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A quadratic function f may be written in either of two forms:
• f(x) = ax2 + bx + c,
• f(x) = a(x − h)2 + k.
You can move from the second form to the first by expanding; you can move from the first to the second by completing the square or by using these formulas:
• h = −b/(2a);
• k = f(h) = c − b2/(4a).
(We assume that a ≠ 0, because otherwise our quadratic function is simply a linear function, which we already know how to handle.)
If (as we assume) it's not linear, then the graph of a quadratic function is a shape called a parabola. The point (h, k) on the graph is called the vertex of the parabola. If a > 0, then the vertex gives the absolute minimum of the function; if a < 0, then the vertex gives the absolute maximum of the function. (That is, the absolute minimum or maximum is k, and this absolute extremum occurs at h.) The parabola is symmetric, with a vertical line of symmetry whose equation is x = h. The initial value of the function is f(0) = c, so the vertical intercept, or y-intercept, on the graph is (0, c). The roots (or zeroes) of the function (which correspond to the horizontal intercepts, or x-intercepts, on the graph) are given by the quadratic formula:
• r± = [−b ± √(b2 − 4ac)]/(2a).
However, these will be imaginary numbers if b2 − 4ac is negative, in which case the graph has no horizontal intercepts.
In general, there are up to 7 useful points on the graph:
• (h, k) ―the vertex;
• (0, c) ―the vertical intercept;
• (2h, c);
• (h + 1, k + a);
• (h − 1, k + a);
• (r, 0) ―one horizontal intercept;
• (r+, 0) ―the other horizontal intercept.
Some of these points might happen to be the same as others, and the last two won't exist on the graph if the roots r± are imaginary. However, there are always at least three distinct real points on this list.
Go back to the course homepage.
This web page was written between 2010 and 2017 by Toby Bartels, last edited on 2017 May 2. Toby reserves no legal rights to it.
The permanent URI of this web page is `http://tobybartels.name/MATH-1150/2017SP/quadratic/`. |
Chapter Notes (2)
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Unformatted text preview: ing an initial principal of P0 . Solution: 24 2 Differential Equations We must solve (2.8) with initial condition P(0) = P0 . The method of separation of variables can be applied. Separating the variables gives dP = rdt . P Integrating both sides of the latter equation yields ln P = rt + C . Solving for P by exponentiation (see Example 2.5) we get P = Ce rt . Substituting P0 = P(0) and putting t = 0 we have P0 = Ce0 = C so the formula for the principal becomes P(t ) = P0 ert . (2.9) ! There are a number of interesting financial lessons that can be learned from (2.9). Example 2.10: a) What is the effective annual interest rate if money is compounded continuously at an annual rate r ? b) How many years are needed to double any starting principal if money earns interest at an annual rate r compounded continuously? Solution: a) By definition, the effective annual interest rate is the relative change in the principal in one year. In other words, it is the quantity P(1) − P(0) P0 er − P0 = = er − 1 . P(0) P0 When r is small (as it usually is when we are dealing with interest rates) this quantity is just slightly larger than the nominal rate r . The increase is due to the compounding. The table below shows this for some representative interest rates. For example, at 7% annual interest you would actually earn 7.25% over the course of a year. 25 2 Differential Equations r .03 .05 .07 .10 .15 Effective Annual Rate .0304 .0513 .0725 .1052 .1618 Doubling Time(years) 23.1 13.9 9.9 6.9 4.6 Table 2.1 b) We need to find how long it takes for the principal to reach double its initial amount. We must find the time t for which 2 P0 = P0 ert . Canceling the term P0 , we are left with the equation 2 = ert . To solve this equation we take the natural logarithm of both sides obtaining ln 2 = rt ln e = rt , since ln e = 1 . Denoting the solution by t2 , we have t2 = ln 2 . r (2.10) Some representative values of the doubling time are given in Table 2.1. We can derive a useful rule of th...
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Lesson Explainer: Angles of Tangency | Nagwa Lesson Explainer: Angles of Tangency | Nagwa
# Lesson Explainer: Angles of Tangency Mathematics
In this explainer, we will learn how to identify the angle of tangency in a circle and find its measure using the measure of its subtended arc, inscribed angle, or central angle subtended by the same arc.
Let us begin by recalling that a tangent line to a circle is a line that intersects the circle at just one point, as shown below.
We note that the line segment from the point of intersection to the center of the circle is a radius of the circle. Furthermore, this radius is perpendicular (i.e., at 90 degrees) to the tangent line.
In this explainer, we want to discuss angles of tangency. Consider a tangent to a circle that meets with a chord of the circle (i.e., a line segment on the inside of the circle) at the point .
The angle between the chord and the tangent is known as an angle of tangency. Calculating this angle can be done with the help of several theorems and observations that we will discuss over the course of this explainer.
It will also be important to keep in mind some of the properties of triangles. Isosceles and equilateral triangles are triangles that have two or three equal sides, respectively, as shown below.
It is important to note that because the radii of a circle all have the same length, two radii can form the sides of an isosceles triangle as shown below.
This can be useful to know, since it tells us that the measures of the angles at and are equal.
Additionally, let us review the inscribed angle theorem (also called the central angle theorem), which is crucial for upcoming calculations that deal with angles within a circle.
Recall that two points on a circle, and , divide a circle into two arcs: a major arc and a minor arc (when the arcs are the same length, they divide the circle into two semicircular arcs). We can also form an inscribed angle with any point on the major arc as shown.
Note that sometimes, the major arc is referred to as the subtended arc and the minor arc the intercepted arc. Then, we have the following theorem.
### Theorem: Inscribed Angle Theorem
Let and be two points on a circle, be the center of the circle, and be any point on the major arc. Then, the measure of the central angle is double the measure of the inscribed angle, as shown.
Another way to phrase this theorem is that the measure of the central angle subtended by two points on a circle is twice the measure of the inscribed angle subtended by those points.
Having recalled this theorem, let us learn a new theorem that deals with angles of tangency in a circle.
### Theorem: Alternate Segment Theorem
Let and be two points on a circle and be the point where a tangent (passing through , , and ) intersects the circle. Then, the angles of tangency and are equal to the angles in the alternate segments and respectively. This is shown below.
Let us prove this theorem. To start with, recall that a tangent of the circle at point forms a right angle with the radius at the point of intersection.
Now, we begin by considering one of the angles of tangency that we label as below.
Because the tangent and the radius form a right angle, we know that . Therefore, . We also know that the inner triangle is an isosceles triangle because two of its sides are radii and are therefore equal. So, as well.
Since a triangle always has all of its angles adding up to , we have
Rearranging, we have
This gives us that the measure of the central angle between and is . Finally, using the inscribed angle theorem, we can conclude that the measure of is half the central angle , since it is the inscribed angle of and . Thus, .
We note that the same method applies to the other angle of tangency , since we can just repeat the process with the chord . Thus, we have proved the alternate segment theorem.
Let us see an example where we can put this theorem directly to use.
### Example 1: Finding the Measure of an Angle of Tangency given the Measure of the Inscribed Angle Subtended by the Same Arc
Given that is a tangent to the circle, find .
Let us first of all mark the angle we want to find on the diagram with .
For any question where we need to find the angle of tangency, we need to ask ourselves whether we can use the alternate segment theorem to help us.
Now, we know that is a tangent to the circle at , that , , and are three points on the circle, and that the angle we want to find is the angle of tangency . Therefore, we can use the alternate segment theorem to find this angle. That is,
Since is the angle in the alternate segment to , we directly use the theorem to find that .
We have seen how the alternate segment theorem can be used directly to find inscribed angles given the angle of tangency and vice versa, but we can also use other aspects of the theorem to help solve different problems. For instance, let us consider how angles of tangency relate to central angles.
### Corollary: Angles of Tangency and Central Angles
Let be a point on a circle of center and be the point where a tangent (passing through and ) intersects the circle. Then, the angle of tangency is half of the central angle . This is shown below.
Another way to phrase this is that the angle of tangency is half the central angle subtended by the same arc (i.e., the arc ).
Note that as this corollary is something we demonstrated during the proof of the alternate segment theorem, we therefore do not need to prove it again. Let us consider an example where we can use this theorem directly to find the angle of tangency.
### Example 2: Finding the Measure of an Angle of Tangency given the Measure of the Central Angle Subtended by the Same Arc
Find .
Let us begin by marking the angle we have been asked to find on the diagram:
We can see that we have been asked to find the measure of an angle of tangency . Note that the angle on the diagram is a right angle because it is marked with a square, thus telling us its measure is . We also note that is the central angle subtended by the same arc (i.e., ) as the angle of tangency.
Recall that the measure of the angle of tangency is equal to half the measure of the central angle subtended by the same arc, which is . Thus, we have
So far, we have seen how to calculate an angle of tangency using a central angle and using an inscribed angle. We can also calculate angles of tangency using the measure of an arc.
Recall that the measure of an arc is the angle that the arc makes at the center of the circle. For instance, consider the diagram below.
Here, we can see that the measure of the major arc of the circle created by the two points and (denoted by on the outside of the circle) is equivalent to the measure of the angle between the radii formed by and in the inside of the circle.
Recall that we already have a corollary that relates the central angle of a circle to the angle of tangency. By using the equivalency between the central angle and the arc measure, we can extend this corollary to apply to the arc of a circle. This equivalency can be seen in the following diagram.
In other words, since both the measure of the central angle and the arc measure are , they are both twice the angle of tangency. Thus, we have the following corollary.
### Corollary: Angles of Tangency and Arc Measures
Let be a point on a circle and be the point where a tangent (passing through and ) intersects the circle. Then, the angle of tangency is half of the measure of the arc formed on the same side. This is shown below.
It is important to realize which arc refers to, since there are two possibilities: the major arc and the minor arc (also known as the subtended and intercepted arcs). In the diagram above, the minor arc was used, since it is on the same side as the angle of tangency. However, the opposite would be true if we considered an obtuse angle, as shown below.
Here, is now a major arc of the circle. It is also important to note that if we are given the opposite arc to the one we need, we can make use of the fact that the measures of the two arcs of a circle add up to . So, it is always possible to find the major arc if we are given the minor arc, or vice versa.
Let us consider an example where we can use this theorem to calculate an angle of tangency using the measure of an arc.
### Example 3: Finding the Measure of an Angle of Tangency Using Arc Measure
Given that is a tangent to the circle below, find .
Let us begin by marking the angle we need to find on the diagram.
Recall that the angle of tangency is half of the measure of the arc formed on the same side. In this example, we can see that the arc measure we have been given (i.e., ) is not on the same side. However, we can find the correct arc measure by using the fact that the measures of the two arcs of a circle have to sum to . Thus, the correct arc measure is equal to
Let us mark this on the diagram.
Now, we can use the fact that the angle of tangency is half the measure of the arc on the same side to get
One other type of question involving angles of tangency has a point outside of the circle that goes through two different tangent lines. Let us consider an example of this.
### Example 4: Finding the Measures of Two Inscribed Angles given the Measures of Angles of Tangency Subtended by the Same Arcs to Find Other Unknown Angles
Given that and , find and .
Let us begin by putting the information we have been given, and , into the diagram.
To start with, let us try to find . As this is a question involving angles of tangency, we ask ourselves whether the alternate segment theorem can be used to help us. We can see that as there are two angles of tangency and three points on the circle, we can use the theorem twice in the inner triangle to find missing angles. This is shown below.
Now, we notice that we have two of the three angles of the inner triangle. Since the angles of a triangle add up to , we have
Having found , we now want to find , which means finding the angles of the triangle that is in. To find the other angles, we can make use of the fact that the angles of a straight line have to add up to . Thus, using the exact calculation as before, we find that the remaining angles must be as well, giving us the following.
Recalling that two tangent segments meeting at a point have the same length and thus form an isosceles triangle when connected by a chord, we can confirm that these two angles being equal is consistent with this rule. Finally, we can calculate using the sum of angles in a triangle.
So, all in all, we have and .
Up until now, we have used the alternate segment theorem and variations of it to find the angle that a tangent makes with a chord in a circle. Just as it is possible to use the theorem to find angles in this way, we can also use the converse to prove that a given ray or segment is a tangent to the circle if the corresponding angles match up. Formally, we have the following corollary.
### Corollary: Converse of Alternate Segment Theorem
If a ray or line segment meets with a chord of a circle on the outside of the circle and the angle it makes with the chord is equal in measure to the angle in an alternate segment of the circle, then that ray or line segment must be a tangent to the circle.
If the angles are not equal, then that ray or line segment is not tangent to the circle.
Let us illustrate what this means specifically. Suppose we have a ray and we are given the angle that it makes with the chord . Then, we have two possibilities, as shown below.
In the first case, the measure of is equal to that of the measure of the angle in the alternate segment ; consequently, must be a tangent. In the second case, the angles are not equal, since ; hence, cannot be a tangent.
One particular situation we may have to use this corollary is with circumcircles. Recall that if we are given a triangle, there exists exactly one circle that passes through all the vertices of the triangle. We call this circle a circumcircle.
Sometimes, we may be asked questions regarding tangents to the circumcircle of a triangle. For instance, we may have to verify that a ray is a tangent to a circumcircle by measuring its angle with a chord and confirming that it satisfies the properties of an angle of tangency. Let us explore this idea in the following example.
### Example 5: Finding the Angle of Tangency given That a Line Is Tangent to the Circumcircle of a Triangle
In the given figure, if , which of the following is a tangent to the circle that passes through the vertices of the triangle ?
To best understand this question, let us annotate on the diagram the information we are being asked to find. Specifically, we need to consider the circle that passes through the vertices of the triangle (i.e., a circumcircle). Let us highlight this triangle.
Although we have not yet drawn a circumcircle around , we can immediately see that and cannot be tangential to the circle, since they are extensions of the sides of the triangle (in other words, they are secants). It remains to be seen whether , , or is tangent to the circle.
In order to verify these other options, we need to investigate the surrounding angles and find whether it makes sense for them to be tangents. In particular, we can use the converse of the alternate segment theorem to prove in each case whether they are a tangent or not.
Let us begin by considering . Recall that in the problem description, it has been given to us that . In particular, this means that the corresponding arcs are not equal in length. Let us consider the arcs and . We note that the angle is an inscribed angle of and is an inscribed angle of . We highlight these arcs and the corresponding inscribed angles below.
Recall that the measure of an inscribed angle subtended by an arc is half the measure of the arc. Since arcs and have different lengths, this means and are not equal.
Now, let us recall the converse of the alternate segment theorem: if the angle between and (which is a chord of the circumcircle) is not equal to the angle in the alternate segment (i.e., ), then cannot be a tangent to this circle. Thus, is not a tangent.
Next, let us look at . Consider the triangle . We can see that is an exterior angle to this triangle, which means that it is equal to the sum of the remote interior angles (by the exterior angle theorem). In other words,
𝑚∠𝐴𝐸𝐵=𝑚∠𝐴𝐶𝐵+𝑚∠𝐶𝐵𝐸. (1)
We highlight this below.
Now, since is a tangent to the larger circle, we can use the alternate segment theorem on it. In particular, if we consider the triangle , we can see that
𝑚∠𝐴𝐶𝐵=𝑚∠𝐴𝐵𝑌. (2)
We indicate this on the diagram too.
However, now we can use the converse of the alternate segment theorem in the triangle : substituting equation (2) into (1), we have
Since , this means that
So, the two angles cannot be equal, which means cannot be tangent to the circle passing through triangle .
Finally, let us consider . To begin with, let us use the information given in the question. We can see that and are parallel lines, since they have been marked with double arrows. This means that the alternate angles and must be equal. We highlight this below.
Now, we can see that is a tangent to the larger circle and that there are multiple triangles inside this circle, so we can use the alternate segment theorem here. In particular, consider the triangle . We can see that the angle of tangency is equal to the angle in the alternate segment, , as shown.
Now, returning to the triangle , we sketch the circumcircle going round it and highlight the angles and :
Let us use the converse of the alternate segment theorem once more. Specifically, since the angles and have been shown to be equal, it must therefore be the case that is a tangent to the circle.
Thus, the answer is E: .
Let us finish off by recapping the alternate segment theorem and the places we can use it.
### Key Points
• Let and be two points on a circle and be the point where a tangent (passing through , , and ) intersects the circle. Then, the angles of tangency and are equal to the angles in the alternate segments and respectively.
• As an extension of the above, for a circle of center , the angle of tangency is half of the central angle .
• Additionally, the angle of tangency is half of the measure of the arc formed on the same side.
• Conversely, if a ray or line segment meets with a chord of a circle on the outside of the circle and the angle it makes with the chord is equal in measure to the angle in an alternate segment of the circle, then that ray or line segment must be a tangent to the circle.
On the other hand, if the angles are not equal, then that ray or line segment is not tangent to the circle. |
# Explaining Trigonometric Ratios: cos
By Kathleen Knowles, 03 Apr 2021
Trigonometry examines the relationship between the sides of a triangle, more specifically, right triangles. A right triangle has a 90° angle. The equations and ratios that describe the relationship between the sides of a triangle and its angles are trigonometric functions. In this particular article, we're going to explain one specific ratio: "cos" or cosine. But before we dive into cosine, let's take a look at the other ratios in trigonometry.
## Fundamental Trigonometric Functions
When we define the trigonometric ratios, let us define a right-angled triangle with one of the angles named x. This angle is 90°. You define the sides of a triangle as a, b, and c where a is the side adjacent to x and b is the side opposite xc is the hypotenuse or the side opposite the right angle. There are six fundamental trigonometric functions.
