written Solution
#3:
Instructions: Solve each equation.
a) $$(12x + 4)^{2} = 400$$
Watch the Step by Step Video Solution
|
View the Written Solution
#4:
Instructions: Solve each equation.
a) $$(9x - 2)^2 = 121$$
Watch the Step by Step Video Solution
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View the Written Solution
#5:
Instructions: Solve each equation.
a) $$(7n - 15)^2 = 10$$
Watch the Step by Step Video Solution
|
View the Written Solution
Written Solutions:
#1:
Solution:
a) $$a = 6$$ or $$a = -6$$
b) $$n = 2\sqrt{11}$$ or $$n = -2\sqrt{11}$$
Watch the Step by Step Video Solution
#2:
Solution:
a) $$r = 5$$ or $$r = -5$$
b) $$v = 8$$ or $$v = -8$$
Watch the Step by Step Video Solution
#3:
Solution:
a) $$x = -2$$ or $$x = \frac{4}{3}$$
Watch the Step by Step Video Solution
#4:
Solution:
a) $$x = -1$$ or $$x = \frac{13}{9}$$
Watch the Step by Step Video Solution
#5:
Solution:
a) $$n = \frac{\sqrt{10} + 15}{7}$$ or $$n = \frac{-\sqrt{10} + 15}{7}$$
Watch the Step by Step Video Solution
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# Extracting the common factor in parentheses - Examples, Exercises and Solutions
Common Factor Extraction Method:
Identify the largest free number that we can extract.
Then, let's move on to the variables and ask what is the least number of times the $X$ appears in any element?
Multiply the free number by the variable the same number of times we have found and we will obtain the greatest common factor.
To verify that you have correctly extracted the common factor, open the parentheses and see if you have returned to the original exercise.
### Suggested Topics to Practice in Advance
1. Factoring using contracted multiplication
## Practice Extracting the common factor in parentheses
### Exercise #1
$2x^{90}-4x^{89}=0$
### Step-by-Step Solution
The equation in the problem is:
$2x^{90}-4x^{89}=0$Beginning Let's pay attention to the left side The expression can be broken down into factors by taking out a common factor, The greatest common factor for the numbers and letters in this case is $2x^{89}$and that is because the power of 89 is the lowest power in the equation and therefore included both in the term where the power is 90 and in the term where the power is 89, any power higher than that is not included in the term where the power of 89 is the lowest, and therefore it is the term with the highest power that can be taken out of all the terms in the expression as a common factor for the letters,
For the numbers, note that the number 4 is a multiple of the number 2, so the number 2 is the greatest common factor for the numbers for the two terms in the expression,
Continuing if so and performing the factorization:
$2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0$Let's continue and refer to the fact that on the left side of the equation that was obtained in the last step there is an algebraic expression and on the right side the number is 0, so, since the only way to get the result 0 from a product is for at least, one of the factors in the product on the left side, must be equal to zero,
Meaning:
$2x^{89}=0 \hspace{8pt}\text{/}:2\\ x^{89}=0 \hspace{8pt}\text{/}\sqrt[89]{\hspace{6pt}}\\ \boxed{x=0}$In the solution of the equation above, we first equated the two sides of the equation by moving the term with the unknown and then we took out a root of order 89 on both sides of the equation.
(In this case taking out an odd-order root on the right side of the equation yields one possibility)
Or:
$x-2=0 \\ \boxed{x=2}$
In summary if so the solution to the equation:
$2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0 \\ \downarrow\\ 2x^{89}=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$And therefore the correct answer is answer a.
$x=0,2$
### Exercise #2
Extract the common factor:
$4x^3+8x^4=$
### Step-by-Step Solution
First, we use the power law to multiply terms with identical bases:
$a^m\cdot a^n=a^{m+n}$It is necessary to keep in mind that:
$x^4=x^3\cdot x$Next, we return to the problem and extract the greatest common factor for the numbers separately and for the letters separately,
For the numbers, the greatest common factor is
$4$and for the letters it is:
$x^3$and therefore for the extraction
$4x^3$outside the parenthesis
We obtain the expression:
$4x^3+8x^4=4x^3(1+2x)$To determine what the expression inside the parentheses is, we use the power law, our knowledge of the multiplication table, and the answer to the question: "How many times do we multiply the common factor that we took out of the parenthesis to obtain each of the terms of the original expression that we factored?
Therefore, the correct answer is: a.
It is always recommended to review again and check that you get each and every one of the terms of the expression that is factored when opening the parentheses (through the distributive property), this can be done in the margin, on a piece of scrap paper, or by marking the factor we removed and each and every one of the terms inside the parenthesis, etc.
$4x^3(1+2x)$
### Exercise #3
Solve the following by removing a common factor:
$6x^6-9x^4=0$
### Step-by-Step Solution
First, we take out the smallest power
$6x^6-9x^4=$
$6x^4\left(x^2-1.5\right)=0$
If possible, we reduce the numbers by a common factor
Finally, we will compare the two sections with: $0$
$6x^4=0$
We divide by: $6x^3$
$x=0$
$x^2-1.5=0$
$x^2=1.5$
$x=\pm\sqrt{\frac{3}{2}}$
$x=0,x=\pm\sqrt{\frac{3}{2}}$
### Exercise #4
$x^4+2x^2=0$
### Video Solution
$x=0$
### Exercise #5
$x^2-x=0$
### Video Solution
$x=0,1$
### Exercise #1
$7x^3-x^2=0$
### Video Solution
$x=0,x=\frac{1}{7}$
### Exercise #2
$7x^{10}-14x^9=0$
### Video Solution
$x=2,x=0$
### Exercise #3
$x^4+x^2=0$
### Video Solution
$x=0$
### Exercise #4
$x^6+x^5=0$
### Video Solution
$x=-1,x=0$
### Exercise #5
$x^6-4x^4=0$
### Video Solution
$x=0,x=\pm2$
### Exercise #1
$x^5-4x^4=0$
### Video Solution
$x=4,x=0$
### Exercise #2
$3x^2+9x=0$
### Video Solution
$x=0,x=-3$
### Exercise #3
$4x^4-12x^3=0$
Solve the equation above for x.
### Video Solution
$x=0,3$
### Exercise #4
$x^7-x^6=0$
### Video Solution
$x=0,1$
### Exercise #5
$x^7-5x^6=0$
### Video Solution
$x=0,5$
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## What is the formula of cross-sectional area of a circle?
The area of a circle is given by the formula πr2, where r is the radius. It therefore makes sense that the volume of a cylinder would be the area of one of the circles forming its base. If the cross-section is parallel to the axis of symmetry, then the area of the cross-section is simply a circle with an area of πr2.
Is cross-sectional area the same as diameter?
Cross section is an area. Diameter is a linear measure. That cannot be the same. The cross section or the cross sectional area is the area of such a cut.
What is area of cross-section?
The cross-sectional area is the area of a two-dimensional shape that is obtained when a three-dimensional object – such as a cylinder – is sliced perpendicular to some specified axis at a point. For example, the cross-section of a cylinder – when sliced parallel to its base – is a circle.
### How do you find the minimum cross-sectional area?
Minimum Cross-sectional Area:
1. R =
2. A = Eq. (
3. A =
4. 4*A =
5. =
6. d =
7. d = 4*0.0000645 3.141592 0.5
What is area of cross section?
How do you solve for cross-sectional area?
Cross-sectional area is determined by squaring the radius and then multiplying by 3.14. For example, if a tree is measured as 10” DBH, the radius is 5”. Multiplying 5 by 5 equals 25, which when multiplied by 3.14 equals 78.5. Thus, the cross-sectional area of a 10” DBH tree is 78.5.
#### What is the formula to calculate the area of a circle?
Area of a circle. The formula for the area of a circle is 2 x π x radius, but the diameter of the circle is d = 2 x r, so another way to write it is 2 x π x (diameter / 2).
How do you find the sector of a circle?
A sector in a circle is the region bound by two radii and the circle. Since it is a fractional part of the circle, the area of any sector is found by multiplying the area of the circle, π × r 2, by the fraction x /360, where x is the measure of the central angle formed by the two radii.
Can we find the exact area of a circle?
With that, we can find the exact area. The formula for the area of a circle is: A = πr². Thus, we need to find the diameter of the circle and divide it by 2 to find the radius. To find the diameter of a circle given the circumference, use the formula: d = C / π. d = 8π / π.
## What is the formula for the areaof a circle?
The area of a circle is the amount of space the circle covers. The formula for calculating the area of a circle is A = π_r_ 2 where pi (π) equals 3.14 and the radius (r) is half the diameter.
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Vibrations
# 4 Amplitude and period on time graphs
It’s easy to read amplitude and period from a time domain graph. This chapter shows how.
### Period and amplitude from time domain graphs
The horizontal “distance” between repetitions on the graph equals the period. The vertical distance from the middle of the vibration to either the “top” or the “bottom.” The diagrams below show period and amplitude for SHOs and complex vibrations.
For complex vibrations, the time between complete repeats of the complex pattern equals the period of the fundamental, which is sometimes called the fundamental period.
You can also use the equations from Chapter 2 in conjunction with the graph to find period or frequency:
$\tau = \frac{amount \: of \: time \: for \: N \: cycles}{N}$
$f = \frac{N}{amount \: of \: time \: for \: N \: cycles}$
Identify two points on the graph and find out the number of cycles (N) and the amount of time in between the two points you’ve identified on the graph and plug the quantities into the appropriate equation.
Example: Time graph 1
Find the period and amplitude of the vibration shown on the graph:
Here’s the diagram again, with a few labels added. The dotted orange line shows the center line (equilibrium) of the vibration:
To find period, find the amount of time between one peak and the next. There are lots of ways to do it. You could read the time axis under the peaks marked A and B and subtract the two times. Peak A occurs at t=5 ms and Peak B occurs at t=10 ms (5 ms later), so the period is 5 ms. You could also notice that there are four cycles and 20 ms between points A and D- so each cycle takes 5 ms.
To find amplitude, you can measure the vertical “distance” from the highest point on the graph to the mid-line. The top of each peak aligns with the vertical axis at 7 Volts and the mid-line lines up with 3 Volts, so the amplitude is 4 Volts. If you don’t want to find the mid-line, you can take the vertical “distance” from top of the vibration to the bottom (8 Volts) and divide by two.
Units spotlight: Get the units from the graph
The axis labels of scientific graphs show the units for each quantity on the graph. For time domain graphs, this means the amplitude will have the same units as the quantity on the vertical axis (y-axis) of the graph. The period will always have the same (time) units as the x-axis.
Finding the fundamental frequency of a sound from a graph
The graph below shows a short sample of a sound played by a plastic recorder. What’s the fundamental frequency of the note?
To find frequency, find out how many cycles happen in how much time. To do that, choose two points on the graph, count the cycles between the two points and measure the amount of time between the two points. I’ve chosen the points indicated by the arrows on the graph below.
In principle, any pair of points will work. I’ve chosen one point at time = 0.00 s and the other so that there’s a whole number of complete cycles between the arrows. That’s mainly for convenience. I’ve also chosen two points about as far apart as I can get. That way, if I’m off by a little bit when I estimate the time, my final answer won’t be affected very much. Count up the number of cycles between the arrows (N=21) and read off the amount of time that those cycles take (about 0.0295 s) and substitute into the definition of frequency:
$f = \frac{N}{amount \: of \: time \: for \: N \: cycles}=\frac{21 \: \text{cycle}}{0.0295 \: \text{s}}= 712 \: \text{Hz}$
This answer seems reasonable- 712 Hz is a little higher than A 440 Hz, which is the note the orchestra tunes to.
### Image credits
1. Period and amplitude of a SHO. Created by David Abbott at Desmos.com.
2. Period and amplitude of a complex vibration. Created by David Abbott at Desmos.com.
3. Graphs for plastic recorder example. Created by David Abbott using Logger Pro.
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# The Method of Cylindrical Shells Help
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## Introduction to The Method of Cylindrical Shells
Our philosophy will now change. When we divide our region up into vertical strips, we will now rotate each strip about the y -axis instead of the x -axis. Thus, instead of generating a disk with each strip, we will now generate a cylinder.
Look at Fig. 8.19. When a strip of height h and thickness Δ x , with distance r from the y -axis, is rotated about the y -axis, the resulting cylinder has surface area 2πr · h and volume about 2πr · h · Δ x . This is the expression that we will treat in order to sum up the volumes of the cylinders.
Fig. 8.19
#### Example 1
Use the method of cylindrical shells to calculate the volume of the solid enclosed when the curve y = x2 , 1 ≤ x ≤ 3, is rotated about the y -axis.
#### Solution 1
As usual, we think of the region under y = x2 and above the x -axis as composed of vertical segments or strips. The segment at position x has height x2. Thus, in this instance, h = x2, r = x , and the volume of the cylinder is 2πx · x2 · Δ x . As a result, the requested volume is
We easily calculate this to equal
#### Example 2
Use the method of cylindrical shells to calculate the volume enclosed when the curve y = x 2 , 0 ≤ x ≤ 3, is rotated about the x -axis ( Fig. 8.20 ).
Fig. 8.20
#### Solution 2
We reverse, in our analysis, the roles of the x - and y -axes. Of course y ranges from 0 to 9. For each position y in that range, there is a segment stretching from to x = 3. Thus it has length . Then the cylinder generated when this segment (thickened to a strip of width Δ y ) is rotated about the x -axis has volume
The aggregate volume is then
You Try It: Use the method of cylindrical shells to calculate the volume enclosed when the region 0 ≤ y ≤ sin x , 0 ≤ x ≤ π/2, is rotated about the y -axis.
## Different Axes
Sometimes it is convenient to rotate a curve about some line other than the coordinate axes. We now provide a couple of examples of that type of problem.
#### Example 1
Use the method of washers to calculate the volume of the solid enclosed when the curve , is rotated about the line y = −1. See Fig. 8.21.
Fig. 8.21
#### Solution 1
The key is to notice that, at position x , the segment to be rotated has height the distance from the point on the curve to the line y = −1. Thus the disk generated has area . The resulting aggregate volume is
You Try It: Calculate the volume inside the surface generated when is rotated about the line y = −3, 1 ≤ x ≤ 4.
#### Example 2
Calculate the volume of the solid enclosed when the area between the curves x = ( y − 2)2 + 1 and x = −( y − 2)2 + 9 is rotated about the line y = −2.
Fig. 8.22
#### Solution 2
Solving the equations simultaneously, we find that the points of intersection are (5, 0) and (5, 4). The region between the two curves is illustrated in Fig. 8.22.
At height y , the horizontal segment that is to be rotated stretches from (( y − 2)2 + 1, y ) to (−( y − 2)2 + 9, y ). Thus the cylindrical shell that is generated has radius y − 2, height 8 − 2( y − 2)2 , and thickness Δ y . It therefore generates the element of volume given by
2 · ( y − 2) · [8 − 2( y − 2)2 ] · Δ y.
The aggregate volume that we seek is therefore
You Try It: Calculate the volume enclosed when the curve y = cos x is rotated about the line y = 4, ≤ x ≤ 3 .
Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.
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# 4.3 Systems of Equations - Elimination Objective: The student will be able to: Solve systems of equations using elimination with addition and subtraction.
## Presentation on theme: "4.3 Systems of Equations - Elimination Objective: The student will be able to: Solve systems of equations using elimination with addition and subtraction."— Presentation transcript:
4.3 Systems of Equations - Elimination Objective: The student will be able to: Solve systems of equations using elimination with addition and subtraction.
Solving Systems of Equations So far, we have solved systems using graphing and substitution. These notes show how to solve the system algebraically using ELIMINATION with addition and subtraction. Elimination is easiest when the equations are in standard form.
Solving a system of equations by elimination using addition and subtraction. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add or subtract the equations. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same (either + or -) coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations.
1) Solve the system using elimination. x + y = 5 3x – y = 7 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! The y’s have the same coefficient. Step 3: Add or subtract the equations. Add to eliminate y. x + y = 5 (+) 3x – y = 7 4x = 12 x = 3
1) Solve the system using elimination. Step 4: Plug back in to find the other variable. x + y = 5 (3) + y = 5 y = 2 Step 5: Check your solution. (3, 2) (3) + (2) = 5 3(3) - (2) = 7 The solution is (3, 2). What do you think the answer would be if you solved using substitution? x + y = 5 3x – y = 7
2) Solve the system using elimination. 4x + y = 7 4x – 2y = -2 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. The x’s have the same coefficient. Step 3: Add or subtract the equations. Subtract to eliminate x. 4x + y = 7 (-) 4x – 2y = -2 3y = 9 y = 3
2) Solve the system using elimination. Step 4: Plug back in to find the other variable. 4x + y = 7 4x + (3) = 7 4x = 4 x = 1 Step 5: Check your solution. (1, 3) 4(1) + (3) = 7 4(1) - 2(3) = -2 4x + y = 7 4x – 2y = -2
Solving a system of equations by elimination using addition and subtraction. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add or subtract the equations. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same (either + or -) coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations.
Solve using elimination. 2x – 3y = -2 x + 3y = 17 1. (2, 2) 2. (9, 3) 3. (4, 5) 4. (5, 4)
3) Solve the system using elimination. y = 7 – 2x 4x + y = 5 Step 1: Put the equations in Standard Form. 2x + y = 7 4x + y = 5 Step 2: Determine which variable to eliminate. The y’s have the same coefficient. Step 3: Add or subtract the equations. Subtract to eliminate y. 2x + y = 7 (-) 4x + y = 5 -2x = 2 x = -1
2) Solve the system using elimination. Step 4: Plug back in to find the other variable. y = 7 – 2x y = 7 – 2(-1) y = 9 Step 5: Check your solution. (-1, 9) (9) = 7 – 2(-1) 4(-1) + (9) = 5 y = 7 – 2x 4x + y = 5
Solving a system of equations by elimination using addition and subtraction. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add or subtract the equations. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same (either + or -) coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations.
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Home / General Studies / Competetive Zone / What is the angle between two sides of a Polygon
# What is the angle between two sides of a Polygon
In our regular life we can see many types of figures/ shapes/structures. A simple closed figure which is formed by three or more lines or more than three lines is called a polygon.
If a polygon has n-sides then it is said to be N-gon. i.e. for example see the below polygon it has 4 sides and it is said to be 4-gon.
Diagonal of a polygon:- A line segment which is joining any two non-consecutive vertices of a polygon is called its diagonal. Observe the Figure-1 Here A and C joins with a line segment AC like wise BD also so here AC,BD are called Diagonal of the polygon.
Here Vertices are A,B,C and D.
To find the No.of diagonal of a n-sided polygon we can use the below formula
[n(n-3)]/2 here n is no of sides of the polygon.
Consider the above figure we have n is 4 here and substitute the n value as 4 here then we get 2.
Polygons are basically Two types those are:
1. Convex polygon
2.concave polygon
Convex polygon: In this type of polygon each angle is measured less than 180 degrees means two right angles such a polygon is called convex polygon.
Concave polygon: In this type of polygon if an angle is measured 180 degrees means two right angles such a polygon is called concave polygon.
Angles of a polygon: Basically polygons having two types of angles those are Exterior and interior angles.
Interior angle Exterior angle:
An angle inside the polygon is called the interior angle (Green color i the picture) and an angle outside of the polygon is called exterior angle(Red color in the Picture).
Note 1:
Sum of the outside angle of polygon is 360 degrees.
Each exterior angle is measured by using 360/n here n is side of the polygon.
Note 2:
Sum of interior angle of a polygon is (n-2)*180 degrees. here n is side of the polygon.
Each interior angle is measured by using [(n-2)*180]/n , here n is side of the polygon.
• For example a polygon has 5 sides then find the each exterior and interior angle and sum of the interior and exterior angle.
• Sum of exterior angle is =360 degrees.
• Sum of interior angle is (n-2)*180 here n is 5
(5-2)*180=540 degrees.
• Each exterior angle is 360/n n is 5 here so 360/5=72 degrees.
• Each interior angle is [(n-2)*180]/n here n is 5 so by using formula we get 108 degrees.
• Apart from this polygons are divide into two those are Equilateral and Equiangular.
• A polygon in which all sides are equal then is called equilateral polygon . example is Rhombus.
• A polygon in which all angles are equal then is called equiangularl polygon . example is rectangle.
Hi, i am Santosh Gadagamma, a tutor in Software Engineering and an enthusiast for sharing knowledge in Computer Science and other domains. I developed this site to share knowledge to all the aspirants of technologies like, Java, C/C++, DBMS/RDBMS, Bootstrap, Big Data, Javascript, Android, Spring, Hibernate, Struts and all levels of software project design, development, deployment, and maintenance. As a programmer I believe that, “The world now needs computers to function.” Hope, this site guides you as a learning tool towards greater heights. I believe that Education has no end points and i wish to learn more in the process of teaching you….,
## What is Antibody
Antibody (plural-antibodies) Antibodies are the proteins which are found in the human body. They are …
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The shortest distance b/w the line $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z= 0, 2x-y+z+3= 0$ is Option: 1 Option: 2 Option: 3 Option: 4
The shortest distance between the lines
$\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1} \text { and } x+y+z+1=0=2x-y+z+3=0$
\begin{aligned} &\text { Line of intersection of planes } \mathrm{x}+\mathrm{y}+\mathrm{z}+1=0=2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3 \text { is }\\ &\text { Eliminating y gives } 3 x+2 z+4=0 \end{aligned}
$\Rightarrow \mathrm{x}=\frac{-2 \mathrm{z}-4}{3}$
\begin{aligned} &\text { Substituting above } \mathrm{x} \text { in } \mathrm{x}+\mathrm{y}+\mathrm{z}+1=0 \text { gives } 3 \mathrm{y}+\mathrm{z}-1=0\\ &\Rightarrow \mathrm{z}=-3 \mathrm{y}+1-(2) \end{aligned}
$\text { From (1) and (2), line equation is } \frac{3 x+4}{-2}=-3 y+1=z$
$\Rightarrow \frac{\mathrm{x}-(-4 / 3)}{-2 / 3}=\frac{\mathrm{y}-(1 / 3)}{-1 / 3}=\frac{\mathrm{z}}{1}$
Given line is $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$
Shortest distance between above two skew lines is $\frac{(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})}{|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|}$
$\begin{array}{l} \overrightarrow{\mathrm{a}}=(1,-1,0) \\\\ \overrightarrow{\mathrm{b}}=\left(-\frac{4}{3}, \frac{1}{3}, 0\right) \\\\ \overrightarrow{\mathrm{c}}=(0,-1,1) \\\\ \overrightarrow{\mathrm{d}}=\left(-\frac{2}{3},-\frac{1}{3}, 1\right) \end{array}$
$\therefore \frac{(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})}{|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}|}=\left|\begin{array}{|ccc|} 7 / 3 & -4 / 3 & 0 \\ 0 & -1 & 1 \\ -2 / 3 & -1 / 3 & 1 \\ \hline \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -1 & 1 \\ -2 / 3 & -1 / 3 & 1 \end{array}\right|=\frac{1}{\sqrt{3}}$
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### MEAP Preparation - Grade 7 Mathematics2.41 Solving Equations
Simplifying with Addition and Subtraction: We can use addition and subtraction to get all the terms with variables on one side of an equation, and all the numeric terms on the other. The equations 3x = 17, 21 = y, and z/12 = 24 each have a variable term on one side of the '=' sign, and a number on the other. The equations x + 3 = 12, 21 = 30 - y, and (z + 2) × 4 = 10 do not. We usually do this after simplifying each side using the distributive rules, eliminating parentheses, and combining like terms. Since addition is associative, it can be helpful to add a negative number to each side instead of subtracting. Examples: For the equation 3x + 4 = 12, we can separate the variable term on the left by subtracting a 4 from both sides: 3x + 4 - 4 = 12 - 4 3x = 8. x = 8/3 For the equation 7y - 200 = 10, if we add 200 to both sides of the equation, the 200 and -200 will cancel each other on the left hand side: 7y + (-200) + 200 = 10 + 200 7y = 210 y = 30 Example: For the equation 8 = 20 - z, we can add z to both sides to get 8 + z = 20 - z + z => 8 + z = 20. Now subtracting 8 from both sides, 8 + z - 8 = 20 - 8 Therefore, z = 12 Simplifying by Multiplication When solving for a variable, we want to get a solution like x = 3 or z = 198 etc. When a variable is divided by some number, we can use multiplication on both sides to solve for the variable. Example: Solve for x in the equation x ÷ 12 = 5. Since the x on the left side is being divided by 12, the equation is the same as x × 1/12 = 5. Multiplying both sides by 12 will cancel the 1/12 on the left side: x × 1/12 × 12 = 5 × 12 ==> x × 1 = 60 ==> x = 60. Simplifying by Division When solving for a variable, we want to get a solution like x = 9 or y = 21 etc. When a variable is multiplied by some number, we can use division on both sides to solve for the variable. Example: Solve for x in the equation 7x = 280. Since the x on the left side is being multiplied by 7, we can divide both sides by 7 to solve for x: 7x ÷ 7 = 280 ÷ 7 (7x)/7 = 280 ÷ 7 x/1 = 40 x = 40. Note: Dividing by 7 is the same as multiplying both sides by 1/7. Directions: Solve the following equations and enter the solution. Also write at least 10 examples of your own.
