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# Complex Number Formula 5/5 - (1 bình chọn) Mục Lục ## Complex Number Complex numbers are helpful in finding the square root of negative numbers. The concept of complex numbers was first referred to in the 1st century by a greek mathematician, Hero of Alexandria when he tried to find the square root of a negative number. But he merely changed the negative into positive and simply took the numeric root value. Further, the real identity of a complex number was defined in the 16th century by Italian mathematician Gerolamo Cardano, in the process of finding the negative roots of cubic and quadratic polynomial expressions. Complex numbers have applications in many scientific research, signal processing, electromagnetism, fluid dynamics, quantum mechanics, and vibration analysis. Here we can understand the definition, terminology, visualization of complex numbers, properties, and operations of complex numbers. ## What are Complex Numbers? A complex number is the sum of a real number and an imaginary number. A complex number is of the form a + ib and is usually represented by z. Here both a and b are real numbers. The value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z). Also, ib is called an imaginary number. Some of the examples of complex numbers are etc. Power of i The alphabet i is referred to as the iota and is helpful to represent the imaginary part of the complex number. Further the iota(i) is very helpful to find the square root of negative numbers. We have the value of i2 = -1, and this is used to find the value of √-4 = √i24 = +2i The value of i2 = -1 is the fundamental aspect of a complex number. Let us try and understand more about the increasing powers of i. • i = √-1 • i2 = -1 • i = i.i2 = i(-1) = -i • i4 = (i2)2 = (-1)2 = 1 • i4n = 1 • i4n + 1 = i • i4n + 2 = -1 • i4n + 3 = -i ## Graphing of Complex Numbers The complex number consists of a real part and an imaginary part, which can be considered as an ordered pair (Re(z), Im(z)) and can be represented as coordinates points in the euclidean plane. The euclidean plane with reference to complex numbers is called the complex plane or the Argand Plane, named after Jean-Robert Argand. The complex number z = a + ib is represented with the real part – a, with reference to the x-axis, and the imaginary part-ib, with reference to the y-axis. Let us try to understand the two important terms relating to the representation of complex numbers in the argand plane. The modulus and the argument of the complex number. ### Modulus of the Complex Number The distance of the complex number represented as a point in the argand plane (a, ib) is called the modulus of the complex number. This distance is a linear distance from the origin (0, 0) to the point (a, ib), and is measured as Further, this can be understood as derived from the Pythagoras theorem, where the modulus represents the hypotenuse, the real part is the base, and the imaginary part is the altitude of the right-angled triangle. ### Argument of the Complex Number The angle made by the line joining the geometric representation of the complex number and the origin, with the positive x-axis, in the anticlockwise direction is called the argument of the complex number. The argument of the complex number is the inverse of the tan of the imaginary part divided by the real part of the complex number. Argz (θ) = ## Polar Representation of a Complex Number With the modulus and argument of a complex number and the representation of the complex number in the argand plane, we have a new form of representation of the complex number, called the polar form of a complex number. The complex number z = a + ib, can be represented in polar form as z = r(Cosθ + iSinθ). Here r is the modulus (r = \sqrt{a^2 + n^2}\), and θ is the argument of the complex number(θ = ## Properties of a Complex Number The following properties of complex numbers are helpful to better understand complex numbers and also to perform the various arithmetic operations on complex numbers. ### Conjugate of a Complex Number The conjugate of the complex number is formed by taking the same real part of the complex number and changing the imaginary part of the complex number to its additive inverse. If the sum and product of two complex numbers are real numbers, then they are called conjugate complex numbers. For a complex number z = a + ib, its conjugate is ¯z = a – ib. The sum of the complex number and its conjugate is z+¯z = ( a + ib) + (a – ib) = 2a, and the product of these complex numbers z.¯z = (a + ib) × (a – ib) = a2 + b2. ### Reciprocal of a Complex Number The reciprocal of complex numbers is helpful in the process of dividing one complex number with another complex number. The process of division of complex numbers is equal to the product of one complex number with the reciprocal of another complex number.. The reciprocal of the complex number z = a + ib is This also shows that z≠z−1. ### Equality of Complex Numbers The equality of complex numbers is similar to the equality of real numbers. Two complex numbers z1=a1+ib1 and z2=a2+ib2 are said to be equal if the rel part of both the complex numbers are equal a1=a2, and the imaginary parts of both the complex numbers are equal b1=b2. Also, the two complex numbers in the polar form are equal, if and only if they have the same magnitude and their argument (angle) differs by an integral multiple of 2π. ### Ordering of Complex Numbers The ordering of complex numbers is not possible. Real numbers and other related number systems can be ordered, but complex numbers cannot be ordered. The complex numbers do not have the structure of an ordered field, and there is no ordering of the complex numbers that are compatible with addition and multiplication. Also, the non-trivial sum of squares in an ordered field is a number , but in a complex number, the non-trivial sum of squares is equal to i2 + 12 = 0. The complex numbers can be measured and represented in a two-dimensional argrand plane by their magnitude, which is its distance from the origin. Euler’s Formula: As per Euler’s formula for any real value θ we have e = Cosθ + iSinθ, and it represents the complex number in the coordinate plane where Cosθ is the real part and is represented with respect to the x-axis, Sinθ is the imaginary part that is represented with respect to the y-axis, θ is the angle made with respect to the x-axis and the imaginary line, which is connecting the origin and the complex number. As per Euler’s formula and for the functional representation of x and y we have ex + iy = ex(cosy + isiny) = excosy + iexSiny. This decomposes the exponential function into its real and imaginary parts. ## Operations on Complex Numbers The various operations of addition, subtraction, multiplication, division of natural numbers can also be performed for complex numbers also. The details of the various arithmetic operations of complex numbers are as follows. Th addition of complex numbers is similar to the addition of natural numbers. Here in complex numbers, the real part is added to the real part and the imaginary part is added to the imaginary part. For two complex numbers of the form and , the sum of complex numbers . The complex numbers follow all the following properties of addition. • Closure Law: The sum of two complex numbers is also a complex number. For two complex numbers and , the sum of is also a complex number. • Commutative Law: For two complex numbers we have . • Associative Law: For the given three complex numbers we have . • Additive Identity: For a complex number z = a + ib, there exists 0 = 0 + i0, such that z + 0 = 0 + z = 0. • Additive Inverse: For the complex number z = a + ib, there exists a complex number -z = -a -ib such that z + (-z) = (-z) + z = 0. Here -z is the additive inverse. ### Subtraction of Complex Numbers The subtraction of complex numbers follows a similar process of subtraction of natural numbers. Here for any two complex numbers, the subtraction is separately performed across the real part and then the subtraction is performed across the imaginary part. For the complex numbers z1 = a + ib, z2=c+id, we have z1−z2 = (a – c) + i(b – d) ### Multiplication of Complex Numbers The multiplication of complex numbers is slightly different from the multiplication of natural numbers. Here we need to use the formula of i2=−1. For the two complex numbers z1 = a + ib, z2 = c + id, the product is z1.z2 = (ca – bd) + i(ad + bc). The multiplication of complex numbers is polar form is slightly different from the above mentioned form of multiplication. Here the absolute values of the two complex numbers are multiplied and their arguments are added to obtain the product of the complex numbers. For the complex numbers z1=r1(Cosθ1+iSinθ1), and z2 = z2=r1(Cosθ2+iSinθ2), the product of the complex numbers is z1.z2=r1.r2(Cos(θ1+θ2)+iSin(θ1+θ2)). ### Division of Complex Numbers The division of complex numbers makes use of the formula of reciprocal of a complex number. For the two complex numbers z1= a + ib, z2 = c + id, we have the division as ## Algebraic Identities of Complex Numbers All the algebraic identities apply equally for complex numbers.The addition and subtraction of complex numbers and with exponents of 2 or 3 can be easily solved using algebraic identities of complex numbers. ### Complex Numbers Tips and Tricks: • All real numbers are complex numbers but all complex numbers don’t need to be real numbers. • All imaginary numbers are complex numbers but all complex numbers don’t need to be imaginary numbers. • The conjugate of a complex number z=a+ib is ¯¯¯z=a−ib. • The magnitude of a complex number z=a+ib is \|z\|=√a2+b2\| ## Complex Numbers in Maths Complex numbers are the numbers that are expressed in the form of a+ib where, a,b are real numbers and ‘i’ is an imaginary number called “iota”. The value of i = (√-1). For example, 2+3i is a complex number, where 2 is a real number (Re) and 3i is an imaginary number (Im). Examples of complex numbers: • 1 + j • -13 – 3i • 0.89 + 1.2 i • √5 + √2i An imaginary number is usually represented by ‘i’ or ‘j’, which is equal to √-1. Therefore, the square of the imaginary number gives a negative value. The main application of these numbers is to represent periodic motions such as water waves, alternating current, light waves, etc., which rely on sine or cosine waves, etc. ## Definition The complex number is basically the combination of a real number and an imaginary number. The complex number is in the form of a+ib, where a = real number and ib = imaginary number. Also, a,b belongs to real numbers and i = √-1. Hence, a complex number is a simple representation of addition of two numbers, i.e., real number and an imaginary number. One part of it is purely real and the other part is purely imaginary. ## Notation An equation of the form z= a+ib, where a and b are real numbers, is defined to be a complex number. The real part is denoted by Re z = a and the imaginary part is denoted by Im z = ib. ### Examples See the table below to differentiate between a real number and an imaginary number. ### Is 0 a complex Number? As we know, 0 is a real number. And real numbers are part of complex numbers. Therefore, 0 is also a complex number and can be represented as 0+0i. ## Graphical representation In the graph below, check the representation of complex numbers along the axes. Here we can see, the x-axis represents real part and y represents the imaginary part. Let us see an example here. If we have to plot a graph of complex number 3 + 4i, then: ## Absolute Value The absolute value of a real number is the number itself. The absolute value of x is represented by modulus, i.e. \|x\|. Hence, the modulus of any value always gives a positive value, such that; \|3\| = 3 \|-3\| = 3 Now, in case of complex numbers, finding the modulus has a different method. Suppose, z = x+iy is a complex number. Then, mod of z, will be: \|z\| = √(x2+y2) This expression is obtained when we apply the Pythagorean theorem in a complex plane. Hence, mod of complex number, z is extended from 0 to z and mod of real numbers x and y is extended from 0 to x and 0 to y respectively. Now these values form a right triangle, where 0 is the vertex of the acute angle. Now, applying Pythagoras theorem, \|z\|2 = \|x\|2+\|y\|2 \|z\|2 = x2 + y2 \|z\| = √(x2+y2) ## Algebraic Operations on Complex Numbers There can be four types of algebraic operation on complex numbers which are mentioned below. Visit the linked article to know more about these algebraic operations along with solved examples. The four operations on the complex numbers include: • Subtraction • Multiplication • Division ## Roots of Complex Numbers When we solve a quadratic equation in the form of ax2 +bx+c = 0, the roots of the equations can be determined in three forms; • Two Distinct Real Roots • Similar Root • No Real roots (Complex Roots) ## Complex Number Formulas While performing the arithmetic operations of complex numbers such as addition and subtraction, combine similar terms. It means that combine the real number with the real number and imaginary number with the imaginary number. (a + ib) + (c + id) = (a + c) + i(b + d) ### Subtraction (a + ib) – (c + id) = (a – c) + i(b – d) ### Multiplication When two complex numbers are multiplied by each other, the multiplication process should be similar to the multiplication of two binomials. It means that the FOIL method (Distributive multiplication process) is used. (a + ib). (c + id) = (ac – bd) + i(ad + bc) ### Division The division of two complex numbers can be performed by multiplying the numerator and denominator by its conjugate value of the denominator, and then applying the FOIL Method. (a + ib) / (c + id) = (ac+bd)/ (c2 + d2) + i(bc – ad) / (c2 + d2) ## Power of Iota (i) Depending upon the power of “i”, it can take the following values; i4k+1 = i.i4k+2 = -1 i4k+3 = -i.i4k = 1 Where k can have an integral value (positive or negative). Similarly, we can find for the negative power of i, which are as follows; i-1 = 1 / i Multiplying and dividing the above term with i, we have; i-1 = 1 / i × i/i × i-1 = i / i2 = i / -1 = -i / -1 = -i Note: √-1 × √-1 = √(-1 × -1) = √1 = 1 contradicts to the fact that i2 = -1. Therefore, for an imaginary number, √a × √b is not equal to √ab. ## Identities Let us see some of the identities. 1. (z+ z2)2 = (z1)+ (z2)2 + 2 z1 × z2 2. (z– z2)2 = (z1)+ (z2)2 – 2 z1 × z2 3. (z1)– (z2)2 = (z+ z2)(z– z2) 4. (z+ z2)3 = (z1)3 + 3(z1)2 z +3(z2)2 z1 + (z2)3 5. (z– z2)3 = (z1)3 – 3(z1)2 z +3(z2)2 z1 – (z2)3 ## Properties The properties of complex numbers are listed below: • The addition of two conjugate complex numbers will result in a real number • The multiplication of two conjugate complex number will also result in a real number • If x and y are the real numbers and x+yi =0, then x =0 and y =0 • If p, q, r, and s are the real numbers and p+qi = r+si, then p = r, and q=s • The complex number obeys the commutative law of addition and multiplication. z1+z2 = z2+z1 z1. z2 = z2. z1 • The complex number obeys the associative law of addition and multiplication. (z1+z2) +z= z1 + (z2+z3) (z1.z2).z= z1.(z2.z3) • The complex number obeys the distributive law z1.(z2+z3) = z1.z+ z1.z3 • If the sum of two complex number is real, and also the product of two complex number is also real, then these complex numbers are conjugate to each other. • For any two complex numbers, say zand z2, then \|z1+z2\| ≤ \|z1\|+\|z2\| • The result of the multiplication of two complex numbers and its conjugate value should result in a complex number and it should be a positive value. ## Modulus and Conjugate Let z = a+ib be a complex number. The Modulus of z is represented by \|z\|. Mathematically, ## Argand Plane and Polar Form Similar to the XY plane, the Argand(or complex) plane is a system of rectangular coordinates in which the complex number a+ib is represented by the point whose coordinates are a and b. We find the real and complex components in terms of r and θ, where r is the length of the vector and θ is the angle made with the real axis. Check out the detailed argand plane and polar representation of complex numbers in this article and understand this concept in a detailed way along with solved examples. ## Complex Numbers Examples Example 1: Can we help Sophia express the roots of the quadratic equation x2+x+1=0 as complex numbers? Solution: Comparing the given equation with ax2+bx+c=0 ,a=1 b=1 c=1 (This equation is as same as the one we saw in the beginning of this page). Substitute these values in the quadratic formula: Thus the roots of the given quadratic equation are: Example 2: Express the sum, difference, product, and quotient of the following complex numbers as a complex number. Solution: Sum: Difference: Quotient: Therefore, we have: Sum = -1 – i Difference = -3 + 3i Product = 5i Division = -4/5 – 3i/5 Example 1: Simplify a) 16i + 10i(3-i) b) (7i)(5i) c) 11i + 13i – 2i Solution: a) 16i + 10i(3-i) = 16i + 10i(3) + 10i (-i) = 16i +30i – 10 i2 = 46 i – 10 (-1) = 46i + 10 b) (7i)(5i) = 35 i2 = 35 (-1) = -35 c) 11i + 13i – 2i = 22i Example 2: Express the following in a+ib form: (5+√3i)/(1-√3i). Solution: Given: (5+√3i)/(1-√3i) ## FAQs on Complex Numbers ### What are Complex Numbers in Math? A complex number is a combination of real values and imaginary values. It is denoted by z = a + ib, where a, b are real numbers and i is an imaginary number. i = √−1−1 and no real value satisfies the equation i2 = -1, therefore, I is called the imaginary number. ### What are Complex Numbers Used for? The complex number is used to easily find the square root of a negative number. Here we use the value of i2 = -1 to represent the negative sign of a number, which is helpful to easily find the square root. Here we have √-4 = √i24 = + 2i. Further to find the negative roots of the quadratic equation, we used complex numbers. ### What is Modulus in Complex Numbers? The modulus of a complex number z = a + ib is the distance of the complex number in the argand plane, from the origin. It is represented by \|z\| and is equal to If a complex number is considered as a vector representation in the argand plane, then the module of the complex number is the magnitude of that vector. ### What is Argument in Complex Numbers? The argument of a complex number is the angle made by the line joining the origin to the geometric representation of the complex number, with the positive x-axis in the anticlockwise direction. The argument of the complex number is the inverse of the tan of the imaginary part divided by the real part of the complex number. Argz (θ) = ### How to Perform Multiplication of Complex Numbers? The multiplication of complex numbers is slightly different from the multiplication of natural numbers. Here we need to use the formula of i2 = -1. For the two complex numbers z1 = a + ib, z2 = c + id, the product is z1.z2 = (ca – bd) + i(ad + bc). ### What is Polar Form in Complex Numbers? The polar form of a complex number is another form of representing and identifying a complex number in the argand plane. The polar form makes the use of the modulus and argument of a complex number, to represent the complex number. The complex number z = a + ib, can be represented in polar form as z = r(Cosθ + iSinθ). Here r is the modulus (r = \sqrt{a^2 + n^2}\), and θ is the argument of the complex number ### What Are Real and Complex Numbers? Complex numbers are a part of real numbers. Certain real numbers with a negative sign are difficult to compute and we represent the negative sign with an iota ‘i’, and this representation of numbers along with ‘i’ is called a complex number. Further complex numbers are useful to find the square root of a negative number, and also to find the negative roots of a quadratic or polynomial expression. ### How To Divide Complex Numbers? The division of complex numbers makes use of the formula of reciprocal of a complex number. For the two complex numbers z1 = a + ib, z2 = c + id, we have the division as ### How to Graph Complex Numbers? The complex number of the form z = a + ib can be represented in the argand plane. The complex number z = a + ib can be presented as the coordinates of a point as (Re(z), Im(z)) = (a, ib). Here the real part is presented with reference to the x-axis, and the imaginary part is presented with reference to the y-axis. ### How to Convert Complex Numbers to Polar Form? The complex number can be easily converted into a polar form. The complex number z = a + ib has a modulus (r = \sqrt{a^2 + n^2}\), and an argument With the help of modulus and argument the polar form of the complex number is written as z = r(Cosθ + iSinθ). ### How to Write Complex Numbers in Standard Form? The standard form of writing a complex number is z = a + ib. The standard form of the complex number has two parts, the real part, and the imaginary part. In the complex number z = a + ib, a is the real part and ib is the imaginary part. ### How Complex Numbers were Invented? The inverse of a complex numbers is helpful in the process of dividing one complex number with another complex number. The process of division of complex numbers is equal to the product of one complex number with the inverse of another complex number.. The inverse of the complex number z = a + ib is This also shows that z≠z−1. ### What are Real Numbers? Any number which is present in a number system such as positive, negative, zero, integer, rational, irrational, fractions, etc. are real numbers. It is represented as Re(). For example: 12, -45, 0, 1/7, 2.8, √5, etc., are all real numbers. ### What are Imaginary Numbers? The numbers which are not real are imaginary numbers. When we square an imaginary number, it gives a negative result. It is represented as Im(). Example: √-2, √-7, √-11 are all imaginary numbers. The complex numbers were introduced to solve the equation x2+1 = 0. The roots of the equation are of form x = ±√-1 and no real roots exist. Thus, with the introduction of complex numbers, we have Imaginary roots. We denote √-1 with the symbol ‘i’, which denotes Iota (Imaginary number). ### What is meant by complex numbers? The complex number is the combination of a real number and imaginary number. An example of a complex number is 4+3i. Here 4 is a real number and 3i is an imaginary number. ### How to divide the complex numbers? To divide the complex number, multiply the numerator and the denominator by its conjugate. The conjugate of the complex number can be found by changing the sign between the two terms in the denominator value. Then apply the FOIL method to simplify the expression. ### Mention the arithmetic rules for complex numbers. The arithmetic rules of complex numbers are: Addition Rule: (a+bi) + (c+di) = (a+c)+ (b+d)i Subtraction Rule: (a+bi) – (c+di) = (a-c)+ (b-d)i Multiplication Rule: (a+bi) . (c+di) = (ac-bd)+(ad+bc)i ### Write down the additive identity and inverse of complex numbers. The additive identity of complex numbers is written as (x+yi) + (0+0i) = x+yi. Hence, the additive identity is 0+0i. The additive inverse of complex numbers is written as (x+yi)+ (-x-yi) = (0+0i). Hence, the additive inverse is -x-yi. ### Write down the multiplicative identity and inverse of the complex number. The multiplicative identity of complex numbers is defined as (x+yi). (1+0i) = x+yi. Hence, the multiplicative identity is 1+0i. The multiplicative identity of complex numbers is defined as (x+yi). (1/x+yi) = 1+0i. Hence, the multiplicative identity is 1/x+yi. Related articles Complex Number Formula Complex Number Division Formula ⭐️⭐️⭐️⭐️⭐ Complex Number Power Formula ⭐️⭐️⭐️⭐️⭐
# Heights and Distances: Theory & Concepts Views:28698 The topic of height and distance in trigonometry is an important topic from competitive examination point of view. Generally you must have seen the problems where the height of a building is given and then from the top of this building the angles of elevation or depression are given for another building and you have to find the height of the second building. In this article we will cover these problems. There are certain terms associated with the heights and distances which are described as follows: 1. Angle of Elevation: Let us consider a situation where a person is standing on the ground and he is looking at an object which is at some height say the top of the building. The line joining the eye of the man with the top of the building is called the line of sight. The angle made by the line of sight with the horizontal line is called angle of elevation. In this figure the line of sight is making an angle θ with the horizontal line. This angle is the angle of elevation. 2. Angle of Depression: Now let us take another situation where the person is standing at some height with respect to the object he is seeing. In this case again the line joining the eye of the man with the bottom of the building is called the line of sight. The angle made by the line of sight with the horizontal line is called angle of depression. In the above figure ‘θ’ is the angle of depression. Note: The angle of elevation is equal to the angle of depression. The questions on this topic require basic knowledge of Trigonometry. You should be aware of the basic trigonometric ratios and their values. Let us recall that the ratios of the sides of a right angled triangle are called trigonometric ratios. These are sine, cosine, tangent, cosecant, secant and cotangent. Let the ΔABC is a right angled triangle. Then Sin θ = AB/AC Cos θ = BC/AC Tan θ = AB/BC Cosec θ = AC/AB Sec θ = AC/BC Cot θ = BC/AB Also you should know the values of these trigonometric ratios of some common angles as given in the following table: 00 300 450 600 900 Sin 0 1/2 1/√2 √3/2 1 Cos 1 √3/2 1/√2 1/2 0 Tan 0 1/√3 1 √3 Not Defined The values in the table will be used while solving the questions on height and distances. Let us now discuss some basic problems on height and distances. After going through these examples you will learn how to measure the height and how to find the distance. Views:28698
Theorem 3.13: Given two circles whose equations are (x - x1)2 + (y - y1)2 = r12 and (x - x2)2 + (y - y2)2 = r22 their points of intersection are given by and where Proof: The points of intersection lie on a line whose equation is given from Theorem 3.10. We will consider two cases. The first is where the centers are not horizontal, and the second is where they are. Case 1: The centers are not horizontal. We can use the simpler formula for the points where a line intersects a circle from Theorem 3.3 and where by Theorem 3.10, and The rational terms are the coordinates of the foot of the center of the circle centered at (x1, y1) with radius r1. The slope of the line between the points of intersection of the two circles is the negative reciprocal of the slope of the line between the centers we know that the line joining the points of intersection is perpendicular to the line between the centers, so the foot of the center in the line joining the two points is the intesection of those lines, and we know from Theorem 3.12 that those coordinates are and so the coordinates of the points of intersection are and We should now work with the radical terms. Let us concentrate on the expression under the radicals (m2+ 1)r12 - (y1 - mx1 - b) First note that When we substitiute for m and b, the expression inside the radical becomes Find common denominators. Remove the parentheses in the second term. Combine like terms in the numerator of the second fraction. This can be rewritten as or We we find common denominators for these two fractions. We can now factor the top as a difference of squares. Remove the inside parentheses, and rearrange the terms. or Both factors on top are differences of squares, and will factor. Which can be written as Rearrange the factors and we find that we have two sum times difference multiplications which we can multiply as We put this back under the radical in the formula for x and using the fact that get for the radical term. Invert and multiply the denominator, and simplify the denominator under the radical. If we put it all together, we get Since we will be taking both the positive and negative square root, we can drop the absolute value signs around y2 - y1. To get the y-coordinate of the points of intersection we can substitute this value of x into the equation from Theorem 3.10. This would give us y = mx + b when you multiply m times x, you will need to multiply m by all the terms, but when you add b, you will only be adding it in once. When you multiply m times the rational terms and add b, you will get the y-coordinate of the foot of the (x1, y1) in the line joining the points of intersection of the circles, but after that, you will simply multiply the radical term by m. Since the y2 - y1's will cancel leaving us with an x2 - x1 on top. The negative sign in front of the slope will change th sign resulting in Case 2: y1 = y2. In this case the line which contains the points of intersection has the equation by Theorem 3.12. Subtitute this value for x into the equation for the first circle (x - x1)2 + (y - y1)2 = r12 and get This will simplify to This is fairly easy to solve for y. Transpose the first term on the left to the right. Take square roots of both sides. or To simplify this, first find common denominators for the terms inside the parentheses under the radical. Under the condition that y1 = y2, \|x2 - x1\| = d, the distance between the centers of the circles.After finding common denominators for the two terms inside the radical, we get The expression under the radical is essentially the same as the expression under the radical in the last computation, so it will simplify to the same thing. which is the formula to which the y coordinates of the points of intersection of the circles, in the statement of the theorem, will simplify, when y1 = y2. top Next theorem (T3.14)
Mathematics # Equilateral Triangle Triangles are closed figures joining with the straight line segments .there are different types of triangles from which the Equilateral Triangle is a regular polygon figure that has three equal-lengths congruent sides (same measure). hence their all three angles should also be equal (equiangular). as the internal angles are acute, the equilateral triangle is itself acute. since, we knew that all the sides and angles are equal in length, so their sum will also be equal to 180º and each interior angle will be 60º. ### What are the Elements of the Equilateral Triangle? from the above figure, the following are elements of such type triangles: Sides: AB, BC, and CA are three equal sides and measures a,b and c respectively. Vertices: A, B, C are called Vertices. Interior Angles: let ∝, β, and γ are three interior angles and are equal to 60º each. Exterior Angles: let x,y,z are three exterior angles that are supleentary to their adjacent interior angles i.e., ∝ + x = β + y = γ + z = 180º. points to remember: • for equilateral triangle, sides a = b = c. • ∝ = β = γ = 60º • x=y=z= 120º or obtuse (greater than 90º) ## Equilateral Triangle Area The area of a flat figure like the Equilateral Triangle represents the size of its surface. the multiplication of base (b) with height (h) and dividing by 2 will give us the area of an equilateral triangle. in other words, the area will be one-half of the base (a) times its height. let suppose we have a height √3a/2. by putting the value of height from the above figure we have the area an equilateral triangle: ## The perimeter of an equilateral triangle as the following triangle have three sides all equal to each other, the perimeter corresponds to the sum of all sides or three times the length of one of its sides ( a ). Mathematically, P = a + a + a = 3a ## equilateral triangle height to find the height of an equilateral triangle we use the Pythagorean theorem. for this purpose, let us divide the triangle into two parts to make a right triangle. now take the sides a, a / 2, and height h. where side a / 2 and h are known as legs while a is called the hypotenuse. hence, we find the height (h) of the triangle = √3a/2. ## What is the Pythagorean theorem? this theorem states that two legs of the right triangle lie on the hypotenuse. the two sides that characteristic a right triangle are known as legs while the longest opposite side to legs is the hypotenuse. according to the theorem: the square of the hypotenuse is equal to the sum of the square of its legs. c² = a² + b² where a and b are two legs and c point to the hypotenuse. ## Properties The following properties fall under the Mentioned triangle: • Angle Bisector: • Segment Bisector: • median is that line that starts from mid of its any side and meets at the vertex of another opposite side. this means, we can draw three medians at a time and they will meet at a single point called the center of gravity. • axes of symmetry: it has three axes of symmetry that pass through the midpoint of each opposite side and its vertex • consists of three equal sides • known as a regular polygon having three sides. • The area is equal to √3a2/ 4 • The perimeter is equal to 3a. • all the angles are 60º and are congruent to each other. • centroid and ortho-center meet at the same point. Questions and Solved Excercise Q. if the side of an equilateral triangle is 12cm side, find its area? Ans: since, A=L² √3/2 A=12² √3/2 A=144 √3/2 A= 72 √3 cm² Q. let the sides of a triangle are equal to 5cm each. what will e it’s area? we know that Area= √3/4.a² Area= √3/4.5² = 10.8253175473 cm². Q. if the sides are sides 24cm long, calculate the area of a triangle. Ans: since, area formula is given by: P = 3. L P = 3.24 P = 72 What is the equilateral triangle? Ans: it is a regular polygon triangle that has three sides of equal lengths as well as equal angles. the internal angles are called congruent and are of 60º each. What is equilateral triangle example? Ans: rack in billiards, a slice of a pizza is an example of such type of triangle. What is the area and perimeter of equilateral triangle? Ans: the area is said to e √3 a2/ 4 while the perimeter is equal to 3a. where a is the side of an equilateral triangle. What are the rules for an equilateral triangle? Ans: basic properties and rules are given below: • it is a regular polygon in geometry. • All three sides are equal in length. • All three angles are congruent and equal to 60º. • The ortho-center and centroid meet at the same point. What are the 3 main types of triangles? Ans: equilateral, isosceles, and scalene are three major types of triangles based on their sides. How many sides does an equilateral triangle have? Ans: it has 3 sides all are equal in length. You May Also Like: Check Also Close
# Use the graph shown to determine whether each system is consistent or inconsistent and if it is independent or dependent. Save this PDF as: Size: px Start display at page: Download "Use the graph shown to determine whether each system is consistent or inconsistent and if it is independent or dependent." ## Transcription 1 Use the graph shown to determine whether each system is consistent or inconsistent and if it is independent or dependent. y = 3x + 1 y = 3x + 1 The two equations intersect at exactly one point, so they are consistent and independent. y = 3x + 1 y =x 3 The two equations intersect at exactly one point, so they are consistent and independent. y =x 3 y =x+3 These two equations do not intersect, so they are inconsistent. y =x+3 x y= 3 Rearrange the first equation into slope intercept form to determine which line it is. The two equations are the same and so intersect in an infinite amount of points. Therefore, they are consistent and dependent. x y= 3 y = 3x + 1 Rearrange the first equation into slope intercept form to determine which line it is. Page 1 2 6-1 Graphing Systemsare of the Equations The two equations same and so intersect in an infinite amount of points. Therefore, they are consistent and dependent. x y= 3 y = 3x + 1 Rearrange the first equation into slope intercept form to determine which line it is. The two equations intersect at exactly one point, so they are consistent and independent. y = 3x + 1 y =x 3 The two equations intersect at exactly one point, so they are consistent and independent. Graph each system and determine the number of solutions that it has. If it has one solution, name it. y =x+4 y= x 4 The graphs intersect at one point, so there is one solution. The lines intersect at ( 4, 0). y =x+3 y = 2x + 4 Page 2 3 6-1 Graphing The graphssystems intersectof at Equations one point, so there is one solution. The lines intersect at ( 4, 0). y =x+3 y = 2x + 4 The graphs intersect at one point, so there is one solution. The lines intersect at ( 1, 2). Write an equation to represent the pages each boy has read. Graph each equation. How long will it be before Alberto has read more pages than Ashanti? Check and interpret your solution. Let y represent the number of pages read and let x represent the number of days. Alberto: y = 20x + 35; Ashanti: y = 10x + 85 b. The solution is (5, 135). To check this answer, enter it into both equations. Page 3 4 Let y represent the number of pages read and let x represent the number of days. Alberto: y = 20x + 35; Ashanti: y = 10x Graphing Systems of Equations b. The solution is (5, 135). To check this answer, enter it into both equations. So, it is a solution for the first equation. Now check the second equation. So it is a solution for the second equation. Alberto will have read more pages than Ashanti after 5 days. Use the graph shown to determine whether each system is consistent or inconsistent and if it is independent or dependent. y =6 y = 3x + 4 The two equations intersect at exactly one point, so they are consistent and independent. y = 3x + 4 Page 4 5 6-1 Graphing Systems So it is a solution foroftheequations second equation. Alberto will have read more pages than Ashanti after 5 days. Use the graph shown to determine whether each system is consistent or inconsistent and if it is independent or dependent. y =6 y = 3x + 4 The two equations intersect at exactly one point, so they are consistent and independent. y = 3x + 4 y = 3x + 4 The two equations intersect at exactly one point, so they are consistent and independent. y = 3x + 4 y = 3x 4 These two equations do not intersect, so they are inconsistent. y = 3x 4 y = 3x 4 The two equations intersect at exactly one point, so they are consistent and independent. 3x y = 4 y = 3x + 4 Rearrange the first equation into slope intercept form to determine which line it is. The two equations are identical, so they are consistent and dependent. 3x y = 4 3x + y = 4 Page 5 6 The two equations are identical, so they are consistent and dependent. 3x y = 4 3x + y = 4 Rearrange the two equations into slope intercept form to determine which lines they are. The two equations intersect at exactly one point, so they are consistent and independent. Graph each system and determine the number of solutions that it has. If it has one solution, name it. y= 3 y =x 3 The graphs intersect at one point, so there is one solution. The lines intersect at (0, 3). y = 4x + 2 y = 2x Page 6 7 6-1 Graphing The graphssystems intersectof at Equations one point, so there is one solution. The lines intersect at (0, 3). y = 4x + 2 y = 2x The graphs intersect at one point, so there is one solution. The lines intersect at. y =x 6 y =x+ These two lines are parallel, so they do not intersect. Therefore, there is no solution. x +y = 4 3x + 3y = Page 7 8 These two lines are parallel, so they do not intersect. Therefore, there is no solution. x +y = 4 3x + 3y = The two lines are the same line. Therefore, there are infinitely many solutions. x y= 2 x +y = Page 8 9 The two lines are the same line. Therefore, there are infinitely many solutions. x y= 2 x +y = The two lines are the same line. Therefore, there are infinitely many solutions. x + 2y = 3 x=5 Rearrange the first equation into slope-intercept form. Page 9 10 The two lines are the same line. Therefore, there are infinitely many solutions. x + 2y = 3 x=5 Rearrange the first equation into slope-intercept form. The graphs intersect at one point, so there is one solution. The lines intersect at (5, 1). 2x + 3y = 12 2x y = 4 Page 10 11 The graphs intersect at one point, so there is one solution. The lines intersect at (5, 1). 2x + 3y = 12 2x y = 4 The graphs intersect at one point, so there is one solution. The lines intersect at (3, 2). 2x + y = 4 y + 2x = 3 Page 11 12 The graphs intersect at one point, so there is one solution. The lines intersect at (3, 2). 2x + y = 4 y + 2x = 3 These two lines are parallel, so they do not intersect. Therefore, there is no solution. 2x + 2y = 6 5y + 5x = 15 Page 12 13 These two lines are parallel, so they do not intersect. Therefore, there is no solution. 2x + 2y = 6 5y + 5x = 15 The two lines are the same line. Therefore, there are infinitely many solutions. Akira and Jen are competing to see who can sell the most tickets for the Winter Dance. On Monday, Akira sold 22 and then sold 30 per day after that. Jen sold 53 one Monday and then sold 20 per day after that. Write equations for the number of tickets each person has sold. Graph each equation. Solve the system of equations. Check and interpret your solution. a. Let x represent the number of days and let y Number of tickets sold by each equals tickets sold per day times the number of days plus the number of tickets sold initially esolutions Manual - Powered by Cognero Page 13 Akira: y = 30x Jen: y = 20x + 53 14 The two lines are the same line. Therefore, there are infinitely many solutions. Akira and Jen are competing to see who can sell the most tickets for the Winter Dance. On Monday, Akira sold 22 and then sold 30 per day after that. Jen sold 53 one Monday and then sold 20 per day after that. Write equations for the number of tickets each person has sold. Graph each equation. Solve the system of equations. Check and interpret your solution. a. Let x represent the number of days and let y Number of tickets sold by each equals tickets sold per day times the number of days plus the number of tickets sold initially Akira: y = 30x Jen: y = 20x + 53 b. Graph y = 30x + 22 and y = 20x Choose appropriate scales for the horizontal and vertical axes. c. The graphs appear to intersect at (3.1, 115) Use substitution to check this. Since the answer works in both equations, (3.1, 115) is the solution to the system of equations. For less than 3 days Jen will have sold more tickets. After about 3 days Jen and Akira will have sold the same If x is the number of years since 2000 and y is the percent of people using travel services, the following equations represent the percent of people using travel agents and the percent of the people using the Internet to plan travel. Travel agents: y = 2x + 30 Internet: y = 6x + 41 Graph the system of equations. Page 14 15 Since the answer works in both equations, (3.1, 115) is the solution to the system of equations. 6-1 Graphing Systems For less than 3 days of JenEquations will have sold more tickets. After about 3 days Jen and Akira will have sold the same If x is the number of years since 2000 and y is the percent of people using travel services, the following equations represent the percent of people using travel agents and the percent of the people using the Internet to plan travel. Travel agents: y = 2x + 30 Internet: y = 6x + 41 Graph the system of equations. Estimate the year travel agents and the Internet were used equally. a. The intersection occurs close to x = 1. So, this means that the year is , or Graph each system and determine the number of solutions that it has. If it has one solution, name it. y =x+2 The graphs intersect at one point, so there is one solution. The lines intersect at ( 4, 2). y = 6x + 6 y = 3x + 6 Page 15 16 6-1 Graphing The graphssystems intersectof at Equations one point, so there is one solution. The lines intersect at ( 4, 2). y = 6x + 6 y = 3x + 6 The graphs intersect at one point, so there is one solution. The lines intersect at (0, 6). y = 2x 17 y = x 10 The graphs intersect at one point, so there is one solution. The lines intersect at (7, 3). 8x 4y = 16 5x 5y = 5 Page 16 17 The graphs intersect at one point, so there is one solution. The lines intersect at (7, 3). 8x 4y = 16 5x 5y = 5 The graphs intersect at one point, so there is one solution. The lines intersect at (1, 2). 3x + 5y = 30 3x + y = 18 Page 17 18 The graphs intersect at one point, so there is one solution. The lines intersect at (1, 2). 3x + 5y = 30 3x + y = 18 The graphs intersect at one point, so there is one solution. The lines intersect at (5, 3). 3x + 4y = 24 4x y = 7 Page 18 19 The graphs intersect at one point, so there is one solution. The lines intersect at (5, 3). 3x + 4y = 24 4x y = 7 The graphs intersect at one point, so there is one solution. The lines intersect at (4, 9). 2x 8y = 6 x 4y = 3 Page 19 20 The graphs intersect at one point, so there is one solution. The lines intersect at (4, 9). 2x 8y = 6 x 4y = 3 The two lines are the same line. Therefore, there are infinitely many solutions. 4x 6y = 12 2x + 3y = 6 Page 20 21 The two lines are the same line. Therefore, there are infinitely many solutions. 4x 6y = 12 2x + 3y = 6 The two lines are the same line. Therefore, there are infinitely many solutions. 2x + 3y = 10 4x + 6y = 12 Page 21 22 The two lines are the same line. Therefore, there are infinitely many solutions. 2x + 3y = 10 4x + 6y = 12 These two lines are parallel, so they do not intersect. Therefore, there is no solution. 3x + 2y = 10 2x + 3y = 10 Page 22 23 These two lines are parallel, so they do not intersect. Therefore, there is no solution. 3x + 2y = 10 2x + 3y = 10 The graphs intersect at one point, so there is one solution. The lines intersect at (2, 2). 3y x= 2 y x=2 Page 23 24 The graphs intersect at one point, so there is one solution. The lines intersect at (2, 2). 3y x= 2 y x=2 These two lines are parallel, so they do not intersect. Therefore, there is no solution. Page 24 25 These two lines are parallel, so they do not intersect. Therefore, there is no solution. The two lines are the same line. Therefore, there are infinitely many solutions. Page 25 26 The two lines are the same line. Therefore, there are infinitely many solutions. These two lines are parallel, so they do not intersect. Therefore, there is no solution. Page 26 27 These two lines are parallel, so they do not intersect. Therefore, there is no solution. The graphs intersect at one point, so there is one solution. The lines intersect at (1, 1). Page 27 28 The graphs intersect at one point, so there is one solution. The lines intersect at (1, 1). The graphs intersect at one point, so there is one solution. The lines intersect at (3, 3). Suppose x represents the number of cameras sold and y represents the number of years since Then the number of digital cameras sold each year since 2000, in millions, can be modeled by the equation y = 12.5x The number of film cameras sold each year since 2000, in millions, can be modeled by the equation y = 9.1x Graph each equation. In which year did digital camera sales surpass film camera sales? In what year will film cameras stop selling altogether? Page 28 29 The graphs intersect at one point, so there is one solution. The lines intersect at (3, 3). Suppose x represents the number of cameras sold and y represents the number of years since Then the number of digital cameras sold each year since 2000, in millions, can be modeled by the equation y = 12.5x The number of film cameras sold each year since 2000, in millions, can be modeled by the equation y = 9.1x Graph each equation. In which year did digital camera sales surpass film camera sales? In what year will film cameras stop selling altogether? What are the domain and range of each of the functions in this situation? a. The digital camera passed the film camera at x = 3, so it was in The film camera will stop selling at x d. these equations is the set of real numbers greater than or equal to 0. Graph each system and determine the number of solutions that it has. If it has one solution, name it. 2y = 1.2x 10 4y = 2.4x Page 29 30 The film camera will stop selling at x d. 6-1 Graphing Systems of Equations these equations is the set of real numbers greater than or equal to 0. Graph each system and determine the number of solutions that it has. If it has one solution, name it. 2y = 1.2x 10 4y = 2.4x These two lines are parallel, so they do not intersect. Therefore, there is no solution. Page 30 31 6-1 Graphing of Equations These two Systems lines are parallel, so they do not intersect. Therefore, there is no solution. The two lines are the same line. Therefore, there are infinitely many solutions. Personal publishing site Lookatme had 2.5 million visitors in Each year after that, the number of visitors rose by 13.1 million. Online auction site Buyourstuff had 59 million visitors in 2009, but each year after that the number of visitors fell by 2 million. Write an equation for each of the companies. Make a table of values for 5 years for each of the companies. Graph each equation. When will Lookatme and Buyourstuff s sites have the same number of visitors? Page 31 32 The two lines are the same line. Therefore, there are infinitely many solutions. Personal publishing site Lookatme had 2.5 million visitors in Each year after that, the number of visitors rose by 13.1 million. Online auction site Buyourstuff had 59 million visitors in 2009, but each year after that the number of visitors fell by 2 million. Write an equation for each of the companies. Make a table of values for 5 years for each of the companies. Graph each equation. When will Lookatme and Buyourstuff s sites have the same number of visitors? Name the domain and range of these functions in this situation. a. Let x represent the number of years past 2009 and y represent the number of visitors in the millions. Lookatme: y = 13.1x + 2.5; Buyourstuff: y = 2x + 59 c. The two sites will have the same number of visitors somewhere between x = 3 and x = 4, meaning that the rd intersection will occur some time in the 3 year past 2009, or during D = {x x y y In this problem, you will explore different methods for finding the intersection of the graphs of two linear equations esolutions Manual - Powered by Cognero.. Page 32 33 The two sites will have the same number of visitors somewhere between x = 3 and x = 4, meaning that the rd intersection will occur some time in the 3 year past 2009, or during D = {x x y y In this problem, you will explore different methods for finding the intersection of the graphs of two linear equations.. Use a graph to solve and y = x How is the equation in part a related to the system in part b? Explain how to use the graph in part b to solve the equation in part a. b. (6, 6) c. Both sides of the equation in part a are set equal to y in the system of linear equations in part b. d. x coordinate of the intersection of the two. Then, you can substitute that value into the equation in part a and see if it checks. Store A is offering a 10% discount on the purchase of all electronics in their store. Store B is offering \$10 off all the electronics in their store. Francisca and Alan are deciding which offer will save them more money. Is either of them correct? Explain your reasoning. Page 33 34 c. Both sides of the equation in part a are set equal to y in the system of linear equations in part b. d. x coordinate of the intersection of the two. Then, you can substitute that 6-1 Graphing Systems of Equations value into the equation in part a and see if it checks. Store A is offering a 10% discount on the purchase of all electronics in their store. Store B is offering \$10 off all the electronics in their store. Francisca and Alan are deciding which offer will save them more money. Is either of them correct? Explain your reasoning. If the item is less than \$100, then \$10 off is better. If the item is more than \$100, then the 10% is better. Consider purchasing an item that cost \$175. The cost they will have to pay at Store A is \$175(1 0.10) or \$ At Store Use graphing to find the solution of the system of equations 2x + 3y = 5, 3x + 4y = 6, and 4x + 5y = 7. One method would be to rearrange the three equations into slope-intercept form. Page 34 35 Notice the fractions in the equations. This will make it a little difficult to graph the lines. Another option would be to plot the x- and y-intercepts of each graph. Substitute 0 in for x and y to identify the intercepts. Equation 1: Equation 2: Page 35 36 Equation 2: Equation 3: Plot each set of points. Draw a line through each set. The intersection of these lines is the solution. All three lines intersect at ( 2, 3) Determine whether a system of two linear equations with (0, 0) and (2, 2) as solutions Page 36 sometimes, always, or never has other solutions. Explain. 37 All three lines intersect at ( 2, 3) Determine whether a system of two linear equations with (0, 0) and (2, 2) as solutions sometimes, always, or never has other solutions. Explain. f the equations are linear and have more than one common solution, they must be consistent and dependent, which means that they have an infinite number of solutions in common. Consider the equations of y = and 2y = 2x. There are WHICH ONE DOESN other three? Explain your reasoning. Which one of the following systems of equations doesn t belong with the Rearrange the equations into slope intercept form. st 1 box nd 2 box rd 3 box th 4 box The third box has two equations that are parallel, which means that it is an inconsistent system. The other three systems are independent and consistent. Write three equations such that they form three systems of equations with y = 5x 3. The three systems should be inconsistent, consistent and independent, and consistent and dependent, respectively. Sample answers: y = 5x + 3; y = 5x 3; 2y = 10x 6 The first equation is parallel to the original equation, so there is no intersection, which makes the system inconsistent. The Manual second- Powered equationbyhas one intersection, which makes the system consistent and independent. The last equation is 37 esolutions Cognero Page the same as the original equation, which makes the system consistent and dependent. 38 th 4 box The third box has two that are parallel, which means that it is an inconsistent system. The other three 6-1 Graphing Systems of equations Equations systems are independent and consistent. Write three equations such that they form three systems of equations with y = 5x 3. The three systems should be inconsistent, consistent and independent, and consistent and dependent, respectively. Sample answers: y = 5x + 3; y = 5x 3; 2y = 10x 6 The first equation is parallel to the original equation, so there is no intersection, which makes the system inconsistent. The second equation has one intersection, which makes the system consistent and independent. The last equation is the same as the original equation, which makes the system consistent and dependent. Describe the advantages and disadvantages to solving systems of equations by graphing. Graphing clearly shows whether a system of equations has one solution, no solution, or infinitely many solutions. However, finding the exact values of x and y from a graph can be difficult. For example, solve the system 2x + y = 5 and 6x + 3y = 12 by graphing. Write each equation in slope-intercept form. y = 2x + 5 and y = 2x 4. Graph each line on the same coordinate plane. It is clear from the graph that the lines are parallel and do not intersect. So, the system has no solutions. For the system 2x + y = 3 and 3x 2y = 4, the equations in slope-intercept form are y = 2x + 3 and y = x and the graphs of these equations yields a pair of intersecting lines. 2 The graph indicates that the system has one solution but it is not possible to determine from the graph that the solution is. Certain bacteria can reproduce every 20 minutes, doubling the population. If there are 450,000 bacteria in a population at 9:00 A.M., how many bacteria will be in the population at 2:00 P.M.? Page 38 39 Describe the advantages and disadvantages to solving systems of equations by graphing. Graphing clearly shows whether a system of equations has one solution, no solution, or infinitely many solutions. However, finding the exact values of x and y from a graph can be difficult. For example, solve the system 2x + y = 5 and 6x + 3y = 12 by graphing. Write each equation in slope-intercept form. y = 2x + 5 and y = 2x 4. Graph each line on the same coordinate plane. It is clear from the graph that the lines are parallel and do not intersect. So, the system has no solutions. For the system 2x + y = 3 and 3x 2y = 4, the equations in slope-intercept form are y = 2x + 3 and y = x and the graphs of these equations yields a pair of intersecting lines. 2 The graph indicates that the system has one solution but it is not possible to determine from the graph that the solution is. Certain bacteria can reproduce every 20 minutes, doubling the population. If there are 450,000 bacteria in a population at 9:00 A.M., how many bacteria will be in the population at 2:00 P.M.? Between 9 A.M. and 2 P.M., there are 5 hours, or 15 twenty-minute sections. That means that the bacteria will double 15 times. 450, , ,800, ,600, ,200, ,400, ,800, ,600, ,200, ,400, ,800, ,600, ,843,200,000 Page 39 40 The graph indicates that the system has one solution but it is not possible to determine from the graph that the solution is. Certain bacteria can reproduce every 20 minutes, doubling the population. If there are 450,000 bacteria in a population at 9:00 A.M., how many bacteria will be in the population at 2:00 P.M.? Between 9 A.M. and 2 P.M., there are 5 hours, or 15 twenty-minute sections. That means that the bacteria will double 15 times. 450, ,000 1,800,000 3,600,000 7,200,000 14,400,000 28,800,000 57,600, ,200, ,400, ,800, ,600,000 1,843,200,000 3,686,400,000 7,372,800,000 14,745,600, Therefore, 14,745,600,000 bacteria will be present at 2:00 P.M. An 84 centimeter piece of wire is cut into equal segments and then attached at the ends to form the edges of a cube. What is the volume of the cube? 294 cm 343 cm cm cm There are twelve edges in a cube. So, if you cut an 84 centimeter piece of wire into 12 equal parts, they are each 7 cm long. To calculate the volume of the cube, multiply length times width times height. So, the correct choice is B. What is the solution of the inequality 9 < 2x + 3 < 15? Page 40 41 13 3,686,400, ,372,800, ,745,600,000 Therefore, 14,745,600,000 bacteria will be present at 2:00 P.M. An 84 centimeter piece of wire is cut into equal segments and then attached at the ends to form the edges of a cube. What is the volume of the cube? 294 cm 343 cm cm cm There are twelve edges in a cube. So, if you cut an 84 centimeter piece of wire into 12 equal parts, they are each 7 cm long. To calculate the volume of the cube, multiply length times width times height. So, the correct choice is B. What is the solution of the inequality 9 < 2x + 3 < 15? x x 0 0 6<x<6 5<x<5 First, express -9 < 2x + 3 < 15 using and. Then solve each inequality. The solution is -6 < x < 6. So, the correct choice is H. What is the solution of the system of equations? x + 2y = 1 2x + 4y = 2 ( 1, 1) (2, 1) Page 41 42 The solution is -6 < x < 6. So, the correct choice is H. What is the solution of the system of equations? x + 2y = 1 2x + 4y = 2 ( 1, 1) (2, 1) no solution infinitely many solutions Solve by graphing. Rewrite each equation in slope-intercept form. Because both equations are the same in slope-intercept form, they are on the same line. Thus it is a true statement, all numbers will work. Therefore, the correct choice is D. Graph each inequality. 3x + 6y >0 Solve for y in terms of x. Page 42 43 Because both equations are the same in slope-intercept form, they are on the same line. Thus it is a true statement, 6-1 Graphing Systems of Equations all numbers will work. Therefore, the correct choice is D. Graph each inequality. 3x + 6y >0 Solve for y in terms of x. Because the inequality involves >, graph the boundary using a dashed line. Choose (1, 1) as a test point. Since 9 is greater than 0, shade the half plane that contains (1, 1). 4x 2y < 0 Solve for y in terms of x. Because the inequality involves >, graph the boundary using a dashed line. Choose (1, 1) as a test point. Page 43 44 4x 2y < 0 Solve for y in terms of x. Because the inequality involves >, graph the boundary using a dashed line. Choose (1, 1) as a test point. Since 2 is not less than 0, shade the half plane that does not contain (1, 1). 3y x 9 Solve for y in terms of x. Since 0 is less than or equal to 9, shade the half plane that contains (0, 0). Page 44 45 3y x 9 Solve for y in terms of x. Since 0 is less than or equal to 9, shade the half plane that contains (0, 0). 4y 3x 12 Solve for y in terms of x. Since 0 is not greater than or equal to 12, shade the half plane that does not contain (0, 0). Page 45 46 4y 3x 12 Solve for y in terms of x. Since 0 is not greater than or equal to 12, shade the half plane that does not contain (0, 0). y < 4x 8 Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. Since 0 is not less than 8, shade the half plane that does not contain (0, 0). Page 46 47 y < 4x 8 Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. Since 0 is not less than 8, shade the half plane that does not contain (0, 0). 3x 1 >y Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. Since 1 is not greater than 0, shade the half plane that does not contain (0, 0). To get a grant from the city s historical society, the number of history books must be within 25 of What is the range of the number of history books that must be in the library? The least number of books is 1500 the range is 1475 to 1525 books. 25 = 1475, while the greatest number of books is = Therefore, Camilla s scores on three math tests are shown in the table. The fourth and final test of the grading an average of at least 92 to receive an A for the grading period. Page 47 period is tomorrow. needs esolutions Manual - Powered byshe Cognero 48 What is the range of the number of history books that must be in the library? 6-1 Graphing Systems Equations The least number of of books is 1500 the range is 1475 to 1525 books. 25 = 1475, while the greatest number of books is = Therefore, Camilla s scores on three math tests are shown in the table. The fourth and final test of the grading period is tomorrow. She needs an average of at least 92 to receive an A for the grading period. If m represents her score on the fourth math test, write an inequality to represent this situation. If Camilla wants an A in math, what must she score on the test? Is your solution reasonable? Explain. a. Therefore, Camilla needs a 94 or higher. c. Yes, the score is attainable and Camilla has scored higher than that before. Write the slope intercept form of an equation for the line that passes through the given point and is perpendicular to the graph of the equation. ( 3, 1), y = x+2 The slope of the line with equation. The slope of the perpendicular line is the opposite reciprocal of, or 3. Page 48 49 Therefore, Camilla needs a 94 or higher. c. Yes, the score is attainable and Camilla has scored higher than that before. Write the slope intercept form of an equation for the line that passes through the given point and is perpendicular to the graph of the equation. ( 3, 1), y = x+2 The slope of the line with equation. The slope of the perpendicular line is the opposite reciprocal of, or 3. Write the equation in slope-intercept form. (6, 2) y = x 4 The slope of the line with equation, or. The slope of the perpendicular line is the opposite reciprocal of. Write the equation in slope-intercept form. (2, 2), 2x + y = 5 Rearrange the equation into slope intercept form. The slope of the line with equation Page The slope of the perpendicular line is the opposite reciprocal of 50 (2, 2), 2x + y = 5 Rearrange the equation into slope intercept form. The slope of the line with equation 2. The slope of the perpendicular line is the opposite reciprocal of 2, or. Write the equation in slope-intercept form. ( 3, 3), 3x + y = 6 Rearrange the equation into slope intercept form. The slope of the line with equation 3. The slope of the perpendicular line is the opposite reciprocal of 3, or. Write the equation in slope-intercept form. Page 50 51 Find the solution of each equation using the given replacement set. f 14 = 8; {12, 15, 19, 22} f f 14 = 8 True or False? = 8 False = 8 False = 8 False = 8 True The solution is {22}. 15(n + 6) = 165; {3, 4, 5, 6, 7} n 15(n + 6) = 165 True or False? 3 15(3 + 6) = 165 False 4 15(4 + 6) = 165 False 5 15(5 + 6) = 165 True 6 15(6 + 6) = 165 False 7 15(7 + 6) = 165 False The solution is {5}. ; {91, 92, 93, 94, 95} d True or False? 91 False 92 True 93 False 94 False 95 False The solution is {92}. Page 51 52 ; {78, 79, 80, 81} t The solution is {81}. True or False? 78 False 79 False 80 False 81 True Evaluate each expression if a = 2, b = 3, and c = 11. a + 6b 7 ab (2c + 3a) 4 Page 52 53 b 2 + (a 3 8)5 Page 53 ### 0-5 Systems of Linear Equations and Inequalities 21. Solve each system of equations. Eliminate one variable in two pairs of the system. Add the first equation and second equations to eliminate x. Multiply the first equation by 3 and add it to the second ### Study Guide and Review - Chapter 4 State whether each sentence is true or false. If false, replace the underlined term to make a true sentence. 1. The y-intercept is the y-coordinate of the point where the graph crosses the y-axis. The ### In this section, we ll review plotting points, slope of a line and different forms of an equation of a line. Math 1313 Section 1.2: Straight Lines In this section, we ll review plotting points, slope of a line and different forms of an equation of a line. Graphing Points and Regions Here s the coordinate plane: ### Writing the Equation of a Line in Slope-Intercept Form Writing the Equation of a Line in Slope-Intercept Form Slope-Intercept Form y = mx + b Example 1: Give the equation of the line in slope-intercept form a. With y-intercept (0, 2) and slope -9 b. Passing ### Linear Equations Review Linear Equations Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The y-intercept of the line y = 4x 7 is a. 7 c. 4 b. 4 d. 7 2. What is the y-intercept ### Solving Equations Involving Parallel and Perpendicular Lines Examples Solving Equations Involving Parallel and Perpendicular Lines Examples. The graphs of y = x, y = x, and y = x + are lines that have the same slope. They are parallel lines. Definition of Parallel Lines ### EQUATIONS and INEQUALITIES EQUATIONS and INEQUALITIES Linear Equations and Slope 1. Slope a. Calculate the slope of a line given two points b. Calculate the slope of a line parallel to a given line. c. Calculate the slope of a line ### The slope m of the line passes through the points (x 1,y 1 ) and (x 2,y 2 ) e) (1, 3) and (4, 6) = 1 2. f) (3, 6) and (1, 6) m= 6 6 Lines and Linear Equations Slopes Consider walking on a line from left to right. The slope of a line is a measure of its steepness. A positive slope rises and a negative slope falls. A slope of zero means ### Lines and Linear Equations. Slopes Lines and Linear Equations Slopes Consider walking on a line from left to right. The slope of a line is a measure of its steepness. A positive slope rises and a negative slope falls. A slope of zero means ### Use the graph to determine whether each system is consistent or inconsistent and if it is independent or dependent. Use the graph to determine whether each system is consistent or inconsistent and if it is independent or dependent. y = 2x 1 y = 2x + 3 The lines y = 2x 1 and y = 2x + 3 intersect at exactly one point ### Section 7.1 Solving Linear Systems by Graphing. System of Linear Equations: Two or more equations in the same variables, also called a. Algebra 1 Chapter 7 Notes Name Section 7.1 Solving Linear Systems by Graphing System of Linear Equations: Two or more equations in the same variables, also called a. Solution of a System of Linear Equations: ### Ordered Pairs. Graphing Lines and Linear Inequalities, Solving System of Linear Equations. Cartesian Coordinates System. Ordered Pairs Graphing Lines and Linear Inequalities, Solving System of Linear Equations Peter Lo All equations in two variables, such as y = mx + c, is satisfied only if we find a value of x and a value ### Section 1.1 Linear Equations: Slope and Equations of Lines Section. Linear Equations: Slope and Equations of Lines Slope The measure of the steepness of a line is called the slope of the line. It is the amount of change in y, the rise, divided by the amount of ### Practice Test - Chapter 4. y = 2x 3. The slope-intercept form of a line is y = mx + b, where m is the slope, and b is the y-intercept. y = 2x 3. The slope-intercept form of a line is y = mx + b, where m is the slope, and b is the y-intercept. Plot the y-intercept (0, 3). The slope is. From (0, 3), move up 2 units and right 1 unit. Plot ### Section 3.4 The Slope Intercept Form: y = mx + b Slope-Intercept Form: y = mx + b, where m is the slope and b is the y-intercept Reminding! m = y x = y 2 y 1 x 2 x 1 Slope of a horizontal line is 0 Slope of a vertical line is Undefined Graph a linear ### Number of Solutions to Simultaneous Equations Worksheet 3.5 Simultaneous Equations Section 1 Number of Solutions to Simultaneous Equations In maths we are sometimes confronted with two equations in two variables and we want to find out which values ### Section 1.4 Graphs of Linear Inequalities Section 1.4 Graphs of Linear Inequalities A Linear Inequality and its Graph A linear inequality has the same form as a linear equation, except that the equal symbol is replaced with any one of,, ### 2.1 Algebraic Expressions and Combining like Terms 2.1 Algebraic Expressions and Combining like Terms Evaluate the following algebraic expressions for the given values of the variables. 