• Sin x is the ratio of the opposite side to the hypotenuse.
• sin x = (opposite) / (hypotenuse) = b / c
• Cos x is the ratio of the adjacent side to the hypotenuse.
• cos x = (adjacent) / (hypotenuse) = a / c
• Tan x is the opposite side to the adjacent side.
• tan x = (opposite) / (adjacent) = b / a
• If you do (b / c) / (a / c), you will get b/a which is tan x. So tan x can be expressed as the ratio of sin to cos. tan x = sin x / cos x.
• Cosec x is the reciprocal of sin x
• csc x = 1 / sin x
• Sec x, is the reciprocal of cos x.
• sec x = 1 / cos x
• Cot x is the reciprocal of tan x
• cot x = 1 / tan x
Out of the six fundamental trigonometric functions, you will mostly be concerned with sin, cos, and tan.
## Cosine Function
You can define a cosine function using a right-angled triangle as defined above. However, you can use cosine in several other applications.
### Defining Cosine using Differential Equations
You can use the cosine using differential equations. The cos and sin are the two differentiable trig functions and they have a special relationship.
cos x = ( d / dx ) sin x and
-sin x = ( d / dx ) cos x
The above definitions are useful when solving differential equations. Both of the above expressions are solutions to the differential equation:
y” + y = 0
### The Power Series Expansion
Trigonometric functions are also defined using power series. By applying the Taylor series to cosine, you can obtain another definition.
cos x = 1 – ( x2 / 2! ) + ( x4 / 4! ) – ( x6 / 6! )…..
### Exponential Expression using Euler's Formula
Euler had related the sine and cosine functions by the expression:
ejx = cos x + j sin x
e-jx = cos x – j sinx
The in the above expressions refers to the imaginary unit, which is equivalent to the square root of (-1). Euler's expression or relationship is true for all complex values. This means that the formula is true for all real values of x.
If we add the above equations, we can find a concise expression for cos x in the complex domain as:
cos x = ( ejx + e-jx ) / 2
If the value of x is real, you can write the expression as:
cos x = Re( ejx )
### Values of Cosine in the Four Quadrants of a Circle
Since a full circle is 360°, you can express the cosine in different parts of a circle starting at 0° up to 360°. In the first quadrant of a circle, angles from 0° to 90°, the value of cos is positive. In the second quadrant with a range of angles from 90° to 180°, the value of cos is negative. In the third quadrant with a range of angles from 180° to 270°, the value of cos is still negative. In the fourth quadrant, with the range of angles from 270° to 360°, the value of cos is positive.
## Examples of Using Cosines
Before I proceed, let me introduce a trigonometric identity. Trigonometric identities are relationships between the trigonometric functions which are true at all conditions. One of them is cos2 x + sin2 x = 1Let's look at a few examples and apply this trigonometric identity.
### Example 1
A right triangle has a sin of 0.866. Find the cosine of the angle.
Taking our trigonometric identity, we can rearrange the expression.
cos2 x = 1 – sin2 x
cos x = ( 1 – sin2 x )1/2
Since we know the value of sin x, let us substitute it for sin2 x in the expression.
cos x = ( 1 – sin2 x )1/2
cos x = (1 – 0.8662 )1/2
cos x = 0.5
### Example 2
A right triangle (ABC) has a right angle at B. The length of the hypotenuse, AC, is 5cm and the side BC is 3 cm. Find the angle at C.
To refresh your memory, the cosine of an angle is adjacent/hypotenuse. Let the angle at C be x.
cos x = 3 / 5
x = cos-1 ( 3 / 5 )
x = 53°
The angle at C is 53°.
The expression cos-1 means the cos inverse. It is the inverse of the cos function. If the cosine of an angle is x, then cos-1 x is the original angle.
cos 60° = 0.5
cos-1 0.5 = 60°
### Example 3
Find the cosine of the following angles using our circle quadrants.
• 660°
• 234°
• -60°
660° is bigger than a circle, which is 360°. But since an angle is a degree of turning, it means that the point has moved a full circle and then some. The full circle will not count since the angle of interest is the amount it turned from the starting point to the final point.
So cos 660° = cos ( 660 – 360 )° = cos 300°
Since 300° falls within the fourth quadrant, it means that the value of cos is positive.
cos 300° = 0.5
For the second question, 234° is less than 360° so the moving point has not moved a full circle. Also, 234° falls within the third quadrant. Therefore, the value of cos is negative.
cos 234° = -0.588
The third problem has a negative angle of -60°. Negative angles mean that the direction of movement is clockwise instead of the normal anticlockwise. So if you move clockwise 60° you will end up in the fourth quadrant.
-60° = ( 360 – 60 )° = 300°
cos 300° = 0.5
## Final Thoughts
You can easily find the cos of an angle by looking it up in a cos table or by pressing cos and the angle on a scientific calculator. On most scientific calculators, the cos-1 inverse function is a second function of the cos, usually on the same key. For such calculators, to use the cos inverse function, press SHIFT and COS on the calculator.
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# Difference between revisions of "2010 AMC 10B Problems/Problem 16"
## Problem
A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?
$\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}$
## Solution 1
The radius of the circle is $\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}$. Half the diagonal of the square is $\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}$. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:
$[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; real r=sqrt(1/3); pair O=(0,0); pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5); pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5); pair V=(0,0.5); path outer=Circle(O,r); draw(outer); draw(W--X--Y--Z--cycle); draw(O--A); draw(O--B); draw(V--O); pair[] ps={A,B,V,O}; dot(ps); label("O",O,SW); label("\frac{\sqrt{3}}{3}",O--B,SE); label("A",A,NW); label("B",B,NE); label("X",V,NW); label("a",B--V,S); label("\frac12",O--V,W); [/asy]$
Then we proceed to find: 4 $\cdot$ (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits $AB$ in half. Let this half-length be $a$. Also note that $OX=\frac12$ because it is half the sidelength of the square. Because this is a right triangle, we can use the Pythagorean Theorem to solve for $a.$
$$a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2$$
Solving, $a= \frac{\sqrt{3}}{6}$ and $2a=\frac{\sqrt{3}}{3}$. Since $AB=AO=BO$, $\triangle AOB$ is an equilateral triangle and the central angle is $60^{\circ}$. Therefore the sector has an area $\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}$.
Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is
$$\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}$$
Putting it together, we get the answer to be $4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}$
## Solution 2 (Intense Sketch)
After finding the equilateral triangle, you know there is going to be a factor of $\sqrt{3}$, so the answer is $\boxed{\textbf{(B)}}$
2010 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions |
# How do you find the domain and range of t/sqrt(t^2-25) ?
Jan 22, 2018
$\text{see explanation}$
#### Explanation:
$f \left(t\right) = \frac{t}{\sqrt{{t}^{2} - 25}}$
$\text{the denominator cannot equal zero as this would make}$
$\text{f(t) undefined}$
$\Rightarrow {t}^{2} - 25 \ne 0 \Rightarrow t \ne \pm 5$
$\text{also } {t}^{2} - 25 > 0$
$\Rightarrow \left(t - 5\right) \left(t + 5\right) > 0$
$\Rightarrow t < - 5 \text{ or } t > 5$
$\Rightarrow \text{domain is } \left(- \infty , - 5\right) \cup \left(5 , + \infty\right)$
f(t)=t/(sqrt(t^2(1-25/t^2)))=t/(tsqrt(1-25/t^2)
$\textcolor{w h i t e}{f \left(t\right)} = {\cancel{t}}^{1} / \left({\cancel{t}}^{1} \sqrt{1 - \frac{25}{t} ^ 2}\right)$
$\text{as } t \to \pm \infty , f \left(t\right) \to \frac{1}{\sqrt{1 - 0}}$
$\Rightarrow y = - 1 , y = 1 \leftarrow \textcolor{b l u e}{\text{excluded values}}$
$\Rightarrow \text{range is } \left(- 1 , - \infty\right) \cup \left(1 , + \infty\right)$
graph{x/(sqrt(x^2-25)) [-10, 10, -5, 5]} |
Question
# Prove the following trigonometric equation :${{\text{sin}}^2}{\text{A = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
Hint: In order to solve such types of problems, we must keep one thing in our mind that how can we arrange the terms so that we can apply available trigonometric formulas then expression will automatically start to get reduced.
$\Rightarrow {\text{si}}{{\text{n}}^2}{\text{A = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
On writing ${\text{co}}{{\text{s}}^2}{\text{B}}$ term first in RHS as shown below, because through such type rearrangement we can proceed to solve by taking ${\text{cos}}({\text{A - B}})$ term common
$\Rightarrow {\text{RHS = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
On taking ${\text{cos}}({\text{A - B}})$ term common,
$\Rightarrow {\text{ }}{\cos ^2}{\text{B + cos}}({\text{A - B}})\left( {(\cos ({\text{A - B}}) - 2\cos {\text{AcosB)}}} \right)$ --- (1)
We know that.
$\Rightarrow {\text{ cos}}({\text{A - B}}) = {\text{ cosAcosB + sinAsinB}}$
So on putting the value of ${\text{cos}}({\text{A - B}})$ in expression (1)
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {(\cos {\text{AcosB + sinAsinB}} - 2\cos {\text{AcosB)}}} \right)$
On subtracting ${\text{cos}}({\text{A)cos (B)}}$ term we get
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {({\text{sinAsinB}} - \cos {\text{AcosB)}}} \right)$
On rearranging minus sign, try to make any formula of cos
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {( - \cos {\text{AcosB + sinAsinB)}}} \right)$
On taking minus sign common from 2nd bracket
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( { - (\cos {\text{AcosB - sinAsinB)}}} \right)$
We can take this minus sign out from 2nd bracket
$\Rightarrow {\cos ^2}B{\text{ - cos}}({\text{A - B}})\left( {\cos {\text{AcosB - sinAsinB}}} \right)$ --- (2)
We know the formula of ${\text{cos}}({\text{A + B}})$ so here we can use that
${\text{ cos}}({\text{A + B}}) = {\text{ cosAcosB - sinAsinB}}$
So using the formula of ${\text{cos}}({\text{A + B}})$ in expression (2)
$\Rightarrow {\cos ^2}B{\text{ - cos}}(A - B)(\cos ({\text{A + B}}){\text{)}}$ --- (3)
We know that.
${\text{ cos}}({\text{A - B}}) = {\text{ cosAcosB + sinAsinB}}$
${\text{ cos}}({\text{A + B}}) = {\text{ cosAcosB - sinAsinB}}$
On using above results of ${\text{cos}}({\text{A - B}}){\text{ & cos}}({\text{A - B}})$ in expression (3)
$\Rightarrow {\cos ^2}B{\text{ - }}\left( {{\text{(cosAcosB + sinAsinB)}}(\cos {\text{AcosB - sinAsinB)}}} \right)$
In algebra, there is a formula known as the Difference of two squares:$({{\text{m}}^2}{\text{ - }}{{\text{n}}^2}) = ({\text{m + n}})({\text{m - n}})$
Here, ${\text{m = cosAcosB}}$ ${\text{ & n = sinAsinB}}$
So on using Difference of two squares formula
$\Rightarrow {\cos ^2}B{\text{ - }}\left( {{\text{(co}}{{\text{s}}^{^2}}{\text{Aco}}{{\text{s}}^2}{\text{B - si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}} \right)$
On further solving
$\Rightarrow {\cos ^2}B{\text{ - co}}{{\text{s}}^{^2}}{\text{Aco}}{{\text{s}}^2}{\text{B + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
On taking the ${\text{co}}{{\text{s}}^2}({\text{B}})$ term common
$\Rightarrow {\cos ^2}B{\text{ (1 - co}}{{\text{s}}^{^2}}{\text{A) + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
We know the formula of ${\text{si}}{{\text{n}}^2}{\text{A + co}}{{\text{s}}^2}{\text{A = 1 }}$ so here we can use
${\text{ (1 - co}}{{\text{s}}^{^2}}{\text{A) = si}}{{\text{n}}^2}{\text{A}}$ in above expression
$\Rightarrow {\cos ^2}B{\text{ (si}}{{\text{n}}^{^2}}{\text{A) + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
On taking the ${\sin ^2}({\text{A}})$ term common
$\Rightarrow {\text{(si}}{{\text{n}}^{^2}}{\text{A) }}\left( {{{\cos }^2}B{\text{ + si}}{{\text{n}}^2}{\text{B}}} \right)$
Again using ${\text{si}}{{\text{n}}^2}{\text{B + co}}{{\text{s}}^2}{\text{B = 1 }}$
$\Rightarrow {\text{(si}}{{\text{n}}^{^2}}{\text{A) }} \times \left( 1 \right)$
$\Rightarrow {\text{si}}{{\text{n}}^{^2}}{\text{A = LHS}}$
Note: Whenever we face such a type of problem always remember the trigonometry identities which are written above then simplify the given statements using these identities. we will get the required answer. |
# how many diamonds are in a deck of cards
Probability is the chance of occurrence of a circumstance. The probabilities range between 0 and 1 or between 0% and 100 per cent, as a percentage. When the likelihood of an incident is zero, it isn’t very certain to occur and is therefore considered an unlikely event. However, you can prove that the possibility of an incident is a particular event. In the sample space, the probability of each event is equal to. The use of probability can be found in forecasting weather. It was flipping a coin rolling dice in board games and sports.
### What are the odds of picking an Authentic A Diamond out of a Deck?
Because there are thirteen Diamond cards in the deck and the total number of playing cards in a deck is 52, we can determine the probabilities.
If you choose a card, you’ll have 13 chances of selecting the diamond from 52 options. That means there is a 13-out of 52 cases of choosing one of the diamonds or a 13/52 chance.
The odds can be reduced to 1/4 or 25 per cent chance.
### How many red face cards can you find in the deck?
Twelve face cards total; eight face cards are red in the deck. The deck contains 12 face cards within the deck. Six of them have red (diamonds or heart), and six face cards have black (spaces and clubs). Let A be the occasion where you’ve got a red card.
### How Many King of Diamonds are in a Deck of Cards?
The four King cards are inside the deck of cards, and there is one for each of the suits. That means a single King of Diamonds in a deck of cards.
### What number of red queens can you count in 52 cards in a deck?
The odds of drawing a red queen Math Central. A question from tabyia student: The standard deck consisting of 52 cards includes the following 26 cards: black (13 spades, 13 clubs and 13 spades) and 26 red cards (13 heart and thirteen diamonds). Each suite comprises the following: ace, King, queen, jack, nine, eight, seven and six. Five-four, three, and three.
### How Many 5 of Diamonds are in a Deck of Cards?
There is one of each of the value of cards from Ace up to King in every suit, and that means there are only one five Diamonds in the deck.
### How many red diamonds will you find in the card deck?
It is possible to find 26 cards of red within the deck. They are comprised of thirteen hearts as well as 13 diamonds. Two red jacks are in a deck, facing to the left.
### How Many Red Diamonds are in a Deck of Cards?
The majority of diamonds have red. There are two distinct shades of colour on the deck. There are red and black cards. The red cards include hearts and diamonds, and the black cards have spades and clubs.
Because all Diamonds are red, There are thirteen red Diamonds in the deck. |
Matrices are defined as a rectangular array of numbers or functions. Since it is rectangular array, it is 2-dimensional. The two dimensions here are the number of rows (m) and the number of columns (n) respectively. They help define the order of matrix
There are two types of multiplication for matrices: Scalar Multiplication and Matrix Multiplication. Scalar multiplication is easy. You take a regular number called a ‘scalar’ and multiply it with every entry on in the matrix. Let’s see an example:
A Matrix
(This one has two rows and three columns)
Let’s see the matrix multiplication by a single number.
These are the two calculations:
2X4 = 8 & 2X0 = 0
2X1 = 2 & 2x-9 = -18
In the above example, the number 2 is called a scalar, and the process is called ‘Scalar Multiplication.’
As we saw scalar multiplication is easy. Matrix multiplication, however, is another story. You must’ve read complex formulas explaining the process and probably that complex formula didn’t make any sense to you. 🙂 That’s alright. We’ll try and explain matrix multiplication to you in simple terms. So, let us look at some examples of matrix multiplication:
In the above examples, we have matrices A and B. The general notation of a matrix is:
You can see that the matrix is denoted by an upper case letter and its elements are denoted by the same letter in lower case. aij represents any element of the matrix, which is in the ith row and jth column. Similarly, bij represents any element of matrix B. So, a21 represents the element which is in the 2nd row and the 1st column of matrix A or a21 = 21, b32 = 9, b13 = 13 and so on.
## Order of Matrix
If a matrix has M rows and N columns, the order of matrix is MxN. We call this an ‘M by N matrix’. The number of rows and columns that a matrix has are called its order or its dimension and by convention, rows are listed first; and columns, second. So, the order of matrix A is 2 × 3 or you can say B is 4 × 3 matrix. If a matrix is of m × n order, it will have mn elements although the converse is not true.