Q 1: Solve: x/7 =4Answer: Q 2: Solve: 5x =45Answer: Q 3: Solve: 7u = 35Answer: Q 4: Solve: d + 16 = 64Answer: Q 5: Solve: 6k = 24Answer: Q 6: Solve: d - 12 =69Answer: Q 7: Solve: u + 16 = 18Answer: Q 8: Solve: f + 16 =100Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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# Point Slope Form Calculator
## How to use the point-slope form calculator?
Point slope form calculator is an easy-to-use tool. Follow the below steps to use this tool.
• Input x1, y1, and the slope m in the required input boxes.
• Click the calculate button
• Press the reset button to enter new values.
Result
Data:
Formula:
Solution:
## Point slope form calculator with steps
Point slope form calculator is an online tool used to find the linear equation of the line by using the coordinate points (x1, y1) and the slope of the line.
There are various methods to find the equation of the straight line such as point-slope form, slope-intercept form, and two-point intercept form
## What is point-slope form?
The point-slope form is a technique used to find the straight-line equation. The point-slope form follows an equation:
(y – y1) = m(x – x1)
• x & y are the fixed points of the line.
• (x1, y1) are the coordinate points of the line.
• & m is the slope of the line
## How to find the point-slope form?
Below are a few solved examples of the point-slope form.
### For positive coordinate points
Example
Find the linear equation of the line if the slope is 4 and the coordinate points are (3, 5).
Solution
Step 1: Identify the given values.
x1 = 3
y1 = 5
slope = m = 4
Step 2: Take the formula of the point-slope form and substitute the given values.
(y – y1) = m(x – x1)
(y – 5) = 4(x – 3)
y – 5 = 4x – 12
y – 5 – 4x + 12 = 0
y – 4x + 7 = 0
Multiply by -1 on both sides of the above expression.
-1(y – 4x + 7) = -1(0)
-y + 4x – 7 = 0
4x – y – 7 = 0
### For negative coordinate points
Example
Find the linear equation of the line if the slope is 3.5 and the coordinate points are (-13, -12).
Solution
Step 1: Identify the given values.
x1 = -13
y1 = -12
slope = m = 3.5
Step 2: Take the formula of the point-slope form and substitute the given values.
(y – y1) = m(x – x1)
(y – (-12)) = 3.5(x – (-13))
(y + 12) = 3.5(x + 13)
y + 12 = 3.5x + 45.5
y + 12 – 3.5x – 45.5 = 0
y – 3.5x – 33.5 = 0
Multiply by -1 on both sides of the above expression.
-1(y – 3.5x – 33.5) = -1(0)
-y + 3.5x + 33.5 = 0
3.5x – y + 33.5 = 0
Try the point slope form calculator to cross-check the above result.
5 months ago
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# Write The Prime Factorization Of 27
Write The Prime Factorization Of 27. 2 × 3 × 5 prime factorization of 31: Therefore, the prime factors of 27 are 3, 3 and 3. When we count the number. The prime factorization of a positive integer is a list of.
Prime number calc prime factorization of 27 what is the prime factorization of 27 [solved] answer the prime factors of 27: Therefore, the factorization of 27 gives us, 1 × 3 × 3 × 3. For calculation, here's how to calculate prime factorization of 275 using the formula above, step by step instructions are given below.
## Prime number calc prime factorization of 27 what is the prime factorization of 27 [solved] answer the prime factors of 27:
Divide the given composite number by its smallest prime factor if the resulting quotient is a composite number, divide it by its smallest prime factor repeat until resulting quotient is a. So, it is possible to draw its prime tree. Prime number calc prime factorization of 27 what is the prime factorization of 27 [solved] answer the prime factors of 27:
## So, It Is Possible To Draw Its Prime Tree.
Thus, the prime factors of 27 are: Consider 36, 756, and 1001. Solution for write the prime factorization of the number. The prime factorization of 27 = 3 3.
### The Prime Factorization Of 27 = 3 3.
To check your answer multiply all the numbers together in the prime factor list and this should match your original number.
### Kesimpulan dari Write The Prime Factorization Of 27.
2 2 × 7 prime factorization of 29: The number 27 is a composite number. 2 × 3 × 5 prime factorization of 31: Factors of 27 the factors of 27 are the numbers, which produce the result as 27 when a pair of factors are multiplied together.
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# A manufacturer produces 80 units of a product at a cost of Rs 22000 and 125 units at a cost of Rs 28750. Assuming the cost curve to be linear, find the equation of the cost curve and then use it to estimate the cost of 95 units.
Given:
The total cost of 80 units of a product at a cost of Rs 22000 and 125 units at a cost of Rs 28750.
To do:
We have to find the cost of 95 units.
Solution:
The cost curve is linear.
Let its equation will be $y = Ax + B$, where $y =$ Total cost and $x =$ Number of units.
This implies,
$22000 = 80A + B$
$80A + B = 22000$.........(i)
$28750 = 125A + B$
$125A + B = 28750$.........(ii)
Solving (i) and (ii), we get,
$125A+B-80A-B=28750-22000$
$45A=6750$
$A=\frac{6750}{45}$
$A=150$
$\Rightarrow 80(150)+B =22000$
$B=22000-12000$
$B=10000$
Therefore, the equation of the cost curve is $y = 150x + 10000$..........(iii)
This implies, the cost of 95 units is,
$y = 150(95) + 10000$
$y = 14250 + 10000$
$y = 24250$
Hence, the cost of 95 units is Rs. 24250.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
67 Views
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# Integrated Reasoning – Two Part Analysis Questions
Let’s continue our series and look at another Integrated Reasoning question type today – two part analysis. As complicated as it sounds, it’s actually the simplest of the IR question types in my opinion. The reason for this is that it tests no new skills; it checks your ability to handle the same old PS and CR questions.
The only reason it is new is that it reduces the probability of guessing correctly and it puts more time pressure on you! Your probability of guessing correctly is 20% in PS/CR questions; it goes down to 4% in two part analysis because you have to guess correctly twice. As for time pressure, you get about 2 mins for every PS and about 1.5 mins for every CR question. For each part of two part analysis, you have only 1.25 mins.
Anyway, let’s look at a sample question to get familiar with this question type.
Question: A grocery store sells fruits in pre packed closed bags such that individual pieces of fruit are not sold. Mangoes are sold at the rate of \$5 per bag (each bag contains two mangoes) and apples at the rate of \$8 per bag (each bag contains five apples). During a particular day, the store started with some mangoes and apples and sold them all by the end of the day. The revenue at the end of the day from selling mangoes and apples that day was \$128. Which of the following could represent the number of mangoes and the number of apples that were in the store at the beginning of that day?
Choose only one from each column:
Solution:
Note that it is a PS question. Only the format of the question is different. Also, the use of the word ‘could’ in the question stem suggests that there could be multiple solutions to this problem. Let’s take a closer look at how to solve it.
Say, number of bags of mangoes is ‘m’ and number of bags of apples is ‘p’.
Then 5m + 8p = \$128 (total revenue)
Each bag of mangoes has 2 mangoes and each bag of apples has 5 apples.
So number of mangoes sold = 2m (to be selected in the first column)
Number of apples sold = 5p (to be selected in the second column)
We need to solve for this equation: 5m + 8p = 128
It is easy to see that one solution to this equation is m = 0, p = 16. The next solution will be m = 8, p = 11. Another will be m = 16, p = 6. Yet another will be m = 24, p = 1. If you are wondering how we are landing on one solution after another so effortlessly, you need to check out a previous post of QWQW – Integral Solutions to Equations in Two Variables.
So there are three solutions possible to our question: Which of the following could represent the number of mangoes and the number of apples that were in the store at the beginning of that day?
There are three different cases possible:
Case 1: Number of mangoes sold could be 16 (= 2m when m is 8). In that case number of apples sold will be 55 ( = 5p when m is 8, p is 11)
Case 2: Number of mangoes sold could be 32 (= 2m when m is 16). In that case number of apples sold will be 30 ( = 5p when m is 16, p is 6)
Case 3: Number of mangoes sold could be 48 ( = 2m when m is 24). In that case number of apples sold will be 5 (= 5p when m is 24, p is 1)
The case we select should be that of which both numbers are included in the options. Case 3 satisfies this condition. So we select 48 in the first column and 5 in the second column.
This is the only ‘exotic’ step of the two part analysis. The rest of the question is just like any other PS question.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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# Need to calculate:The factorization of 3p^{3}-2p^{2}-9p+6.
Need to calculate:The factorization of $3{p}^{3}-2{p}^{2}-9p+6$.
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Formula used:
The factors of a polynomial can be found by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a\left(b+c\right)+d\left(b+c\right)$
$=\left(a+d\right)\left(b+c\right)$
Or,
$ab-ac+bd-cd=a\left(b-c\right)+d\left(b-c\right)$
$=\left(a+d\right)\left(b-c\right)$
Calculation:
Consider the polynomial $3{p}^{3}-2{p}^{2}-9p+6$.
This is a four term polynomial, factorization of this polynomial can be found by factor by grouping as,
$3{p}^{3}-2{p}^{2}-9p+6=\left(3{p}^{3}-2{p}^{2}\right)+\left(-9p+6\right)$
$={p}^{2}\left(3p-2\right)-3\left(3p-2\right)$
As, $\left(3p-2\right)$ is the common factor of the polynomial,
The polynomial can be factorized as,
$3{p}^{3}-2{p}^{2}-9p+6={p}^{2}\left(3p-2\right)—3\left(3p-2\right)$
$=\left(3p-2\right)\left({p}^{2}-3\right)$
Therefore, the factorization of the polynomial is $\left(3p-2\right)\left({p}^{2}—3\right)$.
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# Video: Finding the Distance Covered by a Car Moving with Uniform Acceleration
A body was moving in a straight line accelerating uniformly. If it covered 55 meters in the first 4 seconds and 57 meters in the next 4 seconds, determine the total distance covered in the first 10 seconds of its motion.
05:39
### Video Transcript
A body was moving in a straight line accelerating uniformly. If it covered 55 meters in the first four seconds and 57 meters in the next four seconds, determine the total distance covered in the first 10 seconds of its motion.
In order to solve this problem, we will use some of the equations of motion or SUVAT equations. We will use 𝑠 equals 𝑢𝑡 plus half 𝑎𝑡 squared, 𝑠 equals 𝑣𝑡 minus half 𝑎𝑡 squared, and 𝑣 equals 𝑢 plus 𝑎𝑡, where 𝑠 is the displacement, 𝑢 is the initial velocity, 𝑣 is the final velocity, 𝑎 is the acceleration, and 𝑡 is the time. We can see from the diagram that in the first four seconds the body covered 55 meters. In the next four seconds, it covered 57 meters. We need to calculate the total distance covered in the first 10 seconds.
If we let the velocity of the body at 𝑡 equals four be 𝑣 meters per second, then for the first part of the journey we have a displacement of 55, we have a final velocity of 𝑣 and acceleration of 𝑎 and a time of four seconds. We can substitute these values into the equation 𝑠 equals 𝑣𝑡 minus a half 𝑎𝑡 squared. This gives us 55 is equal to 𝑣 times four minus a half multiplied by 𝑎 multiplied by four squared.
Simplifying this equation gives us 55 is equal to four 𝑣 minus eight 𝑎. This means that during the time period 𝑡 equals zero to 𝑡 equals four, the motion of the body satisfies the equation four 𝑣 minus eight 𝑎 equals 55.
If we now consider the time period between 𝑡 equals four and 𝑡 equals eight, we have a displacement of 57 meters. The initial velocity is equal to 𝑣 as this is the velocity at time equals four. The acceleration is still equal to 𝑎. And our time is four seconds. We can substitute these values into the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. This gives us 57 is equal to 𝑣 multiplied by four plus a half multiplied by 𝑎 multiplied by four squared. Simplifying this gives us 57 is equal to four 𝑣 plus eight 𝑎.
We now have two simultaneous equations: one for the time period 𝑡 equals zero to 𝑡 equals four and one for the time period between 𝑡 equals four and 𝑡 equals eight. We can use these equations to calculate the acceleration and the velocity at 𝑡 equals four. Adding equation one and equation two gives us eight 𝑣 is equal to 112. Dividing both sides of this equation by eight gives us a value of 𝑣 of 14 meters per second. The velocity of the body at 𝑡 equals four is 14 meters per second.
Subtracting equation one from equation two gives us 16𝑎 is equal to two. Dividing both sides of this equation by 16 gives us a value of 𝑎 of one-eighth meter per second squared. The uniform acceleration of the body is one-eighth meter per second squared.
Our next step is to calculate the initial velocity of the body. In order to do this, we’ll consider the time between 𝑡 equals zero and 𝑡 equals four. Our displacement is 55 meters. Our initial velocity is 𝑢. Our final velocity is 14 meters per second. Our acceleration is one-eighth meter per second squared. And our time is four seconds. Using the equation 𝑣 equals 𝑢 plus 𝑎𝑡 gives us 14 is equal to 𝑢 plus an eighth multiplied by four. One-eighth multiplied by four is 0.5. Therefore, 14 equals 𝑢 plus 0.5. Subtracting 0.5 from both sides of this equation gives us an initial velocity 𝑢 of 13.5 meters per second.
We now have enough information to consider the first 10 seconds of the motion. The initial velocity 𝑢 is 13.5. Our acceleration is one-eighth. 𝑡 equals 10 and 𝑠 — the total displacement — is the unknown. Using the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared gives us 𝑠 is equal to 13.5 multiplied by 10 plus a half multiplied by an eighth multiplied by 10 squared. This gives us a value of 𝑠 of 141.25.
Therefore, the total distance covered by the body in the first 10 seconds of its motion is 141.25 meters.
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# Program for Picard’s iterative method | Computational Mathematics
• Difficulty Level : Medium
• Last Updated : 01 Sep, 2022
The Picard’s method is an iterative method and is primarily used for approximating solutions to differential equations. This method of solving a differential equation approximately is one of successive approximation; that is, it is an iterative method in which the numerical results become more and more accurate, the more times it is used. The Picard’s iterative method gives a sequence of approximations Y1(x), Y2(x), …Yk(x) to the solution of differential equations such that the nth approximation is obtained from one or more previous approximations. The Picard’s iterative series is relatively easy to implement and the solutions obtained through this numerical analysis are generally power series. Picard’s iteration method formula:
Picard’s iteration formula.
Steps involved:
• Step 1: An approximate value of y (taken, at first, to be a constant) is substituted into the right hand side of the differential equation: dy/dx= f(x, y).
• Step 2: The equation is then integrated with respect to x giving y in terms of x as a second approximation, into which given numerical values are substituted and the result rounded off to an assigned number of decimal places or significant figures.
• Step 3: The iterative process is continued until two consecutive numerical solutions are the same when rounded off to the required number of decimal places.
Picard’s iteration example: Given that: and that y = 0 when x = 0, determine the value of y when x = 0.3, correct to four places of decimals. Solution: We may proceed as follows: where x0 = 0. Hence: where y0 = 0. which becomes:
• First Iteration: We do not know y in terms of x yet, so we replace y by the constant value y0 in the function to be integrated. The result of the first iteration is thus given, at x = 0.3, by:
• Second Iteration: Now, we use: Therefore, which gives: The result of the second iteration is thus given by: at x=0.3.
• Third Iteration: Now we use: Therefore, which gives: The result of the third iteration is thus given by: at x = 0.3.
• Hence, y = 0.0451, correct upto four decimal places, at x = 0.3.