3 3 3 Simplify the following algebraic expressions by combining like ### 2-4 Writing Linear Equations. Write an equation in slope-intercept form for the line described. 2. passes through ( 2, 3) and (0, 1) SOLUTION: Write an equation in slope-intercept form for the line described 2 passes through ( 2, 3) and (0, 1) Substitute m = 1 and in the point slope form 4 passes through ( 8, 2); Substitute m = and (x, y) = ( ### Chapter 2 Section 4: Equations of Lines. 4.* Find the equation of the line with slope 4 3, and passing through the point (0,2). Chapter Section : Equations of Lines Answers to Problems For problems -, put our answers into slope intercept form..* Find the equation of the line with slope, and passing through the point (,0).. Find ### Lesson 22: Solution Sets to Simultaneous Equations Student Outcomes Students identify solutions to simultaneous equations or inequalities; they solve systems of linear equations and inequalities either algebraically or graphically. Classwork Opening Exercise ### 4-5 Scatter Plots and Lines of Fit Determine whether each graph shows a positive, negative, or no correlation. If there is a positive or negative correlation, describe its meaning in the situation. The graph shows a positive correlation. ### Make sure you look at the reminders or examples before each set of problems to jog your memory! Solve Name Date Make sure you look at the reminders or examples before each set of problems to jog your memory! I. Solving Linear Equations 1. Eliminate parentheses. Combine like terms 3. Eliminate terms by ### What does the number m in y = mx + b measure? To find out, suppose (x 1, y 1 ) and (x 2, y 2 ) are two points on the graph of y = mx + b. PRIMARY CONTENT MODULE Algebra - Linear Equations & Inequalities T-37/H-37 What does the number m in y = mx + b measure? To find out, suppose (x 1, y 1 ) and (x 2, y 2 ) are two points on the graph of ### Chapter 6 Notes. Section 6.1 Solving One-Step Linear Inequalities Chapter 6 Notes Name Section 6.1 Solving One-Step Linear Inequalities Graph of a linear Inequality- the set of all points on a number line that represent all solutions of the inequality > or < or circle ### Slope-Intercept Form of a Linear Equation Examples Slope-Intercept Form of a Linear Equation Examples. In the figure at the right, AB passes through points A(0, b) and B(x, y). Notice that b is the y-intercept of AB. Suppose you want to find an equation ### 3-1 Solving Systems of Equations. Solve each system of equations by using a table. 2. SOLUTION: Write each equation in slope-intercept form. Solve each system of equations by using a table. 2. Write each equation in slope-intercept form. Also: Make a table to find a solution that satisfies both equations. Therefore, the solution of the system ### Study Guide and Review - Chapter 5 State whether each sentence is true or false. If false, replace the underlined term to make a true sentence. Set-builder notation is a less concise way of writing a solution set. The statement is false. ### 5 \$75 6 \$90 7 \$105. Name Hour. Review Slope & Equations of Lines. STANDARD FORM: Ax + By = C. 1. What is the slope of a vertical line? Review Slope & Equations of Lines Name Hour STANDARD FORM: Ax + By = C 1. What is the slope of a vertical line? 2. What is the slope of a horizontal line? 3. Is y = 4 the equation of a horizontal or vertical ### Chapter 9. Systems of Linear Equations Chapter 9. Systems of Linear Equations 9.1. Solve Systems of Linear Equations by Graphing KYOTE Standards: CR 21; CA 13 In this section we discuss how to solve systems of two linear equations in two variables ### The Point-Slope Form 7. The Point-Slope Form 7. OBJECTIVES 1. Given a point and a slope, find the graph of a line. Given a point and the slope, find the equation of a line. Given two points, find the equation of a line y Slope ### Then Now Why? connected.mcgraw-hill.com. Your Digital Math Portal Systems of Linear Equations and Inequalities Then Now Why? You solved linear equations in one variable. In this chapter, you will: Solve systems of linear equations by graphing, substitution, and elimination. ### ( ) # 0. SOLVING INEQUALITIES and Example 1. Example 2 SOLVING INEQUALITIES 9.1.1 and 9.1.2 To solve an inequality in one variable, first change it to an equation and solve. Place the solution, called a boundary point, on a number line. This point separates ### Warm Up. Write an equation given the slope and y-intercept. Write an equation of the line shown. Warm Up Write an equation given the slope and y-intercept Write an equation of the line shown. EXAMPLE 1 Write an equation given the slope and y-intercept From the graph, you can see that the slope is ### Slope-Intercept Equation. Example 1.4 Equations of Lines and Modeling Find the slope and the y intercept of a line given the equation y = mx + b, or f(x) = mx + b. Graph a linear equation using the slope and the y-intercept. Determine ### 6-3 Elimination Using Addition and Subtraction. Use elimination to solve each system of equations. Multiply the second equation by 1. Use elimination to solve each system of equations. 5m p = 7 7m p = 11 Multiply the second equation by 1. Then, add this to the first equation. Now, substitute 2 for m in either equation to find the value ### Math 018 Review Sheet v.3 Math 018 Review Sheet v.3 Tyrone Crisp Spring 007 1.1 - Slopes and Equations of Lines Slopes: Find slopes of lines using the slope formula m y y 1 x x 1. Positive slope the line slopes up to the right. ### 4.1 & Linear Equations in Slope-Intercept Form 4.1 & 4.2 - Linear Equations in Slope-Intercept Form Slope-Intercept Form: y = mx + b Ex 1: Write the equation of a line with a slope of -2 and a y-intercept of 5. Ex 2:Write an equation of the line shown ### Systems of Linear Equations - Introduction Systems of Linear Equations - Introduction What are Systems of Linear Equations Use an Example of a system of linear equations If we have two linear equations, y = x + 2 and y = 3x 6, can these two equations ### Linear Equations and Graphs 2.1-2.4 Linear Equations and Graphs Coordinate Plane Quadrants - The x-axis and y-axis form 4 "areas" known as quadrants. 1. I - The first quadrant has positive x and positive y points. 2. II - The second ### Section P.9 Notes Page 1 P.9 Linear Inequalities and Absolute Value Inequalities Section P.9 Notes Page P.9 Linear Inequalities and Absolute Value Inequalities Sometimes the answer to certain math problems is not just a single answer. Sometimes a range of answers might be the answer. ### 7-1 Multiplication Properties of Exponents Determine whether each expression is a monomial. Write yes or no. Explain your reasoning. 1. 15 15 is a monomial. It is a constant and all constants are monomials.. 3a 3a is not a monomial. There is subtraction ### Linear Equations. Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)} Linear Equations Domain and Range Domain refers to the set of possible values of the x-component of a point in the form (x,y). Range refers to the set of possible values of the y-component of a point in ### Final Graphing Practice #1 Final Graphing Practice #1 Beginning Algebra / Math 100 Fall 2013 506 (Prof. Miller) Student Name/ID: Instructor Note: Assignment: Set up a tutoring appointment with one of the campus tutors or with me. ### Systems of Linear Equations in Two Variables Chapter 6 Systems of Linear Equations in Two Variables Up to this point, when solving equations, we have always solved one equation for an answer. However, in the previous chapter, we saw that the equation ### 2. THE x-y PLANE 7 C7 2. THE x-y PLANE 2.1. The Real Line When we plot quantities on a graph we can plot not only integer values like 1, 2 and 3 but also fractions, like 3½ or 4¾. In fact we can, in principle, plot any real ### Section 2.2 Equations of Lines Section 2.2 Equations of Lines The Slope of a Line EXAMPLE: Find the slope of the line that passes through the points P(2,1) and Q(8,5). = 5 1 8 2 = 4 6 = 2 1 EXAMPLE: Find the slope of the line that passes ### HIBBING COMMUNITY COLLEGE COURSE OUTLINE HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam, ### Section 1.4 Notes Page Linear Equations in Two Variables and Linear Functions., x Section. Notes Page. Linear Equations in Two Variables and Linear Functions Slope Formula The slope formula is used to find the slope between two points ( x, y ) and ( ) x, y. x, y ) The slope is the vertical ### Exam 2 Review. 3. How to tell if an equation is linear? An equation is linear if it can be written, through simplification, in the form. Exam 2 Review Chapter 1 Section1 Do You Know: 1. What does it mean to solve an equation? To solve an equation is to find the solution set, that is, to find the set of all elements in the domain of the ### Algebra I Pacing Guide Days Units Notes 9 Chapter 1 ( , ) Algebra I Pacing Guide Days Units Notes 9 Chapter 1 (1.1-1.4, 1.6-1.7) Expressions, Equations and Functions Differentiate between and write expressions, equations and inequalities as well as applying order ### Pre-AP Algebra 2 Lesson 2-5 Graphing linear inequalities & systems of inequalities Lesson 2-5 Graphing linear inequalities & systems of inequalities Objectives: The students will be able to - graph linear functions in slope-intercept and standard form, as well as vertical and horizontal ### 2.3 Writing Equations of Lines . Writing Equations of Lines In this section ou will learn to use point-slope form to write an equation of a line use slope-intercept form to write an equation of a line graph linear equations using the ### 5. Equations of Lines: slope intercept & point slope 5. Equations of Lines: slope intercept & point slope Slope of the line m rise run Slope-Intercept Form m + b m is slope; b is -intercept Point-Slope Form m( + or m( Slope of parallel lines m m (slopes ### Graphing Linear Equations Graphing Linear Equations I. Graphing Linear Equations a. The graphs of first degree (linear) equations will always be straight lines. b. Graphs of lines can have Positive Slope Negative Slope Zero slope ### Study Resources For Algebra I. Unit 1D Systems of Equations and Inequalities Study Resources For Algebra I Unit 1D Systems of Equations and Inequalities This unit explores systems of linear functions and the various methods used to determine the solution for the system. Information ### 5.1: Rate of Change and Slope 5.1: Rate of Change and Slope Rate of Change shows relationship between changing quantities. On a graph, when we compare rise and run, we are talking about steepness of a line (slope). You can use and ### A synonym is a word that has the same or almost the same definition of Slope-Intercept Form Determining the Rate of Change and y-intercept Learning Goals In this lesson, you will: Graph lines using the slope and y-intercept. Calculate the y-intercept of a line when given ### GRAPHING LINEAR EQUATIONS IN TWO VARIABLES GRAPHING LINEAR EQUATIONS IN TWO VARIABLES The graphs of linear equations in two variables are straight lines. Linear equations may be written in several forms: Slope-Intercept Form: y = mx+ b In an equation ### Chapter 3: Systems of Linear Equalities and Inequalities Chapter 3: Systems of Linear Equalities and Inequalities Chapter 3: Systems of Linear Equations and Inequalities Assignment Sheet Date Topic Assignment Completed 3.1 Solving Systems of Linear Equations ### Algebra I. In this technological age, mathematics is more important than ever. When students In this technological age, mathematics is more important than ever. When students leave school, they are more and more likely to use mathematics in their work and everyday lives operating computer equipment, ### Algebra Chapter 6 Notes Systems of Equations and Inequalities. Lesson 6.1 Solve Linear Systems by Graphing System of linear equations: Algebra Chapter 6 Notes Systems of Equations and Inequalities Lesson 6.1 Solve Linear Systems by Graphing System of linear equations: Solution of a system of linear equations: Consistent independent system: ### Math 10 - Unit 7 Final Review - Coordinate Geometry Class: Date: Math 10 - Unit Final Review - Coordinate Geometry Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Determine the slope of this line segment. ### Name: Class: Date: Does the equation represent a direct variation? If so, find the constant of variation. c. yes; k = 5 3. c. Name: Class: Date: Chapter 5 Test Multiple Choice Identify the choice that best completes the statement or answers the question. What is the slope of the line that passes through the pair of points? 1. ### Algebra Course KUD. Green Highlight - Incorporate notation in class, with understanding that not tested on Algebra Course KUD Yellow Highlight Need to address in Seminar Green Highlight - Incorporate notation in class, with understanding that not tested on Blue Highlight Be sure to teach in class Postive and ### 10.1 Systems of Linear Equations: Substitution and Elimination 726 CHAPTER 10 Systems of Equations and Inequalities 10.1 Systems of Linear Equations: Sustitution and Elimination PREPARING FOR THIS SECTION Before getting started, review the following: Linear Equations ### 3.4. Solving Simultaneous Linear Equations. Introduction. Prerequisites. Learning Outcomes Solving Simultaneous Linear Equations 3.4 Introduction Equations often arise in which there is more than one unknown quantity. When this is the case there will usually be more than one equation involved. ### MATH 60 NOTEBOOK CERTIFICATIONS MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5 ### 5.2. Systems of linear equations and their solution sets 5.2. Systems of linear equations and their solution sets Solution sets of systems of equations as intersections of sets Any collection of two or more equations is called a system of equations. The solution ### REVIEW SHEETS INTERMEDIATE ALGEBRA MATH 95 REVIEW SHEETS INTERMEDIATE ALGEBRA MATH 95 A Summary of Concepts Needed to be Successful in Mathematics The following sheets list the key concepts which are taught in the specified math course. The sheets ### Algebra I: Strand 2. Linear Functions; Topic 2. Formalizing Slope and y-intercept; Task 1 TASK : DESCRIBE STACKING CUPS Solutions Make a scatter plot of the data. Write a symbolic function rule and a verbal description of the situation. 1. 1 1 1 Number of Cups 0.5 1 6.0 2 8.5 11.0 2.5 2.5 ### Sect The Slope-Intercept Form Concepts # and # Sect. - The Slope-Intercept Form Slope-Intercept Form of a line Recall the following definition from the beginning of the chapter: Let a, b, and c be real numbers where a and b are not ### Practice Test - Chapter 3. TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. a. State the independent and dependent variables.explain. b. Determine the C- and K-intercepts and describe what ### Algebraic expressions are a combination of numbers and variables. Here are examples of some basic algebraic expressions. Page 1 of 13 Review of Linear Expressions and Equations Skills involving linear equations can be divided into the following groups: Simplifying algebraic expressions. Linear expressions. Solving linear ### Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers. Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used ### Math 155 (DoVan) Exam 1 Review (Sections 3.1, 3.2, 5.1, 5.2, Chapters 2 & 4) Chapter 2: Functions and Linear Functions 1. Know the definition of a relation. Math 155 (DoVan) Exam 1 Review (Sections 3.1, 3.2, 5.1, 5.2, Chapters 2 & 4) 2. Know the definition of a function. 3. What ### Unit III Practice Questions Unit III Practice Questions 1. Write the ordered pair corresponding to each point plotted below. A F B E D C 2. Determine if the ordered pair ( 1, 2) is a solution of 2x + y = 4. Explain how you know. ### 2-4 Writing Linear Equations. Write an equation in slope-intercept form for the line described. Write an equation in slope-intercept form for the line described 1 slope 15, passes through (0, 5) m = 15 (x, y) = (0, 5) in the equation y = mx + b 3 passes through (3, 5); m = 2 m = 2 (x, y) = (3, 5) ### A coordinate system is formed by the intersection of two number lines, the horizontal axis and the vertical axis. Consider the number line. State whether each sentence is true or false. If false, replace the underlined term to make a true sentence. An exponent indicates the number the base is to be multiplied by. The exponent indicates the ### 7.5 Systems of Linear Inequalities 7.5 Systems of Linear Inequalities In chapter 6 we discussed linear equations and then saw linear inequalities, so since here, in chapter 7, we are talking about systems of linear equations, it makes sense ### 5.1. Systems of Linear Equations. Linear Systems Substitution Method Elimination Method Special Systems 5.1 Systems of Linear Equations Linear Systems Substitution Method Elimination Method Special Systems 5.1-1 Linear Systems The possible graphs of a linear system in two unknowns are as follows. 1. The ### Vocabulary Words and Definitions for Algebra Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms ### Alg2 S4 Summative Review Determine whether the relation is a function. (0, 4), (1, 4), (2, 5), (3, 6), (4, 6) Alg2 S4 Summative Review 15-16 1 Determine whether the relation is a function (0, 4), (1, 4), (2, 5), (3, 6), (4, 6) 2 The boiling temperature of water T (in Fahrenheit) can be modeled by the function ### Solving Systems of Two Equations Algebraically 8 MODULE 3. EQUATIONS 3b Solving Systems of Two Equations Algebraically Solving Systems by Substitution In this section we introduce an algebraic technique for solving systems of two equations in two unknowns ### This is a function because no vertical line can be drawn so that it intersects the graph more than once. Determine whether each relation is a function. Explain. A function is a relation in which each element of the domain is paired with exactly one element of the range. So, this relation is a function. A ### Brunswick High School has reinstated a summer math curriculum for students Algebra 1, Geometry, and Algebra 2 for the 2014-2015 school year. Brunswick High School has reinstated a summer math curriculum for students Algebra 1, Geometry, and Algebra 2 for the 2014-2015 school year. Goal The goal of the summer math program is to help students ### 2x - y 4 y -3x - 6 y < 2x 5x - 3y > 7 DETAILED SOLUTIONS AND CONCEPTS GRAPHICAL REPRESENTATION OF LINEAR INEQUALITIES IN TWO VARIABLES Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. ### 7-2 Division Properties of Exponents. Simplify each expression. Assume that no denominator equals zero. SOLUTION: SOLUTION: SOLUTION: SOLUTION: Simplify each expression. Assume that no denominator equals zero. 1. 2. 3. 4. Page 1 4. 5. 6. 7. Page 2 7. 8. 9. 10. Page 3 10. 11. 12. A value to the zero power is 1. 13. A value to the zero power is ### Chapter 4 SOLVING SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES Chapter 4 SOLVING SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES 4.1 Introduction to Systems of Linear Equations: Solving by Graphing Objectives A Decide whether an ordered pair is a solution of a system ### Lesson 8.3 Exercises, pages Lesson 8. Eercises, pages 57 5 A. For each function, write the equation of the corresponding reciprocal function. a) = 5 - b) = 5 c) = - d) =. Sketch broken lines to represent the vertical and horizontal ### CAHSEE/CCSS: 7 AF 1.1/ A REI.17 Warm-Up y CAHSEE/CCSS: 7 AF 1.1/ A REI.17 Review/CCSS: ALG 6.0/ 8.EE.5 Which system of equations represents the statements below? The sum of two numbers is ten. One number is five times the other. Graph ### Section 1.10 Lines. The Slope of a Line Section 1.10 Lines The Slope of a Line EXAMPLE: Find the slope of the line that passes through the points P(2,1) and Q(8,5). = 5 1 8 2 = 4 6 = 2 1 EXAMPLE: Find the slope of the line that passes through ### Section 8.1 System of Linear Equations: Substitution and Elimination Identifying Linear Systems Linear Systems in two variables Solution, 0 1 1 0 Consistent/ Independent Consistent/ Dependent Inconsistent Exactly one solution Infinitely many solutions No solution Two lines ### Section 2.1 Rectangular Coordinate Systems P a g e 1 Section 2.1 Rectangular Coordinate Systems 1. Pythagorean Theorem In a right triangle, the lengths of the sides are related by the equation where a and b are the lengths of the legs and c is ### Chapter R - Basic Algebra Operations (69 topics, due on 05/01/12) Course Name: College Algebra 001 Course Code: R3RK6-CTKHJ ALEKS Course: College Algebra with Trigonometry Instructor: Prof. Bozyk Course Dates: Begin: 01/17/2012 End: 05/04/2012 Course Content: 288 topics ### Grade 10 Applied Math Unit 3: Linear Equations Lesson: Date: Completed Level Lesson 1: Coordinates Review Lesson 2: Constructing Tables and Sketching Grade 10 Applied Math Unit 3: Linear Equations Lesson: Date: Completed Level Lesson 1: Coordinates Review Lesson 2: Constructing Tables and Sketching Graphs Lesson 3: Interpreting Lines Lesson 4: Write ### x x y y Then, my slope is =. Notice, if we use the slope formula, we ll get the same thing: m = Slope and Lines The slope of a line is a ratio that measures the incline of the line. As a result, the smaller the incline, the closer the slope is to zero and the steeper the incline, the farther the 1.3 LINEAR EQUATIONS IN TWO VARIABLES Copyright Cengage Learning. All rights reserved. What You Should Learn Use slope to graph linear equations in two variables. Find the slope of a line given two points
Mathematics Maxima and Minima For CBSE-NCERT Click for Only Video ### Topic covered star Maxima and Minima star Local maximum value and local minimum value ### Maxima and Minima \color{green} ✍️ Let f be a function defined on an interval I. (a) Then f is said to have a maximum value in I, if there exists a point c in I such that color{green}{f (c) ≥ f (x)} , for all x ∈ I. => The number f(c) is called the maximum value of f in I and the point c is called a point of maximum value of f in I. (b) f is said to have a minimum value in I, if there exists a point c in I such that color{green}{f (c) ≤ f (x)}, for all x ∈ I. => The number f(c), in this case, is called the minimum value of f in I and the point c, in this case, is called a point of minimum value of f in I. \color{green} ✍️ (c) f is said to have an extreme value in I if there exists a point c in I such that f (c) is either a maximum value or a minimum value of f in I. \color{green} ✍️ The number f (c), in this case, is called an extreme value of f in I and the point c is called an extreme point. Q 3135145962 Find the maximum and the minimum values, if any, of the function f given by f (x) = x^2, x ∈ R . Class 12 Chapter 6 Example 26 Solution: From the graph of the given function (Fig 6.10), we have f (x) = 0 if x = 0. Also f (x) ≥ 0, for all x ∈ R. Therefore, the minimum value of f is 0 and the point of minimum value of f is x = 0. Further, it may be observed from the graph of the function that f has no maximum value and hence no point of maximum value of f in R. Note : If we restrict the domain of f to [– 2, 1] only, then f will have maximum value (– 2)^2 = 4 at x = – 2. Q 3165145965 Find the maximum and minimum values of f , if any, of the function given by f (x) = \| x \|, x ∈ R. Class 12 Chapter 6 Example 27 Solution: From the graph of the given function (Fig 6.11) , note that f (x) ≥ 0, for all x ∈ R and f (x) = 0 if x = 0. Therefore, the function f has a minimum value 0 and the point of minimum value of f is x = 0. Also, the graph clearly shows that f has no maximum value in R and hence no point of maximum value in R. Note : (i) If we restrict the domain of f to [– 2, 1] only, then f will have maximum value \|– 2\| = 2. (ii) One may note that the function f in Example is not differentiable at x = 0 Q 3115145969 Find the maximum and the minimum values, if any, of the function given by f (x) = x, x ∈ (0, 1). Class 12 Chapter 6 Example 28 Solution: The given function is an increasing (strictly) function in the given interval (0, 1). From the graph (Fig 6.12) of the function f , it seems that, it should have the minimum value at a point closest to 0 on its right and the maximum value at a point closest to 1 on its left. Are such points available? Of course, not. It is not possible to locate such points. In fact, if a point x_0 is closest to 0, then we find (x_0)/2 < x_0 for all x_0 ∈ (0,1) . Also, if x_1 is closest to 1, then (x_1 +1)/2 > x_1 for all x_1 ∈(0,1) . Therefore, the given function has neither the maximum value nor the minimum value in the interval (0,1). "Remark:" The reader may observe that in Example 28, if we include the points 0 and 1 in the domain of f , i.e., if we extend the domain of f to [0,1], then the function f has minimum value 0 at x = 0 and maximum value 1 at x = 1. Infact, we have the following results (The proof of these results are beyond the scope of the present text) ### Monotonic Functions => By a monotonic function f in an interval I, we mean that f is either increasing in I or decreasing in I. =>color{green}{"Every monotonic function assumes its maximum/minimum value at the"} color{green}{" end points of the domain of definition of the function."} => Every continuous function on a closed interval has a maximum and a minimum value. ### Local/Global maximum value and Local/Global minimum value => Observe that at points A, B, C and D on the graph, the function changes its nature from decreasing to increasing or vice-versa. These points may be called turning points of the given function. => The function has maximum value in some neighbourhood of points B and D which are at the top of their respective hills. For this reason, the points A and C may be regarded as points of local minimum value (or relative minimum value) and points B and D may be regarded as points of local maximum value (or relative maximum value) for the function \color{green} ✍️ Local maxima and local minima are the largest and smallest value of the function, either within a given range (the local or relative extrema) or in an entire domain of a function . \color{green} ✍️ Global maxima and Global minima are the largest and smallest value of the function, either within a given range or entire domain of a function See Fig => ### Concept Of Local minima and local maxima \color{green} ✍️ Let f be a real valued function and let c be an interior point in the domain of f. (a) c is called a point of local maxima if there is an h > 0 such that color{red}{f (c) ≥ f (x)}, for all x in (c – h, c + h) => The value f (c) is called the local maximum value of f. (b) =>c is called a point of local minima if there is an h > 0 such that color{red}{f (c) ≤ f (x),} for all x in (c – h, c + h) => The value f (c) is called the local minimum value of f . color{blue}{"Note :"} The function f is increasing (i.e., (f ′(x) > 0) in the interval (c – h, c) and decreasing (i.e., (f ′(x) < 0) in the interval (c, c + h). =>Similarly, if c is a point of local minima of f , then the graph of f around c will be as shown in fig. Here f is decreasing (i.e., f ′(x) < 0) in the interval (c – h, c) and increasing (i.e., (f ′(x) > 0) in the interval (c, c + h). ### Theorem \color{green} ✍️ Let f be a function defined on an open interval I. Suppose c ∈ I be any point. If f has a local maxima or a local minima at x = c, then either f ′(c) = 0 or color{red}{f \ \ "is not differentiable at" \ \ c.} color{blue}{"Remark-"} The converse of above theorem need not be true, that is, a point at which the derivative vanishes need not be a point of local maxima or local minima. => For example, if f (x) = x^3, then f′(x)= 3x^2 and so f′(0) = 0. But 0 is neither a point of local maxima nor a point of local minima .
kidzsearch.com > wiki Explore:images videos games # Mean (Redirected from Mean (statistics)) In mathematics and statistics, the mean is another name for the average. The mean is calculated by adding all of the values together, then dividing by the number of values. Example: 1, 2, 2, 100, 100 is a set of numbers or scores. If we add all the numbers, the answer is 205. By dividing this number by the number of numbers, we discover that the mean is 41. The difficulty with this particular set of numbers is that no one in this group scored anything like a 41, and the mean does not tell us much about what kind of scores these numbers represent. ## Calculation details To find the average of $N$ numbers, the $N$ numbers are added and the total is divided by $N$. In symbols, if the numbers are $X_1$, $X_2$, $X_3$, ... $X_N$, the total is: $X_1 + X_2 + X_3 + ... + X_N$ The total is divided by $N$ to make the average: ${X_1 + X_2 + X_3 + ... + X_N} \over N$ ### Examples • Lucy is 5 years old. Tom is 6 years old. Emily is 7 years old. To find the average age: • Add the three numbers : $5+6+7=18$ • The total is 18. Divide the total 18 by three: $18/3=6$ • The average of the three numbers is 6. $\frac{5+6+7}{3}$ Therefore, the average age of Lucy, Tom and Emily is 6 years. ## Related calculations The idea behind the mean is to represent a number of measurements, or values, by one value only. But there are different ways to calculate such a representing value. • The median is the number that divides all the samples in such a way that half of the samples are below it, and the other half above. Example: 1, 10, 50, 100, 100 is a set of numbers or scores. If we look at these scores we discover that the number 50 falls in the middle of the range of numbers and tells us that half the numbers or scores are above this number and half the numbers and scores are below this number. This is more information, depending on what you are trying to find out about this group of numbers, to help you find out what you want to know. It is not always possible to make the higher and lower group each exactly half of the total (example: 1, 2, 2), but that is the ideal. • The modus or mode is the number that occurs most often. Example: 1, 2, 2, 100, 200 is a set of numbers or scores. If we look at the numbers we discover that the number 2 recurs most often and would tell us that the number or score of 2 is the most common score or number in the group. • The arithmetic mean is just the average, the value that is the sum of all values, divided by their number. This is what is most often referred to as mean. • The geometric mean is the root of the product of all values. For example, the geometric mean of 4, 6, and 9 is 6, because 4 times 6 times 9 is 216, and the cube root (because there are three values) of 216 is 6. • The harmonic mean is the reciprocal of the arithmetic mean of the reciprocals. It is often used when people want a mean of rates or percentages. • The root mean square is the square root of the arithmetic mean of the squares of the values. The root mean square is at least as high as the arithmetic mean, and usually higher. If people do many different measurements, they will get many different results. Those results have a certain distribution, and they can also be centered around an average value. This average value is what mathematicians call arithmetic mean. Mean can also stand for expected value.es:Promedio it:Promedio
# Comparison of crank based leg mechanism/draw the Jansen linkage First of all, we need a drawing with the constants. Theo Jansen has published the numbers on his website: [1] (Video "The Legsystem", 3:36) or on archived version the webpage: [2] I suggest to name the points. The naming can be arbitrary. I have used Z..S to avoid confusion with the lengths a..m. It can be very helpful to print out the drawing to scribble on. ## Construction by hand Now we have to think about how we could construct the mechanism by hand. We need a non collapsible compass, a ruler, paper and a pencil. We set the origin Z to an arbitrary point. For Y, we are going down l and a to the left. The crank m can be at an arbitrary angle. Drawing the crank results in the point X. From this point, we construct triangles that are defined by two points and two lengths. In geometry, this is the SSS case (constructing a Triangle with three sides given, see also Solution of triangles#Three sides given (SSS)). On paper, this can be easily solved using a compass. Let us start with setting the compass to length b and then putting the compass in point Y. Then we set the compass to the length j and put the compass in point X. The crossing point of the arcs is the point W. Note that when two points and two lengths are given, there are always two solutions, speak crossing points. Given that we already know the general shape of the mechanism, we know which one we need. But keep it mind for later. The rest is more or less "rinse and repeat". ### Geometric Construction Instructions #### Conventions • lengths are written in lower case • point are written in upper case • lines and arcs used for construction have a ' #### Instructions 1. construct point Z 1. draw a horizontal line h 2. draw a vertical line v crossing the horizontal line h 3. mark point Z where the h and v cross • that will be the origin of this construction and crank axis of the leg mechanism 2. construct point Y 1. draw a horizontal line h' distance l down from h 2. draw a vertical line v' distance a left from v 3. label the crossing point of h' and v' as fix point Y 3. construct point X 1. draw a line m' trough Z add an arbitrary angle (the angle is the crank angle) 2. draw an arc from point Z with radius m, the crossing point of arc n and Point Z is crank pivot X draw line m from Z to X 4. construct point W (SSS case) 1. draw an arc j' from point X with radius j 2. draw an arc b' from point Y with radius b 3. the crossing point of arc j' and b' is point W draw line j from X to W draw line b from Y to W 5. construct point V (SSS case) 1. draw an arc e' from point W with radius e 2. draw an arc d' from point Y with radius d draw line e from W to V draw line d from Y to V 1. the crossing point of arc e' and d' is point V 6. construct point U (SSS case) 1. draw an arc c' from point Y with radius c 2. draw an arc k' from point X with radius k draw line c from Y to U draw line k from X to U 1. the crossing point of arc c' and k' is point U 7. construct point T (SSS case) 1. draw an arc f' from point V with radius f 2. draw an arc g' from point U with radius g 3. the crossing point of arc f' and g' is point T draw line f from V to T draw line g from U to T 8. construct point S (SSS case) 1. draw an arc h' from point T with radius h 2. draw an arc i' from point U with radius i 3. the crossing point of arc f' and g' is point S • this is the "foot" of the mechanism draw line h from T to S draw line i from U to S #### given lengths ``` a=38.0 b=41.5 c=39.3 d=40.1 e=55.8 f=39.4 g=36.7 h=65.7 i=49.0 j=50.0 k=61.9 l= 7.8 m=15.0 ``` #### Points ``` Z fix point, origin, crank axis Y fix point X crank axis W V U T S foot ```
# Writing a function rule given a table of ordered pairs: One-step rules In this lesson, we find the function rule given a table of ordered pairs. We first identify the input and the output variables and their values. We find if the function is increasing or decreasing. If the function is increasing, it means there is either an addition or multiplication operation between the two variables. If the function is decreasing, it means there is either a subtraction or division operation between the two variables. Consider the following table − x y 3 15 5 25 6 30 8 40 9 45 We see that the y values are increasing as the x values are increasing. So it is an increasing function. So, the variables x and y must be related either by addition or multiplication operation. We check addition operation on x and y values as follows − 3 + 12 = 15 5 + 12 = 17 We check multiplication operation on x and y values as follows − 3 x 5 = 15 5 x 5 = 25 and so on We see that the relation between x and y is a multiplication operation here and the constant for which all values are satisfied is 5. So the function rule for this table of x and y values is “Multiply by 5”. Consider another table − x y 10 13 15 18 19 22 23 26 28 31 Here we identify the input and output and then see the output y is increasing as input x is increasing. 13 = 10 + 3; 18 = 15 + 3; 22 = 19 + 3 and so on. So, output y = input x + 3 Therefore, we identify the function rule here as “Add 3”. Given the following table of ordered pairs, write a one-step function rule. Input(x) Output(y) 0 3 2 5 4 7 6 9 8 11 ### Solution Step 1: From the table 0 + 3 = 3; 2 + 3 = 5 and so on Step 2: Input + 3 = Output or x + 3 = y Step 3: Therefore the function rule here is ‘Add 3’ to the input to get the output. Given the following table of ordered pairs, write a one-step function rule. Input(x) Output(y) 0 0 1 6 2 12 3 18 4 24 ### Solution Step 1: From the table 0 × 6 = 0; 1 × 6 = 6 and so on Step 2: Input × 6 = Output or x × 6 = y Step 3: Therefore the function rule here is ‘Multiply by 6’ the input to get the output.
# Circular arc A circular sector is shaded in green. Its curved boundary of length L is a circular arc. A circular arc is the arc of a circle between a pair of distinct points. If the two points are not directly opposite each other, one of these arcs, the minor arc, will subtend an angle at the centre of the circle that is less than π radians (180 degrees), and the other arc, the major arc, will subtend an angle greater than π radians. The length (more precisely, arc length) of an arc of a circle with radius r and subtending an angle θ (measured in radians) with the circle center — i.e., the central angle — is and, with α being the same angle measured in degrees, since θ = α/180π, the arc length equals A practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement: For example, if the measure of the angle is 60 degrees and the circumference is 24 inches, then This is so because the circumference of a circle and the degrees of a circle, of which there are always 360, are directly proportional. The area of the sector formed by an arc and the center of a circle (bounded by the arc and the two radii drawn to its endpoints) is The area A has the same proportion to the circle area as the angle θ to a full circle: Using the conversion described above, we find that the area of the sector for a central angle measured in degrees is The area of the shape bounded by the arc and the straight line between its two end points is Using the intersecting chords theorem (also known as power of a point or secant tangent theorem) it is possible to calculate the radius r of a circle given the height H and the width W of an arc: Consider the chord with the same endpoints as the arc. Its perpendicular bisector is another chord, which is a diameter of the circle. The length of the first chord is W, and it is divided by the bisector into two equal halves, each with length W/2. The total length of the diameter is 2r, and it is divided into two parts by the first chord. The length of one part is the sagitta of the arc, H, and the other part is the remainder of the diameter, with length 2r − H. Applying the intersecting chords theorem to these two chords produces
# How do you solve 4(5b +2)-b=2+8(3b-8)? Jun 29, 2017 $b = 14$ #### Explanation: Both brackets neednto be expanded first, where $a \left(b + c\right) \equiv a b + a c$ This means that $4 \left(5 b + 2\right) \equiv \left(4 \cdot 5 b\right) + \left(4 \cdot 2\right) = 20 b + 8$ It also means that $8 \left(3 b - 8\right) \equiv \left(8 \cdot 3 b\right) + \left(8 \cdot - 8\right) = 24 b - 64$ You now have $20 b + 8 - b = 2 + 24 b - 64$. Put all b's on the left hand side and numbers on the right. $20 b + 8 - b = 2 + 24 b - 64$ $20 b - b - 24 b = - 8 + 2 - 64$ $- 5 b = - 70$ Now divide both sides by $- 5$. $\frac{- 5 b}{- 5} = \frac{- 70}{- 5}$ $b = 14$ Jun 29, 2017 $b = 14$ #### Explanation: $4 \left(5 b + 2\right) - b = 2 + 8 \left(3 b - 8\right)$ First open the brackets and simplify by multiplying each term with the number outside the bracket. $20 b + 8 - b = 2 + 24 b - 64$ Simplify each side. $19 b + 8 = 24 b - 62$ Add $62$ to each side. $19 b + 70 = 24 b$ Subtract $19 b$ from each side. $70 = 5 b$ Divide both sides by $5$. $14 = b$
# Fractions on the Number Line Any fractional number can be represented on a number line which is known as fractions on the number line. Here we will learn how to represent fraction on a number line. We will also learn how to add or subtract with the help of a fraction number line. Follow the examples step-by-step how to add or subtract using a fraction number line. 1. Calculate the sum of 3/10 + 4/10 on the fraction number line. 3/10 + 4/10 = (3 + 4)/10 = 7/10 2. Calculate 3/5 + 2/5 on the fraction number line. 3/5 + 2/5 = (3 + 2)/5 = 5/5 = 1/1 = 1. 3. Find the different on the fraction number line 7/8 - 3/8 7/8 - 3/8 = 4/8 ÷ 4/4 = 1/2 4. Find the difference of 9/10 and 7/10 9/10 - 7/10 = 2/10 or 2/10 ÷ 2/2 = 1/5 Related Concept Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Comparison of Numbers \| Compare Numbers Rules \| Examples of Comparison May 18, 24 02:59 PM Rule I: We know that a number with more digits is always greater than the number with less number of digits. Rule II: When the two numbers have the same number of digits, we start comparing the digits… 2. ### Numbers \| Notation \| Numeration \| Numeral \| Estimation \| Examples May 12, 24 06:28 PM Numbers are used for calculating and counting. These counting numbers 1, 2, 3, 4, 5, .......... are called natural numbers. In order to describe the number of elements in a collection with no objects 3. ### Face Value and Place Value\|Difference Between Place Value & Face Value May 12, 24 06:23 PM What is the difference between face value and place value of digits? Before we proceed to face value and place value let us recall the expanded form of a number. The face value of a digit is the digit… 4. ### Patterns in Numbers \| Patterns in Maths \|Math Patterns\|Series Patterns May 12, 24 06:09 PM We see so many patterns around us in our daily life. We know that a pattern is an arrangement of objects, colors, or numbers placed in a certain order. Some patterns neither grow nor reduce but only r…
# Step-by-step Solution ## Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{6x^2-3}{\sqrt{x^2-1}}\right)$ Go 1 2 3 4 5 6 7 8 9 0 x y (◻) ◻/◻ 2 e π ln log lim d/dx Dx > < >= <= sin cos tan cot sec csc asin acos atan acot asec acsc sinh cosh tanh coth sech csch asinh acosh atanh acoth asech acsch ### Videos $\frac{12x\sqrt{x^2-1}-\left(x^2-1\right)^{-\frac{1}{2}}x\left(6x^2-3\right)}{x^2-1}$ ## Step-by-step explanation Problem to solve: $\frac{d}{dx}\left(\frac{6x^2-3}{\sqrt{x^2-1}}\right)$ 1 Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$ $\frac{\sqrt{x^2-1}\cdot\frac{d}{dx}\left(6x^2-3\right)-\left(6x^2-3\right)\frac{d}{dx}\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)^2}$ 2 The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$ $\frac{\sqrt{x^2-1}\cdot\frac{d}{dx}\left(6x^2-3\right)-\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}\left(6x^2-3\right)\frac{d}{dx}\left(x^2-1\right)}{\left(\sqrt{x^2-1}\right)^2}$ $\frac{12x\sqrt{x^2-1}-\left(x^2-1\right)^{-\frac{1}{2}}x\left(6x^2-3\right)}{x^2-1}$ $\frac{d}{dx}\left(\frac{6x^2-3}{\sqrt{x^2-1}}\right)$ Quotient rule ~ 1.24 seconds ### Struggling with math? Access detailed step by step solutions to millions of problems, growing every day!
#### Read Microsoft Word - Document5 text version `Le sson 1 0: Almost a Doub leLesson Level: TWO & THREELe sson Object ive s· · · To use an understanding of doubles relationships to work with near-doubles. To build understanding of the almost-double combinations (that are built upon the doubles anchors) to help solve computation problems. To recognize the difference between even numbers (can be represented as a pair of equal numbers) and odd numbers (paired numbers plus one).Activit y Bac kground and Introd uct ion·Begin by illustrating the following on your demonstration Rekenrek.· ·Ask students to find the double that is within the group of 5 beads on the left. Visually separate the double (two groups of two) from the remaining 5th bead. A pencil may be used to physically separate the beads.· ·Ask students, "Now that you can see the double (i.e., 2 + 2 = 4), what can you say about a double plus one?" Illustrate the thinking by emphasizing the visualization, and also the symbolic equivalent: 2 + 3 = ? means... 2 + (2 + 1) = ? (2 + 2) + 1 = 4 + 1Le sson P rog re ssion· · Continue with additional examples. As students to find the double within the representation, and then use that information to find the sum of the neardouble. Visually compare the doubles with the near doubles, highlighting the differences between even and odd numbers (odd numbers have solo bead).Lesson 10: Almost a Double24681013579· ·You may wish to highlight the doubles that may be found within these "almost double" visualizations as has been illustrated on the examples of 5, 7 and 9. Using the idea of hidden doubles as a way to work with "almost doubles" number relationships is a powerful strategy. It is important to help students transfer this visual strategy to symbolic representations. Take the example of 7. Children should be encouraged to see the "6" that exists within this starting number. 4 +3 3+4 =7 just as (3 + 3) +1 = 7· ·By demonstrating this relationship, students begin to develop a relational view of the equal sign that the equal sign means "the same as" rather than simply a symbol that indicates the answer is approaching. Develop this idea by doing several additional examples with the Rekenrek. Ask students to use the Rekenrek to "prove" whether or not the following are true. Have students visually identify each component of the statements. Does 6 + 7 = 12 + 1? Does 3 + 2 = 4 + 1? Does 4 + 5 = 8 + 1? Does 8 + 9 = 16 + 1?` #### Information ##### Microsoft Word - Document5 2 pages Find more like this #### Report File (DMCA) Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us: Report this file as copyright or inappropriate 269461
6. Applications of Integration # 6.7 Integrals, Exponential Functions, and Logarithms ### Learning Objectives • Write the definition of the natural logarithm as an integral. • Recognize the derivative of the natural logarithm. • Integrate functions involving the natural logarithmic function. • Define the number through an integral. • Recognize the derivative and integral of the exponential function. • Prove properties of logarithms and exponential functions using integrals. • Express general logarithmic and exponential functions in terms of natural logarithms and exponentials. We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key details in the previous discussions. For example, we did not study how to treat exponential functions with exponents that are irrational. The definition of the number is another area where the previous development was somewhat incomplete. We now have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section. For purposes of this section, assume we have not yet defined the natural logarithm, the number , or any of the integration and differentiation formulas associated with these functions. By the end of the section, we will have studied these concepts in a mathematically rigorous way (and we will see they are consistent with the concepts we learned earlier). We begin the section by defining the natural logarithm in terms of an integral. This definition forms the foundation for the section. From this definition, we derive differentiation formulas, define the number and expand these concepts to logarithms and exponential functions of any base. # The Natural Logarithm as an Integral Recall the power rule for integrals: Clearly, this does not work when as it would force us to divide by zero. So, what do we do with Recall from the Fundamental Theorem of Calculus that is an antiderivative of Therefore, we can make the following definition. ### Definition For define the natural logarithm function by For this is just the area under the curve from 1 to For we have so in this case it is the negative of the area under the curve from (see the following figure). Notice that Furthermore, the function for Therefore, by the properties of integrals, it is clear that is increasing for # Properties of the Natural Logarithm Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus. ### Derivative of the Natural Logarithm For the derivative of the natural logarithm is given by ### Corollary to the Derivative of the Natural Logarithm The function is differentiable; therefore, it is continuous. A graph of is shown in (Figure). Notice that it is continuous throughout its domain of ### Calculating Derivatives of Natural Logarithms Calculate the following derivatives: #### Solution We need to apply the chain rule in both cases. Calculate the following derivatives: #### Hint Apply the differentiation formula just provided and use the chain rule as necessary. Note that if we use the absolute value function and create a new function we can extend the domain of the natural logarithm to include Then This gives rise to the familiar integration formula. ### Integral of (1/) du The natural logarithm is the antiderivative of the function ### Calculating Integrals Involving Natural Logarithms Calculate the integral #### Solution Using -substitution, let Then and we have Calculate the integral #### Hint Apply the integration formula provided earlier and use -substitution as necessary. Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here. ### Properties of the Natural Logarithm If and is a rational number, then ## Proof 1. By definition, 2. We have Use on the last integral in this expression. Let Then Furthermore, when and when So we get 3. Note that Furthermore, Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have for some constant Taking we get Thus and the proof is complete. Note that we can extend this property to irrational values of later in this section. Part iii. follows from parts ii. and iv. and the proof is left to you. ### Using Properties of Logarithms Use properties of logarithms to simplify the following expression into a single logarithm: #### Solution We have Use properties of logarithms to simplify the following expression into a single logarithm: #### Hint Apply the properties of logarithms. # Defining the Number Now that we have the natural logarithm defined, we can use that function to define the number ### Definition The number is defined to be the real number such that To put it another way, the area under the curve between and is 1 ((Figure)). The proof that such a number exists and is unique is left to you. (Hint: Use the Intermediate Value Theorem to prove existence and the fact that is increasing to prove uniqueness.) The number can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series in the second volume of this text). Its approximate value is given by # The Exponential Function We now turn our attention to the function Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by Then, The following figure shows the graphs of and We hypothesize that For rational values of this is easy to show. If is rational, then we have Thus, when is rational, For irrational values of we simply define as the inverse function of ### Definition For any real number define to be the number for which Then we have for all and thus for all # Properties of the Exponential Function Since the exponential function was defined in terms of an inverse function, and not in terms of a power of we must verify that the usual laws of exponents hold for the function ### Properties of the Exponential Function If and are any real numbers and is a rational number, then ## Proof Note that if and are rational, the properties hold. However, if or are irrational, we must apply the inverse function definition of and verify the properties. Only the first property is verified here; the other two are left to you. We have Since is one-to-one, then As with part iv. of the logarithm properties, we can extend property iii. to irrational values of and we do so by the end of the section. We also want to verify the differentiation formula for the function To do this, we need to use implicit differentiation. Let Then Thus, we see as desired, which leads immediately to the integration formula We apply these formulas in the following examples. ### Using Properties of Exponential Functions Evaluate the following derivatives: #### Solution We apply the chain rule as necessary. Evaluate the following derivatives: #### Hint Use the properties of exponential functions and the chain rule as necessary. ### Using Properties of Exponential Functions Evaluate the following integral: #### Solution Using -substitution, let Then and we have Evaluate the following integral: #### Hint Use the properties of exponential functions and as necessary. # General Logarithmic and Exponential Functions We close this section by looking at exponential functions and logarithms with bases other than Exponential functions are functions of the form Note that unless we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let’s rectify that here by defining the function in terms of the exponential function We then examine logarithms with bases other than as inverse functions of exponential functions. ### Definition For any and for any real number define as follows: Now is defined rigorously for all values of . This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of It is straightforward to show that properties of exponents hold for general exponential functions defined in this way. Let’s now apply this definition to calculate a differentiation formula for We have The corresponding integration formula follows immediately. ### Derivatives and Integrals Involving General Exponential Functions Let Then, and If then the function is one-to-one and has a well-defined inverse. Its inverse is denoted by Then, Note that general logarithm functions can be written in terms of the natural logarithm. Let Then, Taking the natural logarithm of both sides of this second equation, we get Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base Again, let Then, Let Then, ### Calculating Derivatives of General Exponential and Logarithm Functions Evaluate the following derivatives: #### Solution We need to apply the chain rule as necessary. Evaluate the following derivatives: #### Hint Use the formulas and apply the chain rule as necessary. ### Integrating General Exponential Functions Evaluate the following integral: #### Solution Use and let Then and we have Evaluate the following integral: #### Hint Use the properties of exponential functions and as necessary. ### Key Concepts • The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way. • The cornerstone of the development is the definition of the natural logarithm in terms of an integral. • The function is then defined as the inverse of the natural logarithm. • General exponential functions are defined in terms of and the corresponding inverse functions are general logarithms. • Familiar properties of logarithms and exponents still hold in this more rigorous context. # Key Equations • Natural logarithm function • Z • Exponential function • Z For the following exercises, find the derivative 1. 2. 3. #### Solution For the following exercises, find the indefinite integral. 4. 5. For the following exercises, find the derivative (You can use a calculator to plot the function and the derivative to confirm that it is correct.) 6. [T] 7. [T] 8. [T] 9. [T] 10. [T] 11. [T] 12. [T] 13. [T] #### Solution 14. [T] 15. [T] For the following exercises, find the definite or indefinite integral. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. #### Solution For the following exercises, compute by differentiating 26. 27. 28. 29. 30. 31. 32. 33. 1 34. 35. #### Solution For the following exercises, evaluate by any method. 36. 37. 38. 39. #### Solution 40. For the following exercises, use the function If you are unable to find intersection points analytically, use a calculator. 41. Find the area of the region enclosed by and above #### Solution 42. [T] Find the arc length of from to 43. Find the area between and the -axis from #### Solution 44. Find the volume of the shape created when rotating this curve from around the -axis, as pictured here. 45. [T] Find the surface area of the shape created when rotating the curve in the previous exercise from to around the -axis. #### Solution 2.8656 If you are unable to find intersection points analytically in the following exercises, use a calculator. 46. Find the area of the hyperbolic quarter-circle enclosed by above 47. [T] Find the arc length of from #### Solution 3.1502 48. Find the area under and above the -axis from For the following exercises, verify the derivatives and antiderivatives. 49. 50. 51. 52. 53.
# Algebra/Solving Equations ### Solving Equations Solving equations is very easy. Take any ordinary equation, such as: 2x+7=13 It may look tough, but when broken down, it is very simple. All you have to do is take it step-by-step, and perform opposite operations on each side. Opposite operations are operations that cancel each other out, such as addition and subtraction, and multiplication and division. So, the first step would be to get rid of that annoying "+7", so you subtract seven from both sides, like so: 2x+7-7=13-7 2x=6 (You may leave out the first step shown in the gap above as you get better, but beginners may want to add that step in.) Now you have a fairly simple equation. What number multiplied by two will give you six? Why, three, of course! If you didn't get that, you may want to go backward a few years. But how did you get that? Think. You divided six by two. And that's what you do: 2x/2=6/2 x=3 The goal of solving one equation is leave the variable alone. Solving equation is possible by using 1 step (+, - *, /) or 2 step ( a math operation combination) or more than 2 steps. Once you get good at some basic equations, you may want to tackle equations with variables on both sides, like this: 7x+3=5x+7 To do that, simply bring all the x's over to one side and solve like a normal equation. 7x-5x+3=5x-5x+7 2x+3=7 2x+3-3=7-3 2x=4 2x/2=4/2 x=2 Well, good luck!
# How to find the 3 Pythagorean triplets with only one number? Given: Pythagorean triplets To explain: Here we have to explain how to find Pythagorean triplets with only one number. Solution: If the number is odd: Square the number (n) and then divide it by 2. Take the integer that is immediately before and after that number i.e. $\left(\frac{n^{2}}{2} \ -\ 0.5\right)$ and $\left(\frac{n^{2}}{2} \ +\ 0.5\right)$. Pythagorean triplet = n, $\left(\frac{n^{2}}{2} \ -\ 0.5\right)$ and $\left(\frac{n^{2}}{2} \ +\ 0.5\right)$. Example: Take number n = 3. On squaring the number, we get 9. Now take half of it: $\frac{9}{2}$ = 4.5 The integer immediately before 4.5 = 4 The integer immediately after 4.5 = 5 Therefore, the triplets are 3, 4 and 5. If the number is even: Take the half of that number (n) and then square it. Pythagorean triplet = n, $\left(\frac{n}{2}\right)^{2} \ -\ 1$ and $\left(\frac{n}{2}\right)^{2} \ +\ 1$ Example: Take number n = 8 Half of n = 4. Pythagorean triplet = 8, (42 $-$ 1) and (42 $+$ 1) = 8, 15 and 17. Updated on: 10-Oct-2022 105 Views
Courses Courses for Kids Free study material Offline Centres More Store In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying e.g., a section of Class ${\text{I}}$ will plant 1 tree, a section of Class ${\text{II}}$ will plant 2 trees and so on till Class ${\text{XII}}$. There are three sections of each class. How many trees will be planted by the students? Last updated date: 13th Jul 2024 Total views: 450.3k Views today: 7.50k Verified 450.3k+ views Hint- Convert the problem statement in the form of a series of $n$ natural numbers and then find the sum of first $n$ natural numbers. According to the problem statement, every class has three sections. Number of trees planted from Class ${\text{I}}$$= 3 \times 1$ (1 from each section) Number of trees planted from Class ${\text{II}}$$= 3 \times 2$ (2 from each section) Number of trees planted from Class ${\text{III}}$$= 3 \times 3$ (3 from each section) Number of trees planted from Class ${\text{IV}}$$= 3 \times 4$ (4 from each section) And so on up to Class${\text{XII}}$ Number of trees planted from Class ${\text{XII}}$$= 3 \times 12$ (12 from each section) Therefore, for the total number of trees planted we will add all the trees planted from all the classes. Total number of trees planted$= 3 \times 1 + 3 \times 2 + 3 \times 3 + 3 \times 4 + ...... + 3 \times 11 + 3 \times 12 = 3\left[ {1 + 2 + 3 + 4 + ..... + 11 + 12} \right]$ As we know that the sum of first $n$ natural numbers is given by ${\text{S}} = \dfrac{{n\left( {n + 1} \right)}}{2}$ Therefore, the sum of the first 12 natural numbers can be computed if we put $n = 12$ in the above formula. i.e., $1 + 2 + 3 + 4 + ..... + 11 + 12 = \dfrac{{12\left( {12 + 1} \right)}}{2} = \dfrac{{12 \times 13}}{2} = 78$ Total number of trees planted$= 3\left[ {1 + 2 + 3 + 4 + ..... + 11 + 12} \right] = 3 \times 78 = 234$ Therefore, the total number of trees which are planted by the students are 234 trees. Note- In these types of problems, it is very crucial to interpret the problem statement very carefully and missing any of the minute detail will lead to a wrong answer. If we observe this particular problem carefully, the sum is reduced to an arithmetic progression for which we have used directly the formula of the sum of first $n$ natural numbers because here the common difference is 1 for the series 1,2,3...12.
# Chain Rule of Differentiation in Calculus The chain rule of differentiation of functions in calculus is presented along with several examples and detailed solutions and comments. Also in this site, Step by Step Calculator to Find Derivatives Using Chain Rule Let f(x) = (g o h)(x) = g(h(x)) Let u = h(x) Using the above, function f may be written as: f(x) = g(u) the derivative of f with respect to x, f ' is given by: f '(x) = (df / du) (du / dx) We now present several examples of applications of the chain rule. Example 1: Find the derivative f '(x), if f is given by f(x) = 4 cos (5x - 2) Solution to Example 1 Let u = 5x - 2 and f(u) = 4 cos u, hence du / dx = 5 and df / du = - 4 sin u We now use the chain rule f '(x) = (df / du) (du / dx) = - 4 sin (u) (5) We now substitute u = 5x - 2 in sin (u) above to obtain f '(x) = - 20 sin (5x - 2) Example 2: Find the derivative f '(x), if f is given by f(x) = (x 3 - 4x + 5) 4 Solution to Example 2 Let u = x 3 - 4x + 5 and f(u) = u 4 , hence du / dx = 3 x 2 - 4 and df / du = 4 u3 Use the chain rule f '(x) = (df / du) (du / dx) = (4 u 3) (3 x 2 - 4) We now substitute u = x 3 - 4x + 5 above to obtain f '(x) = 4 (x 3 - 4x + 5) 3 (3 x 2 - 4) Example 3: Find f '(x), if f is given by f(x) = √ (x 2 + 2x -1)Solution to Example 3 Let u = x 2 + 2x -1 and f(u) = √u , hence du / dx = 2x + 2 and df / du = 1 / (2 √u) Use the chain rule f '(x) = (df / du) (du / dx) = () 1 / (2 √u) ) (2x + 2) Substitute u = x 2 + 2x -1 above to obtain f '(x) = (2x + 2) ( 1 / (2 √(x 2 + 2x -1)) ) Factor 2 in numerator and denominator and simplify f '(x) = (x + 1) / (√(x 2 + 2x -1)) Example 4: Find the first derivative of f if f is given by f(x) = sin 2 (2x + 3)Solution to Example 4 Let u = sin (2x + 3) and f(u) = u 2 , hence du / dx = 2 cos(2x + 3) and df / du = 2 u Use the chain rule f '(x) = (df / du) (du / dx) = 2 u 2 cos(2x + 3) Substitute u = sin (2x + 3) above to obtain f '(x) = 4 sin (2x + 3) cos (2x + 3) Use the trigonometric formula sin (2x) = 2 sin x cos x to simplify f '(x) f '(x) = 2 sin (4x + 6) Example 5: Find the first derivative of f if f is given by f(x) = ln(x2 + x)Solution to Example 5 Let u = x2 + x and f(u) = ln u , hence du / dx = 2 x + 1 and df / du = 1 / u Use the chain rule and substitute f '(x) = (df / du) (du / dx) = (1 / u) (2x + 1) = (2x + 1) /(x2 + x) Exercises: Find the first derivative to each of the functions. 1) f(x) = cos (3x -3) 2) l(x) = (3x 2 - 3 x + 8) 4 3) m(x) = sin [ 1 / (x - 2)] 4) t(x) = √ (3x 2 - 3 x + 6) 5) r(x) = sin 2 (4 x + 20) Solutions to the above exercises 1 ) f '(x) = -3 sin (3 x - 3) 2 ) l(x) = 12 (2 x - 1) (3 x 2 - 3 x + 8) 3 3 ) m(x) = - 1 / (x - 2) 2 cos [ 1 / (x - 2)] 4 ) t(x) = (3/2)(2 x - 1) / √ (3 x 2 - 3 x + 6) 5 ) r(x) = 4 sin (8x + 4) More on differentiation and derivatives Solve Rate of Change Problems in Calculus Solve Tangent Lines Problems in Calculus Home Page
# Slope of a Line The slope of a line is a number that describes how steep a line is. The slope formula is the ratio between the vertical change and the horizontal change of the line. If the slope is a large number, then the line will be very steep. If it is a small number, then the line will just be a gentle incline. A perfectly flat horizontal line has a slope of 0. If the slope is a positive number, then that means the line is sloping “uphill” from left to right. If it is a negative number, then that means the line is sloping “downhill”. ## Slope Formula There are two formulas you can use to find the slope of a line. The first one is the most useful when you are finding the slope of a line on a graph. The second one is most useful when you are finding the slope of a line from two points on the line. ## How to Find Slope on a Graph Step 1: Choose two points on the line. When you are finding the slope, you can choose ANY two points on the line. So, technically, you could choose a point like (-0.5,3) on the green line. However, the points that I’ve labeled are easier to work with because they have integer coordinates instead of decimals. Step 2: Find the vertical difference (rise) between the two points you chose. To find the rise of the line, choose one of your points as your starting point. From the starting point, find the vertical distance to the second point. If you go down, the rise will be negative. If you go up, the rise will be positive. Step 3: Find the horizontal difference (run) between the two points you chose. To find the run of the line, find the horizontal distance from the starting point to the second point. If you go left, the run will be negative. If you go right, the run will be positive. Step 4: Divide the rise by the run to find the slope. The rise always comes first in the division problem (or the top of the fraction). If you have a hard time remembering that, you can use this memory tool: “You first have to RISE from your bed before you can go on a RUN.” Also, if the rise or the run is a negative number, make sure to follow the rules for dividing negative numbers when you divide them. It is reasonable that the slope of the green line is +5 because it is an “uphill” line and it is relatively steep. It is reasonable that the slope of the red line is -2 because it is a “downhill” line and it is steep but not as steep as the green line. It is reasonable that slope of the purple line is +1/2 because it is an “uphill” graph and it is not very steep.