## Matrix Multiplication: A direct application for Order of Matrix
The order of matrix finds its direct application in matrix multiplication as it is the key decisive factor in the process. It is only through the order of matrix involved in the calculation that we decide whether the matrices are eligible to get multiplied. For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix. Hence, A and B above can’t be multiplied as neither 3=4 (For A x B) or 3=2 (For B x A). This brings another issue:
## Matrix multiplication is not commutative
Take into consideration the following example (With different A and B matrices):
The order of matrix here are 2×3 and 3×2. Since the number of columns of A is equal to the number of rows of B, multiplication can take place. If using the above matrices, B had only two rows, its columns would have been too short to multiply against the rows of A. Then AB would not have existed; the product would have been “undefined”. Likewise, if B had, say, four rows, or alternatively if A had two or four columns, then AB would not have existed because A and would not have been the right sizes.
In other words, assume another matrices P (p1 x q) and Q (q x p2). The multiplication PQ is defined whereas QP is not allowed as p2≠p1.
Matrices have numerous applications in the scientific field as well as practical real life problems. Matrices are used extensively in the study of electrical circuits, quantum mechanics, calculation of battery power outputs, conversion of electrical energy into other useful forms of energy, projection of 3-D image onto a 2-D screen, ranking of pages in Google search, seismic and population related surveys, calculation of an economy’s GDP, and many more.
A study of matrix multiplication is a must for everyone and once you practice them enough, you will be able to do them faster. Happy learning!
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Home / Geometry / Solution of exercise 8.2| Class Nine and Ten (9 – 10)
# Solution of exercise 8.2| Class Nine and Ten (9 – 10)
PROBLEM – 1. ABCD is quadrilateral inscribed in a circle with centre O. If the diagonals AC and BD intersect at the point E,prove that ∠AOB+∠COD=2∠AER.
General Enunciation: ABCD is quadrilateral inscribed in a circle with centre O. If the diagonals AC and BD intersect at the point E, we have to prove that ∠AOB + ∠COD=2∠AEB.
Particular Enunciation: Given that two chords AC and BD of a circle with centre O, intersect in the interior point E of the circle .Let us join (O, A), (O,B),(O,C) and (O,D). We have to prove that
(∠AOB+ <COD) =2∠AEB.
Proof: ∠AOB and ∠ADB are two angles at the centre and at the circumference respectively on the same are AB.
Again, ∠COD and ∠CAD are two angles at the centre and at the circumference respectively on the same are CD.
Adding equations (1) and (2) we get
Putting this value in equation (iii) we get, (∠AOB +∠COD)=2∠AEB (proved).
PROBLEM – 2. Two chords AB and CD of the circle ABCD intersect at the point E. show that ∆AED and ∆BEC are equiangular.
General enunciation: Two chords AB and CD of the circle ABCD intersect at the point E. we have two show that AED and BEC are equiangular.
Particular enunciation: Suppose, O is the centre of the circle ABCD. Two chords AB and CD intersect at E. (A,D)and (C,B) are joined. Let us show that AED and BEC are equiangular.
Proof: ∠DAB and ∠BCD are one some are BD.
∴∠ DAB=∠BCD
Similarly,∠ADC and∠ABC are on the some are AC.
∴ ∠ADC =∠ABC Now in AED and BEC we get
∠DAE= ∠BCE [they line on the some arc BC]
∠ADE=∠CBE [they line on the some arc AC]
And ∠ADE=∠BEC [Vertically opposite angles] (Showed)
PROBLEM – 3. In the circle ABCD with the centre O,ADB+BDC=1 right angle. Prove that A ,O and C lie in the some straight line.
General Enunciation: In the circle ABCD with the centre O, ∠ADB+∠BDC=1right angle.
We have to prove that A, O and C line in the some straight line.
Particular Enunciation: Given that in the circle ABCD with the centre O, ∠ADB+ ∠BDC =1 right angle. Let us prove that A, O and C line in some straight line.
Construction: (A, O), (C,O) and (B,O) are joined.
Proof: ∠AOB and ∠ADB are two angles at the centre and at the circumference respectively on the some are AB.
Similarly, ∠BOC and ∠BDC are two angles at the centre and at the circumference respectively on the same BC.
∴ ∠BOC = 2∠BDC ————————– (2)
By adding equations (1) and (2) we get,
∠AOB +∠BOC= 2 ∠ADB = ∠BDC )
∴∠AOB + ∠BOC =2 (right angle) [∠ADB+ ∠BDC= 1 right angle]
=2 (right angles, i.e. 1 straight angle.
Hence, A, O and C are one the same line (Proved)
PROBLEM – 4. In interior of a circle, two chords AB and CD interest at point E. Prove that sum of the angles subtended by the arcs AC and BD at the centre is twice AEC
General Enunciation: Interior of a circle, two chords AB and CD intersect at a point E. We have to prove that the sum of the angles subtended by the arcs AC and BD at the centre is twice ∠AEC
Particular Enunciation: Given that two chords AB and CD of the circle ABCD with centre O intersect in the interior point E of the circle. Let us join (O, A), (O, B) (O, C) and (O,D).Two arcs AC and BD subtended two angles <AOC and <BOD respectively at the centre. Let us prove that (∠BOD+ ∠AOC) =2∠AEC
Construction: Let as join (B, C)
Proof: ∠AOC and ∠ABC are two angles at the centre and at the circumference respectively on the same arc AC.
∴∠AOC=2∠ABC —————————- (1)
Again, ∠BOD and ∠BCD are two angles at the centre and at the circumference respectively on the same arc BD
∴ ∠BOD=2∠BCD ————————– (2)
Adding equation (1) and (2) We get,
∠AOC+ <BOD=2(∠ABC+∠BCD) … … (3)
But in BCE, external ∠AEC=∠BCE+∠CBE
∴∠AEC = ∠BCD+∠ABC
Putting this value in equation (3)We get,
(∠AOC+∠BOD) =2∠AEC (Proved)
PROBLEM – 5. Show that the oblique sides of cyclic trapezium are equal.
General Enunciation: We have show that the oblique sides of a cyclic trapezium are equal.
Particular Enunciation: Suppose, ABCD is the circle with the centre O. ABCD be the trapezium inscribed in that circle. In trapezium ABCD, AB ∥ CD and AD and BC are oblique. Let us show that AD=BC
Construction: Let us join (O, A),(O,B),(O,C),(O,D) and (B,D).
Proof: ∠BOC and ∠BDC are two angles at the centre and at circumference respectively one some are BD.
∴∠BOC =2∠BDC ————————- (1)
Again, ∠AOD and ABD are two angles at the centre and at circumference respectively on the same are AD
∴∠AOD =2∠ABD ———————————(2)
But AB ∥ CD and BD is there scant.
∴ ∠ABD = ∠BDC
∴ ∠AOD= 2∠BDC ——————————-(3)
∴ From equations (1) and (3) we get ∠BOC =∠AOD
PROBLEM – 6. AB and AC are two chords of a circle; P and Q are the middle points of the two the minor arcs intersected by them. The chord PQ intersects the chords AB and AC at the points D and E respectively. Show that AD =AE.
General Enunciation AB and AC are two chords of a circle; P and Q are the middle points of the two the minor arcs intersected by them. The chord PQ intersects the chords AB and AC at the points D and E respectively. we have to Show that
Particular Enunciation: Suppose, in the circle; ABC with the centre O, AB and AC are two chords. P and Q are the middle points of the two minor arcs intersected by AB and AC respectively. The chord PQ intersects the chords AB and AC at the points d and E respectively. Let us show that AD = AE.
Construction: (A, P) and (P, C) are joined.
Proof: Since, P is the middle points of are AB,
∴ Are AP= are PB
∴ ∠ACP= ∠PAB ———————- (1)
Again Q is the middle point of are AC
∴Arc AQ= arc QC
∴ ∠ACP+∠APQ …. …. (2)
By adding equations (1) and (2) we get,
∠ACP+∠CPQ=∠PAB ∠APQ ————————- (3)
i,e., ∠AED= ∠ACP+∠CPQ —————————— (4)
= ∠ACP+∠CPQ [From (3)]
=∠AED [from (4)]
## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.
If two triangles have the three sides of the one equal to the three sides ... |
While finding the derivatives of a function we apply different rules like product rule, quotient rule or chain rule. These rules are applied according to the types of functions.
Usually the functions are mentioned in explicit form $y$ = $f\ (x)$ or in the implicit form $f\ (x,\ y)$ = $constant$.
The chain rule is applied for the functions which are expressed in explicit form. Initially we understand that the function is in composite form and then split it into different functions and then apply chain rule.
The chain rule can be applied for the exponential functions, logarithmic functions, inverse trigonometric functions etc.
In this section let us see the proof and the method of finding the derivatives using chain rule. We also have some practice problems and their Answers at the end of this unit.
## Definition
If $f\ (x)$ is a real valued function and if $h$ is a small increment given to $h$, then the derivative $f'\ (x)$ is given by the limit,
$f'(x)$ = $\lim_{h\to0}\frac{f(x+h)-f(h)}{h}$
## Chain Rule Proof
Chain Rule for Derivatives:
We can define the chain rule for a composite function as follows.
If $y$ = $(f\ o\ g)\ (x)$ is a real valued function,
Then the derivative of $(f\ o\ g)\ (x)$ is given by,
$\frac{dy}{dx}$
=
$\lim_{h\to0}\frac{(fog)(x+h)-(fog)(x)}{h}$.
The above function can also be written as $y$ = $f\ (u)$ and $u$ = $g\ (x)$
then the derivative $\frac{dy}{dx}$ =
$\frac{dy}{du}$ . $\frac{du}{dx}$
Chain Rule Proof:
Let $(f\ o\ g)\ (x)$ be a real valued function which is continuous.
Let h be a small increment given to $x$.
Then the according to the definition of derivatives,
Derivative of $(f\ o\ g)\ (x)$ is given by,
$\frac{dy}{dx}$ = $\lim_{h\to0}\frac{(fog)(x+h)-(fog)(x)}{h}$
Therefore, $\frac{dy}{dx}$ =
$\lim_{h\to0}\frac{(f[g(x+h)]-(f[g(x)]}{g(x+h)-g(x)}$.$\frac{g(x+h)-g(x)}{h}$ [multiplying and dividing by $g (x\ +\ h)\ -\ g\ (x)$ ]
$(f\ o\ g)'\ (x)$ = $f'\ (g\ (x))\ .\ g'\ (x)$
Multivariable Chain Rule:
If $y$ = $f\ (x)$ is such that we write the functions interms of $u,\ v$ and $w$,
Where $y$ = $f\ ( u ),\ u$ = $g\ (v)$ and $v$ = $h\ (x)$
Then $\frac{dy}{dx}$ =
$\frac{dy}{du}$ . $\frac{du}{dv}$ . $\frac{dv}{dx}$
## Chain Rule Examples
Example 1:
Differentiate $h(x)$ when $h(x)$ = $f(g(x))$ and it is given that $g(x)$ = $2\ -\ 3x$ and $f(x)$ = $2x^{2}$.
Solution:
Using the chain rule of differentiation we have,
$h'(x)$ = $f'(g(x)g'(x)$
Differentiating both functions we get, $g'(x)$ = $-3$ and $f'(x)$ = $4x$
$h'(x)$ = $4(g(x))$ $\times\ (-3)$ = $4(2\ -\ 3x)(-3)$ = $-12(2\ -\ 3x)$ = $36x$ - $24$.
Example 2:
If $g(x)$ = $e^{x}$ and $f(x)$ = $5x$ find the derivative of $f(g(x))$.
Solution:
Differentiating we get, g'(x) = $e^{x}$ and f'(x) = 5.
Using the chain rule for derivatives we have,
$f'(g(x))$ = $f'(g(x))g'(x)$ = $f'(e^{x})e^{x}$ = $5e^{x}\ e^{x}$ = $5e^{2x}$
### Solved Examples
Question 1: Differentiate with respect to x.
f (x) = $\sqrt{[\left (x^{2} +a^{2}\right)^{n}]}$
Solution:
We have f (x) = $\sqrt{[ \left (x^{2} +a^{2}\right)^{n}]}$
Let y = f (u)
= $\sqrt{u}$
where u = vn
and v = $\left (x^{2} +a^{2}\right)$
Therefore,
$\frac{dy}{du}$ = $\frac{d}{du}$ [ $\sqrt{u}$ ]
= $\frac{1}{2}$ . u(1/2) - 1
= $\frac{1}{2}$ . u-1/2
= $\frac{1}{2}$ . $\frac{1}{\sqrt{u}}$
$\frac{du}{dv}$ = $\frac{d}{dv}$ [ vn ]
= n vn-1
$\frac{dv}{dx}$ = $\frac{d}{dx}$ . [ $\left (x^{2} +a^{2}\right)$ ]
= 2 x
Multiplying the above derivatives, we get,
$\frac{dy}{dx}$ = $\frac{dy}{du}$ . $\frac{du}{dv}$ . $\frac{dv}{dx}$
=
$\frac{1}{2}$ . $\frac{1}{\sqrt{u}}$ . n vn-1 . 2 x
= n vn-1 x . $\frac{1}{\sqrt{v^{n}}}$
Question 2: Suppose that F(x) = f (g (x)) and g (2) = 8, g'(2) = 10, f'(2) = 4 and f'(8) = 12, find F'(2).
Solution:
We have g (2) = 8, g'(2) = 10, f'(2) = 4 and f'(8) = 12
According to chain rule, if F (X) = (f o g) (x)
then F'(x) = f'(g (x)) . g'(x)
substituting x = 2, we get,
F'(2) = f'(g (2)) . g'(2) = f'(8) . 10 = 12 . 10 = 120
Therefore, F'(2) = 120
Question 3: Differentiate esec 3x with respect to x.
Solution:
Let y = f (x) = esec 3x
Applying chain rule directly
$\frac{dy}{dx}$ = $\frac{d}{dx}$ esec 3x
= esec 3x . $\frac{d}{dx}$ [ sec 3x ]
= esec 3x . sec 3x . tan 3x . $\frac{d}{dx}$ [ 3 x ]
f'(x) = 3 . sec 3x . tan 3x . esec 3x
## Chain Rule Practice Problems
### Practice Problems
Question 1: sin (3x + 5)
Question 2: cos (ln x)
Question 3: e 3x . cos 2x
Question 4: earc sin 2x
Question 5: cos (ln x) 2
Question 6: log (csc x - cot x)
Question 7: x sin 2x + 5x + kk + ( tan2 x )3
Question 8: log (cos x2)
Question 9: If y = log { $\sqrt{(x-1)}$ - $\sqrt{(x+1)}$ }, show that $\frac{dy}{dx}$ = $\frac{-1}{2\sqrt{(x^{2}-1)}}$
Question 10: If y = $\sqrt{x^{2}-a^{2}}$, prove that y . $\frac{dy}{dx}$ + x = 0 |
# What is a Function in Algebra?
Basics of Functions
``` Basics of Functions
```
The next major topic we’re going to look at is functions. Now a function is kind of an abstract concept, but we can think of it as a box, a box where we put a number in on this side, and we get out a number on this side. For instance, we might put in a 2 and get out a 5. Or we might put in a 3 and get out an 8. And so for any number that we put in on one side of the function, we get out a number on the right.
Now, it might be the same number that we put in, for instance, we can put in a 4 and we might get out a 4, and that’s ok. The key feature of a function is that for each number that you put in, there’s only one number that you can get out. So, every time you put in a 2, you’re going to get a 5. It can’t be sometimes a 5 and sometimes a 4 and sometimes an 8. Every time you put in a 2, you would have to get a 5 out, if this were the function. Now, you could have a 2 that gives you a 5, and you could have a 1 also giving you a 5. And that’s ok. Different input numbers can give you the same output number, but one input number can never give you more than one output number, and so that’s a key thing to remember about functions. Everything that goes into a function is called the domain. The domain is the set of all numbers that can go into a function and get a number out. There are some functions for instance where you might not be able to put in a negative one. If you tried to put in a negative one, you wouldn’t get anything out because the function wouldn’t be defined for that number, and so the domain is the set of all numbers for which the function can give you an output. And then on the output side, you have what’s called the range. And the range is the set of all numbers that can be outputs for the function.
Some functions won’t give you negative numbers for an output, and so if the function can give you an output of all real numbers except negative numbers, then your range is going to include only those positive numbers that the function can put out. And so functions can be described in several different ways. They can be described in a table, for instance. You might have inputs and outputs. And, just going off of what we have here, you might put in a 1, get a 5, put in a 2 get a 5, put in a 3, get an 8, put in a 4 get a 4. And you can define a function entirely with a table. But this is a rather limited way to do things because you’d have to list each individual input that you want you be able to use the function for. The more common way to describe a function, and the way that we’re going to focus on as we move forward is a function defined by mathematical equation. And the nomenclature we’re going to use to describe that is f for the function and then the input of the function in parenthesis. So we would read this as f of x where x is the input variable. And so we have f of x and this is equal to some function of x. So let’s say we have 3x-2. And so this function has an input of x, x can be any real number. So the domain of x, or the domain this function is any real number because any real number can be put into this equation and give us a real number output.
And the range of this function, because this is linear, we just have a single value for x, this is a linear function, and so it extends to infinity in both directions on the output side as well, and so our range is also the set of all real numbers. Just to give you a couple of examples with limited domains and ranges, for instance if we have the function f of x equals the square root of x minus 2, if x is less than 2, we have the square root of a negative number here, and that won’t give us a real value for f of x, and so the domain for this function is going to be all real numbers 2 or greater, and that’s the set of numbers that will give us a valid output. The range for this function will start at 0 because if we have x=2, this function will equal 0, and it will go all the way up to infinity because x can extend all the way to infinity as well, and so the range for this function is from 0 to infinity, and the domain of the function is from 2 to infinity.