Program for Picard’s iterative method:
## C
`// C program for Picard's iterative method` ` ` `#include ` `#include ` ` ` `// required macros defined below:` `#define Y1(x) (1 + (x) + pow(x, 2) / 2)` `#define Y2(x) (1 + (x) + pow(x, 2) / 2 + pow(x, 3) / 3 + pow(x, 4) / 8)` `#define Y3(x) (1 + (x) + pow(x, 2) / 2 + pow(x, 3) / 3 + pow(x, 4) / 8 + pow(x, 5) / 15 + pow(x, 6) / 48)` ` ` `int` `main()` `{` ` ``double` `start_value = 0, end_value = 3,` ` ``allowed_error = 0.4, temp;` ` ``double` `y1[30], y2[30], y3[30];` ` ``int` `count;` ` ` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``y1[count] = Y1(temp);` ` ``y2[count] = Y2(temp);` ` ``y3[count] = Y3(temp);` ` ``}` ` ` ` ``printf``(``"\nX\n"``);` ` ``for` `(temp = start_value;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error) {` ` ` ` ``// considering all values` ` ``// upto 4 decimal places.` ` ``printf``(``"%.4lf "``, temp);` ` ``}` ` ` ` ``printf``(``"\n\nY(1)\n"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``printf``(``"%.4lf "``, y1[count]);` ` ``}` ` ` ` ``printf``(``"\n\nY(2)\n"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``printf``(``"%.4lf "``, y2[count]);` ` ``}` ` ` ` ``printf``(``"\n\nY(3)\n"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``printf``(``"%.4lf "``, y3[count]);` ` ``}` ` ``return` `0;` `}`
## Java
`// Java program for Picard's iterative method` `import` `java.util.*;` `class` `GFG {` ` ``// required macros defined below:` ` ``static` `double` `Y1(``double` `x)` ` ``{` ` ``return` `(``1` `+ (x) + Math.pow(x, ``2``) / ``2``);` ` ``}` ` ``static` `double` `Y2(``double` `x)` ` ``{` ` ``return` `(``1` `+ (x) + Math.pow(x, ``2``) / ``2` ` ``+ Math.pow(x, ``3``) / ``3` `+ Math.pow(x, ``4``) / ``8``);` ` ``}` ` ``static` `double` `Y3(``double` `x)` ` ``{` ` ``return` `1` `+ (x) + Math.pow(x, ``2``) / ``2` ` ``+ Math.pow(x, ``3``) / ``3` `+ Math.pow(x, ``4``) / ``8` ` ``+ Math.pow(x, ``5``) / ``15` `+ Math.pow(x, ``6``) / ``48``;` ` ``}` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``double` `start_value = ``0``, end_value = ``3``,` ` ``allowed_error = ``0.4``, temp;` ` ``double` `y1[] = ``new` `double``[``30``];` ` ``double` `y2[] = ``new` `double``[``30``];` ` ``double` `y3[] = ``new` `double``[``30``];` ` ``int` `count;` ` ``for` `(temp = start_value, count = ``0``;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``y1[count] = Y1(temp);` ` ``y2[count] = Y2(temp);` ` ``y3[count] = Y3(temp);` ` ``}` ` ``System.out.printf(``"\nX\n"``);` ` ``for` `(temp = start_value; temp <= end_value;` ` ``temp = temp + allowed_error) {` ` ``// considering all values` ` ``// upto 4 decimal places.` ` ``System.out.printf(``"%.4f "``, temp);` ` ``}` ` ``System.out.printf(``"\n\nY(1)\n"``);` ` ``for` `(temp = start_value, count = ``0``;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``System.out.printf(``"%.4f "``, y1[count]);` ` ``}` ` ``System.out.printf(``"\n\nY(2)\n"``);` ` ``for` `(temp = start_value, count = ``0``;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``System.out.printf(``"%.4f "``, y2[count]);` ` ``}` ` ``System.out.printf(``"\n\nY(3)\n"``);` ` ``for` `(temp = start_value, count = ``0``;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``System.out.printf(``"%.4f "``, y3[count]);` ` ``}` ` ``}` `}` `// This code is contributed by phasing17`
## Python3
`# Python3 program for Picard's iterative method` `import` `math` ` ` `# required macros defined below:` `def` `Y1(x): ` ` ``return` `(``1` `+` `(x) ``+` `pow``(x, ``2``) ``/` `2``) ` `def` `Y2(x):` ` ``return` `(``1` `+` `(x) ``+` `pow``(x, ``2``) ``/` `2` `+` `pow``(x, ``3``) ``/` `3` `+` `pow``(x, ``4``) ``/` `8``) ` `def` `Y3(x) :` ` ``return` `(``1` `+` `(x) ``+` `pow``(x, ``2``) ``/` `2` `+` `pow``(x, ``3``) ``/` `3` `+` `pow``(x, ``4``) ``/` `8` `+` `pow``(x, ``5``) ``/` `15` `+` `pow``(x, ``6``) ``/` `48``)` ` ` `start_value ``=` `0` `end_value ``=` `3` `allowed_error ``=` `0.4` `y1 ``=` `[``0` `for` `_ ``in` `range``(``30``)]` `y2 ``=` `[``0` `for` `_ ``in` `range``(``30``)]` `y3 ``=` `[``0` `for` `_ ``in` `range``(``30``)]` `temp ``=` `start_value` `count ``=` `0` `while` `temp <``=` `end_value:` ` ``y1[count] ``=` `Y1(temp);` ` ``y2[count] ``=` `Y2(temp);` ` ``y3[count] ``=` `Y3(temp);` ` ` ` ``temp ``=` `temp ``+` `allowed_error` ` ``count ``+``=` `1` ` ` `print``(``"\nX"``);` `temp ``=` `start_value` `while` `temp <``=` `end_value:` ` ` ` ``# considering all values` ` ``# upto 4 decimal places.` ` ``print``(``round``(temp, ``4``), end ``=` `" "``)` ` ``temp ``=` `temp ``+` `allowed_error` ` ` `print``(``"\n\nY(1)"``);` `temp ``=` `start_value` `count ``=` `0` `while` `temp <``=` `end_value:` ` ``print``(``round``(y1[count], ``4``), end ``=` `" "``)` ` ``temp ``=` `temp ``+` `allowed_error` ` ``count ``+``=` `1` ` ` `print``(``"\n\nY(2)"``);` `temp ``=` `start_value` `count ``=` `0` `while` `temp <``=` `end_value:` ` ``print``(``round``(y2[count], ``4``), end ``=` `" "``)` ` ``temp ``=` `temp ``+` `allowed_error` ` ``count ``+``=` `1` `print``(``"\n\nY(3)"``);` `temp ``=` `start_value` `count ``=` `0` `while` `temp <``=` `end_value:` ` ``print``(``round``(y3[count], ``4``), end ``=` `" "``)` ` ``temp ``=` `temp ``+` `allowed_error` ` ``count ``+``=` `1` ` ` `# This code is contributed by phasing17`
## C#
`// C# program for Picard's iterative method` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG {` ` ``// required macros defined below:` ` ``static` `double` `Y1(``double` `x)` ` ``{` ` ``return` `(1 + (x) + Math.Pow(x, 2) / 2);` ` ``}` ` ``static` `double` `Y2(``double` `x)` ` ``{` ` ``return` `(1 + (x) + Math.Pow(x, 2) / 2` ` ``+ Math.Pow(x, 3) / 3 + Math.Pow(x, 4) / 8);` ` ``}` ` ``static` `double` `Y3(``double` `x)` ` ``{` ` ``return` `1 + (x) + Math.Pow(x, 2) / 2` ` ``+ Math.Pow(x, 3) / 3 + Math.Pow(x, 4) / 8` ` ``+ Math.Pow(x, 5) / 15 + Math.Pow(x, 6) / 48;` ` ``}` ` ``public` `static` `void` `Main(``string``[] args)` ` ``{` ` ``double` `start_value = 0, end_value = 3,` ` ``allowed_error = 0.4, temp;` ` ``double``[] y1 = ``new` `double``[30];` ` ``double``[] y2 = ``new` `double``[30];` ` ``double``[] y3 = ``new` `double``[30];` ` ``int` `count;` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``y1[count] = Y1(temp);` ` ``y2[count] = Y2(temp);` ` ``y3[count] = Y3(temp);` ` ``}` ` ``Console.Write(``"\nX\n"``);` ` ``for` `(temp = start_value; temp <= end_value;` ` ``temp = temp + allowed_error) {` ` ``// considering all values` ` ``// upto 4 decimal places.` ` ``Console.Write(Math.Round(temp, 4) + ``" "``);` ` ``}` ` ``Console.Write(``"\n\nY(1)\n"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``Console.Write(y1[count] + ``" "``);` ` ``}` ` ``Console.Write(``"\n\nY(2)\n"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``Console.Write(Math.Round(y2[count], 4) + ``" "``);` ` ``}` ` ``Console.Write(``"\n\nY(3)\n"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``Console.Write(Math.Round(y3[count], 4) + ``" "``);` ` ``}` ` ``}` `}` `// This code is contributed by phasing17`
## Javascript
`// JavaScript program for Picard's iterative method` ` ` `// required macros defined below:` `function` `Y1(x) ` ` ``{ ``return` `(1 + (x) + Math.pow(x, 2) / 2) }` `function` `Y2(x) ` ` ``{ ``return` `(1 + (x) + Math.pow(x, 2) / 2 + Math.pow(x, 3) / 3 + Math.pow(x, 4) / 8) }` `function` `Y3(x) ` ` ``{ ``return` `(1 + (x) + Math.pow(x, 2) / 2 + Math.pow(x, 3) / 3 + Math.pow(x, 4) / 8 + Math.pow(x, 5) / 15 + Math.pow(x, 6) / 48)}` ` ` `let start_value = 0, end_value = 3,` ` ``allowed_error = 0.4, temp;` `let y1 = ``new` `Array(30);` `let y2 = ``new` `Array(30);` `let y3 = ``new` `Array(30);` `let count;` ` ` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``y1[count] = Y1(temp);` ` ``y2[count] = Y2(temp);` ` ``y3[count] = Y3(temp);` ` ``}` ` ` ` ``console.log(``"\nX"``);` ` ``for` `(temp = start_value;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error) {` ` ` ` ``// considering all values` ` ``// upto 4 decimal places.` ` ``console.log(temp.toFixed(4));` ` ``}` ` ` ` ``console.log(``"\n\nY(1)"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ``console.log((y1[count]).toFixed(4));` ` ``}` ` ` ` ``console.log(``"\n\nY(2)"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``console.log((y2[count]).toFixed(4));` ` ``}` ` ` ` ``console.log(``"\n\nY(3)"``);` ` ``for` `(temp = start_value, count = 0;` ` ``temp <= end_value;` ` ``temp = temp + allowed_error, count++) {` ` ` ` ``console.log((y3[count]).toFixed(4));` ` ``}` `// This code is contributed by phasing17`
Output:
```X
0.0000 0.4000 0.8000 1.2000 1.6000 2.0000 2.4000 2.8000
Y(1)
1.0000 1.4800 2.1200 2.9200 3.8800 5.0000 6.2800 7.7200
Y(2)
1.0000 1.5045 2.3419 3.7552 6.0645 9.6667 15.0352 22.7205
Y(3)
1.0000 1.5053 2.3692 3.9833 7.1131 13.1333 24.3249 44.2335```
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Related Articles
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## What two numbers add to negative?
When you have two negative signs, one turns over, and they add together to make a positive. If you have a positive and a negative, there is one dash left over, and the answer is negative.
## What are the integer rules for adding and subtracting?
To add integers having the same sign, keep the same sign and add the absolute value of each number. To add integers with different signs, keep the sign of the number with the largest absolute value and subtract the smallest absolute value from the largest. Subtract an integer by adding its opposite.
## What is the first step in subtracting integers?
Steps on How to Subtract Integers First, keep the first number (known as the minuend). Second, change the operation from subtraction to addition. Third, get the opposite sign of the second number (known as the subtrahend) Finally, proceed with the regular addition of integers.
## What are the rules to integers?
Multiplication and Division of Integers. RULE 1: The product of a positive integer and a negative integer is negative. RULE 2: The product of two positive integers is positive. RULE 3: The product of two negative integers is positive.
## When two negative integers are added we get *?
Answer: The sum of any integer and its opposite is equal to zero. Summary: Adding two positive integers always yields a positive sum; adding two negative integers always yields a negative sum.
## When two negative integers are added we get a positive integer?
Rule: The sum of any integer and its opposite is equal to zero. Summary: Adding two positive integers always yields a positive sum; adding two negative integers always yields a negative sum.
## When two negative integers are added we get a negative integer True or false?
The correct statement is: When two negative integers are added we get a negative integer. For example, (– 56) + (– 73) = – 129.
## Which integer is added to integer?
Answer. Let the integer to be added is x. x=9. So 9 is the number to be added.
## When two positive integers are added we get a integer?
When two positive integers are added the result will be a positive integer. When two negative integers are added the result will be a negative integer. When a positive integer and a negative integer are added, the result will be a negative or positive integer.
## Is negative 3 a natural number?
−3 is negative so it is not a natural or whole number. Rational numbers are numbers that can be expressed as a fraction or ratio of two integers.
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# Commutative Properties of Multiplication
Save
Simply put, the commutative property of multiplication means that no matter how you order the numbers you are multiplying, you will get the same answer. Addition also shares the commutative property with multiplication, whereas division and subtraction do not. For example, if you multiply 3 by 5 or 5 by 3, you will get the same answer of 15.
## Commutative Property Basics
• The root word for "commutative" is "commute." You can remember the meaning of commutative by thinking of the definition of "commute," which means to move around, change places, travel or interchange. The product will be the same no matter the order of the factors. In the operation of addition, if you add 5 and 3 or 3 and 5, you get the same sum of 8. The same applies in multiplication: The order of factors makes no difference.
## Example Problems
• The examples of 3 x 5 = 15 and 5 x 3 = 15 are numerical examples of the commutative property associated with multiplication. This can also be illustrated by an array. Draw on a piece of paper 15 circles, but arrange them in columns and rows. Whether you created three rows of five circles or five rows of three circles, both arrangements equal 15 circles. The same logic applies for algebraic terms, such as ab = ba or (4x)(2y) = (2y)(4x).
## Word Problems
• Although both addition and multiplication have the commutative property, when you must perform such operations after reading word problems, the interpretations are somewhat different. If you are reading a word problem that involves adding 112 houses with 134 houses, the meaning does not change whatever order you add the numbers. Suppose you are asked to determine the total number of flowers: If the word problem states that there are five groups of four flowers, you should interpret the equation as 5 x 4; if the problem states four groups of five, you should multiply 4 x 5. Although the answers are the same, it is worthwhile to take the time to read a word problem slowly to understand the exact question. You can even draw the groupings before producing your final answer.
## Related Properties
• Some mathematical properties go hand in hand with the commutative property. The associative property also pertains to both addition and multiplication. In multiplication, if you have three or more factors, the order and groupings of the factors does not matter -- the product will always be the same. For example, (2 x 3) x 4 is the same as (3 x 4) x 2, and each equals 24. The distributive property only pertains to multiplication. According to this property, the sum of two numbers multiplied by a third number is the same as multiplying each of the numbers being added by that factor. In algebraic terms, this can be represented by x (y + z) = xy + xz.
## References
• Photo Credit Liquidlibrary/liquidlibrary/Getty Images
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# A US President’s Proof of the Pythagorean Theorem
James Garfield, the 20th president of the United States, came up with an original proof of the Pythagorean Theorem in 1876 when he was still a Congressman. His proof was published in New England Journal of Education.
Recall that the Pythagorean Theorem states that given a right triangle with sides $a$, $b$, and hypotenuse $c$, the following equation is always satisified:
$a^2 + b^2 = c^2$.
President Garfield’s proof is quite simple. We can do this in three steps:
1. Find the area of figure above using the trapezoid
2. Find the area of the same figure using the three triangles
3. Equate the results in 1 and 2.
Proof:
(1) Finding the area of the figure using the trapezoid
$b_1$: $a$
$b_2$: $b$
$h$: $a + b$
$A = \displaystyle\frac{1}{2}h(b_1 + b_2)$
$A = \displaystyle \frac{1}{2}(a+b)(a+b) = \frac{a^2}{2}+ab+\frac{b^2}{2}$.
(2) Finding the area of the figure using the triangles
area of the red triangle = $\displaystyle\frac{ab}{2}$
area of the green triangle = $\displaystyle\frac{c^2}{2}$
area of the blue triangle = $\displaystyle\frac{ab}{2}$
total area $A = \displaystyle\frac{ab}{2} + \frac{c^2}{2} + \frac{ab}{2} = ab + \frac{c^2}{2}$
(3) The areas calculated in (1) and (2) are equal, therefore we can equate (1) and (2).
$\displaystyle\frac{a^2}{2} + ab + \frac{b^2}{2} = ab + \frac{c^2}{2}$
Multiplying both sides by $2$, we have $a^2 + 2ab + b^2 = 2ab + c^2$.
Subtracting $2ab$ from both sides, we have $a^2 + b^2 = c^2$. That completes the proof.
## 4 thoughts on “A US President’s Proof of the Pythagorean Theorem”
1. I find this proof interesting. It is obviously the standard ‘square proof’ – in which the green triangle above is half of what is usually a square. Even though it is of course different to the proof I’m referring to, it is ultimately based on the same idea. Still, great to think that a president was attempting his own proofs!
• You’re right. Doubling the figure and putting them together is the ‘square proof.’ However, what is interesting is that President Garfield did not take up a math-related course.
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# Math photo solver
This Math photo solver supplies step-by-step instructions for solving all math troubles. We will also look at some example problems and how to approach them.
## The Best Math photo solver
One instrument that can be used is Math photo solver. To solve for all values of x by factoring, we need to factor the equation. To do this, we need to find all the factors of the equation and set them equal to zero. Once we have done this, we can solve for x.
We can solve exponential functions using logarithms. Here is an example: To solve an exponential function, we use the power rule: We double the base to the power x, then add 1. This tells us how many times to multiply the original number by itself. The power rule enables us to solve exponential functions by computing two numbers - one for the exponent and a second for the base. We can then use these values to solve for the original number as follows: For example, if we want to solve 4x5^2, we would first compute 5x4^2 and then find 4 in this expression. Similarly, if we want to find 8x5^2, we would first compute 5x8^2 and then find 8 in this expression.
There are a few different ways to solve a 3x3 system of equations. One way is to use substitution. This involves solving for one variable in one equation and substituting that variable into the other equations. Another way is to use elimination. This involves adding or subtracting equations to eliminate one variable at a time. Once all variables are eliminated, the remaining equation will have only one variable and can be solved.
This results in a new equation with two fewer terms: By solving equations like this, we can simplify an expression. For instance, if we multiply 4x + 2y by 6x – 3y, we get 16x + 12y: By multiplying and adding the terms from both sides of this equation, we get 20x + 8y: We can also add or subtract like terms to simplify an expression. For example: By adding like terms and then multiplying, we get 9x + 5y: We can also subtract like terms and then divide by the same number to get a simpler result: And lastly, we can add or subtract like terms and then divide by a smaller number to get a simpler result: When simplifying expressions, it’s important to keep track of units. We don’t want to end up with incorrect numbers that are too small or too large! In other words, we want our final answer to be accurate. To avoid getting confused about units when working with exponents and powers, it’
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## Intermediate Algebra (6th Edition)
$f(x)=-\dfrac{3}{2}x-1$
Using the properties of equality, the given equation, $2x-3y=6 ,$ is equivalent to \begin{array}{l} -3y=-2x+6 \\\\ y=\dfrac{-2}{-3}x+\dfrac{6}{-3} \\\\ y=\dfrac{2}{3}x-2 .\end{array} Using $y=mx+b$, where $m$ is the slope, the slope of the given line is \begin{array}{l} m=\dfrac{2}{3} .\end{array} Using $m= -\dfrac{3}{2}$ (negative reciprocal slope since the lines are perpendicular) and the given point $( -4,5 ),$ then the equation of the line is \begin{array}{l} y-5=-\dfrac{3}{2}(x-(-4)) \\\\ y-5=-\dfrac{3}{2}(x+4) \\\\ y-5=-\dfrac{3}{2}x-6 \\\\ y=-\dfrac{3}{2}x-6+5 \\\\ y=-\dfrac{3}{2}x-1 .\end{array} In function notation, this is equivalent to $f(x)=-\dfrac{3}{2}x-1 .$
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# Subtraction
Subtraction is the arithmetic operation for finding the difference between two numbers, though it can also be generalized to other mathematical objects such as vectors and matrices.[1] The names of the numbers in a subtraction expression are: ${\displaystyle {\textrm {minuend}}-{\textrm {subtrahend}}={\textrm {difference}}}$.[2][3] For example, the expression ${\displaystyle 7-4=3}$ can be read as "seven minus four equals three", "seven take away four leaves three", or "four from seven leaves three".
If the minuend is less than the subtrahend, the difference will be a negative number. For example, ${\displaystyle 17-25=-8}$. This can be read as "seventeen minus twenty-five equals negative eight".
Subtraction is how cash registers determine the change a buyer receives, when the buyer pays with more money than the purchase cost.
## Properties
### Anti-commutativity
Subtraction is anti-commutative, meaning that swapping the numbers around the minus sign will give a number with the same magnitude, but the opposite sign (opposite number):
${\displaystyle a-b=-(b-a)}$
### Non-associativity
Subtraction is 'not' associative, which comes up when one tries to define repeated subtraction. In general, the expression
${\displaystyle a-b-c}$
means ${\displaystyle (a-b)-c}$ or ${\displaystyle a-(b-c)}$, but these two possibilities lead to different answers. To resolve this issue, one must establish an order of operations, with different orders yielding different results.
### Algebraic subtraction
Subtraction in algebra is different than normal arithmetic because you add negative numbers to a value instead of simply subtracting two positive numbers. Both numbers can also be placed inside parentheses. This is usually to make reading the equation easier.
${\displaystyle (a)+(-b)}$
### Predecessor
In the context of integers, subtraction of one also plays a special role: for any integer ${\displaystyle a}$, the integer ${\displaystyle a-1}$ is the largest integer that is smaller than ${\displaystyle a}$, also known as the predecessor of ${\displaystyle a}$.
## References
1. "Comprehensive List of Algebra Symbols". Math Vault. 2020-03-25. Retrieved 2020-08-26.
2. Weisstein, Eric W. "Subtraction". mathworld.wolfram.com. Retrieved 2020-08-26.
3. "Subtraction". www.mathsisfun.com. Retrieved 2020-08-26.
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What’s the probability…
Presentation on theme: "What’s the probability…"— Presentation transcript:
What’s the probability…
That this slideshow is about probability? By Ellen Brown, Kaylee Kline, and Madeline Zoe Hill McNichols
What is probability? Steps for solving probability problems:
1.) Determine all possibilities in a logical manner. Count them. 2.) Determine the number of these possibilities that are “favorable.” Thou shalt call these “winners.”
Example Problem If one of the four points is picked at random, what is the probability that the point lies on the angle? Solution! We follow the two basic steps by listing all possibilities and circle the winners.
Example Problem Suppose there are marbles. 10 marbles. In a pickle jar. If there are four red marbles and the rest are chartruse, what is the probability that when you reach into the pickle jar you will retrieve a red marble? number of red marbles 4 total marbles in jar = 2:5 therefore if I choose a marble from the pickle jar there is a 2:5 chance that I will choose a red marble
Example Problem Problem: If two of the four points are selected at random, the probability that they both lie on YW? W X Solution! We follow the two basic steps by listing all the possibilities in an orderly manner and circling the winners. If three of the six points lie on YW then the probability is 1:2. Z Y WX XY YZ WY XZ WZ
Practice Problem Two dice are thrown. What is the probability of getting two sixes?
SOLUTION! First we list the possibilities: 1,1 2,2 3,3 4,4 5,5 6,6
1,1 2,2 3,3 4,4 5,5 6,6 1,2 2,3 3,4 4,5 5,6 1,3 2,4 3,5 4,6 1,4 2,5 3,6 1,5 2,6 1,6 Now we circle the winner(s). There is only one winner in this problem therefore the answer is 1: 21.
Practice Problem There are six items in my pocket, half of the items are goats and half are kittens. If I take on item out of my pocket and put it on the table and then take out another item what is the probability that both of the items are goats?
SOLUTION! First we list the possibilities:
K1, K2 K2, K3 K3, G1 G1, G2 G2, G3 K1, K3 K2,G1 K3, G2 G1, G3 K1, G1 K2,G2 K3, G3 K1, G2, K2,G3 K1, G3 Then we circle the winner(s). There are three possibilities therefore there is a 3:15 chance that I will choose two goats which simplifies to 1:5.
Works Cited "The Math Forum at Drexel." Mathforum.org. Web. 17 Jan <http://mathforum.org/dr.math/faq/faq.prob.intro.html> "Expert Math Tutoring." Expertmathtutoring.com. Web. 17 Jan <http://www.expertmathtutoring.com/Probabilty-Knowledge-Examples.php>. Geometry for Enjoyment and Challenge. Evanston, Illinois : McDougal, Littell and Company, Print.
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## Precalculus (6th Edition) Blitzer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right]$.
Consider the provided matrix $A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]$. Compute matrix in the form of: $\left[ \left. A \right|I \right]$ Augment matrix with identity matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 3 & 0 & 0 & 1 & 0 & 0 \\ 0 & 6 & 0 & 0 & 1 & 0 \\ 0 & 0 & 9 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, we will use row operations to reduce in row echelon form for the inverse: \begin{align} & {{R}_{3}}\to \frac{1}{9}\times {{R}_{3}}, \\ & {{R}_{2}}\to \frac{1}{6}\times {{R}_{_{2}}}, \\ & {{R}_{1}}\to \frac{1}{3}\times {{R}_{1}} \\ \end{align} The resulting matrix is: \begin{align} & \left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{6} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{9} \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Therefore, the inverse of the matrix is: ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right]$ Where $B={{A}^{-1}}$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} And, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Thus, $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.
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# The value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear is
1. -1
2. 2
3. 1
4. None of these
Option 3 : 1
## Detailed Solution
Concept:
1. If three points (x1, y1), (x2, y2), and (x3, y3) are collinear then the area of the triangle determined by the three points is zero.
$$\left| {\begin{array}{*{20}{c}} {{{\rm{x}}_1}}&{{{\rm{y}}_1}}&1\\ {{{\rm{x}}_2}}&{{{\rm{y}}_2}}&1\\ {{{\rm{x}}_3}}&{{{\rm{y}}_3}}&1 \end{array}} \right|{\rm{\;}} = {\rm{\;}}0$$
2. If three or more points are collinear then the slope of any two pairs of points is the same.
For example, let three points A, B, and C are collinear then
slope of AB = slope of BC = slope of AC
The slope of the line if two points $$({{\rm{x}}_1},{{\rm{y}}_1}){\rm{\;and\;}}\left( {{{\rm{x}}_2},{\rm{\;}}{{\rm{y}}_2}} \right)$$ are given by:
$$⇒ \left( {\bf{m}} \right) = \;\frac{{{{\bf{y}}_2} - {{\bf{y}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}}{\rm{\;}}$$
Calculation:
Given that,
(x, -1), (2, 1) and (4, 5) are collinear
⇒ $$\left| {\begin{array}{*{20}{c}} {{{\rm{x}}}}&{{-1}}&1\\ {{2}}&{{1}}&1\\ {{4}}&{{5}}&1 \end{array}} \right|{\rm{\;}} = {\rm{\;}}0$$
⇒ x[1×1 - 1×5] + 1[2×1 - 4×1] + 1[2×5 - 1×4] = 0
⇒ -4x - 2 + 6 = 0
⇒ -4x = -4
⇒ x = 1
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# How do you find the area of the triangle ABC with a = 7.2, c = 4.1, and B=28 degrees?
Nov 17, 2015
The area of the triangle is approximately $6.93$.