Making a letter the subject how do i do it? How do i do the equation in the picture>? a = (6a - n)/(3 + n) Solve for "a", and then solve for "n", Begin by multiplying each side of the equation by (3 + n). Comments for Making a letter the subject Apr 06, 2011 Making a letter the subject by: Staff The question: a = (6a – n) ÷ (3 + n) how do I do it? The answer: a = (6a – n) ÷ (3 + n) a = (6a – n)/(3 + n) Part I: solve for “a” Part II: solve for “n” Part I: solve for “a” a = (6a – n)/(3 + n) Multiply each side of the equation by (3 + n) a * (3 + n) = [(6a – n)/(3 + n)]* (3 + n) a * (3 + n) = (6a – n)* [(3 + n)/(3 + n)] a * (3 + n) = (6a – n)* 1 a * (3 + n) = 6a – n 3a + an = 6a – n Add -6a to each side of the equation 3a - 6a + an = 6a - 6a – n 3a - 6a + an = 0 – n 3a - 6a + an = – n -3a + an = – n Factor “a” from the expression (-3a + an) on the left hand side of the equation using the distributive law a * (-3 + n) = – n a * (n - 3) = – n Divide each side of the equation by (n – 3) a * (n - 3)/(n – 3) = – n/(n – 3) a * 1 = – n/(n – 3) a = – n/(n – 3) Part II: solve for “n” a = (6a – n)/(3 + n) Multiply each side of the equation by (3 + n) a * (3 + n) = [(6a – n)/(3 + n)]* (3 + n) a * (3 + n) = (6a – n)* [(3 + n)/(3 + n)] a * (3 + n) = (6a – n)* 1 a * (3 + n) = 6a – n Use the distributive law to eliminate the parentheses on the left side of the equation 3a + an = 6a – n Add -3a to each side of the equation 3a – 3a + an = 6a – 3a – n 0 + an = 6a – 3a – n an = 6a – 3a – n an = 3a – n Add +n to each side of the equation an + n = 3a – n + n an + n = 3a + 0 an + n = 3a Factor “n” from the expression (an + n) on the left hand side of the equation using the distributive law n * (a + 1) = 3a Divide each side of the equation by (a + 1) n * (a + 1)/(a + 1) = 3a/(a + 1) n * 1 = 3a/(a + 1) n = 3a/(a + 1) the final answer is: a = – n/(n – 3) n = 3a/(a + 1) Thanks for writing. Staff www.solving-math-problems.com Aug 10, 2020 change to algerbic expressions NEW by: Anonymous I = (E-P)/(R+r) how to make r the subject
# How do you Subtract (-16x^2 - 19x - 10) from (-18x^2 - 13x - 14)? May 2, 2018 "Subtract A from B" is the same as "Find B - A". First, you need to distribute the negative though the terms of the A part, then combine the results for like terms (terms of the same power of x). #### Explanation: $\left(- 18 {x}^{2} - 13 x - 14\right) - \left(- 16 {x}^{2} - 19 x - 10\right) =$ $\left(- 18 {x}^{2} - 13 x - 14\right) + \left(16 {x}^{2} + 19 x + 10\right) =$ $\left(- 18 {x}^{2} + 16 {x}^{2}\right) + \left(- 13 x + 19 x\right) + \left(- 14 + 10\right) =$ $- 2 {x}^{2} + 6 x - 4$ May 2, 2018 $- 2 {x}^{2} + 6 x - 4$ #### Explanation: Subtracting ( $- 16 {x}^{2} - 19 x - 10$ ) from ( $- 18 {x}^{2} - 13 x - 14$ ) is same as subtracting 5 from 49. (I just mean any two real numbers). We do something like this: =49-5 =44 Similarly, we do the subtraction of polynomials but only between numbers with same coefficients. We'll start like this: = $\left(- 18 {x}^{2} - 13 x - 14\right) - \left(- 16 {x}^{2} - 19 x - 10\right)$ = $- 18 {x}^{2} - 13 x - 14 + 16 {x}^{2} + 19 x + 10$ = $\left(- 18 + 16\right) {x}^{2} + \left(- 13 + 19\right) x + \left(- 14 + 10\right)$ = $- 2 {x}^{2} + 6 x - 4$ HAPPY LEARNING !!
Distortion produced by Projection and Rational Equations by Jiyoon Chun Projection and Rational Functions is a combination of . We can get these relationships in projective geometry. We will investigated how the rational function relates to projection, and what is the mathematical facts that we can understand and apply in high school mathematics using projective geometry. What is projection and projective lines? Think about the railroads. they are parallel in Euclidean geometry. However, when we see the rails, It seems like they meet. More precisely, they meet at the horizon. Euclidean geometry works fine in many cases, but when we want to draw this, we have to think differently. If we are cling to Euclidean geometry to draw this, then we will ignore the rule of perspective. Before artists followed the rule of perspective, their pictures could not express which one is nearer and which one is farther. In the picture below, the floor looks like vertical. So, why do we want to deceive ourselves? Although the lines are parallel, they look like meeting at a point on the horizon in the real life. Projective geometry deals with the problem that Euclidean geometry cannot explain (the parallel postulate). Naively saying, projective geometry is assuming that there is an eye which is looking at something. In addition, we want to project the thing what we are watching to another thing such as a line or plane. Basic Concepts of Projective Lines Let's go back to the first picture. It seems like two rails meet at a horizon. Then, what does horizon mean? It is at the sight of the eye. Let's see the picture below. Suppose there is an eye at origin, and we are looking down y=-1. Then,the line passing through origin corresponds to the point of intersection of the line and y=-1. Pay attention to the slopes of the line above. They are getting closer to closer to x-axis. Think about the point on y=-1 which is very far from the origin. The line which connect the point on y=-1 and origin still has an intersection. In other words, the lines through origin always have intersection with y=-1 except when the line is parallel to y=-1. Intuitively, we can see that y=0 meets y= -1 at horizon. y=0 is at the height of the eye. Therefore, we can say that two parallel lines meet at horizon, which is called point of infinity in projective geometry. Distortion Produced by Projection Although there are same sizes of trees are aligned on a railroad, the practical sizes that are observed are different. The nearer, the bigger, the farther, the smaller the trees we see. Thus, distortion happens in projective geometry. Let's say that the eye is at origin, and want to project y=1 to x=1. In the picture above, the projected points on x=1 are not equally spaced whereas the points on y=1 are equally spaced. We can see there is distortion while we are projecting points of y=1 onto x=1. Say the eye can see all 360 degree. Let's mark all the corresponding points on x=1. To see how much it is distorted according to the points of y=1, let's plot points. Since we are projecting points of y=1 onto x=1, the coordinates of the new ordered pairs is (x-coordinated of the point of y=1, y-coordinate of the point x=1 which are projected image of the points on y=1). The new ordered pairs would be plotted as the picture below. These curves are familiar to us. Yes. It is . Proof of Now, we are interested y-coordinate of N' depend on x-coordinate of N. If we make this relation as an ordered pair, we have . This leads to . For the skeptics, I attach the picture when n<0. The proof is the same. Asymptotes and the limits of the graph 1) Asymptote y=0 Since y=0 is parallel to y=1, it does not pass through x=1. As we can see in the railroad picture, the point of infinity located at the height of the eye. Suppose that we have a point (n,1) on y=1, and n goes to infinity. Since the point of infinity locates at the height of the eye, the projected points locates at the height of the eye which is on y=0. Thus, we have . This can also explains why . We assumed that the eye can see 360 degree. When n goes negative infinity, it means that the eye looks back. Although the eye looks back, the height of the eye stays. Therefore, , the height of the eye. The projection gives a lot of idea of the limits. It is not easy to understand why and since n goes to the opposite direction. In projective plane, y=0 is actually connected (there is no disconnected part of the line through origin (because the eye can see back and forth)). Thus, although n goes to negative infinity or positive infinity, they are on the same line. This gives an idea why their limits are the same. 2) Asymptote x=0 When n approaches to 0, we know that the slope of the lines which connect the eye and the points on y=1 gets steeper. However, it cannot give explicit explanation why . It is tricky because is undefined. In projective geometry, it is more clear. The line connecting the eye and (0,1) on y=1 is parallel x=1. It means that we are looking at (0,1) parallel to x=1, which we want to project point on. ntuitively speaking, it is like looking at railroad. Tow parallel lines meet at infinity. Thus, . Symmetry of the Graph If we think origin as the eye, then it is easy to explain why is symmetry over origin (the place where the eye is). See the graph below. Thus, we can say is symmetry over the eye in the way of projective geometry not just an origin. This gives us an insight of the symmetry of other graphs not just by algebraically. Locations of the graphs Now, we understand the place of eye is the key concept of projective line because at the height of the eye, there is an point of infinity. Therefore, origin does not mean a lot in the projective plane. When we refer to quadrants, it will make more sense that the place of eye acts as origin of Cartesian plane. In the picture above, we can see four quadrants with respect to the place of the eye. Since the eye is at origin, it is hard to see the difference. I will deal with the projection line when the eye is not at origin. Look at the 2nd and 4th quadrant respect to the eye. On the 4th quadrant, there is not any points to project because we are projecting the points of y=1, and y=1 does not pass the 4th quadrant. Therefore, there are no points plotted. On the 2nd quadrant, the projected line does not exist since we project y=1 to x=1, and x=1 does not cross the 2nd quadrant. Thus, there are no points plotted. On the 1st quadrant, we see there are two lines: the line we want to project (y=1), and the line on which we want to project the points of y=1 (x=1). Therefore, we know that there would be some points plotted on the 1st quadrant. By the symmetry of the graph, we can see that the graph also is on the 3rd plane. What if x=-1 instead of x=1? Let's project x=-1 onto y=1. See the graph below. It looks like all points lie on . The proof is similar to the previous one. However, I will prove it algebraic way. Now, we are interested y-coordinate of N' depend on x-coordinate of N. If we make this relation as an ordered pair, we have . This leads to . How about in case of x=-1, and y=-1? Or x=1, y=-1? 1) When we project point x=-1 to y=-1 By the symmetry of the projective line, we have the same curve with when x=1 and y=1. We also can conclude that if we project points on x=1 to y=-1, we also have the same curve with . Generalization: When we project the points of x=s onto y=t . Therefore, when we project the point of x=s onto y=t, the set of ordered pairs of the x-coordinate of x=s and the y-coordinate of the projected point onto y=t is . How do the lines of projection x=s, and y=t explain the shape of ? Let's compare the two cases when x=1,y=1 and x=4,y=1. N'=the projected point of Ny=1 onto x=1 N''=the projected point of Ny=1 onto x=4 Since x=4 located on the right side of x=1, the y-coordinate of N'' is bigger that the y-coordinate of N'. Therefore, when we make ordered pairs of the x-coordinate of N and y-coordinate of the projected points, the ordered pairs from N' locates higher than those from N'. Intuitively, we can say the set of ordered pairs locates farther when we project N onto x=4 than x=1 because the line where the images (the projected points) sit on is farther than x=1. See the flash below. In this way, we can explain the locations of the curves. We know that the curves locates on the 1st and 3rd when st>0, and on the 2nd and 4th quadrant when st<0 where . By the property that we found of the location and symmetry of the curve according to x=s and y=t, we know that we have two curves on the 1st and 3rd quadrant when s>0,t>0 and s<0,t<0. With the same sense, the curve locates on the 2nd and the 4th quadrant when s<0,t>0 and s>0,t<0. When the eye is not at origin Let's say the eye is at (1,2), and we want to project the points of y=3 onto x=2. The set of points are . By the property of asymptotes, we know that the intersection of the asymptote is at the place of the eye. Therefore, the asymptotes of the set of the ordered pairs of x-coordinate of a point y=3 and y coordinate of the projected point onto x=2 is x=1 and y=2. Since the distances of the eye to x=2 and y=3 are the same with , the shape of the graph is the same with . Generalization: When the eye is at (p,q), and we project the points of y=t onto x=s 1) Asymptotes of the graph: The point of infinity: x=p, y=q (the eye) The the intersection of the line through the eye and parallel to x=s (which is x=p) project (p,t) to infinity. Therefore, the projected points approaches to x=p vertically. y=q is parallel to y=t, and passing through the eye. When we think that we are looking at the horizon., the point which is infinitely far from the eye projected on the horizon. Thus, when the x-coordinate goes to infinity, it projected on the horizon., which is the height of the eye, q. 2) Symmetry of the graph By the property of the congruent triangles, point A and point B are reflection over the eye. Thus, the graph is symmetry over the eye. 3) The equation of the graph Let's find y-coordinate of k which is the projection point of ny=t onto x=t. Since the yellow and the orange triangles are similar, the slopes of the hypotenuses of the two triangles are the same. The slope of the hypotenuse of the yellow triangle=the slope of the hypotenuse of the orange triangle. This gives Therefore, (x-coordinate of (n,t), y-coordinate of (s,k)=(n,k)=. This gives us that the equation of the set of ordered pair (n,k) is . If we rewrite this as an x-y relationship, this gives the familiar equation such that . 4) The location of the graph As I mentioned above, when (t-q)(s-p)>0, the curves locate on the 1st and the 3rd quadrant respect to the eye. If (t-q)(s-p)<0, the curves are on the 2nd and 4th quadrant. 5) The shape of the graph The bigger \|(t-q)(s-p)\|is, the farther the curves are from the eye. \|(t-q)(s-p)\| is the area of the rectangle of the base (s-p) and the height (t-q). Compare the two pictures above. When the blue rectangle of the area \|(t-q)(s-p)\|, the projected points on x=s is spread out. When the blue rectangle is small, the projected points on x=s are squeezed in for equally spaced points on y=t. See the flash below. Let's fix x=s and move y=t as in the flash above. We can see the slope of the eyesight is getting gentler as it approaches to y=q. For the same spaced points, there will be more projected points when \|t-q\| gets smaller. Therefore, the ordered pairs of (n, k)= would be squeezed in. Although this is not a formal proof, this gives us nice intuitive idea of the shape according to \|(t-q)(s-p)\|. One more interest thing is the case when the lines approach to the eye. The curves disappear because all the points were sent to the point of infinity. The flash above gives us nice visual explanation about the asymptotes If you want to play this on GSP file, click here. 6) The graph movement: Moving the eye for fixed distance between the eye and the two lines ( which means (t-q) and (s-p) are fixed) Since \|(t-q)(s-p)\| affects the shape of the graph, the shape of the curves stay as long as the distance between the eye and x=s, y=t are fixed. Comparison of and projection of y=t onto x=s at the eye (p,q) The way of understanding algebraically and projective geometrically are little bit different. See the table below. projection of y=t onto x=s at the eye (p,q) Equation form Domain reason The denominator cannot be zero. When we connect the eye and (p,n) on y=t, the line is parallel to x=s. Therefore, (p,n) cannot be projected on the line x=s. The line connecting of the eye and (p,n) and x=s meets at infinity. Since infinity does not exist in R, the function is undefined when x=p. Range reason Since cannot be 0, y cannot be . y=q is parallel to y=t. In projective geometry, parallel lines meet at the height of the eye, the horizon. Therefore, the line through the eye and (n,t) on y=t meet y=q when n goes infinity. Since there is no infinity in R, y=q cannot hit any point on y=q. Asymptotes , , reason Since the denominator cannot be zero, there would no y-value corresponding . is an asymptote because . (p,t) corresponds to the infinity since the eye is looking at the point vertically up and parallel to the projected line. (n,t) when (n,t) projected to the horizon, at the height of the eye, y=q Symmetry Over Over (p,q) reason By the congruency of the triangles Location * the 1st and the 3rd quadrant when >0 respect to * the 2nd and the 4th quadrant when <0 respect to * the 1st and 3rd quadrant when (t-q)(s-p)>0 respect to the eye * the 2nd and 4th quadrant when (t-q)(s-p)<0 respect to the eye reason Thus, if there is a curve on the 1st quadrant, there must be the symmetry curve on the 3rd quadrant by the property of the odd functions. When the rectangle of the base (s-p) and the height (t-q) are on the 1st quadrant respect to the eye, the projected points are on the rectangle. Therefore, the ordered pairs are on the 1st quadrant. This counts for every quadrant. Shape The bigger is, the farther the curves from . The bigger \|(t-q)(s-p)\|is, the farther the curves are from the eye. reason Since is multiplied to , f(x) is also multiplied by . When the blue rectangle of the area \|(t-q)(s-p)\|, the projected points on x=s is spread out. Graph movement horizontally k and vertically m Substitute (x-k) for x, and (y-m) for y Move the eye (p,q) horizontally k, vertically m with preserving the distance between the eye and the lines. As we see in the table above, understanding rational function projective geometrically makes more sense than algebraic way. Personal opinion I first learned projective geometry this year, and I was fascinated because I could find the other way to understand the limits. Although I could solve limit problems, it is hard to understand the concept of limits. I remember I just accepted what teachers said, and I also repeated the same thing to my own students. Students learn projective geometry very early in art class. When they draw or paint the picture, they first figure out where the eye is looking at the place that they want to put on the canvas. In addition, they also know where the horizon., and the point of infinity. They already experience the perspective view. Thus, projective geometry is not totally new thing to the students. Rather, it is more familiar than Euclidean geometry in the sense that we always see things in a perspective view. Understanding rational functions in projective geometry way enable us to supply visual reasons about the asymptotes and the concepts of infinity. Some might think projective geometry could be to hard to the students simply because they did not learn it in high school or simply because it is not in the curriculum. When I taught rational functions to the students, I knew that there are so many students who are embarrassed by the strange looking curves. If we teach projective geometry and help them understand rational functions in this way, I think students can understand better since projective geometry is on the extension of common sense. Reference The four pillars geometry by John Stillwel RETURN
# NCERT Solutions Class 10 Maths Chapter 1 : Real Numbers ### Exercise 1.1 1. Use Euclid's division algorithm to find the HCF of: 1. 135 and 225 2. 196 and 38220 3. 867 and 255 Solution I. Since 225 > 135, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=225 and b=135. 225 = 135 × 1 + 90 where r=90 Since r≠0, we will re-apply the algorithm on b = 135, and r = 90 135 = 90×1 + 45 where r1=45 Since r1≠0, we will re-apply the algorithm on b1=90 and r1=45 90 = 45×2 + 0 where r2=0 We have r2=0 Therefore, the HCF of 135 and 225 is 45. II. Since 38220 > 196, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=38220 and b=196. 38220 = 196×195 + 0 We have r=0 Therefore, the HCF of 196 and 38220 is 196. III. Since 867 > 255, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=867 and b=255. 867 = 255 × 3 + 102 where r=102 Since r≠0, we will re-apply the algorithm on b = 255, and r = 102 255 = 102×2 + 51 where r1=51 Since r1≠0, we will re-apply the algorithm on b1=102 and r1=51 102 = 51×2 + 0 where r2=0 We have r2=0 Therefore, the HCF of 867 and 255 is 51. 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. Solution Let there be an odd positive integer a, such that a=6q + r; 0≤ r< 6 (By Euclid's Division Lemma) where q is the quotient and r is the remainder. Since 0≤ r< 6, we can put 0, 1, 2, 3, 4, or 5 in place of r. For each of the cases we get the following: r=0 r=1 r=2 r=3 r=4 r=5 a=6q +0 a=6q +1 a=6q +2 a=6q +3 a=6q +4 a=6q +5 As given previously, a is an odd positive integer. Hence, r cannot be equal to 2 or 4 because a=6q, a=6q+2 and a=6q+4 present a as an even positive integer which is false. Hence, an odd positive integer can be only of the forms 6q+1, 6q+3 or 6q+5. 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution It is clear from the problem that we need to find the maximum number of columns, which is the HCF of 616 and 32. Since 616 > 32, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=867 and b=32. 616 = 32 × 19 + 8 where r=8 Since r≠0, we will re-apply the algorithm on b = 32, and r = 8 32 = 8×4 + 0 where r1=0 We have r1=0 Therefore, the HCF of 616 and 32 is 8. Hence, the maximum number of columns the army contingent of 616 members can march behind band of 32 members is 8. 4. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Solution Let there be a positive integer a. After applying Euclid's division lemma with b=3 we get a=3q + r; 0≤ r< 3 Therefore, a can be of the forms a=3q, a=3q+1 or a=3q+2 Squaring both sides in each equation we get a2=9q2 a2=3(3q2) a2=3m; where m=3q2 a2=9q2+1+6q a2=3(3q2 +2q) +1 a2=3m+1; where m=3q2 +2q a2=9q2 +4+12q a2=3(3q2+4q+1) +1 a2=3m+1; where m=3q2+4q+1 Hence, square of any positive integer is always of the form 3m or 3m+1. 5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Solution Let a be positive integer. After applying Euclid's Division Lemma with b=3 we get a=3q+r; 0≤ r< 3 Therefore, a can be of the forms a=3q, a=3q+1 or a=3q+2 Cubing both sides of each equation we get a3=27q3 a3=9(3q3) a3=9m where m=3q3 a3=27q3 + 1+ 27q3+ 9q a3=9(3q3+3q2 + q) +1 a3=9m+1 where m=3q3+3q2 + q a3=27q3+ 8+ 54q3+36q a3=9(3q3+6q2 + 4q) +8 a3=9m +8 where m=3q3+6q2 + 4q Hence, cube of any positive integer is always of the form 9m, 9m+1, or 9m+8. ### Exercise 1.2 1. Express each number as a product of its prime factors: 1. 140 2. 156 3. 3825 4. 5005 5. 7429 I. By applying prime factorization, II. By applying prime factorization, III. By applying prime factorization, IV. By applying prime factorization, V. By applying prime factorization, 2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 1. 26 and 91 2. 510 and 92 3. 336 and 54 Solution I. Apply prime factorization to find LCM and HCF 26 = 2 × 13 And 91 = 7 × 13 Therefore, HCF(26, 91)=13 and LCM(26, 91) = 2 × 7 × 13 = 182. Product of both numbers = 26 × 91 = 2366 LCM × HCF = 182 × 13= 2366. Hence, product of the two numbers has been verified to be equal to LCM × HCF. II. Apply prime factorization to find LCM and HCF 510 = 2 × 3 × 5 × 17 And 92 = 2 × 2 × 23 Therefore, HCF(510, 92) = 2 and LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460. Product of both numbers = 510 × 92 = 46920 LCM × HCF = 23460 × 2 = 46920. Hence, product of the two numbers has been verified to be equal to LCM × HCF. III. Apply prime factorization to find LCM and HCF 336 = 2 × 2 × 2 × 2 × 3 × 7 And 54 = 2 × 3 × 3 × 3 Therefore, HCF(336, 54) = 2 × 3 = 6 and LCM(336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024. Product of both numbers = 336 × 54 = 18144 LCM × HCF = 3024 × 6 = 18144. Hence, product of the two numbers has been verified to be equal to LCM × HCF. 3. Find the LCM and HCF of the following integers by applying the prime factorization method. 1. 12, 15, and 21 2. 17, 23 and 29 3. 8, 9 and 25 Solution I. By performing prime factorization, we get 12 = 2 × 2 × 3 15 = 3 × 5 21 = 3 × 7 HCF (12, 15, 21) = 3 LCM (12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420 II. By performing prime factorization, we get 17 = 17 × 1 23 = 23 × 1 29 = 29 × 1 HCF (17, 23, 29) = 1 LCM (17, 23, 29) = 17 × 23 × 29 = 11339 III. By performing prime factorization, we get 8 = 2 × 2 × 2 9 = 3 × 3 × 3 25 = 5 × 5 HCF (8, 9, 25) = 1 LCM (8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800 4. Given that HCF (306, 657) = 9, find LCM (306, 657). Solution Since, it is known that Product of two given numbers = LCM × HCF Therefore, 306 × 657 = LCM × 9 Solving the above equation, we get LCM = 22338 Hence, LCM (306, 657) = 22338 5. Check whether 6n can end with the digit 0 for any natural number n. Solution 6n will end with a 0 if and only if 5 is one of the prime factors of 6. By using prime factorization, we have 6 = 2 × 3 Since, 5 is not a prime factor of 6, 6n can never end with the digit 0. 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Solution Let 7 × 11 × 13 + 13 = a and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = b. a = 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 2 × 3 × 13 (Prime Factorization of 78) Since a can be expressed as a product of primes, it is a composite number. b = 7 × 6 × 5 × 4 × 3 × 2 + 5 = 5 × (7 × 6 × 1 × 4 × 3 × 2 + 1) =5 × (1008 + 1) = 5 × 1009 Since b can be expressed as a product of primes, it is a composite number. Hence, 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are both composite numbers. 7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Solution Sonia takes 18 minutes while Ravi takes 12 minutes. In order to find the time elapsed (in minutes) when they meet again at the starting point, we need to calculate the LCM of 18 and 12 (time taken by each of them to complete a round). By Prime factorization, 18 = 2 × 3 × 3 12 = 2 × 2 × 3 LCM = 2 × 2 × 3 × 3 = 36 Hence, it will take them 36 minutes to meet again at the starting point. ### Exercise 1.3 1. Prove that √5 is irrational. Solution Let us assume that √5 is a rational. Since, √5 is rational, there exists two coprime integers a and b; b ≠ 0 such that √5= a/b b√5= a Squaring both sides 5b2 = a2 … (I) From the equation, it can be deduced that a2 is divisible by 5, which further implies that a is also divisible by 5. Therefore, we can write a = 5c where c is an integer. If we substitute this value of a in (I), we get 5b2 = 25c2 b2 = 5c2 From the equation, it can be deduced that b2 is divisible by 5, which further implies that b is also divisible by 5. Hence, a and b are both divisible by 5. This contradicts the fact that a and b are coprime integers as they both have 5 as a common factor. The contradiction has arisen due to our wrong assumption that √5 is rational. Hence, √5 is irrational. 2. Prove that 3 + 2√5 is irrational. Solution Let us assume that 3 + 2√5 is rational. Since, 3 + 2√5 is rational, there exists two coprime integers a and b; b ≠ 0 such that But √5 is irrational because of the following proof: #### (NOTE: It is not necessary for students to add the following proof when answering the questions unless the question awards more than 3 marks) Let us assume that √5 is a rational. Since, √5 is rational, there exists two coprime integers a and b; b ≠ 0 such that √5= a/b b√5= a Squaring both sides 5b2 = a2 … (I) From the equation, it can be deduced that a2 is divisible by 5, which further implies that a is also divisible by 5. Therefore, we can write a = 5c where c is an integer. If we substitute this value of a in (I), we get 5b2 = 25c2 b2 = 5c2 From the equation, it can be deduced that b2 is divisible by 5, which further implies that b is also divisible by 5. Hence, a and b are both divisible by 5. This contradicts the fact that a and b are coprime integers as they both have 5 as a common factor. The contradiction has arisen due to our wrong assumption that √5 is rational. Hence, √5 is irrational. From the above proof it is clear that √5 is irrational. This contradicts the assumption that 3 + 2√5 is rational. Hence, 3 + 2√5 is irrational. 3. Prove that the following are irrationals: 1. 1/√2 2. 7√5 3. 6 + √2 Solution I. Let us assume that √2 is rational. Since, √2 is rational, there exist two coprimes a and b; b ≠ 0 such that √2 = a/b. √2b = a Squaring both sides, we get 2b2 = a2 …(I) We can deduce that a2 is divisible by 2, hence a is also divisible by 2. Therefore, a = 2c for some integer c. Substituting value of a in (I) we get 2b2 = 4c2 b2 = 2c2 This implies that b2 is divisible by 2, hence b is also divisible by 2. Therefore, a and b have 2 as a common factor. This contradicts the fact that a and b are coprimes. The contradiction has arisen due to our wrong assumption that √2 is rational. Hence, √2 is irrational. This implies that 1/√2 is also irrational. II. Let us assume that √5 is a rational. Since, √5 is rational, there exists two coprime integers a and b; b ≠ 0 such that √5= a/b b√5= a Squaring both sides 5b2 = a2 … (I) From the equation, it can be deduced that a2 is divisible by 5, which further implies that a is also divisible by 5. Therefore, we can write a = 5c where c is an integer. If we substitute this value of a in (I), we get 5b2 = 25c2 b2 = 5c2 From the equation, it can be deduced that b2 is divisible by 5, which further implies that b is also divisible by 5. Hence, a and b are both divisible by 5. This contradicts the fact that a and b are coprime integers as they both have 5 as a common factor. The contradiction has arisen due to our wrong assumption that √5 is rational. Hence, √5 is irrational. This implies that 7√5 is also irrational. III. Let us assume that 6 + √2 is a rational. Since, 6 + √2 is rational, there exists two coprime integers a and b; b ≠ 0 such that 6 + √2 = a/b. Therefore, √2 is also rational. But √2 is irrational because of the following proof: #### (NOTE: It is not necessary for students to add the following proof when answering the questions unless the question awards more than 3 marks) Let us assume that √2 is a rational. Since, √2 is rational, there exists two coprime integers a and b; b ≠ 0 such that √2= a/b b√2= a Squaring both sides 2b2 = a2 … (I) From the equation, it can be deduced that a2 is divisible by 2, which further implies that a is also divisible by 2. Therefore, we can write a = 2c where c is an integer. If we substitute this value of a in (I), we get 2b2 = 4c2 b2 = 2c2 From the equation, it can be deduced that b2 is divisible by 2, which further implies that b is also divisible by 2. Hence, a and b are both divisible by 2. This contradicts the fact that a and b are coprime integers as they both have 2 as a common factor. The contradiction has arisen due to our wrong assumption that √2 is rational. Hence, √2 is irrational. From the above proof it is clear that √2 is irrational. This contradicts the assumption that 6 + √2 is rational. Hence, 6 + √2 is irrational. ### Exercise 1.4 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 1. 13/3125 2. 17/8 3. 64/455 4. 15/1600 5. 29/343 6. 23/2352 7. 129/225775 8. 6/15 9. 35/50 10. 77/210 Solution Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a terminating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a non-terminating repeating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a non-terminating repeating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a non-terminating repeating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a terminating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2n5m. Therefore, the given number has a non-terminating repeating decimal expansion. 2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. Solution I, II, IV, VI, VIII, IX in question 1 have a terminating decimal expansion. These can be calculated as follows: Multiply and divide by 53 Multiply and divide by 54 Multiply and divide by 2 3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q what can you say about the prime factors of q? 1. 43.123456789 2. 0.120120012000120000. . . 3. 43.123456789 Solution I. Since, the given decimal expansion is terminating, it is a rational. The denominator q can be expressed in the form of 2n5m. II. Since, the given decimal expansion is non-terminating non-repeating, it is irrational and hence the denominator cannot be represented as 2n5m. III. Since, the given decimal expansion is non-terminating repeating, it is rational but the denominator cannot be expressed in the form of 2n5m.
# How do you evaluate the limit (sqrtx-2)/(x-4) as x approaches 4? Nov 23, 2016 ${\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}$ #### Explanation: The denominator can be viewed as the difference of two squares, so we can write: ${\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{{\sqrt{x}}^{2} - {2}^{2}}$ $\therefore {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{\left(\sqrt{x} - 2\right) \left(\sqrt{x} + 2\right)}$ $\therefore {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = {\lim}_{x \rightarrow 4} \frac{1}{\left(\sqrt{x} + 2\right)}$ $\therefore {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{\left(\sqrt{4} + 2\right)}$ $\therefore {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{\left(2 + 2\right)}$ $\therefore {\lim}_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}$ This is confirmed visually from the graph $y = \frac{\sqrt{x} - 2}{x - 4}$ graph{(sqrt(x)-2)/(x-4) [-0.166, 4.835, -1.01, 1.49]}
# AP Statistics Curriculum 2007 Distrib Dists ## General Advance-Placement (AP) Statistics Curriculum - Geometric, HyperGeometric, Negative Binomial Random Variables and Experiments ### Geometric • Definition: The geometric distribution is the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set {1, 2, 3, ...}. • Mass function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is $$P(X = x) = (1 - p)^{x-1} \times p$$, for x = 1, 2, 3, 4,.... • Expectation: The expected value of a geometrically distributed random variable X is $${1\over p}.$$ • Variance: The variance is $${1-p\over p^2}.$$ ### HyperGeometric The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. An experimental design for using Hypergeometric distribution is illustrated in this table: Type Drawn Not-Drawn Total Defective k m-k m Non-Defective n-k N+k-n-m N-m Total n N-n N • Explanation: Suppose there is a shipment of N objects in which m are defective. The hypergeometric distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective. • Mass function: The random variable X follows the hypergeometric distribution with parameters N, m and n, then the probability of getting exactly k successes is given by $P(X=k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}.$ This formula for the hypergeometric mass function may be interpreted as follows: There are $${{N}\choose{n}}$$ possible samples (without replacement). There are $${{m}\choose{k}}$$ ways to obtain k defective objects and there are $${{N-m}\choose{n-k}}$$ ways to fill out the rest of the sample with non-defective objects. #### Examples • SOCR Activity: The SOCR Ball and Urn Experiment provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can be varied with scroll bars. • A lake contains 1,000 fish; 100 are randomly caught and tagged. Suppose that later we catch 20 fish. Use SOCR Hypergeometric Distribution to: • Compute the probability mass function of the number of tagged fish in the sample of 20. • Compute the expected value and the variance of the number of tagged fish in this sample. • Compute the probability that this random sample contains more than 3 tagged fish. ### Negative Binomial • Definition: The family of negative binomial distributions is a two-parameter family; p and r with 0 < p < 1 and r > 0. • Mass Function: The probability mass function of a Negative Binomial random variable (X~NegBin(r, p))is give by: $P(X=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k \!$, for k = 0,1,2,.... #### Application Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and 70% chance that she'll fail. • What's the probability mass function that the last 6th state where she succeeds to secure all electoral votes happens to be the at the nth state she campaigns in? NegBin(r, p) distribution describes the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is n=k+6. The random variable we are interested in is X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}. So, k = n-6, and $$X\sim NegBin(6, 0.3)$$. Thus, for $$n \geq 6$$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is: $P(X=n) = {(n-6) + 6 - 1 \choose 6-1} \; 0.3^6 \; 0.7^{n-6} = {n-1 \choose 5} \; \frac{3^6 \; 7^{n-6} \; }{10^n}$ • What's the probability that Jane finishes her campaign in the 10th state? $P(X=10) = 0.022054$ • What's the probability that Jane finishes campaigning on or before reaching the 8th state? $P(X\leq 8) = 0.011292$ Error creating thumbnail: File missing • Suppose the success of getting all electoral votes within a state is reduced to only 10%, then X~NegBin(r=6, p=0.1). Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below)! What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)? $P(X\geq 50) = 0.632391$
# How do you write (-28)^(7/5) in radical form? Jun 6, 2017 ${\left(\sqrt[5]{- 28}\right)}^{7}$ #### Explanation: ${m}^{\frac{1}{n}} \equiv \sqrt[n]{m}$ ${\left(- 28\right)}^{\frac{5}{7}}$ $= {\left({\left(- 28\right)}^{\frac{1}{5}}\right)}^{7}$ $= {\left(\sqrt[5]{- 28}\right)}^{7}$ Jun 6, 2017 See a solution process below: #### Explanation: We can rewrite this expression as: ${\left(- 28\right)}^{7 \times \frac{1}{5}}$ We can now use this rule of exponents to rewrite the expression again: ${x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}} = {\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}}$ ${\left(- 28\right)}^{\textcolor{red}{7} \times \textcolor{b l u e}{\frac{1}{5}}} = {\left(- {28}^{\textcolor{red}{7}}\right)}^{\textcolor{b l u e}{\frac{1}{5}}}$ We can now use this rule for exponents and radicals to put the expression into radical form: ${x}^{\frac{1}{\textcolor{red}{n}}} = \sqrt[\textcolor{red}{n}]{x}$ ${\left(- {28}^{7}\right)}^{\frac{1}{\textcolor{red}{5}}} = \sqrt[\textcolor{red}{5}]{- {28}^{7}}$
PUMPA - THE SMART LEARNING APP Helps you to prepare for any school test or exam The inverse operation of a cube is cube root. The symbol used to represent the cube root is $$\sqrt[3]{}$$. A cube root is a unique value that gives us the original number when we multiply itself by three times. The cube root of $$a$$ is denoted by $$\sqrt[3]{a}$$ or $$a^{\frac{1}{3}}$$. Example: Find the cube root of $$64$$. Solution: $$\sqrt[3]{64} = \sqrt[3]{4 \times 4 \times 4}$$ $$= \sqrt[3]{4^3}$$ $$= 4$$ Thereforethe cube root of $$64$$ is $$4$$. By the observation of the above example, we can conclude that: The cube of $$4$$ is $$64$$. The cube root of $$64$$ is $$4$$. The following table consist of cube and cube roots of the first $$20$$ numbers. Number Cube number Cube root Number Cube number Cube root 1 $$1^3 = 1$$ $$\sqrt[3]{1} = 1$$ 11 $$11^3 = 1331$$ $$\sqrt[3]{1331} = 11$$ 2 $$2^3 = 8$$ $$\sqrt[3]{8} = 2$$ 12 $$12^3 = 1728$$ $$\sqrt[3]{1728} = 12$$ 3 $$3^3 = 27$$ $$\sqrt[3]{27} = 3$$ 13 $$13^3 = 2197$$ $$\sqrt[3]{2197} = 13$$ 4 $$4^3 = 64$$ $$\sqrt[3]{64} = 4$$ 14 $$14^3 = 2744$$ $$\sqrt[3]{2744} = 14$$ 5 $$5^3 = 125$$ $$\sqrt[3]{125} = 5$$ 15 $$15^3 = 3375$$ $$\sqrt[3]{3375} = 15$$ 6 $$6^3 = 216$$ $$\sqrt[3]{216} = 6$$ 16 $$16^3 = 4096$$ $$\sqrt[3]{4096} = 16$$ 7 $$7^3 = 343$$ $$\sqrt[3]{343} = 7$$ 17 $$17^3 = 4913$$ $$\sqrt[3]{4913} = 17$$ 8 $$8^3 = 512$$ $$\sqrt[3]{512} = 8$$ 18 $$18^3 = 5832$$ $$\sqrt[3]{5832} = 18$$ 9 $$9^3 = 729$$ $$\sqrt[3]{729} = 9$$ 19 $$19^3 = 6859$$ $$\sqrt[3]{6859} = 19$$ 10 $$10^3 = 1000$$ $$\sqrt[3]{1000} = 10$$ 20 $$20^3 = 8000$$ $$\sqrt[3]{8000} = 20$$
# GED Math Practice Test 1 This is the first of our free GED Math practice tests, and it has been fully updated to reflect the new GED Test that was rolled out in 2014. To prepare for your GED Math test you will want to work through as many practice questions as possible. After you answer each question, the correct answer will be provided along with a very detailed explanation. Get started on your test prep right now with this GED Math practice test. Directions: Solve each problem and select the best answer. You may use your calculator. You may also refer to the GED Formula Sheet as needed. Congratulations - you have completed . You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%% Question 1 Devon owns a house cleaning company and has to give price quotes to potential customers. He figures out his price by assuming a \$25 base charge and then adding \$8 for each bathroom and \$4 for each other room. If he uses P to represent the price, B for bathroom, and R for other rooms, which of the following represents his price quote formula? A P = 25 + 12(BR) B P = 25 + 8B + 4R C P = 25(4R + 8B) D P = (4)(8)(R + B) + 25 Question 1 Explanation: The price quote, P, will be equal to the total amount of charges. It’s given that there is a \$25 base charge, so we can begin by writing the price quote as P = \$25. To this base charge of \$25, \$8 per each bathroom, represented by the variable B, is added; our price quote is now P = \$25 + \$8B. Lastly, \$4 per each other room, R, is added, and the actual price quote is P = \$25 + \$8B + \$4R, or P = 25 + 8B + 4R. Question 2 What are the coordinates of the point shown on this coordinate plane? A (5, −5) B (4, −5) C (−5, 4) D (5, −4) Question 2 Explanation: The coordinate plane is formed by a horizontal x-axis and a vertical y-axis that intersect at the origin, represented as (0, 0). Points on the plane have both an x coordinate and a y coordinate and are written in the form (x, y). Negative x values are to the left of the origin, positive x values are to the right of the origin; negative y values are below the origin, and positive y values are above the origin. In this case, the point is to the left and above the origin. The only answer choice containing a negative x coordinate must be correct. Question 3 Given ƒ(x) = 2x2 − 3x + 7, find ƒ(2.5) A 5 B 12 C 19 D 24.5 Question 3 Explanation: This is a function problem. A function relates an input to an output. To find ƒ(2.5), substitute 2.5 for every x in the original function and evaluate the expression: ƒ(2.5) = 2(2.5)2 − 3(2.5) +7 ƒ(2.5) = 2(6.25) − 7.5 + 7 ƒ(2.5) = 12.5 − 7.5 + 7 ƒ(2.5) = 12 Question 4 Solve 5x + 8 < 3(x + 2) A x < −1 B x < −3 C x > 1 D x > −¾ Question 4 Explanation: In order to solve inequalities in one variable, follow the order of operations (PEMDAS) to simplify the expressions, then combine like terms, then isolate the variable. Inequalities can be treated much like equations, but with one major difference: when multiplying or dividing either side of an inequality by a negative value, the direction of the inequality symbol must switch. In this case, begin by multiplying the 3 through the parentheses, before then isolating the variable: 5x + 8 < 3(x + 2) 5x + 8 < 3x + 6 2x + 8 < 6 2x < −2 x < −1 Question 5 Simplify (x6)(x5) A 2x11 B 2x30 C x11 D x30 Question 5 Explanation: Recall that the multiplication of terms sharing the same base can be simplified by adding the exponents: xa * xb = x(a + b). If this rule is forgotten, you can quickly derive it by considering the simple case of 21 * 22, which can be rewritten as 2 * 2 * 2, or 23. The expression in this case, then, can be simplified as: (x6) (x5) = x5+6 = x11 Question 6 The area of the circle shown below is 100π. What is the diameter (D) of the circle? A 10 B 20 C 30 D 40 Question 6 Explanation: Begin by writing out the formula for the area of a circle, which is one of the formulas provided on the GED Formula Sheet. Next, substitute the given value for the area, and then solve for the radius. Recall that the square root of both sides must be evaluated in order to isolate r. Multiplying the value of r by 2 gives the diameter: Question 7 Max struggled with his math class early in the year, but he has been working hard to improve his scores. There is one test left, and he is hoping that his final average test score will be 75. What score will he need to get on Test 6 to finish the year with an average score of 75? A 75 B 85 C 92 D 100 Question 7 Explanation: The average of a set of data points is the sum of the data points divided by the total number of data points. In this case, we are given 5 out of 6 data points, the number of data points, and the desired average. Substitute the given values into the formula for the average, using a variable to represent the unknown test score, and then solve for the variable: Average = sum of data points ÷ number of data points 75 = (50 + 52 + 77 + 88 + 91 + x) ÷ 6 75 = (358 + x) ÷ 6 Next, eliminate the denominator by multiplying both sides by 6: 450 = 358 + x The last step is to subtract 358 from both sides: x = 92 Question 8 If a = −4, and b = 3, what is \|a − b\|? A 1 B 3 C 4 D 7 Question 8 Explanation: The vertical bars surrounding ab designate the absolute value of the difference. The absolute value of an expression is the distance from the simplified expression to 0; absolute value is always positive. Substitute the given values of a and b into the expression, evaluate the difference, and then apply the absolute value to make it positive. \|a − b\| = \|−4 − 3\| = \|−7\| = 7 Recall that −4 − 3 is equivalent to −4 + −3 = −7 Question 9 Which of the following graphs represents the equation y = 2x + 1? A B C D Question 9 Explanation: An equation of the form y = ax + b, where a and b are constants, designates a straight line and is known as the slope-intercept form of the line because it explicitly provides both the slope and y-intercept. The variable a represents the slope of the line, which can be negative, zero, positive, or undefined; the variable b represents the y-intercept, or the point along the y-axis at which the line crosses. In this case, the slope of the line, also known as the “rise over run”, is positive 2, indicating that the line rises from left to right. The y-intercept is positive 1, indicating that the line passes through the point (0, 1). Lines can always be easily graphed by forming an (x, y) table, substituting values of either x or y, and solving for the other variable. Once two points on the line are known, the line can be accurately graphed. Question 10 What is the probability of selecting a male from a group of 4 males and 8 females? A B C D Question 10 Explanation: The probability of an event taking place is the ratio of the number of successful events to the total number of possible events. In this case, a successful event is randomly selecting a male from the group. The total number of possible events is the number of people in the group. The ratio is then: Once you are finished, click the button below. Any items you have not completed will be marked incorrect. There are 10 questions to complete. ← List →
What is 1/458 as a decimal? Solution and how to convert 1 / 458 into a decimal 1 / 458 = 0.002 The basis of converting 1/458 to a decimal begins understanding why the fraction should be handled as a decimal. Both are used to handle numbers less than one or between whole numbers, known as integers. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals. 1/458 is 1 divided by 458 The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (1) by the denominator (458). Here's 1/458 as our equation: Numerator: 1 • Numerators represent the number of parts being taken from a denominator. Small values like 1 means there are less parts to divide into the denominator. The bad news is that it's an odd number which makes it harder to covert in your head. Smaller numerators doesn't mean easier conversions. Now let's explore the denominator of the fraction. Denominator: 458 • Denominators represent the total number of parts, located below the vinculum or fraction bar. Larger values over fifty like 458 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Ultimately, don't be afraid of double-digit denominators. Let's start converting! Converting 1/458 to 0.002 Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 458 \enclose{longdiv}{ 1 }$$ We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once. Step 2: Extend your division problem $$\require{enclose} 00. \\ 458 \enclose{longdiv}{ 1.0 }$$ Because 458 into 1 will equal less than one, we can’t divide less than a whole number. Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 458 into 1 + 0 or 10. Step 3: Solve for how many whole groups you can divide 458 into 10 $$\require{enclose} 00.0 \\ 458 \enclose{longdiv}{ 1.0 }$$ Now that we've extended the equation, we can divide 458 into 10 and return our first potential solution! Multiple this number by our furthest left number, 458, (remember, left-to-right long division) to get our first number to our conversion. Step 4: Subtract the remainder $$\require{enclose} 00.0 \\ 458 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$ If you hit a remainder of zero, the equation is done and you have your decimal conversion. If there is a remainder, extend 458 again and pull down the zero Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 1/458 fraction into a decimal is long division just as you learned in school. Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 1/458 and 0.002 bring clarity and value to numbers in every day life. Here are examples of when we should use each. When you should convert 1/458 into a decimal Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions. When to convert 0.002 to 1/458 as a fraction Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct. Practice Decimal Conversion with your Classroom • If 1/458 = 0.002 what would it be as a percentage? • What is 1 + 1/458 in decimal form? • What is 1 - 1/458 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.002 + 1/2?
# How do you evaluate (2x ^ { 5} ) ( - 3x ^ { 9} ) ( 9x ^ { 9} )? Nov 29, 2017 you should be able to do it in your head - it's not difficult: #### Explanation: 1. Multiply the constants 2, -3, and 9. (-54) 2. Multiply the powers of 'x' terms, by writing x, raised to a power equal to the SUM of the powers of the original terms. This is ${x}^{5 + 9 + 9} = {x}^{23}$ 1. Put it all together: $- 54 {x}^{23}$ GOOD LUCK Nov 29, 2017 See a solution process below: #### Explanation: First, rewrite the expression as: $\left(2 \times - 3 \times 9\right) \left({x}^{5} \times {x}^{9} \times {x}^{9}\right) \implies$ $- 54 \left({x}^{5} \times {x}^{9} \times {x}^{9}\right)$ Now, use this rule for exponents to multiply the $x$ terms: ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$ $- 54 \left({x}^{\textcolor{red}{5}} \times {x}^{\textcolor{b l u e}{9}} \times {x}^{\textcolor{g r e e n}{9}}\right) \implies$ $- 54 {x}^{\textcolor{red}{5} + \textcolor{b l u e}{9} + \textcolor{g r e e n}{9}} \implies$ $- 54 {x}^{23}$
Which interval notation represents the set of all numbers greater than and less than? Which interval notation represents the set of all numbers greater than and less than? Consider a variable x and two numbers A and B. Using the rules listed above the interval notation that represents the set of all numbers greater than or equal to 5 and less than 12 is option 1. The correct interval notation is [5, 12) which is option 1. What is the set of all real numbers greater than or equal to? You can also say, “The domain is the set of all real numbers, and the range is the set of all real numbers greater than or equal to negative two.” Sometimes you have restrictions because of a square root. How do you write a set of real numbers in interval notation? The interval “all real numbers greater than −5” is written as (−5,∞), and “all real numbers less than or equal to 7” is written as (−∞,7]. This does not mean that ∞ is a number; it is just a convenient shorthand. How do you write all real numbers greater than 0? {x \| x ≥ 0}. If the domain of a function is all real numbers (i.e. there are no restrictions on x), you can simply state the domain as, ‘all real numbers,’ or use the symbol to represent all real numbers. How do you write greater than or equal to in interval notation? With interval notation, we use use round parentheses, ( or ). With inequalities, we use “less than or equal to”: ≤ or “greater than or equal to”: ≥ to include the endpoint of the interval. With interval notation, we use use square brackets, [ or ]. Enter DNE for an empty set. What is the symbol for all real numbers? symbol R Symbol of Real Numbers Since the set of real numbers is the collection of all rational and irrational numbers, real numbers are represented by the symbol R. What does interval notation look like? Intervals are written with rectangular brackets or parentheses, and two numbers delimited with a comma. The two numbers are called the endpoints of the interval. The number on the left denotes the least element or lower bound. The number on the right denotes the greatest element or upper bound. How do you show all real numbers in set notation? *Note that “the set of all real numbers” can be written as a script upper case R. In handwriting we usually make a double line in the left down stroke of the R to indicate this. How do you write a set of all real numbers? Is 3 less than or equal to 3? Inequalities x≠3 x is not equal to 3 x<3 x is less than 3 x>3 x is greater than 3 x≤3 x is less than or equal to 3 x≥3 x is greater than or equal to 3
# How Do You Describe A Set In Math? ## What are the 3 ways to describe a set? There are three main ways to identify a set: • A written description, • List or Roster method, • Set builder Notation, ## How do you represent a set in math? Sets, in mathematics, are an organized collection of objects and can be represented in set -builder form or roster form. Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set. ## What do you call the method of describing a set in words? Answer: A method of describing a set in words is called Rule Method. Step-by-step explanation: A set can be described by writing a description of its elements. ## What is the best description of set? A set is a group or collection of objects or numbers, considered as an entity unto itself. Sets are usually symbolized by uppercase, italicized, boldface letters such as A, B, S, or Z. Each object or number in a set is called a member or element of the set. ## How do you describe set? A set in mathematics is a collection of well defined and distinct objects, considered as an object in its own right. The most basic properties are that a set “has” elements, and that two sets are equal (one and the same) if and only if every element of one is an element of the other. You might be interested: Question: What Is A Converse Statement In Math? ## What is proper set example? A proper subset of a set A is a subset of A that is not equal to A. In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not in B. For example, if A={1,3,5} then B={1,5} is a proper subset of A. ## How do you write the elements of a set? Generally, the elements of a set are written inside a pair of curly (idle) braces and are represented by commas. The name of the set is always written in capital letter. Here ‘A’ is the name of the set whose elements (members) are v, w, x, y, z. ## What is rule method? (2) Set – builder method or Rule method: In this method, a set is described by a characterizing property P(x) of its elements x. In such a case the set is described by {x: P(x) holds} or {x \| P(x) holds}, which is read as ‘the set of all x such that P(x) holds’. The symbol ‘\|’ or ‘:’ is read as ‘such that’. ## How do you name a set? The objects in the set are called its elements. Set notation uses curly braces, with elements separated by commas. The following conventions are used with sets: 1. Capital letters are used to denote sets. 2. Lowercase letters are used to denote elements of sets. 3. Curly braces { } denote a list of elements in a set. ## How many ways can you name a set? Answer. There are three ways to represent a set. ## How do you describe a set example? Two methods of describing sets are the roster method and set -builder notation. Example: B = {1, 2, 3, 4, 5} Example: C = {x\| x ∈ N where x > 4} Example: Write B = {1, 4, 9, 16, …} in set builder notation. Example: Given C = {2x \| x > 10 and x is a natural number}, list the elements of the set. You might be interested: Quick Answer: What Is Prime Factorization In Math? ## What is verbal description method? Verbal Descriptions: a verbal description of a set uses an English sentence to state a rule that allows us to determine the class of objects being discussed and to determine for any particular object whether or not it is in the set. We sometimes name the set using a capital letter. ## What is equivalent set? What are Equivalent Sets? To be equivalent, the sets should have the same cardinality. Definition 2: Two sets A and B are said to be equivalent if they have the same cardinality i.e. n(A) = n(B). In general, we can say, two sets are equivalent to each other if the number of elements in both the sets is equal.
# How to Multiply a Polynomial and a Monomial Learn how to multiply monomials to polynomials using the distributive property and the multiplication property of exponents. ## Step by step guide to Multiplying a Polynomial and a Monomial • When multiplying monomials, use the product rule for exponents. • When multiplying a monomial by a polynomial, use the distributive property. $$\color{blue}{a×(b+c)=a×b+a×c}$$ ### Multiplying a Polynomial and a Monomial – Example 1: Multiply expressions. $$2x(-2x+4)=$$ Solution: Use Distributive Property: $$\color{blue}{a \ (b \ + \ c)=ab \ + \ ac}$$ Then: $$2x(-2x+4)=-4x^2+8x$$ ### Multiplying a Polynomial and a Monomial – Example 2: Multiply expressions. $$-2x(3x^2+4y^2 )=$$ Solution: Use Distributive Property: $$\color{blue}{a \ (b \ + \ c)=ab \ + \ ac}$$ Then: $$-2x(3x^2+4y^2 )=-6x^3-8xy^2$$ ### Multiplying a Polynomial and a Monomial – Example 3: Multiply expressions. $$-4x(5x+9)=$$ Solution: Use Distributive Property: $$-4x(5x+9)=-20x^2-36x$$ ### Multiplying a Polynomial and a Monomial – Example 4: Multiply expressions. $$2x(6x^2-3y^2 )=$$ Solution: Use Distributive Property: $$2x(6x^2-3y^2 )=12x^3-6xy^2$$ ## Exercises for Multiplying a Polynomial and a Monomial ### Find each product. 1. $$\color{blue}{5 (3x – 6y)}$$ 2. $$\color{ blue }{9x (2x + 4y)}$$ 3. $$\color{blue}{8x (7x – 4)}$$ 4. $$\color{blue}{12x (3x + 9)}$$ 5. $$\color{blue}{11x (2x – 11y)}$$ 6. $$\color{blue}{2x (6x – 6y)}$$ 1. $$\color{blue}{15x – 30y}$$ 2. $$\color{blue}{18x^2 + 36xy}$$ 3. $$\color{blue}{56x^2 – 32x}$$ 4. $$\color{blue}{36x^2 + 108x}$$ 5. $$\color{blue}{22x^2 – 121xy}$$ 6. $$\color{blue}{12x^2 – 12xy}$$ 36% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment
# Question 4 and 5, Exercise 10.2 Solutions of Question 4 and 5 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. If $\cos \theta =-\dfrac{3}{7}$ and terminal ray of $\theta$ is in the third quadrant, then find $\sin \dfrac{\theta }{2}$. Given: $\cos \theta =-\dfrac{3}{7}$ and terminal ray of $\theta$ is in the third quadrant, that is, \begin{align}&\pi < \theta < \dfrac{3\pi}{2} \\ \implies &\frac{\pi}{2} < \frac{\theta}{2} < \dfrac{3\pi}{4}\end{align} This gives $\frac{\theta}{2}$ lies in 2nd quadrant. By using the half angle identity: $$\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos \theta }{2}}$$ As $\frac{\theta}{2}$ lies in 2nd quadrant and $\sin$ is positive in 2nd quadrant, therefore \begin{align}\sin\dfrac{\theta }{2}&=-\sqrt{\dfrac{1-\cos \theta }{2}}\\ &=-\sqrt{\dfrac{1-\left( -\dfrac{3}{7} \right)}{2}}\\ &=-\sqrt{\dfrac{1+\dfrac{3}{7}}{2}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin\dfrac{\theta}{2}=-\sqrt{\dfrac{5}{7}}}$$ Use double angle identities to evaluate exactly $\sin \dfrac{2\pi }{3}$. Given: $\sin \dfrac{2\pi }{3}$. By using double angle identities, we have \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ \implies \sin 2\left(\dfrac{\pi }{3}\right)&=2\sin \dfrac{\pi }{3}\cos \dfrac{\pi}{3}\\ &=2\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{2} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin\dfrac{2\pi}{3}={\dfrac{\sqrt{3}}{2}}}$$ Use the double angle identities to evaluate exactly $\cos \dfrac{2\pi }{3}$. Given: $\cos \dfrac{2\pi }{3}$ By using double angle identities, we have \begin{align}\cos 2\theta &=2{{\cos }^{2}}\theta -1\\ \implies \cos 2\left(\dfrac{\pi }{3}\right)&=2{{\cos }^{2}}\dfrac{\pi }{3}-1\\ &=2{{\left( \dfrac{1}{2} \right)}^{2}}-1\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}}$$ • fsc-part1-kpk/sol/unit10/ex10-2-p4
1. Introduction Mathematical formulas are often made more accessible by using parenthesis. However, in computers, expressions with multiple parentheses can be inefficient. So, mathematicians have created different notations such as infix, prefix, and postfix expressions to reduce computational work. In this tutorial, we’ll explore these different ways of writing and evaluating expressions. 2. Infix Expressions Infix expressions are the most usual type of expression. This notation is typically employed when writing arithmetic expressions by hand. Moreover, in the infix expression, we place the operator between the two operands it operates on. For example, the operator “+” appears between the operands A and B in the expression “A + B”. The following figure depicts the example: Furthermore, infix expressions can also include parentheses to indicate the order of operations. In this way, we should observe the operator precedence rules and use parentheses to clarify the order of operations in expressions in infix notation. Operator precedence rules specify the operator evaluation order in an expression. So, in an expression, operators with higher precedence are evaluated before operators with lower precedence. Some operator precedence rules follow: • Parentheses: expressions inside parentheses are evaluated first • Exponentiation: exponents are evaluated next • Multiplication and division: multiplication and division are evaluated before addition and subtraction However, if an expression has multiple operators with the same precedence, the evaluation of those operators occurs from left to right. 3. Prefix Expressions Prefix expressions, also known as Polish notation, place the operator before the operands. For example, in the expression “+ A B”, we place the “+” operator before the operands A and B, as demonstrated in the image next: We should consider that prefix expressions are evaluated from right to left. Thus, we apply each operator to its operands as it is encountered. 4. Postfix Expressions Postfix expressions, also known as reverse Polish notation, where we place the operator after the operands. For instance, in the expression “A B +”, the “+” we place the operator after the operands A and B. The figure next depicts the example: Hence, we can evaluate postfix expressions from left to right, with each operator being applied to its operands as encountered. 5. Comparison of the Expression Notations The infix notation is the simplest notation for humans to read and write, but it requires more complex parsing algorithms for computers due to parentheses and operator precedence rules. The prefix and postfix notations are computationally efficient and do not require parentheses or operator precedence tracking. Furthermore, the prefix notation can easily handle unary operators, while infix and postfix notations require special handling. The infix notation uses parentheses for function arguments, while the prefix and postfix notations can use other delimiters. The infix notation is the most usual notation for writing mathematical expressions, while the prefix and postfix notations are appropriate for particular applications. Examples of these applications are stack-based algorithms and programming languages. 6. Conversion of Infix to Postfix One of the applications of postfix notation is to build a calculator or evaluate expressions in a programming language. In addition, we can evaluate postfix expressions efficiently using a stack data structure. Therefore, postfix notation is effective for implementing algorithms such as postfix notation evaluation and expression parsing. The process of converting an infix expression to a postfix expression involves the following steps: 1. First, we create an empty stack and an empty postfix expression 2. Next, we iterate through the infix expression from left to right and append operands to the postfix expression 3. If an operator is encountered, we pop operators from the stack and append them to the postfix expression until an operator with lower or equal precedence is found 4. The current operator is then pushed onto the stack 5. If a left parenthesis is encountered, we push it onto the stack 6. If a right parenthesis is encountered, we pop operators from the stack and append them to the postfix expression until a left parenthesis is found 7. Finally, we pop any remaining operators from the stack and append them to the postfix expression Considering the previously defined steps, we can convert an infix expression like “5 + 6 * 2 – 3 / 2” into the postfix expression “5 6 2 * + 3 2 / -“. This notation facilitates a computer to evaluate the expression. 7. Conclusion The infix, prefix, and postfix notations are three different ways of writing and evaluating expressions. While infix expressions are common and intuitive for humans to read and write, prefix and postfix notations are computationally efficient and valuable for creating computer programs that manipulate expressions. In particular, we can easily evaluate the postfix expressions with a stack data structure, making them appropriate for implementing some algorithms, such as parsers. Comments are open for 30 days after publishing a post. For any issues past this date, use the Contact form on the site.