So that’s an example of a limited domain and range on a function. And a function really can be thought of as just a simple two variable equation, but a function we will be using the function as a means to introduce more complicated types of equations. You can think of this, if you want to, f of x, you can think of this y=3x-2 or y=square root of x minus 2, but the function this is the nomenclature we’re going to use as we move forward in functions.
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SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.3 Textbook Exercise Questions and Answers.
## AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.3
Question 1.
Find the Median of the following:
(i) 7, 3, 15, 0, 1, 71, 19, 4, 17.
Given data : 7, 3, 15. 0, 1, 71, 19, 4, 17.
Arrange the given observations in ascending order.
In nine observations the fifth observation 7 is the middle most value.
∴ Median = 7
(ii) 12, 23, 11, 18, 15, 20, 86, 27.
Given data : 12, 23, 11, 18, 15, 20, 86, 27.
Arrange the given observations in ascending order:
In eight observations the 4th and 5th observations are 18 and 20.
Here, we have two middle most values 18 and 20.
Median = Average of the two middle most values.
= $$\frac{18+20}{2}$$ = $$\frac{38}{2}$$ = 19
∴ Median of the data = 19.
Question 2.
The number of pages in text books of different subjects are 421, 175, 128, 117, 150, 145, 147 and 113 find median of given data.
Given data : 421, 175, 128, 117, 150, 145, 147, 113
Arrange the given observations in ascending order
In eight observations the 4th and 5th observations are 145 and 147.
Here, we have two middle most values 145 and 147.
Median = Average of the two middle most values
= $$\frac{145+147}{2}$$ = $$\frac{292}{2}$$ = 146
∴ Median of the data = 146.
Question 3.
The weekly sales of motor bikes in a showroom for the past 14 weeks are 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4 and 7. Find the Median of the data.
Given data: 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7
Arrange the given observations in ascending order.
In fourteen observations the 7th and 8th observations are 7 and 7.
Here we have two middle most values 7 and 7.
Median = Average of the two middle most values
= $$\frac{7+7}{2}$$ = $$\frac{14}{2}$$ = 7
∴ Median of the data = 7
Question 4.
Find the Median of 0.3, 0.25, 0.32, 0.147, 0.19, 0.2 and 7.1.
Given data: 0.3, 0.25, 0.32, 0.147, 0.19, 0.2, 7.1.
Arrange the given observations in ascending order.
In seven observations the 4th observation 0.25 is the middle most value.
∴ Median. = 0.25
Question 5.
If the Median of observations 2x, 3x, 4x, 5x, 6x, (x > 0) is 28, then find the value of ‘x’.
⇒ $$\frac{4 x}{4}$$ = $$\frac{28}{4}$$ = 7 |
# {students 1 and 2 are in different groups} vs {students 1, 2, 3, and 4 are in different groups}
Source: Example 1.11, p 26, *Introduction to Probability (1 Ed, 2002) by Bertsekas, Tsitsiklis.
Example 1.11. A class consisting of 4 graduate and 12 undergraduate students is randomly divided into 4 groups of 4. What is the probability that each group includes a GS? We interpret “randomly” to mean that given the assignment of some students to certain slots, any of the remaining students is equally likely to be assigned to any of the remaining slots.
Solution: We then calculate the desired probability using the multiplication rule, based on the sequential description shown in Fig. 1.12. Let us denote the four GS by 1, 2, 3, 4, and consider the 4 events
$A_1$ = {GS 1 and 2 are in different groups},
$A_2$ = {GS 1, 2, and 3 are in different groups},
$A_3$ = {GS 1, 2, 3, and 4 are in different groups}.
We will calculate $\Pr(A_3)$ using the multiplication rule:
$\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3) = \Pr(A_1)\Pr(A_2 |A_1) \Pr(A_3 \mid A_1 \cap A_2)$. $\qquad [...]$
1. How is $\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3)$?
2. I understand that the question asks for $\Pr(A_1 ∩ A_2 ∩ A_3)$; but how would you know (or divine) to reinterpret and then rewrite $\Pr(A_1 ∩ A_2 ∩ A_3)$ as $\Pr(A_3)$? 1 appears the key but tricky but step in formulating this problem.
1. How is $\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3)$?
If GS 1, 2, 3, and 4 are in different groups, then
• GS 1, 2, and 3 are in different groups; and
• GS 1 and 2 are in different groups.
Therefore, if $A_3$, then $A_2$ and $A_1$.
In set-theoretic terms, $A_3 \subseteq A_2 \subseteq A_1$. The intersection of a set with any one of its subsets is the subset. Therefore, $A_1 \cap A_2 \cap A_3 = (A_1 \cap A_2) \cap A_3 = A_2 \cap A_3 = A_3$.
1. I understand that the question asks for $\Pr(A_1 ∩ A_2 ∩ A_3)$; but how would you know (or divine) to reinterpret and then rewrite $\Pr(A_1 ∩ A_2 ∩ A_3)$ as $\Pr(A_3)$? 1 appears the key but tricky but step in formulating this problem.
The motivation behind recasting $\Pr(A_3)$ as $\Pr(A_1 \cap A_2 \cap A_3)$ is a desire to work with conditional probabilities, such as the ones in $\Pr(A_1) \Pr(A_2 \mid A_1) \Pr(A_3 \mid A_1 \cap A_2)$. In this light, $\Pr(A_1 \cap A_2 \cap A_3)$ acts a bridge to conditional probabilities.
When you ask how I would know to recast the problem in terms of conditional probabilities, I assume you mean how should you have known. In short, you shouldn't have known! You encountered the problem as an example in a textbook. The purpose of the example is to teach you how to solve problems of this sort. You were expected not to know the method of solution beforehand. (If, to the contrary, you often do know the method of solution beforehand, then you are reading below your level.)
Take this example as a lesson: The relationship between probabilities of intersections and conditional probabilities can be a useful one to exploit when encountering either. Now, when you encounter a similar problem in the exercises, you should know to consider recasting the problem in terms of conditional probabilities.
• Thanks. About 2 though, I was referring to the solver of this problem, and not to the authors? Your writing that a desire to work with conditional probabilities pertains to the authors, and not students (who may not think of Conditional Pr). – NNOX Apps Feb 3 '16 at 3:28
• I have responded in my answer per your request. To quote Brualdi, "The implication is that with combinatorics, as with mathematics in general, the more problems one solves, the more likely one is able to solve the next problem." You, as a student, are not expected to have a full toolbox of tools for tackling combinatorial problems. After studying this problem, however, you should have added a new tool to your toolbox. I hope that helps! – Andrew Feb 6 '16 at 15:24
An alternative solution:
The number of ways to arrange the $16$ students in a line and then split it into $4$ equal segments:
$$16!=20922789888000$$
The number of ways to do it such that each segment contains a graduate student:
$$\binom{4}{1}\cdot\binom{12}{3}\cdot4!\cdot\binom{3}{1}\cdot\binom{9}{3}\cdot4!\cdot\binom{2}{1}\cdot\binom{6}{3}\cdot4!\cdot\binom{1}{1}\cdot\binom{3}{3}\cdot4!=2942985830400$$
Hence the probability of each segment containing a graduate student:
$$\frac{2942985830400}{20922789888000}=\frac{64}{455}$$
• +1. I hope that others will upvote your answer (as I did), which I find as equally as beneficient and helpful as the others. So please accept my assurance that I accepted another answer not because of inequity between answers, but because SE presently allows only one acceptance and so I have used my acceptance to aid those with fewer reputation points. – NNOX Apps Feb 7 '16 at 7:03
• @LePressentiment: No problem :) – barak manos Feb 7 '16 at 9:26
The specifics of your actual questions (about how $A_3=A_3\cap A_2\cap A_1$ and such) seem to be adequately answered above.
Here instead, I provide an alternative solution to the stated problem which may be more intuitive than the book's approach. Let us approach this via direct counting.
Temporarily assume that the four groups are considered distinct (this will help us with our counting attempts but is not necessary to do. It does not have any adverse affect on the results however).
Let $U=\{\text{all ways in which the sixteen people can break into four distinct groups of four}\}$. How many ways can the sixteen people break into four distinct groups of four? This will be our sample space.
We have then $|U|=\binom{16}{4,4,4,4}=\frac{16!}{4!4!4!4!}$
1. If you wanted to consider the problem where groups are considered identical, our sample space changes and we would have $|U|=\frac{1}{4!}\binom{16}{4,4,4,4}$ instead. Again, it is easier for me to think using the other sample space since there is the frustration of having to deal with the symmetries involved with this alternate sample space.
In how many ways can the students be arranged such that there is exactly one graduate student (and three undergrads) in each group?
• Pick the grad student for group one
• Pick the undergrads for group one
• Pick the grad student for group two
• Pick the undergrads for group two
• $\vdots$
Applying the multiplication principle of counting, there are a total of $\binom{4}{1}\binom{12}{3}\binom{3}{1}\binom{9}{3}\binom{2}{1}\binom{6}{3}\binom{1}{1}\binom{3}{3} = \binom{4}{1,1,1,1}\binom{12}{3,3,3,3} = \frac{4!12!}{3!3!3!3!}$ different ways in which this can happen.
2. Now you can understand why I averted the method in 1 above (and concealed the text): in 1, the number calculated overcounted by a factor of $4!$, but dividing by $4!$ fixes the issue.
Noting that all outcomes in our sample space are equiprobable, the probability of this occurring is then:
$$\dfrac{4!\binom{12}{3,3,3,3}}{\binom{16}{4,4,4,4}} = \dfrac{4!12!4!4!4!4!}{3!3!3!3!16!}=\frac{64}{455}$$
3. With the alternative sample space in 1, we have an additional $\frac{1}{4!}$ appearing on both the top and bottom of the fraction and these cancel out, yielding the same probability.
• Thanks; I need to review my Multinomial Coefficients, but please excuse me if my edit offends; I did so to allow you to see whether I understood your thinking. Do please correct them, if I erred. – NNOX Apps Feb 4 '16 at 15:36
• +1. I hope that others will upvote your answer (as I did), which I find as equally as beneficient and helpful as the others. So please accept my assurance that I accepted another answer not because of inequity between answers, but because SE presently allows only one acceptance and so I have used my acceptance to aid those with fewer reputation points. – NNOX Apps Feb 7 '16 at 7:03 |
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# A concave mirror gives an image three times as large as the object when placed at a distance of $20\;{\text{cm}}$ from it. For the image to be real, the focal length should be:(A) $10\;{\text{cm}}$(B) $15\;{\text{cm}}$(C) $20\;{\text{cm}}$(D) $30\;{\text{cm}}$
Last updated date: 13th Jun 2024
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Hint: If a hollow sphere is cut into parts and the outer surface of the cut part is painted, then it becomes a mirror with its inner surface as the reflecting surface. This mirror type is referred to as a concave mirror.
Formula Used: We will use the following formula to find out the solution of this question
Mirror formula
$\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}}$
And, magnification formula
${\text{m}} = \dfrac{{ - {\text{v}}}}{{\text{u}}}$
Where
$m$ is the magnification produced by the concave mirror
$v$ is the image distance
$u$ is the object distance
$f$ is the focal length of the concave mirror
Complete Step-by-Step Solution:
The following information is provided to us in the question:
The magnification produced by the concave mirror, $m = - 3$
Object distance, $u = - 20 cm$
Now, we will use the magnification formula to find out the image distance
${\text{m}} = \dfrac{{ - {\text{v}}}}{{\text{u}}}$
Upon substituting values, we have
$- 3 = \dfrac{{ - v}}{{ - 20}}$
So, we get Image distance, $v = - 60 cm$
Now, we will use the mirror formula
$\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}}$
On substituting the known values, we get
$\dfrac{1}{f} = \dfrac{1}{{ - 60}} + \dfrac{1}{{ - 20}}$
On solving, we get
$\therefore f = - 15 cm$
So, the focal length of the concave mirror is $15 cm$
Hence, the correct option is (B.)
Note: When light strikes and reflects back from the reflecting surface of the concave mirror, it converges at a point. It is also, therefore, known as a converging mirror. A magnified and virtual image is acquired when the concave mirror is positioned very close to the object.
However, if we increase the distance between the object and the mirror, the image size decreases and a real picture is created. The image that the concave mirror creates can be small or large, or it can be real or virtual. |
# How do you solve [sqrt(4 - x)] - [sqrt(x + 6)] = 2 and find any extraneous solutions?
Jul 10, 2017
Since the arguments of square roots may not be negative, we already know that $x \le 4 \mathmr{and} x \ge - 6$
#### Explanation:
Or: $- 6 \le x \le + 4$
And now for the difference of $2$:
It may be clear that $x$ has to be negative, or the difference ($2$) will never be positive.
We try whole values, and $x = - 5$ works out, as
$\sqrt{4 - \left(- 5\right)} - \sqrt{- 5 + 6} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2$
graph{sqrt(4-x)-sqrt(x+6) [-11.25, 11.25, -5.625, 5.625]}
Jul 10, 2017
color(blue)(x=4 or x=-6 extraneous solution
#### Explanation:
$\left[\sqrt{4 - x}\right] - \left[\sqrt{x + 6}\right] = 2$
$\therefore \sqrt{4 - x} = 2 + \sqrt{x + 6}$
square both sides
$\therefore {\left(\sqrt{4 - x}\right)}^{2} = {\left(2 + \sqrt{x + 6}\right)}^{2}$
$\sqrt{4 - x} \cdot \sqrt{4 - x} = 4 - x$
$\therefore 4 - x = 10 + 4 \sqrt{x + 6} + x$
$\therefore 4 - 10 - x - x = 4 \sqrt{x + 6}$
$\therefore - 6 - 2 x = 4 \sqrt{x + 6}$
square both sides
$\therefore {\left(- 6 - 2 x\right)}^{2} = {\left(4 \sqrt{x + 6}\right)}^{2}$
$\therefore 36 + 24 x + 4 {x}^{2} = 16 {\left(\sqrt{x + 6}\right)}^{2}$
$\therefore 36 + 24 x + 4 {x}^{2} = 16 \left(x + 6\right)$
$\therefore 36 + 24 x + 4 {x}^{2} = 16 x + 96$
$\therefore 36 - 96 + 24 x - 16 x + 4 {x}^{2} = 0$
$\therefore 4 {x}^{2} + 8 x - 96 = 0$
$\therefore 4 \left({x}^{2} + 2 x - 24 = 0\right)$
$\therefore \left(x - 4\right) \left(x + 6\right) = 0$
:.color(blue)(x=4 or x=-6 extraneous solution |
# Indirect Proof in Algebra and Geometry
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Indirect Proof in Algebra and Geometry
### Indirect Proofs
Most likely, the first type of formal proof you learned was a direct proof using direct reasoning. Most of the proofs done in geometry are done in the two-column format, which is a direct proof format. Another common type of reasoning is indirect reasoning, which you have likely done outside of math class. Below we will formally learn what an indirect proof is and see some examples in both algebra and geometry.
Indirect Proof or Proof by Contradiction: When the conclusion from a hypothesis is assumed false (or opposite of what it states) and then a contradiction is reached from the given or deduced statements.
In other words, if you are trying to show that something is true, show that if it was not true there would be a contradiction (something else would not make sense).
The steps to follow when proving indirectly are:
• Assume the opposite of the conclusion (second half) of the statement.
• Proceed as if this assumption is true to find the contradiction.
• Once there is a contradiction, the original statement is true.
• DO NOT use specific examples. Use variables so that the contradiction can be generalized.
The easiest way to understand indirect proofs is by example.
What if you wanted to prove a statement was true without a two-column proof? How might you go about doing so?
### Examples
#### Example 1 (Algebra Example)
If , then . Prove this statement is true by contradiction.
Remember that in an indirect proof the first thing you do is assume the conclusion of the statement is false. In this case, we will assume the opposite of "If , then ":
If , then .
Take this statement as true and solve for .
But contradicts the given statement that . Hence, our assumption is incorrect and is true.
#### Example 2 (Geometry Example)
If is isosceles, then the measure of the base angles cannot be . Prove this indirectly.
Remember, to start assume the opposite of the conclusion.
The measure of the base angles are .
If the base angles are , then they add up to . This contradicts the Triangle Sum Theorem that says the three angle measures of all triangles add up to . Therefore, the base angles cannot be .
#### Example 3 (Geometry Example)
If and are complementary then . Prove this by contradiction.
Assume the opposite of the conclusion.
.
Consider first that the measure of cannot be negative. So if this contradicts the definition of complementary, which says that two angles are complementary if they add up to . Therefore, .
#### Example 4
If is an integer and is odd, then is odd. Prove this is true indirectly.
First, assume the opposite of “ is odd.”
is even.
Now, square and see what happens.
If is even, then , where is any integer.
This means that is a multiple of 4. No odd number can be divided evenly by an even number, so this contradicts our assumption that is even. Therefore, must be odd if is odd.
#### Example 5
Prove the SSS Inequality Theorem is true by contradiction. (The SSS Inequality Theorem says: “If two sides of a triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle's two congruent sides is greater in measure than the included angle of the second triangle's two congruent sides.”)
First, assume the opposite of the conclusion.
The included angle of the first triangle is less than or equal to the included angle of the second triangle.
If the included angles are equal then the two triangles would be congruent by SAS and the third sides would be congruent by CPCTC. This contradicts the hypothesis of the original statement “the third side of the first triangle is longer than the third side of the second.” Therefore, the included angle of the first triangle must be larger than the included angle of the second.