#### Explanation:
For this problem, we will use the following:
The law of cosines:
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right)$
Heron's formula:
$A r e a = \sqrt{p \left(p - a\right) \left(p - b\right) \left(p - c\right)}$ where $p = \frac{a + b + c}{2}$
Applying the law of cosines gives us
${b}^{2} = {a}^{2} + {c}^{2} - 2 a \mathcal{o} s \left(B\right)$
$\implies {b}^{2} = {7.2}^{2} + {4.1}^{2} - 2 \left(7.2\right) \left(4.1\right) \cos \left({28}^{\circ}\right)$
$\implies b = \sqrt{51.84 + 16.81 - 2 \left(7.2\right) \left(4.1\right) \cos \left({28}^{\circ}\right)} \approx 4.065$
(Note that for greater accuracy, we can perform the calculation of the square root at the very end, however for convenience we will use this approximation here)
Now, let $p = \frac{a + b + c}{2} \approx 7.6825$
By Heron's formula, then, we get the area $A$ of the triangle as
$A = \sqrt{p \left(p - 7.2\right) \left(p - 4.065\right) \left(p - 4.1\right)}$
Plugging this in gives us the final result
$A \approx 6.93$
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Math 101 Intermediate Algebra
Solving Systems of Linear Equations
Chapter 4, Section 1
Idea
2 equations in 2 unknowns (variables)
• Can be solved for both unknowns
An example of a system of equations in 2 unknowns:
2x + 3y = 7 (1) 3x + 2y = 8 (2)
Observe that for the above example:
1. (2, 1) satifies both equations.
• Plug x = 2 and y = 1 into equations (1) and (2), and both equations hold true.
2. (5, -1) does not satisfy both equations.
• Plug x = 5 and y = -1 into equations (1) and (2), and equation (1) holds true but equation (2) is false.
3. Conclusion:
• (2, 1) is a solution to the example system of equations.
• (5, -1) is not a solution to the example system of equations.
Notice that equations (1) and (2) are each equations of lines in standard form.
Three methods (techniques) of solution studied in this section
1. Substitution
2. Elimination
3. Graphical
Solution Types
Consistent 1 solution Lines intesect in 1 point Inconsistent No solutions Lines are parallel Dependent Infinitely many solutions Lines are the same
Substitution Method
Steps to solution...
1. Solve one of the equations for one of the variables.
2. Substitute in the other equation: the expression for the variable found in step 1.
3. Solve the resulting equation (that is an equation in one variable after step 2).
4. Substitute the value of the variable found in step 3 into either of the original equations and solve for the value of the other variable.
5. You now have values for the 2 variables (i.e., you have the solution). Check that the solution satisfies both equations.
Elimination Method--Also known as the Addition Method
Idea: Multiply one, or both, equations by a number, or numbers, so that one of the variables dissappears when you add the equations together. Then you can solve the resulting equation for a value of a variable and substitute that value back into one of the original equations to find the other variable.
Steps to solution...
1. Write both given equations in standard form (best to get rid of fractions in this step!).
2. Multiply one, or both, equations by a number, or numbers, so that one of the variables dissappears when you add the equations together.
3. Add the equations together. That means, add the left-hand-sides to get the left-hand-side of the resulting equation and add the right-hand-sides to get the right-hand-side of the resulting equation.
4. You now have an equation in 1 variable--solve it for the value of that variable.
5. Substitute the value of the variable found in step 4 into either of the original equations and solve for the value of the other variable.
6. You now have values for the 2 variables (i.e., you have the solution). Check that the solution satisfies both equations.
Graphical Method
Steps to solution...
1. Graph the 2 equations.
2. The solution is the intersection of the 2 lines.
• The lines may intersect in 1 point meaning they are consistent (i.e., the system of equations has 1 solution).
• The lines may be the same (colinear) meaning they are dependent (i.e., the system of equations has infinitely many solutions).
• The lines may be parallel meaning they are inconsistent (i.e., the system of equations has no solutions).
3. Look at the intesection and write down the solution.
4. Check that the solution satisfies both equations.
The problem with this method is that it is often difficult to accurately write down the solution by looking at the point of intersection. For example, it is hard to look at a graph and tell that a point is (1.625, 3.875).
The usefulness of this method is that you can look at the intersection of two lines and tell if the solution point you found, by the substitution or elimination method, is close to correct.
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## Introducing the fraction notation
It is not uncommon to see fraction notation introduced as a way of recording a double count. First we count the number of parts shaded, next we count the total number of parts and then record the first count over the second count as a description of a fraction, as in the following figure.
Two-thirds resulting from a double count
Developing fraction notation from two counts relies upon an additive interpretation as the fixed nature of the whole is ignored. If the whole is ignored in naming fractions, the logical errors associated with fraction operations are reinforced.
An additive interpretation of fractions due to lack of reference to the whole
The emphasis in teaching fractions on counting the number of parts has meant that some students have developed concept images for fractions that are solely dependent on the number of parts represented. In the following example, the student has made three equal parts to represent one-third and six equal parts to represent one-sixth. That is, the number of parts corresponding to the denominator appears to represent the fraction.
For some students, fractions are defined solely by the number of parts without attention to the equality of all of the parts. In the following example, the student has represented fractional quantities as the number of parts out of the total number of parts. This is not a comparison of areas but rather a comparison of the number of parts.
Incomplete fraction concepts can be formed from activities associated with common parts-of-a-whole models. In the above examples the number of parts formed is the defining characteristic of the fraction representations rather than the area of the parts. A similar focus appears when students use equidistant parallel partitioning to form sub-units.
The circle below, shaded by a student to represent one-sixth of the circle, appears to have been subdivided by equidistant parallel partitioning.
Equidistant partitioning
As using vertical parallel lines works in creating fractions of a rectangular region, some students also attempt to use them in a circular region to produce thirds, fourths or fifths. Parallel partitioning can result in a number of parts that are treated additively, rather than a relationship between areas.
The dominance of the vertical and horizontal directions when forming parts of shapes suggests that some students may only attend to linear distance when dividing regional models, even circles. Equidistant parallel partitioning can also be described as linear partitioning, as attending to length rather than area forms the parts. Linear partitioning is also associated with students using ‘half-way points’ for one-half. The use of linear partitioning can persist well into high school with one major study in New South Wales (Gould, 2008, p. 145) reporting approximately 10% of students in Years 4–8 using parallel partitioning to attempt to represent one-third of a circle.
Instead of seeing the relationship between the parts and the whole, some students see:
• parts from parallel partitions,
• a number of parts (not equal parts), and
• a number of equal parts (not a fraction of the whole).
Although it is a common practice, introducing the standard fraction notation a/b as a ‘count’ interpretation of the regional ‘parts of a whole’ model is very limiting. A more useful introduction of the fraction notation arises from the idea of accumulation of length.
That is, instead of introducing the notation 3/4 as 3 parts out of 4 parts, it is better introduced as adding related units of length: 3/4 of the length is 1/4 of the length and 1/4 of the length and 1/4 of the length.
This also addresses the limitation of the parts of a whole interpretation of fractions; namely, how can you have 5 parts out of 4 parts? Developing a multiplicative interpretation of fractions and their notation is important to understanding why we form common denominators to add or subtract fractions.
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# Residual Standard Deviation/Error: Guide for Beginners
The residual standard deviation (or residual standard error) is a measure used to assess how well a linear regression model fits the data. (The other measure to assess this goodness of fit is R2).
But before we discuss the residual standard deviation, let’s try to assess the goodness of fit graphically.
Consider the following linear regression model:
Y = β0 + β1X + ε
Plotted below are examples of 2 of these regression lines modeling 2 different datasets:
Just by looking at these plots we can say that the linear regression model in “example 2” fits the data better than that of “example 1”.
This is because in “example 2” the points are closer to the regression line. Therefore, using a linear regression model to approximate the true values of these points will yield smaller errors than “example 1”.
In the plots above, the gray vertical lines represent the error terms — the difference between the model and the true value of Y.
Mathematically, the error of the ith point on the x-axis is given by the equation: (Yi – Ŷi), which is the difference between the true value of Y (Yi) and the value predicted by the linear model (Ŷi) — this difference determines the length of the gray vertical lines in the plots above.
Now that we developed a basic intuition, next we will try to come up with a statistic that quantifies this goodness of fit.
## Residual standard deviation vs residual standard error vs RMSE
The simplest way to quantify how far the data points are from the regression line, is to calculate the average distance from this line:
Where n is the sample size.
But, because some of the distances are positive and some are negative (certain points are above the regression line and others are below it), these distances will cancel each other out — meaning that the average distance will be biased low.
In order to remedy this situation, one solution is to take the square of this distance (which will always be a positive number), then calculate the sum of these squared distances for all data points and finally take the square root of this sum to obtain the Root Mean Square Error (RMSE):
We can take this equation one step further:
Instead of dividing by the sample size n, we can divide by the degrees of freedom df to obtain an unbiased estimation of the standard deviation of the error term ε. (If you’re having trouble with this idea, I recommend these 4 videos from Khan Academy which provide a simple explanation mainly through simulations instead of math equations).
The quantity obtained is sometimes called the residual standard deviation (as referred to it in the textbook Data Analysis Using Regression and Multilevel Hierarchical Models by Andrew Gelman and Jennifer Hill). Other textbooks refer to it as the residual standard error (for example An Introduction to Statistical Learning by Gareth James, Daniela Witten, Trevor Hastie and Robert Tibshirani).
In the statistical programming language R, calling the function summary on the linear model will calculate it automatically.
The degrees of freedom df are the sample size minus the number of parameters we’re trying to estimate.
For example, if we’re estimating 2 parameters β0 and β1 as in:
Y = β0 + β1X + ε
Then, df = n – 2
If we’re estimating 3 parameters, as in:
Y = β0 + β1X1 + β2X2 + ε
Then, df = n – 3
And so on…
Now that we have a statistic that measures the goodness of fit of a linear model, next we will discuss how to interpret it in practice.
## How to interpret the residual standard deviation/error
Simply put, the residual standard deviation is the average amount that the real values of Y differ from the predictions provided by the regression line.
We can divide this quantity by the mean of Y to obtain the average deviation in percent (which is useful because it will be independent of the units of measure of Y).
Here’s an example:
Suppose we regressed systolic blood pressure (SBP) onto body mass index (BMI) — which is a fancy way of saying that we ran the following linear regression model:
SBP = β0 + β1×BMI + ε
After running the model we found that:
• β0 = 100
• β1 = 1
• And the residual standard error is 12 mmHg
So we can say that the BMI accurately predicts systolic blood pressure with about 12 mmHg error on average.
More precisely, we can say that 68% of the predicted SBP values will be within ∓ 12 mmHg of the real values.
Why 68%?
Remember that in linear regression, the error terms are Normally distributed.
And one of the properties of the Normal distribution is that 68% of the data sits around 1 standard deviation from the average (See figure below).
Therefore, 68% of the errors will be between ∓ 1 × residual standard deviation.
For example, our linear regression equation predicts that a person with a BMI of 20 will have an SBP of:
SBP = β0 + β1×BMI = 100 + 1 × 20 = 120 mmHg.
With a residual error of 12 mmHg, this person has a 68% chance of having his true SBP between 108 and 132 mmHg.
Moreover, if the mean of SBP in our sample is 130 mmHg for example, then:
12 mmHg ÷ 130 mmHg = 9.2%
So we can also say that the BMI accurately predicts systolic blood pressure with a percentage error of 9.2%.
The question remains: Is 9.2% a good percent error value? More generally, what is a good value for the residual standard deviation?
The answer is that there is no universally acceptable threshold for the residual standard deviation. This should be decided based on your experience in the domain.
In general, the smaller the residual standard deviation/error, the better the model fits the data. And if the value is deemed unacceptably large, consider using a model other than linear regression.
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A Visual Approach to Simplifying Radicals (A Get Out of Jail Free Card)
The radical sign is like a prison. Twelve can be expressed as a product of prime factors so √12 = √(2×2×3). The 2’s pair up and try to break out. Sadly, only one of them survives the escape. √12 becomes 2√3.
That’s how I was taught to simplify radicals. No joke.
I imagined the numbers yelling “All in the name of liberty! Got to be free! JAlLBREAK!” as they scaled the prison walls. To this day, I can’t get this song out of my head when I teach this topic.
Many students are shown this method, albeit without the prison imagery. Write the prime factorization of the number. Circle the pairs. Write/multiply circled numbers outside the radical sign. There is real math behind this procedure. By definition, √2 × √2 = 2. However, I found that students who were taught this method couldn’t tell me why √(2×2×3) = 2√3. Where did the other 2 go?
Instead, I asked students to evaluate √12, then 2√3, using their calculators. Why are they equivalent? Students factored √12 as √4 × √3 (with some scaffolding for some). They understood where the 2 came from. Some began by factoring √12 as √6 × √2. Correct, but not helpful. The importance of finding factors that are perfect squares was discussed.
Marc Garneau shared with me his visual approach to simplifying radicals.
Consider a square with an area of 24. The side has length √24.
This square can be divided into 4 smaller squares, each with an area of 6. The sides of these smaller squares have length √6. Two of these lengths make up the side length of the large square, so √24 = 2√6.
24 can also be divided into 3 rectangles, each with an area of 8. Again, correct, but not helpful. How to simplify √45 as 3√5 and √72 as 6√2 are also shown above. Again, factors that are perfect squares are key.
I think it would be interesting to try this out. Some students may prefer this method, but most students will likely move towards simplifying radicals without drawing pictures. But by drawing pictures as they are learning this skill, students will be connecting mathematical ideas and building conceptual understanding. New learning (simplifying radicals in Math 10) will be connected to prior learning (concept of a square root introduced in Math 8). Students will have a more solid understanding of why perfect squares are used.
Turn it Around
In More Good Questions: Great Ways to Differentiate Secondary Mathematics Instruction, Dr. Marian Small discusses the turn-around strategy to create open questions.
Instead of asking “The legs of a right triangle are 3 cm and 6 cm long. What is the hypotenuse?” the teacher can ask “The answer is √45. What could the question be?”
There are many possible questions. For example,
Determine the length of the hypotenuse.
Determine the length of x.
A square has an area of 45 cm². What is the side length?
What is an example of a square root that has a value between 6 and 7?
Which number is the greatest: √37, 6, 6½, √45?
Students will come up with a variety of questions. However, at first, I imagine the response to open questions such as “The answer is √45. What could the question be?” will be silence. Students are used to being asked questions where there is one correct answer. In math, you either get it or you don’t. It’s not just questions that need turning around. This black and white view of mathematics also needs turning around. With time and practice, class discussions about open questions can help change this attitude.
“Under the M… the square root of 12”
On this blog, sometimes I share my thoughts about transforming math education. This is not one of those times.
Here, I’m using my blog as a digital filing cabinet.
One activity that my students enjoyed was MATHO (and its variations FACTO and TRIGO).
Have students select and place answers from the bottom of each column to fill up their MATHO cards. In some versions, I pulled prepared questions from a hat. In other versions, I translated answers to questions on the fly. For example, if I grabbed 2√3, I called out “Under the M… the square root of 12”. After a student shouts “MATHO!” ask potential winners to read aloud their numbers. (Remember to keep track of answers you have called.)
Nothing revolutionary here – just a fun way to review content.
By the way, if you are looking to read about changing things, please check out Sam Shah’s recent post, The Messiness of Trying Something New.
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# Difference Between Factors and Multiples
Factors and multiples are two basic concepts in mathematics. Most of the time, people need clarification and to consider the difference between factors and multiples. They have distinct meanings and serve different purposes in different fields. In this article, we will see the difference between factors and multiples.
## What are the Factors?
Any number that divides a particular number leaving no remains behind, is a Factor. For example, the factors of 32 are 1, 2, 4, 8, and 16 because they can divide 32 without any remainder. Here point to be noted is that factors always come in pairs. For example, if 4 is a factor of 12, 3 must also be a factor of 12 because 4 × 3 = 12. Similarly, if 6 is a factor of 12, then two must also be a factor of 12 because 6 × 2 = 12.
## What are Multiples?
Multiples come when a number is multiplied by a whole number. For example, the multiples of 5 are 5, 10, 15, 20, 25, 30, 35, and so on because each of these numbers can be obtained by multiplying five by another whole number (1, 2, 3, 4, 5, 6, 7, and so on). Multiples can be infinite in number, and they become more extensive as we multiply the given number by more significant whole numbers.
## Difference Between Factors and Multiples
Although factors and multiples are interrelated, they have some essential points of differences. The main difference between factors and multiples is the way they are calculated. Factors are obtained by dividing a given number by other numbers, whereas multiples are obtained by multiplying a given number. Hence, factors and multiples are obtained by division and multiplication, respectively.
Another difference between factors and multiples is their relationship to the original number. Factors are always smaller than or equal to the number provided, whereas multiples are always greater than or equal to the actual number. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12, which are all smaller than or equal to 12. On the other hand, the multiples of 12 include 12, 24, 36, 48, and so on, all greater than or equal to 12.
Factors are limited in number, whereas multiples are unlimited. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12 only. But, the multiples of 12 are 12, 24, 36, 48, and so on.
Finally, factors and multiples are used in different areas of mathematics. Factors are used mainly in number theory and algebra, whereas multiples are used in arithmetic and geometry. In geometry, multiples are used to find the scale factor between two similar figures: the ratio of the corresponding sides.
## Some Examples Based on Factors
Example 1: Find the greatest common factor (GCF) of 24 and 36.
Solution: To find the GCF of 24 and 36, we must find the largest number to divide 24 and 36 without leaving any remainder. First, we must find all the factors of both numbers and then detect the greatest common factor.
The factors of 24 include 1, 2, 3, 4, 6, 8, 12, and 24.
The factors of 36 have 1, 2, 3, 4, 6, 9, 12, 18, and 36.
The common factors of 24 and 36 are 1, 2, 3, 4, 6, and 12.
Therefore, the GCF of 24 and 36 is 12.
Example 2: Find all the factors of 24.
Solution: To find the factors of 24, we need to find all the numbers that can divide 24 without leaving any remainder. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
We can find the factors of 24 by dividing 24 by each number from 1 to 24 and checking if the division leaves no remainder. For example, 24 ÷ 3 = 8, so 3 and 8 are factors of 24.
Example 3: Determine whether 15 is a factor of 75.
Solution: To determine whether 15 is a factor of 75, we need to check if 15 can divide 75 without leaving any remainder. We can do this by dividing 75 by 15. If the division leaves no remainder, then 15 is a factor of 75.
75 ÷ 15 = 5, which means that 15 is a factor of 75. We can also say that 75 is a multiple of 15 since 15 multiplied by 5 gives 75.
We can use the prime factorization or common factor method to find the greatest common factor (GCF) of 48 and 72. Here are both methods:
### Method 1: Prime factorization method
Step 1: Find the prime factorization of 48 and 72.
Prime factorization of 48: 2 × 2 × 2 × 2 × 3 = 2^4 × 3
Prime factorization of 72: 2 × 2 × 2 × 3 × 3 = 2^3 × 3^2
Step 2: Identify the common prime factors and their smallest exponents.
The common prime factors of 48 and 72 are 2 and 3. The smallest exponent for 2 is 3, and for 3 is 1.
Step 3: Multiply the common prime factors raised to their smallest exponents to find the GCF.
GCF of 48 and 72 = 2^3 × 3^1 = 24
Therefore, the greatest common factor of 48 and 72 is 24.
### Method 2: Common factor method
Step 1: List factors 48 and 72.
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Step 2: Identify the common factors of 48 and 72.
The common factors of 48 and 72 are 1, 2, 3, 4, 6, 8, 12, and 24.
Step 3: Find the greatest common factor from the common factors.
The greatest common factor of 48 and 72 is 24.
Therefore, the greatest common factor of 48 and 72 is 24, the same as the answer we got using the prime factorization method.
## Some Examples Based on Multiples
Example 1: List the first 5 multiples of 6.
Solution: To find the multiples of 6, we must multiply 6 by all the natural numbers. The first 5 multiples of 6 are:
6 × 1 = 6
6 × 2 = 12
6 × 3 = 18
6 × 4 = 24
6 × 5 = 30
Example 2: Determine whether 50 is a multiple of 3
Solution: No, 50 is not a multiple of 3.
A multiple of 3 is a number that can be evenly divided by 3. In other words, a number is a multiple of 3 if the remainder of the division by 3 is 0. When we divide 50 by 3, we get a remainder of 2. Therefore, 50 is not a multiple of 3.
Example 2: Determine whether 24 is a multiple of 3.
Solution: To determine whether 24 is a multiple of 3, we need to check if 3 can divide by 24 without leaving any remainder. We can do this by dividing 24 by 3. If the division leaves no remainder, 24 is a multiple of 3.
24 ÷ 3 = 8, meaning 24 is a multiple of 3. We can also say that 8 is the factor of 24, since 3 multiplied by 8 gives 24.
Example 3: Find the least common multiple (LCM) of 12 and 18.
Solution: To find the LCM of 12 and 18, we need to find the smallest number, a multiple of both 12 and 18. One way to do this is to list the multiples of both numbers until we find the first common multiple.
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ...
The multiples of 18 are 18, 36, 54, 72, 90, 108, 126, 144, ...
The first common multiple of 12 and 18 is 36, so the LCM of 12 and 18 is 36.
We can also use the prime factorization method to find the LCM of 12 and 18. The prime factorization of 12 is 2 × 2 × 3, and the prime factorization of 18 is 2 × 3 × 3. To find the LCM, we take the product of all the prime factors and raise it to the greatest exponent. In this case, the prime factors are 2 and 3, and the greatest exponent for 2 is 2, and for 3 is 2. Therefore, the LCM of 12 and 18 is 2 × 2 × 3 × 3 = 36.
Factors Multiples
Obtained by division Obtained by multiplication
Always equals to or smaller than the actual number Always equals to or larger than the actual number
Example: factors of 12 are 1, 2, 3, 4, 6, and 12. Example: first five multiples of 4 are 4, 8, 12, 16, and 20.
Finite in number. Infinite.
## Conclusion
Factors and multiples are two essential topics in mathematics. Although they are related, they have different applications and meanings and serve different purposes in various fields. Factors are used primarily in number theory, whereas multiples are used in arithmetic and geometry. By understanding the difference between factors and multiples, we can better understand the properties of numbers and their relationships to each other.
Next TopicDifference between
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# Additive Identity Vs Multiplicative Identity
Additive Identity and Multiplicative Identity are two different identity properties of numbers. When additive identity is added to a number, it returns the original number. Similarly, when multiplicative identity is multiplied by any number, it returns the original number. Both these properties are applicable to all real numbers. Additive identity vs multiplicative identity describes here the difference between the two properties.
Zero (0) is the additive identity and one (0) is the multiplicative identity for all the numbers such as whole numbers, natural numbers, integers, etc.
Additive identity of numbers, as the name suggests, is a property of numbers that are applied when carrying out addition operations. The property states that when a number is added to zero it will give the same number. “Zero” is called the identity element, (also known as additive identity) If we add any number with zero, the resulting number will be the same number. This is true for any real numbers, complex numbers and even for imaginary numbers.