# If θ = 30°, verify that Question: If $\theta=30^{\circ}$, verify that (i) $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$ (ii) $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$ (iii) $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ (iv) $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$ Solution: (i) Given:' $\theta=30^{\circ}$....(1) To verify: $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$....(2) Now consider LHS of the expression to be verified in equation (2) Therefore, LHS $=\tan 2 \theta$ Now by substituting the value of $\theta$ from equation (1) in the above expression We get, $\mathrm{LHS}=\tan 2 \times\left(30^{\circ}\right)$ $=\tan 60^{\circ}$ $=\sqrt{3}$ Now by substituting the value of $\theta$ from equation (1) in the expression $\frac{2 \tan \theta}{1-\tan ^{2} \theta}$ We get, $\mathrm{RHS}=\frac{2 \tan \left(30^{\circ}\right)}{1-\tan ^{2}\left(30^{\circ}\right)}$....(4) $\mathrm{RHS}=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}$ $=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1^{2}}{(\sqrt{3})^{2}}}$ $=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}$ $=\sqrt{3}$ Now by comparing equation (3) and (4) We get, $\mathrm{LHS}=\mathrm{RHS}=\sqrt{3}$ Hence $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$ (ii) Given: $\theta=30^{\circ} \ldots \ldots(1)$ To verify: $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$...(2) $\sin 2 \theta=\sin 2 \times 30$ $=\sin 60$ $=\frac{\sqrt{3}}{2}$ Now consider right hand side $\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\frac{2 \tan 30}{1+\tan ^{2} 30}$ $=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$ $=\frac{\sqrt{3}}{2}$ Hence it is verified that, $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$ (iii) Given: $\theta=30^{\circ}$....(1) To verify: $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$....(2) Now consider left hand side of the equation (2) Therefore, $\cos 2 \theta=\cos 2 \times 30$ $=\cos 60$ $=\frac{1}{2}$ Now consider right hand side of equation (2) Therefore, $\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{1-(\tan 30)^{2}}{1+(\tan 30)^{2}}$ $=\frac{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$ $=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}$ $=\frac{1}{2}$ Hence it is verified that, $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ (iv) Given: $\theta=30^{\circ}$...(1) To verify: $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \ldots \ldots(2)$ Now consider left hand side of the expression in equation (2) Therefore $\cos 3 \theta=\cos 3 \times 30$ $=\cos 90$ $=0$ Now consider right hand side of the expression to be verified in equation (2) Therefore, $4 \cos ^{3} \theta-3 \cos \theta=4 \cos ^{3} 30-3 \cos 30$ $=4 \times\left(\frac{\sqrt{3}}{2}\right)^{3}-3 \times \frac{\sqrt{3}}{2}$ $=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}$ $=0$ Hence it is verified that, $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$
# Balance puzzle facts for kids Kids Encyclopedia Facts An animation of a solution to the a false coin problem involving ten coins. In this example, the false coin is lighter than the others. A balance puzzle or weighing puzzle is a logic puzzle about balancing items—often coins—to determine which holds a different value, by using balance scales a limited number of times. These differ from puzzles that assign weights to items, in that only the relative mass of these items is relevant. Known Goal Maximum Coins for n weighings Number of Weighings for c coins Whether target coin is lighter or heavier than others Identify coin $3^n$ $\lceil\log_3(c)\rceil$ Target coin is different from others Identify coin $\frac{3^n-1}2$ $\lceil\log_3(2c+1)\rceil$ Target coin is different from others, or all coins are the same Identify if unique coin exists, and whether it is lighter or heavier $\frac{3^n-1}2-1$ $\lceil\log_3(2c+3)\rceil$ For example, in detecting a dissimilar coin in three weighings (n = 3), the maximum number of coins that can be analyzed is 33 − 12 = 13. Note that with 3 weighs and 13 coins, it is not always possible to determine the identity of the last coin (whether it is heavier or lighter than the rest), but merely that the coin is different. In general, with n weighs, you can determine the identity of a coin if you have 3n − 12 - 1 or less coins. In the case n = 3, you can truly discover the identity of the different coin out of 12 coins. ## Nine-coin problem Solution to the balance puzzle for 9 coins in 2 weighings, where the odd coin is lighter than the others – if the odd coin were heavier than the others, the upper two branches in each weighing decision are swapped A well-known example has up to nine items, say coins (or balls), that are identical in weight except one, which is lighter than the others—a counterfeit (an oddball). The difference is perceptible only by weighing them on scale—but only the coins themselves can be weighed. How can one isolate the counterfeit coin with only two weighings? ### Solution To find a solution, we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one, we can compare any two coins, leaving the third out. If the two coins weigh the same, then the lighter coin must be one of those not on the balance. Otherwise, it is the one indicated as lighter by the balance. Now, imagine the nine coins in three stacks of three coins each. In one move we can find which of the three stacks is lighter (i.e. the one containing the lighter coin). It then takes only one more move to identify the light coin from within that lighter stack. So in two weighings, we can find a single light coin from a set of 3 × 3 = 9. By extension, it would take only three weighings to find the odd light coin among 27 coins, and four weighings to find it from 81 coins. ## Twelve-coin problem A more complex version has twelve coins, eleven or twelve of which are identical. If one is different, we don't know whether it is heavier or lighter than the others. This time the balance may be used three times to determine if there is a unique coin—and if there is, to isolate it and determine its weight relative to the others. (This puzzle and its solution first appeared in an article in 1945.) The problem has a simpler variant with three coins in two weighings, and a more complex variant with 39 coins in four weighings. ### Solution This problem has more than one solution. One is easily scalable to a higher number of coins by using base-three numbering: labeling each coin with a different number of three digits in base three, and positioning at the n-th weighings all the coins that are labeled with the n-th digit identical to the label of the plate (with three plates, one on each side of the scale and one off the scale). Other step-by-step procedures are similar to the following. It is less straightforward for this problem, and the second and third weighings depend on what has happened previously, although that need not be the case (see below). • Four coins are put on each side. There are two possibilities: 1. One side is heavier than the other. If this is the case, remove three coins from the heavier side, move three coins from the lighter side to the heavier side, and place three coins that were not weighed the first time on the lighter side. (Remember which coins are which.) There are three possibilities: 1.a) The same side that was heavier the first time is still heavier. This means that either the coin that stayed there is heavier or that the coin that stayed on the lighter side is lighter. Balancing one of these against one of the other ten coins reveals which of these is true, thus solving the puzzle. 1.b) The side that was heavier the first time is lighter the second time. This means that one of the three coins that went from the lighter side to the heavier side is the light coin. For the third attempt, weigh two of these coins against each other: if one is lighter, it is the unique coin; if they balance, the third coin is the light one. 1.c) Both sides are even. This means that one of the three coins that was removed from the heavier side is the heavy coin. For the third attempt, weigh two of these coins against each other: if one is heavier, it is the unique coin; if they balance, the third coin is the heavy one. 2. Both sides are even. If this is the case, all eight coins are identical and can be set aside. Take the four remaining coins and place three on one side of the balance. Place 3 of the 8 identical coins on the other side. There are three possibilities: 2.a) The three remaining coins are lighter. In this case you now know that one of those three coins is the odd one out and that it is lighter. Take two of those three coins and weigh them against each other. If the balance tips then the lighter coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is lighter. 2.b) The three remaining coins are heavier. In this case you now know that one of those three coins is the odd one out and that it is heavier. Take two of those three coins and weigh them against each other. If the balance tips then the heavier coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is heavier. 2.c) The three remaining coins balance. In this case you just need to weigh the remaining coin against any of the other 11 coins and this tells you whether it is heavier, lighter, or the same. ## Variations Given a population of 13 coins in which it is known that 1 of the 13 is different (mass) from the rest, it is simple to determine which coin it is with a balance and 3 tests as follows: 1) Subdivide the coins in to 2 groups of 4 coins and a third group with the remaining 5 coins. 2) Test 1, Test the 2 groups of 4 coins against each other: a. If the coins balance, the odd coin is in the population of 5 and proceed to test 2a. b. The odd coin is among the population of 8 coins, proceed in the same way as in the 12 coins problem. 3) Test 2a, Test 3 of the coins from the group of 5 coins against any 3 coins from the population of 8 coins: a. If the 3 coins balance, then the odd coin is among the remaining population of 2 coins. Test one of the 2 coins against any other coin; if they balance, the odd coin is the last untested coin, if they do not balance, the odd coin is the current test coin. b. If the 3 coins do not balance, then the odd coin is from this population of 3 coins. Pay attention to the direction of the balance swing (up means the odd coin is light, down means it is heavy). Remove one of the 3 coins, move another to the other side of the balance (remove all other coins from balance). If the balance evens out, the odd coin is the coin that was removed. If the balance switches direction, the odd coin is the one that was moved to the other side, otherwise, the odd coin is the coin that remained in place. ## Generalizations The generalization of this problem is described in Chudnov. Let $\mathbb{R}^n$ be the $n$-dimensional Euclidean space, $[\mathrm{e}^1, \mathrm{e}^2]$ be the inner product of vectors $\mathrm{e}^1$ and $\mathrm{e}^2$ from $\mathbb{R}^n,$ For vectors $\mathrm{e} = (e_1, \dots,e_n) \in \mathbb{R}^n$ and subsets $E = \{\mathrm{e}^j\} \subseteq \mathbb{R}^n,$ the operations $(\cdot)^{}$ and $(\cdot)^{+}$ are defined, respectively, as $\mathrm{e}^{} = (sign(e_i))_i$ ; $E^{} = \{(\mathrm{e}^j) ^{}\}$, $\mathrm{e}^{+} = (\|sign(e_i)\|)_i$, $E^{+} = \{(\mathrm{e}^j)^{+}\}.$ By $I^n$ we shall denote the discrete [−1; 1]-cube in $\mathbb{R}^n$; i.e., the set of all sequences of length $n$ over the alphabet $I = \{ -1,0, 1 \}$ The set $I^n_t = \{\mathrm{x} \in I^n \|w(\mathrm{x}) \le t \} \subseteq I^n$ is the discrete ball of radius $t$ (in the Hamming metric $w()$ ) with centre at the point $\mathrm{0}.$ Relative weights of $n$ objects are given by a vector $\mathrm{x} = (x_1, \dots, x_n) \in I^n ,$ which defines the configurations of weights of the objects: the $i$th object has standard weight if $x_i = 0;$ the weight of the $i$th object is greater (smaller) by a constant (unknown) value if $x_i = 1$ (respectively, $x_i = -1$). The vector $\mathrm{x}^{+}$ characterizes the types of objects: the standard type, the non-standard type (i.e., configurations of types), and it does not contain information about relative weights of non-standard objects. A weighing (a check) is given by a vector $\mathrm{h} \in I^n ;$ the result of a weighing for a situation $\mathrm{x} \in I^n$ is $s(\mathrm{x}; \mathrm{h}) = sign([\mathrm{x}; \mathrm{h}]).$ The weighing given by a vector $\mathrm{h} = (h_1, \dots, h_n)$ has the following interpretation: for a given check the $i$th object participates in the weighing if $h_i \ne 0$; it is put on the left balance pan if $h_i < 0$ and is put on the right pan if $h_i > 0.$ For each weighing $\mathrm{h}$, both pans should contain the same number of objects: if on some pan the number of objects is smaller than as it should be, then it receives some $r(\mathrm{h}) = [\mathrm{h}; 1 ,\dots, 1]$ reference objects. The result of a weighing $s(\mathrm{x}; \mathrm{h})$ describes the following cases: the balance if $s(\mathrm{x}; \mathrm{h}) = 0$, the left pan outweighs the right one if $s(\mathrm{x}; \mathrm{h}) = -1$, and the right pan outweighs the left one if $s(\mathrm{x}; \mathrm{h}) = 1.$ The incompleteness of initial information about the distribution of weights of a group of objects is characterized by the set of admissible distributions of weights of objects $Z \subseteq I^n,$ which is also called the set of admissible situations, the elements of $z \in Z$ are called admissible situations. Each weighing $\mathrm{h}$ induces the partition of the set $I^n$ by the plane (hyperplane ) $[\mathrm{x}; \mathrm{h}] = 0$ into three parts $W(s\|I^n ; \mathrm{h}) = \{\mathrm{x} \in I^n \|s(\mathrm{x}; \mathrm{h}) = s\}$, $s \in I,$ and defines the corresponding partition of the set $Z = W(0\|Z, \mathrm{h}) + W(1\|Z, \mathrm{h}) + W(-1\|Z, \mathrm{h}),$ where $W(s\|Z, \mathrm{h}) = W(s\|I^n , \mathrm{h}) \cap Z.$ Definition 1. A weighing algorithm (WA) $\mathcal{A}$ of length $m$ is a sequence $\mathcal{A} = < \mathrm {A}_1, \dots, \mathrm {A}_m >,$ where $\mathrm {A}_j : I^{j-1} \to I^n$ is the function determining the check $\mathrm{h}^j = \mathrm{A}_j(s^{j-1}); \mathrm{h}^j \in I^n,$ at each $j$th step, $j = 1, 2, \dots, m,$ of the algorithm from the results of $\mathrm{s}^{j-1} = (s_1, \dots, s_{j-1}) \in I^{j-1}$ weighings at the previous steps ( $\mathrm{h}^1 = \mathrm {A}_1()$ is a given initial check). Let $S(Z, \mathcal{A})$ be the set of all $(Z, \mathcal{A})$-syndromes and $W(s\|\mathcal{A}) \subseteq I$ be the set of situations with the same syndrome $s$; i.e., $W(s\|\mathcal{A}) = \{\mathrm{z} \in I^m \|s(z\|\mathcal{A}) = s \}$; $W(s\|Z; \mathcal{A}) =W(s\|\mathcal{A}) \cap Z.$ Definition 2. A WA $\mathcal{A}$ is said to: a) identify the situations in a set $Z$ if the condition $\|W(s\|Z, \mathcal{A}) \| = 1$ is satisfied for all $s \in S(Z\mathcal{A});$ b) identify the types of objects in a set $Z$ if the condition $\|W^{+}(s\|Z\mathcal{A})\| = 1$ is satisfied for all $s \in S(Z \mathcal{A}).$ It is proved in that for so-called suitable sets $Z$ an algorithm of identification the types identifies also the situations in $Z.$ As an example the perfect dynamic (two-cascade) algorithms with parameters $n = 11, m = 5, t = 2$ are constructed in which correspond to the parameters of the perfect ternary Golay code (Virtakallio-Golay code). At the same time, it is established that a static WA (i.e. weighting code) with the same parameters does not exist. Each of these algorithms using 5 weighings finds among 11 coins up to two counterfeit coins which could be heavier or lighter than real coins by the same value. In this case the uncertainty domain (the set of admissible situations) contains $1 + 2 C_{11} ^ 1 + 2 ^ 2 C_{11} ^ 2=3^5$ situations, i.e. the constructed WA lies on the Hamming bound for $t=2$ and in this sense is perfect. To date it is not known whether there are other perfect WA that identify the situations in $I_t^n$ for some values of $n, t$. Moreover, it is not known whether for some $t> 2$ there exist solutions for the equation $\sum_ {i = 0}^t 2^iC_n^i = 3^m$ (corresponding to the Hamming bound for ternary codes) which is, obviously, necessary for the existence of a perfect WA. It is only known that for $t = 1$ there are no perfect WA, and for $t = 2$ this equation has the unique nontrivial solution $n=11, m=5$ which determines the parameters of the constructed perfect WA. ## Original parallel weighings puzzle Konstantin Knop invented this puzzle. There are N indistinguishable coins, one of which is fake (it is not known whether it is heavier or lighter than the genuine coins, which all weigh the same). There are two balance scales that can be used in parallel. Each weighing lasts one minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes? Balance puzzle Facts for Kids. Kiddle Encyclopedia.
Understanding Dot Product (with example) -- \|\| Introduction to the problem Hi and welcome, I should preface this by saying that the mathematics going into this post has been simplified for easy understanding and proving formulae and theory is not going to be talked over. So first off, what are we dealing with here? The Dot Product is a mathematical operations that we can use to get outputs regarding the relationship between the vectors we are using them on. If you’re already familiar, you can skip the next few sections and get to the part where I use it in a code example. What is a vector / scalar? To help understand what a vector is, we should start by describing what a scalar is. A scalar is an object with a value or magnitude (size). An example would be a float variable that determines the value for a character in a game. Speed = 100. Vectors build upon this, they are objects that have not only just a magnitude, but also a direction. Below is an example of a vector U. Its pointing in a diagonal direction due to a positive x and y (i,j) components and has a magnitude (\|U\|) which we can figure out using vector. We can calculate the magnitude of this 2D Vector is by using Pythagoras Theorem. We take the value of x² and y² and square root the addition of them both (both x and y are variables of any given value, an example in this scenario would be [5,8]). What is Dot Product? Now that we’ve gone over the fundamentals of vectors, we can now start to use them with operations such as Dot Product. As stated in the introduction, Dot Product is a mathematical operation used to find the relationship between 2 vectors. The result is a scalar value. There are 2 equations used to find the Dot Product Equation 1: U • V = \|U\|\|V\|Cos(θ) Equation 2: U • V = UxVx + UyVy + (UzVz)* * Only needed if the Vector is 3D It is somewhat easy to tell which one seems less complex than the other however I’ll try to breifly explain what each equation is doing. Equation 1: If you recall earlier, I did mention the \|U\| symbol and this is a fancy way of saying ‘the magnitude of a (given) vector’. The equation is asking for the magnitude of both vectors and then the Cosine of the angle between the vectors. Here’s a diagram explaining these visually. With this particular equation, the Cos(θ) is what’s most intriguing and arguably allows you to more or less guess what a relative value of the Dot Product could be. I say this because Cosine of any angle will produce a value in the range between -1 and 1. From trigonometric knowlege, Cos(90) = 0. Therefore when the angle between 2 vectors is 90 degrees, a right angle, both vectors are perpendicular and anything multiplied by 0 is 0. This means that when the vectors are perpendicular the Dot Product is 0. Additionally, when the angle is > 90 degrees, as the angle approaches 0 from 90, the cosine value will go from 0 to 1. Therefore the Dot Product will have a positive sign. Alternatively if the angle is <90, the sign will be negative. (However this statement isnt entirely true as it only applies to the range of angles from 0–180, beyond this range the less obtuse angle will need to be used to calculate the Dot Product) Equation 2: This equation is much simpler on paper, purely because we are multiplying the vectors by their relative components and then adding them together. This is also the equation that is more commonly used as its much easier to work with as we don’t have to deal with an angle. Both equations ultimately give you the same scalar at the end of the calculations. Now let’s get on with the fun stuff! What are some use cases in Game Development? It’s most commonly used when comparing vectors to see if they meet certain criteria to trigger events, some examples are, if an AI is looking at a player or not, if the player is looking directly at an object such as a cup or a table, if something is behind the player, if the player is moving towards or away from a direction, detecting whether or not a player is either facing downhill or uphill on a slope, etc. There’s also a very well thought out example from a Naughty Dog GDC where Dot Products are used to calculate clipping for wheels: Better yet, I’ll give you an example of which I am using in my own project, Primal Dominion: Aftermath. In this example the function is used to determine how much a character should lean given by its velocity. Let’s break the following 2 lines down first. Line 1 I am using multiple operations to calculate the sign (-1/1) of a Dot Product. I am using 2 frames of reference for my vectors, they’re both the velocity of the character but the ‘old’ velocity is 1 frame old, this can give me a different direction and speed when comparing between ticks and thus can tell me which direction the character is going. 0 for idle, -1 for decel, +1 for accel. On Line 2 I use the sign to determine whether or not to use an accel value or decel value. The syntax is a bit fancy but ultimately means, if Sign > 0 then use Accel, if not use Decel. Following the code, I then unrotate the character’s worldspace rotation to a scaled local space vector The last couple lines is then producing the final leaning value which is interpolated to produce a smoother rate of change awell as caching the current frame velocity for the next frame. Alternatively you can extend this by adding a variable interpolation speed to signify ‘tiredness’ as the character will be faster to slouch into the direction. Within unreals animation graph, I use that variable and do some small adjustments as the characters in question will lean different amounts. I drive this as an addititive rotation transform on the target bones such as Tail and Spine. Heres an exaggerated version of the result:
# 5.08 Equivalent number sentences Lesson ## Ideas Do you remember how to make a target number ? ### Examples #### Example 1 Find the missing number to equal the number 98. a 89 + ⬚ = 98 Worked Solution Create a strategy Use the number line below to count how many jumps are between the starting number and the target number. Apply the idea Locate where 89 is. Jump to the right of 89 and count the number of spaces until we reach 98. We have jumped 9 spaces to the right of 89. So 89 + 9 = 98 b 20 + ⬚ = 98 Worked Solution Create a strategy Use the number line below to count how many jumps are between the starting number and the target number. Apply the idea Locate where 20 is. Jump to the right of 20 and count the number of spaces until we reach 98. We have jumped 78 spaces to the right of 20. So 20 + 78 = 98 c 28 + ⬚ = 98 Worked Solution Create a strategy Use the number line below to count how many jumps are between the starting number and the target number. Apply the idea Locate where 28 is. Jump to the right of 28 and count the number of spaces until we reach 98. We have jumped 70 spaces to the right of 28. So 28 + 70 = 98 Idea summary We can use number lines to make a target number by plotting the beginning number and the target number, and counting how many spaces are between them. ## Equivalent number sentences For our number sentences to be equivalent, both sides need to be equal. Let's see how to check if this is true, as well as how we can make it true. ### Examples #### Example 2 Complete the number sentence: 31 - ⬚ = 16 Worked Solution Create a strategy Use a number line. Apply the idea Locate 31 in a number line. Jump back from 31 and count the number of units until we reach 16. We have jumped back 15 units from 31, so the complete number sentence would be: 31 - 15 = 16 Idea summary We can use number lines to make a target number using subtraction by plotting the beginning number and the target number, and counting how many spaces are between them. ## Find a missing number If we need to find a missing number, we can use things such as bridge to 10, and number lines to help us. Let's see how we can do this for an addition and a subtraction problem, to make a sentence equal. ### Examples #### Example 3 Complete the number sentence: 48 - ⬚ = 19 + 16 Worked Solution Create a strategy Find the right hand side of the equation first then use a number lin to complete the number sentence. Apply the idea Add the numbers in the right hand side of the equation in a vertical algorithm. \begin{array}{c} & &1 &9 \\ &+ &1 &6 \\ \hline \\ \hline \end{array} Add the ones column first: 9 + 6 = 15. Bring down the 5 and carry the 1 to the tens column. \begin{array}{c} & &\text{}^1 1 &9 \\ &+ &1 &6 \\ \hline & & &5 \\ \hline \end{array} Add the tens column: 1 + 1 + 1 = 3. \begin{array}{c} & &\text{}^1 1 &9 \\ &+ &1 &6 \\ \hline & &3 &5 \\ \hline \end{array} The simplified number sentence would be: 48 - ⬚ = 35 Locate 48 in the number line. Jump back from 48 and count the number of units until we reach 35. We have jumped back 13 units from 48. So the complete number sentence would be: 48 - 13 = 19 + 16 Idea summary For number sentences to be equivalent, or balanced, both sides must equal the same amount. ### Outcomes #### MA2-5NA uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers
# Rd Sharma Solutions Class 6 Chapter 7 ## Rd Sharma Solutions Class 6 Chapter 7 Decimals Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Rd Sharma Class 6 Mathematics, Chapter 7, Decimals. Here students can easily find Exercise wise solution for chapter 7, Decimals. Students will find proper solutions for Exercise 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8 and 7.9. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum. Decimals Exercise 7.1 Solution Question no – (1) Solution : Given number, (i) 52.5; (ii) 12.57 (iii) 15.05; (iv) 74.059 (v) 0.503 Tens Ones Tenths Hundredths Thousands (i) 5 2 5 (ii) 1 2 5 7 (iii) 1 5 0 5 (iv) 7 4 0 5 9 (v) 0 5 0 3 Question no – (2) Solution : (i) 307. 12 (ii) 9543.025 (iii) 12.503 Question no – (3) Solution : (i) 175.04 = One Hundred Seventy five and four hundredths. (ii) 0.21 = Zero and twenty one hundredths. (iii) 9.004 = Nine and four thousandths. (iv) 0.459 = Zero and four hundred fifty nine thousandths. Question no – (4) Solution : (i) 65 + 2/10 + 7/100 = 56.27 (ii) 45 + 9/100 = 45.09 (iii) 88 + 5/10 + 2/1000 = 88. 502 (iv) 3/10 + 7/1000 = 0.307 Question no – (5) Solution : (i) Five and four tenths = 5 + 4/10 = 5.4 So, Five and four tenths is 5.4 (ii) Twelve and four hundredths = 12 + 4/100 = 12.04 So, Twelve and four hundredths = 12.04 (iii) 9 + 705/100 = 9.705 So, Nine and Seven hundred five thousandths = 9.705 (iv) Zero point five two six = 0.526 (v) = 47 + 6/1000 = 47.006 Therefore, Forty seven and six thousandths = 47.006 (vi) 8/1000 = 0.008 So, Eight thousandths = 0.008 Decimals Exercise 7.2 Solution Question no – (1) Solution : (i) 3/10 = 0.3 So, Three tenths = 0.3 (ii) = 2 + 5/10 = 2.5 So, two ones and five tenths = 2.5 (iii) 30 + 1/10 = 30.1 So, thirty and one tenths = 30.1 (iv) 22 + 6/10 = 22.6 So, twenty tow and six tenths = 22.6 (v) 100 + 2 + 3/10 = 102.3 Thus, One hundred, two ones and three tenths = 102.3 Question no – (2) Solution : (i) 30 + 6 + 2/10 = 36.2 So, decimal of 30 + 6 + 2/10 is 36.2 (ii) 700 + 5 + 7/10 = 705.7 So, decimal of 700 + 5 + 7/10 is 705.7 (iii) 200 + 6 + 5 1/10 = 265.1 So, decimal of 200 + 6 + 5 1/10 is 265.1 (iv) 200 + 70 + 9 + 5/10 = 279.5 So, decimal of 200 + 70 + 9 + 5/10 is 279.5 Question no – (3) Solution : (i) 22/10 = 2.2 (ii) 3/2 = 1.5 (iii) 2/5 = 0.4 Question no – (4) Solution : (i) 40 2/5 = 40 + 0.4 = 40.4 Therefore, decimal of 40 2/5 is 40.4 (ii) 39 2/10 = 39 + 0.2 = 39.2 Hence, decimal of 39 2/10 is 39.2 (iii) 4 3/5 = 4 + 0.6 = 4.6 Thus, decimal of 4 3/5 is 4.6 Question no – (5) Solution : (i) 3.8 = 3.8 = 3 + 8/10 = 38/10 = 19/5 (ii) 21.2 = 21.2 = 21 + 2/10 = 212/10 = 106/5 Question no – (6) Solution : Given numbers, (i) 0.2; (ii) 1.9 (iii) 1.1; (iv) 2.5 Now on number line : Question no – (7) Solution : Between Numbers : (i) 0.8 = 0 and 1 (ii) 5.1 = 5 and 6 (iii) 2.6 = 2 and 3 (iv) 6.4 = 6 and 7 (v) 9.0 = 9 (vi) 4.9 Nearer to the number: (i) 1 (ii) 5 (iii) 3 (iv) 6 (v) 9 (vi) 5 Question no – (8) Solution : A = 0.8 C = 1.8 B = 1.3 D = 2.9 Decimals Exercise 7.3 Solution Question no – (1) Solution : (i) Five hundred twenty five and forty hundredths. = 525. 40 (ii) Twelve and thirty five thousandths = 12.035 (iii) Fifteen and seventeen thousandths, = 15.017 (iv) Eighty eight and forty eight-hundredths = 88.48 Question no – (2) Solution : (i) Given 137 + 5/100 = 137.05 So, decimal of 137 + 5/100 is 137.05 (ii) 20 + 9 + 4/100 = 29.04 So, decimal of 20 + 9 + 4/100 is 29.04 Question no – (3) Solution : (i) 8/100 = 0.008 So, 8/100 as decimal = 0.008 (ii) 300/1000 = 0.3 So, 300/1000 as decimal = 0.3 (iii) 18/1000 = 0.018 So, 18/1000 as decimal = 0.018 (iv) 208/100 = 2.08 So, 208/100 as decimal = 2.08 (v) 888/1000 = 0.888 So, 888/1000 as decimal = 0.888 Question no – (4) Solution : (i) 12 1/4 = 12 1/4 = 50/4 = 12.25 (ii) 7 1/8 = 7 1/8 = 57/8 = 7.125 (iii) 5 1/20 = 5 1/20 = 101/20 = 5.05 Question no – (5) Solution : (i) Given, 0.04 = 4/100 .. (as fraction) = 2/50 = 1/25..(reduce form) (ii) Given, 2.34 = 234/100 … (as fraction) = 117/50 …(reduce form) (iii) Given, 0.342 = 342/100 … (as fraction) = 117/50 …(reduce form) (iv) Given, 1.20 = 120/100… (as fraction) = 12/10 = 6/5…(reduce form) (v) Given, 17.38 = 1738/100 … (as fraction) = 869/50 …(reduce form) Question no – (6) Solution : (i) 20 + 9 + 4/10 + 1/000 = 29.41 (ii) 30 + 4/10 + 8/100 3/1000 = 30.483 (iii) 137 + 5/100 = 137 + 5/100 = 137.05 (iv) 7/10 + 6/100 + 4/1000 = 7/10 + 6/100 + 4/100 = 0.764 (v) 23 + 2/10 + 6/1000 = 23.206 (vi) 700 + 20 + 5 + 9/100 = 725.09 Decimals Exercise 7.4 Solution Question no – (1) Solution : (i) 23/10 = 23/10 = 2.3 (ii) 139/100 = 1.39 (iii) 4375/1000 = 4.375 (iv) 12 1/2 = 25/2 = 12.5 (v) 75 1/4 = 75.25 (vi) 25 1/8 = 25.125 (vii) 18 3/24 = 18.125 (viii) 39 7/35 = 1372/35 = 39.2 (ix) 15 1/25 = 15.4 (x) 111/250 = 111/250 = 0.444 Question no – (2) Solution : (i) 0.5 = 5/10 = 1/2 (ii) 2.5 = 25/10 = 5/2 (iii) 0.60 = 60/100 = 6/10 = 3/5 (iv) 0.18 = 18/100 = 9/50 (v) 5.25 = 525/100 = 21/4 (vi) 7.125 = 7125/100 = 201/8 (vii) 15.004 = 15004/1000 = 3751/280 (viii) 20.375 = 20.375 = 20375/100 = 163/8 (x) 59.48 = 5948/100 = 1487/25 Decimals Exercise 7.5 Solution Question no – (1) Solution : (i) 25.35 > 8.47 (ii) 20.695 < 20.93 (iii) 0.39 < 0.72 (iv) 0.019 < 0.23 (v) 0.2306 > 0.201 (vi) 0.93 < 0.99 Question no – (2) Solution : (i) 1.800 > 1.008 (ii) 3.8 = 3.300 (ii) 5.64 > 5.603 (iv) 1.5 = 1.50 (v) 1.431 < 1.439 (vi) 0.5 > 0.005 Decimals Exercise 7.6 Solution Question no – (1) Solution : As we know that, 100 paise = 1 rupee. (i) 15 paisa = 0.15 Rs (ii) 5 paisa = 0.5 Rs (iii) 350 paisa = 3.50 Rs (iv) 2 rupees 60 paisa = 2.60 Rs Question no – (2) Solution : As we know, 100 cm = 1 m (i) 15 cm = 15/100 = 0.15 m (ii) 8 cm = 8/100 = 0.8 m (iii) 135 cm = 135/100 = 1.35 m (iv) 3 m 65 cm = 3m + 65 cm = 3.65 m Question no – (3) Solution : As we know that, 10 mm = 1 cm (i) 5 mm = 5/10 = 0.5 cm (ii) 60 mm = 60/10 = 6 cm (iii) 175 mm = 175/10 = 17.5 cm (iv) 4 cm 5 mm = 4 cm + 5 mm = 4.5 cm Question no – (4) Solution : As we know, 1000 m = 1 km (i) 5 m = 5/1000 = 0.005 km (ii) 55 m = 55/1000 0.055 km (iii) 555 m = 55/1000 = 0.555 km (iv) 5555 m = 5555/1000 = 5.555 km (v) 15 km 35 m = 15 km + 35 m = 15.035 km Question no – (5) Solution : As we know, 1000 g = 1 kg (i) 8 g = 0.008 kg (ii) 150 g = 150/1000 = 0.150 kg (iii) 2750g = 2750/1000 = 2.750 kg (iv) 5 kg 750 g = 5750/1000 = 5.750 kg (v) 36 kg 50 g = 36 kg + 50 g = 36.050 kg Question no – (6) Solution : (i) Rs 5.25 = 525/100 = 21/4 Rs (ii) 8.354 kg = 8354/100 kg (iii) 3.5 cm = 35/10 cm = 7/2 cm (iv) 3.05 km = 305/100 km = 61/20 km (v) 7.54 m = 754/100 m (vi) 15.005 = 15005/1000 kg (vii) 12.05 = 12.05 m (viii) 0.2 cm = 1/5 cm Decimals Exercise 7.7 Solution Question no – (1) Solution : (i) 0.8 (0, 80, 0.85, 0,800,0.08) = 0.85, 0.08 (ii) 25.1 (25, 01, 25.10 25.100, 25.001) = 25.01, 25.001 (iii) 45.05 (45. 050, 45. 005, 45.0500) = 45.005, 45.500 Question no – (2) Solution : Number (i) and Number (iii) are lie decimals because equal no of digits are present after the decimal point. Question no – (3) Solution : (i) 8.05 and 7.95 are like decimals. = Correct (ii) 0.95, 0.306, 7.10 are unlike decimals. = correct. (iii) 3.70 and 3.7 are like decimals. = Incorrect. (iv) 13.59, 1.359, 135.9 are like decimals. = Incorrect. (v) 5.60, 3.04, 0.45 are like decimals. = Correct. Question no – (4) Solution : (i) 7.8, 7.85 Like decimal = 7.80, 7.85 (ii) 2.02, 3.2 Like decimal = 2.02, 3.20 (iii) 0.6, 5.8, 12.765 Like decimal = 0.600, 5.800, 12.765 (iv) 5.296, 5.2, 5.29 Like decimal = 5.296, 5.200, 5.290 (v) 4.329, 43.29, 432.94 Like decimal = 4.329, 43.2900, 432.94000 Decimals Exercise 7.8 Solution Question no – (2) Solution : (i) Given, 41.8, 39.24, 5.01 and 62.6 = 41.8 + 39.24 + 5.01 + 62.6 = 148.65 (ii) Given, 4.702, 4.2, 6.02 and 1.27 = 4.702 + 4.2 + 6.02 + 1.27 = 16.192 (iii) Given, 18.03, 146.3, 0.829 and 5.324 = 18.03 + 146.3 + 0.829 + 5.324 = 170.483 Question no – (3) Solution : (i) 0.007 + 8.5 + 30.08 —————————- = 38.587 (ii) 280.69 + 25.2 + 38 —————————- = 343.89 (iii) 25.65 + 9.005 + 3.7 —————————- = 38.355 (iv) 27.076 + 0.55 + 0.004 —————————- = 27.630 Question no – (4) Solution : In the question, Mother gave her = Rs 10.50 Father gave her = Rs 15.80 Total amount given by her presents, = (10.50 + 15.80) Rs = 26.30 Rs Therefore, the total amount 26.30 Rs. Question no – (5) Solution : In the question, Apples = 4 kg 90 g, Grapes = 2 kg 60 g Mangoes = 5 kg 300 g Weight of all fruits, = (4.090 + 2.060 + 5.300) kg = 11.450 kg Therefore, the weight of the fruits is 11.450 kg. Question no – (6) Solution : Given in the question, Shirt = 3 m 20 cm cloth Skit = 2 m 5 cloth Total cloth bought by her, = (3.20 + 2.05) m = 5.25 m Therefore, the total cloth bought by her is 5.25 m. Question no – (7) Solution : In the question, By bus = 15 km 268 m, By car = 7 km 7m By foot = 500 m Distance from residence to school, = (15.26.8 + 7.007 + 0.500) km = 22.775 km Thus, the school is 22.775 kilometers away from his residence. Decimals Exercise 7.9 Solution Question no – (1) Solution : (i) 46.2 – 37.5 ———————— = 8.73 (ii) 128.4 – 53.05 ———————— = 75.35 (iii) 45.03 – 27.80 ———————— = 17.23 (iv) 23.93 – 5.946 ———————— = 17.984 Question no – (2) Solution : (i) 9.756 – 628 = 3.476 The value is 3.476 (ii) 21.05 – 15.27 = 5.78 The value is 5.78 (iii) 18.5 – 6.79 = 11.71 The value is 11.71 (iv) 48.1 – 0.37 = 47.73 The value is 47.73 (v) 108.032 – 86.8 = 21.232 The value is 21.232 (vi) 91.001 – 72.9 = 18.101 The value is 18.101 (vii) 32.7 – 25.86 = 6.84 The value is 6.84 (viii) 100 – 26.32 = 73.68 The value is 73.68 Question no – (3) Solution : In the question, Sum of two numbers = 100 One of them = 78.01 Other number is, = (100 – 78.01) = 21.99 Therefore, the other number will be 21.99 Question no – (4) Solution : Distance of school = 5 km 350 m Travels on foot = 1 km 70 m Distance travel by bus, = (5.350 – 1.070) = (5.350 – 1.070) km = 4.280 km Hence, she 4.280 km travel by bus. Question no – (5) Solution : We know, Book price = Rs 35.65. He gave shopkeeper = Rs 50 Balance return, = (50 – 35.65) Rs = 14.35 Rs Therefore, he get back Rs 14.35 from the shopkeeper. Question no – (6) Solution : As we know, Ruby bought watermelon = 5 kg 200 g. she gave to her neighbor = 2 kg 750 g Weight of watermelon left with Ruby, = (5.200 – 2.750) kg = 2.450 kg Therefore, weight of the watermelon left with Ruby is 2.450 kg. Question no – (7) Solution : From the given data we know, Victor drove on Saturday = 89.050 km Victor drove on Sunday = 73.9 km Distance travel more by victor, = (89.050 – 7.3.9) = 15.25 km Hence, he 15.25 km more drive on Saturday. Question no – (9) Solution : According to the question, Gopal travelled by bus = 125.5 km By pony = 14.25 km Total distance = 150 km Distance travelled by Gopal on foot, = (150 – 125.5 – 14.25) km = 10.25 km Therefore, he travel 10.25 km on foot. Question no – (10) Solution : According to the question, Tina had = 20 m 5 cm long cloth. She cuts = 4 m 50 cm length of cloth Length of cloth left her, = (20.05 – 4.50) = 15.55 m Therefore, 15.55 m cloth is left with her. Question no – (11) Solution : 1st we need to find the total cost, = (18.90 + 8.50 + 5.05) = 32.45 Rs Now, balance return back, = (50 – 32.45) Rs = 17.55 Thus, Vineeta will get back 17.55 Rs. Question no – (12) Solution : According to the question, Tanuj walked on Monday = 8.62 km Tuesday = 7.05 km Distance walked by Tanuj, = (21.01 – 8.62 – 7.05) km = 5.34 km Therefore, Tanuj walk 5.34 km on Wednesday Previous Chapter Solution : Updated: June 7, 2023 — 3:55 pm
FRACTIONS TO DECIMALS What Are Fractions? A fraction is really a division statement. The number ½ says that we divided 1 whole thing into 2 equal parts. Similarly, ¾ indicates that we divided something -- say a pizza -- into 4 equal parts and we ate 3 of them. It could also mean that we have 4 things of equal size or value and 3 of them are of interest because of some quality, say they're red or they've been eaten. So if we ordered 4 pizzas and ate 3 of them, we'd have eaten ¾ of all the pizzas. . Changing Fractions to Decimal Equivalents by Division Here's some fraction images Since a fraction indicates division, that's exactly how we change it into a decimal. We do the division. So, to get the decimal equivalent of ¾, we divide 4 into 3 like this: However, if we're smart, we'll just remember that three "quarters" = 75 cents so ¾ = 0.75. To find the decimal equivalent of a fraction, divide the numerator by the denominator. Repeating Decimals When we use division to find the decimal equivalent of fractions with 3, 7, 9, 11 etc.in the denominator, the quotient is a repeating digit. These are called repeating decimals and as always in math, we use a precise symbol to denote them. Let's use division to find the decimal equivalent of . ... We must make sure the "bar" is over ALL THE REPEATING DIGITS in the pattern! For instance, and Note that the bar covers all 6 repeating digits. The decimal equivalent of 1/19 = 0.052631578947368421 052631578947368, with 18 repeating digits in the pattern. All 18 must be covered by the bar. Changing the Denominator to a Power of 10 Since the denominator of any decimal's fraction equivalent is a power of 10, when we have to find a fraction equivalent to a decimal value, and it's easy to make the denominator a power of 10 through multiplication by a fraction equal to 1, we can proceed like this. We know that 5 × 20 = 100 so we multiplied by 5/5 which equals 1 -- therefore we didn't change the value of our fraction -- we just wrote an equivalent fraction with a denominator of 100. In the second case, we knew that 8 × 125 = 1000, so we multiplied by 8/8. When we have mixed numbers -- part whole number, part fraction, we change the fraction part to its decimal equivalent and write the whole number as usual. So 1¼ = 1.25 Now get a pencil, an eraser and a note book, copy the questions, do the practice exercise(s), then check your work with the solutions. If you get stuck, review the examples in the lesson, then try again. . Practice 1) Multiply by 1 to make the denominator a power of 10, then write the decimal equivalent. a) b) c) d) e) . 2) Find the decimal equivalent by division. Round to 3 decimal places. Indicate repeats. a) b) c) d) e) . 3) Write the fraction equivalent of these decimals. Reduce to lowest terms. a) 0.36 = b) 0.012 = c) 0.6 = d) 0.237 = e) 0.0027 = . Solutions 1) Change each denominator to a power of 10, then write the decimal equivalent. a) b) c) d) e) . 2) Use division to find the decimal equivalent of these fractions. Round to 3 decimal places. a) 0.267 b) 0.417 c) 0.714 d) 0.727 e) 0.444 . 3) Write the fraction equivalent of these decimals. Reduce to lowest terms. a) 0.36 = b) 0.012 = c) 0.6 = d) 0.237 = e) 0.0027 = .
PSAT Math Practice Questions: Linear Equations Linear equations and linear graphs are some of the most common test questions on the PSAT Math Test. Linear Equations can be used to model relationships and changes such as those concerning time, temperature, or population. The graphs of these equations are as important as the equations themselves. Let’s review an example. PSAT Math Practice Question: Linear Equations Work through the Kaplan Method for Math step-by-step to solve the following question. The table below shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right. 8/7(x – 101/220) + 4(x +8/9) = 38 Which approximate value of x satisfies the equation shown? A) 4.29 B) 4.65 C) 6.6 D) 6.8 Strategic Thinking Math Scratchwork Step 1: Read the question, identifying and organizing important information as you go. Notice a pattern? Again, the question is asking you to solve for x. 8/7(x − 101/220) + 4(x + 8/9) = 38 Step 2: Choose the best strategy to answer the question How can you quickly solve this problem? Clearing the fraction outside the parentheses is a smart move. Multiply both sides by 7. How are they different from the answers in the last problem? The presence of decimals means your calculator will be a great asset here. Don’t worry about common denominators. Divide the fractions. Because the answer choices are only written to two decimal places, write your intermediate steps to two places as well. Now distribute the numbers outside the parentheses, collect like terms, and solve for x. 8(x − 101/220) + 28(x + 8/9) = 266 8(x – 0.46) + 28(x + 0.89) = 266 8x – 3.68 + 28x + 24.92 = 266 36x = 244.76 x = 6.8 Step 3: Check that you answered the right question Double-check the question stem. Choice (D) is correct. x = 6.8 Many graphing calculators have a built-in function that will let you input and solve algebraic equations like the previous one. Consider learning how to use it before Test Day by reading the instruction manual or searching online. Notice in the previous question that careful use of your calculator can eliminate the need to complete time-consuming tasks by hand. Be conscious of the format of the answer choices—decimal answers are a great clue that you can use your calculator.
## + (855) 550-0571 POTENTIAL AND CREATIVITY WITH A FREE TRIAL CLASS DEVELOP TECHNICAL, SOFT, & ENTREPRENEURIAL SKILLS AGE 7-16 YEARS CARD BY ATTENDING A FREE TRIAL CLASS BOOK A FREE TRIAL # Vedic Math Sutra: Sutra-9 Chalana-Kalanabhyam ## \| Chalana-Kalanabhyam” is a Sanskrit phrase used in Vedic Mathematics. The term translates to “by movement and by position.” It’s one of the sixteen Sutras, or word formulae, proposed by Sri Bharati Krishna Tirthaji in his book “Vedic Mathematics,” each of which he claimed represented a mathematical principle. The Chalana-Kalanabhyam Sutra mainly applies to simplify specific algebraic equations, especially quadratic and cubic ones. Recommended Reading: HOW VEDIC MATH CAN HELP GEN Z TO OVERCOME MATH FEAR Let’s take an example of how to apply the Chalana-Kalanabhyam Sutra for solving cubic equations. #### Consider the cubic equation x^3 – 6x^2 + 11x – 6 = 0 #### This equation can be solved using Chalana-Kalanabhyam as follows: 1. First, note the coefficients of the equation: 1 (for x^3), -6 (for x^2), 11 (for x), and -6 (the constant). 2. Start by assuming the root of the equation to be ‘x.’ Consider ‘x’ and start by subtracting the coefficient x^2 from it, that is, x – (-6) = x + 6. 3. Now, multiply this root with the coefficient of x (11) and subtract the constant term (-6). This gives (x + 6)*11 – 6. 4. If the result of this operation is zero, then we have found the root of the cubic equation. However, in this case, you will find that for x = 1, the result is indeed zero, making x = 1 a root of this cubic equation. 5. The remaining roots can be found by dividing the original cubic equation by (x – 1), resulting in a quadratic equation that can be solved using standard methods. This example illustrates the use of Chalana-Kalanabhyam for solving cubic equations. However, remember that the applicability of this sutra is limited and does not extend to all types of equations. Moonpreneur understands the needs and demands this rapidly changing technological world is bringing with it for our kids. Our expert-designed Advanced Math course for grades 3rd, 4th, 5th, and 6th will help your child develop math skills with hands-on lessons, excite them to learn, and help them build real-life applications. Register for a free 60-minute Advanced Math Workshop today! #### Moonpreneur Moonpreneur is an ed-tech company that imparts tech entrepreneurship to children aged 6 to 15. Its flagship offering, the Innovator Program, offers students a holistic learning experience that blends Technical Skills, Power Skills, and Entrepreneurial Skills with streams such as Robotics, Game Development, App Development, Advanced Math, Scratch Coding, and Book Writing & Publishing. Subscribe Notify of Inline Feedbacks Emily Smith 7 months ago What are the 5 rules of Vedic Maths? Editor Simran Chawla 6 months ago There are 5 basic rules of Vedic Maths; Nikhilam Sutra, Gyarasguna Sutra, Ekanunena Purneva Sutra, Antyaordasake Pi, and Navamguna Sutra. These rules deal with the multiplication operation. rose 6 months ago How powerful is Vedic Maths? Editor Simran Chawla 6 months ago Vedic math offers quicker mental calculation methods, subject to individual preference. Adoption in mainstream education varies, with opinions on its effectiveness differing among individuals. ## MOST POPULAR ### Chain Rule: Guide with Theorem and Examples JOIN A FREE TRIAL CLASS ### MATH QUIZ FOR KIDS - TEST YOUR KNOWLEDGE Start The Quiz
004 - 320 CHAPTER 5 NONLINEAR SYSTEMS which at(0 0 is equal... • Notes • 1 This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: 320 CHAPTER 5 NONLINEAR SYSTEMS which at (0. 0) is equal to (‘3?) so the linearized system at (0, 0) is (we could also see this by “dropping the higher order terms“). (c) The eigenvalues of the linearized system at the origin are -l and 1. so the origin is a saddle. The linearized system decouples, so solutions approach the origin along the x-axis and tend away form the origin along the y-axis. 5. (a) Using separation of variables (or simple guessing). we have x0) = 10:" . (b) The equation dy _=_ 3 dr 4x +y is a ﬁrst-order. linear equation. We write the equation as Therefore. the integrating factor is e". Multiplying both sides of the equation by 9" yields ‘0 -r 3 -r ( d! y) e -— 41: e . Note that the left-hand side is just the derivative of ye" and the right-hand side is —4x3e"3’ e“ since x(t) = we" . Therefore we have i 9", dry -4r ‘3'e" = “use . = -4X3€ After integrating and simplifying, we have y(r) = x3e'3' + (yo — xgk‘. (c) The general solution of the system is x0) = x0e" y(t) = x3?” + (yo — x3)e‘. (d) For all solutions, x(r) —> 0 as t -> 00. For a solution to tend to the origin as r —> 00, we must have 370‘) —> 0, and this can happen only if yo - x3 = 0. (e) Since 1: = we", we see that a solution will tend toward the origin as r —> -00 only if x0 = 0. In that case, y(t) = yoe‘. and y(!) -v 0 as t -> —oo. ... View Full Document • Spring '08 • JU {[ snackBarMessage ]} What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern
### R.3-Estimating, R.4 ```R.3 - ROUNDING NUMBERS, ESTIMATING R.4 – SUBTRACTION WITH WHOLE NUMBERS Math 081 Catherine Conway Rounding Steps for Rounding Numbers • Locate the digit just to the right of the place you are to round to. • If that digit is less than 5, replace it and all digits to its right with zeros. • If that digit is 5 or greater, replace it and all digits to its right with zeros, and add 1 to the digit to its left. Examples Round 5,482 to the nearest hundred. Change to zeros 5,482 = 5,500 Greater than 5 You Try: pg 28 #42, 43, 46 Pg 28 #42, 43, 46 5,950 6,000 5,900 6,000 6,000 6,000 11,110 11,100 11,000 Practice The pie chart shows how a family of 3 earning \$52,402 a year spends their money SPENDING Taxes, 7,248, Miscellaneous, 6,987 Entertainment, 3,278 Car Expenses, 8,413 Food, 6,423 House Payment, 17,904 Savings Account, 2,149, 1. To the nearest ten dollars, what is the total amount spent on food and entertainment? \$9,700 2. To the nearest thousand dollars, how much of their income is spent on items other than taxes and savings? \$43,000 Estimating • Estimation is used to simplify the problem so that an Example: Estimate the answer to the following problem by rounding to the nearest thousand. 4,621 + 1,687 + 888 + 4,210 Round each number to the nearest thousand 5,000 + 2,000 + 1,000 + 4,000 12,000 You Try: Pg 30 #60, 62 Pg 30 #60, 62 Subtraction with Whole Numbers • Common words/phrases that represent subtraction: • difference, subtract, take away, decrease, fewer, less than, decline, below, minus • See Table 1 on page 33 Best way to Subtract numbers: Align the numbers vertically by place values and subtract in each column. Practice: pg 34 & 35 -Example 1 and 2 Subtraction with Borrowing • Subtraction must involve borrowing when the bottom digit in any column is larger than the digit above it. 83 = 8 tens + 3 ones 35 = 3 tens + 5 ones 83 = 8 tens + 3 ones Borrow - 7 tens + 13 ones Practice: 35 = 3 tens + 5 ones pg 39 # 38, 4 tens + 8 ones = 48 40 Example: 83 – 35 Practice: pg 39 # 38, 40 Correction on the homework •R.3: Question 4 and 9 It says to round to the nearest hundred, but it really wants you to round to the nearest thousand. ```
# Question #d0859 Feb 1, 2018 $h = 15$ #### Explanation: $4 \left(h + 3\right) = 6 \left(h - 3\right)$ $4 h + 12 = 6 h - 18$ $18 + 12 = 6 h - 4 h$ $2 h = 30$ $h = 15$ -Sahar:) Feb 1, 2018 $h = 15$ #### Explanation: Lets start with using the distributive property. You are going to multiply the number on the outside of the parenthesis with everything that is inside of the parenthesis. The steps for the first part of the equation which is $4 \left(h + 3\right)$ is going to be red, and the steps for the second part of the equation which is $6 \left(h - 3\right)$ is going to be blue. $\textcolor{red}{4 \cdot h = 4 h ,}$ and $\textcolor{red}{4 \cdot 3 - 12}$. This is going to give you the first part of the equation which is $\textcolor{red}{4 h + 12}$. Now lets do the next part. $\textcolor{b l u e}{6 \cdot h = 6 h}$. and $\textcolor{b l u e}{6 \cdot - 3 = - 18}$. This gives you the second part of your equation. Now your equation looks like this... $\textcolor{red}{4 h + 12}$ $\textcolor{g r e e n}{=}$ $\textcolor{b l u e}{6 h - 18}$. The next step is to add 18 to both sides. $\textcolor{red}{4 h + 12 + 18 = 4 h + 30}$, and $\textcolor{b l u e}{6 h - 18 + 18 = 6 h}$ This will now give you the equation of $\textcolor{red}{4 h + 30}$ $\textcolor{g r e e n}{=}$ $\textcolor{b l u e}{6 h}$ The next step is to subtract $4 h$ from both sides. $\textcolor{red}{4 h + 30 - 4 h}$, and $\textcolor{b l u e}{6 h - 4 h}$ This will give you the answer of $\textcolor{red}{30}$ $\textcolor{g r e e n}{=}$ $\textcolor{b l u e}{2 h}$. The last step is to divide both sides by 2. $\textcolor{red}{30 \div 2}$ $\textcolor{g r e e n}{=}$ $\textcolor{b l u e}{2 h \div 2}$ $\textcolor{red}{15}$ $\textcolor{g r e e n}{=}$ $\textcolor{b l u e}{h}$ So the answer to the problem is $\textcolor{p u r p \le}{h = 15}$
How do you find the vertical, horizontal or slant asymptotes for f(x)=x/(x-1)^2? Aug 9, 2017 $\text{vertical asymptote at } x = 1$ $\text{horizontal asymptote at } y = 0$ Explanation: The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. $\text{solve "(x-1)^2=0rArrx=1" is the asymptote}$ $\text{horizontal asymptotes occur as}$ ${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{( a constant)}$ Divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$ $f \left(x\right) = \frac{\frac{x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{\frac{1}{x}}{1 - \frac{2}{x} + \frac{1}{x} ^ 2}$ as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 - 0 + 0}$ $\Rightarrow y = 0 \text{ is the asymptote}$ Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes. graph{x/(x-1)^2 [-10, 10, -5, 5]}
# Thales’ Theorem ### New York State Common Core Math Geometry, Module 5, Lesson 1 Student Outcomes • Using observations from a pushing puzzle, explore the converse of Thales’ theorem: If triangle ABC is a right triangle, then A, B, and C are three distinct points on a circle with segment AC a diameter. • Prove the statement of Thales’ theorem: If A, B, and C are three different points on a circle with segment AC a diameter, then angle ABC is a right angle. Thales’ Theorem Classwork Opening Exercise a. Mark points 𝐴 and 𝐵 on the sheet of white paper provided by your teacher. b. Take the colored paper provided, and push that paper up between points 𝐴 and 𝐵 on the white sheet. c. Mark on the white paper the location of the corner of the colored paper, using a different color than black. Mark that point 𝐶. See the example below. d. Do this again, pushing the corner of the colored paper up between the black points but at a different angle. Again, mark the location of the corner. Mark this point 𝐷. e. Do this again and then again. Exploratory Challenge Choose one of the colored points (𝐶, 𝐷, …) that you marked. Draw the right triangle formed by the line segment connecting the original two points 𝐴 and 𝐵 and that colored point. Take a copy of the triangle, and rotate it 180° about the midpoint of 𝐴𝐵. Label the acute angles in the original triangle as 𝑥 and 𝑦, and label the corresponding angles in the rotated triangle the same. Todd says 𝐴𝐶𝐵𝐶′ is a rectangle. Maryam says 𝐴𝐶𝐵𝐶′ is a quadrilateral, but she is not sure it is a rectangle. Todd is right but does not know how to explain himself to Maryam. Can you help him out? a. What composite figure is formed by the two triangles? How would you prove it? i. What is the sum of the measures of 𝑥 and 𝑦? Why? ii. How do we know that the figure whose vertices are the colored points (𝐶, 𝐷, …) and points 𝐴 and 𝐵 is a rectangle? b. Draw the two diagonals of the rectangle. Where is the midpoint of the segment connecting the two original points 𝐴 and 𝐵? Why? c. Label the intersection of the diagonals as point 𝑃. How does the distance from point 𝑃 to a colored point (𝐶, 𝐷, …) compare to the distance from 𝑃 to points 𝐴 and 𝐵? d. Choose another colored point, and construct a rectangle using the same process you followed before. Draw the two diagonals of the new rectangle. How do the diagonals of the new and old rectangle compare? How do you know? e. How does your drawing demonstrate that all the colored points you marked do indeed lie on a circle? Example In the Exploratory Challenge, you proved the converse of a famous theorem in geometry. Thales’ theorem states the following: If 𝐴, 𝐵, and 𝐶 are three distinct points on a circle, and 𝐴𝐵 is a diameter of the circle, then ∠𝐴𝐶𝐵 is right. Notice that, in the proof in the Exploratory Challenge, you started with a right angle (the corner of the colored paper) and created a circle. With Thales’ theorem, you must start with the circle and then create a right angle. Prove Thales’ theorem. a. Draw circle 𝑃 with distinct points 𝐴, 𝐵, and 𝐶 on the circle and diameter 𝐴𝐵. Prove that ∠𝐴𝐶𝐵 is a right angle. b. Draw a third radius (𝑃𝐶). What types of triangles are △ 𝐴𝑃𝐶 and △ 𝐵𝑃𝐶? How do you know? c. Using the diagram that you just created, develop a strategy to prove Thales’ theorem. d. Label the base angles of △ 𝐴𝑃𝐶 as 𝑏° and the base angles of △ 𝐵𝑃𝐶 as 𝑎°. Express the measure of ∠𝐴𝐶𝐵 in terms of 𝑎° and 𝑏°. e. How can the previous conclusion be used to prove that ∠𝐴𝐶𝐵 is a right angle?8 Exercises 1. 𝐴𝐵 is a diameter of the circle shown. The radius is 12.5 cm, and 𝐴𝐶 = 7 cm. a. Find 𝑚∠𝐶. b. Find 𝐴𝐵. c. Find 𝐵𝐶. 2. In the circle shown, 𝐵𝐶 is a diameter with center 𝐴. a. Find 𝑚∠𝐷𝐴𝐵. b. Find 𝑚∠𝐵𝐴𝐸. c. Find 𝑚∠𝐷𝐴𝐸. Lesson Summary Theorems: THALES’ THEOREM: If 𝐴, 𝐵, and 𝐶 are three different points on a circle with a diameter ̅𝐴𝐵̅̅̅ , then ∠𝐴𝐶𝐵 is a right angle. CONVERSE OF THALES’ THEOREM: If △ 𝐴𝐵𝐶 is a right triangle with ∠𝐶 the right angle, then 𝐴, 𝐵, and 𝐶 are three distinct points on a circle with a diameter 𝐴𝐵. Therefore, given distinct points 𝐴, 𝐵, and 𝐶 on a circle, △ 𝐴𝐵𝐶 is a right triangle with ∠𝐶 the right angle if and only if 𝐴𝐵 is a diameter of the circle. • Given two points 𝐴 and 𝐵, let point 𝑃 be the midpoint between them. If 𝐶 is a point such that ∠𝐴𝐶𝐵 is right, then 𝐵𝑃 = 𝐴𝑃 = 𝐶𝑃. Relevant Vocabulary CIRCLE: Given a point 𝐶 in the plane and a number 𝑟 > 0, the circle with center 𝐶 and radius 𝑟 is the set of all points in the plane that are distance 𝑟 from the point 𝐶. RADIUS*: May refer either to the line segment joining the center of a circle with any point on that circle (a radius) or to the length of this line segment (the radius). DIAMETER: May refer either to the segment that passes through the center of a circle whose endpoints lie on the circle (a diameter) or to the length of this line segment (the diameter). CHORD: Given a circle 𝐶, and let 𝑃 and 𝑄 be points on 𝐶. 𝑃𝑄 is called a chord of 𝐶. CENTRAL ANGLE: A central angle of a circle is an angle whose vertex is the center of a circle. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
## The effect of adding b to the linear function y = ax + b In the previous post, we have learned about the effects of a in the linear function with equation y = ax. In this post, we learn about the effects of adding b to that equation. That is, we want to learn the effects of b in the linear function with equation y = ax + b. Consider the graph of the functions y = x, y = x + 2 and y = x – 3. The table of values (click figure to enlarge) below shows the corresponding y values of the three linear functions. The effect of adding 2 to the function y = x adds 2 to all the y values of y = x. This implies that in the graph, all the points with corresponding x values are moved 2 units above the graph of y = x. In addition, in the graph of y = x – 3, the -3 subtracts 3 from all the y values of y = x. In effect, all the points with corresponding x values are moved 3 units below the graph of y = x. In addition, for y = x, if x = 0, y = 0. That means that the graph passes through the origin. On the other hand, for y = x + 2, when x = 0, then y = 0 + 2 = 2. This means that the graph passes through y = 2. Further, for y = 0 – 3 = -3. This means that the graph passes through y = -3. These are shown both in the table above and in the graph below. » Read more
Albert Teen powered by YOU ARE LEARNING: Algebraic Fractions: Division # Algebraic Fractions: Division ### Algebraic Fractions: Division Algebraic fractions can also be divided together to form new fractions containing algebra. 1 Dividing algebraic fractions has one key first step we need to do that before anything else 2 We want to divide these fractions $\dfrac{3x}{2} \div \dfrac{5}{7x}$ . What is the first step we need to do? 3 Which fraction do we need to flip here? $\dfrac{3x}{2} \div \dfrac{5}{7x}$ 4 This means that our division becomes a multiplication: $\dfrac{3x}{2} \div \dfrac{5}{7x}=\dfrac{3x}{2} \times\dfrac{7x}{5}$ 5 We then multiply to get our answer. What is $\dfrac{3x}{2} \times\dfrac{7x}{5}$? 6 We've found our final answer! $\dfrac{3x}{2} \div \dfrac{5}{7x}=\dfrac{21x^2}{10}$ 1 Let's try another example. $\dfrac{8a}{3}\div \dfrac{a^2}{6}$ 2 What is the first step we need to take? $\dfrac{8a}{3}\div \dfrac{a^2}{6}$ 3 We flip, or invert, the second fraction and the division becomes multiplication. $\dfrac{8a}{3}\div \dfrac{a^2}{6}=\dfrac{8a}{3}\times \dfrac{6}{a^2}$ 4 Now we multiply - first let's cancel all the common factors. 5 Now we have cancelled our common factors, what is the final answer? $\dfrac{8{\cancel{a}}}{{\cancel{3}}1}\times \dfrac{{\cancel{6}}2}{a^{\cancel{2}}}$ 6 We could have chosen to multiply first. $\dfrac{8a}{3}\times \dfrac{6}{a^2}=\dfrac{48a}{3a^2}$ 7 We then need to simplify $\dfrac{48a}{3a^2}$. What common factor is there in the numerator and denominator? 8 Simplify $\dfrac{48a}{3a^2}$ using the common factor $3a$. 9 Both methods gave the same answer! 😎 $\dfrac{8a}{3}\div \dfrac{a^2}{6}=\dfrac{16}{a}$ 1 Give this one a try. We'll go through it step by step. $\dfrac{3a}{7}\div \dfrac{5a^2}{14}$ 2 First we invert the second fraction. $\dfrac{3a}{7}\div \dfrac{5a^2}{14}$ 3 Now we have $\dfrac{3a}{7}\times \dfrac{14}{5a^2}$, what common factors are there? 4 Now we have the common factors, $7$ and $a$, what is $\dfrac{3a}{7}\times \dfrac{14}{5a^2}$ in its simplest terms? Try this one - remember to make sure your answer is fully simplified. $\dfrac{3y^2}{4}\div \dfrac{4y}{3}$ This one is a little harder $\dfrac{12x-8}{3x}\div \dfrac{4x}{3}$. 1 Which option gives the correct first step? 2 We now have $\dfrac{12x-8}{3x}\times \dfrac{3}{4x}$. Our next step is to factorise the numerator of the first fraction. 3 Factorise means to find a common factor and add in brackets. What is the common factor in the terms $12x-8$? 4 The common factor in $12x-8$ is $4$. We can now write this as $4(3x-2)$. 5 Our fraction is now $\dfrac{4(3x-2)}{3x}\times \dfrac{3}{4x}$. One common factor is $3$, what is the other? 6 We can cancel our common factors $3$ and $4$. What answer do we get? $\dfrac{{\cancel{4}}(3x-2)}{{\cancel{3}}x}\times \dfrac{{\cancel{3}}}{{\cancel{4}}x}$ 7 Well done! There were lots of steps in this one. $\dfrac{12x-8}{3x}\div \dfrac{4x}{3}=\dfrac{3x-2}{x^2}$. Try another one! Simplify $\dfrac{2x-6}{3}\div \dfrac{2}{x}$. 1 In summary, to divide algebraic fractions flip the second fraction and multiply. 2 It must be the second fraction that you flip, or invert $\dfrac{3x}{10}\div {\color{orange}{\dfrac{2}{5x}}}=\dfrac{3x}{10}\times {\color{orange}{\dfrac{5x}{2}}}$ 3 Multiply the numerators and the denominators $\dfrac{3x}{10} \times \dfrac{5x}{2}=\dfrac{15x^2}{20}$ 4 Remember to check if your answer can simplify. $\dfrac{15x^2}{20}=\dfrac{3x^2}{4}$ 5 A trickier one might be: $\dfrac{4}{3x}\div \dfrac{2}{3x-12}$ 6 First we must invert the second fraction. $\dfrac{4}{3x}\div \dfrac{2}{3x-12}=\dfrac{4}{3x}\times \dfrac{3x-12}{2}$ 7 Then we factorise the numerator in the second fraction. $\dfrac{4}{3x}\times \dfrac{3x-12}{2}=\dfrac{4}{3x}\times \dfrac{3(x-4)}{2}$ 8 Then we can see our common factors. $\dfrac{{\cancel{4}}2}{{\cancel{3}}x}\times \dfrac{{\cancel{3}}(x-4)}{{\cancel{2}}}$ 9 Finally we multiply to give our answer: $\dfrac{{\cancel{4}}2}{{\cancel{3}}x}\times \dfrac{{\cancel{3}}(x-4)}{{\cancel{2}}}=\dfrac{2x-8}{x}$
# Dilations on the Coordinate Plane Videos and solutions to help Grade 8 students learn how to describe the effect of dilations on two-dimensional figures using coordinates. Related Topics: Lesson Plans and Worksheets for Grade 8 Lesson Plans and Worksheets for all Grades ## New York State Common Core Math Module 3, Grade 8, Lesson 6 ### Lesson 6 Student Outcomes • Students describe the effect of dilations on two-dimensional figures using coordinates. ### Lesson 6 Summary Dilation has a multiplicative effect on the coordinates of a point in the plane. Given a point (x, y) in the plane, a dilation from the origin with scale factor r moves the point (x, y) to (rx, ry) For example, if a point (3, -5) in the plane is dilated from the origin by a scale factor of r = 4, then the coordinates of the dilated point are (4 × 3, 4 × (-5)) = (12, - 20) ### NYS Math Module 3 Grade 8 Lesson 6 Classwork Example 1 Students learn the multiplicative effect of scale factor on a point. Note that this effect holds when the center of dilation is the origin. In this lesson, the center of dilation will always be assumed to be (0, 0) Example 2 Students learn the multiplicative effect of scale factor on a point. Example 3 The coordinates in other quadrants of the graph are affected in the same manner as we have just seen. Based on what we have learned so far, given point A = (-2. 3) predict the location of A' when A is dilated from a center at the origin, (0, 0) by scale factor r = 3. Exercises 1 - 5 1. Point A = (7, 9) is dilated from the origin by scale factor r = 6. What are the coordinates of point A'? 2. Point B = (-8, 5) is dilated from the origin by scale factor r = 1/2. What are the coordinates of point B'? 3. Point C = (6, -2) is dilated from the origin by scale factor r = 3/4. What are the coordinates of point C'? 4. Point D = (0, 11) is dilated from the origin by scale factor r = 4. What are the coordinates of point D'? 5. Point E = (-2, -5) is dilated from the origin by scale factor r = 3/2. What are the coordinates of point E'? Example 4 Students learn the multiplicative effect of scale factor on a two dimensional figure. Example 5 Students learn the multiplicative effect of scale factor on a two-dimensional figure. Exercises 6 - 8 6. The coordinates of triangle ABC are shown on the coordinate plane below. The triangle is dilated from the origin by scale factor r = 12. Identify the coordinates of the dilated triangle A'B'C'. 7. Figure DEFG is shown on the coordinate plane below. The figure is dilated from the origin by scale factor r = 2/3. Identify the coordinates of the dilated figure D'E'F'G', then draw and label figure D'E'F'G' on the coordinate plane. 8. The triangle ABC has coordinates (3, 2) (12, 3) and (9, 12). Draw and label triangle ABC on the coordinate plane. The triangle is dilated from the origin by scale factor r = 1/3. Identify the coordinates of the dilated triangle A'B'C', then draw and label triangle A'B'C' on the coordinate plane.