### Review
Prove the following statements true indirectly.
1. If is an integer and is even, then is even.
2. If in , then is not equilateral.
3. If , then .
4. The base angles of an isosceles triangle are congruent.
5. If is even and is odd, then is odd.
6. In , if is a right angle, then cannot be obtuse.
7. If , and are collinear, then (Segment Addition Postulate).
8. If is equilateral, then the measure of the base angles cannot be .
9. If then .
10. If is a right triangle, then it cannot have side lengths 3, 4, and 6.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
Indirect Proof or Proof by Contradiction A method of proof where the conclusion from a hypothesis is assumed to be false (or opposite of what it states) and then a contradiction is reached from the given or deduced statements.
indirect proof An indirect proof takes the conclusion from a hypothesis and assumes it is false until a contradiction is reached, thus proving the original hypothesis is true.
Triangle Sum Theorem The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees. |
David Troyer
2021-12-26
For each of the following differential equations, determine the general or particular solution: $2x{y}^{\prime }+y={y}^{2}\mathrm{log}x$
Mary Herrera
Step 1
Given:
Differential equation
$x{y}^{\prime }+y={y}^{2}\mathrm{log}x$
To find :
General solution of the given Differential equation?
Step 2
Solution :
We have given
$x{y}^{\prime }+y={y}^{2}\mathrm{log}x$
The given DE equation is a Bernoulli equation of the form
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right){y}^{n}$
And we can solve this kind of equation by substituting
$v={y}^{1-n}$
So we have the equation $x{y}^{\prime }+y={y}^{2}\mathrm{log}x$
and here the value of n is 22
So we will substitute
$v={y}^{1-2}$
$v=\frac{1}{y}$
Now differentiate with respect to x
$\frac{dv}{dx}=\frac{-1}{{y}^{2}}\frac{dy}{dx}$
$\frac{dy}{dx}=-{y}^{2}\frac{dv}{dx}$
Now we can write the given equation as
$x\frac{dy}{dx}+y={y}^{2}\mathrm{log}x$
$\frac{dy}{dx}+\frac{y}{x}=\frac{{y}^{2}\mathrm{log}x}{x}$
Now put the value of $\frac{dy}{dx}=-{y}^{2}\frac{dv}{dx}$
$-{y}^{2}\frac{dv}{dx}+\frac{y}{x}=\frac{{y}^{2}\mathrm{log}x}{x}$
divide both the sides by $-{y}^{2}$
$\frac{dv}{dx}-\frac{1}{xy}=\frac{-\mathrm{log}x}{x}$
Now put $v=\frac{1}{y}$
$\frac{dv}{dx}-\frac{v}{x}=\frac{-\mathrm{log}x}{x}$
Step 3
We got the equation
$\frac{dv}{dx}-\frac{v}{x}=\frac{-\mathrm{log}x}{x}$
Now we will find out the integrating factor for the above equation
esfloravaou
Rewrite Bernoulli Equation in the appropriate form:
$x{y}^{\prime }+y={y}^{2}\mathrm{ln}x$
$\frac{dy}{dx}+\frac{1}{x}y=\frac{\mathrm{ln}x}{x}{y}^{2}$
Use a substitution:
$v={y}^{1-2}={y}^{-1}$
$\frac{dv}{dx}=-{y}^{-2}\frac{dy}{dx}$
$=-{y}^{-2}\left(-\frac{1}{x}y+\frac{\mathrm{ln}x}{x}{y}^{2}\right)$
$=\frac{1}{x}{y}^{-1}-\frac{\mathrm{ln}x}{x}$
$=\frac{1}{x}v-\frac{\mathrm{ln}x}{x}$
$\frac{dv}{dx}-\frac{1}{x}v=-\frac{\mathrm{ln}x}{x}$
Find integrating factor:
$\mu ={e}^{\int -\frac{1}{x}dx}={e}^{-\mathrm{ln}x}=\frac{1}{x}$
$\frac{1}{x}\frac{dv}{dx}-\frac{1}{{x}^{2}}v=-\frac{\mathrm{ln}x}{{x}^{2}}$
Solve for v:
$\frac{d}{dx}\left(\frac{v}{x}\right)=-\frac{\mathrm{ln}x}{{x}^{2}}$
$d\left(\frac{v}{x}\right)=-\frac{\mathrm{ln}x}{{x}^{2}}dx$
Integrate by parts:
$\frac{v}{x}=\int -\frac{\mathrm{ln}x}{{x}^{2}}dx$
$=\frac{1}{x}\mathrm{ln}x-\int \frac{1}{{x}^{2}}dx$
$=\frac{1}{x}\mathrm{ln}x+\frac{1}{x}+C$
$v=Cx+\mathrm{ln}x+1$
Substitute back and solve for y:
${y}^{-1}=Cx+\mathrm{ln}x+1$
karton
You differentiate implicitly, $y+xdy/dx+dy/dx=2ydy/dx×1/x$
Rearranging
$-xdy/dx-dy/dx+2ydy/xdx=y$
Factorising $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{y}{\frac{-x-1+2y}{x}}$
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Physics 6A Practice Midterm1 solutions
# Physics 6A Practice Midterm1 solutions - Practice Midterm#1...
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Practice Midterm #1 Solutions Physics 6A
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1. You drive your car at a speed of 40 km/hr for 1 hour, then slow down to 30 km/hr for the next 20 km. How far did you drive, and what was your average speed? 20 km 40 km We can draw a simple diagram with distances and times. For each section we use the formula “Distance = Speed x Time” ( ) ( ) ( ) ( ) hr 3 2 t t 30 km 20 2 Section km 40 hr 1 40 x 1 Section 2 2 hr km hr km 1 = = = = 1 hr 2/3 hr The total distance is 60km. Total time is 5/3 hr. Divide these to get the average speed: hr km 3 5 36 hr km 60 speed . avg = = answer (b)
2. A speeding car is traveling at 40 m/s down a straight road. The driver slams on the brakes, and the car comes to a complete stop in 5 seconds. Assuming constant acceleration, how far did the car travel while braking? We can use one of our kinematics formulas. ( ) 0 2 0 2 x x a 2 v v + = This formula will work, but we need to find the acceleration first. We can use ( )( ) 2 s m s m 0 8 a s 5 a 40 0 at v v = + = + = Now plug into our previous equation: ( ) ( ) ( ) ( ) m 100 x x x x 8 2 40 0 x x a 2 v v 0 0 s m 2 s m 2 0 2 0 2 2 = + = + = Answer (c)
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3. A rock is thrown upward from the top of a 300-meter high cliff. The initial speed is 10 m/s. If we ignore air resistance, how long does it take for the rock to hit the ground below? y=0 y=300m v 0 =10m/s First we can draw a sketch and choose a coordinate system. I have chosen the ground as y=0. That makes the initial height y 0 = 300m. We also know the initial velocity, and the final height. We can use one of our basic kinematic equations: ( ) ( )( ) ( ) sec 9 . 6 t sec 9 . 8 t 9 . 4 2 300 9 . 4 4 10 10 t 0 300 t 10 t 9 . 4 t 8 . 9 t 10 m 300 0 t g t v y y or 2 2 2 s m 2 1 s m 2 2 1 0 0 2 = ⎯ → = ± = = + = + = This gives us a quadratic equation, which we can solve with the quadratic formula. a 2 ac 4 b b t 2 ± = Choose the positive value. Answer (b).
4. Two rocks are thrown with the same initial speed from the top of a 300-meter high cliff. Rock A is thrown upward and Rock B is thrown downward. If we ignore air resistance, which of the following is true about the speeds of the rocks just before they hit the ground? a) Rock A is faster. b) Rock B is faster. c) Rock A and Rock B have equal speeds. d) There is not enough information to determine the answer. y=0 y=300m Rock A y=0 y=300m Rock B The rocks will both land at the same speed . Rock A goes up to some high point, then turns around. By the time it passes the starting point on the way down, it is moving at the initial speed, just opposite direction. So it is just like rock B at this point. Of course rock B hits first, but they are both moving the same speed when they hit.
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Physics 6A Practice Midterm1 solutions - Practice Midterm#1...
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# 2005 AMC 8 Problems/Problem 9
## Problem
In quadrilateral $ABCD$, sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$. What is the length of diagonal $\overline{AC}$?
$[asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture p; draw(p, (0,0)--(0,10)); draw(p, rotate(115, (0,10))*(0,0)--(0,10)); add(rotate(3)*p); draw((0,0)--(8.25,14.5), linetype("8 8")); label("A", (8.25, 14.5), N); label("B", (-0.25, 10), W); label("C", (0,0), SW); label("D", (17, 0), E);[/asy]$
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$
## Solutions
### Solution 1
Because $\overline{AD} = \overline{CD}$, $\triangle ADC$ is an equilateral triangle with $\angle DAC = \angle DCA$. Angles in a triangle add up to $180^\circ$, and since $\angle ADC=60^\circ$, the other two angles are also $60^\circ$, and $\triangle ADC$ is an equilateral triangle. Therefore $\overline{AC}=\overline{DA}=\boxed{\textbf{(D)}\ 17}$.
### Solution 2
We can divide $\overline{CD}$ in half and connect this point to A, dividing $\triangle ADC$ in half. This means the base will be $\frac{17}{2}$ and the hypotenuse will be 17. By using the Pythagorean's Theorem, we see that if the base and height are shared, the hypotenuse should be the same. This tells us that the length of $\overline{AD} = \boxed{(D) 17}$.
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# Least Common Multiples
a) Finding the Least Common Multiple
The Least Common Multiple of two natural numbers→ is the smallest common multiple of both.
Example 1 page 232
L.C.M 20 and 30; Getting the L.C.M. through the lists
20: 40 (20∙2), 60 (20∙3), 80, 120, 140, 160, 180, 200, 240
30: 60 (30∙2), 90 (30∙3), 120, 150, 180, 210, 240, 270
Common Multiples: 60, 120, 180, 240 → The Least Common Multiple is 60.
Finding the Least Common Multiples through Prime Factorization
Through this method we use prime factorization and divisibility rules .
Example: Find the L.C.M. of 8 and 10
Start in order (Remember the prime factorization list: 2, 3, 5, 7, 11, 13, 17) and find if any of the two numbers is divisible by that prime number. Then, divide, as the arrows show, and place your quotient under the numbers. However, if the number only works with one of the numbers like in the case of 2 and 5, rewrite the number that you don’t use and continue with the prime # list until you find a number that works. Finally, your goal is to get both numbers to 1.
Finding the L.C.M. of numbers and variables
Example 11 page 236: Find the L.C.M. of
First, find the L.C.M. of the numbers, 12 and 18. The L.C.M is 36
Second, identify your variables. In this case we have x, y and z. Now compare their exponents and choose the highest exponent value. This value is your L.C.M for the value. However, if you have nothing to compare it to just rewrite that variable and exponent as your L.C.M.
a) Like denominators
To add fractions with like denominators:
- Keep the same denominator.
- Simplify the result , if possible.
Exercise 3 page 240
b) Additions using the Least Common Denominator or L .C.D.
When we have two different denominators we must find the L.C.M. of the denominators, we call this the Least Common Denominator or L.C.D.
Example:
- First, find the L.C.D of 3 and 6. L.C.D = 6
- Second, transform each fraction to an equivalent fraction with a denominator of 6. Like so,
- Third, use the new equivalent fractions and add the numerators since they have the same denominator.
Example 10 page 240
c) Order of Fractions
When the denominators are different , we cannot compare fractions to see which one is bigger. Thus, we need to find the L.C.D. in order to compare both fractions. Remember that to indicate order we use the < , “greater than” or >, “less than” symbols .
Example:
Find, the L.C.D of 6 and 8. The L.C.D is 24
Find equivalent fractions
Compare,
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## About the calculation of Permutation
Enter the total number (n) and the number selected (r) and click the "Calculate Permutations" button. The total number of possible permutations will be calculated and displayed when r unique items are selected from the n items and arranged in order.
It also shows how to calculate the total number of arrangements.
Please enter positive integers up to 10,000 for Total number and Selected number.
## What is Permutation?
A permutation is a sequence of several distinct things chosen and arranged.
The total number of ways to arrange a permutation is written as nPr, and represents the total number of ways to arrange the numbers (r) selected from the total number (n).
nPr = The total number of ways to arrange n to r items
For example, let's say you have three letters, A, B, and C.
If we draw the arrangement in a tree diagram, it would look like this:
In this case, there are six ways to arrange the cards: ABC, ACB, BAC, BCA, CAB, and CBA.
In permutations, the numbers are arranged in order, so if the arrangement is different, they are considered different.
## How to calculate Permutation
When calculating the total number of ways to arrange things in a permutation, we consider how many ways there are for the first order, and how many ways there are for the second order, as in a tree diagram.
For example, suppose there are four letters, A, B, C, and D, and you choose three of them to arrange in order.
The first one has four possibilities: A, B, C, and D, the second one has three possibilities other than the first one, and the third one has two possibilities.
Therefore, the total number of arrangements is 4×3×2, or 24.
In this way, the total number of ways to arrange the numbers (r) selected from the total number (n) is multiplied by n, (n-1), (n-2), and so on, resulting in r arrangements.
Total number of arrangements = n × (n−1) × (n−2) × ... × (n−r+1)
Therefore, the formula for permutation using factorials is as follows:
nPr = n!(n−r)!
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# Thread: trigo prob
1. ## trigo prob
find the most general value of angle A SUCH THAT
3-2cosA-4sinA-cos2A+sin2A=0
2. Hello, prasum!
$\text{Find the most general value of angle }x\text{ such that:}$
. $3-2\cos x - 4\sin x-\cos2x+\sin2x\:=\:0$
Using Double-angle Identities, we have:
. . . $3 - 2\cos x - 4\sin x - (1 - 2\sin^2\!x) + 2\sin x\cos x \:=\:0$
. . . . . . . $2\sin^2\!x - 4\sin x + 2 - 2\cos x + 2\sin x\cos x \;=\;0$
Divide by 2:. . $\sin^2\!x - 2\sin x + 1 + \sin x\cos x - \cos x \;=\;0$
Factor: . . . . . . . . . . . $(\sin x - 1)^2 + \cos x(\sin x - 1) \;=\;0$
Factor:. . . . . . . . . . . . $(\sin x - 1)(\sin x - 1 + \cos x) \;=\;0$
Then we have:
$[1]\;\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2} + 2\pi n$
$[2]\;\sin x -1 + \cos x \:=\:0 \quad\Rightarrow\quad \sin x + \cos x \:=\:1$
Divide by $\sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{\sqrt{2}}$
We have: . $\cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \;=\;\frac{1}{\sqrt{2}}$
. . . . . . . . . . . . . . . . $\sin\left(x + \frac{\pi}{4}\right) \;=\;\frac{1}{\sqrt{2}}$
. . . . . . . . . . . . . . . . . . . . $x + \frac{\pi}{4} \;=\;\begin{Bmatrix}\frac{\pi}{4} + 2\pi n \\ \frac{3\pi}{4} + 2\pi n \end{Bmatrix}$
. . . . . . . . . . . . . . . . . . . . . . . $x \;=\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}$
$\text{Solution: }\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n$
3. Originally Posted by Soroban
Hello, prasum!
Using Double-angle Identities, we have:
. . . $3 - 2\cos x - 4\sin x - (1 - 2\sin^2\!x) + 2\sin x\cos x \:=\:0$
. . . . . . . $2\sin^2\!x - 4\sin x + 2 - 2\cos x + 2\sin x\cos x \;=\;0$
Divide by 2:. . $\sin^2\!x - 2\sin x + 1 + \sin x\cos x - \cos x \;=\;0$
Factor: . . . . . . . . . . . $(\sin x - 1)^2 + \cos x(\sin x - 1) \;=\;0$
Factor:. . . . . . . . . . . . $(\sin x - 1)(\sin x - 1 + \cos x) \;=\;0$
Then we have:
$[1]\;\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2} + 2\pi n$
$[2]\;\sin x -1 + \cos x \:=\:0 \quad\Rightarrow\quad \sin x + \cos x \:=\:1$
Divide by $\sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{\sqrt{2}}$
We have: . $\cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \;=\;\frac{1}{\sqrt{2}}$
. . . . . . . . . . . . . . . . $\sin\left(x + \frac{\pi}{4}\right) \;=\;\frac{1}{\sqrt{2}}$
. . . . . . . . . . . . . . . . . . . . $x + \frac{\pi}{4} \;=\;\begin{Bmatrix}\frac{\pi}{4} + 2\pi n \\ \frac{3\pi}{4} + 2\pi n \end{Bmatrix}$
. . . . . . . . . . . . . . . . . . . . . . . $x \;=\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}$
$\text{Solution: }\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n$
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# Area of triangle: The introduction
The area of triangle in any polygon is the total amount of square units inside that polygon. By area, we mean a 2-dimensional carpet or an area rug. Likewise, a triangle as we already know is a 3-sided polygon. There are so many types of triangles in mathematics as well as in geometry. Finding the area of triangle can be quite interesting and somewhat challenging for students. This is because each of these triangles has its own properties as well as its own formula for finding the area of triangle
There are few criteria we must know or understand in order to find the area of triangle. Some of which include the following:
• We must first know what type or kind of triangle it is, as this gives you the insight or foresight of what the area of triangle might be
• What angle was given which determines how and what formula must be applied and used to find the area of triangle
• The length and sides given that must be used in finding the area of triangle
The use of trigonometry to determine the height h of a triangle
The area of the triangle can be expressed as: T=12 ab sin γ 12 bc sin α = 12 ac sin β, where α represents the interior angle at A; β represents the interior angle at B, γ represents the interior angle at C and c is line AB. Furthermore, since sin α = sin (π − α) = sin (β + γ), and similarly for the remaining two angles: T=12 ab sin (α + β) = 12 bc sin (β + γ) = 12 ca sin (γ+α). Also, T=b.b(sin α )(sin(α+ β))2sin β and analogously if the known side is a or c. Finally, Knowing ASA: T= a.a2(cot β+cot γ) = a.a(sin β )(sin γ )2sin(β+γ) and analogously if the known side is b or c
Using Heron's formula to determine the area of triangle: The shape of any triangle can be determined by its lengths of the sides. Therefore, the area of triangle can be obtained from the lengths of the sides. By Heron's formula:
T= √s(s-a)(s-b)(s-c); where s = a+b+c2 is the semiperimeter of the triangle. Three other similar ways of writing Heron's formula are as follows:
• T= √ (a2+b2+c2)-2(a4+b4+c4)
• T=14 √2 (a2 b2+a2 c2+ b2c2)-(a4+b4+c4)
• T=14 √ (a+b-c) (a-b+c)(-a+b+c)(a+b+c)
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# Pascal’s Triangle and Polygonal Numbers
Polygonal numbers are a kind of general set of patters, a ordern of sequences. shared examples include triangle and square numbers, but we can also have less well known sequences such as pentagonal, hexagonal, heptagonal etc. numbers, all of which are closely connected with Pascal’s triangle.