Suppose, ‘a’ is any real number, then
a + 0 = a = 0 + a
For example, 120 + 0 = 120, illustrates identity property of addition, where 0 is the additive identity.
## What is Multiplicative Identity?
Multiplicative identity of numbers, as the name suggests, is a property of numbers which is applied when carrying out multiplication operations Multiplicative identity property says that whenever a number is multiplied by the number
$$\begin{array}{l}1\end{array}$$
(one) it will give that number as product. “
$$\begin{array}{l}1\end{array}$$
” is the multiplicative identity of a number. It is true if the number being multiplied is
$$\begin{array}{l}1\end{array}$$
itself. The multiplicative identity property is represented as:
a × 1 = a = 1 × a (a is any real number)
Some examples:
−1 + 0 = −1 (−1 here is the number on which the operation is carried out and “0” is additive identity.
0 + 259 = 259
−1 × 1 = −1 (−1 here is the number on which the operation is carried out and “1” is a multiplicative identity)
Note: (−1) × (−1) = 1 (proves that −1 is not a multiplicative identity)
## Difference Between Additive Identity and Multiplicative Identity
Additive Identity Multiplicative Identity Additive identity for any real number is 0. Multiplicative identity for any real number is 1. It is denoted by a + 0 = a, where a is any real number It is denoted by a x 1 = a, where a is any real number It is used in addition It is used in the multiplication operation Example: 7 + 0 = 7 Example: 7 x 0 = 7
## Solved Examples
Q.1: Which of the following illustrates the multiplicative identity and additive identity?
1. 45 + 1 = 46
2. 50 × 2 = 100
3. 14 × 1 = 14
4. −54 + 0 = −54
Solution:
According to the identity property of multiplication, the product of any number multiplied by 1 is the number itself.
Here, only 14 × 1 = 14 satisfies the property.
Therefore, 14 × 1= 14 illustrates the Multiplicative identity.
According to the identity property of addition, the sum of any number added to 0 is the number itself.
Here, only −54 + 0 = −54 satisfied the property.
Therefore, −54 + 0 = −54 illustrates the additive identity.
Q.2: If a + 0 = -11, then what is the value of a?
Solution: Given, a + 0 = -11
By additive identity, if 0 is added to a number, it returns the original number.
Thus, a = -11
Q.3: If n x 1 = 1100, then what is the value of n?
Solution: Given, n x 1 = 1100
By multiplicative identity, if 1 is multiplied by a number, it returns the original number.
Thus, n = 1100
To solve more problems on the topic, download BYJU’S – The Learning App from Google Play Store and watch interactive videos.
## Frequently Asked Questions – FAQs
Additive identity is the value when added to a number, results in the original number. When we add 0 to any real number, we get the same real number. For example, 5 + 0 = 5. Therefore, 0 is the additive identity of any real number.
### Which is the multiplicative identity, 0 or 1?
Multiplicative identity of a real number is 1. When we multiply 1 to any real number then we get the same number. Examples are 3 x 1 = 3, -9 x 1 = -9, ½ x 1 = ½.
### Is -1 also a multiplicative identity?
-1 is not a multiplicative identity, because if we multiply -1 to any real number, the sign of that number gets changed. Examples are:
4 x -1 = -4
-4 x -1 = 4
### What is the multiplicative identity of 5?
Multiplicative identity if 5 is 1 only, since 5 x 1 = 5.
### What is the additive identity of x?
Additive identity of x is 0, since x+0 = x.
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# 9.6: Supplementary and Complementary Angles
Difficulty Level: At Grade Created by: CK-12
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Practice Supplementary and Complementary Angle Pairs
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Paul is fixing a house. The wood frames have already been placed and he is almost done with one of the rooms. He looks around the room and sees that along each wall, the boards are placed at a 90 degree angle. There are pairs of 90 degree angles next to each other. He knows that there is a special name for the angle pair, but he can't remember the name. What type of angle pair is formed by two 90 degree angles?
In this concept, you will learn about complementary and supplementary angles.
### Supplementary and Complementary Angles
Sometimes, two angles are a part of each other or are connected to each other. These angles are called angle pairs.
Let's look at two special types of angle pairs, supplementary angles or complementary angles.
Supplementary angles are two angles whose sum is equal to . In other words when you add the measure of one angle in the pair with the other angle in the pair, they equal 180 degrees.
These two angles are supplementary because together they form a straight line. You can also tell that they are supplementary because when you add their angle measures the sum is equal to 180 degrees.
Complementary angles are a pair of angles whose sum is . Here is an example of complementary angles.
If you add up the two angle measures, the sum is equal to 90 degrees. Therefore, the two angles are complementary.
You can find missing angle measures by using information about supplementary and complementary angles.
Let's look at an example.
Find the measure of .
First, identify that these two angles are supplementary. They form a straight line. The total number of degrees in a straight line is 180. Therefore, you can write the following equation to solve.
The missing angle is equal to .
Here the two angles are complementary. Therefore the sum of the two angles is equal to 90 degrees. Write an equation and solve for the missing angle measure.
The missing angle measure is equal to .
### Examples
#### Example 1
Earlier, you were given a problem about Paul and the house.
He has an angle pair that is formed by two 90 degree angles. What type of angle pair is formed when two 90 degree angles are combined?
First,add the two angles to calculate their sum.
90 + 90 = 180
Then, identify which angle pair adds up to 90 degrees.
Supplementary angles
The answer is that the pair are supplementary angles.
#### Example 2
Identify the angle pair as complementary or supplementary.
First, add the two angles to calculate their sum.
75 + 105 = 180
Then, identify which angle pair adds up to 180 degrees.
Supplementary angles
Write whether each pair is complementary or supplementary.
#### Example 3
Identify the angle pair as complementary or supplementary.
If the sum of the angles is equal to 180 degrees.
First, identify which angle pair adds up to 180 degrees.
Supplementary angles
#### Example 4
Identify the angle pair as complementary or supplementary.
If one angle is 60 degrees and the other angle is 120 degrees.
First, add the two angles together to find the sum.
60 + 120 = 180
Then, identify which angle pair adds up to 180 degrees.
Supplementary angles
#### Example 5
Identify the angle pair as complementary or supplementary.
First, add the two angles to calculate their sum.
30 + 60 = 90
Then, identify which angle pair adds up to 90 degrees.
Complementary angles
The answer is that the pair are complementary angles.
### Review
Identify each angle pair as supplementary or complementary angles.
Use what you have learned about complementary and supplementary angles to answer the following questions.
1. If two angles are complementary, then their sum is equal to _________ degrees.
2. If two angles are supplementary, then their sum is equal to ________ degrees.
3. True or false. If one angle is , then the second angle must be equal to for the angles to be supplementary.
4. True or false. If the angles are supplementary and one angle is equal to , then the other angle must be equal to .
5. True or false. The sum of complementary angles is .
6. True or false. The sum of supplementary angles is .
Identify whether the angles are supplementary, complementary or neither based on the angle measures.
1. One angle is 50 degrees. The other angle is 130 degrees.
2. One angle is 30 degrees. The other angle is 60 degrees.
3. One angle is 112 degrees. The other angle is 70 degrees.
4. One angle is 110 degrees. The other angle is 50 degrees.
5. One angle is 35 degrees. The other angle is 55 degrees.
To see the Review answers, open this PDF file and look for section 9.6.
### Notes/Highlights Having trouble? Report an issue.
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# Retail Math Formulas by cch21045
VIEWS: 287 PAGES: 20
• pg 1
``` Chapter 2 Section 4: Formulas
In this section, we will…
Use formulas to
solve word problems
Solve a formula
for a given variable
2.4 Formulas
We will use the following formulas from business: simple interest
formula
interest = principal rate time OR i prt Don’t forget to
to a decimal before
profit = revenue - cost OR p r c using it in a formula!
retail price = cost + markup OR r cm
We will use the following formulas from science:
distance = rate time OR d rt
5( F 32)
C Where C and F are the temperature in degrees
9 Celsius and Fahrenheit respectively.
2.4 Use Formulas to Solve Word Problems
Example: Find the markup on a CD Example: After expenses of \$55.15
player whose wholesale cost is \$219 were paid, a Rotary Club donated
and whose retail price is \$395. \$875.85 in proceeds from a pancake
breakfast to a local health clinic.
How much did the breakfast gross?
2.4 Use Formulas to Solve Word Problems
Example: Cryobiologists freeze living matter to preserve it for future use.
They can work with temperatures as low as 270 C. Change this to
degrees Fahrenheit.
2.4 Use Formulas to Solve Word Problems
Example: Three years after opening an account that paid 6.45% annually,
a depositor withdrew the \$3,483 in interest earned. How much money was
left in the account?
2.4 Use Formulas to Solve Word Problems
Example: Rose Parade floats travel down the 5.5 mile-long parade route at
a rate of 2.5 miles per hour. How long will it take a float to complete the
parade if there are no delays?
2.4 Use Formulas to Solve Word Problems
Geometry Formulas to Know
Rectangle Rectangular Box
Perimeter = sum of all sides Volume l w h in 3
units
or in
units
Perimeter 2l 2w
in 2
Area = l w units
2.4 Use Formulas to Solve Word Problems
Geometry Formulas to Know
Triangle Parallelogram in
units
Perimeter = sum of all sides Perimeter = sum of all sides
Area = 1
2 bh Area bh
in 2
units
For these formulas:
The height meets
the base-side at a
90-degree angle
2.4 Use Formulas to Solve Word Problems
Geometry Formulas to Know
Trapezoid Circle in
units
Perimeter = sum of all sides Circumference d
a+b
Area = h Area r 2
2
3.14159...
227
*To minimize rounding-
errors we will use the
button on our TI
2.4 Use Formulas to Solve Word Problems
Geometry Formulas to Know
base
Pyramid
Cone
Volume hB 1 *
Volume 1 r 2 h
3
* 3
B is the area of the base
2.4 Use Formulas to Solve Word Problems
Geometry Formulas to Know
Sphere Cylinder
Volume 4 r 3
3 Volume r 2 h
2.4 Use Formulas to Solve Word Problems
Example: For each of the following scenarios, indicate which geometric
concept (perimeter, circumference, area or volume) should be used and
which unit of measurement (units, units 2 , units 3 ) would be appropriate.
• The amount of storage in a freezer
• The distance around the outside of a building lot
• The amount of land making up the Sahara Desert
• How far a bike tire rolls in one revolution
2.4 Use Formulas to Solve Word Problems
Example: A horse trots in a perfect Example: Find the perimeter and
circle around its trainer at the end of area of the truss below:
a 28 foot-long rope. How far does
the horse travel as it circles the
6 ft
the nearest tenth.) 16 ft
perimeter:
area:
2.4 Use Formulas to Solve Word Problems
Example: Find the amount of space Example: Find the exact area of the
in the hamster tube below. Round square road sign below:
12 in
3 in
2.4 Use Formulas to Solve Word Problems
Example: Find the volume of a spherical-shaped pumpkin that has a
diameter of 9 inches. Round your answer to the nearest hundredth.
2.4 Use Formulas to Solve Word Problems
Solving Linear Equations in One Variable:
1. If the equation contains fractions, multiply both sides by the
magic number (LCM of the denominators) to clear fractions
2. Use the Distributive Property to remove the parentheses
“undoing” the operations:
Isolate the variable by
(then combine the like-terms on each side of the equation)
3. Use the Addition and Subtraction Properties to get all of the
{ variables on one side of the equation together
(and all of the numbers on the other side of the equation)
4. Use the Multiplication and Division Properties to make the
coefficient of the variable equal to 1
Remember that solving an equation means that we
isolate the variable we are solving for!
2.4 Solve a Formula for a Given Variable
Example: Solve each formula for the given variable.
d rt for t I prt for r
2.4 Solve a Formula for a Given Variable
Example: Solve each formula for the given variable.
P 2l 2w for l K 1 mv 2 for m
2
2.4 Solve a Formula for a Given Variable
Example: Solve each formula for the given variable.
5 y 25 x for y 6 y 12 5x for y
We will need to
have this skill for
chapters 3 and 7.
2.4 Solve a Formula for a Given Variable
Independent Practice
You learn math by doing math. The best way to learn math is to practice,
practice, practice. The assigned homework examples provide you with an
opportunity to practice. Be sure to complete every assigned problem (or more
if you need additional practice). Check your answers to the odd-numbered
problems in the back of the text to see whether you have correctly solved each
problem; rework all problems that are incorrect.
|
# Ordinal Numbers
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
Show that the right distributive law $\left( {\alpha + \beta } \right) \cdot \gamma = a \cdot \gamma + \beta \cdot \gamma$ does not hold for all ordinals.
### Example 2
Show that the following rule does not hold for all ordinals: ${\text{If } \alpha + \gamma = \beta + \gamma,\;}\kern0pt{\text{then } \alpha = \beta .}$
### Example 3
Show that the following rule does not hold for all ordinals: ${\left( {\alpha \cdot \beta } \right)^2} = {\alpha ^2} \cdot {\beta ^2}.$
### Example 4
Show that the following rule does not hold for all ordinals:
$$\text{If } \gamma \gt 0 \text{ and } \alpha \cdot \gamma = \beta \cdot \gamma,\;$$ $$\text{then } \alpha = \beta .$$
### Example 5
Simplify the ordinal expression $\left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right).$
### Example 1.
Show that the right distributive law $\left( {\alpha + \beta } \right) \cdot \gamma = a \cdot \gamma + \beta \cdot \gamma$ does not hold for all ordinals.
Solution.
We know that $$2 \cdot \omega = \omega.$$ Then we have
$2 \cdot \omega = \left( {1 + 1} \right) \cdot \omega = \omega \ne 1 \cdot \omega + 1 \cdot \omega = \omega + \omega.$
### Example 2.
Show that the following rule does not hold for all ordinals: ${\text{If } \alpha + \gamma = \beta + \gamma,\;}\kern0pt{\text{then } \alpha = \beta .}$
Solution.
Let $$\alpha = 1,$$ $$\beta = 2,$$ $$\gamma = \omega.$$ We see that
$\alpha + \gamma = 1 + \omega = \omega,\;\;\beta + \gamma = 2 + \omega = \omega,$
so $$\alpha + \gamma = \beta + \gamma = \omega,$$ though
$\alpha = 1 \ne \beta = 2.$
### Example 3.
Show that the following rule does not hold for all ordinals: ${\left( {\alpha \cdot \beta } \right)^2} = {\alpha ^2} \cdot {\beta ^2}.$
Solution.
Let $$\alpha = \omega$$ and $$\beta = 2.$$ Check that
$\left( {\omega \cdot 2} \right)^2 \ne {\omega^2} \cdot {2^2}.$
Ordinal multiplication is associative. Since $$n \cdot \omega = \omega,$$ the left-hand side ($$LHS$$) of this expression is written as
${\left( {\omega \cdot 2} \right)^2} = {\left( {\omega \cdot 2} \right)^{1 + 1}} = \left( {\omega \cdot 2} \right) \cdot \omega \cdot 2 = \omega \cdot \left( {2 \cdot \omega} \right) \cdot 2 = \omega \cdot \omega \cdot 2 = {\omega^2} \cdot 2.$
We see that $$LHS \ne RHS.$$
### Example 4.
Show that the following rule does not hold for all ordinals:
$$\text{If } \gamma \gt 0 \text{ and } \alpha \cdot \gamma = \beta \cdot \gamma,\;$$ $$\text{then } \alpha = \beta .$$
Solution.
Let $$\alpha = 2,$$ $$\beta = 3,$$ and $$\gamma = \omega.$$ For these ordinal numbers, we have
$\alpha \cdot \gamma = 2 \cdot \omega = \omega,\;\;\beta \cdot \gamma = 3 \cdot \omega = \omega.$
So $$\alpha \cdot \gamma = \beta \cdot \gamma = \omega,$$ but
$\alpha = 2 \ne \beta = 3.$
### Example 5.
Simplify the ordinal expression $\left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right).$
Solution.
We denote this expression by $$\alpha.$$ Given that $$n \cdot \omega = \omega,$$ we can substitute $$4\omega = \omega$$ and $$2\omega = \omega:$$
$\alpha = \left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right) = \left( {1 + \omega} \right)\left( {3 + \omega + {\omega^2}} \right).$
Now recall that for limit ordinals, $$n + \omega = \omega.$$ Therefore,
$\alpha = \left( {1 + \omega} \right)\left( {3 + \omega + {\omega^2}} \right) = \omega\left( {\omega + {\omega^2}} \right).$
Using left distributivity and the exponentiation property $${\omega^{\beta + \gamma }} = {\omega^\beta } \cdot {\omega^\gamma },$$ we get
$\alpha = \omega\left( {\omega + {\omega^2}} \right) = \omega \cdot \omega + \omega \cdot {\omega^2} = {\omega^2} + {\omega^3}.$
|
# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8
These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8
Question 1.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8, x ∈ [- 4, 2]
Solution:
Now f(x) = x² + 2x – 8 is a polynomial
∴ It is continuous and derivable in its domain x ∈ R.
Hence it is continuous in the interval [- 4, 2] and derivable in the interval (- 4, 2)
f(-4) = (- 4)² + 2(- 4) – 8 = 16 – 8 – 8 = 0,
f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied.
f'(x) = 2x + 2
∴ f’ (c) = 2c + 2 = 0
or c = – 1, c = – 1 ∈ [- 4, 2]
Thus f’ (c) = 0 at c = – 1.
Question 2.
Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?
(i) f(x) = [x] for x ∈ [5, 9]
(ii) f (x) = [x] for x ∈ [-2, 2]
(iii) f (x) = x² – 1 for x ∈ [1, 2]
Solution:
(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6, 7, 8 Hence Rolle’s theorem is not applicable
(ii) f (x) = [x] is not continuous and derivable at – 1, 0, 1. Hence Rolle’s theorem is not applicable.
(iii) f(x) = (x² – 1), f(1) = 1 – 1 = 0,
f(2) = 22 – 1 = 3
f(a) ≠ f(b)
Though it is continous and derivable in the interval [1,2].
Rolle’s theorem is not applicable.
In case of converse if f (c) = 0, c ∈ [a, b] then conditions of rolle’s theorem are not true.
(i) f (x) = [x] is the greatest integer less than or equal to x.
∴ f(x) = 0, But fis neither continuous nor differentiable in the interval [5, 9].
(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [- 2, 2].
(iii) f (x) = x² – 1, f'(x) = 2x. Here f'(x) is not zero in the [1, 2], So f (2) ≠ f’ (2).
Question 3.
If f: [- 5, 5] → R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).
Solution:
For Rolle’s theorem
If (i) f is continuous in [a, b]
(ii) f is derivable in [a, b]
(iii) f (a) = f (b)
then f’ (c) = 0, c ∈ (a, b)
∴ f is continuous and derivable
i.e., f'(c) ≠ 0. Hence $$\frac{f(5)-f(-5)}{10}$$
but f (c) ≠ 0 ⇒ f(a) ≠ f(b)
⇒ f(-5) ≠ f(5)
Question 4.
Verify Mean Value Theorem, if
f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1, 4] and derivable in (1,4), So all the condition of mean value theorem hold.
then f’ (x) = 2x – 4,
f'(c) = 2c – 4
f(4) = 16 – 16 – 3 = – 3,
f(1) = 1 – 4 – 3 = – 6
∴ f'(c) = 0 $$\frac{f(b)-f(a)}{b-a}$$ = $$\frac{f(4)-f(1)}{4-1}$$
⇒ 2c – 4 = $$\frac{-3-6}{4-1}$$
⇒ 2c – 4 = 1 ⇒ c = $$\frac{5}{2}$$ ∈ (1, 4)
∴ Mean Value Theorem is verified for f(x) on (1, 4)
Question 5.
Verify Mean Value Theorem, if f (x) = x3 – 5x2 – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f’ (c) = 0.
Solution:
f (x) = x3 – 5x2 – 3x
f'(x) = 3x² – 10x – 3
Since f'(x’) exists, f(x) is continous on [1, 3]
f(x) is differentiable on (1, 3)
f'(c) = 3c² – 10c – 3
f(b) = f(3) = – 27
f(a) = f(1) = – 7
∴ f'(c) = 0 $$\frac{f(b)-f(a)}{b-a}$$
⇒ 3c² – 10c – 3 = $$\frac{-27-7}{3-1}$$
⇒ 3c² – 10c – 3 = – 10
⇒ 3c² – 10c + 7 = 0
⇒ (c – 1)(3c – 7) = 0 ⇒ c = 1 or c = $$\frac{7}{3}$$
$$\frac{7}{3}$$ ∈ (1, 3)
∴ Mean Value Theorem is verified for f(x) on (1, 3)
Question 6.
Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.
Solution:
(i) F (x)= [x] for x ∈ [5, 9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.
(ii) f (x) = [x], for x ∈ [-2, 2],
Again f (x) = [x] in the interval [-2, 2] is neither continous, nor differentiable.
(iii) f(x) = x² – 1 for x ∈ [1,2], It is a polynomial.
Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
f (x) = 2x, f(1) = 1 – 1 = 0 ,
f(2) = 4 – 1 = 3, f'(c) = 2c
∴ f'(c) = 0 $$\frac{f(b)-f(a)}{b-a}$$
2c = $$\frac{3-0}{2-1}$$ = $$\frac{3}{1}$$
∴ c = $$\frac{3}{2}$$ which belong to (1, 2)
error: Content is protected !!
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# 5.01 Name and identify fractions
Lesson
We've used fraction bars to help us name and identify fractions before. Let's try this problem to help us remember.
### Examples
#### Example 1
Which of the following shows \dfrac{1}{10} of the area of the shape shaded?
A
B
C
D
Worked Solution
Create a strategy
The numerator tells us how many parts should be shaded. The denominator tells us how many parts to divide the shape into.
Apply the idea
The fraction \dfrac{1}{10} is asking for one part of the shape to be shaded out of 10 total parts. The shape in option C has 10 total parts with 1 shaded part.
Idea summary
• The numerator (top number) is the number of parts shaded to represent the fraction.
• The denominator (bottom number) is the number of equal parts the shape is divided into.
## Area models
This video shows how to name and identify fractions using area models.
### Examples
#### Example 2
What fraction of the square is shaded blue?
Worked Solution
Create a strategy
Write the fraction as: \,\, \dfrac{\text{Number of shaded parts}}{\text{Total number of parts}}.
Apply the idea
There are 9 squares shaded blue and 16 squares in total.
So, the fraction shaded blue is \,\dfrac{9}{16}.
Idea summary
A fraction from an area model can be written as:\,\, \dfrac{\text{Number of shaded parts}}{\text{Total number of parts}}
## Identify fractions using number lines
This video shows how to name and identify fractions using number lines.
### Examples
#### Example 3
Plot \dfrac{1}{10} on the number line.
Worked Solution
Create a strategy
Apply the idea
Since the number line is already divided into 10 spaces, we just need to move right 1 space.
Idea summary
When plotting a fraction on a number line:
• the denominator (bottom number) shows how many parts there should be between each whole number.