# How to compare fractions Sometimes we need to compare two fractions to discover which is larger or smaller. There are two main ways to compare fractions: using decimals, or using the same denominator. ## The Decimal Method of Comparing Fractions Convert each fraction to decimals, and then compare the decimals. • Convert each fraction to a decimal. • We can use a calculator (3÷8 and 5÷12), or the method on Converting Fractions to Decimals. • Anyway, these are the answers I get: 3 8 = 0.375, and 5 12 = 0.4166… So 5 12 is bigger. ### The Same Denominator Method 1. The denominator is the bottom number in a fraction. 2. It shows how many equal parts the item is divided into 3. When two fractions have the same denominator they are easy to compare: is less than 4 9 5 9 But when the denominators are not the same we need to make them the same (using Equivalent Fractions). Look at this: • When we multiply 8 × 3 we get 24, • and when we multiply 12 × 2 we also get 24, so let's try that (important: what we do to the bottom we must also do to the top): × 3 3 = 9 8 24 × 3 and × 2 5 = 10 12 24 × 2 We can now see that 9 24 is smaller than 10 24 (because 9 is smaller than 10). so 5 12 is the larger fraction. is less than 3 8 5 12 ### Making the Denominators the Same There are two main methods to make the denominator the same: • Common Denominator Method, or the • Least Common Denominator Method They both work, use which one you prefer! Using the Common Denominator method we multiply each fraction by the denominator of the other: × 15 5 = 75 6 90 × 15 and × 6 11 = 66 15 90 × 6 We can see that 75 90 is the larger fraction (because 75 is more than 66) so 5 6 is the larger fraction. is more than 5 6 11 15 Copyright © 2020 MathsIsFun.com ## Using the Number Line to Compare Fractions In previous posts, we’ve learned how to place different numbers on the number line. Today we’re going to learn how to represent fractions on a number line. This is very useful when we need to compare them. ### Fractions with the same denominator As we know, it’s easy to compare fractions with the same denominator. In these cases, the fraction with the larger numerator is the bigger fraction. For example, eight-thirds is bigger than three thirds. ### Fractions with the same numerator Something that seems a bit more complicated, but only on the surface, is comparing fractions with the same numerator. In this situation, the fraction with the smaller denominator is always the bigger fraction. For example, fifteen-thirds is bigger than fifteen fourths. See also: Dangerous and disgusting medicines of centuries past ### Using the number line for fractions with different numerators and denominators The comparison becomes more complicated when the fractions we’re comparing don’t have the same numerator or denominator. Which fraction is greater: eleven thirds or thirteen fourths? This is where the number line is particularly useful. We can use it to give us a visual representation of the fractions, which we can then compare without having to do any mathematical calculations. Although the number of subdivisions of each unit will be different on the number line for each fraction, it’s important to make sure the unit is the same size on each one. Then we can compare the fractions visually. Once we’ve placed the two fractions with different numerators and denominators on the number line, we can tell which of the two is the greater simply by looking. Without having to make any calculations or operate with fractions, we can see that thirteen-fourths is less than eleven thirds. The first fraction is placed close to the three, while the second fraction is closer to the four. • It’s easy, right? • If you want to learn more about the number line and other mathematical content, log in to Smartick and try it for free. • Learn More: • Learn and Practice How to Find Sums of Fractions • Learning How to Subtract Fractions • How to Find a Sum of Fractions • Learn and Practice How to Multiply Fractions • Learn How to Subtract Fractions ## Comparing Fractions with Different Denominators • As we've begun our study of fractions, we've learned how to identify equivalent fractions. • We used the equal fractions property to create equivalent fractions. • We will also use this same property to help us compare fractions with different denominators. If the numerator and denominator of a fraction are multiplied (or divided) by the same nonzero number, then the resulting fraction is equivalent to the original fraction. Remember how we multiplied the numerator and denominator by the same number in order to create equivalent fractions? We will use the same process to compare fractions. Take a look…. ### Example 1 – Are these two fractions equal? If you are confused, please take a look at the video lesson below that will fully explain Example 1. That example was pretty easy since 8 is a factor of 24. Now let's take a look at another example that is a little harder. In this example we will use our knowledge of least common multiple. See also: Vinegar + baking soda = the ultimate cleanser? ### Example 2 – Which fraction is greater? Take a look at this example on video! You must always make sure that the fractions have the same denominator. Let's take a look at one more example for comparing fractions with different denominators. This time we'll compare more than two fractions. ### Example 3 – Ordering from least to greatest Hopefully you now feel comfortable comparing and ordering proper fractions. 1. Home 2. > 3. Fractions 4. > 5. Comparing Fractions ## 8 Ways to Compare Fractions in the Late Elementary Grades In this post, I want to share a collection of methods that you can use with late elementary students, in Grades 3 to 5, to compare fractions. Plus, download a FREE set of problems in which students are asked to use and to think about different strategies for comparing fractions. When I was younger, I only learned one way to compare fractions. It wasn’t until I was tutoring math as a community college student that I started to explore the different ways of representing and comparing fractions. There have been a few times as a teacher when I have had “Aha!” moments that have given me even deeper insights into fractions. Students’ ability to justify their choice of strategy for comparing fractions is part of Standard for Mathematical Practice 3, “construct viable arguments and critique the reasoning of others.” ### Eight Ways To Compare Fractions In Grades 3–5 Below I've outlined eight ways to compare fractions. Once you develop these different strategies for comparing fractions with your students, check their reasoning skills using the download for this post. My printable Explaining Your Reasoning to Compare Fractions Activity asks students to compare fractions using given strategies, by choosing between strategies, and by creating a problem of their own. ### 1. Equivalent Denominators This is the easiest situation in which to compare fractions. If two fractions have equivalent denominators, then compare the numerators to determine which faction is greater. Students at the earliest stages of learning about fractions should be able to do this. ## COMPARING FRACTIONS Example 1. 23 is to 58 as 2 × 8 is to 3 × 5 as 16 is to 15. 16 and 15 are the numerators we would get if we expressed 23 and 58 with the common denominator 24. as And since 16 is larger than 15, we would know that 23 is larger than 58 . Example 2. Which is larger, 47 or 59 ? • Answer. On cross-multiplying, • as • 36 is to 35. • 36 is larger than 35. Therefore, • Note: We must begin multiplying with the numerator on the left: • 4 × 9. See also: How to divide fractions using invert and multiply Example 3. 14 is to 12 as which whole numbers? 1. Answer. On cross-multiplying, 2. as 3. 2 is to 4. 4. That is, Example 4. What ratio has 2½ to 3? Answer. First, express 2½ as the improper fraction 52 . Then, treat the whole number 3 as a numerator, and cross-multiply: 2½ is five sixths of 3. Equivalently, since 3 = 62 (Lesson 21, Question 2), then 52 is to 62 as 5 is to 6. In general: To express the ratio of a fraction to a whole number, multiply the whole number by the denominator. 67 is to 3 as 6 is to 21. For an application of this, see Lesson 26. Example 5. On a map, 34 of an inch represents 60 miles. How many miles does 2 inches represent? Solution. Proportionally, 34 of an inch is to 2 inches as 60 miles is to ? miles. Therefore: 3 is to 8 as 60 miles is to ? miles. • Since 20 × 3 = 60, then 20 × 8 = 160 miles. • The theorem of the same multiple. • Or, inversely: 8 is to 3 as ? miles is to 60 miles. 1. Now, 2. 8 is two and two thirds times 3. 3. (Lesson 18, Example 5.) Therefore, the missing term will be Two and two thirds times 60 = Two times 60 + two thirds of 60 (Lesson 16) = 120 + 40 = 160 miles. More than or less than ½ ## Comparing fractions This is a more difficult task the two comparisons above. Depending on your child’s level, this maybe a step you skip and then come back to later. These fractions are compared by changing the denominators to a common number. This can be done by multiplying the top and bottom of the fraction by the same number since this will give a fraction with an equivalent value. Do this as shown below to both fractions to get a common denominator and then compare them. So, in the example above, now that the denominators are equal, the fraction with the greatest numerator is the largest. Some students may respond to questions about which fraction is bigger by saying it depends on what size the whole is and, depending on the wording of the question, they may actually be correct! Encourage them to think in terms which is the larger or smaller share and be precise with your questions. e.g. “which is the largest fraction?” not just “which is the largest?”
# When do we use the Chain Rule? ## The Formula If s(x)=f(g(x)), and if g is differentiable at x and f is differentiable at g(x) then $$s'(x)=f'(g(x))g'(x)$$ At this point we have only found the derivative of functions built with arithmetic operations (addition, subtraction, multiplication, division). If two simpler functions were combined using multiplication we used the product rule to find the derivative, if they were combined using division we used the quotient rule, and if addition or subtraction was used the addition/subtraction rule implemented. # What is composition? Answer: We will now look at another operation, composition , which is when one function is embedded in another (f(g(x))). In the imagine below g associates g(x) with x and the function f associates f(g(x)) with g(x). Therefore, the composition of f with g associates f(g(x)) with each element x in the domain of g for which g(x) is in the domain of f. This figure shows g(x) evalualted in f(x) . Example of Compositon Suppose $f(x)=\sqrt{x}$ and $g(x)=2x-3$. Then: $$f(g(x))=\sqrt{2x-3}$$ Similarly, $$g(f(x))=2\sqrt{x-3}$$ Now the question is can we find the derivative of $s(x)$ given $s(x)=f(g(x))$? More specifically can we express the first derivative of s in terms of the first derivatives of f and g? The answer is yes, and the \textit{chain rule} help us accomplish this. In fact, the chain rule will make the differentiation of many functions a lot easier. # Identifying the composition of two functions However, before we begin taking the derivative of functions using the chain rule, if is helpful for us to practice identifying the composition of functions. We will start off with one function inside another (f(gx)) and then progress into the composition of three functions h(f(g(x))). Once we become comfortable recognizing the order of the composition (inner most, next inner most, $\cdots$, outer most), taking the derivative with the chain rule becomes rather straight forward. ## Practice Identifying Composition of functions Function Inside Outisde Composition $sin(2x)$ $2x$ $sin(x)$ $f(g(x))$ $e^{5x+1}$ $(5x+1)$ $e^x$ $f(g(x))$ $\sqrt{3x^5}$ $3x^5$ $\sqrt{x}$ $f(g(x))$ $(5x^2+ln(x))^{10}$ $5x^2+ln(x)$ $x^{10}$ $f(g(x))$ $ln\big(\frac{x}{sin(x)}\big)$ $\frac{x}{sin(x)}$ $ln(x)$ $f(g(x))$ $\big(\frac{2x-5}{1-x^2}\big)^-5$ $\frac{2x-5}{1-x^2}$ $x^-5$ $f(g(x))$ $tan^2(x)$ $tan(x)$ $x^2$ $f(g(x))$ ## Identifying the composition of three functions h(f(g(x))) g(x) f(x) h(x) Composition $ln(\sqrt{4x})$ $4x$ $\sqrt{x}$ $ln(x)$ $h(f(g(x))$ $cos^4(3x+2)$ $(3x+2)$ $cos(x)$ $x^4$ $h(f(g(x))$ $tan\big((6x)^{10}\big)$ $6x$ $x^{10}$ $tan(x)$ $h(f(g(x))$ $e^{e^{\pi x}}$ $\pi x$ $e^x$ $e^x$ $h(f(g(x))$ $csc(\sqrt[3]{1-sin(x)})$ $sin(x)$ $\sqrt[3]{x}$ $csc(x)$ $h(f(g(x)))$ # What is the chain rule? If s(x)=f(g(x)), and if g is differentiable at x and f is differentiable at g(x) then $$s'(x)=f'(g(x))g'(x)$$ ## The Chain Rule in Words To find the derivative of s(x), find the derivative of the outside function $(f')$ evaluated at the inside function left alone $f'(g(x))$ times the derivative of the inside function $f'(g(x))g'(x)$. $$\frac{d}{dx}[f(g(x))]=f'(g(x))$$ Basically, we start with the outer most function take it derivative, leave everything else the same, multiplied by the derivative of the next outer most function. If there are multipliable functions, we keep work from outside to inside using the same technique . ## Examples of the Chain Rule Example 1 $$s(x)=(3x^2+1)^5$$ Example 2 $$s(x)=sin^3(x)$$ Example 3 $$s(x)=\sqrt{ln(x)}$$ $$s(x)=e^{x^2}$$ $$s(x)=\big(\frac{3x+4}{6x-1}\big)^3$$
# Solve linear system graphically In this blog post, we will show you how to Solve linear system graphically. Our website will give you answers to homework. ## Solving linear system graphically There are also many YouTube videos that can show you how to Solve linear system graphically. The roots of the equation are then found by solving the Quadratic Formula. The parabola solver then plots the points on a graph and connecting them to form a parabola. Finally, the focus and directrix of the parabola are found using the standard form of the equation (y = a(x-h)^2 + k). Solve slope intercept form is an algebraic equation that can be used to find the y-intercept of a line. It uses the slope of two points on a graph and the y-intercet to find the y-intercept. It is used in algebra classes and in statistics. To solve it, first find the equation of the line: b>y = mx + c/b> where b>m/b> is the slope and b>c/b> is the y-intercept. Add them up for both sides: b>y + mx = c/b>. Solve for b>c/b>: b>c = (y + mx) / (m + x)/b>. Substitute into your original equation: b>y = mx + c/b>. Finally, take your original data points and plug them into this new equation to find the y-intercept: b>y = mx + c/b>. In words, solve "for c" by plugging your data into both sides of your equation as you would solve any algebraic equation. Then solve for "y" by adjusting one side until you get "c" back on top. Example 1: Find the y-intercept if this line is graphed below. A math answer scanner is a tool that can be used to check answers to math problems. This can be a useful tool for students who want to check their work, or for teachers who want to check answers to problems before giving them to their students. In the case of separable differential equations, it is possible to solve the system by separating it into several smaller sub-models. This approach has the advantage that it allows for a more detailed analysis of the source of error. In addition, it can be used to implement model validation and calibration. Furthermore, the problem can also be solved in parallel using different approaches (e.g., different solvers). In addition, since each sub-model treats only a small part of the overall system, it is possible to use a very limited computer memory and computational power. Separable differential equations solvers are divided into two main groups: deterministic and stochastic. Stochastic solvers are based on probability models, which simulate the relative frequencies of system events as they occur. The more frequently an event occurs, the higher its probability of occurring; therefore, a stochastic solver will tend to converge faster than a deterministic solver when used in parallel. Deterministic solvers are based on probabilistic models that estimate the probability of each state transition occurring so that they can predict what the next state will be given any input data. Both types of solvers can be classified further into two major categories: explicit and implicit. Explicit models have explicit equations describing how to go from one state to another; implicit models do not have explicit equations but instead rely ## We solve all types of math problems I am loving the app! Been using it for about a month now and all of the things I don’t understand in math are solved with this app. It's great and I really recommend it because it's simple and provides you with all the steps in solving a problem and has many ways of solving them. Harlie Hughes Perfect, it not only gives you answers, but step by step methods. If the camera function isn’t working, you can choose to use the type function where you can just type in the sum. Love the app I'm 10 and don't really know math but this helps me so much!! amazing app Thalia Garcia
Linear_Transformations # 09 kilometers herea0andb1609 This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: each quiz X = old grade X* = new grade X* = x + 5 a = 5, b = 1 We changed the origin ( 0 = 5 ) Examples Examples • If a distance x is measured in miles, the same distance in kilometers is: x* = 1.609(x) – That is 10 miles is the same as 16.09 kilometers. – Here a = 0 and b = 1.609. – This transformation changes the units without changing the origin a distance of 0 miles is the same as a distance of 0 kilometers. Example Example • A temperature x measured in degrees Fahrenheit (°F) must be reexpressed in degrees Celsius (°C) to align with every other country in the world. – The transformation is: X* = 5/9( X – 32 ) • What does “a” and “b” equal? • a = 160/9 • b = 5/9 – This changes both the unit size and the origin of the measurements. 0º C is 32º F. Problem #1 Problem #1 • The average salary for an employee at Acme Corporation is \$30,000 per year. This year, management awards the following bonuses to every employee. – A Christmas bonus of \$500 – An incentive bonus equal to 10% of the employee’s salary. • What is the mean bonus received by employees? – (A) \$500 (B) \$3,000 (C) \$3,500 (D) None of the above. (E) There is not enough information to answer this question. Problem #1 Continued Problem #1 Continued • The correct answer is C. To compute the bonus, management applies the following linear transformation to the each employee's salary: – x* = bx + a Y = 0.10 * x + 500 – where x* is the transformed variable (the bonus), x is the original variable (the salary), b is the multiplicative constant 0.10, and a is the additive constant 500. • Since we know that the mean salary is \$30,000, we can compute the mean bonus from the following equation: – Y = bx + a Y = 0.10 * \$30,000 + \$500 = \$3,500 Homework Homework • Textbook pp. 90 – 99 – # 23, 24, 35, 36, 46 • Chapter 3 Project Due Monday, September 12th!! • Unit 1 Test – Thursday, September 15th!... View Full Document ## This document was uploaded on 03/30/2014 for the course MATH AP Statist at Richard Montgomery High. Ask a homework question - tutors are online
# Sketching inequalities involving complex numbers ## Homework Statement Sketch all complex numbers z which satisfy the given condition: $\|z-i\|\geq\|z-1\|$ ## Homework Equations $z=a+bi$ $\|z\|=\sqrt{a^{2}-b^{2}}$ ## The Attempt at a Solution First I find the boundary between the regions where the inequality holds and does not hold by replacing the inequality sign with an equality sign. Then I substitute z=a+bi into the equation and solve for b: $\|(a+bi)-i\|=\|(a+bi)-1\|$ $\sqrt{(a)^{2}+(bi-i)^{2}}=\sqrt{(a-1)^{2}-b^{2}}$ $a=-b$ So the boundary cuts through the origin and has a slope of -1. To find which side of the boundary the equality holds, I can plug random values from different sides of the boundary, in this case z=-1-i and z=1+i. Doing this, I get $\sqrt{-3}\geq\sqrt{3}$ and $\sqrt{1}\geq\sqrt{-1}$ What do I do from here? $\|z\|=\sqrt{a^{2}-b^{2}}$ This is incorrect. $\|z\| = \sqrt{a^2+b^2}$ You can also solve this without using a and b, in a rather simple way. This how I would try, by drawing the graph. $\|z-i\|$ represents all circles(vectors) with center at (0,1) and $\|z-1\|$ represents circles with center (1,0). The perpendicular bisector of the line joining these two points gives the condition when $\|z-i\|=\|z-1\|$. So, which side of the curve would you sketch z, such that $\|z-i\|>\|z-1\|$? Try drawing a graph, and it should become really simple to figure out Ah, I thought i was included in the formula. I re-did the entire thing with the correct form, and I got it now. Thanks! :) Ah, I thought i was included in the formula. I re-did the entire thing with the correct form, and I got it now. Thanks! :) Graphically, modulus/absolute value represents the magnitude of distance between the point and origin. Keeping that in mind, its easy to see why it comes out to be $\sqrt{a^2+b^2}$
Verify commutativity of addition of rational numbers pairs of rational numbers -4 and $$\dfrac{4}{-7}$$ Asked by Sakshi \| 1 year ago \| 57 ##### Solution :- Firstly we need to convert the denominators to positive numbers. $$\dfrac{ 4}{-7} = \dfrac{(4 × -1)}{ (-7 × -1)} = \dfrac{-4}{7}$$ By using the commutativity law, the addition of rational numbers is commutative. $$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{c}{d} + \dfrac{a}{b}$$ In order to verify the above property let us consider the given fraction $$\dfrac{ -4}{1} and \dfrac{-4}{7}$$ as $$\dfrac{ -4}{1}+ \dfrac{-4}{7}$$ and $$\dfrac{-4}{7}+ \dfrac{ -4}{1}$$ The denominators are 1 and 7 By taking LCM for 1 and 7 is 7 We rewrite the given fraction in order to get the same denominator Now, $$\dfrac{-4}{1} = \dfrac{(-4 × 7)}{ (1×7)} = \dfrac{-28}{7}$$ $$\dfrac{-4}{7} = \dfrac{(-4 ×1) }{ (7 ×1)} =\dfrac{ -4}{7}$$ Since the denominators are same we can add them directly $$\dfrac{ -28}{7} +\dfrac{ -4}{7} = \dfrac{(-28 + (-4))}{7}$$ $$= \dfrac{(-28-4)}{7} = \dfrac{-32}{7}$$ $$\dfrac{ -4}{7} +\dfrac{ -4}{1}$$ The denominators are 7 and 1 By taking LCM for 7 and 1 is 7 We rewrite the given fraction in order to get the same denominator Now, $$\dfrac{ -4}{7}= \dfrac{(-4 ×1) }{ (7 ×1) }= \dfrac{ -4}{7}$$ $$\dfrac{ -4}{1} =\dfrac{ (-4 × 7) }{ (1×7)} =\dfrac{ -28}{7}$$ Since the denominators are same we can add them directly $$\dfrac{ -4}{7} + \dfrac{ -28}{7} = \dfrac{(-4 + (-28))}{7}$$ $$= \dfrac{(-4-28)}{7} = \dfrac{-32}{7}$$ $$\dfrac{ -4}{1} +\dfrac{ -4}{7} = \dfrac{-4}{7} + \dfrac{-4}{1}$$ is satisfied. Answered by Sakshi \| 1 year ago ### Related Questions #### By what number should 1365 be divided to get 31 as quotient and 32 as remainder? By what number should 1365 be divided to get 31 as quotient and 32 as remainder? #### Which of the following statement is true / false? Which of the following statement is true / false? (i) $$\dfrac{ 2 }{ 3} – \dfrac{4 }{ 5}$$ is not a rational number. (ii) $$\dfrac{ -5 }{ 7}$$ is the additive inverse of $$\dfrac{ 5 }{ 7}$$ (iii) 0 is the additive inverse of its own. (iv) Commutative property holds for subtraction of rational numbers. (v) Associative property does not hold for subtraction of rational numbers. (vi) 0 is the identity element for subtraction of rational numbers. #### If x = 4 / 9, y = – 7 / 12 and z = – 2 / 3, then verify that x – (y – z) ≠ (x – y) – z If x = $$\dfrac{4 }{ 9}$$, y =$$\dfrac{-7 }{ 12}$$ and z = $$\dfrac{-2 }{ 3}$$, then verify that x – (y – z) ≠ (x – y) – z If x = $$\dfrac{ – 4 }{ 7}$$ and y = $$\dfrac{2 }{ 5}$$, then verify that x – y ≠ y – x Subtract the sum of $$\dfrac{ – 5 }{ 7} and\dfrac{ – 8 }{ 3}$$ from the sum of $$\dfrac{5 }{ 2} and \dfrac{– 11 }{ 12}$$
Edit Article # wikiHow to Solve for X There are a number of ways to solve for x, whether you're working with exponents and radicals or if you just have to do some division or multiplication. No matter what process you use, you always have to find a way to isolate x on one side of the equation so you can find its value. Here's how to do it: ### Method 1 Using a Basic Linear Equation 1. 1 Write down the problem. Here it is: • 22(x+3) + 9 - 5 = 32 2. 2 Resolve the exponent. Remember the order of operations: PEMDAS, which stands for Parentheses, Exponents, Multiplication/Division, and Addition/Subtraction. You can't resolve the parentheses first because x is in the parentheses, so you should start with the exponent, 22. 22 = 4 • 4(x+3) + 9 - 5 = 32 3. 3 Do the multiplication. Just distribute the 4 into (x +3). Here's how: • 4x + 12 + 9 - 5 = 32 4. 4 Do the addition and subtraction. Just add or subtract the remaining numbers. Here's how: • 4x+21-5 = 32 • 4x+16 = 32 • 4x + 16 - 16 = 32 - 16 • 4x = 16 5. 5 Isolate the variable. To do this, just divide both sides of the equation by 4 to find x. 4x/4 = x and 16/4 = 4, so x = 4. • 4x/4 = 16/4 • x = 4 6. 6 Check your work. Just plug x = 4 back into the original equation to make sure that it checks out. Here's how: • 22(x+3)+ 9 - 5 = 32 • 22(4+3)+ 9 - 5 = 32 • 22(7) + 9 - 5 = 32 • 4(7) + 9 - 5 = 32 • 28 + 9 - 5 = 32 • 37 - 5 = 32 • 32 = 32 ### Method 2 With Exponents 1. 1 Write down the problem. Let's say you're working with this problem where the x term includes an exponent: • 2x2 + 12 = 44 2. 2 Isolate the term with the exponent. The first thing you should do is combine like terms so that all of the constant terms are on the right side of the equation while the term with the exponent is on the left side. Just subtract 12 from both sides. Here's how: • 2x2+12-12 = 44-12 • 2x2 = 32 3. 3 Isolate the variable with the exponent by dividing both sides by the coefficient of the x term. In this case, 2 is the x coefficient, so divide both sides of the equation by 2 to get rid of it. Here's how: • (2x2)/2 = 32/2 • x2 = 16 4. 4 Take the square root of each side of the equation. Taking the square root of x2 will cancel it out. So, take the square root of both sides. You'll get x left over on one side and the square root of 16, 4, on the other side. Therefore, x = 4. 5. 5 Check your work. Just plug x = 4 back in to the original equation to make sure it checks out. Here's how: • 2x2 + 12 = 44 • 2 x (4)2 + 12 = 44 • 2 x 16 + 12 = 44 • 32 + 12 = 44 • 44 = 44 ### Method 3 Using Fractions 1. 1 Write down the problem. Let's say you're working with the following problem:[1] • (x + 3)/6 = 2/3 2. 2 Cross multiply. To cross multiply, simply multiply the denominator of each fraction by the numerator of the other fraction. You'll be essentially multiplying in two diagonal lines. So, multiply the first denominator, 6, by the second numerator, 2, to get 12 on the right side of the equation. Multiply the second denominator, 3, by the first numerator, x + 3, to get 3 x + 9 on the left side of the equation. Here's how it will look: • (x + 3)/6 = 2/3 • 6 x 2 = 12 • (x + 3) x 3 = 3x + 9 • 3x + 9 = 12 3. 3 Combine like terms. Combine the constant terms in the equation to subtract 9 from both sides of the equation. Here's what you do: • 3x + 9 - 9 = 12 - 9 • 3x = 3 4. 4 Isolate x by dividing each term by the x coefficient. Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you're left with x = 1. 5. 5 Check your work. To check your work, just plug x back in to the original equation to make sure that it works. Here's what you do: • (x + 3)/6 = 2/3 • (1 + 3)/6 = 2/3 • 4/6 = 2/3 • 2/3 = 2/3 ### Method 4 Using Radical Signs 1. 1 Write down the problem. Let's say you're solving for x in the following problem:[2] • √(2x+9) - 5 = 0 2. 2 Isolate the square root. You have to move the part of the equation with the square root sign to one side of the equation before you can proceed. So, you'll have to add 5 to both sides of the equation. Here's how: • √(2x+9) - 5 + 5 = 0 + 5 • √(2x+9) = 5 3. 3 Square both sides. Just as you would divide both sides of an equation by a coefficient that is being multiplied by x, you would square both sides of an equation if x appears under the square root, or the radical sign. This will remove the radical sign from the equation. Here's how you do it: • (√(2x+9))2 = 52 • 2x + 9 = 25 4. 4 Combine like terms. Combine like terms by subtracting both sides by 9 so that all of the constant terms are on the right side of the equation while x remains on the left side. Here's what you do: • 2x + 9 - 9 = 25 - 9 • 2x = 16 5. 5 Isolate the variable. The last thing you have to do to solve for x is to isolate the variable by dividing both sides of the equation by 2, the coefficient of the x term. 2x/2 = x and 16/2 = 8, so you're left with x = 8. 6. 6 Check your work. Plug 8 back in to the equation for x to see if you get the right answer: • √(2x+9) - 5 = 0 • √(2(8)+9) - 5 = 0 • √(16+9) - 5 = 0 • √(25) - 5 = 0 • 5 - 5 = 0 ### Method 5 Using Absolute Value 1. 1 Write down the problem. Let's say you're trying to solve for x in the following problem:[3] • \|4x +2\| - 6 = 8 2. 2 Isolate the absolute value. The first thing you have to do is to combine like terms and get the terms inside the absolute value sign on one side. In this case, you would do it by adding 6 to both sides of the equation. Here's how: • \|4x +2\| - 6 = 8 • \|4x +2\| - 6 + 6 = 8 + 6 • \|4x +2\| = 14 3. 3 Remove the absolute value and solve the equation. This is the first and easiest step. You'll have to solve for x twice whenever you work with absolute value. Here's how you do it the first time: • 4x + 2 = 14 • 4x + 2 - 2 = 14 -2 • 4x = 12 • x = 3 4. 4 Remove the absolute value and change the sign of the terms on the opposite side of the equal sign before you solve. Now, do it again, except set the first part of the equation equal to -14 instead of 14. Here's how: • 4x + 2 = -14 • 4x + 2 - 2 = -14 - 2 • 4x = -16 • 4x/4 = -16/4 • x = -4 5. 5 Check your work. Now that you know that x = (3, -4), just plug both numbers back in to the equation to see that it works. Here's how: • (For x = 3): • \|4x +2\| - 6 = 8 • \|4(3) +2\| - 6 = 8 • \|12 +2\| - 6 = 8 • \|14\| - 6 = 8 • 14 - 6 = 8 • 8 = 8 • (For x = -4): • \|4x +2\| - 6 = 8 • \|4(-4) +2\| - 6 = 8 • \|-16 +2\| - 6 = 8 • \|-14\| - 6 = 8 • 14 - 6 = 8 • 8 = 8 ## Community Q&A Search • How I solve this problem: 15 : 1000 : x : 100? Write it as one fraction equal to another: 15 / 1000 = x / 100. Solve for x either by inspection or by cross-multiplication. • How do you solve (2x(x-1)=4)? wikiHow Contributor You would have to factor it out into a monomial equation using either the f.o.i.l. method, or generic squares. • Solve for x; Peter is twice as old as his son. Twelve years ago he was four times older than him. How old are they now? Let x be the son's current age. Then 2x is Peter's current age. Twelve years ago the son's age was (x-12), and Peter's age was (2x-12). Set up an equation stating that Peter's age 12 years ago was four times the son's age then. Solve for x and 2x. • How do I set up an equation for x from a word problem? wikiHow Contributor If you are told to find the value of something, let that "something" be x. For example, suppose there are two trees next to the beach. George planted a number of trees there. Now the total number of trees is 4. How many trees did George plant? Since you do not know the number of trees George planted, that number is x in your equation. Now you have x + 2 = 4. Solve for x, and you get x = 2. Thus the answer is George planted two trees. Use the same approach for other word problems. • How do I solve for x in the following equation? (x-3)(x-5)=9 wikiHow Contributor Multiply them together to get x^2 +2x-15=9, then subtract 9 to get x^2+2x-24=0. The only numbers that add up to 2 and equal 24 when multiplied are -4 and 6. Write it as (x-4)(x+6) = 0. So, x must be 4 or -6. • How do I draw y = (x)+3? Choose some small values for x, calculate the corresponding values for y, and plot the x-y pairs on a graph. It will be a straight line at a 45° angle (lower left to upper right), elevated three units vertically above the origin of the coordinate plane. • How do I solve 3 to the power of x is equals 81? The only practical way of solving this is trial and error: just try various convenient numbers for x until you solve the equation. As it happens, the square root of 81 is 9, and the square root of 9 is 3. Therefore, 3 raised to the fourth power is 81. • For the following problem: d = 4/5 x 8, the solution is d = 6.4. How did they get this answer? Multiply the numerator by the whole number, and then divide by the denominator. Thus, 4 x 8 = 32, divided by 5 = 6.4. 200 characters left ## Tips • Radicals, or roots, are another way of representing exponents. The square root of x = x^1/2. • To check your work, plug the value of x back into the original equation and solve. ## Article Info Categories: Algebra In other languages: Español: encontrar el valor de X, Italiano: Trovare la X, Português: Achar o Valor de X em uma Equação, Deutsch: Gleichungen nach X auflösen, Français: trouver X, 中文: 求X, Nederlands: X oplossen in een vergelijking, Русский: решать уравнение с одним неизвестным, Bahasa Indonesia: Mencari Nilai X Thanks to all authors for creating a page that has been read 121,077 times.
# What is the sum of the geometric sequence -1, 6, -36, ... if there are 7 terms? Sep 2, 2015 ${S}_{7} = - 39991$ #### Explanation: In this sequence we have: ${a}_{1} = - 1$, $q = - 6$, $n = 7$. If we apply the formula for ${S}_{n}$ we get: ${S}_{7} = \left(- 1\right) \cdot \frac{1 - {\left(- 6\right)}^{7}}{1 - \left(- 6\right)}$ ${S}_{7} = \left(- 1\right) \cdot \frac{1 + 279936}{7}$ ${S}_{7} = - 39991$ Aug 20, 2016 ${S}_{n} = - 39 , 991$ #### Explanation: In the given GP, we have the following: ${a}_{1} = - 1 , \mathmr{and} n = 7$ We can find $r = \frac{6}{-} 1 = - \frac{36}{6} = - 6$ There are 2 formulae for ${S}_{n}$ depending on whether r is a proper fraction or not. ${S}_{n} = \frac{a \left({r}^{n} - 1\right)}{r - 1} \text{ substitute with the values}$ ${S}_{n} = \frac{- 1 \left({\left(- 6\right)}^{7} - 1\right)}{- 6 - 1}$ ${S}_{n} = \frac{- 1 \left({\left(- 6\right)}^{7} - 1\right)}{- 6 - 1}$ ${S}_{n} = \frac{- 1 \left(- 279 , 937\right)}{- 7}$ ${S}_{n} = - 39 , 991$
The fraction 8/11 is same to 0.72727272727273 as soon as converted to a decimal. See listed below detalis on exactly how to convert the portion 8/11 come a decimal value. You are watching: 8/11 as a decimal ### Fraction to Decimal Converter Enter a portion value:Ex.: 1/2, 2 1/2, 5/3, etc. Note that 2 1/2 means two and fifty percent = 2 + 1/2 = 2.5 Fraction to decimal explained: ## How to transform from fraction to decimal? To easily transform a portion to a decimal, division the numerator (top number) through the denominator (bottom number). ### Example 1: exactly how to convert 4/8 come a decimal? Step 1:Divide 4 by 8: 4 ÷ 8 = 1 ÷ 2 = 0.5Step 2:Multiply the an outcome by 100 and add the decimal sign: 0.5 × 100%Answer: 4/8 = 50% ### Example 2: just how to transform 1 1/3 come a decimal? Step 1:Divide 1 by 3: 1 ÷ 3 = 0.3333Step 2:Add this worth to the the integer part: 1 + 0.3333 = 1.3333Step 3:Multiply the an outcome by 100 and add the decimal sign: 1.3333 × 100%Answer: 1 1/3 = 133.33% Note: the result was rounded come 2 decimal places. fractioninchesmm 1/640.01560.3969 1/320.03130.7938 3/640.04691.1906 1/160.06251.5875 5/640.07811.9844 3/320.09382.3813 7/640.10942.7781 1/80.12503.1750 9/640.14063.5719 5/320.15633.9688 11/640.17194.3656 3/160.18754.7625 13/640.20315.1594 7/320.21885.5563 15/640.23445.9531 1/40.25006.3500 fractioninchesmm 17/640.26566.7469 9/320.28137.1438 19/640.29697.5406 5/160.31257.9375 21/640.32818.3344 11/320.34388.7313 23/640.35949.1281 3/80.37509.5250 25/640.39069.9219 13/320.406310.3188 27/640.421910.7156 7/160.437511.1125 29/640.453111.5094 15/320.468811.9063 31/640.484412.3031 1/20.500012.7000 fractioninchesmm 33/640.515613.0969 17/320.531313.4938 35/640.546913.8906 9/160.562514.2875 37/640.578114.6844 19/320.593815.0813 39/640.609415.4781 5/80.625015.8750 41/640.640616.2719 21/320.656316.6688 43/640.671917.0656 11/160.687517.4625 45/640.703117.8594 23/320.718818.2563 47/640.734418.6531 3/40.750019.0500 See more: 3 Best Methods To Remember Debits On The Left Credits On The Right Song fractioninchesmm 49/640.765619.4469 25/320.781319.8438 51/640.796920.2406 13/160.812520.6375 53/640.828121.0344 27/320.843821.4313 55/640.859421.8281 7/80.875022.2250 57/640.890622.6219 29/320.906323.0188 59/640.921923.4156 15/160.937523.8125 61/640.953124.2094 31/320.968824.6063 63/640.984425.0031 11.000025.4000
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Basic Algebra Concepts Go to the latest version. # Chapter 12: Rational Equations and Functions Difficulty Level: Basic Created by: CK-12 ## Introduction The final chapter of this text introduces the concept of rational functions; that is, equations in which the variable appears in the denominator of a fraction. A common rational function is the inverse variation model, similar to the direct variation model you studied in a Concept in chapter 4. We finish the chapter with solving rational equations and using graphical representations to display data. ## Summary This chapter begins by talking about inverse variation models, the graphs of rational functions, and the division of polynomials. It then moves on to discuss rational expressions in detail, including the multiplication, division, addition, and subtraction of rational expressions. Next, instruction is given on solving rational equations by using proportions and by clearing denominators. Finally, surveys and samples in statistics are highlighted. ### Rational Equations and Functions; Statistics Review Define the following terms used in this chapter. 1. Inverse variation 2. Asymptotes 3. Hyperbola 4. Points of discontinuity 5. Least common multiple 6. Random sampling 7. Stratified sampling 8. Biased 9. Cherry picking 10. What quadrants are the branches of a hyperbola located if k<0\begin{align}k<0\end{align}? Are the following examples of direct variation or inverse variation? 1. The number of slices n\begin{align}n\end{align} people get from sharing one pizza 2. The thickness of a phone book given n\begin{align}n\end{align} telephone numbers 3. The amount of coffee n\begin{align}n\end{align} people receive from a single pot 4. The total cost of pears given that nectarines cost \$0.99 per pound For each variation equation: 1. Translate the sentence into an inverse variation equation. 2. Find k\begin{align}k\end{align}, the constant of variation. 3. Find the unknown value. 1. y\begin{align}y\end{align} varies inversely as x\begin{align}x\end{align}. When x=5,y=215\begin{align}x=5, y=\frac{2}{15}\end{align}. Find y\begin{align}y\end{align} when x=12\begin{align}x=- \frac{1}{2}\end{align}. 2. y\begin{align}y\end{align} is inversely proportional to the square root of y\begin{align}y\end{align}. When x=16,y=0.5625\begin{align}x=16, y=0.5625\end{align}. Find y\begin{align}y\end{align} when x=18\begin{align}x=\frac{1}{8}\end{align}. 3. Habitat for Humanity uses volunteers to build houses. The number of days it takes to build a house varies inversely as the number of volunteers. It takes eight days to build a house with twenty volunteers. How many days will it take sixteen volunteers to complete the same job? 4. The Law of the Fulcrum states the distance you sit to balance a seesaw varies inversely as your weight. If Gary weighs 20.43 kg and sits 1.8 meters from the fulcrum, how far would Shelley sit, assuming she weighs 36.32 kilograms? For each function: 1. Graph it on a Cartesian plane. 2. State its domain and range. 3. Determine any horizontal and/or vertical asymptotes the function may have. 1. y=4x\begin{align}y=\frac{4}{x}\end{align} 2. f(x)=24x\begin{align}f(x)=\frac{2}{4-x}\end{align} 3. g(x)=1x+1\begin{align}g(x)=\frac{-1}{x+1}\end{align} 4. y=63x+12\begin{align}y=\frac{6}{3x+1}-2\end{align} 5. f(x)=3x5\begin{align}f(x)=\frac{3}{x}-5\end{align} Perform the indicated operation. 1. 5a65b4b\begin{align}\frac{5a}{6}-\frac{5b}{4b}\end{align} 2. 43m+4m5\begin{align}\frac{4}{3m}+\frac{4m}{5}\end{align} 3. 3x2xy+43\begin{align}\frac{3x}{2xy}+\frac{4}{3}\end{align} 4. 25n2+2n2\begin{align}\frac{2}{5n-2}+\frac{2n}{2}\end{align} 5. 2x+13x+9x+53x+9\begin{align}\frac{2x+1}{3x+9}-\frac{x+5}{3x+9}\end{align} 6. 5m+n30n44m+n30n4\begin{align}\frac{5m+n}{30n^4}-\frac{4m+n}{30n^4}\end{align} 7. r64r212r+8r+64r212r+8\begin{align}\frac{r-6}{4r^2-12r+8}-\frac{r+6}{4r^2-12r+8}\end{align} 8. 216x3y2+x2y16x3y2\begin{align}\frac{2}{16x^3 y^2}+\frac{x-2y}{16x^3 y^2}\end{align} 9. n6n+2+2n5\begin{align}\frac{n-6}{n+2}+\frac{2n}{5}\end{align} 10. 84x+5x+8\begin{align}\frac{8}{4}-\frac{x+5}{x+8}\end{align} 11. 3x2(x+1)+67x6\begin{align}\frac{3x}{2(x+1)}+\frac{6}{7x-6}\end{align} 12. 11820x22\begin{align}\frac{11}{8} \cdot \frac{20x^2}{2}\end{align} 13. 17r167r416\begin{align}\frac{17r}{16} \cdot \frac{7r^4}{16}\end{align} 14. 15181417t\begin{align}\frac{15}{18} \cdot \frac{14}{17t}\end{align} 15. 2(b11)14bb+5(b+5)(b11)\begin{align}\frac{2(b-11)}{14b} \cdot \frac{b+5}{(b+5)(b-11)}\end{align} 16. 17w2w+418(w+4)17w2(w9)\begin{align}\frac{17w^2}{w+4} \cdot \frac{18(w+4)}{17w^2 (w-9)}\end{align} 17. 10s330s230s210s3s38\begin{align}\frac{10s^3-30s^2}{30s^2-10s^3} \cdot \frac{s-3}{8}\end{align} 18. 1f5÷f+3f2+6f+9\begin{align}\frac{1}{f-5} \div \frac{f+3}{f^2+6f+9}\end{align} 19. (a+8)(a+3)4(a+3)÷10a2(a+10)4\begin{align}\frac{(a+8)(a+3)}{4(a+3)} \div \frac{10a^2 (a+10)}{4}\end{align} 20. 1(h10)(h+7)÷(h4)4h(h10)\begin{align}\frac{1}{(h-10)(h+7)} \div \frac{(h-4)}{4h(h-10)}\end{align} 21. 2(5x8)4x2(85x)÷64x2\begin{align}\frac{2(5x-8)}{4x^2 (8-5x)} \div \frac{6}{4x^2}\end{align} 22. 2(q7)40q(q+1)÷140q(q+1)\begin{align}\frac{2(q-7)}{40q(q+1)} \div \frac{1}{40q(q+1)}\end{align} Solve each equation. 1. 33x2=1x+13x2\begin{align}\frac{3}{3x^2}=\frac{1}{x}+\frac{1}{3x^2}\end{align} 2. 25x2=12x3\begin{align}\frac{2}{5x^2}=-\frac{12}{x-3}\end{align} 3. 7xx6=34x+16\begin{align}\frac{7x}{x-6}=\frac{3}{4x+16}\end{align} 4. 4c2=3c+4\begin{align}\frac{4}{c-2}=\frac{3}{c+4}\end{align} 5. \begin{align}\frac{d-4}{4d^2}=\frac{1}{4d^2}+\frac{1}{4d}\end{align} 6. \begin{align}\frac{1}{2}=\frac{2z-12}{z} - \frac{z+1}{4z}\end{align} 7. \begin{align}\frac{1}{n}=\frac{1}{n^2} +\frac{6}{n}\end{align} 8. \begin{align}\frac{1}{2a}=\frac{1}{2a^2}+\frac{1}{a}\end{align} 9. \begin{align}\frac{k+4}{k^2} =\frac{5k-30}{3k^2}+\frac{1}{3k^2}\end{align} 10. It takes Jayden seven hours to paint a room. Andie can do it in five hours. How long will it take to paint the room if Jayden and Andie work together? 11. Kiefer can mow the lawn in 4.5 hours. Brad can do it in two hours. How long will it take if they worked together? 12. Melissa can mop the floor in 1.75 hours. With Brad’s help, it took only 50 minutes. How long would it take Brad to mop it alone? 13. Working together, it took Frankie and Ricky eight hours to frame a room. It would take Frankie fifteen hours doing it alone. How long would it take Ricky to do it alone? 14. A parallel circuit has \begin{align}R_1=50 \Omega\end{align} and \begin{align}R_t=16 \Omega\end{align}. Find \begin{align}R_2\end{align}. 15. A parallel circuit has \begin{align}R_1=6 \Omega\end{align} and \begin{align}R_2=9 \Omega\end{align}. Find \begin{align}R_T\end{align}. 16. A series circuit has \begin{align}R_1=200 \Omega\end{align} and \begin{align}R_t=300 \Omega\end{align}. Find \begin{align}R_2\end{align}. 17. A series circuit has \begin{align}R_1=11 \Omega\end{align} and \begin{align}R_2=25 \Omega\end{align}. Find \begin{align}R_T\end{align}. 18. Write the formula for the total resistance for a parallel circuit with three individual resistors. 19. What would be the bias in this situation? To determine the popularity of a new snack chip, a survey is conducted by asking 75 people walking down the chip aisle in a supermarket which chip they prefer. 20. Describe the steps necessary to design and conduct a survey. 21. You need to survey potential voters for an upcoming school board election. Design a survey with at least three questions you could ask. How will you plan to conduct the survey? 22. What is a stratified sample? Name one case where a stratified sample would be more beneficial. ### Rational Equations and Functions; Statistics Test 1. True or false? A horizontal asymptote has the equation \begin{align}y=c\end{align} and represents where the denominator of the rational function is equal to zero. 2. A group of SADD members wants to find out about teenage drinking. The members conduct face-to-face interviews, wearing their SADD club shirts. What is a potential bias? How can this be modified to provide accurate results? 3. Name the four types of ways questions can be biased. 4. Which is the best way to show data comparing two categories? 5. Consider \begin{align}f(x)= -\frac{4}{x}\end{align}. State its domain, range, asymptotes, and the locations of its branches. 6. \begin{align}h\end{align} varies inversly as \begin{align}r\end{align}. When \begin{align}h=-2.25, r=0.125\end{align}. Find \begin{align}h\end{align} when \begin{align}r=12.\end{align} 7. Name two types of visual displays that could be used with a frequency distribution. 8. Tyler conducted a survey asking the number of pets his classmates owned and received the following results: 0, 2, 1, 4, 3, 2, 1, 0, 0, 0, 0, 1, 4, 3, 2, 3, 4, 3, 2, 1, 1, 1, 5, 7, 0, 1, 2, 3, 2, 1, 4, 3, 2, 1, 1, 0 1. Display this data with a frequency distribution chart. 2. Use it to make a histogram. 3. Find its five-number summary. 4. Draw a box-and-whisker plot. 5. Make at least two conclusions regarding Tyler’s survey. 1. Find the excluded values, the domain, the range, and the asymptotes of: Perform the indicated operation. 1. \begin{align}\frac{4}{21r^4}+\frac{4r+5t}{21r^4}\end{align} 2. \begin{align}\frac{a-v}{12a^3}-\frac{a+5v}{12a^3}\end{align} 3. \begin{align}\frac{8}{g+8}+\frac{g-3}{g-5}\end{align} 4. \begin{align}\frac{4t}{5t-8}+\frac{24}{12}\end{align} 5. \begin{align}\frac{4}{5} \cdot \frac{80}{48m}\end{align} 6. \begin{align}\frac{1}{d-8} \div \frac{d+7}{2d+14}\end{align} 7. \begin{align}\frac{1}{u-3} \div \frac{u-4}{2u-6}\end{align} Solve. 1. \begin{align}\frac{7w}{w-7}=\frac{7w}{w+5}\end{align} 2. \begin{align}\frac{p-6}{3p^2-6p}=\frac{7}{3}\end{align} 3. \begin{align}\frac{2}{x^2} =\frac{1}{2x^2}-\frac{x+1}{2x^2}\end{align} 4. \begin{align}\frac{1}{2}-\frac{1}{4r}=\frac{3}{4}\end{align} 5. \begin{align}\frac{y-5}{3y^2}=-\frac{1}{3y}+\frac{1}{y^2}\end{align} 6. Working together, Ashton and Matt can tile a floor in 25 minutes. Working alone, it would take Ashton two hours. How long would it take Matt to tile the floor alone? 7. Bethany can paint the deck in twelve hours. Melissa can paint the deck in five hours. How long would it take the girls to paint the deck, working together? 8. A parallel circuit has \begin{align}R_t=115 \ \Omega\end{align} and \begin{align}R_2=75 \ \Omega\end{align}. Find \begin{align}R_1\end{align}. 9. A series circuit has \begin{align}R_1=13 \ \Omega\end{align} and \begin{align}R_t=21 \ \Omega\end{align}. Find \begin{align}R_2\end{align}. #### Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9622. Basic 8 , 9 Feb 24, 2012
# Lesson 15 Putting All the Solids Together • Let’s calculate volumes of prisms, cylinders, cones, and pyramids. ### 15.1: Math Talk: Volumes Evaluate the volume of each solid mentally. ### 15.2: Missing Measurements 1. Answer the questions for each of the two solids shown. 1. Which measurement that you need to calculate the volume isn’t given? 2. How can you find the value of the missing measurement? 3. What volume formula applies? 4. Calculate the volume of the solid, rounding to the nearest tenth if necessary. 2. Calculate the volume of each solid, rounding to the nearest tenth if necessary. For a sphere with radius $$r$$, its volume is $$\frac43 \pi r^3$$ and its surface area is $$4 \pi r^2$$. Here is a half-sphere bowl pressed out of a piece of sheet metal with area 1 square foot. What is the volume of the bowl? ### 15.3: Spinning into Three Dimensions Suppose this two-dimensional figure is rotated 360 degrees using the vertical axis shown. Each small square on the grid represents 1 square inch. 1. Draw the solid that would be traced out. Label the dimensions of the solid. 2. Find the volume of the solid. Round your answer to the nearest tenth if needed. ### Summary Before computing volume, it’s important to select the right formula and find all the dimensions represented in the formula. For example, consider a company that makes two chew toys for dogs. One toy is in the shape of a cylinder with radius 9 cm and height 2.5 cm. The other looks like the cone in the image. The company wants to know which toy uses more material. The toys are solid, not hollow. To calculate the cylinder toy’s volume, use the expression $$Bh$$. The radius measures 9 cm, so the area of the base, $$B$$, is $$81\pi$$ cm2. The volume is $$202.5\pi$$, or approximately 636 cm3, because $$81\pi \boldcdot 2.5 = 202.5\pi$$. For the cone, the height is unknown. A right triangle is formed by the radius 6 cm and the height $$h$$, with hypotenuse 16 cm. By the Pythagorean Theorem, $$6^2+h^2=16^2$$. Solving, we get $$h=\sqrt{220}$$. Since this is a cone, use the expression $$\frac13 Bh$$. The area of the base, $$B$$, is $$36\pi$$ cm2. The volume is approximately 559 cm3 because $$\frac13 \boldcdot 36 \pi \boldcdot \sqrt{220} \approx 559$$. The cylinder-shaped toy uses more material. ### Glossary Entries • apex The single point on a cone or pyramid that is the furthest from the base. For a pyramid, the apex is where all the triangular faces meet.