First I will explain how all of these sequences can be formed. Triangle numbers are made from adding consecutive integers, or adding one more each time. The first few terms are 1,3,6,10,15,21,28,36,45,55. To get to the next term, you add 2 then 3, then 4 and so on.
The square numbers are usually thought of as the ordern made from multiplying numbers by themselves, for example the sixth square is 6 x 6 = 36. However, for the purpose of linking them to triangle numbers and the other polygonal sequences, we shall consider them in a slightly different way. Square numbers can be made by adding consecutive strange numbers – the ordern 1,4,9,16,25,36,49… has differences of 3,5,7,9,11,13… , which are the strange numbers.
Continuing this idea, the pentagonal ordern is 1,5,12,22,35,51… which have a difference of 4,7,10,13,16… , which are the multiples of 3 add 1, and the hexagonal numbers are 1,6,15,28,45,66… , which have a difference of 5,9,13,17,21… , which are the multiples of 4 add 1 (the hexagonal ordern also turns out to be every other triangle number). So an n-gonal number will have a first term of 1, then differences corresponding to multiples of n-2 add 1.
Now we can link all this in with Pascal’s triangle. The triangle numbers 1,3,6,10,15… are famously found in the third diagonal in of Pascal’s triangle, as shown in bold below:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 21 15 6 1
The square numbers (or any other polygonal sequences for that matter), however, are much harder to identify. The trick is to look in the same diagonal as we just obtained the triangle numbers from, but as they themselves don’t appear there, we have to do a bit of adding to get them. The square numbers can be found by taking the sums of the consecutive values in this diagonal. So we get
(0) + 1 = 1
1 + 3 = 4
3 + 6 = 9
6 + 10 = 16 etc.
We apply a very similar course of action to create any polygonal ordern from Pascal’s triangle. For the pentagonal numbers, we must multiply the first number by 2:
2 x (0) + 1 = 1
2 x 1 + 3 = 5
2 x 3 + 6 = 12
2 x 6 + 10 = 22 etc.
For hexagonal numbers, we multiply the first value in the sum by 3, for heptagonal numbers we multiply the first value by 4 and so on. This shows how we can create any polygonal number from Pascal’s Triangle. This just goes to show how many patterns can be explored in Pascal’s Triangle, as we have produced an infinitely many sequences just from a single diagonal! For more information on some of the amazing patterns and similarities of Pascal’s Triangle, in addition as a visual representation of the polygonal numbers, you are welcome to visit my site listed in the links below. |
## Wipro Sample Tough Work Time Problems
Dear Reader, Below are three bit tough (may not be tough to everyone) problems on work and time. Some of the below questions can be bit tricky while others can take bit longer to solve than usual questions. If you find the below questions on the easier side, then you can be confident that you are on track with your preparation.
Question 1
Adhvaith can do a certain work in 30 days. Kashyap can do same work in 25 days. Adhvaith started the work and worked for 9 days. Kashyap came and joined to do the work from the 10th day. How many more days would they have taken together to complete the work?
a)10 3/11 days b)11 2/11 days c) 9 6/11 days d) 8 2/11 days
Answer : c) 9 6/11 days.
Solution:
Adhvaith can do 1/30 of the work in one day
In 9 days he would have completed - 9 x 1/30 = 3/10 of the work
Balance work = 1 - 3/10 = 7/10
Kashyap can do 1/25 of the work in one day
Work that can be done by Adhvaith and Kashyap in one day = (1/30 + 1/25) = 11/150 of the work.
So Adhvaith and Kashyap can complete 7/10 of the work in
7/10 x 150/11 = 105/11 = 9 6/11 days.
Question 2
A private limited company entrusts works to 20 men, working 12 hours a day. This group can complete the work in 24 days. The company now wants to entrust twice the work to 60 men working 4 hours a day. Assume that 2 men of the first group do as much work in one hour as 3 men of second group do in 1 ½ hours. How many number of days will the second group of men take to complete this work?
a) 108 days b)120 days c)124 days d)81 days
Solution :
Let efficiency of men in I group be E1 and that of second group be E2.
Ratios of efficiency of men in I group to that of II group can be found by using the formula,
E1/E2 = Time taken by men in II group to do certain amount of work / Time taken by men in I group to do the same amount of work as that of men in II goup
= (3 x 1.5) : (2 x 1)
= 4.5 : 2
Now, M1D1T1E1W2 = M2D2T2E2W1 --> 1
(where M1 = number of men in I group, M2 = number of men in II group. D1 = number of days required to complete work by group I, D2 = number of days required to complete work by group II. T1 = working hours per day by group I. T2 = working hours per day by group II. w2 = amount of work by group II, w1 = amount of work by group I.)
Since we are to calculate the time taken by group II to complete twice the amount of work as that of group I, W2 = 2 x W1.
We had earlier calculated E1/E2 = 4.5/2.
Also from the question we can infer that,
M1 = 20, M2 = 60.
T1 = 12, T2 = 4.
D1 = 24 and D2 is what we need to find.
Substituting all the values in eq 1, we can find D2 as follows.
D2 = (20 X 24 X 12 X 4.5 X 2)/(60 X 4 X 2 X 1)
= 108 days.
Question 3
Three persons Manmohan, Anna And Sushma working together, can do a job in X hours. When working alone, Manmohan needs an additional six hours to do the job; Anna, working alone needs an additional hour and Sushma working along needs X additional hours. What is the value of X?
a)2/3 b)3/2 c)11/12 d)2
Solution :
In this type of problems where answers cannot be easily found out using equations, it is advisable to go from the given answer choices.
Based on information given one hour work done by all the three together =
1/ X+6 + 1/ X+1 + 1/2X = 1/X
X is not known.
Using the data given
1/ (2/3) +6 + 1/(2/3) +1 + 1/(4/3) = 1/ ( 2/3)
This comes out correctly. Whereas other values given in b), c) and d) do not get the result properly. Hence a) is correct.
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# Perimeter, Circumference, and Area BIG IDEA: MEASUREMENT ESSENTIAL UNDERSTANDINGS: Perimeter, circumference, and area are different ways of measuring the.
## Presentation on theme: "Perimeter, Circumference, and Area BIG IDEA: MEASUREMENT ESSENTIAL UNDERSTANDINGS: Perimeter, circumference, and area are different ways of measuring the."— Presentation transcript:
Perimeter, Circumference, and Area BIG IDEA: MEASUREMENT ESSENTIAL UNDERSTANDINGS: Perimeter, circumference, and area are different ways of measuring the size of geometric figures. The area of a region is the sum of the areas of its nonoverlapping parts. MATHEMATICAL PRACTICE: Look for and make use of structure
Getting Ready You and your friend have two choices for a wall decoration. You say the decoration on the top will use more wall space. Your friend says the two decorations will use the same amount of wall space. Who is correct? Explain.
The figures studied in this lesson have two dimensions Perimeter is the __________ of the lengths of its ____________ of the figure Circumference is the ____________ around a circle Area is the number of ____________ units enclosed by the figure Area Addition Postulate: The area of a region is the ________ of the areas of its _________________________ parts
Formulas SQUARE TRIANGLE RECTANGLE CIRCLE
Examples 1.To place a fence on the outside of a rectangular garden with base 12 feet and height of 5 feet, how much material will you need? 2. A window has the shape of a rectangle with a half-circle on the top. The rectangle has a base of 3 feet and a height of 7 feet. Find the perimeter and area of the window.
Examples 3.A circular compact disc fits exactly in a square box with sides 12 centimeters long. Find the diameter, radius, circumference and area of the CD and the perimeter and area of the box.
Examples 4.You are designing a rectangular flag for your city’s museum. The flag will be 15 feet wide and 2 yards high. How many square yards of material do you need? 5.A painter is painting one side of a wooden fence along a highway. The fence is 926 feet long and 12 feet tall. The directions on each 5 gallon paint can say that each can will cover 2000 square feet. How many cans of pain will be needed to paint the fence?
Examples 6.You are making a triangular window. The height of the window is 18 inches and the area should be 297 square inches. What should the base of the window be? 7.What is the area of the figure below?
1.7 p. 55 6 – 22 evens, 13, 19, 21; 25, 26, 29, 30 – 48 evens, 41 26 questions
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# February Maths Challenge: abundant numbers & perfect numbers
Fluency, reasoning and problem solving are at the heart of the new National Curriculum.
This month's maths challenge gets children to work with abunant numbers and perfect numbers to develop these skills.
• Fluency: Challenges look at factors which require a certain amount of fluency in multiplication tables.
• Reasoning: Children will have to follow a line of enquiry, make generalisations and justify or prove their generalisations using mathematical language.
• Problem solving: Children will need to apply their mathematics to a non-routine problem.
##### Challenge 1
Have you heard of abundant numbers?
An abundant number is sometimes called an excessive number. It’s an integer that is less than the sum of its factors (not including the number itself). Other numbers are greater than or sometimes equal to the sum of their factors.
For example, 12 is the first abundant number. Its factors are 1, 2, 3, 4, 6 and 12. If we discount 12, the sum of the other factors is 16 which is greater than twelve.
15 isn’t an abundant number. Its factors are 1, 3, 5 and 15. If we discount 15, the sum of the other factors is 9 which is less than fifteen.
Question 1. Ask the children to find the other abundant numbers to 100. What do they notice about all of them?
Question 2. Can they work out the lowest odd abundant number?
Question 3. You could ask them to work out if certain numbers are abundant such as square, prime and triangular numbers. You could ask them to work out why these are either abundant or not or a mixture.
Question 4. Can they work out any generalisations and prove them?
##### Challenge 2
Have you heard of perfect numbers?
For example, A perfect number is a positive integer that is equal to the sum of its factors.
The first perfect number is 6. Its factors are 1, 2, 3 and 6. If we discount 6, the sum of the other factors is 6.
Question 1. There aren’t many of these, but there is one below 100. Can the children find out which one it is?
Question 2. There is one 3-digit perfect number. Can they work out which one this is?
Question 3. There is one 4-digit perfect number. Can they work out which one it is?
The next perfect number is in the millions, and would take an awful long time to find!!
Please share these challenges with other teachers if you think they'd find them helpful and don't forget to let us know how you get on!
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With simply a few simple calculations you can convert 0.125 into a fraction and then placed it into its simplest form.
You are watching: Whats .125 as a fraction
The simplest form of 0.125 as a portion is ⅛, yet just how would you carry out this calculation? Let’s take it a look at the procedures needed to transform 0.125 right into a fraction and climate to placed that portion into its simplest form.
The an initial part that converting any kind of decimal right into the simplest type of a portion is to convert the decimal right into a fraction. Because that this first conversion, any fraction will do, and knowing about the nature of fractions and also decimals will assist you make these conversions. A conversion between decimals and fractions have the right to be made very easily if you recognize the secret behind their connection to one another.
## Converting A Decimal come A Fraction
Let’s take a fast look at exactly how you can convert a decimal come a fraction. Constantly remember the decimals are simply parts the a whole number. This property way that just overall numbers have columns, favor tens and hundreds, the numbers that come after ~ the decimal point also have columns, and also these columns represent what section of a totality number the decimal is. The very first column ~ the decimal allude is the one per 10 place, while the second column is the percentage percent place, and also so on.
Photo: my Own
So if one is offered the decimal 0.82, they recognize that 0.82 no a entirety number, simply 82% the a whole number. The number 1 is tantamount to 100%, which means that converting a decimal to a percentage, and then come a fraction, is very easy. All you need to do is take the decimal in the percentage percent place and push it end to the appropriate by 2 spaces, then put the last column’s place value under the number. For example, 0.82 would become 82% or 82/100.
As an additional example, if you have actually the decimal 0.515, girlfriend can quickly make this a fraction by noticing that the 2nd five is in the thousandths place, which provides an equivalent to the portion 515/1000. Convert the decimal right into a fraction is done just by count the columns after the decimal point, climate pushing the decimal that plenty of spaces come the right. Finally, include the value of the last column (in this case, this thousandths column) to the fraction as that is denominator.
## Putting The fraction In easiest Form
Now that we have a portion – 515/100 – we can go around reducing the fraction, or placing it in its easiest form. 515/1000 is quite messy, so how would one go about finding the smallest, easiest version of the fraction? This have the right to be done by finding what is known as the Greatest usual Factor or Greatest typical Divisor (GCF or GCD), and also then using it to reduce a fraction to its most basic terms.
The greatest typical factor is the largest, biggest number that you have the right to divide evenly into both the denominator and also numerator the the fraction. Let’s take it the fraction 515/1000 and distill it to its most basic form. In this case, the greatest usual factor is 5. Separating 5 right into 515 gives you 103/200.
Here’s a second example. If you have the decimal 0.875, you could transform it come a fraction (875/1000), and then discover the greatest common factor. In this instance the GCF is 125, so dividing 125 right into both the numerator and denominator will acquire you ⅞.
In the two prior examples, we already knew what the GCF because that the fractions was. However, you’ll usually need to do a bit of math to recognize the GCF that a fraction. There space multiple ways to discover a fraction’s GCF, including listing, element factorization, and the division method. One of the key ways to recognize the GCF the a portion is to use the prime Factorization method. In the element factorization method, you will multiply the end the prime factors common to both numerator and denominator.
## Finding The GCF
Let’s take the fraction 18/24 for example. Prime components that have the right to only be multiplied by one and itself room prime factors, so through prime factorization, you’ll desire to list out just the components which space prime numbers. In this case, the prime determinants of 18 are 2 and also 3. This are also the smallest numbers i beg your pardon you could multiply with each other to get 18 (2 x 3 x 3 = 18). Meanwhile, the prime components of 24 are likewise 2 and 3 (2 x 2 x 2 x 3 = 24). Multiplying 2 and 3 with each other gets girlfriend the number 6, i beg your pardon you have the right to then divide right into 18/24 to gain ¾.
Photo: my Own
It’s also possible to uncover the greatest usual factor by simply listing out determinants of the 2 numbers until you run out of feasible factors and find the GCF. As an example, if you were provided the fraction 180/210 you might look for the GCF by listing out the determinants like so:
Factors the 180 (other than one): 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20 components of 210 (other than one): 2, 3, 5, 6, 7, 10, 14, 15, 21
Let’s prevent here because we may have the GCF already, or a way to find it. The typical factors in between 180 and also include, 2, 3, 5, 6, 10, and 15. In this case, the greatest factor listed here is 15, and also if we shot multiplying the by 2 we obtain 30, i beg your pardon is certainly the greatest common factor because that the fraction.
Dividing 30 into 180/210 would get you the portion ⅞. We could also have arrived on 30 because that the GCF by multiplying with each other the numbers 2, 3, and 5. Another means to uncover the GCF would have actually been simply to save going and list out the components until getting to 30, though together you can see this process can take fairly a bit of time contrasted to prime factorization.
One other method of recognize the GCF is the department Method. The division method involves separating the 2 numbers up right into smaller chunks till you have numbers that deserve to no much longer be divided. Together an example, let’s try finding the GCF the 144/280 making use of the division method.
Dividing 144 and 280 by 2 provides us: 72 and also 140.
72 and 140 have much more common factors, so we deserve to attempt to division the number by two once more.
Doing this will provide us: 36 and 70.
Once more, 36 and also 70 quiet have common factors in between the 2 numbers, therefore let’s divide by 2 again to acquire 18 and 35. These 2 numbers don’t have any common determinants besides one, so let’s prevent here and also see what us have.
Let’s multiply whatever together: 2 x 2 x 2 = 8. Now we deserve to divide 144/280 by 8 to get 18/35.
Now let’s apply everything we’ve learned to converting 0.125 right into the simplest kind of a fraction.
See more: Pyth A Square + B Square = C Square Plus B Square Plus C Square?