• the numerator (top number) shows the number of parts to move to the right from the previous whole number.
## Mixed numbers and improper fractions
This video shows how to change a fraction written as a mixed number to an improper fraction, and also going the other way.
### Examples
#### Example 4
Rewrite \dfrac{17}{4} as a mixed number.
Worked Solution
Create a strategy
Divide the numerator by the denominator. The remainder will be the numerator of the mixed fraction.
Apply the idea
17 divided by 4 is 4 remainder 1. This is because 4\times 4=16 and 16+1=17.
So, \dfrac{17}{4} is made up of 4 wholes and 1 out of 4 remaining.\dfrac{17}{4}=4\dfrac{1}{4}
Idea summary
To convert an improper fraction to a mixed number:
• Divide the numerator by the denominator.
• The number of times the denominator goes into the numerator is the whole part of the mixed number.
• The remainder is the numerator of the mixed number.
• The denominator stays the same.
To convert a mixed number to an improper fraction:
• Multiply the denominator and whole part.
• Add the numerator to the result.
• The final result is the new numerator.
• The denominator stays the same.
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# Glossary (Version 8.4)
In algebra, a parabola is the graph of a function of the general form $$y=ax^2+bx+c$$, where a, b and c are real numbers and a≠0. Two examples of parabolas are shown below.
Parallel lines are lines in a plane which do not intersect or touch each other at any point. Parallel lines can never intersect, even if they were to continuously extend toward infinity.
Two lines are parallel, if they have the same gradient (or slope).
The lines below are parallel to one another, as indicated by the use of the arrow signs. In text, the symbol ∥ is used to denote parallel lines; for example, a∥b is read as “line a is parallel to line b”.
A parallelogram is a quadrilateral whose opposite sides are parallel.
Thus the quadrilateral ABCD shown below is a parallelogram because AB∥DC and AD∥BC.
Properties of a parallelogram
• The opposite angles of a parallelogram are equal.
• The opposite sides of a parallelogram are equal.
• The diagonals of a parallelogram bisect each other.
Partitioning means dividing a quantity into parts. In the early years, it commonly refers to the ability to think about numbers as made up of two parts, such as, 10 is 8 and 2. In later years it refers to dividing both continuous and discrete quantities into equal parts.
A percentage is a fraction whose denominator is 100; for example, 6 percent (written as 6%) is the percentage whose value is $$\frac6{100}$$.
Similarly, 40 as a percentage of 250 is $$\frac{40}{250}\times100=16\%$$.
Percentiles are the 99 values that divide an ordered data set into 100 (approximately) equal parts. It is only possible to divide a data set into exactly 100 equal parts when the number of data values is a multiple of one hundred.
Within the above limitations, the first percentile divides off the lower 1% of data values. The second, the lower 2% and so on. In particular, the lower quartile (Q1) is the 25th percentile, the median is the 50th percentile and the upper quartile is the 75th percentile.
Percentiles are often used to report comparative test results. A student who scores in the 90th percentile for a given test has scored higher than 90% of other students who took the test. A student who scores in the 10th percentile would have scored better than only 10% of students who took the test.
The perimeter of a plane figure is the length of its boundary. The perimeter of a figure can be calculated by adding the lengths of all its sides.
In geometry, two lines are said to be perpendicular to each other, if they meet at a right angle (90 degrees).
Pi is the name of the Greek letter π that is used to denote the ratio of the circumference of any circle to its diameter. The number $$\pi$$ is irrational, but $$\frac{22}7$$ is a rational approximation. The decimal expansion of $$\pi$$ begins:
$$\pi=3.141\text{ }592\text{ }653\text{ }589\text{ }79\dots$$
A picture graph is a statistical graph for organising and displaying categorical data.
Place value refers to the value of a digit as determined by its position in a number, relative to the ones (or units) place. For integers, the ones place is occupied by the rightmost digit in the number; for example, in the number 2 594.6 the 4 denotes 4 ones, the 9 denotes 90 ones or 9 tens, the 5 denotes 500 ones or 5 hundreds, the 2 denotes 2000 ones or 2 thousands, and the 6 denotes $$\frac6{10}$$ of a one or 6 tenths.
A point marks a position, but has no size.
A polygon is a plane figure bounded by three or more line segments. The word derives from Greek polys “many” and gonia “angle”.
The regular pentagon shown below is an example of a polygon. It is called a pentagon because it has five sides (and five angles). It is called regular because all sides have equal length and all interior angles are equal.
A polyhedron is a three-dimensional object, or a solid, which consists of a collection of polygons, joined at their edges and making up the faces of the solid. The word derives from Greek polys “many” and hedra “base” or “seat”.
The solid below is an example of a polyhedron (called an icosahedra and consisting of 20 faces).
A polynomial in one variable $$x$$ is a finite sum of terms of the form $$ax^k$$, where $$a$$ is a real number and $$k$$ is a non-negative integer.
A non-zero polynomial can be written in the form $$a_0+a_1x+a_2x^2+\cdots+a_nx^n,$$, where $$n$$ is a non-negative integer and $$a_n\neq0$$.
The term that contains the variable $$x$$ raised to the highest power, that is $$a_nx^n$$, is called the leading term.
The numbers $$a_0,a_1,\dots,a_n$$ are called the coefficients of the terms. Coefficients include the preceding sign.
For example, in the polynomial $$3x^2-5x+2$$, the leading term is $$3x^2$$ and the coefficient of the second term is $$-5$$.
A population is the complete set of individuals, objects, places etc. about which we want information.
A census is an attempt to collect information about the whole population.
A positive integer is an integer that excludes negative numbers and zero. Positive integers are 1, 2, 3, 4, 5, 6, ….
Primary data is original data collected by the user. Primary data might include data obtained from interviews the user has conducted herself, or observations the user has made during an experiment.
A prime factor of a number is a factor of that number which is prime.
A prime number is a natural number greater than 1 that has no factor other than 1 and itself.
A prism is a polyhedron that has two congruent and parallel faces and all its remaining faces are parallelograms.
A right prism is a polyhedron that has two congruent and parallel faces and all its remaining faces are rectangles. A prism that is not a right prism is often called an oblique prism.
Some examples of prisms are shown below.
The probability of an event is a number between 0 and 1 that indicates the chance of that event happening; for example, the probability that the sun will come up tomorrow is 1, the probability that a fair coin will come up ‘heads’ when tossed is 0.5, while the probability of someone being physically present in Adelaide and Brisbane at exactly the same time is zero.
A product is the result of multiplying together two or more numbers or algebraic expressions; for example, 36 is the product of 9 and 4, and $$x^2-y^2$$ is product of $$x-y$$ and $$x+y$$.
A proof is a rigorous mathematical argument that demonstrates the truth of a given proposition. A mathematical statement that has been established by means of a proof is called a theorem.
Two quantities are in proportion, if there is a constant ratio between them.
A protractor is an instrument for measuring angles. It uses degrees as the unit of measurement and is commonly in the shape of a semi-circle (180°) or circle (360°).
A pyramid is a polyhedron with a polygonal base and triangular sides that meet at a point called the vertex. The pyramid is named according to the shape of its base.
Pythagoras’ theorem states that for a right-angled triangle:
The square of the hypotenuse of a right-angled triangle equals the sum of the squares of the lengths of the other two sides.
In symbols, $$c^2=a^2+b^2$$.
The converse
If $$c^2=a^2+b^2$$ in a triangle ABC, then ∠C is a right angle.
|
# What are all the factors, the prime factorization, and factor pairs of 90100?
To find the factors of 90100, divide 90100 by each number starting with 1 and working up to 90100
## What is a factor in math ?
Factors are the numbers you multiply together to get another number. For example, the factors of 15 are 3 and 5 because 3 × 5 = 15.
The factors of a number can be positive or negative, but they cannot be zero.
The factors of a number can be used to find out if the number is prime or not.
A prime number is a number that has only two factors: itself and 1. For example, the number 7 is prime because its only factors are 7 and 1.
## List all of the factors of 90100 ?
To calculate the factors of 90100 , you can use the division method.
1. Begin by dividing 90100 by the smallest possible number, which is 2.
2. If the division is even, then 2 is a factor of 90100.
3. Continue dividing 90100 by larger numbers until you find an odd number that does not divide evenly into 90100 .
4. The numbers that divide evenly into 90100 are the factors of 90100 .
Now let us find how to calculate all the factors of Ninety thousand one hundred :
90100 ÷ 1 = 90100
90100 ÷ 2 = 45050
90100 ÷ 4 = 22525
90100 ÷ 5 = 18020
90100 ÷ 10 = 9010
90100 ÷ 17 = 5300
90100 ÷ 20 = 4505
90100 ÷ 25 = 3604
90100 ÷ 34 = 2650
90100 ÷ 50 = 1802
90100 ÷ 53 = 1700
90100 ÷ 68 = 1325
90100 ÷ 85 = 1060
90100 ÷ 100 = 901
90100 ÷ 106 = 850
90100 ÷ 170 = 530
90100 ÷ 212 = 425
90100 ÷ 265 = 340
90100 ÷ 340 = 265
90100 ÷ 425 = 212
90100 ÷ 530 = 170
90100 ÷ 850 = 106
90100 ÷ 901 = 100
90100 ÷ 1060 = 85
90100 ÷ 1325 = 68
90100 ÷ 1700 = 53
90100 ÷ 1802 = 50
90100 ÷ 2650 = 34
90100 ÷ 3604 = 25
90100 ÷ 4505 = 20
90100 ÷ 5300 = 17
90100 ÷ 9010 = 10
90100 ÷ 18020 = 5
90100 ÷ 22525 = 4
90100 ÷ 45050 = 2
90100 ÷ 90100 = 1
As you can see, the factors of 90100 are 1 , 2 , 4 , 5 , 10 , 17 , 20 , 25 , 34 , 50 , 53 , 68 , 85 , 100 , 106 , 170 , 212 , 265 , 340 , 425 , 530 , 850 , 901 , 1060 , 1325 , 1700 , 1802 , 2650 , 3604 , 4505 , 5300 , 9010 , 18020 , 22525 , 45050 and 90100 .
## How many factors of 90100 are there ?
The factors of 90100 are the numbers that can evenly divide 90100 . These numbers are 1 , 2 , 4 , 5 , 10 , 17 , 20 , 25 , 34 , 50 , 53 , 68 , 85 , 100 , 106 , 170 , 212 , 265 , 340 , 425 , 530 , 850 , 901 , 1060 , 1325 , 1700 , 1802 , 2650 , 3604 , 4505 , 5300 , 9010 , 18020 , 22525 , 45050 and 90100.
Thus, there are a total of 36 factors of 90100
## What are the factor pairs of 90100 ?
Factor Pairs of 90100 are combinations of two factors that when multiplied together equal 90100. There are many ways to calculate the factor pairs of 90100 .
One easy way is to list out the factors of 90100 :
1 , 2 , 4 , 5 , 10 , 17 , 20 , 25 , 34 , 50 , 53 , 68 , 85 , 100 , 106 , 170 , 212 , 265 , 340 , 425 , 530 , 850 , 901 , 1060 , 1325 , 1700 , 1802 , 2650 , 3604 , 4505 , 5300 , 9010 , 18020 , 22525 , 45050 , 90100
Then, pair up the factors:
(1,90100),(2,45050),(4,22525),(5,18020),(10,9010),(17,5300),(20,4505),(25,3604),(34,2650),(50,1802),(53,1700),(68,1325),(85,1060),(100,901),(106,850),(170,530),(212,425) and (265,340) These are the factor pairs of 90100 .
## Prime Factorisation of 90100
There are a few different methods that can be used to calculate the prime factorization of a number. Two of the most common methods are listed below.
1) Use a factor tree :
1. Take the number you want to find the prime factorization of and write it at the top of the page
2. Find the smallest number that goes into the number you are finding the prime factorization of evenly and write it next to the number you are finding the prime factorization of
3. Draw a line under the number you just wrote and the number you are finding the prime factorization of
4. Repeat step 2 with the number you just wrote until that number can no longer be divided evenly
5. The numbers written on the lines will be the prime factors of the number you started with
For example, to calculate the prime factorization of 90100 using a factor tree, we would start by writing 90100 on a piece of paper. Then, we would draw a line under it and begin finding factors.
The final prime factorization of 90100 would be 2 x 2 x 5 x 5 x 17 x 53.
2) Use a factorization method :
There are a few different factorization methods that can be used to calculate the prime factorization of a number.
One common method is to start by dividing the number by the smallest prime number that will divide evenly into it.
Then, continue dividing the number by successively larger prime numbers until the number has been fully factorised.
For example, to calculate the prime factorization of 90100 using this method, we keep dividing until it gives a non-zero remainder.
90100 ÷ 2 = 45050
45050 ÷ 2 = 22525
22525 ÷ 5 = 4505
4505 ÷ 5 = 901
901 ÷ 17 = 53
53 ÷ 53 = 1
So the prime factors of 90100 are 2 x 2 x 5 x 5 x 17 x 53.
## Frequently Asked Questions on Factors
### What are all the factors of 90100 ?
The factors of 90100 are 1 , 2 , 4 , 5 , 10 , 17 , 20 , 25 , 34 , 50 , 53 , 68 , 85 , 100 , 106 , 170 , 212 , 265 , 340 , 425 , 530 , 850 , 901 , 1060 , 1325 , 1700 , 1802 , 2650 , 3604 , 4505 , 5300 , 9010 , 18020 , 22525 , 45050 and 90100.
### What is the prime factorization of 90100 ?
The prime factorization of 90100 is 2 x 2 x 5 x 5 x 17 x 53 or 22 x 52 x 171 x 531, where 2 , 5 , 17 , 53 are the prime numbers .
### What are the prime factors of 90100 ?
The prime factors of 90100 are 2 , 5 , 17 , 53 .
### Is 90100 a prime number ?
A prime number is a number that has only two factors 1 and itself.
90100 it is not a prime number because it has the factors 1 , 2 , 4 , 5 , 10 , 17 , 20 , 25 , 34 , 50 , 53 , 68 , 85 , 100 , 106 , 170 , 212 , 265 , 340 , 425 , 530 , 850 , 901 , 1060 , 1325 , 1700 , 1802 , 2650 , 3604 , 4505 , 5300 , 9010 , 18020 , 22525 , 45050 and 90100.
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Interpreting Quadratic Functions in Vertex Form
Quadratic function rules can be expressed in different ways. Each form is useful because it highlights different characteristics of the parabola. As the name implies, vertex form gives the vertex of the parabola.
Vertex Form
Vertex form is an algebraic format used to express quadratic function rules.
In this form, gives the direction of the parabola. When the parabola faces upward and when it faces downward. The vertex of the parabola lies at and the axis of symmetry is Consider the graph of
From the graph, we can connect the following characterisitcs to the function rule.
Notice that although the factor in the function rule shows is actually equal to This coincides with a horizontal translation of a quadratic function.
Graphing a Quadratic Function in Vertex Form
Writing quadratic functions in vertex form, is advantageous because it clearly presents three characteristics of the function. It is possible to graph a quadratic function using its characteristics. Consider the function
1
Identify and plot the vertex
To begin, identify the vertex, from the function rule. Since the rule is Thus, the vertex is Next, plot the vertex on a coordinate plane.
2
Draw the axis of symmetry
The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two mirror images. Since the axis of symmetry is here it is
3
Determine and plot the -intercept
Next, determine the -intercept. Recall that for all -intercepts, the -coordinate is Thus, substitute into the function rule and solve for
Substitute for and evaluate
Thus, the -intercept of the parabola is This point can be added to the graph.
4
Reflect the -intercept across the axis of symmetry
The axis of symmetry divides the parabola into two mirror images. Thus, points on one side of the parabola can be reflected across the axis of symmetry. Notice that the -intercept is units away from the axis of symmetry.
There exists another point directly across from the -intercept that is also units from the axis of symmetry.
5
Draw the parabola
Now, with three points plotted, the general shape of the parabola can be seen. It appears that the parabola faces upward. Since in the given function rule, this should be expected. To draw the parabola, connect the points with a smooth curve.
fullscreen
Exercise
The characteristics of two functions, and are described below. Determine which of the function rules could represent and
Number Function rule
Show Solution
Solution
When a quadratic function is given in vertex form, , both the vertex and the axis of symmetry can be easily seen. It is given that the axis of symmetry for is and for it is Since the axis of symmetry is given as this gives for and for Thus, we can begin to write the function rules for and in vertex form as follows. Note that becomes We can use the given vertices to determine the values of The vertex of is This gives Similarly, 's vertex gives Thus, we can write the rules as Comparing these rules to the table above, it can be seen that rule number and can represent and and can represent
Number Function rule
There are two candidates remaining for each function. The rules with correspond to a downward parabola, whereas the rules with correspond to an upward parabola. To determine if any option is completely correct, we'd need more information about the parabola, specifically its direction or if its vertex is a minimum or a maximum.
fullscreen
Exercise
Write the rule for the quadratic function that has the vertex and passes through the point .
Show Solution
Solution
It's possible to write the rule of a quadratic function in vertex form using two points. Let's start with the vertex. In vertex form, is the coordinate of the vertex. Knowing this, we can substitute and into the rule. Notice that the rule is incomplete. We can use the other point to determine the value of Since lies on the parabola, and can be substituted into the rule. Then, we can solve for
Thus, we can write the complete function rule as To verify that passes through and we can graph the parabola and mark the points.
Since the parabola passes through the given points, the created rule is correct.
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# If α and β are zeroes of the quadratic polynomial x2 − 6x + a; find the value of ‘a’ if 3α + 2β = 20.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
If α and β are zeroes of the quadratic polynomial x2 − 6x + a; the value of ‘a’ if 3α + 2β = 20 is -16. A quadratic equation in x is a second-degree polynomial, or one of the following forms: ax2 + bx + c = 0, where a, b, and c are real values. Before attempting to solve polynomial equations, candidates should write them in standard form. Once the factors have all been calculated and reached zero, set each variable factor to zero. The answers to the derived equations are the solutions to the original equations.
## x2 − 6x + a. Find the Value of ‘a’ if 3α + 2β = 20.
The question states If α and β are zeroes of the quadratic polynomial x2 − 6x + a; find the value of ‘a’ if 3α + 2β = 20. When a variable term in the polynomial expression has the highest power of 2, the polynomial is said to be quadratic. Only the variable’s exponent is considered when determining a polynomial’s degree.
It is not taken into account how strong a coefficient or constant term is. The roots or zeros of the quadratic equation are the names given to the solutions of such an equation. The steps to solve the given quadratic polynomial are as follows:
Given that: α and β are the zeroes of the quadratic polynomial x2 − 6x + a and 3α + 2β = 20, then we must determine a’s value.
α + β = – (-6)/1
In simplification, we get the:
α + β = 6 ….. (1)
αβ = a/1
The above equation can be written as:
αβ = a….. (2)
3α + 2β = 20
Taking common
2(α + β) + α = 20
Substituting the values from equation (1) we get:
2 x 6 + α = 20
α = 8
Now, we are putting the value α in equation (1), and we get:
8 + β = 6
β = -2
From equation (2), we have:
a = 8 x (-2)
a = -16
Hence, the value of a is −16.
Summary:
## If α and β are zeroes of the quadratic polynomial x2 − 6x + a; find the value of ‘a’ if 3α + 2β = 20.
The value of ‘a’ if 3α + 2β = 20 is -16. If α and β are zeroes of the quadratic polynomial x2 − 6x + a. A quadratic equation or function is created when a quadratic polynomial equals 0. Any polynomial can be easily solved using fundamental algebraic ideas and factorization techniques. The first step in solving the polynomial equation is to set the value of the right-hand side to zero.
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## Engage NY Eureka Math 7th Grade Module 5 Lesson 16 Answer Key
### Eureka Math Grade 7 Module 5 Lesson 16 Example Answer Key
Example 2: Using Random Numbers to Select a Sample
The histogram indicates the differences in the number of words in the collection of 150 books. How many words are typical for a best-selling children’s book? Answering this question would involve collecting data, and there would be variability in those data. This makes the question a statistical question. Think about the 150 books used to create the histogram on the previous page as a population. How would you go about collecting data to determine the typical number of words for the books in this population?
How would you choose a random sample from the collection of 150 books discussed in this lesson?
The data for the number of words in the 150 best-selling children’s books are listed below. Select a random sample of the number of words for 10 books.
Sample response: I would add up all of the words in the 150 books and divide by 150. This would be the mean number of words for the 150 books. As the data distribution is not symmetrical, I could also find the median of the 150 books, as it would be a good description of the typical number of words. (Note: Discuss with students that using data for all 150 books is very tedious. As a result, students may indicate that selecting a random sample of the 150 books might be a good way to learn about the number of words in these children’s books.)
Sample response: I would make 150 slips of paper that contained the names of the books. I would then put the slips of paper in a bag and select 10 or 15 books. The number of pages of the books selected would be my sample.
If necessary, explain how to use ten numbers selected from a bag that contains the numbers from 1 to 150 to select the books for the sample.
If students need more direction in finding a random sample, develop the following example: Consider the following random numbers obtained by drawing slips from a bag that contained the numbers 1 to 150: {114,65,77,38,86,105,50,1,56,85}. These numbers represent the randomly selected books. To find the number of words in those books, order the random numbers {1,38,50,56,75,77,85,86,105,114}. Count from left to right across the first row of the list of the number of words, then down to the second row, and so on. The sample consists of the 1st element in the list, the 38th, the 50th, and so on.
Use the above example of random numbers to help students connect the random numbers to the books selected and to the number of words in those books.
→ What number of words corresponds to the book identified by the random number 1?
59,635; the 1st children’s book listed has 59,635 words.
→ What number of words corresponds to the book identified by the random number 38?
3,252; the 38th children’s book listed has 3,252 words.
### Eureka Math Grade 7 Module 5 Lesson 16 Exercise Answer Key
Exercises 1–2
Exercise 1.
From the table, choose two books with which you are familiar, and describe their locations in the data distribution shown in the histogram.
Answers will vary. Sample response: I read The Mouse and the Motorcycle, and that has 22,416 words. It is below the median number of words and may be below the lower quartile. Harry Potter and the Chamber of Secrets has 84,799 words. It is one of the books with lots of words but may not be in the top quarter for the total number of words.
Exercise 2.
Put dots on the number line below that you think would represent a random sample of size 10 from the number of words distribution above.
Answers will vary. The sample distribution might have more values near the maximum and minimum than in the center.
Exercises 3–6
Exercise 3.
Follow your teacher’s instructions to generate a set of 10 random numbers. Find the total number of words corresponding to each book identified by your random numbers.
Answers will vary. Sample response: I generated random numbers 123, 25, 117, 119, 93, 135, 147, 69, 48, 46, which produces the sample 94505, 80798, 92037, 83149, 90000, 8245, 89239, 89167, 3101, 22476.
Exercise 4.
Choose two more different random samples of size 10 from the data, and make a dot plot of each of the three samples.