#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (x) maths textbook solution. Answer : $x e^{y}=\tan y+C$ Give : $e^{-y} \sec ^{2} y d y=d x+x d y$ Hint : Using $\int \sec ^{2} x d x$ Explanation: $e^{-y} \sec ^{2} y d y=d x+x d y$ \begin{aligned} &=e^{-y} \sec ^{2} y d y-x d y=d x \\ &=\left(e^{-y} \sec ^{2} y-x\right) d y=d x \\ &=\frac{d x}{d y}=e^{-y} \sec ^{2} y-x \\ &=\frac{d y}{d x}+x=e^{-y} \sec ^{2} y \end{aligned} This is a first order linear differential equation of the form \begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=1 \text { and } Q=e^{-y} \sec ^{2} y \end{aligned} The integrating factor $If$ of this differential equation is \begin{aligned} &I f=e^{\int 1 d y} \\ &=e^{\int d y} \\ &=e^{y} \; \; \; \; \; \; \; \; \; \; \quad\left[\int d y=y+C\right] \end{aligned} Hence, the solution of different equation is \begin{aligned} &x I f=\int Q I f d y+C \\ &=x e^{y}=\int e^{-y} \sec ^{2} y e^{y} d y+C \end{aligned} \begin{aligned} &=x e^{y}=\int \sec ^{2} y d y+C \; \; \; \; \; \; \quad\left[e^{-y} e^{y}=e^{-y+y}=e^{0}=1\right] \\ &=x e^{y}=\tan y+C \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sec ^{2} x d x=\tan x+C\right] \end{aligned}
How do you simplify 3^3 - 8 times 3 ÷ 12 using PEMDAS? Jul 29, 2016 $25$ Explanation: Count the number of terms first. Each term must end in an answer. You can work in different in the same step. In the last step all the answers will be added or subtracted. Do powers and roots first because they are the strongest operations. Multiplication and division are equally strong and can be done together. $\textcolor{b l u e}{{3}^{3}} - \textcolor{red}{8 \times 3 \div 12} \text{ there are 2 terms}$ =$\textcolor{b l u e}{27} - \textcolor{red}{\frac{8 \times 3}{12}}$ =$\textcolor{b l u e}{27} - \textcolor{red}{\frac{24}{12}}$ =$\textcolor{b l u e}{27} - \textcolor{red}{2}$ =$25$
Sections: # Multiplying Integers Test #3 About: Additional Resources: In this section, we continue to learn about the topic of integers. Now, we turn to multiplying integers and dividing integers. Let's start by discussing integer multiplication. How do we multiply two or more integers together? When we are multiplying integers, we use the follwing procedure. Multiplying Integers: 1. Determine the sign of the product: • Same signs produce a positive result: (+) • (+) = + and (-) • (-) = + • Different signs produce a negative result: (+) • (-) = - and (-) • (+) = - 2. Mutliply absolute values 3. Attach the sign to the answer Example 1: Multiply (-5) • (-2) 1) Determine the sign, since we have the same signs the result will be positive 2) Multiply the absolute values: 5 • 2 = 10 3) Attach the sign to the answer: Our answer is already positive, so we end up with 10 as the answer (-5) • (-2) = 10 Example 2: Multiply (-6) • 3 1) Determine the sign, since we have different signs the result will be negative 2) Multiply the absolute values: 6 • 3 = 18 3) Attach the sign to the answer: Our answer needs to be negative, so we end up with -18 as the answer (-6) • 3 = -18 We think about dividing integers in the same manner: Dividing Integers: 1. Determine the sign of the product: • Same signs produce a positive result: (+) ÷ (+) = + and (-) ÷ (-) = + • Different signs produce a negative result: (+) ÷ (-) = - and (-) ÷ (+) = - 2. Divide absolute values 3. Attach the sign to the answer Example 3: Divide (-15) ÷ 5 1) Determine the sign, since we have different signs the result will be negative 2) Divide the absolute values: 15 ÷ 5 = 3 3) Attach the sign to the answer: Our answer needs to be negative, so we end up with -3 as the answer (-15) ÷ 5 = -3 + Show More +
# Mensuration class 8 worksheets In this page we have Mensuration class 8 worksheets Chapter 9 . Hope you like them and do not forget to like , social share and comment at the end of the page. Question 1 Find the area of a trapezium whose parallel sides are 12 cm and 20 cm and the distance between them is 15 cm. Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=12 cm , b=20 cm h =15 cm Therefore $A=\frac {1}{2} [12 +20] \times 15 =240 \; cm^2$ Question 2 Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 18 cm. Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=38.7 cm , b=22.3 cm h =18 cm Therefore $A=\frac {1}{2} [38.7 +22.3] \times 18 =549 \; cm^2$ Question 3 The area of a trapezium is 1440 cm2. If the lengths of its parallel sides are 54.6 cm and 35.4 cm, find the distance between them. Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=54.6 cm , b=35.4 cm, h =?, A=1440 cm2 Therefore $1440=\frac {1}{2} [38.7 +22.3] \times h$ $h= 32$ cm Question 4 The area of a trapezium is 1586 cm2 and the distance between its parallel sides is 26 cm. If one of the parallel sides is 84 cm, find the other. Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=84 cm , b=?, h =26 cm., A=1586 cm2 Therefore $1586=\frac {1}{2} [84 +b] \times 26$ $h= 38$ cm Question 5 The area of a trapezium is 384cm2. Its parallel sides are in the ratio 2: 6 and the perpendicular distance between them is 12 cm. Find the length of each of the parallel sides. Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=2x cm , b=6x, h =12 cm., A=384 cm2 Therefore $384=\frac {1}{2} [2x +6x] \times 12$ $x= 8$ cm So parallel sides are 16 cm and 48 cm Question 6 Mitesh wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. let x be the length on the road, then 2x is the length on the river side Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=2x m , b=x, h =100 m., A=10500 m2 Therefore $10500=\frac {1}{2} [2x +x] \times 100$ $x= 70$ m So parallel sides are 70 m and 140 m Question 7 The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides. let x be one side, then x+6 is the length of another parallel side Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the $A = \frac {1}{2} [a +b] \times h$ Here a=x m , b=x+6, h =9 cm., A=180 cm2 Therefore $180=\frac {1}{2} [x +x+6] \times 9$ $x= 17$ cm So parallel sides are 17 cm and 23 cm Question 8 The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium. Let ABCD is the trapezium with AB and CD are the parallel sides. Now AB=10 cm, CD=20 cm, BC=13 cm Now draw line BE \|\| AD and draw a perpendicular from B on EC Now ABED is a parallelogram, then BE=13 cm In triangle BEC, BE =BC, So Isoceles triangle,So perpendicular will bisect the EC Hence EF=FC Now EC= DC -DE = 20 -10 =10 cm Therefore EF=FC= EC/2 = 5 cm Now in Triangle BEF, it is right angle at F,So by pythagorous theorem $BE^2 = BF^2 + EF^2$ $169 = BF^2 + 25$ $BF=12 cm$ This is also the perpendicular distance between the parallel sides.So now coming back to Area of trapezium $A = \frac {1}{2} [a +b] \times h$ Here a=10 cm , b=20 cm, h =12 cm., A=? Therefore $A=\frac {1}{2} [10 +20] \times 12$ $A= 180 \; cm^2$ ## True and False Question 9 (i) Amount of region occupied by a solid is called its surface area (ii)The areas of any two faces of a cube are equal. (iii)The areas of two oppossite faces of a cuboid are equal. (iv) 2.5 litres is equal to 0.0025 cubic meters (v) Ratio of area of a circle to the area of a square whose side equals radius of circle is 1 : π. (i) False (ii) True (iii) True (iv) True (v) false ## Fill in the blanks Question 10 (i) Curved surface area of a cylinder of radius 2b and height 2a is _______. (ii) Volume of a cylinder with radius p and height q is __________. (iii) 6.55m2 = ______cm2 (iv)The volume of a cylinder becomes __________ the original volume if its radius becomes doubles of the original radius and height becomes half of its original values (v)The surface area of a cylinder which exactly fits in a cube of side b is____ (i) $S=2\pi r H= 2\pi 2b 2a= 8\pi ab$ (ii)$V=\pi r^2 H= \pi p^2 q$ (iii) 6.55m2 =6.55 * 104 cm2 (iv) $V_1=\pi r^2 H$ ,$V_2= \pi (2r)^2 (1/2)H=2 pi r^2 H$, So it becomes double (v) Radius of cylinder =(1/2)b and height =b, So Surface area=2\pi b^2\$ Question 11 (p) -> (ii) (q) -> (iii) (r) -> (i) Question 12 (p) -> (iv) (q) -> (i) (r) -> (ii) (s) -> (iii) ## Summary This mensuration class 8 worksheets is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.
# TangentDetermine the equation of the tangent to the curve y=3x-x^3 at x=2 . sciencesolve \| Certified Educator You need to use the equation of tangent line to the curve, at a given point `(x_0,y_0)` , such that: `y - y_0 = f'(x_0)(x - x_0)` The problem provides `x_0 = 2` , hence, you may evaluate `y_0` replacing 2 for x in equation of the function, such that: `y_0 = f(x_0) = f(2) => y_0 = 6 - 2^3 => y_0 = 6 - 8 => y_0 = -2` You need to differentiate the function with respect to x, such that: `f'(x) = 3 - 3x^2 => f'(x_0) = f'(2) => f'(2) = 3 - 12 = -9` You need to replace -2 for `y_0` , -9 for `f'(x_0)` and 2 for `x_0` in tangent equation, such that: `y + 2 = -9(x - 2) => y = -9x + 18 - 2 => y = -9x + 16` Hence, evaluating the equation of the tangent line to the given curve, at` x = 2` , yields `y = -9x + 16.` giorgiana1976 \| Student To determine the equation of the tangent line to the graph of y, we'll have to determine the derivative of y at x = 2. f'(2) = lim [f(x) - f(2)]/(x-2) f(2) = 32 - 2^3 f(2) = 6 - 8 f(2) = -2 lim [f(x) - f(2)]/(x-2) = lim (3x - x^3 + 2)/(x-2) We'll substitute x by 2: lim (3x - x^3 + 2)/(x-2) = (6-8+2)/(2-2) = 0/0 Since we've obtained in indeterminacy acse, we'll apply L'Hospital rule: lim (3x - x^3 + 2)/(x-2) = lim (3x - x^3 + 2)'/(x-2)' lim (3x - x^3 + 2)'/(x-2)' = lim (3-3x^2)/1 We'll substitute x by 2: lim (3-3x^2)/1 = 3 - 34 = 3-12 = -9 f'(2) = -9 But the derivative of y at x = 2 is the slope of the tangent line to the curve y. m = -9 Now, we'll write the equation of the tangent line, whose slope is m=-9 and it passes through the point that has x coordinate = 2.We'll compute the y coordinate of this point: f(2) = 3x - x^3 f(2) = 6 - 8 f(2) = -2 The equation of the tangent line is: y - (-2) = m(x - 2) y + 2 = -9(x - 2) We'll remove the brackets: y + 2 = -9x + 18 y + 9x + 2 - 18 = 0 y + 9x - 16 = 0
# Polynomials • An expression of the form p(x) = a0 + a1x + a2x2 + a3x3 + ... + anxn , where p(x), is called a polynomial in x of degree n. Here, a0, a1, a2, a3, ... a are real numbers and each power of x is a non-negative integer. • The exponent of the highest degree term in a polynomial is known as its degree. • A polynomial of degree 0 is called a constant polynomial. • A polynomial of degree 1 is called a linear polynomial. A linear polynomial is of the form p(x) = ax + b, where a ≠ 0. • A polynomial of degree 2 is called a quadratic polynomial. A quadratic polynomial is of the form p(x) = ax2 + bx + c, where a ≠ 0,. • A polynomial of degree 3 is called a cubic polynomial. A cubic polynomial is of the form p(x) = ax3 + bx2 + cx + d, where a ≠ 0. • A polynomial of degree 4 is called a biquadratic polynomial. A biquadratic polynomial is of the form p(x) = ax4 + bx3 + cx2 + dx + e, where a ≠ 0,. • If p(x) is a polynomial in x and if α is any real number, then the value obtained by putting x = α in p(x) is called the value of p(x) at x = α . The value of p(x) at x = α is denoted by p(α) . • A real number α is called a zero of the polynomial p(x), if p(α) = 0. • A polynomial of degree n can have at most n real zeroes. • Geometrically the zeroes of a polynomial p(x) are the x-coordinates of the points, where the graph of p(α) = 0. intersects x-axis. • Zero of the linear polynomial ax + b is $\dpi{120} \frac{-b}{a}=\frac{- \emph{constant term}}{\emph{coefficient of }x}$ • If α and β are the zeroes of a quadratic polynomial p(x) = ax2 + bx + c, a ≠ 0,, then $\dpi{120} \alpha +\beta =\frac{-b}{a}=\frac{- \text{coefficient of }x}{\text{coefficient of }x^2}$ $\dpi{120} \alpha \beta =\frac{c}{a}=\frac{ \text{constant term }}{\text{coefficient of }x^2}$ • If α , β and γ are the zeroes of a cubic polynomial p(x) = ax3 + bx2 + cx + d, a ≠ 0, then $\dpi{120} \alpha +\beta + \gamma =\frac{-b}{a}=\frac{- \text{coefficient of }x^2}{\text{coefficient of }x^3}$ $\dpi{120} \alpha\beta +\beta\gamma + \gamma\alpha =\frac{c}{a}=\frac{\text{coefficient of }x}{\text{coefficient of }x^3}$ $\dpi{120} \alpha \beta \gamma =\frac{-d}{a}=\frac{- \text{constant term }}{\text{coefficient of }x^3}$ • A quadratic polynomial whose zeroes are α , β is given by p(x) = x2 – (α + β)x + αβ = x2 – (sum of the zeroes) x + product of the zeroes. • A cubic polynomial whose zeroes are α, β , γ is given by p(x) = x3 – (α + β + γ) x2 + (αβ + βγ + γα)x + αβγ = x3 – (sum of the zeroes) x2 + (sum of the products of the zeroes taken two at a time) x – product of the zeroes. • The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomial q(x) and r(x) such that p(x) = g(x)q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x). # Important Questions 1. For what value of k, (–4) is a zero of the polynomial x2 – x – (2k + 2)? 2. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a and a + b, fnd the values of a and b. 3. Find a quadratic polynomial whose zeroes are 3 + √5 and 3–√5. 4. α, β are the roots of the quadratic polynomial p(x) = x2 – (k – 6) x + (2k + 1). Find the value of k, if α + β = αβ . 5. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4 respectively. Find g(x). 6. Find the zeroes of 4√5x2 + 17x + 3√5 and verify the relation between the zeroes and coeffcients of the polynomial. 7. Find the zeroes of the quadratic polynomial p(x) = x2 – (√3 +1)x + 3 and verify the relationship between the zeroes and its coeffcients. 8. If α and β are the zeroes of the quadraticpolynomial f(x) = x2 – px + q, prove that $\dpi{120} \frac{\alpha^2 }{\beta^2 }+\frac{\beta^2 }{\alpha^2 }=\frac{p^4}{q^2}-\frac{4p^2}{q}+2.$ 9. If α and β are the zeroes of the quadraticpolynomial p(s) = 3s2 – 6s + 4, fnd the value of $\dpi{120} \frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2\left ( \frac{1}{\alpha }+\frac{1}{\beta } \right )+3\alpha \beta.$ 10. If α and β are zeroes of the quadratic polynomial f(x) = kx2 + 4x + 4 such that α2 + β2 = 24, find the value of k. 11. If one zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, find the value of k. 12. If the sum of the zeroes of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product, find the value of k. 13. If the square of the difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, find the value of p. 14. If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, find the value of a. 15. If α, β, γ are zeroes of the polynomial 6x3 + 3x2 – 5x + 1, then fnd the value of α–1 + β–1 + γ–1. 16. Find a cubic polynomial with the sum, sum of the products of its zeroes taken two at a time and product of its zeroes as 3, –1 and –3 respectively. 17. If α, β are the zeroes of the polynomial 21y2 – y – 2, find a quadratic polynomial whose zeroes are 2α and 2β. 18. If α, β are the zeroes of the polynomial 6y2 – 7y + 2, find a quadratic polynomial whose zeroes are $\dpi{120} \frac{1}{\alpha } \text{ and } \frac{1}{\beta}.$ 19. If α and β are the zeroes of the quadratic polynomial f(x) = x2 + px + q, form a polynomial whose zeroes are (α + β)2 and (α – β)2. 20. If α and β are the zeroes of the polynomial f(x) = x2 – 2x + 3, find a polynomial whose zeroes are α + 2 and α + β. 21. Divide 30x4 + 11x3 – 82x2 – 12x + 48 by (3x2 + 2x – 4) and verify the result by division algorithm. 22. If α and β are zeroes of the quadratic polynomial x2 – 6x + a; fnd the value of a if 3α + 2β = 20. 23. Find the value of p for which the polynomial x3 + 4x2 – px + 8 is exactly divisible by x – 2. 24. If the remainder on division of x3 + 2x2 + kx + 3 by x – 3 is 21, fnd the quotient and the value of k. Hence, fnd the zeroes of the cubic polynomial x3 + 2x2 + kx – 18. 25. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be (x + a), fnd the values of k and a. 26. Find all other zeroes of the polynomial p(x) = 2x3 + 3x2 – 11x – 6, if one of its zero is –3. 27. Find all the zeroes of the polynomial 2x4 – 3x3 – 5x2 + 9x – 3, it being given that two of its zeros are √3 and –√3. 28. Find all the zeroes of the polynomial 2x4 – 10x3 + 5x2 + 15x – 12, if it is given that two of its zeroes are $\dpi{100} \small \sqrt{\frac{3}{2}} \text{ and} -\sqrt{\frac{3}{2}}$ . 29. Find all the zeroes of the polynomial x4 + x3 – 34x2 – 4x + 120, if two of its zeroes are 2 and –2. 30. If two zeroes of p(x) = x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3 , fnd the other zeroes. 31. If the polynomial 6x4 + 8x3 – 5x2 + ax + b is exactly divisible by the polynomial 2x2 – 5, then fnd the values of a and b. 32. What must be added to the polynomial P(x) = 5x4 + 6x3 – 13x2 – 44x + 7 so that the resulting polynomial is exactly divisible by the polynomial Q(x) = x2 + 4x + 3 and the degree of the polynomial to be added must be less than degree of the polynomial Q(x). 33. What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x2 – 4x + 3?
# Important Questions Class 11 Maths Chapter 4 ## Important Questions Class 11 Mathematics Chapter 4 ### Important Questions for CBSE Class 11 Mathematics Chapter 4 – Principle of Mathematical Induction Important Questions Class 11 Mathematics Chapter 4 prepared by Extramarks, will help students gain a better understanding of concepts and terminologies related to the chapter, Principle of Mathematical Induction. Students can get answers to their doubts as these notes are comprehensive. These notes are compiled by subject matter experts to give a brief overview of all the important questions in this chapter. Chapter 4 Class 11 Mathematics Important Questions are made in accordance with the CBSE Syllabus These notes are beneficial for students while preparing for the exams, as they cover all the main topics of this chapter. Students can go through these notes before their exams so that they can brush up on all the important formulas and concepts. ### CBSE Class 11 Mathematics Chapter 4 Important Questions Students can preview the Class 11 Mathematics Chapter 4 Important Questions given below. They can also click the link to view additional questions. Q1. Prove that 102n-1+1 is divisible by 11. Ans. Using the method of P.M.I., Given, P(n):102n-1+1 Checking if the statement is true or not for n=1 For n=1 P(1):10(21)-1+1=11. Which is divisible by 11. Therefore, P(1) is true. Let, P(n):102n-1+1 is true for n=k. Q2. Prove 2n+7<n+32. Ans. Using the method of P.M.I., Given, P(n):2n+7<n+32. Checking if the statement is true or not for n=1. So, for n=1. P(1): 9<42 9<16 which is true. Thus, P(1) is true. Let, P(n) : 2n+7<n+32 is true for n=k. That is, P(k):2k+7<k+32 …………(1) Now, we have to show that the given statement P(n):2(k+1)+7<k+1+32 is true for n=k+1 So, P(k+1):(2k+9)<k+42 Now, (2k+7+2)<k+42 k+32+2<k+42 from Equation (1) k2+9+6k+2<k+42 k2+6k+11<k+42 k2+6k+11+2k-2k+5-5<k+42 k2+8k+162k+5<k+42 k+422k+5<k+42 Which is true. Hence, P(k+1):(2k+9)<k+42 is true. Thus, P(k+1) is true when Pk is true. Therefore, by P.M.I. the statement (2n+7)<n+32 is true for all n N. Q3. Using induction, prove that 10n+34n+2+5 is divisible by 9. Ans. 10n+34n+2+5 is divisible by 9 by using the method of PMI. Given, P(n): 10n+34n+2+5. Checking if the statement is true or not for n=1. So, For n=1. P(1):101+341+2+5=207. Which is divisible by 9. Thus, P(1) is true. Let, P(n):10n+34n+2+5 is true for n=k. That is, P(k): 10k+34k+2+5=9 , where N …..(1) Now, we have to show that the given statement P(n):10n+34n+2+5 is true for n=k+1 P(k+1):10k+1+34k+1+2+5 10k+1+3 4k+1+2+5 10k10+3 4k43+5 (9 -34k+2-5)10+34k43+5 {from equation(1)} 90 -304k+2-50+34k43+5 90 -304k+2-45+34k+24 90 – 304k+2-45+124k+2 90 – 184k+2-45 910 -24k+2-5 Hence, P(k+1):10k+1+34k+1+2+5 is divisible by 9. Thus, P(k+1) is true when P(k) is true. Therefore, by P.M.I. the given statement is true for every positive integer n. Q4. Prove that n(n+1)(n+5) is a multiple of 3. Ans: nn+1n+5 is a multiple of 3 by using the method of PMI. Given, P(n):nn+1n+5 Checking if the statement is true or not for n=1. So, For n=1 P(1):11+11+5=12 Which is multiple of 3. Thus, P(1) is true. Let, P(n):nn+1n+5 is true for n=k. That is, P(k):kk+1k+5=3 , where N ……(1) Now, we have to show that the given statement P(n):nn+1n+5 is true for n=k+1. P(k+1):(k+1)(k+2)(k+6) (k+1)(k+2)(k+6)=[(k+1)(k+2)](k+6) k(k+1)(k+2)+6(k+1)(k+2) k(k+1)(k+5-3)+6(k+1)(k+2) k(k+1)(k+5)+3k(k+1) +6(k+1)(k+2) k(k+1)(k+5)+ (k+1) 6(k+2)-3k k(k+1)(k+5)+ (k+1)3k+12 k(k+1)(k+5)+ 3k+1k+4 3 + 3k+1k+4 from equation (1) 3 +k+1k+4 Hence, P(k+1):k+1k+2k+6 is multiple of 3. Thus, P(k+1) is true when P(k) is true. Therefore, by P.M.I. the given statement n(n+1)(n+5) is multiple of 3. Q5. Show that the sum of the first n odd natural number is n2. Ans. Using the PMI method to prove 1+3+5+…..+(2n-1)=n2 Given, P(n):1+3+5+…..+(2n-1)=n2 Checking if the statement is true or not for n=1. So, For n=1 P(1):1=1 Which is true Thus, P(1) is true. Let P(n):1+3+5+……+(2n-1)=n2 is true for n=k. That is, P(k):1+3+5+……+(2k-1)=k2 …………………….(1) Now, we have to show that the given statement P(n):1+3+5+……+(2n-1)=n2 is true for n=k+1 So, P(k+1):1+3+5+……+(2(k+1)-1)=(k+1)2 P(k+1):1+3+5+……+(2k+1)=(k+1)2 Now, L.H.S =1+3+5+……+(2k-1)+(2k+1) =k2+(2k+1) …………………..{from equation} (k+1)2= R.H.S Which is true. Hence, P(k+1):1+3+5+……+(2(k+1)-1)=(k+1)2 is true. Thus, P(k+1) is true when P(k) is true. Therefore, by P.M.I. the statement 1+3+5+……+(2n-1)=n2 is true. Q6. Prove x2n-1 is divisible by x-1. Ans. x2n-1 is divisible by x-1 with the method of PMI. Given, P(n):x2n-1. Checking if the statement is true or not for n=1. So, For n=1 P(1):x2-1=(x-1)(x+1) Which is divisible by (x-1). Thus, P(1) is true. Let, P(n):x2n-1 is true for n=k. That is, P(k):x2k-1=x-1 , where N ………(1) Now, we have to show that the given statement P(n):x2n-1 is true for n=k+1. P(k+1):x2k+2-1 =x2kx2-1 ={[(x-1) +1]x2-1} from equation (1) = {(x-1) x2 +x2-1} =(x-1) x2+(x-1)(x+1) =(x-1) x2+(x+1) Hence, P(k+1):x2k+2-1 is divisible by (x-1). Thus, P(k+1) is true when P(k) is true. Therefore, by P.M.I. the given statement is true for every positive integer n Q7. The sum of the cubes of three consecutive natural numbers is divisible by 9. Ans. [n3+n+13+n+23]. Checking if the statement is true or not for n=1. So, For n=1. P(1):13+23+33=243 divisible by 9. Thus, P(1) is true. Let, P(n): [n3+n+13+n+23] is true for n=k. P(k): [k3+k+13+k+23]=9 , where N ……………1 Now, we have to show that the given statement P(n):[n3+n+13+n+23] is true for n=k+1. P(k+1):[k+13+k+23+k+33] =[k+13+k+23+k3+9k2+27k+27] =9 +9k2+27k+27 from equation (1) =9[ +k2+3k+3] divisible by 9. Hence, P(k+1):[k+13+k+23+k+33] is divisible by 9. Thus, P(k+1) is true when Pk is true. Therefore, by P.M.I. the given statement is true for every positive integer n. Q.1 Let P(n) be the statement ?n3 + n is divisible by 3?, then the incorrect statement is P(3). P(6). P(9). P(11). Marks:1 Ans P(11) is the statement 113 + 11 is divisible by 3 i.e., 1342 is divisible by 3, that is not true Therefore, the statement P(11) is false. Q.2 10n + 3.22n + 4 + 5 (for n = 4) is divisible by 7. 9. 11. 13. Marks:1 Ans P(n) = 10n + 3.22n + 4 + 5 Now we check for the value of n = 4 and we get, P(4) = 104 + 3.212 + 5 = 10000 + 12288 + 5 = 22293 = 9 × 2477 Therefore, we can observe that ?10n + 3.22n + 4 + 5 (for n = 4) is divisible by 9. Q.3 Using the principle of mathematical induction,prove that13.7+17.11+111.15+ ? +1(4n-1)(4n+3)=n3(4n+3) for?all?n?N Marks:4 Ans Let the given statement bePn:13.7+17.11+111.15+14n-14n+3=n34n+3 .1Test for n=1LHS= 13.7=121RHS=n34n+3=134+3=121Hence, P1 is true. Let Pn is true for n=k then 1 becomes13.7+17.11+111.15++14k-14k+3=k34k+3 2Now, we shall test for n=k+11 becomes13.7+1711+111.15+14k+34k+7=k+134k+73 LHS=13.7+17.11+111.15+14k-14k+3+14k+34k+7=k34k+3+14k+34k+7 Using 2=14k+3k3+14k+7=14k+34k2+7k+334k+7 =14k+34k+3k+134k+7 =k+134k+7=RHSPk+1 is true. Hence, Pn is true for all nN. Q.4 Prove that : 2n > n for all positive integers n. Marks:3 Ans When n=1,21>1. Hence P1 is TrueAssuming Pk is True Pk:2k>kWe shall prove Pk+1 is Truei.e., Pk+1=2k+1>k+1 Pk+1=2k+1=2k 21=2k 22k.2>2k=k+k>k+1 2k+1>2k>k+1Pk+1 is TrueHence by principle induction Pn is true, for all nN. Q.5 Prove that: 2.7n + 3.5n ? 5 is divisible by 24. Marks:6 Ans Let P n=2.7n+3.5n-5P1 =2.71+3.51-5 =14+15-5 =24P1 is true Assuming Pk is truePk =2.7k+3.5k-5 is divisible by 24We shall prove Pk+1 is truei.e. Pk+1=2.7k+1+3.5k+1-5 is divisible by 242.7k+1+3.5k+1-5=24g let 1Pk+1=2.7k+1+3.5k+1-5 =2.7k.7+3.5k.5-5 =72.7k+3.5k-5-3.5k+5+3.5k.5-5 =724g-3.5k-5-3.5k-5+15.5k-5 [From equation 1] =724g-21.5k+35+15.5k-5 =724g-6.5k+30 =724g-65K-5 =724g-6(4P) 5k-5 is a multiple of 45k-5=4P =724g-24 PWhich is divisible by 24 Pk+1 is true Hence by principle of mathematical induction is true. For all n N ## Please register to view this section ### 1. What is ‘Mathematical Induction’? Why do we use it? Mathematical Induction is a mathematical technique that is used to prove a statement, a formula, or a theorem is true for every natural number. Mathematical induction is usually used in case of all natural numbers to state whether a given statement holds true. ### 2. Name the types of mathematical induction. The types of mathematical induction are as follows: • First principle of mathematical induction • Second principle of mathematical induction • Second principle of mathematical induction(variation) ### 3. How is Important Questions Class 11 Mathematics Chapter 4 beneficial for Students? Important Questions Class 11 Mathematics Chapter 4 made by subject matter of Extramarks is very useful. It has many benefits such as: • This questionnaire consists of important questions which are categorised on the basis of marks distribution. • These questions are comprehensive and concise. • These questions are constructed by keeping in mind the CBSE guidelines. • Chapter 4 Class 11 Maths Important Questions cover all important types of questions that can come for the exams. • These questions are best for revision before exams. ### 4. Prove that 1+xn1+nx, for all natural numbers n, where x＞-1. Let P(n) be the given statement, i.e., P(n): 1+xn1+nx, for x＞-1. We note that P(n) is true when n=1, since 1+x1+x for x＞-1. Assume that P(k): 1+xk1+kx, x＞-1 is true. … (1) We want to prove that P(k+1) is true for x＞-1 whenever P(k) is true. … (2) Consider the identity 1+xk= 1+xk1+x Given that x＞-1, so 1+x > 0. Therefore, by using 1+xk1+kx we have 1+xk+11+kx1+x i.e.1+xk+11+x+kx+kx2. … (3) Here k is a natural number and x20 so that kx2 ≥ 0. Therefore 1+x+kx+kx21+x+kx and so we get 1+xk+11+x+kx i.e. 1+xk+11+1+kx Thus, the statement in (2) is established. Hence, as per the principle of mathematical induction, P(n) is true for all natural numbers.
# Math Homework No Homework Individuals: Clear your desk! Wednesday, April 1 st. ## Presentation on theme: "Math Homework No Homework Individuals: Clear your desk! Wednesday, April 1 st."— Presentation transcript: Math Homework No Homework Individuals: Clear your desk! Wednesday, April 1 st Math Homework Study for Test! Individuals: Homework out ready to check Wednesday, April 1 st Homework Check Unit Review Target/Objective: I can find the perimeter and area of rectangles, parallelograms, and triangles. Math Vocabulary Review Perimeter: the distance around a shape. – To find the perimeter of a shape, add the lengths. Area: the measure of a surface inside a shape. – Area is commonly measured in square units. – The area of a surface is the number of unit squares and fractions of unit squares needed to cover the surface without overlaps and/or gaps. – Different ways of writing square units: Square inch = in² Square meter = m² Square centimeter = cm² Finding the Area for a Rectangle: If the length of the base and the width of a rectangle are known, the area can be found by multiplying. Formula: Area of a rectangle = length x width A = l x w Finding the Area for a Parallelogram: Height is the distance perpendicular to the base of a figure. Any side of the parallelogram can be the base. The choice of the base determines the height. Finding the Area for a Triangle: The area of a triangle is half of the base times the height. Formulas: Area of a ∆ = ½ x Base x Height Area of a ∆ = (Base x Height) ÷ 2 Find the perimeter of each shape! Add the lengths a. c. b. Find the area of each rectangle! Formula: (l x w) a. c. b. 16 cm 9 cm 23 in 7 in Find the area of each parallelogram! Formula: (b x h) a. c. b. Find the area of each triangle! Formula: (b x h) ÷ 2 a. c. b. Journal Work Complete ALL journal pages in Unit 8!! Show your work. Early Finishers: Show Ms. Smith that your journal work is completed; then you may read a book. Time: End of Class Voice Level: 0 Wrap Up! Did we achieve our target/objective? – I can find the perimeter and area of rectangles, parallelograms, and triangles. Individuals: Get ready to move classes!
## How do you find the leading coefficient? How To: Given a polynomial expression, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree. 2. Identify the term containing the highest power of x to find the leading term. 3. Identify the coefficient of the leading term. ## What is the leading coefficient in a graph? In other words, the leading term is the term that the variable has its highest exponent. Basically, the leading coefficient is the coefficient on the leading term. would be – 4. The degree of a term of a polynomial function is the exponent on the variable. ## What is the leading coefficient of 2x 7? The leading coefficient in a polynomial is the coefficient of the leading term. In this case, the leading term is 2×7 2 x 7 and the leading coefficient is 2. You might be interested: Often asked: What is biomechanics? ## What is the leading coefficient of number 10? Answer. The leading coefficient is the coefficient of the leading term, i.e the number in front of the y, so that’s −10. ## What is an example of a leading coefficient? Leading coefficients are the numbers written in front of the variable with the largest exponent. For example, in the equation -7x^4 + 2x^3 – 11, the highest exponent is 4. The coefficient for that term is -7, which means that -7 is the leading coefficient. ## Can pi be in a polynomial? Since π and e are transcendental, neither can be the root of a polynomial with rational coefficients. However, it is easy to construct a polynomial transcendental coefficients (with π or e as one of it’s roots), namely (x−π) and (x−e). ## How do you tell if a graph has a positive leading coefficient? The degree of a polynomial and the sign of its leading coefficient dictates its limiting behavior. In particular, If the degree of a polynomial f(x) is even and the leading coefficient is positive, then f(x) → ∞ as x → ±∞. If f(x) is an even degree polynomial with negative leading coefficient, then f(x) → -∞ as x →±∞. ## How do you use the leading coefficient test? Use the Leading Coefficient Test to determine the end behavior of the graph of the polynomial function f(x)=−x3+5x. Case End Behavior of graph When n is even and an is positive Graph rises to the left and right When n is even and an is negative Graph falls to the left and right You might be interested: Often asked: What are similes? ## How do you know if a graph will cross or bounce? If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. The sum of the multiplicities is n. ## How do you know if a leading coefficient is positive or negative? Use the degree of the function, as well as the sign of the leading coefficient to determine the behavior. 1. Even and Positive: Rises to the left and rises to the right. 2. Even and Negative: Falls to the left and falls to the right. 3. Odd and Positive: Falls to the left and rises to the right. ## What is the degree of 2x 7? the variable can be written x^1 so the expression 2x + 7 is of degree 1. ## What does a degree mean in math? more The largest exponent the variable has in a polynomial with one variable. Example: 4x3 + 2x2 − 7 is degree 3. For more than one variable: add the variables’ exponents for each term and find the highest such value. ## What is a leading term? In a polynomial function, the leading term is the term containing the highest power of x. The coefficient of the leading term is called the leading coefficient. ## How do you find the leading coefficient of a quadratic function? Looking at the general form, y = ax 2 + bx + c, the leading coefficient will always be the a value. ## What is the degree of each product? The degree of the product of two or more polynomials with one variable is the sum of the degrees of each polynomial. For example, the degree of the product of x2+1 and 4×3+5x+1 is 5. This is because the degree of x2+1 is 2, and the degree of 4×3+5x+1 is 3, so the total degree is 2+3=5.
Courses Courses for Kids Free study material Offline Centres More Store # 12% as a Fraction - Converting Percentage into Fraction Last updated date: 23rd Apr 2024 Total views: 196.5k Views today: 3.96k ## What is 12 as a Fraction? What do you do when you and your sibling have to share a large pizza? To avoid a fight between you two, you divide it into several slices of the same size, isn’t it? But how do you know that each slice is of the same size? And how do you divide the pizza? Well, here Maths comes into play. The topic of fraction in Maths defines the portion or section of a whole object or quantity. Fractions are represented in the form of a numerator and a denominator. For example, consider this circle. The circle is divided into 12 equal parts and one part is coloured darker than the others. This darker part is called the twelfth part of the circle. In fractions, it can be written as$\frac{1}{12}$, where 1 = numerator and 12 = denominator. Every integer is also considered a fraction with denominator 1. A circle divided into 12 equal parts. ## Convert 12% into a Fraction 12% is a term written in percentage form but we need to convert it into the fraction form. But how to do it? To express 12 as a fraction, follow the given step-by-step procedure. Step 1: Divide the given number in percentage form by 100 like this: 12% = $\frac{12}{100}$ Step 2: Now we need to multiply both the numerator and denominator by 10 for every number after the decimal point. Since 12 is an integer with no decimal point, we can just skip to the next step. Step 3: In this step, we need to simplify the obtained fraction, i.e. $\frac{12}{100}$ into the simplest form. To simplify any fraction, we find the greatest common divisor (GCD) of the number and then divide both the numerator and denominator by the GCD to obtain our simplified fraction. In this case, the GCD of 12 and 100 is 4. Dividing 12/100 by 4, (12 ÷ 4)/(100 ÷ 4) = 3/25 12 percent when written as a fraction in its simplest form is 3/25. ## Quiz Time! 1.Convert these percentages into the simplest form of fractions. 1. 18% 2. 25% 3. 76% 4. 55% 2. The given picture shows Sylvia’s garden in which she has grown flowers. Can you tell which flowers she grows the most? Express the part of those flowers as a fraction in its simplest form. Fraction exercise for kids 1. 9/50 2. 1/4 3. 19/25 4. 11/20 1. Sylivia grows red flowers the most, which are 8 in number out of a total 12 flowers. Fraction of the field that contains red flowers = 8/12 In simplest form = 2/3 ## Conclusion Now you understand the importance of fractions and how you can express percentages as fractions. Using our detailed explanation of how to write 12 as a fraction, you can practice other similar questions yourself. Also, if you want to learn more about such mathematical concepts, explore our website and enhance your learning experience! ## FAQs on 12% as a Fraction - Converting Percentage into Fraction 1. What are the real-life uses of fractions? Fractions are used in almost every part of our daily lives, such as • We read the time in terms of fractions from an analogue clock or watch. • Splitting the restaurant bill among your friends. • Dividing food into equal portions for the entire family. • Comparing the prices of items in a supermarket when on sale or discounts. 2. How can you teach percentage and fractions to children? Fraction is a fun topic to teach your young ones. You can experiment with different visual methods and hands-on activities to explain fractions to kids easily. Ask them to cut a sandwich into half or four equal parts when demonstrating the fractions$\frac{1}{2}$ or $\frac{1}{4}$. You can use slices of pizza or playdough to explain the concept of simplified fractions.
# (Solution) Math 225N Week 5 Assignment Understanding the Empirical Rule Week 5 Assignment Understanding the Empirical Rule Question A random sample of CO2 levels in a school has a sample mean of x¯=598.4 ppm and sample standard deviation of s=86.7 ppm. Use the Empirical Rule to determine the approximate percentage of CO2 levels that lie between 338.3 and 858.5 ppm. Question A random sample of small business stock prices has a sample mean of x¯=\$54.82 and sample standard deviation of s=\$8.95. Use the Empirical Rule to estimate the percentage of small business stock prices that are more than \$81.67. Question A random sample of vehicle mileage expectancies has a sample mean of x¯=169,200 miles and sample standard deviation of s=19,400 miles. Use the Empirical Rule to estimate the percentage of vehicle mileage expectancies that are more than 188,600 miles. Question A random sample of hybrid vehicle fuel consumptions has a sample mean of x¯=53.2 mpg and sample standard deviation of s=4.8 mpg. Use the Empirical Rule to estimate the percentage of hybrid vehicle fuel consumptions that are less than 43.6 mpg. # Solution: Question A random sample of CO2 levels in a school has a sample mean of x¯=598.4 ppm and sample standard deviation of s=86.7 ppm. Use the Empirical Rule to determine the approximate percentage of CO2 levels that lie between 338.3 and 858.5 ppm. To use the Empirical Rule, we need to know how many standard deviations from the mean are the given values 338.3 and 858.5. Since the mean is 598.4, we see that the value 338.3 is 598.4−338.3=260.1 ppm below the mean. This is 3 standard deviations, since 260.1=3×86.7, so 338.3 is 3 standard deviations less than the mean. Similarly, the value 858.5 is 858.5−598.4=260.1 ppm above the mean. Again, this is 3 standard deviations, since 260.1=3×86.7, so 858.5 is 3 standard deviations greater than the mean. The Empirical Rule states that approximately 99.7% of the data is within 3 standard deviations of the mean. So by the Empirical Rule, we can say that approximately 99.7% of CO2 levels in the school are between 338.3 and 858.5 ppm. Please click the icon below to purchase full answer at only \$5
# Geometric Constructions A series of free, online High School Geometry Video Lessons and solutions. Videos, worksheets, and activities to help Geometry students. In these lessons, we will learn • how to define and construct the altitude of a triangle Click here • how to define and construct the median of a triangle Click here The following diagram shows how to construct the altitude of a triangle from a vertex. Scroll down the page for more examples and solutions. ### Construct an Angle Bisector An angle is formed by two rays with a common endpoint. The angle bisector is a ray or line segment that bisects the angle, creating two congruent angles. To construct an angle bisector you need a compass and straightedge. Bisectors are very important in identifying corresponding parts of similar triangles and in solving proofs. How to define and construct an angle bisector. This video shows how to bisect an acute angle with a compass and straight edge. The process is the same for an obtuse angle. ### Constructing Parallel Lines Constructing parallel lines with a compass and straightedge uses the converse of the parallel lines theorem. Creating congruent corresponding angles (or congruent AIA or AEA) guarantees parallel lines. The first step in constructing parallel lines is to draw a transversal through the given point to intersect the given line. Last, duplicate an angle created by the transversal and the given line. using compass and straightedge to construct parallel lines Instructions on how to construct a line parallel to a given line through a given point. ### Constructing Triangle Altitude Altitudes are defined as perpendicular line segments from the vertex to the line containing the opposite side. In each triangle, there are three triangle altitudes, one from each vertex. In an acute triangle, all altitudes lie within the triangle. In a right triangle, the altitude for two of the vertices are the sides of the triangle. In an obtuse triangle, the altitude from the largest angle is outside of the triangle. Use a compass and a straight edge to create the altitudes of an obtuse triangle. One of the altitude will lie inside the triangle and other two altitudes will lie outside the triangle. ### Constructing a Median A median is a line segment from the vertex to the midpoint of the opposite side in a triangle. In every type of triangle, the median will be contained within the polygon, unlike altitudes which can lie outside the triangle. When constructing a median, we first find the midpoint of the side opposite the desired vertex, then use a straightedge to connect the midpoint and the vertex. How to define a median in a triangle. This video will show you how to construct a midpoint using a compass. It will show you how to construct the three medians of an acute triangle. Using a straight edge and compass constructing a median of a triangle. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
## Linear Algebra and Its Applications, Exercise 3.4.16 Exercise 3.4.16. Given the matrix $A$ whose columns are the following two vectors $a_1$ and $a_3$ [sic]: $a_1 = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \quad a_3 = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$ factor $A$ as $A = QR$. If there are $n$ vectors $a_j$ with $m$ elements each, what are the dimensions of $A$, $Q$, and $R$? Answer: With $a_1$ and $a_3$ as the two columns of $A$, we first choose $a_1' = a_1 = (1, 2, 2)$. We then have $a_3' = a_3 - \frac{(a_1')^Ta_3}{(a_1')^Ta_1'}a_1' = a_3 - \frac{1 \cdot 1 + 2 \cdot 3 + 2 \cdot 1}{1^2 + 2^2 + 2^2}a_1' = a_3 - \frac{9}{9}a_1' = a_3 - a_1'$ $= \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}$ Now that we have calculated the orthogonal vectors $a_1'$ and $a_3'$ we can normalize them to create the orthonormal vectors $q_1$ and $q_3$. We have $\\|a_1'\\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$ $\\|a_3'\\| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$ so that $q_1 = a_1' / \\|a_1'\\| = \frac{1}{3} \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix}$ $q_3 = a_3' / \\|a_3'\\| = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}$ The matrix $Q$ is the 3 by 2 matrix with columns $q_1$ and $q_3$: $Q = \begin{bmatrix} \frac{1}{3}&0 \\ \frac{2}{3}&\frac{1}{\sqrt{2}} \\ \frac{2}{3}&-\frac{1}{\sqrt{2}} \end{bmatrix}$ The matrix $R$ is the 2 by 2 matrix calculated as follows: $R = \begin{bmatrix} q_1^Ta_1&q_1^Ta_3 \\ 0&q_3^Ta_3 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \cdot 1 + \frac{2}{3} \cdot 2 + \frac{2}{3} \cdot 2&\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot 3 + \frac{2}{3} \cdot 1 \\ 0&0 \cdot 1 + \frac{1}{\sqrt{2}} \cdot 3 + (-\frac{1}{\sqrt{2}}) \cdot 1 \end{bmatrix}$ $= \begin{bmatrix} \frac{9}{3}&\frac{9}{3} \\ 0&\frac{2}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} 3&3 \\ 0&\sqrt{2} \end{bmatrix}$ If there are $n$ vectors $a_j$ with $m$ elements each, since they form the columns of $A$ the shape of $A$ will be $m$ by $n$. The matrix $Q$ contains one orthonormal column for each column in $A$, and its orthonormal columns have the same number of elements as the columns of $A$, so $Q$ is also $m$ by $n$. Finally, $R$ is a square matrix with one column for each column of $Q$, so it is $n$ by $n$. NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang. If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books. This entry was posted in linear algebra and tagged , . Bookmark the permalink.
# How to analyse pie charts Written by linda donahue • Share • Tweet • Share • Pin • Email Analysing a pie chart is really about analysing the data; the pie chart itself is actually the visual analysis of data. So, in reality, you analyse the data then create the pie chart as a representation of your analysis. If, however, as a student, you've been given a drawn pie chart without data analysis on it, then you can reverse the process and look at the chart and calculate backwards for the data analysis. For this you'll need a protractor and will have to do some simple maths calculations. • Pie chart • Protractor ## Instructions 1. 1 Look at the pie chart for your initial analysis. The larger the section the more important or greater value that piece represents. Your chart is probably labelled with what the slice represents. For example, say the chart represents the types of horses you have at your farm. So one slice is for pintos, another for palominos, one for thoroughbreds, and so forth. Each slice represents what per cent of the herd is that type of horse. 2. 2 Measure the interior angle of a slice with the protractor. Start with the biggest slice. Work your way around the chart, measuring every angle of every slice. In the end, all the angles together should add up to 360 degrees. If they don't, remeasure. 3. 3 Convert the angle to what fraction of the circle it represents. Take each of your angle measurements and place them over 360. For example, if one slice was 90 degrees, you would create the fraction 90/360. Now the fraction. 90/360 reduces to 1/4. Do this for each section of the pie chart. 4. 4 Convert the fractions to percentages. For this, all you have to do is divide and multiply by 100. If you don't like working with decimals, you can multiply first (as the operations have the same hierarchy.) So, to convert 1/4 to a percentage, you can multiply by 100 getting 100/4 then doing the division. 4 divides into 100 exactly 25 times. Your answer is 25 per cent; that is, that slice of the pie chart represents 25 per cent. So if that was your per cent of pintos, then pintos comprise 25 per cent of your total herd. Do the same analysis for all slices. ### Don't Miss • All types • Articles • Slideshows • Videos ##### Sort: • Most relevant • Most popular • Most recent
What is Infinity? One of the common errors made by calculus learners is to consider infinity as a number. True, there are number systems such as the extended real number system in which infinity is successfully treated as a number, and you can ponder on them if you wish, but for our purposes at this level of learning calculus, infinity is decidedly not a number. Numbers satisfy various properties, and infinity does not satisfy these properties. For example, it might seem reasonable to state that $\infty + 1 = \infty$ (how could this be any different?), but then by the usual properties of numbers, we should be able to subtract $\infty$ from both sides of the equation to obtain $1 = 0$, which is nonsense. This is enough to rule out infinity as a number, but you can have fun deriving all kinds of contradictions based on the assumption that it is a number, for your own amusement. (It won’t take you long to prove that all real numbers are equal, starting from the crazy assumption that $1 = 0$, for example, which is further reinforcement that the assumption that infinity is a number is not valid.) Be alert to the use of the symbol $\infty$ in various arguments in calculus textbooks for various purposes. Recognize that it is a time-saving symbol that represents various facts or processes. While you are striving to understand the facts and processes it represents, remind yourself regularly that infinity, while a very useful concept, is not a number. So what is infinity, then? Let’s start by thinking about the natural numbers. A basic property of the natural numbers is that you can always add the number $1$ to a natural number, and the result is another natural number. For example, $7$ is a natural number, and it is possible to add $1$ to $7$, with the result being $8$, another natural number. Now this is true no matter how large the natural number you choose, because this is a property of all natural numbers. What, then is the largest natural number? You will be able to understand that based on this property of natural numbers, there is no largest natural number. Suppose someone proposes to you that some natural number, no matter how large, is the largest natural number. You could counter this proposal by simply adding $1$ to the proposed largest natural number to produce a natural number that is even larger. But that new natural number is not the largest either, because you can also add $1$ to it to obtain an even larger one. So there is no largest natural number. One could say that there is an unlimited number of natural numbers. Another way to say this is that there is an infinite number of natural numbers. This usage of the word infinite summarizes the fact that there is an unlimited number of natural numbers. There are many cars on Earth, but if you had to do so, you could count all of them. You could take a super-snapshot of Earth at a particular time, and then you could carefully examine this photograph and count all of the cars on Earth. No doubt the number is very large, but it is a natural number. The number of cars on Earth is finite, because in principle you could count the number and the result is a natural number. Similarly, you could (in principle) count all of the atoms on Earth, at a particular time, and this too is a natural number, so we say the number of atoms on Earth is finite.1 Similarly, there are collections of numbers that are finite. For example, consider the collection of odd numbers that are between $1$ and $100$ inclusive. There are $50$ such numbers, right? We could say that the set of odd natural numbers less than or equal to $100$ is finite. You could easily construct any number of finite sets, such as the set of prime natural numbers that are less than $1000$, the set of even numbers between $5000$ and $9000$ inclusive, and so on. So we have finite sets, and then we have infinite sets, such as the set of all natural numbers. In order to be able to have some way of talking about the number of elements of a set in a unified way, whether the set is finite or infinite, mathematicians have coined the term cardinality. It doesn’t make sense to speak about the number of elements in the set of all natural numbers, because there is no such number. It does make sense to speak about the number of elements in the set of odd numbers between $1$ and $100$ inclusive; this number is $50$. We can say that the cardinality of the set described in the previous sentence is $50$, and the cardinality of the set of all natural numbers is infinite. Thus, the concept of cardinality gives us a way of speaking about the “size” of a set, whether the number is finite or infinite. Introducing the concept of cardinality may seem unnecessary, but let’s discuss something that is potentially shocking. It certainly shocked numerous mathematicians when they learned about it from Georg Cantor about a century ago: There are different infinities, of different “sizes.” If you prefer, there are different “levels” of infinity. Is this not mind-boggling?? Are there really infinite sets that have different cardinalities? To understand this amazing fact about infinities (yes, we should use the plural now), we’ll first have to think about how to compare the cardinalities of two sets that are infinite. For example, imagine the set of all natural numbers (we’ll call it $A$), and then imagine the set that includes all natural numbers and that also includes the number $0$ as well; we’ll call this set $B$. Now it seems reasonable to say that set $B$ is larger than set $A$; after all, set $B$ includes everything in set $A$, and set $B$ also includes one number that is not in $A$. In the language of set theory, we would say that set $A$ is a proper subset of set $B$. However, this is not the way we currently understand infinite sets, as you will see in the next few paragraphs. Cantor came up with a criterion for comparing infinite sets that led him to his revolutionary understanding of infinities. He said that two sets have the same cardinality if you could set up a one-to-one correspondence between the two sets. That is, you have to be able to pair the elements of the two sets, so that each pairing matches an element of one set with an element of the other set, no element of either set is included in more than one pairing, and each element of each set is included in some pairing. Think about a sports stadium with $50,000$ seats. Now imagine that the stadium is full of people, so that each seat is occupied by one person, no seats are empty, and each person in the stadium is in a seat. You don’t need to count the people to determine how many of them are in the stadium; you can immediately conclude that there are $50,000$ people in the stadium, because they are in one-to-one correspondence with the seats, and you know how many seats there are. Now let’s apply this concept of comparing cardinalities to infinite sets. In particular, consider the sets $A$ and $B$ described a few paragraphs ago. There exists a one-to-one correspondence between the sets $A$ and $B$, and so they have the same cardinality! Even though we have some sort of sense that we should be considering $B$ to be bigger than $A$, this sense is not applicable to infinite sets. According to Cantor’s definition of equal cardinalities, these two sets have the same cardinality! Can you come up with a formula for a one-to-one correspondence that confirms this? A good way to intuitively understand this is through the story of Hilbert’s Hotel. David Hilbert was one of the greatest mathematicians of about a century ago, and he constructed a series of “thought experiments” involving a hypothetical hotel that has an infinite number of hotel rooms. I imagine the hotel rooms all in a (very long!) row, rather like a motel, to model the natural numbers as we would normally plot them along a number line. Suppose that all of the infinite number of rooms in Hilbert’s Hotel are occupied at the moment, so that there are no vacant rooms. If a new prospective guests walks into the hotel’s lobby desperately asking for a room for the night, is he or she out of luck? Well, not necessarily, according to the clever front desk clerk. The clerk merely asks each guest in the hotel to vacate their room and shift one room over. That is, the person in Room 1 moves to Room 2, the person in Room 2 moves to Room 3, and so on. Each existing guest is perfectly well-accommodated in their new room, but now Room 1 has been vacated, and so it is available for the new guest. Problem solved! Once you have let this remarkable solution sink in, you might then understand why there is a temptation among some people to express this shifty business as $$\infty + 1 = \infty$$ Resist the temptation to do this! This kind of equation encourages us to treat $\infty$ as if it were a number, but we have already argued that this is not so! So avoid such nonsensical equations. Nevertheless, you can also understand the temptation to write such an equation, because it does (in a way) capture an important property of Hilbert’s Hotel; even if all of the infinite number of rooms is occupied, space can always be made available for one more guest. Mind-boggling! Infinity sure is unusual. Does this story help you to feel a bit better about the fact that the sets $A$ and $B$, described earlier, can be placed in one-to-one correspondence, and therefore have the same cardinality? Would it help more if you could find an explicit formula for such a one-to-one correspondence? Here’s one, perhaps the simplest one, that does the trick: $f(n) = n + 1$. The same formula describes the way existing guests must shift rooms: The guest in Room $n$ must shift to Room $n + 1$. You can iterate the front desk clerk’s shifty technique to accommodate two new guests, three new guests, and indeed, any finite number of new guests. (What is the shifting formula in such cases if there are $m$ new guests?) But this kind of shifting clearly won’t work if an infinite number of new guests arrive, right? For example, let’s suppose that there is a neighbouring hotel that is much like Hilbert’s Hotel, in that there are an infinite number of rooms, all currently occupied by guests. There is a power outage at the other hotel; is there any way that this infinite number of guests can be squeezed into Hilbert’s Hotel, with each guest having his or her own room? If there were only $5$ new guests, we could just shift each existing guest five rooms over. But with an infinite number of new guests, shifting in this way doesn’t work. What does it mean to shift each existing guest an infinite number of rooms over? This is meaningless! Where does the guest currently in Room $37$ get moved? Because $\infty$ is not a number, saying that the guest in Room $37$ should be moved to Room $37 + \infty$ has no meaning! But the front desk clerk is very clever, and decides that if for each $n$, the guest in Room $n$ shifts to Room $2n$, then each existing guest will still be accommodated (in all the even-numbered rooms), and yet an infinite number of rooms (the odd-numbered ones) will have been vacated, allowing all of the guests displaced from the other hotel to be accommodated also! Isn’t this amazing? The previous paragraph shows that the cardinality of the even natural numbers is the same as the cardinality of the natural numbers. In some intuitive way, we would wish to say that there are half as many even numbers as natural numbers, but no, this intuition is not helpful when it comes to infinite sets. Similarly, the cardinality of the odd numbers is also the same as the cardinality of the natural numbers. What is a simple formula for a one-to-one correspondence demonstrating this latest fact? Once again, one might be tempted to write $$\infty + \infty = \infty \quad \quad \textrm{or} \quad \quad 2\infty = \infty$$ to express this strange fact, but one should really avoid doing so, as $\infty$ is not a number, and therefore can’t be combined in an equation like this according to the usual rules for manipulating numbers. But you can certainly see why such nonsensical equations are written in some places; they are attempts to express strange and wonderful properties of infinity in a form that is not appropriate for communicating such facts. What if there were two or three other copies of the Hilbert Hotel, whose occupants all had to be squeezed into the Hilbert Hotel? Would you be able to do so if you were the desk clerk? Which formula proves that such redistributions of guests are possible? What if there were $m$ total copies of the Hilbert Hotel (including the HH); can you do the redistribution? What is a formula that proves that such a redistribution is possible? If you were able to complete the tasks in the previous paragraph, you will now be convinced that the cardinality of $m$ copies of the natural numbers, taken as one giant set, is the same as the cardinality of one copy of the natural numbers by itself. Remarkable! In the previous paragraph, $m$ is a finite number. What if you had an infinite number of hotels like the Hilbert Hotel? Would you be able to fit all of the guests in all of these infinite number of hotels into just one Hilbert Hotel by redistributing all of the guests? Surely this is impossible, right? At least it’s not possible using the method of the previous paragraphs for a finite number of copies of the natural numbers. It’s worth pausing right now, turning away, and mulling this over for some time. Return to your reading only after you have mulled things over for a while, and after having writing your thoughts in your research notebook. After mulling it over, what do you think? In fact, it is indeed possible! The cardinality of an infinite number of copies of the natural numbers is the same as the cardinality of the natural numbers! Wow! It is a little more challenging to come up with an explicit formula for a one-to-one correspondence in this case. It may help you to sketch a diagram, where each row of the diagram corresponds to a copy of the natural numbers. Then ask yourself if there is a systematic way to step your way through the entire (infinite) array of numbers, such that you are certain to eventually step on each number in each row. Doing this may help you to understand that this is possible, and provided your pathway is simple enough, you may also be able to write a formula for the correspondence. This is a challenging task, but have fun with it! We stated earlier on that there were different levels of infinity, but so far we have only encountered one, the cardinality of the natural numbers. Each of the infinite sets we have constructed so far has the same cardinality. It turns out that the cardinality of the real numbers is greater than the cardinality of the natural numbers. The proof that this is so is due to Cantor, again, and it is based on a beautiful idea nowadays called Cantor’s diagonal argument, which I’ll now describe. Consider the real numbers between $0$ and $1$. Cantor showed that the cardinality of this set is not equal to the cardinality of the natural numbers by proving that it is not possible to place the two sets into one-to-one correspondence. He did this by using a proof by contradiction, which is to assume that it is possible and then demonstrate a contradiction, showing that the original assumption is false. So, let’s retrace Cantor’s steps by assuming that it is possible to construct a one-to-one correspondence between the natural numbers and the set of real numbers between $0$ and $1$. In effect, this assumption is that you can place the entire set of real numbers between $0$ and $1$ in a list in some way. For example, here is a partial list: 0.3715682… 0.4931657… 0.1153267… 0.0474749… 0.9535360… 0.0088841… 0.5583322 Clearly we can’t display the entire list, nor can we even show the complete decimal expansion of each number in the list, but the assumption is that this can be done. Cantor then argued that this assumption is incorrect by constructing a number that is not in the list. You can do this by constructing a number that differs from the first number in the first decimal place, differs from the second number in the second decimal place, differs from the third number in the third decimal place, and so on. You can do this according to some rule to make it easier; for example, if the given digit is a $3$, then make it a $5$, and if the digit is not a $3$, then make it a $3$. Look at the list of numbers above, and apply this rule to the red digits to construct a new number: $$0.5333533…$$ The particular rule used is not essential; many other rules would work just as well. Consider the new number just constructed and note that it is not in the original list of numbers. You can tell it is not in the original list, because it is not the first number in the list (it differs in the first decimal digit), it is not the second number in the list (it differs in the second decimal digit), it is not the $47$-th number in the list (it differs in the $47$-th digit), and so on. Therefore, it is not in the list, and the assumption that we had a complete list of all real numbers between $0$ and $1$ is false. Can you obtain a complete list of all real numbers between $0$ and $1$ by just including this new number at the top of the list? No. You can see that this attempt will not work by applying Cantor’s diagonal argument again to the new list to construct yet another real number between $0$ and $1$ that is not in the new list either. No matter how many newly constructed numbers you add to the top of the list, it will never be a complete list of all real numbers between $0$ and $1$. The same argument can be applied to any proposed complete list of real numbers whatsoever. Isn’t this an ingenious argument? And isn’t the result absolutely remarkable? Thus, it is not possible to list all of the real numbers between $0$ and $1$. Another way to say this is that it is not possible to place the real numbers between $0$ and $1$ in one-to-one correspondence with the natural numbers, and therefore the cardinality of the real numbers between $0$ and $1$ is different from the cardinality of the natural numbers. It turns out that the cardinality of the set of all real numbers is the same (!!) as the cardinality of the set of real numbers between $0$ and $1$. Can you argue that this must be true? Hint: If you can construct a one-to-one function that maps the entire real line into the interval of real numbers from $0$ to $1$, then this would be an explicit proof. A function that maps the other way would work just as well. Search your memory banks for a graph from high school that will do the trick! Isn’t it mind-boggling that the number of real numbers between $0$ and $1$ is the same (in the sense of one-to-one correspondence) as the number of all real numbers? These two sets have the same cardinality. Because the set of natural numbers is contained within the set of real numbers, and these two sets cannot be placed in one-to-one correspondence, we say that the cardinality of the real numbers is greater than the cardinality of the natural numbers. Thus, we have established the existence of two levels of infinity. Here is some standard terminology: Sets that either contain a finite number of elements or can be placed in one-to-one correspondence with the natural numbers (such as the even numbers, the odd numbers, the integers,2 and so on) are called countable sets. Infinite sets that are countable are also called countably infinite. Infinite sets that are not countable are called uncountable. Thus, we have (so far) two levels of infinite sets, sets that are countably infinite (such as the natural numbers) and sets that are uncountable (such as the real numbers). Are there any levels of infinity that are between the cardinality of the natural numbers and the cardinality of the real numbers? Cantor conjectured in 1878 that the answer to this question is no, in what is now called the continuum hypothesis. Attempts were made for many years to either prove the continuum hypothesis or to discover a counterexample, which culminated in a publication by Kurt Gödel in 1940 in which he showed that it is impossible to disprove the continuum hypothesis within standard set theory. Paul Cohen showed in 1963 that the continuum hypothesis cannot be proved within standard set theory either! This remarkable set of results shows that the continuum hypothesis is independent from standard set theory. To learn more about this very strange result, look up Gödel’s incompleteness theorem. We stated earlier that there is a whole hierarchy of infinities, but so far we have only seen examples of two levels of infinity, that of the natural numbers and that of the real numbers. How can one construct higher levels of infinity? What about the cardinality of the set of points in the plane that you used so much in high school to study functions? Surely the cardinality of the number of points in the plane is greater than the cardinality of the real line? But no, the cardinalities are the same! Proving this is more challenging, though, than Cantor’s diagonal argument. (Look up the Schröder-Bernstein theorem if you are curious about this.) Similarly, the cardinality of the points in three-dimensional space is also the same as the cardinality of the real number line. Thus, simply moving to higher-dimensional spaces does not give us a greater level of infinity. How does one construct sets with cardinalities at higher levels of infinity? We shall leave this discussion for another time, but if you are curious you can consult a work on mathematical analysis or set theory. Before concluding this discussion, it is worth mentioning that the ancient Greeks already distinguished between what they called actual infinity and potential infinity, and this is a useful distinctions. Actual infinity is reserved to describe an infinite set in its entirety, such as the set of natural numbers taken as a whole, or the set of real numbers taken as a whole. Potential infinity is reserved for the idea of a quantity that is increasing without bound, so that the quantity gets larger and larger with each step of the process, with no limitations on how large it gets. Our discussion of limits as $x$ “approaches infinity” fits this sense of potential infinity. In fact, we discuss limits as $$x \rightarrow \infty \qquad \textrm{and} \qquad x \rightarrow -\infty$$ where we typically envision $x$ “moving” to the right indefinitely in the first case, and “moving” to the left indefinitely in the second case. It’s worth emphasizing again that in both of these cases “infinity” is not a place, but rather this is a process of imagining what happens when a quantity ($x$ in this case) either “moves” to the right indefinitely or “moves” to the left indefinitely. 1. It appears that some of the most serious problems we have on Earth is that we humans collectively treat some of our limited resources as if they were infinite instead of finite. 2. Can you prove that the integers can be placed in one-to-one correspondence with the natural numbers by constructing a suitable formula?