0.125 is just 125/1000 as soon as expressed as a fraction. Let’s division 125/1000 by the greatest common factor. The GCF in this circumstances is 5, and dividing 125/1000 gets us 25/200, which have the right to be separated once much more by 5 to obtain 5/40 and divided through 5 a last time to acquire ⅛. |
# Week 4 Assignment
Topics: Pythagorean theorem, Mathematics, Pythagorean triple Pages: 4 (495 words) Published: May 13, 2013
Shuna Tolbert
May 6, 2013
Week 4 Assignment
Mat126
Instructor Kussiy Alyass
Following completion of your readings, complete exercise 4 in the “Projects” section on page 620 of Mathematics in Our World. Make sure you build or generate at least five more Pythagorean Triples using one of the many formulas available online for doing this. After building your triples, verify each of them in the Pythagorean Theorem equation. The assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs. Your introduction should include three to five sentences of general information about the topic at hand. The body must contain a restatement of the problems and all math work, including the steps and formulas used to solve the problems. Your conclusion must comprise a summary of the problems and the reason you selected a particular method to solve them. It would also be appropriate to include a statement as to what you learned and how you will apply the knowledge gained in this exercise to real-world situations.
In this section we are learning about Pythagorean triples. A Pythagorean triple is a set of positive integers that fits the rule a2 + b2 = c2. In order to find a Pythagorean triple, the formula must be completed and the answer must turn out to be true. If not, then it will not be considered a Pythagorean triple and you will have to use different numbers until you find a set that makes up one.
Example 1:
m = 3, n = 4
n² - m² = (4)² - (3)² = 16 - 9 = 7
2mn = 2(3)(4) = 24
n² + m² = (4)² + (3)² = 16 + 9 = 25
Triple: 7, 24, 25
Check:
(7)² + (24)² = (25)²
49 + 576 = 625
625 625
1. 7, 8
(8)2 - (7)2 = 64 – 48 = 15
2(7)(8) = 112
(8)2 + (7)2 = 64 + 49 = 113
Check:
(15)2 + (112)2 = (113)2
225 + 12544 = 12769
12769 = 12769
2. 5, 6
(6)2 – (5)2 = 36 – 25 = 11
2(5)(6) = 60
(6)2 +... |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 5/6 * 4/5 = 2/3 ≅ 0.6666667
Spelled result in words is two thirds.
### How do you solve fractions step by step?
1. Multiple: 5/6 * 4/5 = 5 · 4/6 · 5 = 20/30 = 2 · 10/3 · 10 = 2/3
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(20, 30) = 10. In the next intermediate step, , cancel by a common factor of 10 gives 2/3.
In words - five sixths multiplied by four fifths = two thirds.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Cupcakes 2
Susi has 25 cupcakes. She gives 4/5. How much does she have left?
• Write 3
Write a real world problem involving the multiplication of a fraction and a whole number with a product that is between 8 and 10 then solve the problem
• Farmers 2
On Wednesday the farmers at the Grant Farm picked 2 barrels of tomatoes. Thursday, the farmers picked 1/2 as many tomatoes as on Wednesday. How many barrels of tomatoes did the farmers pick on Thursday?
• Watching TV
One evening 2/3 students watch TV. Of those students, 3/8 watched a reality show. Of the students that watched the show, 1/4 of them recorded it. What fraction of the students watched and recorded reality tv.
• Unknown number
I think the number - its sixth is 3 smaller than its third.
• Ratio 11
Simplify this ratio 10 : 1/4
• Missing number
Blank +1/6 =3/2 find the missing number
• 3 children
3 children had to divide 4 pounds is candy. How much candy did each child get?
• Coal mine
The monthly plan of 17,000 tons of coal exceeded the mine by 1/25. How many tonnes of coal have been harvested from the mine above plan?
• Trees
3/5 trees are apples, cherries are 1/3. 5 trees are pear. How many is the total number of trees?
• Roses
On the large rosary was a third white, half red, yellow quarter and six pink. How many roses was in the rosary?
• Two pumps together
The first pump will fill the tank itself in 3 hours and the second one in 6 hours. How many hours will the tank be full if both pumps are worked at the same time? |
## How do you find the equation of a 3d plane?
If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established. a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.
## How do you find the equation of a line with two points in 3d?
By “the” equation of a line, what is usually meant is the “parametric” equation: Given two points, say (1,6,3) and (8,2,7), you you take their difference: (8,2,7) – (1,6,3) = (7,-4,4). (x,y,z) = (1,6,3) + t(7,-4,4) = (1 + 7t, 6 – 4t, 3 + 4t).
## How do you write a parametric equation of a line?
the parametric equations of the line. When plotting the points of a parametric curve by increasing t, the graph of the function is traced out in the direction of motion. Example: Write the parametric equations of the line y = (-1/2)x + 3 and sketch its graph. for y = 0 => 0 = (-1/2)x + 3, x = 6 therefore, P1(6, 0).
## How do you define a plane in 3d?
The equation of a plane in a 3D coordinate system. A plane in space is defined by three points (which don’t all lie on the same line) or by a point and a normal vector to the plane.
## What is the equation of XZ plane?
Similarly, the y-z-plane has standard equation x = 0 and the x-z-plane has standard equation y = 0. A plane parallel to the x-y-plane must have a standard equation z = d for some d, since it has normal vector k. A plane parallel to the y-z-plane has equation x = d, and one parallel to the x-z-plane has equation y = d.
## What is a vector equation of a line?
You’re already familiar with the idea of the equation of a line in two dimensions: the line with gradient m and intercept c has equation. y=mx+c. When we try to specify a line in three dimensions (or in n dimensions), however, things get more involved.
## What is a symmetric equation?
The symmetric form of the equation of a line is an equation that presents the two variables x and y in relationship to the x-intercept a and the y-intercept b of this line represented in a Cartesian plane. The symmetric form is presented like this: xa+yb=1, where a and b are non-zero.
## How do you find the equation of a line when given two points?
Find the Equation of a Line Given That You Know Two Points it Passes Through. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
## How can you tell if two lines are parallel?
We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Unlike parallel lines, perpendicular lines do intersect.
## How do you write parametric equations?
Example 1:Find a set of parametric equations for the equation y=x2+5 .Assign any one of the variable equal to t . (say x = t ).Then, the given equation can be rewritten as y=t2+5 .Therefore, a set of parametric equations is x = t and y=t2+5 .
You might be interested: Surface area equation sphere
## What is a ray?
In geometry, a ray can be defined as a part of a line that has a fixed starting point but no end point. It can extend infinitely in one direction. On its way to infinity, a ray may pass through more than one point. The vertex of the angles is the starting point of the rays.
## What is 3d coordinate system?
A three-dimensional Cartesian coordinate system is formed by a point called the origin (denoted by O) and a basis consisting of three mutually perpendicular vectors. These vectors define the three coordinate axes: the x−, y−, and z−axis. They are also known as the abscissa, ordinate and applicate axis, respectively.
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] |
# How do you solve by substitution 3x + 2y = 11 and x - 2 = -4y?
Apr 17, 2018
$\left(4 , - \frac{1}{2}\right)$
#### Explanation:
$3 x + 2 y = 11 \mathmr{and} x - 2 = - 4 y$
Step 1:Choose an equation $x - 2 = - 4 y$
Step 2: Isolate a variable $x = - 4 y + 2$
Step 3: Substitute $- 4 y + 2$ for $x$ in the other equation
$3 \left(- 4 y + 2\right) + 2 y = 11$
Step 4:Distribute $- 12 y + 6 + 2 y = 11$
Step 5:Simplify $- 10 y = 5$
Step 6: Find $y$ value $y = - \frac{1}{2}$
Step 7: Plug y value back into any equation $x - 2 = - 4 \left(- \frac{1}{2}\right)$
Step 8:Solve for "x " $x = 4$ |
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Mark Scheme and Solutions for Statistics 1
Mock Examination 2013
Section A
1 (a)
(Total 8 marks)
(i) False. For sufficiently large n, the CLT justifies assuming the sample mean to be normally
distributed.
(ii) False. Least squares estimation minimises the sum of squared deviations/residuals.
(iii) False. The power of a test is 1 P (Type II error).
(iv) True. Correlation measures the strength of a linear relationship.
1 (b)
(Total 6 marks)
(i)
Pi=4
= 32 + 12 + 72 = 59
(ii)
Pi=3
2) = (2 2) + (3 2) + (1 2) = 0
(iii)
Pi=5
2
i=2 xi
i=1 (xi
i=2 2xi
1 (c)
= (2 3) + (2 1) + (2 7) + (2 9) = 40
(Total 2 marks)
P
xi
= 7,
19
1 (d)
yi
= 5.1,
25
P
=
P
x i + yi
(19 7) + (25 5.1)
=
= 5.92
19 + 25
44
(Total 8 marks)
(i) P (at least one odd #) = 1 P (no odd #s) = 1 P ((2, 2) (2, 4) (2, 6) (4, 2) (4, 4)
2
9
(4, 6) (6, 2) (6, 4) (6, 6)) = 1 36
= 1 632 = 27
36 = 0.75
Extending the pattern, P (at least one odd #) = 1 P (no odd #s) = 1
(ii) Let D = defective. Hence P (D) = P (D|A) P (A) + P (D|B) P (B)
So P (D) = (0.1 0.1) + (0.05 0.9) = 0.055
34
64
= 0.9375
Using Bayes Theorem to obtain correct answer:
P (D|B) P (B)
P (D|B) P (B) + P (D|A) P (A)
0.05 0.9
=
0.055
= 0.8182
P (B|D) =
1 (e)
(Total 8 marks)
(i) A test which is significant at the 1% level can be thought of as highly significant and the
data provide strong evidence to support rejection of H0 .
(ii) When significant at the 10% level but not the 5% level there is, at best, weak evidence in
support of rejection of H0 . Re-sampling/increasing n would be recommended.
(iii) H0 : = 0.01, H1 : > 0.01 need H0 and H1
Test statistic: Z =
p=
3
144
qp0.01
(1)
n
N (0, 1) since n is large.
= 0.0208
z = 1.306
Only 1.306 is weakly significant at the 10% level, so insufficient evidence to reject H0 ,
hence no evidence to suggest that more than 1% of transcations contain errors.
1 (f )
(Total 4 marks)
The 90% confidence interval for is x
1.645/ n (correct z-value)
We require 1.645/ n = 1.645 8.5/ n 1.5
Solving for n yields n =
1.6458.5 2
1.5
= 86.89
n must be an integer, so n = 87
1 (g)
(Total 6 marks)
1222
6 )
3722
6 )
= P (Z > 2.5) = 0.0062. Hence percentage of students = 0.62%
N 22, 36 . P (21 X
23) = P (2.5 Z 2.5) = (2.5) (2.5) =
(iii) X
225
0.9938 0.0062 = 0.9876
1 (h)
(Total 4 marks)
- Quota sampling is a form of non-probablitiy/non-random sampling, i.e. the probability of an
individuals selection is not known.
- Interviewers are given quota controls to interview by sex, social class, age etc.
- Basic rules of inference do not apply
- Use when in a hurry to collect sample data
- Use when no available list/sampling frame
- Use to reduce survey cost
- Use when detailed accuracy of results not important
1 (i)
(Total 4 marks)
- Use stratified random sampling to reduce standard errors
- Stratify study population to ensure a representative sample
- Unlike quota sampling, there is a known non-zero (not necessarily equal) chance of being
selected
- Usual inference can be applied
- Survey cost higher than with quota sampling
Section B
2 (a)
(Total 13 marks)
i. H0 : No association between gender and party affiliation
H1 : Association between gender and party affiliation
Compute expected values for each cell (row total column total / N ):
Oij /Eij
Gender
Male
Female
Total
Test statistic, T =
Party Identification
Democrat
Independent
Republican
279 / 261.4163 73 / 70.6531 225 / 244.9306
165 / 182.5837 47 / 49.3469 191 / 171.0694
444
120
416
i,j
(Oij Eij )2
Eij
Total
577
403
980
and t = 7.01
Degrees of freedom = (c 1)(r 1) = 2 1 = 2
At = 0.05, critical value is 5.991, hence reject H0 and conclude there is an
association between gender and party affilitation
Then test at = 0.01, critical value is 9.210, hence we do not reject H0 and conclude
there is no association at the 1% level
Comment: Rejection of H0 depends on the significance level used as the result is
significant at 5% but not 1% thus indicating some support for H1 , but not overwhelmingly
so
ii. Obtain sample proportions for all cases: pDEM = 0.37, pIN D = 0.39, pREP = 0.46 and
pALL = 0.41
Since Republican proportion is highest, we should test H0 : REP = 0.41 vs. H1 : REP >
0.41
z=
q p
(1)
n
0.460.41
q
0.410.59
416
= 2.07
At the 5% level, the critical value is 1.645, hence reject H0 .
Therefore women are more likely to identify with Republicans than other parties.
2 (b)
(Total 12 marks)
Guideline suggestions:
Random sample design specified in question, so the student should mention random
sampling techniques.
Should specify that each unit in a random sample has a known (though not necessarily equal)
4
probability of being selected.
Mention of potential sampling frame / list, e.g. electoral register, national list of addresses
etc.
An interviewer survey is required, hence recommend using a pilot survey e.g. to assess
clarity of survey questions/interviewer.
Asked to provide at least three stratification and clustering factors, hence candidates should
define stratification and cluster sampling and give examples of suitable strata, e.g. gender, social
status and employment status etc. and cluster factors
Individual respondents selected in a random manner. As this is an interviewer survey,
cluster sampling helps to reduce costs as the interviewer is limited to a small geographical
area. May have to sacrifice accuracy due to intra-class correlation. Clusters themselves might
represent constituencies
Discussion of contact method, i.e.
face-to-face interview or telephone interview, with
Potential questions to include in the survey relevant to researching political affiliation.
Minimise non-response by offering incentives, e.g. enter into prize draw
3 (a)
(Total 12 marks)
Award marks for histogram as follows:
Informative title
Frequency density axis label
x-axis label
Sensible number of classes
Plotting of frequency densities
Accuracy
10
8
6
4
0
Frequency density
12
Histogram of Growth Rate Forecasts
Forecast of economists, %
Median = 3.35, lower quartile = 2.55, upper quartile = 4.25
Population mean forecast > median forecast if the sample distribution is positively/right-skewed,
as it is here, mean affected by extreme values, unlike median (sample mean = 3.46)
3 (b)
(Total 13 marks)
(i) Correct calculation of difference between the means: 7.0 5.1 = 1.9. Do not accept
5.1 7.0 = 1.9 as question specifies x y
Correct standard error computation: 0.4501
Use t distribution with nx + ny 2 = 42 degrees of freedom
Confidence interval formula for observed sample means x
and y:
s
1
1
+
x
y t/2,nx +ny 2 s
nx ny
Correct t value: t42 value = 2.018, rounded to t40 = 2.021 from tables
Correct C.I. : (0.9903, 2.8097) if t40 used
Since the confidence interval does not include 0, this does not support the view that
there is no true difference between the population means.
t=
y
1
s2p n + n1
1
q 7.05.1
1
1
2.187( 19
+ 25
)
= 4.22
p-value = 2 P (Z > 4.22) 0
Hence reject H0 at any sensible significance level
This agrees with our original deduction that there was no evidence of no difference between
the means.
4 (a)
(Total 10 marks)
i. H0 : A = Z vs. H1 : A 6= Z
Compute sample proportions: pA = 0.6, pZ = 0.55
Test statistic value: z = 0.4529
For = 0.05, two-tailed critical z-value = 1.96
Hence result is not significant and we do not reject H0 . There is no evidence to
suggest there exists a difference in the population proportions
ii. If nA = nZ = 100, only the standard error is affected
Standard error now 0.0698 resulting in a test statistic value of 0.05/0.0698 = 0.7161,
again not statistically significant
iii. Standard error is 0.0349, hence z = 0.05/0.0349 = 1.432, still not significant
4 (b)
(Total 15 marks)
i. Award scatter diagram marks for:
Informative title
Axis labels
Accuracy
Divorce rate
x
x
x
x
x
40
45
50
55
Mobility rate
ii. Calculation of least squares regression line: y = + x +
P
xy n
xy
= P 2
= 0.1685
x n
x2
= y x
= 2.4893
Correct parameter estimates
Hence the estimated regression line is y = 2.4893 + 0.1685x
For x = 40, the expected divorce rate is 2.4893 + 0.1685(40) = 4.25
iii. Use of divorce rate as the response variable is reasonable due to the likely disruptive effect
moving home may have on relationships. (Or similar)
iv. Calculation of sample correlation coefficient:
P
P P
n xy x y
r=p P
P
P
P
(n x2 ( x)2 )(n y 2 ( y)2 )
Compute r = 0.8552
r = 0.8552 suggests a strong positive correlation/linear relationship between divorce rate
and mobility rate
8
v. Using mobility as the dependent variable would imply that the divorce rate is the driver
of mobility.
Some sensible comment about the feasibility of this argument. |
## ________________
It’s Coach Newton’s birthday this week and the team celebrates by offering him some birthday math trivia and puzzles.
## ________________
### Some Difference
If you add up the two-digit year that somebody was born, the number of the month of their birthday from 1 to 12, and the day of the month of their birthday, we’ll call that their “birthday sum.” For example, somebody born in 1968 on Feb. 18 would have a birthday sum of 68 + 2 + 18 = 88.
What’s the largest possible difference between birthday sums of people born on consecutive days?