Answers will vary. One possible response is displayed below.
Exercise 5.
If your teacher randomly chooses 10 books from your summer vacation reading list, would you be likely to get many books with a lot of words? Explain your thinking using statistical terms.
Answers will vary. Sample response: From my samples, it looks like I probably would get at least one book that had over 90,000 words because the maximum in each of the samples approached or exceeded 90,000 words. The three samples vary a lot, probably because the sample size is only 10. The median numbers of words for the three samples were about 86,000, 35,000, and 70,000, respectively, so it seems like at least half of the books would contain about 50,000 or more words.
Exercise 6.
If you were to compare your samples to your classmates’ samples, do you think your answer to Exercise 5 would change? Why or why not?
Answers will vary. Sample response: The sample size is pretty small, so different samples might be different. I still think that I would have to read some books with a lot of words because of the shape of the population distribution.
Exercises 7–9: A Statistical Study of Balance and Grade
Exercise 7.
Is the following question a statistical question: Do sixth graders or seventh graders tend to have better balance?
Yes, this is a statistical question because the data collected would vary between sixth and seventh graders.
Exercise 8.
Berthio’s class decided to measure balance by finding out how long people can stand on one foot.
a. How would you rephrase the question from Exercise 7 to create a statistical question using this definition of balance? Explain your reasoning.
a. Answers will vary. Sample response: What is the typical length of time that a seventh grader can balance on one foot? The data collected to answer this question will have some variability—1 min., 2 min., 2 min. 10 sec., and so on. So, it is also a statistical question.
b. What should the class think about to be consistent in how they collect the data if they actually have people stand on one foot and measure the time?
Sample response: Would it make a difference if students stood on their right feet or on their left? How high do they have to hold their feet off the ground? Can they do it barefoot or with shoes on? Would tennis shoes be better than shoes with higher heels? What can we use to measure the time?
Exercise 9.
Work with your class to devise a plan to select a random sample of sixth graders and a random sample of seventh graders to measure their balance using Berthio’s method. Then, write a paragraph describing how you will collect data to determine whether there is a difference in how long sixth graders and seventh graders can stand on one foot. Your plan should answer the following questions:
a. What is the population? How will samples be selected from the population? Why is it important that they be random samples?
Sample response: The populations will be all of the sixth graders and all of the seventh graders in our school. To get a random sample, we will find the number of sixth graders, say 62, and generate a list of 15 random numbers from the set 1 to 62, that is, {4,17,19,25,…}. Then, we will go into one classroom and count off the students beginning with 1 and use student 4, 17, and 19. Then we will go into the next classroom and count off the students beginning where we left off in the first room and so on. We will do the same for the seventh graders. This will give random samples because it offers every sixth and seventh grader the same chance of being selected (if using this plan with both grades).
b. How would you conduct the activity?
Sample response: Students will stand for as long as they can using whichever foot they choose in their stocking or bare feet with their eyes open. We will time them to the nearest second using stopwatches from our science class. We will have students do the activity one at a time out in the hall so they cannot see each other.
c. What sample statistics will you calculate, and how will you display and analyze the data?
Sample response: The sample statistics will be the mean time (in seconds) standing on one foot for the sixth graders and seventh graders. We will make a dot plot of the times for the sixth graders and for the seventh graders using parallel number lines with the same scale.
d. What would you accept as evidence that there actually is a difference in how long sixth graders can stand on one foot compared to seventh graders?
We will compare the shape, center, and spread of the sample distributions of times for the sixth graders and do the same for the seventh graders. If the mean times are fairly close together and the spreads not that different, there is not really evidence to say one group of students has better balance.
### Eureka Math Grade 7 Module 5 Lesson 16 Problem Set Answer Key
Question 1.
The suggestions below for how to choose a random sample of students at your school were made and vetoed. Explain why you think each was vetoed.
a. Use every fifth person you see in the hallway before class starts.
Students who are not in the hallway because they have a class in another part of the building would not have a chance to be selected, so the sample would not be a random sample.
b. Use all of the students taking math the same time that your class meets.
The students not taking math at that time would not have a chance of being selected, so the sample would not be a random sample.
c. Have students who come to school early do the activity before school starts.
The sample would be not be a random sample because some students would not be able to get to school early, so they could not be selected.
d. Have everyone in the class find two friends to be in the sample.
Choosing people that members of the class know would not be a random sample because people that members of the class do not know have no chance to be chosen.
Question 2.
A teacher decided to collect homework from a random sample of her students rather than grading every paper every day.
a. Describe how she might choose a random sample of five students from her class of 35 students.
Sample response: You could assign each student a number from 1 to 35, generate five random numbers from 1 to 35, and choose the corresponding students.
b. Suppose every day for 75 days throughout an entire semester she chooses a random sample of five students. Do you think some students will never get selected? Why or why not?
Sample response: Over that many days, it should almost even out. If you think about 375 numbers generated all together with each number from 1 to 35 having an equal chance of showing up each time, then each number should be in the overall set about 10 or 11 times. I generated 75 random samples of numbers from 1 to 35 and looked at how the numbers showed up. Every number from 1 to 35 showed up at least three times, and most of the numbers showed up about 10 or 11 times.
Question 3.
Think back to earlier lessons in which you chose a random sample. Describe how you could have used a random number generator to select a random sample in each case.
a. A random sample of the words in the poem “Casey at the Bat”
Sample response: You could have generated the random numbers from 1 to 26 for the block of words and the random numbers 1 to 20 to choose a word in the block. Or you could number all of the words from 1 to 520 and then generate random numbers between 1 and 520 to choose the words.
b. A random sample of the grocery prices on a weekly flyer
Sample response: Instead of cutting out all of the prices and putting them in a bag, you could just number them on the flyer and use the random number generator to select numbers to identify the items in the sample and use the price of those items.
Question 4.
Sofia decided to use a different plan for selecting a random sample of books from the population of 150 top-selling children’s books from Example 2. She generated ten random numbers between 1 and 100,000 to stand for the possible number of pages in any of the books. Then, she found the books that had the number of pages specified in the sample. What would you say to Sofia?
Sample response: She would have to reject the numbers in the sample that referred to pages that were not in her list of 150 books. For example, if she gets the random numbers 4 or 720, she would have to generate new numbers because no books on the list had either 4 or 720 pages. She would have to throw out a lot of random numbers that did not match the number of pages in the books in the list.
It would take her a long time. But if there were no two books that had the same total number of words in the population, it would be a random sample if she wanted to do it that way. However, because there are quite a few books that have the same number of words as other books in the population, this method would not work for selecting a random sample of the books.
Question 5.
Find an example from a newspaper, a magazine, or another source that used a sample. Describe the population, the sample, the sample statistic, how you think the sample might have been chosen, and whether or not you think the sample was random.
Responses will vary depending on the articles students find. For example, “an estimated 60% of the eligible children in Wisconsin did not attend preschool in 2009.” The population would be all of the children in Wisconsin eligible for preschool in 2009, and the sample would be the ones selected by the study. The sample statistic would be 60%. The article did not tell how the sample was chosen but said the source was from the Census Bureau, so it was probably a random sample.
### Eureka Math Grade 7 Module 5 Lesson 16 Exit Ticket Answer Key
Question 1.
Name two things to consider when you are planning how to select a random sample.
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# A sum of Rs. 10,000 is borrowed at 8% per annum compounded annually. If the amount is to be paid in three equal installments, what will be the annual installment?
Last updated date: 20th Sep 2024
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Hint: In his type of question use the concept that the amount is borrowed it has to pay in three equal installments so this is equal to (first installment / (1 + rate of interest)) + (second installment / (1 + rate of interest)$^2$) + (third installment / (1 + rate of interest)$^3$), so use this concept to reach the solution of the question.
Given data:
Sum of money which is borrowed = Rs. 10,000
Rate of interest at which it is borrowed = 8% per annum compounded annually.
Now it is given that the amount is paid in three equal installments.
Now we have to calculate the annual installment such that the amount is paid in three equal annual installments.
Let the annual installment be Rs. X.
So the amount borrowed = (first installment / (1 + rate of interest)) + (second installment / (1 + rate of interest)$^2$) + (third installment / (1 + rate of interest)$^3$).
As all the installments are the same, the first installment = second installment = third installment = X.
Now substitute the values we have,
$\Rightarrow 10000 = \dfrac{X}{{\left( {1 + \dfrac{8}{{100}}} \right)}} + \dfrac{X}{{{{\left( {1 + \dfrac{8}{{100}}} \right)}^2}}} + \dfrac{X}{{{{\left( {1 + \dfrac{8}{{100}}} \right)}^3}}}$
Now simplify this equation we have,
$\Rightarrow 10000 = \dfrac{X}{{\left( {\dfrac{{108}}{{100}}} \right)}} + \dfrac{X}{{{{\left( {\dfrac{{108}}{{100}}} \right)}^2}}} + \dfrac{X}{{{{\left( {\dfrac{{108}}{{100}}} \right)}^3}}}$
$\Rightarrow 10000 = \dfrac{{25X}}{{27}} + \dfrac{{{{\left( {25} \right)}^2}X}}{{{{\left( {27} \right)}^2}}} + \dfrac{{{{\left( {25} \right)}^3}X}}{{{{\left( {27} \right)}^3}}}$
$\Rightarrow 10000 = X\left( {\dfrac{{25}}{{27}} + \dfrac{{{{\left( {25} \right)}^2}}}{{{{\left( {27} \right)}^2}}} + \dfrac{{{{\left( {25} \right)}^3}}}{{{{\left( {27} \right)}^3}}}} \right)$
$\Rightarrow 10000 = X\left( {\dfrac{{25{{\left( {27} \right)}^2} + {{\left( {25} \right)}^2}27 + {{\left( {25} \right)}^3}}}{{{{\left( {27} \right)}^3}}}} \right)$
$\Rightarrow 10000 = X\left( {\dfrac{{18225 + 16875 + 15625}}{{19683}}} \right)$
$\Rightarrow 10000 = X\left( {\dfrac{{50725}}{{19683}}} \right)$
$\Rightarrow X = \left( {\dfrac{{19683}}{{50725}}} \right)10000 = 3880.335$ Rs.
So this is the required installment he has to pay in each year such that the amount is paid in three equal annual installments.
Note: Whenever we face such types of questions if he has to pay a single amount in after three years then the amount he has to pay is equal to $P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$, where P is the amount which he borrowed, r is the rate of interest and n is the period of time.
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## Engage NY Eureka Math Geometry Module 4 Lesson 2 Answer Key
### Eureka Math Geometry Module 4 Lesson 2 Example Answer Key
Example 1.
→ Name three points inside the rectangular region.
Answers will vary. Possible answers include (2, 2), (5, 4), and (14, 6).
→ Name three points on the boundary of the rectangular region.
Answers will vary. Possible answers include (3, 2), (15, 6), and (7, 7).
→ What if we want to know all of the points in the region and on the boundary? How can I describe those points?
Let students brainstorm and do a 30 – second Quick Write; then, share ideas with the class before asking the following.
→ Can you say anything about the possible value of x?
The x – values are to the right of 1 but to the left of 15; the x – values are greater than or equal to 1; the x – values are less than or equal to 15; x ≥ 1 and x ≤ 15; 1 ≤ x ≤ 15.
→ Can you say anything about the possible value of y?
The y – values are above 2 but below 7; the y – values are greater than or equal to 2; the y – values are less than or equal to 7; y ≥ 2 and y ≤ 7; 2 ≤ y ≤ 7.
→ What can you say about the coordinates of points that lie on the left side of this rectangular region?
The points on the left border of this rectangle all have x – coordinates of 1. On the top side?
The points on the top border of this rectangle all have y – coordinates of 7.
→ The region is a rectangle. Let’s review the properties of a rectangle. (Have students share ideas on the board.)
All rectangles have opposite sides parallel, opposite sides congruent, 4 right angles, and diagonals congruent.
→ What is the length of a diagonal of the rectangular region?
The diagonal is approximately 14.9 units.
→ Does it matter which of the two diagonals you work with?
No, the diagonals of a rectangle are the same length.
→ Can you give the coordinates of a point within the rectangular region that lies on the diagonal that connects (1, 2) to (15, 7)?
There are an infinite number of points on the diagonal. Students may find the point lying in the middle by averaging the values and get (8, 4.5), or they could use the rise over run triangle and find other points such as (3.8, 3).
Example 2.
Draw the triangular region in the plane given by the triangle with vertices (0, 0), (1, 3), and (2, 1). Can we write a set of inequalities that describes this region?
y ≤ 3x
y ≥ $$\frac{1}{2}$$x
y ≤ – 2x + 5
### Eureka Math Geometry Module 4 Lesson 2 Exercise Answer Key
Opening Exercises
Graph each system of inequalities.
a. y ≥ 1
x ≤ 5
i. Is (1, 2) a solution? Explain.
Yes, (1, 2) is inside the region.
ii. Is (1, 1) a solution? Explain.
Yes, (1, 1) is on the border of the region and included in the region.
iii. The region is the intersection of how many half – planes? Explain how you know.
It is the intersection of 2 half planes, x ≤ 5 and y ≥ 1. x ≤ 5 splits the plane in half at the vertical line x=5 and to the left. y ≥ 1 cuts the plane in half horizontally from the line y=1 and above.
b. y < 2x + 1
y ≥ – 3x – 2
i. Is ( – 2, 4) in the solution set?
No, it does not belong to the overlapping shaded region.
ii. Is (1, 3) in the solution set?
No, (1, 3) is not in the solution set because it is on the boundary, but that boundary is not included.
iii. The region is the intersection of how many half – planes? Explain how you know.
It is the intersection of 2 half planes and includes the portion of the line y<2x + 1 above the intersection, and the portion of the line y≥ – 3x – 2 below the intersection.
Exercises 1–3
Exercise 1.
Given the region shown to the right:
a. Name three points in the interior of the region.
Answers will vary but could include (4, 3), (3.5, 7), and (4.5, 19).
b. Name three points on the boundary.
Answers will vary but could include (4, 2), (3, 15), and (5, 18).
c. Describe the coordinates of the points in the region.
All x – coordinates are greater than or equal to 3 and less than or equal to 5, and all y – coordinates are greater than or equal to 2 and less than or equal to 20.
d. Write the inequality describing the x – values.
x ≥ 3, x ≤ 5 or 3 ≤ x ≤ 5
e. Write the inequality describing the y – values.
y ≥ 2, y ≤ 20 or 2 ≤ y ≤ 20
f. Write this as a system of equations.
{(x, y)│3 ≤ x ≤ 5, 2 ≤ y ≤ 20}
g. Will the lines x=4 and y=1 pass through the region? Draw them.
x=4 vertically cuts the region in half. y=1 is below the region and horizontal.
Exercise 2.
Given the region that continues unbound to the right as shown to the right:
a. Name three points in the region.
Answers will vary but could include (2, 2), (4, 3), and (5, 4).
b. Describe in words the points in the region.
The region is above y=1 and below y=5. It starts at x=0 and continues to the right without bound.
c. Write the system of inequalities that describe the region.
{(x, y)│x ≥ 0, 1 ≤ y ≤ 5}
d. Name a horizontal line that passes through the region.
Answers will vary but may include y=4.
Exercise 3.
Given the region that continues down without bound as shown to the right:
a. Describe the region in words.
The region is to the right of x= – 2 and to the left of x=3. It starts at y=0 and continues down without bound.
b. Write the system of inequalities that describe the region.
{(x, y)│ – 2 ≤ x ≤ 3, y ≤ 0}
c. Name a vertical line that passes through the region.
Answers will vary but may include x= – 1.
Exercises 4–5
Exercise 4.
Given the triangular region shown, describe this region with a system of inequalities.
x ≥ 0
y ≥ $$\frac{2}{3}$$x – 4
y ≤ – $$\frac{2}{3}$$ x + 4
Exercise 5.
Given the trapezoid with vertices ( – 2, 0), ( – 1, 4), (1, 4), and (2, 0), describe this region with a system of inequalities.
y ≥ 0
y ≤ 4
y ≥ 4x + 8
y ≤ – 4x + 8
### Eureka Math Geometry Module 4 Lesson 2 Problem Set Answer Key
Question 1.
Given the region shown:
a. How many half – planes intersect to form this region?
2
b. Name three points on the boundary of the region.
( – 1, – 1), ( – 3, 4), and (5, – 1)
c. Describe the region in words.
The region is above and includes y= – 1, is to the right of and includes x= – 3, and extends without bound to the top and right.
Question 2.
Region T is shown to the right.
a. Write the coordinates of the vertices.
( – 1, – 4), ( – 1, 6), (3, 6), and (3, – 4)
b. Write an inequality that describes the region.
{(x, y)│ – 1 ≤ x ≤ 3, – 4 ≤ y ≤ 6}
c. What is the length of the diagonals?
The diagonal is approximately 10.8 units long.
d. Give the coordinates of a point that is both in the region and on one of the diagonals.
Answers will vary. (1, 1) lies on the diagonal and in the interior of region T.
Question 3.
Jack wants to plant a garden in his backyard. His yard is 120 feet wide and 80 feet deep. He wants to plant a garden that is 20 feet by 30 feet.
a. Set up a grid for the backyard, and place the garden on the grid. Explain why you placed your garden in its place on the grid.
Answers will vary, but the backyard should be on the grid with length 120 feet in the x – direction and 80 feet in the y – direction, or this can be set up in the other direction with 80 feet in the x – direction and 120 feet in the y – direction. The garden should form a rectangle somewhere on the grid.
b. Write a system of inequalities to describe the garden.
c. Write the equation of three lines that would go through the region that he could plant on, and explain your choices.
Question 4.
Given the trapezoidal region shown to the right:
a. Write the system of inequalities describing the region.
y ≤ 4x + 12
y ≤ 4
y ≤ – 4x + 16
y ≥ 0
b. Translate the region to the right 3 units and down 2 units. Write the system of inequalities describing the translated region.
y ≤ 4x – 2
y ≤ 2
y ≤ – 4x + 26
y ≥ – 2
Challenge Problems:
Question 5.
Given the triangular region shown with vertices A( – 2, – 1), B(4, 5), and C(5, – 1):
a. Describe the systems of inequalities that describe the region enclosed by the triangle.
y ≤ x + 1
y ≤ – 6x + 29
y ≥ – 1
b. Rotate the region 90° counterclockwise about Point A. How will this change the coordinates of the vertices?
A( – 2, – 1), B( – 8, 5), and C( – 2, 6)
c. Write the system of inequalities that describe the region enclosed in the rotated triangle.
y ≥ – x – 3
y ≤ $$\frac{1}{6}$$x + $$\frac{19}{3}$$
x ≤ – 2
Question 6.
Write a system of inequalities for the region shown.
y ≤ |x|
y ≥ 0
|x| ≤ 3
### Eureka Math Geometry Module 4 Lesson 2 Exit Ticket Answer Key
Question 1.
Given the region shown:
a. Name three points in the region.
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GCF of 10 15 and 25
Let us know what is GCF of 10 15 and 25. The greatest common factor (GCF) of some numbers is the largest number that is equally divisible by all numbers. For example, the GCF of 10, 15 and 25 is 5 .
Similarly, what is GCF 6? The greatest common factor (GCF or GCD or HCF) of the set of whole numbers is the greatest positive integer that divides evenly into all numbers with zero remainder. For example, GCF = 6 for the set of numbers 18, 30 and 42. (GCF of 10 15 and 25)
What are the factors of 21? factors of 21
• 21: Factors 1, 3, 7 and 21.
• Negative factors of 21: -1, -3, -7 and -21.
• 21: 3, prime factors of 7
• Prime factorization of 21: 3 × 7 = 3 × 7.
• The sum of the factors of 21:32.
What is the GCF of 20 and 15? The GCF of 15 and 20 is 5 . To calculate the GCF (Greatest Common Factor) of 15 and 20, we need to factor each number (15 = 1, 3, 5, 15; 20 = 1, 2, 4, 5, 10, 20) and the most Choose the greater factor that exactly divides both 15 and 20, i.e. 5.
Second what is the GCF of 20? The prime factors of 20 are 2 x 2 x 5. No prime factors are the same in 9 and 20. The greatest common factor is 1 .
What is the GCF of 12 and 15?
Answer: GCF of 12 and 15 is 3 .
So what is the GCF of 12 and 21? Answer: GCF of 12 and 21 is 3 .
What is a factor of 15? Factor 15 by prime factorization
The factors of 15 are 1, 3, 5, and 15 .
What is GCF for 21?
The GCF of 21 and 30 is 3 . To calculate the greatest common factor of 21 and 30, we need to factor each number (21 = 1, 3, 7, 21; 30 = 1, 2, 3, 5, 6, 10, 15, 30). must be factored. And choose the greatest factor that exactly divides both 21 and 30, i.e. 3.
How do you make 21? We know that multiplying 1 by 21 gives 21, so they are both factors. 2 is not a factor of 21, because we cannot multiply it by any other whole number. 21.3 is a factor of 21, because it multiplies by 7 to make 21. So the factors of 21 are 1, 3, 7 and 21 .
How do you find GCF?
To find the GCF of a set of numbers, list all the factors of each number . The largest factor that appears in each list is GCF. For example, to find the GCF of 6 and 15, first list all the factors of each number. Because 3 is the largest factor that appears in both lists, 3 is the GCF of 6 and 15.
What is the GCF of 21 and 35? Answer: The GCF of 21 and 35 is 7 .
What is the fastest way to find GCF?
What is the GCF of 21?
The GCF of 21 and 30 is 3 . To calculate the greatest common factor of 21 and 30, we need to factor each number (21 = 1, 3, 7, 21; 30 = 1, 2, 3, 5, 6, 10, 15, 30). must be factored. And choose the greatest factor that exactly divides both 21 and 30, i.e. 3.
How do you do GCF? Here’s how to find the GCF of a set of numbers using prime factors:
1. List the prime factors of each number.
2. Circle each common prime factor – that is, every prime factor that is a factor of every number in the set.
3. Multiply all rounded numbers. The result is GCF.
What is the common factor of 21 and 35? 2 and 21 have 35 common factors, which are 1 and 7 . Therefore, the greatest common factor of 21 and 35 is 7.
What are the factors of 21 and 18?
The factors of 18 and 21 are 1, 2, 3, 6, 9, 18 and 1, 3, 7, 21 respectively.
What are the factors of 20? factors of 20
• 20: Factors 1, 2, 4, 5, 10 and 20.
• Negative factors of 20: -1, -2, -4, -5, -10 and -20.
• 20: prime factors of 2, 5
• Prime factorization of 20: 2 × 2 × 5 = 2 × 5.
• The sum of the factors of 20:42.
What is the GCF of 14 and 21?
Answer: The GCF of 14 and 21 is 7 .
What is the HCF of 20 and 15? Answer: HCF of 15 and 20 is 5 .
Which of the following number is not a factor of 21?
Answer: 6 is not a factor of 21.