# Relation Formulas for Class 11 Home » CBSE Class 11 Maths » Formulas for Class 11 Maths » Relation Formulas for Class 11 ## Relation Formulas for Class 11 Here we are providing Relation Formula For Class 11 Maths. Practice Relation problems based on these formulas. Important terms and definitions are also included so that you can revise them in a very short time. List of Relation Formulas for Class 11 (1) Ordered Pair: An ordered pair consists of two objects or elements grouped in a particular order. Two numbers a and b listed in a specific order and enclosed in parentheses form an ordered pair (a, b) . Here a is the first component and b is the second component. In general, (a, b)≠(b, a) . (2) Equality of Ordered Pairs: Two ordered pairs (a1, b1) and (a2, b2) are equal iff a1=a2 and b1=b2 (a, b) = (c, d) ⇔ a = c and b = d. (3) Cartesian (or Cross) Product of Sets: For two non-empty sets A and B, the set of all ordered pairs (a, b) such that a ∈ A and b ∈ B is called Cartesian product A × B, i.e. A × B = {(a, b) : a ∈ A and b ∈ B} If A = Φ or B = Φ then A x B = Φ (4) Ordered Triplet: Three numbers a, b, c listed in a specific order and enclosed in parentheses form an ordered triplet (a, b, c). (a, b, c) ≠ (b, a, c) ≠ (c, a, b), etc. A× B×C = {(a, b, c) : a ∈ A, b ∈ B and c ∈ C} (5) For any nonempty sets A, B, C, we have (A x B) x C = A x (B x C), each denoted by A x B x C. (6) For any sets A, B and C, we have: (i) A x (B ∪ C) = (A x B) ∪ (A x C) (ii) A x (B ∩ C) = (A x B) ∩ (A x C) (iii) A x (B – C) = (A x B) – (A x C) (iv) (A x B) ∩ (B x A) = (A ∩ B) x (B ∩ A) = (A ∩ B) x (A ∩ B) (v) A x B = A x C ⇒ B = C (vi) A ⊂ B ⇒ (A x A) ⊂ (A x B) ∩ (B x A) (vii) A ⊂ B ⇒ (A x C) ⊂ (B x C) (viii) A ⊂ B and C ⊂ D ⇒ (A x C) ⊂ (B x D) (ix) A x B = B x A ⇔ A = B (7) Let A and B be two nonempty sets and let R ⊆ A x B. Then, R is called a relation from A to B. >> If (a, b) ∈ R, we say that ‘a is related to b‘ and we write, a R b. >> If (a, b) ∉ R, we say that ‘a is not related to b‘ >> Dom (R) = {a : (a, b) ∈ R}, range (R) = {b : (a, b) ∈ R} . (8) We define, R-1 = {(b, a) : (a, b) ∈ R}, dom (R) = range (R-1) and range (R) = dom (R-1). (9) Let A be a nonempty set. Then, every subset of A x A is called a binary relation on A. (10) If either A or B is an infinite set, then A × B is an infinite set. (11) n(A × B) = n(A) × n(B) (12) If A and B be any two non-empty sets having n elements in common, then A × B and B × A have n2 elements in common. Announcements Join our Online JEE Test Series for 499/- Only (Web + App) for 1 Year Join our Online NEET Test Series for 499/- Only for 1 Year
# Tangents and Cotangents of Multiples or Submultiples We will learn how to solve identities involving tangents and cotangents of multiples or submultiples of the angles involved. We use the following ways to solve the identities involving tangents and cotangents. (i) Starting step is A + B + C = π (or, A + B + C = $$\frac{π}{2}$$) (ii) Transfer one angle on the right side and take tan (or cot) of both sides. (iii) Then apply the formula of tan (A+ B) [or cot (A+ B)] and simplify. 1. If A + B + C = π, prove that: tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C Solution: Since, A + B + C = π ⇒ 2A + 2B + 2C = 2π ⇒ tan (2A + 2B + 2C) = tan 2π ⇒ $$\frac{tan 2A+ tan 2B + tan 2C - tan 2A tan 2B tan 2C}{1 - tan 2A tan 2B - tan 2B tan 2C - tan 2C tan 2A}$$ = 0 ⇒ tan 2A + tan 2B + tan 2C - tan 2A tan 2B tan 2C = 0 ⇒ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C. Proved. 2. If A + B + C = π, prove that: $$\frac{cot A + cot B}{tan A + tan B}$$ + $$\frac{cot B + cot C}{tan B + tan C}$$ + $$\frac{cot C + cot A}{tan C + tan A}$$ = 1 Solution: A + B + C = π ⇒ A + B = π - C Therefore, tan (A+ B) = tan (π - C) ⇒ $$\frac{tan A+ tan B}{1 - tan A tan B}$$ = - tan C ⇒ tan A + tan B = - tan C + tan A tan B tan C ⇒ tan A + tan B + tan C = tan A tan B tan C. ⇒ $$\frac{tan A + tan B + tan C}{tan A tan B tan C}$$ = $$\frac{ tan A tan B tan C}{tan A tan B tan C}$$, [Dividing both sides by tan A tan B tan C] ⇒ $$\frac{1}{tan B tan C}$$ + $$\frac{1}{tan C tan A}$$ + $$\frac{1}{tan A tan B}$$ = 1 ⇒ cot B cot C + cot C cot A + cot A cot B = 1 ⇒ cot B cot C($$\frac{tan B + tan C}{tan B + tan C}$$) + cot C cot A ($$\frac{tan C + tan A}{tan C + tan A}$$) + cot A cot B ($$\frac{tan A + tan B}{tan A + tan B}$$) = 1 ⇒ $$\frac{cot B + cot C}{tan B + tan C}$$ + $$\frac{cot C + cot A}{tan C + tan A}$$ + $$\frac{cot A + cot B}{tan A + tan B}$$ = 1 ⇒ $$\frac{cot A + cot B}{tan A + tan B}$$ + $$\frac{cot B + cot C}{tan B + tan C}$$ + $$\frac{cot C + cot A}{tan C + tan A}$$ = 1 Proved. 3. Find the simplest value of cot (y - z) cot (z - x) + cot (z - x) cot (x - y) + cot (x - y) cot(y - z). Solution: Let, A = y - z, B = z - x, C = x - y Therefore, A + B + C = y - z + z - x + x - y = 0 ⇒ A + B + C = 0 ⇒ A + B = - C ⇒ cot (A + B) = cot (-C) ⇒ $$\frac{cot A cot B - 1}{cot A + cot B}$$ = - cot C ⇒ cot A cot B - 1 = - cot C cot A - cot B cot C ⇒ cot A cot B + cot B cot C + cot C cot A = 1 ⇒ cot (y - z) cot (z - x) + cot (z - x) cot (x - y) + cot (x - y) cot (y - z) = 1. Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Successor and Predecessor \| Successor of a Whole Number \| Predecessor May 24, 24 06:42 PM The number that comes just before a number is called the predecessor. So, the predecessor of a given number is 1 less than the given number. Successor of a given number is 1 more than the given number… 2. ### Counting Natural Numbers \| Definition of Natural Numbers \| Counting May 24, 24 06:23 PM Natural numbers are all the numbers from 1 onwards, i.e., 1, 2, 3, 4, 5, …... and are used for counting. We know since our childhood we are using numbers 1, 2, 3, 4, 5, 6, ……….. 3. ### Whole Numbers \| Definition of Whole Numbers \| Smallest Whole Number May 24, 24 06:22 PM The whole numbers are the counting numbers including 0. We have seen that the numbers 1, 2, 3, 4, 5, 6……. etc. are natural numbers. These natural numbers along with the number zero 4. ### Math Questions Answers \| Solved Math Questions and Answers \| Free Math May 24, 24 05:37 PM In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students
# Plane Shapes 2: Perimeter – Perimeter of a Rectangle and Square Perimeter is the outside boundary or edges of a plane shape. Example 1. Irregular shapes. Measure the perimeter of the quadrilateral ABCD in the diagram below. Solution AB = 22mm = 2.2cm BC= 27mm = 2.7cm CD= 10mm =1.0cm DA= 35mm = 3.5cm Total 94mm = 9.4cm • Regular shapes: Find in cm, the perimeter of the regular hexagon ABCDEF below. Solution Length AB= 1.6cm , there are 6 equal sides Perimeter = 6 x 1.6cm = 9.6cm Using Formulae to Calculate Perimeters. Examples 1. A rectangular piece of land measures 57m by 42m. What is the perimeter of the land. Solution Perimeter of land = 2(L + B) = 2(57 +42) m = 2(99) m = 198m. To calculate Length = perimeter of rectangle – breadth 2 To calculate Breadth = perimeter of rectangle/2 – length • A rectangle has a perimeter of 74m. Find • a) the length of the rectangle if its breadth is 17. • b) the breadth of the rectangle if its length is 25m. Solution a) Length = perimeter of rectangle/2 – breadth , 74m/2 -17m = 37m – 17m = 20m b) Breadth= perimeter of rectangle/2 – length, 74m/2 – 25m = 37m – 25m = 12m Squares A square is a regular four sided shape. If the length of one side of a square is L. Perimeter of square = L x 4 = 4L Length of side of square = perimeter of square/4 Example 1. Calculate the perimeter of a square of side 12.3cm. Solution Perimeter of square = 4L, 12.3 x 4 = 49.2cm • A square assembly area has a perimeter of 56m. Find the length of the side of the assembly area. Solution Length of square = perimeter/ 4 , 56m/4 = 14m Assignment 1. A rectangular field measure 400m by 550m. What is the perimeter of the field. 2. Calculate the perimeter of a square of side 15cm. 3. A length of wire is bent to make a square of side 3.2cm. a) what is the length of the wire in the square. b)if the wire is silver, costing ₦50 per centimeter, what is the cost of the square.
# Find a, b and n in the expansion of Question: Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively. Solution: We have : $T_{1}=729, T_{2}=7290$ and $T_{3}=30375$ Now, ${ }^{n} C_{0} a^{n} b^{0}=729$ $\Rightarrow a^{n}=729$ $\Rightarrow a^{n}=3^{6}$ ${ }^{n} C_{1} a^{n-1} b^{1}=7290$ ${ }^{n} C_{2} a^{n-2} b^{2}=30375$ Also, $\frac{{ }^{n} C_{2} a^{n-2} b^{2}}{{ }^{n} C_{1} a^{n-1} b^{1}}=\frac{30375}{7290}$ $\Rightarrow \frac{n-1}{2} \times \frac{b}{a}=\frac{25}{6} \quad \ldots(\mathrm{i})$ $\Rightarrow \frac{(n-1) b}{a}=\frac{25}{3}$ And, $\frac{{ }^{n} C_{1} a^{n-1} b^{1}}{{ }^{n} C_{0} a^{n} b^{0}}=\frac{7290}{729}$ $\Rightarrow \frac{n b}{a}=\frac{10}{1}$ ...(ii) On dividing $($ ii $)$ by $(i)$, we get $\frac{\frac{n b}{a}}{\frac{(n-1) b}{a}}=\frac{10 \times 3}{25}$ $\Rightarrow \frac{n}{n-1}=\frac{6}{5}$ $\Rightarrow n=6$ Since, $a^{6}=3^{6}$ Hence, $a=3$ Now, $\frac{n b}{a}=10$ $\Rightarrow b=5$
# Math picture app This Math picture app helps to fast and easily solve any math problems. We will give you answers to homework. ## The Best Math picture app Best of all, Math picture app is free to use, so there's no reason not to give it a try! To solve for the hypotenuse of a right angled triangle, you can use the Pythagorean Theorem. This theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, in order to solve for the hypotenuse, you would need to square the other two sides and then add them together. Afterwards, you would need to take the square root of the result in order to find the value of the hypoten When you take logs of the numbers in your equation, you will get a number that looks like log(y - y0). You can then subtract this number from the original y value to get y - log(y0) = log(y) + log(y0) This gives you the solution for x. It is as simple as that! Just take logs of each value in your equation and subtract them from one another to get the solution of x. The system of equations is the mathematical representation of a set of related equations. It is an ordered list of equations with and without solutions. The solution of a system of equations is the set of values that satisfies all the given equations. To solve system of equations, first we need to identify all the variables involved in the given system. Then we need to add all unknowns and solve for them individually. Once all unknowns are known, we can add all knowns and solve for them individually. This way, we get a single solution from a set of individual solutions. We use algebra to find a solution or to solve a system of linear equations or inequalities. Algebra is used to simplify, manipulate and evaluate expressions and questions involving variables. Algebra is also used for solving more complicated problems such as quadratic equations, polynomial equations, rational expressions, exponential expressions etc. Algebra can be used to solve systems with several variables or when there are different types of questions (such as multiple choice, fill-in-the-blank). There are various methods one can use to solve system of linear equations like substitution method, elimination method and combination method etc. In this article, we will discuss several approaches on solving systems of linear equation i.e substitution method etc. If you're struggling to solve a word problem, there's no need to agonize over it for hours - there are plenty of online resources that can help. A quick Google search can lead you to websites, videos, and articles that offer step-by-step guidance on solving all kinds of word problems. Many of these resources are free, so you can get help without spending any money. And even if you do choose to use a paid resource, it's still likely to ## We solve all types of math troubles This app is just a lit thank out so much for creating this app it has all the problems even some problems I can’t find it, it's fine as long as I get something it's just amazing. No ads just a perfect app I thought this app will also be like brainly but man this is way better than that no stupid confusions just simple and easy Yana Campbell The free version is amazing. I can’t do math tutors because I get upset easily and it’s much less stressful to teach myself at home and the walkthrough, they have been amazing! I love this so much and thank you for making math so much better and more fun to learn!!! Breanna Green
# Write an inequality of the graph shown above describes To solve a word problem with two unknowns find two equations that show a relationship between the unknowns. Thus the plane extends indefinitely in all directions. Then in the bottom line y we will place the corresponding value of y derived from the equation. Since we have already solved the second equation for x in terms of y, we may use it. The check is left up to you. Since 3,2 checks in both equations, it is the solution to the system. Determine the equations and solve the word problem. The point 0,0 is not in the solution set, therefore the half-plane containing 0,0 is not the solution set. We now locate the ordered pairs -3,9-2,7-1,50,31,12,-13,-3 on the coordinate plane and connect them with a line. To do this, however, we must change the form of the given equation by applying the methods used in section Step 5 If we check the ordered pair 4,-3 in both equations, we see that it is a solution of the system. Since the point 0,0 is not in the solution set, the half-plane containing 0,0 is not in the set. The example above was a system of independent equations. Note that we could solve this system by the substitution method, by solving the first equation for y. In this case we simply multiply each side by Check this point x,y in both equations. Independent equations The two lines intersect in a single point. A system of two linear inequalities consists of linear inequalities for which we wish to find a simultaneous solution. Even though the topic itself is beyond the scope of this text, one technique used in linear programming is well within your reach-the graphing of systems of linear inequalities-and we will discuss it here. This fact will be used here even though it will be much later in mathematics before you can prove this statement. You could have chosen any values you wanted. Check these values also. To graph the solution to this system we graph each linear inequality on the same set of coordinate axes and indicate the intersection of the two solution sets. Systems of Equations and Inequalities In previous chapters we solved equations with one unknown or variable. Step 3 Solve for the unknown. To solve a system of two linear inequalities by graphing, determine the region of the plane that satisfies both inequality statements. In this case there will be infinitely many common solutions. Determine the common solution of the two graphs. Compare your solution with the one obtained in the example. Therefore, the system has as its solution set the region of the plane that is in the solution set of both inequalities. Which graph would be steeper: The line indicates that all points on the line satisfy the equation, as well as the points from the table. Notice that once we have chosen a value for x, the value for y is determined by using the equation. To graph a linear inequality: Locate these points on the Cartesian coordinate system and connect them with a line. There are many types of graphs, such as bar graphs, circular graphs, line graphs, and so on.How Do You Graph an Inequality or an Infinite Set on a Number Line? Number lines are really useful in visualizing an inequality or a set. In. Apr 21, · Best Answer: PROBLEM 1: Just look at the line first. It has a slope of You can use the two intercept points to figure the slope, or Status: Resolved. Lesson 7 Write and Graph Inequalities 27 Main Idea Write and graph inequalities. Write and Graph Inequalities FAIR Jessica is trying to County State Write and graph an inequality to describe the possible ages of a colt. The graph shows ACTIVITIES Basketball Tennis Orchestra Baseball Softball Drama Band. Fit an algebraic two-variable inequality to its appropriate graph. Write the equation of a line in slope-intercept form. Use the y-intercept and the slope to draw the graph, as shown in example 8. Note that this equation is in the form y = mx + b. This gives us a convenient method for graphing linear inequalities. To graph a linear inequality 1. Replace the inequality symbol with an equal sign and. Solving Absolute Value Equations and Inequalities SOLVING EQUATIONS AND INEQUALITIES Solving Absolute Value Equations and Inequalities 51 An absolute value inequality such as \|x º 2 solutions in the original inequality. The graph is shown . Write an inequality of the graph shown above describes Rated 4/5 based on 14 review
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis : # Decimal Number Comparison - Through Graphical And Analytical Method The concept of the decimal is used in each and every field of mathematics. For example measurement of weight, length, money etc. And in each of these sections, there are instances where we need to compare two numbers or quantities. Let us say, we go to a local market to purchase pulse, where one of the vendors is selling the pulse at Rs 35.50/kg and the other is selling it at Rs 35.70/kg. In order to judge which shop we should buy the pulse from, we need to compare both the prices. These prices, as we saw, are in decimal numbers, so it is necessary to learn how decimal numbers are compared. ## Comparing Decimal Numbers There are two methods involved in the decimal number comparisons. Let us discuss the two different methods here in detail. ### Decimal Comparison Method 1 Example: Comparing 0.25 and 0.13 • We take a piece of paper and divide it into rows and columns equivalent to the number of places after decimal we are dealing with. For example, if we are dealing with numbers with one place after the decimal, we draw 10 blocks, 52 or 101; for two places after the decimal we draw 100 blocks with 10 rows and 10 columns. • Since the decimals we are dealing with have maximum two places after the decimal, we draw a 10*10 box. • Upon conversion into fraction, we can see, 0.25 = 25/100, so we fill 25 blocks out of 100 and for 0.13 we fill 13 blocks out of 100. • By observation, we can say that 0.25 is greater than 0.13 as the coloured region in 0.25 is greater than 0.13. The above method is time taking and not suitable for numbers with more than two places after the decimal. For such numbers method, 2 is beneficial and efficient. ### Decimal Comparison Method 2 Example :Â Comparing 7.345 and 7.38 • The whole number present before the decimal is compared first. The number with a greater whole number is considered bigger. In this example, the whole number present before the decimal is the same for both the cases. • The digit at the tenth place is compared. The number with a bigger digit at the tenth place is considered greater. In this example, both the numbers have the same digit in tenth place. • The digit at the hundredth place after the decimal is compared. In this example, the digit at the hundredths place for the number 7.38 is greater than that for 7.345. Hence, the number 7.38 is greater than 7.345. ## Video Lesson on Numbers ### Decimal Number Comparison Examples Letâ€™s Work Out: Example: Compare the given decimal numbers and state whether the given decimals are greater than, less than or equal to each other. (a) 4.1245 or 4.124 (b) 7.524 or 7.5240 (c) 254.913 or 254.15 (d) 1.102 or 1.120 Solution: (a) 4.1245 or 4.124 4.125 > 4.124 (b) 7.524 or 7.5240 7.524 = 7.5240 (c) 254.913 or 254.15 254.913 > 254.15 (d) 1.102 or 1.120 1.102 <.1.120
# Factors of 42 With Examples ## Factors of 42 With Examples Factors of number 42 are the pair of numbers which, when multiplied together gives the original number. There are other composite numbers such as 48, 60, 70, 420, 36, 45, 30, etc. whose factors and prime factorisation can be determined. We can find the factors of a number of 42 in a very simple and easy method. It involves the multiplication of a pair of numbers to get another number. In this article, we will learn the factors of 42, and the pair factors and the prime factors of 42 using the prime factorization method with many solved examples. ## What are the Factors of Forty Two? The factors of Forty Two are the numbers that divide 42 exactly without remainder. As the number 42 is a composite number, it has many factors other than one and the number itself. The factors of Forty Two can be positive or negative, but it cannot be in decimal or fraction form. Therefore, the factors of Forty Two are 1, 2, 3, 6, 7, 14, 21 and 42. Factors of 42: 1, 2, 3, 6, 7, 14, 21 and 42. Prime Factorization of 42: 2 × 3 × 7 ## Pair Factors of Forty Two We can find the factors in pairs for 42, by multiplying two numbers in a combination to get the result as 42. This can be done as follows: 1 × 42 = 42 2 × 21 = 42 3 × 14 = 42 6 × 7 = 42 Therefore, the factors of 42 in pairs are (1, 42), (2, 21), (3, 14) and (6, 7). ### Prime Factorization of 42 The number 42 is a composite number. Now let us find the prime factors of 42. • The first step is to divide the number 42 with the smallest prime factor, i.e. 2. 42 ÷ 2 = 21 • Now, check whether 21 can be further divided by 2 or not. 21 ÷ 2 = 10.5 But, a factor cannot be a fraction. Therefore, we will proceed to the next prime number, which is 3. • Now, dividing 21 by 3 we get, 21 ÷ 3 = 7 • Again, dividing 7 by 3 will give a fraction number. Therefore, taking the next prime number, 5, into consideration. 21 ÷ 5 = 4.2, again a fraction number. So the next prime number is 7. 7 ÷ 7 = 1 • We have received 1 at the end and further, we cannot proceed with the division method, as we know the multiple of 1 is 1 only. So, the prime factors of Forty Two are 2 × 3 × 7, where 2, 3 and 7 are prime numbers. ### Examples Example 1: Find the common factors of Forty Two and 41. Solution: The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42. The factors of 41 are 1 and 41. As the number 41 is a prime number, the common factor of Forty Two and 41 is 1. Example 2: Find the common factors of Forty Two and 43. Solution: Factors of Forty Two = 1, 2, 3, 6, 7, 14, 21 and 42. Factors of 43 = 1 and 43. Thus, the common factor of Forty Two and 43 is 1. ## Frequently Asked Questions on Factors of Forty Two #### Q1 What are the factors of Forty Two? Ans. The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42. #### Q2 What is the prime factorization of 42? Ans. The prime factorization of 42 is 2 × 3 × 7. #### Q3 Write down the positive pair factors of Forty Two. Ans. The positive pair factors of Forty Two are (1, 42), (2, 21), (3, 14), and (6, 7). #### Q4 What are the negative pair factors of Forty Two? Ans. The negative pair factors of Forty Two are (-1, -42), (-2, -21), (-3, -14), and (-6, -7). #### Q5 Is 21 a factor of Forty Two? Ans. Yes, 21 is a factor of Forty Two. As 21 divides 42 exactly, and hence 21 is a factor of Forty Two. Category: Maths
Math Courses / Course / Chapter # Understanding Bar Graphs and Pie Charts Lesson Transcript Chad has taught Math for the last 9 years in Middle School. He has an M.S. in Instructional Technology and Elementary Education. Bar graphs and pie charts are some of the most used graphical ways to present data. Learn how to read bar graphs and pie charts, and explore some examples to understand how they are interpreted. Updated: 08/22/2021 ## Understanding Bar Graphs & Pie Charts One of the more fascinating ways to record data is to put the data in a graph. Two of the more popular graphs are bar graphs and pie graphs. In this lesson, we will examine how to read and interpret the information that can be found on these graphs. An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: How to Calculate Percent Increase with Relative & Cumulative Frequency Tables ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:06 Understanding Bar… • 0:20 Bar Graphs • 2:07 Reading a Bar Graph • 4:05 Pie Charts • 5:10 Reading a Pie Chart • 7:21 Examples and Practice Problems Save Timeline Autoplay Autoplay Speed Speed ## Bar Graphs A bar graph is a graph that can be used to compare the amounts or frequency of occurrence of different types of data. Bar graphs are helpful when comparing groups of data and comparing data quickly. Let's look at the important parts of a bar graph. All bar graphs have a graph title, which gives the reader a brief overview of the type of data that the graph contains. The graph title in this graph is 'Percentage of U.S. Population That Is Foreign-Born.' Bar graphs also contain two axes that are labeled and also have a scale. The axis that is labeled at the bottom of the bars is referred to as the group data axis. This axis details information about the type of data that is displayed. The other axis in a bar graph is called the frequency data axis. This axis has a range of values that measures the frequency that the data occurs. The most important part of a bar graph is the bars. These bars provide an instant comparison of data in a graph. The reader can easily compare the data to see which measure occurs the most often and if any data points have similar frequencies. Let's look at a different bar graph to sharpen our skills on labeling its parts. Looking at this new graph, there is one obvious difference. This bar graph is going horizontal instead of vertical like the first example. The title of this graph is 'Sports Shoes Sales' and is located at the top of our graph. The group data axis is located along the left side, which is directly under the bars. The title of the axis is 'Types of shoes.' The frequency data axis is located at the bottom of the graph and is labeled 'Average cost.' The bars of this graph are running horizontally. When reading a bar graph, it's important to pay attention to the intervals used on the frequency data axis. An interval is the amount of data that occurs between each section or tick mark. Looking at the graph 'Sports Shoes Sales,' we can see that the frequency data axis is located at the bottom of the graph. Along this axis, we can see each tick mark represents \$10. That is important, because there is data between each of these tick marks. Data located between two marks is often estimated to the closest value. For example, look at the bar for 'Gym Shoes.' The bar for 'Gym Shoes' ends between the \$20 and \$30 marks. The bar appears to be more than halfway between these two points, so a good estimate would be either \$26 or \$27. To read a bar graph, find the point where each bar would meet the frequency data axis. Let's look at the 'Favorite Syndicated Programs, 1999 - 2000.' This graph is also displayed horizontally, so the frequency data axis is located at the bottom of the graph. Which program had almost nine percent of TV households viewing? The answer is Jeopardy. Looking at the Jeopardy bar, it is slightly more than nine percent, so a good estimate would be about 9.5%. Which program had the least percentage of TV households viewing? The answer is The Jerry Springer Show. The Jerry Springer Show had the shortest bar and lies between three percent and six percent. A good estimate would be 4.5%. Approximately what percent of TV-viewing households watched Friends? Friends had slightly more than six percent of viewers. A good estimate for this would be 6.2% of viewers. ## Pie Charts Shifting gears slightly, let's look at another popular type of graph used to display and compare data. Pie charts (also referred to as a circle graph) are circular graphs used to show the relationship of a part to a whole. A pie chart displays its data in sectors, which are parts of the circle and are proportional to the other parts displayed in the graph. Pie chart values are represented by percentages, with each chart representing 100%. Let's examine a pie chart and identify the important parts. Looking at this pie chart, the first important part is the graph title. The graph that we are looking at is titled 'What Tops Pizza.' It compares the most popular pizza toppings in relationship to one another. The title and the creative picture used for the pie chart give the reader information about the type of data displayed in the chart. Another important part of this pie chart is the sectors. These sectors are different-sized areas that divide up the circle into proportional parts. This is such a tasty graph, let's now examine how to read the information that it displays. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? Back ### Resources created by teachers for teachers Over 30,000 video lessons & teaching resources‐all in one place. 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Short and Sweet Calculus ## 2.2One-sided limits Consider the function $$F(x)$$ whose graph is shown in Figure 2.2. If we take $$x$$ values closer and closer to 2.5, but less than 2.5, $$F(x)$$ gets closer and closer to 5. In other words, when $$x$$ approaches 2.5 through the values less than 2.5, $$F(x)$$ approaches 5. We express this by saying that “the limit of $$F(x)$$ as $$x$$ approaches 2.5 from the left is 5” or “the left-hand limit of $$F(x)$$ as $$x$$ approaches 2.5 is 5.” The notation for this is $\lim_{x\to2.5^{-}}F(x)=5$ The minus sign that is written after 2.5 means $$x$$ approaches 2.5 from the left. Now consider the case in which $$x$$ takes on the values close to 2.5 but larger than 2.5. As $$x$$ approaches 2.5 from the right, $$F(x)$$ approaches 2. Symbolically we write $\lim_{x\to2.5^{+}}F(x)=2,$ and say “the limit of $$F(x)$$ as $$x$$ approaches 2.5 from the right is 2” or “the right-hand limit of $$F(x)$$ as $$x$$ approaches 2.5 is 2.” In this example, $$F(x)$$ is defined at $$x=2.5$$, but the value of $$F(2.5)$$ has no bearing on the left-hand or right-hand limit of $$F(x)$$. Even if we remove $$x=2.5$$ from the domain of $$F(x)$$ (that is, if $$F(x)$$ were not defined at $$x=2.5$$), the left-hand and right-hand limits would remain the same. 2.2. (left-hand limit) If we can make the values of $$f(x)$$ as close as we please to a number $$L$$ by taking $$x$$ sufficiently close to $$a$$ with $$\boldsymbol{x<a}$$, we say “the limit of $$f(x)$$ as $$x$$ approaches $$a$$ from the left is $$L$$” or “the left-hand limit of $$f(x)$$ as $$x$$ approaches $$a$$ is $$L$$” and write $\lim_{x\to a^{-}}f(x)=L.$ Similarly 2.3. (right-hand limit) If we can make the values of $$f(x)$$ as close as we please to a number $$L$$ by taking $$x$$ sufficiently close to $$a$$ with $$\boldsymbol{x>a}$$, we say “the limit of $$f(x)$$ as $$x$$ approaches $$a$$ from the left is $$L$$” or “the left-hand limit of $$f(x)$$ as $$x$$ approaches $$a$$ is $$L$$” and write $\lim_{x\to a^{-}}f(x)=L.$ Example 2.1. Consider the function $$y=\text{sgn}(x)$$ defined by ${\rm sgn}(x)=\left\{ \begin{tabular}{ll} \ensuremath{1} & if \ensuremath{x>0}\\ 0 & if \ensuremath{x=0}\\ \ensuremath{-1} & if \ensuremath{x<0} \end{tabular}\right.$ (a) Determine $${\displaystyle \lim_{x\to0^{-}}\text{sgn}(x)}$$. (b) Determine $${\displaystyle \lim_{x\to0^{+}}\text{sgn}(x)}$$. Solution The graph of $$y=\text{sgn}(x)$$ is shown below. (a) When $$x$$ is any negative number, the value of $$\text{sgn}(x)$$ is $$-1$$. Therefore $${\displaystyle \lim_{x\to0^{-}}\text{sgn}(x)}=-1$$. (b) When $$x$$ is positive, the value of $$\text{sgn}(x)$$ is $$1$$. Therefore, $${\displaystyle \lim_{x\to0^{+}}\text{sgn}(x)=1}$$. By comparing the definitions of one-sided limits and regular (or two-sided) limits, we see the following is true. 2.1. $${\displaystyle \lim_{x\to a}f(x)}$$ exists and is equal to $$L$$ if and only if $${\displaystyle \lim_{x\to a^{-}}f(x)}$$ and $${\displaystyle \lim_{x\to a^{+}}f(x)}$$ both exist and are equal to $$L$$. That is, $\lim_{x\to a}f(x)=L\quad\Longleftrightarrow\quad\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=L.$ For instance, in the above example$${\displaystyle \lim_{x\to0^{-}}\text{sgn}(x)}\neq{\displaystyle \lim_{x\to0^{+}}\text{sgn}(x)}$$, so $${\displaystyle \lim_{x\to0}\text{sgn}(x)}$$ does not exist.
1. linear differential equation Hello im stuck on the following differential equation. I understand since it linear we find the general solution by finding the complementary function and adding it together with a particular integral. Im stuck because in this example the RHS has both a linear function and an exponential and im unsure of how to go about doing the trail function subsitution. Find a solution to the following differential equation $\displaystyle dy/dx + 4y = 4x + e^x$ with $\displaystyle y(0)=0$ iv gotten as far as the CF = $\displaystyle Ae^-4x$ any help most appreciated 2. Hello, jordanrs! Find a solution to the following differential equation: . . $\displaystyle \frac{dy}{dx} + 4y \:=\: 4x + e^x\;\;\text{ with }y(0)=0$ You already found the homogeneous solution: .$\displaystyle y \:=\:ke^{-4x}$ We can conjecture that the particular solution has the form: . . $\displaystyle y \;=\;Ax + B + Ce^x$ . . $\displaystyle \begin{array}{ccccc}\text{Then: }& \dfrac{dy}{dx} &=& A + Ce^x \\ \\[-4mm] \text{and:} & 4y &=& 4Ax + 4B + 4Ce^x \end{array}$ Add the equations: .$\displaystyle \frac{dy}{dx} + 4y \;=\;4Ax + (A+4B) + 5Ce^x$ This is supposed to equal $\displaystyle 4x + e^x$ . . So we have: .$\displaystyle 4Ax + (A+4B) + 5Ce^x \;=\;4x+ 0+ 1\!\cdot\!e^x$ Equate the coefficients: .$\displaystyle \begin{Bmatrix}4A &=& 4 \\ A+4B &=& 0 \\ 5C &=& 1 \end{Bmatrix}$ . . and we get: .$\displaystyle A = 1,\;B = \text{-}\tfrac{1}{4},\;C = \tfrac{1}{5}$ Hence, the particular solution is: .$\displaystyle y \;=\;x - \tfrac{1}{4} + \tfrac{1}{5}e^x$ So far, the solution is: .$\displaystyle y \;=\;ke^{-4x} + x - \tfrac{1}{4} + \tfrac{1}{5}e^x$ Since $\displaystyle y(0) = 0$, we have: .$\displaystyle ke^0 + 0 - \tfrac{1}{4} + \tfrac{1}{5}e^0 \;=\;0 \quad\Rightarrow\quad k - \tfrac{1}{4} + \tfrac{1}{5} \:=\:0 \quad\Rightarrow\quad k \:=\:\tfrac{1}{20}$ Therefore, the solution is: .$\displaystyle \boxed{y \;=\;\frac{1}{20}e^{-4x} + x - \frac{1}{4} + \frac{1}{5}e^x}$ 3. yep that makes alot of sense using the 2 combined trial solutions in one and then differentiating and solveing through really appreciate the help thanks alot
The key difference in between area and also surface area is that area is provided to describe the an are occupied by a two-dimensional shape, whereas, surface ar area is the area inhabited by the external surface of a three-dimensional shape. Geometry is the branch of math that faces shapes and also solid figures, their area, and surface area and volume. In genuine life, over there are plenty of uses of area and also surface area. For example, area is supplied as a measure up to calculation the space occupied through a component of a soil or if over there is a have to paint a certain part of the wall; surface ar area is provided to calculate the full square units of material needed come cover or plunder a cuboidal box. You are watching: How does surface area differ from area 1 What isArea? 2 What is surface ar Area? 3 Difference in between Area and also Surface Area 4 Solved Examples 5 Practice Questions 6 FAQs on Difference in between Area and also Surface Area Area is a term associated with two-dimensional figures and also shapes favor triangles, squares, and rectangles, and also it is measure up in square units. It deserve to be imagined as the total an are or region occupied through a two-dimensional shape on a level surface. Different figures have different areas. Because that example, a square has 4 same sides and also the area of a square is the total number of square units inhabited by the square and is calculated with the aid of the formula: next × side. The area of a triangle is calculated making use of the formula: 1/2 × base × height. Similarly, different shapes have various formulas to calculation the area. Surface area is a term the is linked with three-dimensional figures and shapes. In other words, surface area applies to a polyhedron, wherein a solid shape has level faces. A cube is a form that has all its deals with in the shape of a square. Therefore, we can say the the surface area that a cube is the sum of the area covered by all the faces of a cube and it is calculated by the formula: surface ar area that a cube = 6a2. The formula supplied to calculate the full surface area of a cuboid = 2(lw+wh+lh), where 'l' is the length, 'w' is the width, and 'h' is the elevation of the cuboid. In a three-dimensional shape choose a cone, the surface ar area is provided by the sum of the curved surface ar area and the area that its base, i m sorry is a circle. The curved surface ar area is the area that the curved part in a hard shape, whereas, the lateral surface is the difference between the full surface area and the area of the base and top the the shape. That is come be provided that the curved surface area uses only to shapes that have curved surfaces. Observe the following figure which mirrors the areas and also surface areas of 2D and 3D shapes. The table shown listed below outlines the vital differences between area and surface area. AreaSurface Area Area is connected with a two-dimensional shape.Surface area is connected with a three-dimensional shape. Area is the room or the region occupied by any kind of flat two-dimensional shape.Surface area is the area or the room occupied by the lateral surface and the area of all the deals with of a three-dimensional shape. A few examples of forms that have actually an area are square, rectangle, and triangle.A couple of examples that solid shapes that have surface area room cube, cuboid, cone, cylinder. Example: Area of a square of next 's' is s × s or s2.Example: surface Area of a cube through side 'a' is 6a2. This is because all the deals with of a cube are in the type of a square. A cube has actually 6 faces, therefore, the surface area is 6a2. The calculate of area involves only 2 dimensions or values. Because that example, a rectangle's area is calculate by acquisition its length and width.The calculate of surface area entails three size or values. Because that example, the surface ar area that a cuboid is calculated by taking its length, width, and also height. In real life, the area is supplied to calculation the an are or an ar to it is in painted ~ above a wall or to calculation the total space occupied by a flat piece that land.In real life, the surface area is offered to calculation the price of spanning a cuboid-shaped box or to repaint the surface of a cubical box. ### Topics related to Difference between Area and also Surface Area Check out some interesting posts related come area and also surface area. Example 1: find the area that the complying with two-dimensional forms whose dimensions space given. a) A triangle with a base of 10 units and a height of 8 units.b) A rectangle through a size of 7 units and a width of 5 units.Solution:a) provided base = 10 units and height = 8 units.Area the a triangle = 1/2 × base × heightTherefore, the area the the triangle = 1/2 × 10 × 8= 40 square units.b) Given, size = 7 units and width = 5 unitsArea that a rectangle = size × widthTherefore, area the the rectangle = 7 × 5= 35 square units. Example 2: discover the surface ar area the the adhering to solid shapes.a) A cube of side 5 units.b) A cuboid of size 5 units, width 6 units, and height 7 units. Solution: a) surface area the a cube = 6a2Given, next of the cube = 5 units.Therefore, surface area of the cube = 6 × 52= 6 × 25= 150 square units.b) total Surface area that a cuboid = 2(lw+wh+lh) provided length = 5 units, width = 6 units and also height = 7 units.Therefore, full surface area the the cuboid = 2<(5 × 6) + (6 × 7) +(5 × 7)>= 2 (30 + 42 + 35)= 214 square units. Therefore, the complete surface area of the cube = 214 square units. See more: What Does “Thus Far” Mean? With Use Thus Far In A Sentence What Does “Thus Far” Mean Indulging in rote learning, you are likely to forget concepts. Through dearteassociazione.org, friend will learn visually and also be surprised through the outcomes.
Translations and Vectors Graphical introduction to image translations Estimated7 minsto complete % Progress Practice Translations and Vectors MEMORY METER This indicates how strong in your memory this concept is Progress Estimated7 minsto complete % Translations Karen looked at the image below and stated that the image was translated thirteen units backwards. Is she correct? Explain. Translations In geometry, a transformation is an operation that moves, flips, or changes a shape to create a new shape. A translation is a type of transformation that moves each point in a figure the same distance in the same direction. Translations are often referred to as slides. If you look at the picture below, you can see that the square is moved 10 units to the right. All points of the square have been moved 10 units to the right to make the translated image . The original square is called a preimage. The final square is called the image. Let's describe the following translations: 1. The preimage is the brown pentagon and the image is the purple pentagon. The pentagon is translated down 8 and over 11 to the right. 1. The preimage is the light blue triangle and the image is the green triangle. The blue triangle moves up 3 units and over 2 units to the left to make the green triangle image. 1. The preimage is the purple shape and the image is the yellow shape. The original shape is translated down 2 and over 7 to the left. Examples Example 1 Earlier, you were told that Karen looked at the image below and stated that the image was translated thirteen units backwards. Is she correct? Explain. Karen is somewhat correct in that the translation is moving to the left (backwards). The proper way to describe the translation is to say that the image has moved 13 units to the left and 2 units up. Example 2 Describe the translation of the pink triangle in the diagram below. The pink triangle is translated down 4 and over 2 to the left. Example 3 Describe the translation of the purple polygon in the diagram below. The purple polygon is translated up 2 and over 12 to the right. Example 4 Describe the translation of the blue hexagon in the diagram below. The blue hexagon is translated down 2 and over 10 to the left. Review Describe the translation of the purple original figures in the diagrams: Use the diagram below to describe the following translations: 1. A onto B 2. A onto C 3. A onto D 4. A onto E 5. A onto F On a piece of graph paper, plot the points and to form . 1. Translate the triangle 3 units to the right and 2 units down. Label this . 2. Translate 3 units to the left and 4 units down. Label this . 3. Describe the translation necessary to bring to . On a piece of graph paper, plot the points and to form . 1. Translate the triangle 2 units to the left and 4 units down. Label this . 2. Translate 5 units to the right and 2 units up. Label this . 3. Describe the translation necessary to bring to . To see the Review answers, open this PDF file and look for section 10.1. Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English TermDefinition Image The image is the final appearance of a figure after a transformation operation. Preimage The pre-image is the original appearance of a figure in a transformation operation. Transformation A transformation moves a figure in some way on the coordinate plane. Translation A translation is a transformation that slides a figure on the coordinate plane without changing its shape, size, or orientation. Rigid Transformation A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure.
# CLASS-4ADDITION - HORIZONTAL OR VERTICAL Vertical addition is a method of adding two or more numbers that involves aligning them vertically and adding the digits in each column. Here are some properties of vertical addition: 1. Commutativity: The order in which the numbers are added does not affect the result. For example, 2 + 3 is the same as 3 + 2. Or Sum of the two numbers remains the same even if the order , in which they are added is changed. Example :- 384475 + 29580 = 414055. Also, 29580+ 384475 = 414055. Thus , 384475 + 29580 = 29580+ 384475 2. Associativity: The grouping of numbers being added does not affect the result. For example, (2 + 3) + 4 is the same as 2 + (3 + 4). In what ever order three or more numbers are grouped for the purpose of addition, their sum remains the same . 3. Identity element: The number 0 is the identity element for addition, which means that adding 0 to any number does not change the value of the number. Addition of Zero – If ‘0’ is added to a number, we get the number itself. Example:1) 892347 + 0 = 8923472. Example.2) 3754093 + 0 = 3754093 4. Carrying: When adding two numbers vertically, carrying may be necessary when the sum of the digits in a column is greater than 9. This involves carrying the "tens" place digit over to the next column and adding it to the sum in that column. 5. Regrouping: If the sum of the digits in a column is greater than or equal to 10, regrouping may be necessary to correctly align the digits in the next column. 6. Place value: The position of a digit in a number indicates its place value, which is the value that the digit contributes to the overall value of the number. When adding numbers vertically, it is important to align digits according to their place values to ensure the correct result. Example.1) 2 5 7 3 4 + 5 3 2 4 5 ----------------------------- 7 8 9 7 9 Example.2) 8 2 7 9 4 3 3 4 9 4 5 6 + 5 3 8 7 3 7 ----------------------------- 1 7 1 6 1 3 6
# The real number √2 as a set In set theory, all kinds of mathematical objects are hereditary sets: Numbers, relations, functions, etc. Often, these mathematical objects can be constructed with sets in different ways and there is no canonical way to do it - so you may encounter different definitions in different textbooks. If you wonder how √2 may look like as a set, here is one example. One of the ways to contruct the real numbers are Dedekind cuts of the rational numbers, so that a real number is defined as the (infinite) set of all rational numbers on one half of a Dedekind cut (one half determines the other). We are using the right half of a Dedekind cut where the left half may have a biggest element and the right half does never have a smallest element. This can be understood as the set of all rational numbers that are bigger than the real number in question. The set of all such cuts then is the set of all real numbers. Klick on a rational number to see how it may be defined as a set. √2 = { 2/1, 3/1, 3/2, 4/1, 5/1, 5/2, 5/3, 6/1, 7/1, 7/2, 7/3, 7/4, 8/1, 8/3, 8/5, 9/1, 9/2, 9/4, 9/5, 10/1, 10/3, 10/7, 11/1, 11/2, 11/3, 11/4, 11/5, 11/6, 11/7, 12/1, 12/5, 12/7, 13/1, 13/2, 13/3, 13/4, 13/5, 13/6, 13/7, 13/8, 13/9, 14/1, 14/3, 14/5, 14/9, 15/1, 15/2, 15/4, 15/7, 15/8, ... } For real numbers R1, R2 as defined above, R1 < R2 is defined as (R2 ⊂ R1 and R1 != R2). R1 + R2 is reduced to addition of rational numbers by defining it as the set of all Q1 + Q2 with Q1 ∈ R1 and Q2 ∈ R2. Here is a way to enumerate all rational numbers a/c that are elements of of √2: ```Let a be 2 Repeat: Let c be 1 While a·a > 2·c·c: Enumerate a/c Increase c by 1. Increase a by 1 ``` This enumeration may list the same rational number several times. To avoid this, you can compare the new number a/c with all already enumerated numbers a1/c1. If a/c = a1/c1 (or easier: a·c1 = a1·c), then the number was already listed. The enumeration as Python function with a limiting parameter k: ```def print_sqrt2(k): dedekind=set() for a in range(2,k+1): c=1 while aa > 2cc: found=False for a1,c1 in dedekind: if ac1 == a1*c: found=True break print str(a)+'/'+str(c)+', ' c+=1 print print "..." ```
Home » Archives for April 2019 # Find the distance from a Speed time graph Today’s post is about an exam question about a speed time graph I was prepping for my students yesterday. I was surprised to see this question because it covers two points I associate more with A-Level work. ### What can we find from the area under a curve? a) Area under the curve as the distance b) Using the ‘ trapezium rule’ Before I say more, here is the question The question asks for the distance the car travels, but the graph is of speed over time. This is where the first of my ‘surprise facts’ comes in – You can find the distance travelled by measuring the ‘area under the curve’. I won’t here show why that is the case – that is for A-Level students (OK I might post on that later). For now though, I will just apply the rule. The area ‘under the curve’ is the area of the shape made by the curve at the top and the line ‘speed = 0’ at the bottom, bound to the left and right by the times we are interested in : In this case t=0 and t=20. ### The Trapezium Rule We can only estimate the area of this shape, given how irregular it is – the top side anyway – and this is where the ‘Trapezium rule’ comes in. To apply this we have to divide the area into strips, each one a trapezium (OK, the first one is a triangle, but we can apply a similar rule. To be fair, the question does give a clue to what to do by saying we should use ‘4 strips of equal width’ We divide up like this. The first area is a triange with area 1/2 x 5 x 22* The value of the speed at t=5 The other areas are trapeziums with area 5 x 1/2(Height at start + height at end). The ‘height’ in each case is the speed values at t= 10, 15 and 20. If we look at the calculation closely we can save time by noting the middle times are included twice each, 1/2 each time. so the area of the trapeziums are 5 x (22 + 28 + 32 + 35 x 1/2) This is a quicker way that fiddling about with all the 1/2s. The 35 – value at t=20 – still needs the 1/2 because it is a measurement only on one trapezium. We get the value 5 x 99.5. This is where I took into account that we are only estimating, so using a value of 100 instead of 99.5 is justifiable. This gives us an estimate of 500m There is a part(b) for this question on the next page. ### Over and under estimates “Is this an over-estimate or an under-estimate for the distance. Give a reason for you answer.’ The answer in this case is that it is an underestimate, because all those trapeziums still left a small space under the curve. I don’t think our rounding up to 500m stops this being the answer. More about the trapezium rule can be found here # A simple algebraic proof As we get close to the exams, my students mostly want to do previous exam questions with me, so today I am sharing one which I found rather interesting – Algebraic proof. Perhaps because it foreshadows the kind of Maths one does at A-Level and degree level but its aimed at GCSE students. The skills required are not that different to solving equations, but the level of thinking required is perhaps one level further on. The question is… Prove that the square of any odd number is one more than a multiple of 4. ### The idea of an algebraic proof The first thing to notice are those words ‘prove that’.. for students who are used to Maths being about doing a sum and finding an answer, this could seem quite new. We are told what to expect. We just need to show the statement is true. One of my students said ‘does this mean I have to look at every odd number’. That would be quite difficult because there is a never -ending list of them. But its not a bad way to start, to convince yourself it might be true. 52 = 25 – and that is 1 more than 24 which is 4 time 6. 112 = 121 and that is 1 more than 120 which is 4 times 30 I could keep going like that – and you should do maybe a couple of your own – but I will never prove its true for ALL numbers like that. ### What is easier : prove its true or prove it isn’t? Its easier in Maths – and in life – to prove something isn’t true. We only need to find ONE odd number where the square is not 1 more than a multiple of 4 to prove that statement is not true. A person only needs to show he was in Birmingham on Tuesday evening to show he didn’t commit the crime in Paris. (You could try but don’t waste your time looking, it is true!) Proving things are true is harder. But we can still do it. We are going to do this with algebra, looking at the ‘general case’. That is why we call this algebraic proof. We need to think of numbers more generally. What is the ‘general’ odd number. Take a number, any number. Will it be odd? Well there is a 50-50 chance. But if we double the number we know it will be even (131 doubled is 262), then add 1. We then know the answer we get will be odd. And in fact, all odd numbers can be found using 2n + 1 – That is a general odd number!! So take our general odd number, and square it. (2n+1)(2n + 1). I could have written this with a 2 but I find its easier to multiply brackets if you can see both. We get Factorise just part of this you get 4(n2 + n) + 1 And that is actually what we are looking for! Let me explain. In 2n + 1, our ‘general’ odd number, n is just any whole number, and so n2 + n is also just a whole number. So 4(n2 + n) must be a multiple of 4 [ 4 times (n2 + n)] and 4(n2 + n) + 1 is 1 more! This completes our example of algebraic proof. # Finding a turning point For today’s post I’m going to look at an exam question I saw posted on Facebook. I confess this is the sort of stuff I get nerdy about! I love the way that equations can describe pictures, and vice versa. Before we start, lets make sure we understand the terms used. What exactly is a ‘turning point’? Well, it is the point where the line stops going down and starts going up (see diagram below). That point at the bottom of the smile. There are three different ways to find that but in all cases, we need to start by finding the equation – finding out the values of ‘a’ and ‘b’. ‘b’ is the easier. take the case where x=0 and the two terms with x can both be disregarded and so b = -5 = the point where the curve crosses the y-axis (also known as the line ‘X = 0’. That is similar to the reason why the term ‘c’ is often called ‘the y intercept’ for straight lines, equation y = mx+c The other point we know is (5,0) so we can create the equation 25 + 5a – 5 = 0 (By substituting the value of 5 in for x) We can solve this for a giving a=-4 The full equation is y = x2 – 4x – 5 I usually check my work at this stage 52 – 4 x 5 – 5 = 0 – as required Now, I said there were 3 ways to find the turning point. I will give all three here bu, be warned, the third does require some A-Level maths. This diary is more aimed at GCSE students but method 3 is actually the way I would usually find the turning point, so I will give a brief description here. Method 1 – The ‘parabola’ is symmetrical. A Parabola is the name of the shape formed by an x2 formula On this version of the graph I’ve marked the turning point with an X and the line of symmetry in green. This tells us the value of x on the turning point lies halfway between the two places where y=0 (These are solutions, or roots, of x2 – 4x – 5 = 0 We know one of these is is x=5. We can get the other by factorising to give (x-5)(x+1) = 0. So x = -1 is the other solution. Halfway between x = -1 and x = 5 is x = 2. when x = 2, y = 22 – 4 x 2 – 5 = -9. So the turning point is (2,-9). Method 2 Complete the Square If we ‘complete the square’ on this equation we get x2 – 4x + 4 – 4 – 5 – I’ve added in 4 and taken 4 away which looks like an eccentric thing to do, but this means we can factorise the first part into a square (x – 2)2 – 9 (also combined the -4 and – 5 to make -9) This where we use our knowledge that a ‘square’ is never less than 0, but it can be 0. So its minimum value is when that square = 0 – so x – 2 = , so x = 2. And we can also see the value of the whole thing there is 0 – 9 = -9 Method 3 – By differentiation This is A-Level stuff really, so I’ll only give an overview. This is the way I would usually do this, but then I have studied Maths through A-level (and beyond!) Differentiation is one branch of Calculus, the mathematics of measuring change. By a rule you will learn if and when you first study calculus, the equation of how much x2 – 4x – 5 is changing is given by 2x – 4. The turning point is where the line isn’t changing, so 2x – 4 = 0 (Zero change) so 2x = 4 and x = 2. y = 9 can be as above. Summary I’ve given three methods here. Fortunately they all give the same answer! (They wouldn’t be good alternatives otherwise!) Choose which of the first two you feel most comfortable with. The third method is the easiest to extend to where we have equations of than an a x2 # Some tricky algebra – and how to make it easier Let’s consider this question from a Higher GCSE paper. First thing I think is – Wow, that looks complicated. [Actually, the first thing I should think is, what sign is that. It is a divide sum in the middle, not an add sign as I first thought] Algebra fractions can stump the brighter students, but its worth remembering to rules are just the same as fractions with numbers. We could do this divide sum by flipping the second and multiplying the numerators together and then the denominators. But given that we are told the final answer is simple, there is certain to be a lot of simplifying we can we first. The first thing I noitices was that the bottom of the second fraction was going to factorise. It looks so close to the ‘difference of 2 squares’ rules. Indeed it does factorise to x(x+5)(x-5) And immediately we see there is some cancelling to be done with the top to give 1/x(x-5) for the second fraction What about the bottom of the first fraction. Looks like there could be some factorising to be done there. And indeed, taking my cue from the (x-5) we have already seen, this factorises to (x – 5)(x + 2) Something I missed before because I was so busy factorising the higher powers of x, but the top can be factorised as 3(x+2) and so we can cancel to 3/(x-5) NOW we can do the divide by flipping the 2nd fraction, and the operation is now much easier than before 3/(x – 5) x(x – 5) [Using an * for multiply here to stop confusion with the x’s] which is then 3x = so a = 3. ; # Circles and Squares – The Answer In my last post I gave the following exam question which I had looked at with two of my students And now, as promised, the answer. I’ve added a few coloured lines to help me talk this through. The width of the square is the width of two of the circles. I have marked this with the two parallel grey lines of different shades. As we are told each of the circles has a radius of 24cm, that makes the width of the square 4 x this – 96cm! The height of the rectangle is a bit less, because there is an overlap of the circle diameters in that direction. We do know the height includes two radiuses* – shown as the parallel red/pink lines. But there is also the height of the green triangle, shown in blue on the height of the rectangle. We can find that using Pythagoras Theorem – By splitting the green triangle in half, we have two right angled triangle with sides 48cm (the hypotenuse) and 24cm… so h2 + 242 = 482. h = sqrt(1728) = 41.57cm (to 4 s.f.)** It can be easy after doing the most complex part of a calculation to forget we haven’t yet finished the question. Let’s not fall into that trap. We still need to find the total height of the rectangle, and thus the area of the rectangle. Total height is 41.57 + 2 x 24 = 89.57cm Area = 89.57 x 96 = 8600cm2 (To 3 s.f) A couple of notes about my solution * Yes, the official plural is radii – from the Latin! I always think that looks a bit old fashioned so I tend to say radiuses! **Its often a judgement call how one rounds ones answer. Exam questions will sometimes say (Round to 3 s.f. or 1d.p) but if they don’t, show what rounding you have done. It is not appropriate to give more figures than any rounding given in the data provided…. BUT if you know you are going to round to 3 s.f. at the end, keep one extra figure in the intermediate numbers. That’s why I use 4.s.f for h, even though I know I am going to give a less precise answer at the end
How do you graph f(x) = 4 sin(x - pi/2 ) + 1? 1 Answer Mar 13, 2016 You have to identify a few important elements of the graph first. Explanation: Amplitude: in a function of the for $y = a \sin b \left(x + c\right) + d$, the amplitude is at $\| a \|$. So, the amplitude is at 4. The amplitude is the distance between the maximum and minimum points and the horizontal line of rotation. This line of rotation would be 0 if d = 0, but since there is a vertical displacement of +1, the line is y = 1. Period: the period is the distance before the function's movement repeats itself. It can be found by $\frac{2 \pi}{\|} b \|$. In this case, b = 1, so the period is $2 \pi$. Phase shift: The phase shift is the horizontal displacement. It can be found by solving the equation $x + c = 0$. $x - \frac{\pi}{2} = 0$ $x = \frac{\pi}{2}$ Since $\frac{\pi}{2}$ is positive, the horizontal displacement is $\frac{\pi}{2}$ units right. Now, we have enough information to graph. If there was no phase shift, you would start on the line of rotation at (0, 1). However, since there is a phase shift of $\frac{\pi}{2}$ we start at $\left(\frac{\pi}{2} , 1\right)$. Usually, you will only be asked to do one complete cycle, so that's what I'll do. In a Sine function, there are 4 parts on the x value in one cycle, so the first point can be found by dividing the period by 4: $\frac{\frac{\pi}{2}}{4} = \frac{\pi}{8}$. The point will be $\left(5 \frac{\pi}{8} , - 3\right)$, since the amplitude is 4 and the line of rotation at y = 1. The next point will be at $\left(3 \frac{\pi}{4} , 5\right)$. The final point in our cycle is $\left(2 \frac{\pi}{2} , 1\right)$, since we have finished our period. Hopefully this helps!
Divisibility Rules: Simplify GMAT Prep March 4, 2017 Time to start the Prep: Application Deadlines to watch out for June 16, 2017 As the Quantitative part of the GMAT focuses on the basics of mathematics, it’s important to know divisibility rules. So let’s discus the divisibility rules for some numbers that you need to be known while preparing for the GMAT. Divisible by 2: All numbers ending with 2 or its multiples (0, 4, 6 and 8) are divisible by 2 Example: 124 is divisible by 2 as it is ending with multiples of 2 Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3 Example: 165 divided by 3 as its sum (1+6+5= 12) is multiple of 3 Divisible by 4: A number is divisible by 4 if its last two digits form a number that is divisible by 4 Example: 2116, as 16 is divisible by 4 so 2116 is divisible by 4 Divisible by 5: A number is divisible by 5 if the number ends in 5 or 0 Example: 100 and 125 are divisible by 5 Divisible by 6: A number is divisible by 6 if it is divisible by both 2 and 3 Example: 948 is even so it is divisible by 2 and the sum of its digits is 21 that is also divisible by 3 Divisible by 7: To determine if a number is divisible by 7, take the last digit of the number, double it and subtract the doubled number from the remaining number. If the result is evenly divisible by 7 (e.g. 14, 7, 0, -7, etc.), then the number is divisible by seven. This may need to be repeated several times. Example: 3101, last digit is 1 so double it (1 x 2=2) then subtract 2 from 310,310 – 2 =308 then again take last digit that is 8 , double it (8 x 2=16) then from 30, that is 30-16=14. Hence 14 is multiple of 7 so 3101 is divisible by 7 Divisible by 8: A number is divisible by 8 if its last three digits form a number that is divisible by 8 Example: 31248, as 248 is divisible by 8 so 31248 is divisible by 8 Divisible by 9: A number is divisible by 9 if the sum of its digits is divisible by 9 Example: 567 is divisible by 9 as its sum (5+6+7=18) is multiple of 9 Divisible by 10: A number is divisible by 10 if it ends in 0. Example: 120, 1000 etc. Divisible by 11: If you sum every second digit and then subtract the result from the sum of all other digits and the answer is 0 or multiples of 11, then the number is divisible by 11. Example: 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9) =-11, is divisible by 11, hence 9,488,699 is divisible by 11 Divisible by 12: If the number is divisible by both 3 and 4, it is also divisible by 12. Example: 324 as it is divisible both 3 and 4 Divisible by 13: To determine if a number is divisible by 13, take the last digit of the number , multiply by 4 and add the result to remaining number, if the result is evenly divisible by 13 (e.g. 26, 0, -13, etc.), then the number is divisible by 13. This may need to be repeated several times. Example: 50661 Step 1: take the last digit i.e. 1, multiply it by 4(1×4=4) then add the result to remaining digits 5066. 5066+4=5070 Step 2: Take the last digit i.e. 0, multiply it by 4(0x4=0) then add the result to remaining digits 507. 507+0=507 Step 3: Take the last digit i.e.7, multiply it by 4(7×4=28) then add the result to remaining digits 50. 50+28=78 78 is multiple of 13, so 50661 is divisible by 13. Simplify your prep exercise by focusing on such simple but vital details of math. Attention to detail does in the long run facilitate preparation and contribute immensely to achieving a good score. At Option GMAT DUBAI we focus significantly on making the prep interesting, enlightening and in the long run effective. 1. 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Sei sulla pagina 1di 4 # CBSE Worksheet-1 CLASS –VII Mathematics (Perimeter and Area) ## 1. Find the area of following triangle: a. 6 cm2 b. 5 cm2 c. 4 cm2 d. 3 cm2 2. A door frame of dimensions 4 m × 5 m is fixed on the wall of dimension 11 m × 11 m. Find the total labour charges for painting the wall if the labour charges for painting 1m2 of the wall is Rs 2.50. a. Rs. 200 b. Rs. 252.50 c. Rs. 300 d. Rs. 350 3. What is the circumference of a circle of diameter 10cm? a. 30 cm b. 35 cm c. 31.4 cm d. none of these 4. Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22m. a. 5 m b. 10 m c. 15 m d. 20 m ## 5. The ______ is the distance around a given two-dimensional object. 6. If we cut a square along one of its diagonals, two triangles are obtained. Area of each triangle obtained = __________. 7. Length of rectangle = ## 8. State true or false: All triangles equal in area are congruent. 9. A rectangular garden is 66 cm long and 50 cm wide. Two cross paths each 2 m wide are to be constructed parallel to the sides. If these paths pass through the centre of the garden, find the cost of constructing the paths at the rate Rs. 69 per m2. 10. The figure given below, shows two circles with the same centre. The radius of the larger circle is 10 cm and the radius of the smaller circle is 4 cm. Find: a. the area of the larger circle, b. the area of the smaller circle, c. the shaded area between the two circles (Take ) 11. A wire is in the shape of a square of side 10 cm. If the wire is bent again into a rectangle of length 12 cm, find its breadth. Which encloses more area - the square or the rectangle? CBSE Worksheet-1 CLASS –VII Mathematics (Perimeter and Area) 1. a Explanation: Area of the triangle = 2. b ## Area of the portion to be painted = 121 - 20 = 101 m2 Cost of painting = 101 × 2.50 = Rs. 252.50 3. c Explanation: Circumference = × diameter = 3.14 × 10 = 31.4 cm 4. d ## Explanation: Area of the rectangle = length × breadth = 440 m2 5. perimeter 6. × Area of the square 7. Area of rectangle 8. False 9. Rs. 15732 Explanation: Area of the path = area of the rectangle EFGH + area of the rectangle PQRS - area of the square XYZW ## Area of the rectangle EFGH = 50 × 2 = 100 m2 Area of the rectangle PQRS = 66 × 2 = 132 m2 ## Area of the path = 100 + 132 - 4 = 228 m2 Cost of constructing the path = 228 × 69 = Rs. 15732 ## c. Area of shaded region = (314 - 50.24) cm2 = 263.76 cm2 Explanation: Area of the larger circle = × radius2 = 3.14 × 102 = 3.14 × 100 = 314cm2 ## Area of the shaded portion = 314 - 50.224 = 263.76 cm2 11. Area of square is greater than the area of rectangle Explanation: Since the same wire is used to make and square and rectangle, the perimeter of both the shapes will be same Perimeter of the square = 4 × side = 4 × 10 = 40 cm Perimeter of the rectangle = 2 ( length + breadth) = 2 (12 + breadth) = 40
# 7.11: Box-and-Whisker Plots Difficulty Level: Basic Created by: CK-12 Estimated8 minsto complete % Progress Practice Box-and-Whisker Plots MEMORY METER This indicates how strong in your memory this concept is Progress Estimated8 minsto complete % Estimated8 minsto complete % MEMORY METER This indicates how strong in your memory this concept is You're a meteorologist and you're collecting temperature data from various locations in your state for the month of February. You've collected over 2,000 temperatures at the same time each day and want to organize them to see if there are patterns. You want to find out the lowest temperature, the highest temperature, the median temperature, the median of the first half of the month and the median of the second half of the month. How would you organize your data to get these answers? ### Watch This First watch this video to learn about box-and-whisker plots. Then watch this video to see some examples. Watch this video for more help. ### Guidance In traditional statistics, data is organized by using a frequency distribution. The results of the frequency distribution can then be used to create various graphs, such as a histogram or a frequency polygon, which indicate the shape or nature of the distribution. The shape of the distribution will allow you to confirm various conjectures about the nature of the data. To examine data in order to identify patterns, trends, or relationships, exploratory data analysis is used. In exploratory data analysis, organized data is displayed in order to make decisions or suggestions regarding further actions. A box-and-whisker plot (often called a box plot) can be used to graphically represent the data set, and the graph involves plotting 5 specific values. The 5 specific values are often referred to as a five-number summary of the organized data set. The five-number summary consists of the following: 1. The lowest number in the data set (minimum value) 2. The median of the lower quartile: \begin{align}Q_1\end{align} (median of the first half of the data set) 3. The median of the entire data set (median) 4. The median of the upper quartile: \begin{align}Q_3\end{align} (median of the second half of the data set) 5. The highest number in the data set (maximum value) The display of the five-number summary produces a box-and-whisker plot as shown below: The above model of a box-and-whisker plot shows 2 horizontal lines (the whiskers) that each contain 25% of the data and are of the same length. In addition, it shows that the median of the data set is in the middle of the box, which contains 50% of the data. The lengths of the whiskers and the location of the median with respect to the center of the box are used to describe the distribution of the data. It's important to note that this is just an example. Not all box-and-whisker plots have the median in the middle of the box and whiskers of the same size. Information about the data set that can be determined from the box-and-whisker plot with respect to the location of the median includes the following: a. If the median is located in the center or near the center of the box, the distribution is approximately symmetric. b. If the median is located to the left of the center of the box, the distribution is positively skewed. c. If the median is located to the right of the center of the box, the distribution is negatively skewed. Information about the data set that can be determined from the box-and-whisker plot with respect to the length of the whiskers includes the following: a. If the whiskers are the same or almost the same length, the distribution is approximately symmetric. b. If the right whisker is longer than the left whisker, the distribution is positively skewed. c. If the left whisker is longer than the right whisker, the distribution is negatively skewed. The length of the whiskers also gives you information about how spread out the data is. A box-and-whisker plot is often used when the number of data values is large. The center of the distribution, the nature of the distribution, and the range of the data are very obvious from the graph. The five-number summary divides the data into quarters by use of the medians of the upper and lower halves of the data. Many data sets contain values that are either extremely high values or extremely low values compared to the rest of the data values. These values are called outliers. There are several reasons why a data set may contain an outlier. Some of these are listed below: 1. The value may be the result of an error made in measurement or in observation. The researcher may have measured the variable incorrectly. 2. The value may simply be an error made by the researcher in recording the value. The value may have been written or typed incorrectly. 3. The value could be a result obtained from a subject not within the defined population. A researcher recording marks from a math 12 examination may have recorded a mark by a student in grade 11 who was taking math 12. 4. The value could be one that is legitimate but is extreme compared to the other values in the data set. (This rarely occurs, but it is a possibility.) If an outlier is present because of an error in measurement, observation, or recording, then either the error should be corrected, or the outlier should be omitted from the data set. If the outlier is a legitimate value, then the statistician must make a decision as to whether or not to include it in the set of data values. There is no rule that tells you what to do with an outlier in this case. One method for checking a data set for the presence of an outlier is to follow the procedure below: 1. Organize the given data set and determine the values of \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align}. 2. Calculate the difference between \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align}. This difference is called the interquartile range (IQR): \begin{align}IQR = Q_3-Q_1\end{align}. 3. Multiply the difference by 1.5, subtract this result from \begin{align}Q_1\end{align}, and add it to \begin{align}Q_3\end{align}. 4. The results from Step 3 will be the range into which all values of the data set should fit. Any values that are below or above this range are considered outliers. #### Example A For each box-and-whisker plot, list the five-number summary and describe the distribution based on the location of the median. a. Minimum value \begin{align}\rightarrow 4\end{align} \begin{align}Q_1 \rightarrow 6\end{align} Median \begin{align}\rightarrow 9\end{align} \begin{align}Q_3 \rightarrow 10\end{align} Maximum value \begin{align}\rightarrow 12\end{align} The median of the data set is located to the right of the center of the box, which indicates that the distribution is negatively skewed. b. Minimum value \begin{align}\rightarrow 225\end{align} \begin{align}Q_1 \rightarrow 250\end{align} Median \begin{align}\rightarrow 300\end{align} \begin{align}Q_3 \rightarrow 325\end{align} Maximum value \begin{align}\rightarrow 350\end{align} The median of the data set is located to the right of the center of the box, which indicates that the distribution is negatively skewed. c. Minimum value \begin{align}\rightarrow 60\end{align} \begin{align}Q_1 \rightarrow 70\end{align} Median \begin{align}\rightarrow 75\end{align} \begin{align}Q_3 \rightarrow 95\end{align} Maximum value \begin{align}\rightarrow 100\end{align} The median of the data set is located to the left of the center of the box, which indicates that the distribution is positively skewed. #### Example B The numbers of square feet (in 100s) of 10 of the largest museums in the world are shown below: 650, 547, 204, 213, 343, 288, 222, 250, 287, 269 Construct a box-and-whisker plot for the above data set and describe the distribution. The first step is to organize the data values as follows: \begin{align}20,400 \quad 21,300 \quad 22,200 \quad 25,000 \quad 26,900 \quad 28,700 \quad 28,800 \quad 34,300 \quad 54,700 \quad 65,000\end{align} Now calculate the median, \begin{align}Q_1\end{align}, and \begin{align}Q_3\end{align}. \begin{align}20,400 \quad 21,300 \quad 22,200 \quad 25,000 \quad \boxed{26,900 \quad 28,700} \quad 28,800 \quad 34,300 \quad 54,700 \quad 65,000\end{align} \begin{align}\text{Median} \rightarrow \frac{26,900+28,700}{2} = \frac{55,600}{2} = 27, 800 \end{align} \begin{align}Q_1 = 22,200\end{align} \begin{align}Q_3 = 34,300\end{align} Next, complete the following list: Minimum value \begin{align}\rightarrow 20,400\end{align} \begin{align}Q_1 \rightarrow 22,200\end{align} Median \begin{align}\rightarrow 27,800\end{align} \begin{align}Q_3 \rightarrow 34,300\end{align} Maximum value \begin{align}\rightarrow 65,000\end{align} The right whisker is longer than the left whisker, which indicates that the distribution is positively skewed. #### Example C Using the procedure outlined above, check the following data sets for outliers: a. 18, 20, 24, 21, 5, 23, 19, 22 b. 13, 15, 19, 14, 26, 17, 12, 42, 18 a. Organize the given data set as follows: \begin{align}& 18, \ 20, \ 24, \ 21, \ 5, \ 23, \ 19, \ 22\\ & 5, \ 18, \ 19, \ 20, \ 21, \ 22, \ 23, \ 24\end{align} Determine the values for \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align}. \begin{align}5, \ \boxed{18, \ 19}, \ 20, \ 21, \ \boxed{22, \ 23}, \ 24\end{align} \begin{align}Q_1 = \frac{18+19}{2} = \frac{37}{2}= 18.5 \qquad Q_3=\frac{22+23}{2}=\frac{45}{2}=22.5\end{align} Calculate the difference between \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align}: \begin{align}Q_3-Q_1=22.5-18.5=4.0\end{align}. Multiply this difference by 1.5: \begin{align}(4.0)(1.5)=6.0\end{align}. Finally, compute the range. \begin{align}Q_1-6.0=18.5-6.0=12.5 \end{align} \begin{align}Q_3+6.0=22.5+6.0=28.5\end{align}. Are there any data values below 12.5? Yes, the value of 5 is below 12.5 and is, therefore, an outlier. Are there any values above 28.5? No, there are no values above 28.5. b. Organize the given data set as follows: \begin{align}& 13, \ 15, \ 19, \ 14, \ 26, \ 17, \ 12, \ 42, \ 18\\ & 12, \ 13, \ 14, \ 15, \ 17, \ 18, \ 19, \ 26, \ 42\end{align} Determine the values for \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align}. \begin{align}12, \ \boxed{13, \ 14}, \ 15, \ \boxed{17}, \ 18, \ \boxed{19, \ 26}, \ 42\end{align} \begin{align}Q_1=\frac{13+14}{2} = \frac{27}{2}=13.5 \qquad Q_3 = \frac{19+26}{2} = \frac{45}{2} = 22.5\end{align} Calculate the difference between \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align}: \begin{align}Q_3-Q_1=22.5-13.5=9.0\end{align}. Multiply this difference by 1.5: \begin{align}(9.0)(1.5)=13.5\end{align}. Finally, compute the range. \begin{align}Q_1-13.5=13.5-13.5=0\end{align} \begin{align}Q_3+13.5=22.5+13.5=36.0\end{align} Are there any data values below 0? No, there are no values below 0. Are there any values above 36.0? Yes, the value of 42 is above 36.0 and is, therefore, an outlier. Points to Consider • Are there still other ways to represent data graphically? • Are there other uses for a box-and-whisker plot? • Can box-and-whisker plots be used for comparing data sets? ### Guided Practice a. For the following data sets, determine the five-number summaries: i. 12, 16, 36, 10, 31, 23, 58 ii. 144, 240, 153, 629, 540, 300 b. Use the data set for part i of the previous question and the five-number summary to construct a box-and-whisker plot to model the data set. a. i. The first step is to organize the values in the data set as shown below: \begin{align}&12, \ 16, \ 36, \ 10, \ 31, \ 23, \ 58\\ &10, \ 12, \ 16, \ 23, \ 31, \ 36, \ 58\end{align} Now complete the following list: Minimum value \begin{align}\rightarrow 10\end{align} \begin{align}Q_1 \rightarrow 12\end{align} Median \begin{align}\rightarrow 23\end{align} \begin{align}Q_3 \rightarrow 36\end{align} Maximum value \begin{align}\rightarrow 58\end{align} ii. The first step is to organize the values in the data set as shown below: \begin{align}&144, \ 240, \ 153, \ 629, \ 540, \ 300\\ &144, \ 153, \ 240, \ 300, \ 540, \ 629\end{align} Now complete the following list: Minimum value \begin{align}\rightarrow 144\end{align} \begin{align}Q_1 \rightarrow 153\end{align} Median \begin{align}\rightarrow 270\end{align} \begin{align}Q_3 \rightarrow 540\end{align} Maximum value \begin{align}\rightarrow 629\end{align} b. The five-number summary can now be used to construct a box-and-whisker plot for part i. Be sure to provide a scale on the number line that includes the range from the minimum value to the maximum value. Minimum value \begin{align}\rightarrow 10\end{align} \begin{align}Q_1 \rightarrow 12\end{align} Median \begin{align}\rightarrow 23\end{align} \begin{align}Q_3 \rightarrow 36\end{align} Maximum value \begin{align}\rightarrow 58\end{align} It is very visible that the right whisker is much longer than the left whisker. This indicates that the distribution is positively skewed. ### Practice 1. Which of the following is not a part of the five-number summary? 1. \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align} 2. the mean 3. the median 4. minimum and maximum values 2. What percent of the data is contained in the box of a box-and-whisker plot? 1. 25% 2. 100% 3. 50% 4. 75% 3. What name is given to the horizontal lines to the left and right of the box of a box-and-whisker plot? 1. axis 2. whisker 3. range 4. plane 4. What term describes the distribution of a data set if the median of the data set is located to the left of the center of the box in a box-and-whisker plot? 1. positively skewed 2. negatively skewed 3. approximately symmetric 4. not skewed 5. What 2 values of the five-number summary are connected with 2 horizontal lines on a box-and-whisker plot? 1. Minimum value and the median 2. Maximum value and the median 3. Minimum and maximum values 4. \begin{align}Q_1\end{align} and \begin{align}Q_3\end{align} 6. For the following data sets, determine the five-number summaries: 1. 74, 69, 83, 79, 60, 75, 67, 71 2. 6, 9, 3, 12, 11, 9, 15, 5, 7 7. For each of the following box-and-whisker plots, list the five-number summary and comment on the distribution of the data: 8. The following data represents the number of coins that 12 randomly selected people had in their piggy banks: \begin{align}35 \quad 58 \quad 29 \quad 44 \quad 104 \quad 39 \quad 72 \quad 34 \quad 50 \quad 41 \quad 64 \quad 54\end{align} Construct a box-and-whisker plot for the above data. 9. The following data represent the time (in minutes) that each of 20 people waited in line at a local book store to purchase the latest Harry Potter book: \begin{align}& 15 \quad 8 \quad 5 \quad \ 10 \quad 14 \quad 17 \quad 21 \quad 23 \quad 6 \quad 19 \quad 31 \quad 34 \quad 30 \quad 31\\ & 3 \quad 22 \quad 17 \quad 25 \quad 5 \quad 16\end{align} Construct a box-and-whisker plot for the above data. Are the data skewed in any direction? 10. Firman’s Fitness Factory is a new gym that offers reasonably-priced family packages. The following table represents the number of family packages sold during the opening month: \begin{align}& 24 \quad 21 \quad 31 \quad 28 \quad 29\\ & 27 \quad 22 \quad 27 \quad 30 \quad 32\\ & 26 \quad 35 \quad 24 \quad 22 \quad 34\\ & 30 \quad 28 \quad 24 \quad 32 \quad 27\\ & 32 \quad 28 \quad 27 \quad 32 \quad 23\\ & 20 \quad 32 \quad 28 \quad 32 \quad 34\end{align} Construct a box-and-whisker plot for the data. Are the data symmetric or skewed? 11. Shown below is the number of new stage shows that appeared in Las Vegas for each of the past several years. Construct a box-and-whisker plot for the data and comment of the shape of the distribution. \begin{align}31 \quad 29 \quad 34 \quad 30 \quad 38 \quad 40 \quad 36 \quad 38 \quad 32 \quad 39 \quad 35\end{align} 12. The following data represent the average snowfall (in centimeters) for 18 Canadian cities for the month of January. Construct a box-and-whisker plot to model the data. Is the data skewed? Justify your answer. Name of City Amount of Snow(cm) Calgary 123.4 Charlottetown 74.5 Edmonton 80.6 Fredericton 73.8 Halifax 64.0 Moncton 82.4 Montreal 63.6 Ottawa 48.9 Quebec City 53.8 Regina 35.9 St. John’s 97.5 Sydney 44.2 Toronto 21.8 Vancouver 12.8 Victoria 8.3 Winnipeg 76.2 1. Using the procedure outlined in this concept, check the following data sets for outliers: 1. 25, 33, 55, 32, 17, 19, 15, 18, 21 2. 149, 123, 126, 122, 129, 120 ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish TermDefinition box-and-whisker plot A box-and-whisker plot is a graph based upon medians. It shows the minimum value, the lower median, the median, the upper median, and the maximum value of a data set. It is also known as a box plot. five-number summary The numbers needed to construct a box-and-whisker plot are called the five-number summary. The five-number summary are: the minimum value, $Q_1$, the median, $Q_2$, and the maximum value. Extremes The extremes are the maximum and minimum values in a data set. five point summary The numbers needed to construct a box-and-whisker plot are called the five-point-summary. The five points are the minimum, the lower median (Q1), the median, the upper median (Q3), and the maximum. line of fit A line of fit is a straight or continuously curved line representing the trend of changes in the comparison of two data sets (or one set of bivariate data). Median The median of a data set is the middle value of an organized data set. observed data Observed data are the values that result from computations performed on the input variable. Outlier In statistics, an outlier is a data value that is far from other data values. Quartile A quartile is each of four equal groups that a data set can be divided into. skewed As with the horizontal skewing of a histogram, stem plots with a obvious skew toward one end or the other tend to indicate an increased number of outliers either lesser than or greater than the mode. statistical correlation Statistical correlation is a representation of possible related changes in values between the two sets of data. trends Trends in data sets or samples are indicators found by reviewing the data from a general or overall standpoint uniform A uniform shaped histogram indicates data that is very consistent; the frequency of each class is very similar to that of the others. 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Worksheet on Multiplying Monomial and Polynomial will assist students in learning the multiplication of algebraic expressions involving Monomial and Polynomial. Learn the necessary math skill so that you will find it easy while solving complex algebraic expressions. Multiplication of Monomial and Polynomial Worksheet will offer step by step explanation for all the problems within so that you will get to know the algebraic methods used in solving the polynomial and monomial multiplication expressions. Multiplying Monomial and Polynomial Worksheet with Answers will guide you in having an in-depth understanding of the concept as well get grip on the rules associated with it. Attain and Fluency by answering the questions from the Free Printable Math Multiplication of Monomial and Polynomial Worksheet on a frequent basis. Do Check: ## Multiplication of a Monomial and Polynomial Worksheet I. Multiply monomial by polynomial: (i) 4x and (2x – 3y + 5z) (ii) (-5m) and (3m – 4n + 2p) (iii) 6xyz and (-7xy – 2yz – zx) (iv) 7a3b2c2 and (4a2b – 2a3c2 – b3c) (v) (-2x3y2z4) and (4x4y3 – 3x3y2z3 – 6xy2z2y) Solution: (i) Given 4x and (2x – 3y + 5z) Multiply the monomial with every term of the polynomial =4x(2x)-3y(4x)+5z(4x) =8x2-12xy+20xz Hence, By multiplying 4x and (2x – 3y + 5z) is 8x2-12xy+20xz. (ii) Given (-5m) and (3m – 4n + 2p) Multiply the monomial with every term of the polynomial =(-5m)(3m) -4n(-5m)+ 2p(-5m) =-15m2+20mn-10pm Hence, By multiplying (-5m) and (3m – 4n + 2p) we get -15m2+20mn-10pm. (iii) Given 6xyz and (-7xy – 2yz – zx) Multiply the monomial with every term of the polynomial =6xyz(-7xy)-2yz(6xyz)-zx(6xyz) =-42x2y2z-12xy2z2-6x2yz2 Hence, By multiplying 6xyz and (-7xy – 2yz -zx) we get -42x2y2z-12xy2z2-6x2yz2. (iv) Given 7a3b2c2 and (4a2b – 2a3c2 – b3c) Multiply the monomial with every term of the polynomial =4a2b(7a3b2c2)-2a3c2(7a3b2c2 )-b3c(7a3b2c2) =28a5b3c2-14a6b2c4-7a3b5c3 Hence, By multiplying 7a3b2c2 and (4a2b – 2a3c2 – b3c) we get 28a5b3c2-14a6b2c4-7a3b5c3 . (v) Given (-2x3y2z4) and (4x4y3 – 3x3y2z3 – 6xy2z2) Multiply the monomial with every term of the polynomial =4x4y3((-2x3y2z4) -3x3y2z3(-2x3y2z4) -6xy2z2(-2x3y2z4) =-8x7y5z4+6x6y4z7+12x4y4z6 Hence, By multiplying (-2x3y2z4) and (4x4y3 – 3x3y2z3 – 6xy2z2) we get -8x7y5z4+6x6y4z7+12x4y4z6. II. Multiply polynomial by monomial: (i) (m+ m4 + 1) and 5m (ii) (ax2 + bx3 + 5x) and x2 (iii) (2x + xz + z3) and 7z (iv) (m – 2mn + 8n) and (–m7) (v) (a + 2bc + ca) and (-a) Solution: (i) Given, (m+ m4 + 1) and 5m Multiply the monomial with every term of the polynomial =m(5m) + m4 (5m) + 1(5m) =5m2+5m5+5m Therefore, By multiplying (m+ m4 + 1) and 5m we get 5m2+5m5+5m. (ii) Given, (ax2 + bx3 + 5x) and x2 Multiply the monomial with every term of the polynomial =ax2(x2) + bx3(x2) + 5x(x2) =ax4+bx5+5x3 Therefore, By multiplying (ax2 + bx3 + 5x) and x2 we get ax4+bx5+5x3. (iii) Given, (2x + xz + z3) and 7z Multiply the monomial with every term of the polynomial =2x(7z) + xz(7z) + z3 (7z) =14xz + 7xz2 + 7z4 Therefore, By multiplying (2x + xz + z3) and 7z is 14xz + 7xz2 + 7z4. (iv) Given, (m – 2mn + 8n) and (–m7) Multiply the monomial with every term of the polynomial =m(–m7) – 2mn(–m7) + 8n(–m7) =-m8+2m8n-8m7n Therefore, By multiplying (m – 2mn + 8n) and (–m7) we get -m8+2m8n-8m7n. (v) Given,(a + 2bc + ca) and (-a) Multiply the monomial with every term of the polynomial =a(-a)+2bc(-a)+ca(-a) =-a2-2abc-a2c Therefore, By multiplying (a + 2bc + ca) and (-a) we get -a2-2abc-a2c. III. Find the product of the following: (i) 3ab(2ab + b2c + 5ca) (ii) (-13m2)(5 + mx + ny) (iii) 2m2n(mn + n – n2) (iv) mn(m2+n2) (v) -6a2bc(3ab + bc – 7ca) (vi) (x+3y)(2x+6y) Solution: (i) Given, 3ab(2ab + b2c + 5ca) Multiply the monomial with every term of the polynomial =3ab(2ab)+b2c(3ab) + 3ab(5ca) =6a2b2 + 3ab3c + 15a2bc Hence, By multiplying 3ab(2ab + b2c + 5ca) we get 6a2b2 + 3ab3c + 15a2bc. (ii) Given, (-13m2)(5 + mx + ny) Multiply the monomial with every term of the polynomial =(-13m2)5 + mx((-13m2) + ny(-13m2) =-65m2-13m3x-13m2ny Hence, By multiplying (-13m2)(5 + mx + ny) we get -65m2-13m3x-13m2ny. (iii) Given, 2m2n(mn + n – n2) Multiply the monomial with every term of the polynomial =2m2n(mn) + n(2m2n)-n2(2m2n) =2m3n2+2m2n2-2m2n3 Hence, By multiplying 2m2n(mn + n – n2) we get 2m3n2+2m2n2-2m2n3. (iv) Given, mn(3m2+2n2) Multiply the monomial with every term of the polynomial =mn(3m2) + mn(2n2) =3m3n + 2mn3 Hence, By multiplying mn(m2+n2) we get 3m3n + 2mn3. (v) Given, -6a2bc(3ab + bc – 7ca) Multiply the monomial with every term of the polynomial =-6a2bc(3ab) + bc(-6a2bc) -7ca(-6a2bc) =-18a3b2c-6a2b2c2+42a3bc2 Hence, By multiplying -6a2bc(3ab + bc – 7ca) we get -18a3b2c-6a2b2c2+42a3bc2. (vi) Given, (x+3y)(2x+6y) By using the distributive property, multiply the polynomials, =x(2x+6y) + 3y(2x+6y) =2×2+6xy+6yx+18y2 =2×2+12xy+18y2 Hence, By multiplying (x+3y)(2x+6y) we get 2×2+12xy+18y2. IV. The product of two numbers is 3m4 if one of them is 1/4m2. Find the other? Solution: Given, The product of two numbers is =3m4 one of the number=1/4m2 Let the other number be x. 1/4m2 × x=3m4 x=3m4.4m2 x=12m6 Therefore, the other number is 12m6. V. If P=5x2+2x, Q=2x, R=24. Find the value of (P × R)/Q. Solution: Given P=5x2+2x, Q=2x+2, R=24 P × R=24(5x2+2x) =120x2 + 48x (P × R)/Q= (120x2 + 48x)/2x =60x+24 Hence, the value of (P × R)/Q is 60x+24. VI. If the length and width of the rectangle are (-4a2+7) and (2a + 5) respectively. Find the area of the rectangle? Solution: Given, The length of the rectangle=(-4a2+7) The breadth of the rectangle=(2a + 5) Area of the rectangle=length * breadth =(-4a2+7) * (2a + 5) =-4a2(2a+5) + 7(2a+5) =-8a3-20a2+14a+35
Calculus Volume 3 # 5.4Triple Integrals Calculus Volume 35.4 Triple Integrals ## Learning Objectives • 5.4.1 Recognize when a function of three variables is integrable over a rectangular box. • 5.4.2 Evaluate a triple integral by expressing it as an iterated integral. • 5.4.3 Recognize when a function of three variables is integrable over a closed and bounded region. • 5.4.4 Simplify a calculation by changing the order of integration of a triple integral. • 5.4.5 Calculate the average value of a function of three variables. In Double Integrals over Rectangular Regions, we discussed the double integral of a function $f(x,y)f(x,y)$ of two variables over a rectangular region in the plane. In this section we define the triple integral of a function $f(x,y,z)f(x,y,z)$ of three variables over a rectangular solid box in space, $ℝ3.ℝ3.$ Later in this section we extend the definition to more general regions in $ℝ3.ℝ3.$ ## Integrable Functions of Three Variables We can define a rectangular box $BB$ in $ℝ3ℝ3$ as $B={(x,y,z)\|a≤x≤b,c≤y≤d,e≤z≤f}.B={(x,y,z)\|a≤x≤b,c≤y≤d,e≤z≤f}.$ We follow a similar procedure to what we did in Double Integrals over Rectangular Regions. We divide the interval $[a,b][a,b]$ into $ll$ subintervals $[xi−1,xi][xi−1,xi]$ of equal length $Δx=b-al,Δx=b-al,$ divide the interval $[c,d][c,d]$ into $mm$ subintervals $[yj−1,yj][yj−1,yj]$ of equal length $Δy=d-cm,Δy=d-cm,$ and divide the interval $[e,f][e,f]$ into $nn$ subintervals $[zk−1,zk][zk−1,zk]$ of equal length $Δz=f-en.Δz=f-en.$ Then the rectangular box $BB$ is subdivided into $lmnlmn$ subboxes $Bijk=[xi−1,xi]×[yj−1,yj]×[zk−1,zk],Bijk=[xi−1,xi]×[yj−1,yj]×[zk−1,zk],$ as shown in Figure 5.40. Figure 5.40 A rectangular box in $ℝ3ℝ3$ divided into subboxes by planes parallel to the coordinate planes. For each $i,j,andk,i,j,andk,$ consider a sample point $(xijk,yijk,zijk)(xijk,yijk,zijk)$ in each sub-box $Bijk.Bijk.$ We see that its volume is $ΔV=ΔxΔyΔz.ΔV=ΔxΔyΔz.$ Form the triple Riemann sum $∑i=1l∑j=1m∑k=1nf(xijk,yijk,zijk)ΔxΔyΔz.∑i=1l∑j=1m∑k=1nf(xijk,yijk,zijk)ΔxΔyΔz.$ We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum. ## Definition The triple integral of a function $f(x,y,z)f(x,y,z)$ over a rectangular box $BB$ is defined as $liml,m,n→∞∑i=1l∑j=1m∑k=1nf(xijk,yijk,zijk)ΔxΔyΔz=∭Bf(x,y,z)dVliml,m,n→∞∑i=1l∑j=1m∑k=1nf(xijk,yijk,zijk)ΔxΔyΔz=∭Bf(x,y,z)dV$ (5.10) if this limit exists. When the triple integral exists on $B,B,$ the function $f(x,y,z)f(x,y,z)$ is said to be integrable on $B.B.$ Also, the triple integral exists if $f(x,y,z)f(x,y,z)$ is continuous on $B.B.$ Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, $ff$ is bounded on $BB$ and continuous except possibly on the boundary of $B.B.$ The sample point $(xijk,yijk,zijk)(xijk,yijk,zijk)$ can be any point in the rectangular sub-box $BijkBijk$ and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections. Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s thereom for triple integrals exists. ## Theorem5.9 ### Fubini’s Theorem for Triple Integrals If $f(x,y,z)f(x,y,z)$ is continuous on a rectangular box $B=[a,b]×[c,d]×[e,f],B=[a,b]×[c,d]×[e,f],$ then $∭Bf(x,y,z)dV=∫ef∫cd∫abf(x,y,z)dxdydz.∭Bf(x,y,z)dV=∫ef∫cd∫abf(x,y,z)dxdydz.$ This integral is also equal to any of the other five possible orderings for the iterated triple integral. For $a,b,c,d,e,a,b,c,d,e,$ and $ff$ real numbers, the iterated triple integral can be expressed in six different orderings: $∫ef∫cd∫abf(x,y,z)dxdydz=∫ef(∫cd(∫abf(x,y,z)dx)dy)dz=∫cd(∫ef(∫abf(x,y,z)dx)dz)dy=∫ab(∫ef(∫cdf(x,y,z)dy)dz)dx=∫ef(∫ab(∫cdf(x,y,z)dy)dx)dz=∫ce(∫ab(∫eff(x,y,z)dz)dx)dy=∫ab(∫ce(∫eff(x,y,z)dz)dy)dx.∫ef∫cd∫abf(x,y,z)dxdydz=∫ef(∫cd(∫abf(x,y,z)dx)dy)dz=∫cd(∫ef(∫abf(x,y,z)dx)dz)dy=∫ab(∫ef(∫cdf(x,y,z)dy)dz)dx=∫ef(∫ab(∫cdf(x,y,z)dy)dx)dz=∫ce(∫ab(∫eff(x,y,z)dz)dx)dy=∫ab(∫ce(∫eff(x,y,z)dz)dy)dx.$ For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside). ## Example 5.36 ### Evaluating a Triple Integral Evaluate the triple integral $∫z=0z=1∫y=2y=4∫x=−1x=5(x+yz2)dxdydz.∫z=0z=1∫y=2y=4∫x=−1x=5(x+yz2)dxdydz.$ ## Example 5.37 ### Evaluating a Triple Integral Evaluate the triple integral $∭Bx2yzdV∭Bx2yzdV$ where $B={(x,y,z)\|−2≤x≤1,0≤y≤3,1≤z≤5}B={(x,y,z)\|−2≤x≤1,0≤y≤3,1≤z≤5}$ as shown in the following figure. Figure 5.41 Evaluating a triple integral over a given rectangular box. ## Checkpoint5.23 Evaluate the triple integral $∭BzsinxcosydV∭BzsinxcosydV$ where $B={(x,y,z)\|0≤x≤π,3π2≤y≤2π,1≤z≤3}.B={(x,y,z)\|0≤x≤π,3π2≤y≤2π,1≤z≤3}.$ ## Triple Integrals over a General Bounded Region We now expand the definition of the triple integral to compute a triple integral over a more general bounded region $EE$ in $ℝ3.ℝ3.$ The general bounded regions we will consider are of three types. First, let $DD$ be the bounded region that is a projection of $EE$ onto the $xyxy$-plane. Suppose the region $EE$ in $ℝ3ℝ3$ has the form $E={(x,y,z)\|(x,y)∈D,u1(x,y)≤z≤u2(x,y)}E={(x,y,z)\|(x,y)∈D,u1(x,y)≤z≤u2(x,y)}$ for two functions $z=u1(x,y)z=u1(x,y)$ and $z=u2(x,y),z=u2(x,y),$ such that $u1(x,y)≤u2(x,y)u1(x,y)≤u2(x,y)$ for all $(x,y)(x,y)$ in $DD$ as shown in the following figure. Figure 5.42 We can describe region $EE$ as the space between $u1(x,y)u1(x,y)$ and $u2(x,y)u2(x,y)$ above the projection $DD$ of $EE$ onto the $xyxy$-plane. ## Theorem5.10 ### Triple Integral over a General Region The triple integral of a continuous function $f(x,y,z)f(x,y,z)$ over a general three-dimensional region $E={(x,y,z)\|(x,y)∈D,u1(x,y)≤z≤u2(x,y)}E={(x,y,z)\|(x,y)∈D,u1(x,y)≤z≤u2(x,y)}$ in $ℝ3,ℝ3,$ where $DD$ is the projection of $EE$ onto the $xyxy$-plane, is $∭Ef(x,y,z)dV=∬D[∫u1(x,y)u2(x,y)f(x,y,z)dz]dA.∭Ef(x,y,z)dV=∬D[∫u1(x,y)u2(x,y)f(x,y,z)dz]dA.$ Similarly, we can consider a general bounded region $DD$ in the $xyxy$-plane and two functions $y=u1(x,z)y=u1(x,z)$ and $y=u2(x,z)y=u2(x,z)$ such that $u1(x,z)≤u2(x,z)u1(x,z)≤u2(x,z)$ for all $(x,z)(x,z)$ in $D.D.$ Then we can describe the solid region $EE$ in $ℝ3ℝ3$ as $E={(x,y,z)\|(x,z)∈D,u1(x,z)≤y≤u2(x,z)}E={(x,y,z)\|(x,z)∈D,u1(x,z)≤y≤u2(x,z)}$ where $DD$ is the projection of $EE$ onto the $xzxz$-plane and the triple integral is $∭Ef(x,y,z)dV=∬D[∫u1(x,z)u2(x,z)f(x,y,z)dy]dA.∭Ef(x,y,z)dV=∬D[∫u1(x,z)u2(x,z)f(x,y,z)dy]dA.$ Finally, if $DD$ is a general bounded region in the $yzyz$-plane and we have two functions $x=u1(y,z)x=u1(y,z)$ and $x=u2(y,z)x=u2(y,z)$ such that $u1(y,z)≤u2(y,z)u1(y,z)≤u2(y,z)$ for all $(y,z)(y,z)$ in $D,D,$ then the solid region $EE$ in $ℝ3ℝ3$ can be described as $E={(x,y,z)\|(y,z)∈D,u1(y,z)≤x≤u2(y,z)}E={(x,y,z)\|(y,z)∈D,u1(y,z)≤x≤u2(y,z)}$ where $DD$ is the projection of $EE$ onto the $yzyz$-plane and the triple integral is $∭Ef(x,y,z)dV=∬D[∫u1(y,z)u2(y,z)f(x,y,z)dx]dA.∭Ef(x,y,z)dV=∬D[∫u1(y,z)u2(y,z)f(x,y,z)dx]dA.$ Note that the region $DD$ in any of the planes may be of Type I or Type II as described in Double Integrals over General Regions. If $DD$ in the $xyxy$-plane is of Type I (Figure 5.43), then $E={(x,y,z)\|a≤x≤b,g1(x)≤y≤g2(x),u1(x,y)≤z≤u2(x,y)}.E={(x,y,z)\|a≤x≤b,g1(x)≤y≤g2(x),u1(x,y)≤z≤u2(x,y)}.$ Figure 5.43 A box $EE$ where the projection $DD$ in the $xyxy$-plane is of Type I. Then the triple integral becomes $∭Ef(x,y,z)dV=∫ab∫g1(x)g2(x)∫u1(x,y)u2(x,y)f(x,y,z)dzdydx.∭Ef(x,y,z)dV=∫ab∫g1(x)g2(x)∫u1(x,y)u2(x,y)f(x,y,z)dzdydx.$ If $DD$ in the $xyxy$-plane is of Type II (Figure 5.44), then $E={(x,y,z)\|c≤y≤d,h1(y)≤x≤h2(y),u1(x,y)≤z≤u2(x,y)}.E={(x,y,z)\|c≤y≤d,h1(y)≤x≤h2(y),u1(x,y)≤z≤u2(x,y)}.$ Figure 5.44 A box $EE$ where the projection $DD$ in the $xyxy$-plane is of Type II. Then the triple integral becomes $∭Ef(x,y,z)dV=∫y=cy=d∫x=h1(y)x=h2(y)∫z=u1(x,y)z=u2(x,y)f(x,y,z)dzdxdy.∭Ef(x,y,z)dV=∫y=cy=d∫x=h1(y)x=h2(y)∫z=u1(x,y)z=u2(x,y)f(x,y,z)dzdxdy.$ ## Example 5.38 ### Evaluating a Triple Integral over a General Bounded Region Evaluate the triple integral of the function $f(x,y,z)=5x−3yf(x,y,z)=5x−3y$ over the solid tetrahedron bounded by the planes $x=0,y=0,z=0,x=0,y=0,z=0,$ and $x+y+z=1.x+y+z=1.$ Just as we used the double integral $∬D1dA∬D1dA$ to find the area of a general bounded region $D,D,$ we can use $∭E1dV∭E1dV$ to find the volume of a general solid bounded region $E.E.$ The next example illustrates the method. ## Example 5.39 ### Finding a Volume by Evaluating a Triple Integral Find the volume of a right pyramid that has the square base in the $xyxy$-plane $[−1,1]×[−1,1][−1,1]×[−1,1]$ and vertex at the point $(0,0,1)(0,0,1)$ as shown in the following figure. Figure 5.46 Finding the volume of a pyramid with a square base. ## Checkpoint5.24 Consider the solid sphere $E={(x,y,z)\|x2+y2+z2≤9}.E={(x,y,z)\|x2+y2+z2≤9}.$ Write the triple integral $∭Ef(x,y,z)dV∭Ef(x,y,z)dV$ for an arbitrary function $ff$ as an iterated integral. Then evaluate this triple integral with $f(x,y,z)=1.f(x,y,z)=1.$ Notice that this gives the volume of a sphere using a triple integral. ## Changing the Order of Integration As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here. ## Example 5.40 ### Changing the Order of Integration Consider the iterated integral $∫x=0x=1∫y=0y=x2∫z=0z=y2f(x,y,z)dzdydx.∫x=0x=1∫y=0y=x2∫z=0z=y2f(x,y,z)dzdydx.$ The order of integration here is first with respect to z, then y, and then x. Express this integral by changing the order of integration to be first with respect to x, then z, and then $y.y.$ Verify that the value of the integral is the same if we let $f(x,y,z)=xyz.f(x,y,z)=xyz.$ ## Checkpoint5.25 Write five different iterated integrals equal to the given integral $∫z=0z=4∫y=0y=4−z∫x=0x=yf(x,y,z)dxdydz.∫z=0z=4∫y=0y=4−z∫x=0x=yf(x,y,z)dxdydz.$ ## Example 5.41 ### Changing Integration Order and Coordinate Systems Evaluate the triple integral $∭Ex2+z2dV,∭Ex2+z2dV,$ where $EE$ is the region bounded by the paraboloid $y=x2+z2y=x2+z2$ (Figure 5.48) and the plane $y=4.y=4.$ Figure 5.48 Integrating a triple integral over a paraboloid. ## Average Value of a Function of Three Variables Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid. ## Theorem5.11 ### Average Value of a Function of Three Variables If $f(x,y,z)f(x,y,z)$ is integrable over a solid bounded region $EE$ with positive volume $V(E),V(E),$ then the average value of the function is $fave=1V(E)∭Ef(x,y,z)dV.fave=1V(E)∭Ef(x,y,z)dV.$ Note that the volume is $V(E)=∭E1dV.V(E)=∭E1dV.$ ## Example 5.42 ### Finding an Average Temperature The temperature at a point $(x,y,z)(x,y,z)$ of a solid $EE$ bounded by the coordinate planes and the plane $x+y+z=1x+y+z=1$ is $T(x,y,z)=(xy+8z+20)°C.T(x,y,z)=(xy+8z+20)°C.$ Find the average temperature over the solid. ## Checkpoint5.26 Find the average value of the function $f(x,y,z)=xyzf(x,y,z)=xyz$ over the cube with sides of length $44$ units in the first octant with one vertex at the origin and edges parallel to the coordinate axes. ## Section 5.4 Exercises In the following exercises, evaluate the triple integrals over the rectangular solid box $B.B.$ 181. $∭B(2x+3y2+4z3)dV,∭B(2x+3y2+4z3)dV,$ where $B={(x,y,z)\|0≤x≤1,0≤y≤2,0≤z≤3}B={(x,y,z)\|0≤x≤1,0≤y≤2,0≤z≤3}$ 182. $∭B(xy+yz+xz)dV,∭B(xy+yz+xz)dV,$ where $B={(x,y,z)\|1≤x≤2,0≤y≤2,1≤z≤3}B={(x,y,z)\|1≤x≤2,0≤y≤2,1≤z≤3}$ 183. $∭B(xcosy+z)dV,∭B(xcosy+z)dV,$ where $B={(x,y,z)\|0≤x≤1,0≤y≤π,−1≤z≤1}B={(x,y,z)\|0≤x≤1,0≤y≤π,−1≤z≤1}$ 184. $∭B(zsinx+y2)dV,∭B(zsinx+y2)dV,$ where $B={(x,y,z)\|0≤x≤π,0≤y≤1,−1≤z≤2}B={(x,y,z)\|0≤x≤π,0≤y≤1,−1≤z≤2}$ In the following exercises, change the order of integration by integrating first with respect to $z,z,$ then $x,x,$ then $y.y.$ 185. $∫ 0 1 ∫ 1 2 ∫ 2 3 ( x 2 + ln y + z ) d x d y d z ∫ 0 1 ∫ 1 2 ∫ 2 3 ( x 2 + ln y + z ) d x d y d z$ 186. $∫ 0 1 ∫ −1 1 ∫ 0 3 ( z e x + 2 y ) d z d x d y ∫ 0 1 ∫ −1 1 ∫ 0 3 ( z e x + 2 y ) d z d x d y$ 187. $∫ −1 2 ∫ 1 3 ∫ 0 4 ( x 2 z + 1 y ) d x d y d z ∫ −1 2 ∫ 1 3 ∫ 0 4 ( x 2 z + 1 y ) d x d y d z$ 188. $∫ 1 2 ∫ −2 −1 ∫ 0 1 x + y z d x d y d z ∫ 1 2 ∫ −2 −1 ∫ 0 1 x + y z d x d y d z$ 189. Let $F,G,andHF,G,andH$ be continuous functions on $[a,b],[c,d],[a,b],[c,d],$ and $[e,f],[e,f],$ respectively, where $a,b,c,d,e,andfa,b,c,d,e,andf$ are real numbers such that $a Show that $∫ a b ∫ c d ∫ e f F ( x ) G ( y ) H ( z ) d z d y d x = ( ∫ a b F ( x ) d x ) ( ∫ c d G ( y ) d y ) ( ∫ e f H ( z ) d z ) . ∫ a b ∫ c d ∫ e f F ( x ) G ( y ) H ( z ) d z d y d x = ( ∫ a b F ( x ) d x ) ( ∫ c d G ( y ) d y ) ( ∫ e f H ( z ) d z ) .$ 190. Let $F,G,andHF,G,andH$ be differential functions on $[a,b],[c,d],[a,b],[c,d],$ and $[e,f],[e,f],$ respectively, where $a,b,c,d,e,andfa,b,c,d,e,andf$ are real numbers such that $a Show that $∫ a b ∫ c d ∫ e f F ′ ( x ) G ′ ( y ) H ′ ( z ) d z d y d x = [ F ( b ) − F ( a ) ] [ G ( d ) − G ( c ) ] [ H ( f ) − H ( e ) ] . ∫ a b ∫ c d ∫ e f F ′ ( x ) G ′ ( y ) H ′ ( z ) d z d y d x = [ F ( b ) − F ( a ) ] [ G ( d ) − G ( c ) ] [ H ( f ) − H ( e ) ] .$ In the following exercises, evaluate the triple integrals over the bounded region $E={(x,y,z)\|a≤x≤b,h1(x)≤y≤h2(x),e≤z≤f}.E={(x,y,z)\|a≤x≤b,h1(x)≤y≤h2(x),e≤z≤f}.$ 191. $∭E(2x+5y+7z)dV,∭E(2x+5y+7z)dV,$ where $E={(x,y,z)\|0≤x≤1,0≤y≤−x+1,1≤z≤2}E={(x,y,z)\|0≤x≤1,0≤y≤−x+1,1≤z≤2}$ 192. $∭E(ylnx+z)dV,∭E(ylnx+z)dV,$ where $E={(x,y,z)\|1≤x≤e,0≤y≤lnx,0≤z≤1}E={(x,y,z)\|1≤x≤e,0≤y≤lnx,0≤z≤1}$ 193. $∭E(sinx+siny)dV,∭E(sinx+siny)dV,$ where $E={(x,y,z)\|0≤x≤π2,−cosx≤y≤cosx,−1≤z≤1}E={(x,y,z)\|0≤x≤π2,−cosx≤y≤cosx,−1≤z≤1}$ 194. $∭E(xy+yz+xz)dV,∭E(xy+yz+xz)dV,$ where $E={(x,y,z)\|0≤x≤1,−x2≤y≤x2,0≤z≤1}E={(x,y,z)\|0≤x≤1,−x2≤y≤x2,0≤z≤1}$ In the following exercises, evaluate the triple integrals over the indicated bounded region $E.E.$ 195. $∭E(x+2yz)dV,∭E(x+2yz)dV,$ where $E={(x,y,z)\|0≤x≤1,0≤y≤x,0≤z≤5−x−y}E={(x,y,z)\|0≤x≤1,0≤y≤x,0≤z≤5−x−y}$ 196. $∭E(x3+y3+z3)dV,∭E(x3+y3+z3)dV,$ where $E={(x,y,z)\|0≤x≤2,0≤y≤2x,0≤z≤4−x−y}E={(x,y,z)\|0≤x≤2,0≤y≤2x,0≤z≤4−x−y}$ 197. $∭EydV,∭EydV,$ where $E={(x,y,z)\|−1≤x≤1,−1−x2≤y≤1−x2,0≤z≤1−x2−y2}E={(x,y,z)\|−1≤x≤1,−1−x2≤y≤1−x2,0≤z≤1−x2−y2}$ 198. $∭ExdV,∭ExdV,$ where $E={(x,y,z)\|−2≤x≤2,-4−x2≤y≤4−x2,0≤z≤4−x2−y2}E={(x,y,z)\|−2≤x≤2,-4−x2≤y≤4−x2,0≤z≤4−x2−y2}$ In the following exercises, evaluate the triple integrals over the bounded region $EE$ of the form $E={(x,y,z)\|g1(y)≤x≤g2(y),c≤y≤d,e≤z≤f}.E={(x,y,z)\|g1(y)≤x≤g2(y),c≤y≤d,e≤z≤f}.$ 199. $∭Ex2dV,∭Ex2dV,$ where $E={(x,y,z)\|1−y2≤x≤y2−1,−1≤y≤1,1≤z≤2}E={(x,y,z)\|1−y2≤x≤y2−1,−1≤y≤1,1≤z≤2}$ 200. $∭E(sinx+y)dV,∭E(sinx+y)dV,$ where $E={(x,y,z)\|−y4≤x≤y4,0≤y≤2,0≤z≤4}E={(x,y,z)\|−y4≤x≤y4,0≤y≤2,0≤z≤4}$ 201. $∭E(x−yz)dV,∭E(x−yz)dV,$ where $E={(x,y,z)\|−y6≤x≤y,0≤y≤1,−1≤z≤1}E={(x,y,z)\|−y6≤x≤y,0≤y≤1,−1≤z≤1}$ 202. $∭EzdV,∭EzdV,$ where $E={(x,y,z)\|2−2y≤x≤2+y,0≤y≤1,2≤z≤3}E={(x,y,z)\|2−2y≤x≤2+y,0≤y≤1,2≤z≤3}$ In the following exercises, evaluate the triple integrals over the bounded region $E={(x,y,z)\|g1(y)≤x≤g2(y),c≤y≤d,u1(x,y)≤z≤u2(x,y)}.E={(x,y,z)\|g1(y)≤x≤g2(y),c≤y≤d,u1(x,y)≤z≤u2(x,y)}.$ 203. $∭EzdV,∭EzdV,$ where $E={(x,y,z)\|−y≤x≤y,0≤y≤1,0≤z≤1−x4−y4}E={(x,y,z)\|−y≤x≤y,0≤y≤1,0≤z≤1−x4−y4}$ 204. $∭E(xz+1)dV,∭E(xz+1)dV,$ where $E={(x,y,z)\|0≤x≤y,0≤y≤2,0≤z≤1−x2−y2}E={(x,y,z)\|0≤x≤y,0≤y≤2,0≤z≤1−x2−y2}$ 205. $∭E(x−z)dV,∭E(x−z)dV,$ where $E={(x,y,z)\|−1−y2≤x≤0,0≤y≤12,0≤z≤1−x2−y2}E={(x,y,z)\|−1−y2≤x≤0,0≤y≤12,0≤z≤1−x2−y2}$ 206. $∭E(x+y)dV,∭E(x+y)dV,$ where $E={(x,y,z)\|0≤x≤1−y2,0≤y≤1,0≤z≤1−x}E={(x,y,z)\|0≤x≤1−y2,0≤y≤1,0≤z≤1−x}$ In the following exercises, evaluate the triple integrals over the bounded region $E={(x,y,z)\|(x,y)∈D,u1(x,y)≤z≤u2(x,y)},E={(x,y,z)\|(x,y)∈D,u1(x,y)≤z≤u2(x,y)},$ where $DD$ is the projection of $EE$ onto the $xyxy$-plane. 207. $∬D(∫12(x+z)dz)dA,∬D(∫12(x+z)dz)dA,$ where $D={(x,y)\|x2+y2≤1}D={(x,y)\|x2+y2≤1}$ 208. $∬D(∫13x(z+1)dz)dA,∬D(∫13x(z+1)dz)dA,$ where $D={(x,y)\|x2−y2≥1,1≤x≤5}D={(x,y)\|x2−y2≥1,1≤x≤5}$ 209. $∬D(∫010−x−y(x+2z)dz)dA,∬D(∫010−x−y(x+2z)dz)dA,$ where $D={(x,y)\|y≥0,x≥0,x+y≤10}D={(x,y)\|y≥0,x≥0,x+y≤10}$ 210. $∬D(∫04x2+4y2ydz)dA,∬D(∫04x2+4y2ydz)dA,$ where $D={(x,y)\|x2+y2≤4,y≥1,x≥0}D={(x,y)\|x2+y2≤4,y≥1,x≥0}$ 211. The solid $EE$ bounded by $y2+z2=9,z=0,x=0,y2+z2=9,z=0,x=0,$ and $x=5x=5$ is shown in the following figure. Evaluate the integral $∭EzdV∭EzdV$ by integrating first with respect to $z,z,$ then $y,and thenx.y,and thenx.$ 212. The solid $EE$ bounded by $y=x,y=x,$ $x=4,x=4,$ $y=0,y=0,$ $z=-2z=-2$, and $z=1z=1$ is given in the following figure. Evaluate the integral $∭ExyzdV∭ExyzdV$ by integrating first with respect to $x,x,$ then $y,y,$ and then $z.z.$ 213. [T] The volume of a solid $EE$ is given by the integral $∫−20∫x0∫0x2+y2dzdydx.∫−20∫x0∫0x2+y2dzdydx.$ Use a computer algebra system (CAS) to graph $EE$ and find its volume. Round your answer to two decimal places. 214. [T] The volume of a solid $EE$ is given by the integral $∫−10∫−x20∫01+x2+y2dzdydx.∫−10∫−x20∫01+x2+y2dzdydx.$ Use a CAS to graph $EE$ and find its volume $V.V.$ Round your answer to two decimal places. In the following exercises, use two circular permutations of the variables $x,y,andzx,y,andz$ to write new integrals whose values equal the value of the original integral. A circular permutation of $x,y,andzx,y,andz$ is the arrangement of the numbers in one of the following orders: $y,z,andxorz,x,andy.y,z,andxorz,x,andy.$ 215. $∫ 0 1 ∫ 1 3 ∫ 2 4 ( x 2 z 2 + 1 ) d x d y d z ∫ 0 1 ∫ 1 3 ∫ 2 4 ( x 2 z 2 + 1 ) d x d y d z$ 216. $∫ 1 3 ∫ 0 1 ∫ 0 −y + 1 ( 2 y + 5 z + 7 x ) d z d y d x ∫ 1 3 ∫ 0 1 ∫ 0 −y + 1 ( 2 y + 5 z + 7 x ) d z d y d x$ 217. $∫ 0 1 ∫ − y y ∫ 0 1 − x 4 − y 4 e x d z d x d y ∫ 0 1 ∫ − y y ∫ 0 1 − x 4 − y 4 e x d z d x d y$ 218. $∫ −1 1 ∫ 0 1 ∫ − y 6 y ( x + y z ) d x d y d z ∫ −1 1 ∫ 0 1 ∫ − y 6 y ( x + y z ) d x d y d z$ 219. Set up the integral that gives the volume of the solid $EE$ bounded by $y2=x2+z2y2=x2+z2$ and $y=a,y=a,$ where $a>0a>0$ and 220. Set up the integral that gives the volume of the solid $EE$ bounded by $x=y2+z2x=y2+z2$ and $x=a2,x=a2,$ where $a>0.a>0.$ 221. Find the average value of the function $f(x,y,z)=x+y+zf(x,y,z)=x+y+z$ over the parallelepiped determined by $x=0,x=1,y=0,y=3,z=0,x=0,x=1,y=0,y=3,z=0,$ and $z=5.z=5.$ 222. Find the average value of the function $f(x,y,z)=xyzf(x,y,z)=xyz$ over the solid $E=[0,1]×[0,1]×[0,1]E=[0,1]×[0,1]×[0,1]$ situated in the first octant. 223. Find the volume of the solid $EE$ that lies under the plane $x+y+z=9x+y+z=9$ and whose projection onto the $xyxy$-plane is bounded by $x=y−1,x=0,x=y−1,x=0,$ and $x+y=7.x+y=7.$ 224. Find the volume of the solid E that lies under the plane $2x+y+z=82x+y+z=8$ and whose projection onto the $xyxy$-plane is bounded by $x=sin−1y,y=0,x=sin−1y,y=0,$ and $x=π2.x=π2.$ 225. Consider the pyramid with the base in the $xyxy$-plane of $[−2,2]×[−2,2][−2,2]×[−2,2]$ and the vertex at the point $(0,0,8).(0,0,8).$ 1. Show that the equations of the planes of the lateral faces of the pyramid are $4y+z=8,4y+z=8,$ $4y−z=−8,4y−z=−8,$ $4x+z=8,4x+z=8,$ and $−4x+z=8.−4x+z=8.$ 2. Find the volume of the pyramid. 226. Consider the pyramid with the base in the $xyxy$-plane of $[−3,3]×[−3,3][−3,3]×[−3,3]$ and the vertex at the point $(0,0,9).(0,0,9).$ 1. Show that the equations of the planes of the side faces of the pyramid are 2. Find the volume of the pyramid. 227. The solid $EE$ bounded by the sphere of equation $x2+y2+z2=r2x2+y2+z2=r2$ with $r>0r>0$ and located in the first octant is represented in the following figure. 1. Write the triple integral that gives the volume of $EE$ by integrating first with respect to $z,z,$ then with $y,y,$ and then with $x.x.$ 2. Rewrite the integral in part a. as an equivalent integral in five other orders. 228. The solid $EE$ bounded by the equation $9x2+4y2+z2=19x2+4y2+z2=1$ and located in the first octant is represented in the following figure. 1. Write the triple integral that gives the volume of $EE$ by integrating first with respect to $z,z,$ then with $y,y,$ and then with $x.x.$ 2. Rewrite the integral in part a. as an equivalent integral in five other orders. 229. Find the volume of the prism with vertices $(0,0,0),(2,0,0),(2,3,0),(0,0,0),(2,0,0),(2,3,0),$ $(0,3,0),(0,0,1),and(2,0,1).(0,3,0),(0,0,1),and(2,0,1).$ 230. Find the volume of the prism with vertices $(0,0,0),(4,0,0),(4,6,0),(0,0,0),(4,0,0),(4,6,0),$ $(0,6,0),(0,0,1),and(4,0,1).(0,6,0),(0,0,1),and(4,0,1).$ 231. The solid $EE$ bounded by $z=10−2x−yz=10−2x−y$ and situated in the first octant is given in the following figure. Find the volume of the solid. 232. The solid $EE$ bounded by $z=1−x2z=1−x2$ and $y=5y=5$ and situated in the first octant is given in the following figure. Find the volume of the solid. 233. The midpoint rule for the triple integral $∭Bf(x,y,z)dV∭Bf(x,y,z)dV$ over the rectangular solid box $BB$ is a generalization of the midpoint rule for double integrals. The region $BB$ is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum $∑i=1l∑j=1m∑k=1nf(xi–,yj–,zk–)ΔV,∑i=1l∑j=1m∑k=1nf(xi–,yj–,zk–)ΔV,$ where $(xi–,yj–,zk–)(xi–,yj–,zk–)$ is the center of the box $BijkBijk$ and $ΔVΔV$ is the volume of each subbox. Apply the midpoint rule to approximate $∭Bx2dV∭Bx2dV$ over the solid $B={(x,y,z)\|0≤x≤1,0≤y≤1,0≤z≤1}B={(x,y,z)\|0≤x≤1,0≤y≤1,0≤z≤1}$ by using a partition of eight cubes of equal size. Round your answer to three decimal places. 234. [T] 1. Apply the midpoint rule to approximate $∭Be−x2dV∭Be−x2dV$ over the solid $B={(x,y,z)\|0≤x≤1,0≤y≤1,0≤z≤1}B={(x,y,z)\|0≤x≤1,0≤y≤1,0≤z≤1}$ by using a partition of eight cubes of equal size. Round your answer to three decimal places. 2. Use a CAS to improve the above integral approximation in the case of a partition of $n3n3$ cubes of equal size, where $n=3,4,…,10.n=3,4,…,10.$ 235. Suppose that the temperature in degrees Celsius at a point $(x,y,z)(x,y,z)$ of a solid $EE$ bounded by the coordinate planes and $x+y+z=5x+y+z=5$ is $T(x,y,z)=xz+5z+10.T(x,y,z)=xz+5z+10.$ Find the average temperature over the solid. 236. Suppose that the temperature in degrees Fahrenheit at a point $(x,y,z)(x,y,z)$ of a solid $EE$ bounded by the coordinate planes and $x+y+z=5x+y+z=5$ is $T(x,y,z)=x+y+xy.T(x,y,z)=x+y+xy.$ Find the average temperature over the solid. 237. Show that the volume of a right square pyramid of height $hh$ and side length $aa$ is $v=ha23v=ha23$ by using triple integrals. 238. Show that the volume of a regular right hexagonal prism of edge length $aa$ is $3a3323a332$ by using triple integrals. 239. Show that the volume of a regular right hexagonal pyramid of edge length $aa$ is $a332a332$ by using triple integrals. 240. If the charge density at an arbitrary point $(x,y,z)(x,y,z)$ of a solid $EE$ is given by the function $ρ(x,y,z),ρ(x,y,z),$ then the total charge inside the solid is defined as the triple integral $∭Eρ(x,y,z)dV.∭Eρ(x,y,z)dV.$ Assume that the charge density of the solid $EE$ enclosed by the paraboloids $x=5−y2−z2x=5−y2−z2$ and $x=y2+z2−5x=y2+z2−5$ is equal to the distance from an arbitrary point of $EE$ to the origin. Set up the integral that gives the total charge inside the solid $E.E.$

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