What is the least to greatest factor of 15? 15 has four factors, i.e. 1, 3, 5 and 15. Therefore, the smallest factor is 1 and the largest factor is 15, itself . The sum of all the factors of 15 is equal to 24.
What are the multiples of 15?
The multiples of 15 are 15, 30, 45 60, 75, 90, 105 .
|
Basic Division Lesson 2: Number 2
Here's an easy example:
4
÷
2
=
2
÷
=
divided by equals
We start with four boxes. We divide these four boxes by two. This means: how many groups of two can we make with the four boxes that we have? Answer: 2.
Here's another example:
10
÷
2
=
5
÷
=
divided equals
We start with ten boxes. We divide these ten boxes by two. This means: how many groups of two can we make with the ten boxes that we have? Answer: 5.
EX21TextBox1 EX21TextBox2 EX21TextBox3 EX21TextBox4 EX21TextBox5
Try this problem.
22
÷
2
=
÷
=
divided equals
We start with twenty-two boxes. We divide these twenty-two boxes by two. This means: how many groups of two can we make with the twenty-two boxes that we have? Answer: 11.
EX31TextBox1 EX31TextBox2 EX31TextBox3 EX31TextBox4 EX31TextBox5 EX31TextBox6 EX31TextBox7 EX31TextBox8 EX31TextBox9 EX31TextBox10 EX31TextBox11
In order to do basic division, you must memorize some equations but fear not: FoxyMath will help you! When you are finished with this lesson, you will know the following equations:
2 ÷ 2 = 1
4 ÷ 2 = 2
6 ÷ 2 = 3
8 ÷ 2 = 4
10 ÷ 2 = 5
12 ÷ 2 = 6
14 ÷ 2 = 7
16 ÷ 2 = 8
18 ÷ 2 = 9
20 ÷ 2 = 10
22 ÷ 2 = 11
24 ÷ 2 = 12
The following exercise will help you learn the above equations and what they mean. Please take your time: take whatever time you need to learn the material.
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# Estimation and Rounding Off
This chapter will introduce the concept of estimation maths or to students. Also, they will learn about overestimation and underestimation and the uses of estimation in real life.
From this learning concept; students will learn to:
• Apply the estimated weight, height and volume
• To use the concept of estimation in daily life to make calculations easier.
The concept is explained to class 3 students using examples, illustrations and concept maps. At the end of the page, two printable worksheets of estimates with solutions are attached for the students.
What Is Estimation?
• Estimate means to find something close to the correct answer.
• To write the estimated value, we use the squiggly equal sign.“ ~”
• Estimation is helpful in both math and real life.
Estimation Math
• To estimate the numbers of any object or animal, it is not necessary to guess the exact number.
• Estimation is to find a number that is close enough to the exact count.
Let us discuss with an example.
Examples: Guess the number of the fish in the picture.
Overestimation and Underestimation
• As estimating numbers are always dependent on guessing, so the estimated number won’t match the exact count often.
• When the estimated number is higher than the exact count, we call it overestimated.
• When the estimated number is lower than the exact count, we call it underestimated.
Let us discuss with an example.
Examples:
(i) A restaurant serves idli to 24 adults and 13 children. Each person will eat at least two idlis. They will have to make
a)50 b) 80 c) 60 d) 100
But the exact count is 24 × 2 + 13 × 2 = 74, which is close to 70.
Hence, the value is underestimated.
(ii) Soumya drove his motorcycle 300 miles on Friday, 105 miles on Saturday and 265 miles on Sunday. which of the below number of miles is the best estimate of the number of miles he drove over 3 days?
a) 500 miles b) 800 miles c) 600 miles d) 100 miles
Answer:The estimated number should be 600 miles.
Now, 300 + 105 + 265 = 670 miles
Hence, the number is overestimated.
Estimation of Weights
• To estimate the weight of some object one needs to compare it with some familiar object.
• In order to do this, we should understand some common units of weight measurement.
• For lighter objects, we use gram.
• For heavier objects, we use kilograms.
Let us discuss with an example.
Examples:
The weight of watermelon is almost equal to 20-24 large tomatoes. How many tomatoes can balance the weight of a half watermelon? Estimate the weight of each large tomato?
Half of the watermelon will be measured around 10-15 large tomatoes. A watermelon weighs around 2 kg. Then large tomatoes will be weighed around 100 - 150 grams.
Estimation of Heights
• To estimate the height of some object one needs to compare it with some familiar object.
• In order to do this, we should understand some common units of height measurement.
• For shorter lengths, we use centimeters or millimeter.
• For longer lengths, we use kilometers, meters.
Let us discuss with an example.
Examples:
The length of a saree is around 12-14 arm length. Estimate the length of two sarees?
The length of the saree is 12-14 arms long. Two sarees will be almost 24-28 arms long. The length of each arm is 60-65 cm. So, the length of each saree is 600-650 cm long. Clearly, length of two sarees will be 1.2-1.5 meters long.
Estimation of Volume
• To estimate the volume of some object one needs to compare it with the volume of some familiar object.
• In order to do this, we should understand some common units that we use to measure volume.
• For a smaller amount of volume, we use millilitre (mL).
• For a larger amount of volume, we use Liter.
Let us discuss with an example.
Examples: A bucket is filled with 5-6 jugs of water. Find how much water is needed to fill half of the bucket.
Answer: A bucket is filled with 5-6 jugs of water. then to fill half of the bucket one will need 2-3 jugs of water. A jug contains almost 1-1.5 liters of water. Hence a bucket must have 5-8 liters of water.
Use of Estimation in real life
• To quickly guess the number of any object.
• Estimation helps us to save time.
• Estimation is used in real life. Like we estimate the number of people coming to any festival. The amount of food or drink, required for a meal.
|
# Get Answers to all your Questions
### Answers (1)
Answer:
$\left ( 3e^{3x} \right )$
Hint:
Use first principle to find the differentiation of $\left ( e^{3x} \right )$
Given:
$\left ( e^{3x} \right )$
Solution:
Let
$f\left ( x \right )= e^{3x}$
$f\left ( x+h \right )= e^{3\left ( x+h \right )}$
So, now we will use formula of differentiation by first principle
$\frac{d}{dx}f\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}\left ( e^{3x} \right )=\lim_{h\rightarrow 0}\frac{e^{3\left ( x+h \right )}-e^{3x}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{3x}\times e^{3h}-e^{3x}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{3x}\left ( e^{3h}-1 \right )}{h}$
Multiply and divide by 3
$=\lim_{h\rightarrow 0}e^{3x}\frac{\left ( e^{3h}-1 \right )}{3h}\times 3$
$=\lim_{h\rightarrow 0}e^{3x}\times 1\times 3$ $\left [ \because \lim_{h\rightarrow 0} \left ( \frac{e^{x}-1}{x} \right )=1\right ]$
$=3e^{3x}$
Hence, the differentiation of $e^{3x}$ is $3e^{3x}$
View full answer
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Q:
# How do you add mixed numbers?
A:
Mixed numbers can be added either by first converting them to improper fractions and then adding, or by adding the fractional part and the whole-number part of the mixed numbers separately and carrying or regrouping as necessary. In either case, the fractions can only be added after a common denominator has been found.
## Keep Learning
Any mixed number can be written as an improper fraction. For example, the mixed number 3 1/8 is equal to three times 8/8 plus 1/8, or 25/8. When adding 3 1/8 to 2 5/6, first convert 3 1/8 to 25/8 and 2 5/6 to 17/6. Using the common denominator of 24, convert 25/8 to 75/24 and 17/6 to 68/24, and add the numbers, producing a sum of 143/24. Since 143/24 is irreducible, the only step remaining is to convert the improper fraction to a mixed number. Since 24 divides into 143 five times with a remainder of 23, the result is 5 23/24.
The other way to add mixed numbers is to add the parts separately and regroup. For example, to add 3 2/3 to 4 3/4, first add the whole numbers 3 and 4 for a sum of 7. Then, using the common denominator of 12, convert 2/3 to 8/12 and 3/4 to 9/12, and add, producing the sum of 17/12. Since 17/12 equals 12/12 plus 5/12, or 1 5/12, regroup by adding the one to the whole number seven, producing a final sum of 8 5/12.
Sources:
## Related Questions
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# 624 and Level 3
Since 624 is one less than 625 which is 25 squared, it should be obvious that 24 x 26 = 624.
624 is the hypotenuse of the Pythagorean triple 240-576-624. What is the greatest common factor of those three numbers?
624 is the sum of consecutive odd numbers and twin primes 311 and 313.
Print the puzzles or type the solution on this excel file: 12 Factors 2015-09-21
—————————————————————————————————
• 624 is a composite number.
• Prime factorization: 624 = 2 x 2 x 2 x 2 x 3 x 13, which can be written 624 = (2^4) x 3 x 13
• The exponents in the prime factorization are 4, 1 and 1. Adding one to each and multiplying we get (4 + 1)(1 + 1)(1 + 1) = 5 x 2 x 2 = 20. Therefore 624 has exactly 20 factors.
• Factors of 624: 1, 2, 3, 4, 6, 8, 12, 13, 16, 24, 26, 39, 48, 52, 78, 104, 156, 208, 312, 624
• Factor pairs: 624 = 1 x 624, 2 x 312, 3 x 208, 4 x 156, 6 x 104, 8 x 78, 12 x 52, 13 x 48, 16 x 39 or 24 x 26
• Taking the factor pair with the largest square number factor, we get √624 = (√16)(√39) = 4√39 ≈ 24.97999199.
—————————————————————————————————
A Logical Approach to solve a FIND THE FACTORS puzzle: Find the column or row with two clues and find their common factor. (None of the factors are greater than 12.) Write the corresponding factors in the factor column (1st column) and factor row (top row). Because this is a level three puzzle, you have now written a factor at the top of the factor column. Continue to work from the top of the factor column to the bottom, finding factors and filling in the factor column and the factor row one cell at a time as you go.
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# How do you find the other five trigonometric values given csc b = -6/5 and cot b > 0?
##### 1 Answer
Jul 3, 2016
$\sin b = - \frac{5}{6}$, $\cos b = - \frac{\sqrt{11}}{6}$, $\tan b = \frac{5}{\sqrt{11}}$,
$\cot b = \frac{\sqrt{11}}{5}$ and $\sec b = - \frac{6}{\sqrt{11}}$.
#### Explanation:
As $\csc b = - \frac{6}{5}$ $\implies$ ${\csc}^{2} b = \frac{36}{25}$ and using ${\csc}^{2} b = 1 + {\cot}^{2} b$, we have
$1 + {\cot}^{2} b = \frac{36}{25}$ and $\cot b = \sqrt{\frac{36}{25} - 1} = \frac{\sqrt{11}}{5}$ (as $\cot b > 0$)
As $\cot b = \frac{\sqrt{11}}{5}$, $\tan b = \frac{1}{\cot} b = \frac{5}{\sqrt{11}}$
As $\csc b = - \frac{6}{5}$, $\sin b = \frac{1}{\csc} b = - \frac{5}{6}$
and $\cos b = \cos \frac{b}{\sin} b \times \sin b = \cot b \times \sin b$
= $\frac{\sqrt{11}}{5} \times \left(- \frac{5}{6}\right) = - \frac{\sqrt{11}}{6}$
and $\sec b = \frac{1}{\cos} b = - \frac{6}{\sqrt{11}}$.
Hence, we have $\sin b = - \frac{5}{6}$, $\cos b = - \frac{\sqrt{11}}{6}$, $\tan b = \frac{5}{\sqrt{11}}$,
$\cot b = \frac{\sqrt{11}}{5}$ and $\sec b = - \frac{6}{\sqrt{11}}$.
|
# What is the general solution of the differential equation y'' +4y =0?
The differential equation $y ' ' + 4 y = 0$ is what we call second order differential equation.
So we need to find its characteristic equation which is
${r}^{2} + 4 = 0$
This equation will will have complex conjugate roots, so the final answer would be in the form of y=e^(αx)*(c_1*sin(βx)+c_2*cos(βx)) where α equals the real part of the complex roots and β equals the imaginary part of (one of) the complex roots.
We need to use the quadratic formula
r=[−b±sqrt(b^2−4ac)]/[2*a]
when $a \cdot {r}^{2} + b \cdot r + c = 0$
In this equation $a = 1$, $b = 0$, and $c = 4$
Hence the roots are ${r}_{1} = 2 i$ and ${r}_{2} = - 2 i$
Now the form of y=e^(αx)*(c_1*sin(βx)+c_2*cos(βx))
where $a = 0$ and β=2
becomes
$y = {e}^{0 \cdot x} \cdot \left({c}_{1} \cdot \sin \left(2 \cdot x\right) + {c}_{2} \cdot \cos \left(2 \cdot x\right)\right)$
Finally
$y = \left({c}_{1} \cdot \sin \left(2 \cdot x\right) + {c}_{2} \cdot \cos \left(2 \cdot x\right)\right)$
The coefficients ${c}_{1} , {c}_{2}$ can be determined if we have initial conditions for the differential equation.
Jun 15, 2017
$y = A \cos 2 x + B \sin 2 x$
#### Explanation:
We have:
$y ' ' + 4 y = 0$
This is a second order linear Homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.
${m}^{2} + 4 = 0$
This has two distinct complex solutions:
$m = \pm 2 i$, or $m = 0 \pm 2 i$
And so the solution to the DE is;
$\setminus \setminus \setminus \setminus \setminus y = {e}^{0} \left(A \cos 2 x + B \sin 2 x\right)$ Where $A , B$ are arbitrary constants
$\therefore y = A \cos 2 x + B \sin 2 x$
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A cylinder can be considered a circular prism, so finding the surface area is a lot like a regular prism:
1. Consider how many sides the shape will have
2. Find the area of each side
3. Find the sum of the areas
Resources
In-Class Assignment
• p186 #3, 5, 6, 8, 9, 10, 11, 12, 13
Things you should be able to do after today:
• determine the area of a cylinder using its net
• determine the area of a cylinder using a formula
• determine the area of a 3d shape that uses parts of cylinders
Every 3d shape has a surface. The total area of that shape is the surface area. If the 3d shape is a polyhedron (something with flat faces) then it is usually pretty easy to find the total area of the surface:
Resources:
• What is a prism? Watch the Prism Video on Youtube if you're not sure. All the prisms we will be looking at in this course are "right prisms"
Notes:
• Watch the 4 videos shown below and complete your notes before coming to next class.
In Class Assignment:
• HW 5.3 p180 #3-8, 10, 12, 13, 15
Things you should be able to do after today:
• find the area of a rectangle
• find the area of a triangle
• label the measurements on a net of a rectangular or triangular prism
• determine the total surface area of a rectangular or triangular prism
What are other ways to represent a 3d shape in two dimensions? We saw last class that you can use one of several different views, but we can also take the shape and look at its "skin". When you cut a polyhedron along some of its edges to lay it flat, we get a net:
Resources:
• Notes: 5.2 Nets
• Triangular Prism
• Rectangular Prism
Fun Ideas with Nets and Networks
Pre-Class Assignment:
• Before next class, read through the notes and copy them down into your notebook.
• Think about the following 2 questions:
• Do you think a net is more or less effective at representing a 3d shape two dimensionally than the 3 views?
• Is there a way to figure out how many faces, vertices and edges a shape will have by looking at its net? Use the examples of the triangular prism and the rectangular prism to help you.
In-Class Assignment:
• P173 #3, 4, 6, 7, 8, 9, *12 *13
• Koninsberg Network Problem
Attachments:
FileDescriptionFile size
5.02 Notes.pdf 312 kB
You will be introduced to 3d objects as we begin looking at properties of surface area and volume.
Many people who work with blueprints, design or models have to try and show what a 3 dimensional object looks like using only picutres or 2 dimensional diagrams. Consider this computer generated model of a cartoon bird:
This design artist felt that it was very important to have 4 different pictures of the bird when working on it. Think about why there might need to be more than 1 view required to see this 3d object.
Resources:
Assignment to be completed during class:
• 5.1 Practice: p168 #3, 4, 5, 6, 7, 9
• Worksheet on 3d Views
Things you should know after today:
• vocabulary:
• face
• vertex
• edge
• isometric
• the 3 main views for a 3d object
• how to represent a 3d object by drawing the 3 main views and the isometric view
• how to draw an isometric view from the 3 main views
• how to draw the 3 main views from the isometric view
• what the 3 main views will change to look like if the 3d object is rotated in some direction
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# Lesson Video: Finding Absolute Value Mathematics
In this video we learn how to find the absolute value of a number, and to evaluate expressions involving the absolute values of terms, which means considering the order of operations.
05:35
### Video Transcript
What is the absolute value? And how do we find it? Absolute value of a number is the distance between that number and zero on a number line. Let me show you what I mean. The distance between zero and four on the number line shown here is four units. The distance between negative four and zero on this number line is also four units. So we say the absolute value of four is four. The straight lines on either side of the four is the symbol for the absolute value. The absolute value of negative four is four. When we talk about absolute value, we’re talking about distance. We’re talking about how far a number is from zero on a number line. And distance is represented by positive integers. Let’s look at two examples.
Example one, what is the absolute value of negative 53?
Remember, we’re trying to answer the question, how far is a negative 53 from zero? The answer is 53 units. The absolute value of negative 53 is 53.
Onto our next example.
The absolute value of 200 equals what?
The distance between 200 and zero on a number line, 200 units.
Example three, evaluate the following.
Before we move forward, I wanted to take a second and note that anytime you see the absolute value bars around something, you read it like this, the absolute value of whatever is inside the absolute value bars. In this case, we say the absolute value of negative three. Reading this would sound like the absolute value of negative three plus three minus the absolute value of negative seven. We’re going to use order of operations to solve this problem. Solving for absolute value occurs in the step for parentheses or grouping. For us, that means we’re going to need to solve the absolute value of negative three and the absolute value of negative seven before we can do anything else.
The absolute value of negative three or the distance from negative three to zero on a number line is three. The absolute value of negative seven or the distance from negative seven to zero on a number line is seven. At this point, we need to copy down the rest of the equation exactly how it was written. We finished all of our grouping. There are no exponents in this problem. There is no multiplication or division; we’re free to add or subtract from left to right. Three plus three is six. Bring down the rest of the equation; six minus seven is negative one.
Here’s another example.
Calculate the following: negative absolute value of negative nine times the negative of the absolute value of negative five.
This problem might sound a little bit more complicated than the last one, but we’ll follow the same steps. Order of operation tells us that we would do the parentheses — the grouping first. Remember that finding the absolute value falls in this step. So we need to start here. The absolute value of negative nine, the distance between negative nine and zero, is nine. The absolute value of negative five is five. The next step is just to copy down the problem exactly how it was written in the first line. Be careful that you copy down the problem exactly how it was in the first line, with the exception of adding in the absolute values that you’ve already solved for. There’s no more parentheses or grouping; there are no exponents in this problem. We can multiply and divide from left to right. In this case, we’ll only be multiplying: negative nine times negative five equals positive 45.
Let’s review the keys to absolute value. The absolute value of a number is the distance from that number to zero. When solving equations that have absolute value in them, absolute value is calculated in the first step — the parentheses step of the order of operations. These are the keys that you need to work with absolute value.
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# How do you find the limit of x^x as x>0^-?
Dec 14, 2017
${\lim}_{x \to {0}^{+}} {x}^{x} = 1$
#### Explanation:
Write the function as:
${x}^{x} = {\left({e}^{\ln} x\right)}^{x} = {e}^{x \ln x}$
Note now that:
${\lim}_{x \to {0}^{+}} x \ln x$
is in the indeterminate form $0 \cdot \infty$. We can reconduce it to the form $\frac{\infty}{\infty}$ and then apply l'Hospital's rule:
${\lim}_{x \to {0}^{+}} x \ln x = {\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{x}} = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \frac{1}{x}} = {\lim}_{x \to {0}^{+}} \frac{\frac{1}{x}}{- \frac{1}{x} ^ 2} = {\lim}_{x \to {0}^{+}} - x = 0$
As ${e}^{x}$ is a continuous function near $0$:
${\lim}_{x \to {0}^{+}} {e}^{x \ln x} = {e}^{{\lim}_{x \to {0}^{+}} x \ln x} = {e}^{0} = 1$
graph{x^x [-10, 10, -5, 5]}
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Lesson: Comparing Numbers and Counting Patterns, Day 3
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Lesson Objective
SWBAT use the inequality symbol when comparing numbers. *Note: builds to comparing 3-digit numbers.
Lesson Plan
Edward W. Brooke Charter School
Unit: Comparing and Ordering Numbers / Counting Patterns
Lesson: 3.3
Teaching Point: Mathematicians use the inequality symbol when comparing numbers.
Materials:
• 3.3 Independent Practice Sheet
• Problem Solving Kits
• Slates and Markers
• You might want to make a greater than / less than symbol that is magnetic so you can use it on the board.
Skip Counting (15 minutes)
• Refer to 3.3 Problem Solving Task
• This problem solving task is going to require some instruction before the students go off to work. On the top of the task, there is an explanation of the inequality symbol (We are going to start by referring to it as the greater or less than symbol). Go through the explanation of the symbol before sending the students off to work. First explain the symbol and then refer to the picture below which is a piranha. The piranha always wants to eat the bigger number so it’s mouth is always facing the bigger number.
• Send students off to work independently.
• Provide students with problem solving kits in case they want to use base-10 blocks when comparing the numbers.
Teaching (20 minutes)
• Have students turn and talk with their partners about the problem solving task.
• Have a whole group discussion about the problem solving task. Important points to include in the discussion:
• 63 and 36 are not the same number so we can’t say 63 = 36. We need a symbol that can show that one number is more or less than another number.
• < and> is the greater than/less than symbol. Mathematicians use this symbol when they are comparing numbers that are not the same.
• The symbol shows which number is more and which number is less depending on which way it is facing. Think of it like a hungry piranha’s mouth. The piranha always wants to eat the larger number so its mouth is always facing the larger number.
• Oftentimes this symbol is taught as two separate symbols (a greater than symbol and a less than symbol). We teach it as one symbol that is facing two different directions.
• 63 > 36. If you were to say this using words you would say “63 is greater than 36.”
• 36 < 63. If you were to say this using words you would say “36 is less than 63.”
• Make sure to go through each example on the board and draw the inequality symbol facing the correct direction.
Let’s try this out together when comparing other numbers.
• Put two numbers up on the board.
• As a class, use place value to decide which number is more/less.
• Have a student come up and draw the greater than/less than symbol facing the correct direction. You might even want to sketch the piranha attached to the symbol to help the students understand.
• As a class say the number sentence using words.
Independent Practice (10 minutes)
• Refer to 3.3 Independent Practice Sheet
• Send students off to work independently
• Since this can potentially be a tricky concept, bring students back to the rug when they are finished working and go over the sheet as a whole class.
Slate Math (10 minutes)
Summary
• By the end of lesson 3.4, all students should be able to use the inequality symbol when comparing numbers.
Lesson Resources
3.3 independent practice sheet.docx Activity 162 3.3 problem solving task.docx Activity 118
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