One of the students jokes with the coach, “You know, you’re not really in the prime of your life.” “What do you mean?” asks Coach Newton. “Well, we noticed that the greatest number of prime ages that can occur in one round decade of your life is four, such as 2, 3, 5, and 7 in the first decade of your life, and 11, 13, 17, and 19 in your teens. Although you are near the beginning of a new decade in your life, it’s not one of those ‘prime decades’ with four prime ages.“ ”Oh, that’s OK with me,” chuckles the Coach.
What is the next prime decade after one’s teen years?
## Solutions to week 74
Frequent Figures. The problem is asking for a number which represented in one base has the form 22, 333, 4444, 55555, or so on, and in another base has a different instance of that same form. (The form “1” always represents the number one in any base, so it is not helpful.) To make the answer as small as possible, we want to use as few digits as possible. So what is 22 in base b? It represents 2×b + 2. Since b must be at least three in order to use the digit “2,” we can represent any even number greater than six as 22 in some base. So what’s the smallest even number that can be represented in another of these forms? Could it be 333 in some base? No, because if the base b is even, then 330 in base b is 3×b² + 3×b is even, so 333 is odd, being three more. And if b is odd, then 330 is 3×b² + 3×b, an odd plus an odd, so again even, and 333 is again odd. So 333 always represents an odd number in any base, and hence can’t represent the same number as 22 in some other base, since that’s always even.
So we have to try 4444. That numeral has the smallest value in base five, and it will certainly be even, so it will equal 22 in some other base. Therefore, the answer is 4444 in base five, which is 4×125 + 4×25 + 4×5 + 4 = 624.
Doubled Digit. It’s well known that any number has the same remainder when divided by nine as the sum of its digits in base ten does. For exactly the same reason, any number has the same remainder when divided by eight as the sum of its digits in base nine does. Therefore the sum of the digits of seven to the eleventh power in base nine has the same remainder when divided by eight as seven to the eleventh does. But checking the powers of seven (7, 49, 343, 2401, …) we see that their remainders when divided by eight alternate between seven and one. Therefore the remainder when seven to the eleventh power is divided by eight is seven. Thus, the sum of its digits in base nine must also have remainder seven when divided by eight. But the sum of the digits zero through eight is 36, with remainder four when divided by eight. Hence the only doubled digit that will bring the remainder to exactly seven is three.
## Recent Weeks
Links to all of the puzzles and solutions are on the Complete Varsity Math page.
Come back next week for answers and more puzzles.
[asciimathsf] |
# 10th STD- MATHS- Chapter 2- Numbers and Sequences Ex 2.2 (All Sums in English)
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### (All Sums in English)
Question 1.
For what values of natural number n, 4n can end with the digit 6?
Solution:
4n = (2 × 2)n = 2n × 2n
2 is a factor of 4n.
So, 4n is always even and end with 4 and 6.
When n is an even number say 2, 4, 6, 8 then 4n can end with the digit 6.
Example:
42 = 16
43 = 64
44 = 256
45 = 1,024
46 = 4,096
47 = 16,384
48 = 65, 536
49 = 262,144
Question 2.
If m, n are natural numbers, for what values of m, does 2n × 5m ends in 5?
Solution:
2n is always even for any values of n.
[Example. 22 = 4, 23 = 8, 24 = 16 etc]
5m is always odd and it ends with 5.
[Example. 52 = 25, 53 = 125, 54 = 625 etc]
But 2n × 5m is always even and end in 0.
[Example. 23 × 53 = 8 × 125 = 1000
22 × 52 = 4 × 25 = 100]
∴ 2n × 5m cannot end with the digit 5 for any values of m.
Question 3.
Find the H.C.F. of 252525 and 363636.
Solution:
To find the H.C.F. of 252525 and 363636
Using Euclid’s Division algorithm
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0.
∴ Again by division algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0.
∴ Again by division algorithm.
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0.
∴ Again by division algorithm
30303 = 20202 × 1 + 10101
The remainder 10101 ≠ 0.
∴ Again using division algorithm
20202 = 10101 × 2 + 0
The remainder is 0.
∴ 10101 is the H.C.F. of 363636 and 252525.
Question 4.
If 13824 = 2a × 3b then find a and b.
Solution:
If 13824 = 2a × 3b
Using the prime factorization tree
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 29 × 33 = 2a × 3b
∴ a = 9, b = 3.
Question 5.
If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1, p2, p3, p4 are primes in ascending order and x1, x2, x3, x4 are integers, find the value of P1, P2, P3, P4 and x1, x2, x3, x4.
Solution:
If p1x1 × p2x2 × p3x3 × p4x4 = 113400
p1, p2, p3, P4 are primes in ascending order, x1, x2, x3, x4 are integers.
using Prime factorization tree.
113400 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7
= 23 × 34 × 52 × 7
= p1x1 × p2x2 × p3x3 × p4x4
∴ p1= 2, p2 = 3, p3 = 5, p4 = 7, x1 = 3, x2 = 4, x3 = 2, x4 = 1.
Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.
Solution:
408 and 170.
408 = 23 × 31 × 171
170 = 21 × 51 × 171
∴ H.C.F. = 21 × 171 = 34.
To find L.C.M, we list all prime factors of 408 and 170, and their greatest exponents as follows.
∴ L.C.M. = 23 × 31 × 51 × 171
= 2040.
Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?
Solution:
To find L.C.M of 24, 15, 36
24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
∴ L.C.M = 23 × 32 × 51
= 8 × 9 × 5
= 360
If a number has to be exactly divisible by 24, 15, and 36, then it has to be divisible by 360. Greatest 6 digit number is 999999.
Common multiplies of 24, 15, 36 with 6 digits are 103680, 116640, 115520, …933120, 999720 with six digits.
∴ The greatest number consisting 6 digits which is exactly divisible by 24, 15, 36 is 999720.
Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Solution:
Find the L.C.M of 35, 56, and 91
35 – 5 × 7 56
56 = 2 × 2 × 2 × 7
91 = 7 × 13
L.C.M = 23 × 5 × 7 × 13
= 3640
Since it leaves remainder 7
The required number = 3640 + 7
= 3647
The smallest number is = 3647
Question 9.
Find the least number that is divisible by the first ten natural numbers.
Solution:
The least number that is divisible by the first ten natural numbers is 2520.
Hint:
1,2, 3,4, 5, 6, 7, 8,9,10
The least multiple of 2 & 4 is 8
The least multiple of 3 is 9
The least multiple of 7 is 7
The least multiple of 5 is 5
∴ 5 × 7 × 9 × 8 = 2520.
L.C.M is 8 × 9 × 7 × 5
= 40 × 63
= 2520 |
# Strategy for integration
## Strategy for integration
• Up to now, we have been learning methods of integration one by one:
• substitution
• trigonometric integration
• partial fractions
• Now, we'll practice figuring out how to calculate an integral without knowing beforehand what method you will have to use.
• As I said before, integration is not an algorithmic process.
• Moreover, the same integral can be calculated via different methods.
• But there's a few steps you can take which can help you figure out how to calculate an integral.
• As a zeroth step, you absolutely have to know by heart the basic integration formulas. For example, you can find a table in the textbook, on the first page of section 7.5.
## I. Simplify the integrand if possible.
• The integrand might be of a different form then one to which one of the methods can be readily applied.
• That is, if you don't immediately know what to do with the integrand, you can try algebraic manipulations.
• Example. 7.5.4. Since we have $\frac{\sin^3x}{\cos x}=\frac{\sin x(1-\cos^2x)}{\cos x}=\tan x-\sin x\cos x=\tan x-\frac12\sin2x,$ we get $\int\frac{\sin^3x}{\cos x}\,\mathrm dx=\int\tan x-\frac12\sin2x\,\mathrm dx=-\ln|\cos x|+\frac14\cos2x+C.$
## II. Look for an obvious substitution.
• You should always be on the lookout for whether the integrand can be recognized as the product of the form $$kf(g(x))g'(x)$$.
• Example. 7.5.32. In the integral $$\int_1^3\frac{e^{3/x}}{x^2}\,\mathrm dx$$, we see a product of the composite function $$\exp(\frac{3}{x})$$ and the function $$x^{-2}$$. Therefore, we can interpret the integrand as $$kf(g(x))g'(x)$$ where $$f(x)=e^x$$, $$g(x)=\frac{3}{x}$$, and $$k=-\frac16$$. Therefore, we have $\int_1^3\frac{e^{3/x}}{x^2}\,\mathrm dx\stackrel{u=\frac{3}{x}}{=}-\frac16\int_3^1e^u\,\mathrm du=\frac{e^3-e}{6}.$
## III. Classify the integrand according to its form.
• If steps 1 or 2 didn't yield a result, based on the form of the integrand $$f(x)$$, you should try to figure out which method of integration you should use.
• Trigonometric functions. If $$f(x)$$ is a product of powers of $$\sin x$$ and $$\cos x$$, of $$\tan x$$ and $$\sec x$$, or of $$\cot x$$ and $$\csc x$$, then you should apply trigonometric integration.
• Example. 7.5.13. Since the integral $$\int\sin^5t\cos^4t\,\mathrm dt$$ has an odd power of $$\sin t$$, we leave one $$\sin t$$ and convert the rest: $\begin{multline*} \int\sin^5t\cos^4t\,\mathrm dt=\int\sin t(1-\cos^2t)^2\cos^4t\,\mathrm dt=\int\sin t(1-2\cos^2t+\cos^4t)\cos^4t\\ \stackrel{u=\cos t}{=}-\int u^4-2u^6+u^8\,\mathrm du=-\frac{\cos^5t}{5}+\frac{2\cos^7t}{7}-\frac{\cos^9t}{9}+C. \end{multline*}$
## III. Classify the integrand according to its form.
• Rational functions. If $$f(x)$$ is a rational function, we use partial fractions.
• Example. 7.5.26. $$\int\frac{3x^2+1}{x^3+x^2+x+1}\,\mathrm dx$$. Since one can check that -1 is a root of the denominator, we can divide in by $$(x+1)$$: $$x^3+x^2+x+1=(x+1)(x^2+1)$$.
• Since $$x^2+1$$ is an irreducible quadratic polynomial, we need to solve $\frac{3x^2+1}{x^3+x^2+x+1}=\frac{A}{x}+\frac{Bx+c}{x^2+1},$ that is \begin{align*} 3x^2+1&=A(x^2+1)+(Bx+C)x\\ &=x^2(A+B)+xC+A. \end{align*}
• We get $$C=0$$, $$A=1$$, and thus $$B=2$$.
• Therefore, we have $\int\frac{3x^2+1}{x^3+x^2+x+1}\,\mathrm dx=\int\frac{1}{x}+\frac{2x}{x^2+1}\,\mathrm dx=\ln|x|+\ln|x^2+1|+C.$
## III. Classify the integrand according to its form.
• Integration by parts. If $$f(x)$$ is a product of a polynomial and a transcendental function (ie.~trigonometric, exponential or logarithmic), then you should use integration by parts.
• Example. 7.5.15. $$\int x\sec x\tan x\,\mathrm dx$$. Since this is a product of a polynomial and a trigonometric function, we can try to derivate the polynomial, and integrate the trigonometric function: $\int x\sec x\tan x\,\mathrm dx=x\sec x-\int\sec x\,\mathrm dx=x\sec x-\ln|\tan x+\sec x|+C.$
• Example 7.5.3. $$\int\sqrt y\ln y\,\mathrm dy$$. We haven't talked about this before: when applying integration by parts to the product to a polynomial and a logarithmic function, unlike the other cases, you can try to integrate the polynomial, and derivate the logarithmic function: $\int\sqrt y\ln y\,\mathrm dy=\frac23\left(\left[y^{3/2}\ln y\right]_1^4-\int_1^4\sqrt y\,\mathrm dy\right)=\frac{32}{3}\ln2-\frac49[y^{3/2}]_1^4=\frac{32}{3}-\frac{31}{9}.$
## III. Classify the integrand according to its form.
• Radicals. If $$\sqrt\pm x^2\pm a^2$$ occurs, use the appropriate trigonometric substitution.
• Example. 7.5.11. $$\int\frac{1}{x^3\sqrt{x^2-1}}\,\mathrm dx$$. We see a $$\sqrt{x^2-1}$$, so we should substitute $$x=\sec\theta$$: $\begin{multline*} \int\frac{1}{x^3\sqrt{x^2-1}}\,\mathrm dx\stackrel{x=\sec\theta}{=}\int\frac{\tan\theta\sec\theta}{\sec^3\theta\tan\theta}\,\mathrm d\theta=\int\cos^2\theta\,\mathrm d\theta\\ =\frac12\int1+\cos2\theta\,\mathrm d\theta=\frac12\sec^{-1}x+\frac14\sin2\theta+C=\frac12\sec^{-1}x+\frac12\sqrt{1-(\cos^{-1}x^{-1})^2}\cos^{-1}x^{-1}+C. \end{multline*}$
• Rationalizing substitution. If a function $$g(x)$$ with a radical occurs, you can try $$u=g(x)$$.
• Example. 7.5.23. $$\int_0^1(1+\sqrt x)^8\,\mathrm dx$$. Let's try substituting $$u=1+\sqrt x$$, which gives $$\mathrm du=\frac12x^{-1/2}\,\mathrm dx$$, that is $$2(u-1)\mathrm du=\mathrm dx$$: $\int_0^1(1+\sqrt x)^8\,\mathrm dx=2\int_1^2(u-1)u^8\,\mathrm du=\left[\frac15u^{10}-\frac29u^9\right]_1^2=\frac{2^{10}-1}{5}-\frac{2^{10}-2}{9}.$
## IV. Try again.
• If none of the first three steps yielded a solution, remember that there are basically only two methods: substitution and parts.
• Try substitution. You can basically try various subtitutions.
• Example. 7.5.21. $$\int\tan^{-1}\sqrt x\,\mathrm dx$$. Here, the function with the radical in inside another function. We can still try $$u=\sqrt x$$: $\int\tan^{-1}\sqrt x\,\mathrm dx=2\int u\tan^{-1}u.$
• Since we have $$(\tan^{-1}u)'=\frac{1}{1+u^2}$$, we can use integration by parts, with integrating $$u$$, and derivating $$\tan^{-1}u$$: $=u^2\tan^{-1}u-\int\frac{u^2}{1+u^2}\,\mathrm du=x\tan^{-1}\sqrt x-\int 1-\frac{1}{1+u^2}\,\mathrm du=x\tan^{-1}\sqrt x-\sqrt x+\tan^{-1}\sqrt x+C.$
• Try parts. Recall that there are cases where you can find $$\int f(x)\,\mathrm dx$$ via integration by parts with $$u=f(x)$$ and $$\mathrm dv=\mathrm dx$$.
• Example. 7.5.14. $\ln(1+x^2)\,\mathrm dx=x\ln(1+x^2)-\int\frac{2x^2}{1+x^2}\,\mathrm dx=x\ln(1+x^2)-2x+2\tan^{-1}x+C.$
## IV. Try again.
• Manipulate the integrand. You can try performing various algebraic manipulations on the integrand, for example expanding it so that you can apply some algebraic or trigonometric formula.
• Example. $$\int\frac{\cos x}{1-\sin x}\,\mathrm dx$$. Since we have $\frac{\cos x}{1-\sin x}=\frac{\cos x}{1-\sin x}\frac{1+\sin x}{1+\sin x}=\frac{\cos x+\sin x\cos x}{1-\sin^2x}=\frac{\cos x}{\cos^2x}+\frac{\sin x\cos x}{\cos^2x}=\sec x+\tan x,$ we get $\int\frac{\cos x}{1-\sin x}\,\mathrm dx=\int\sec x+\tan x\,\mathrm dx=\ln|\tan x+\sec x|-\ln|\cos x|+C.$
• Exercises. I'm sorry, but I'm going to say you should do as many problems as you can from this section. This is one of the most difficult parts of this course. You should get in as much practice as possible.
## Can we integrate all continuous functions?
• One might ask if it is possible to calculate the integral of any continuous function.
• In Calculus I, you have seen that every continuous function is integrable.
• So you might ask: is it possible to calculate the integral of every continuous function?
• Of course, this invites the question: what does it mean to calculate the integral of a continuous function? What does it mean to describe a function?
• The functions we have been dealing with are called elementary function. An elementary function is one you can obtain via addition, multiplication, and composition from polynomial, exponential, trigonometric functions, and their inverses.
• So we should rather ask: is the integral of every elementary function an elementary function.
• The answer is no. The most striking example is that the antiderivative of the *density function of a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$: $\phi(x)=\frac{1}{\sqrt{2\sigma^2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ is not an elementary function. This is one of the most commonly used density functions that can create bell curves. |
# If three dice are rolled, what is the probability of getting a total of 18?
Jan 23, 2018
$\frac{1}{216}$
#### Explanation:
to obtain a total of $18$ we need all three dice to show a $6$
because they are independent events we can simply multiply the separate probabilities
$P \left(18\right) = P \left(6 \cap 6 \cap 6\right)$
$= P \left(6\right) \times P \left(6\right) \times P \left(6\right) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{5}$
$= \frac{1}{216}$ |
## College Algebra (6th Edition)
$\{ (\displaystyle \frac{1}{2},\frac{1}{3},-1) \}$
Plan (see p.533): 1. Eliminate one variable and arrive at a system of two equations in two variables. 2. Solve the system of two equations in two variables. 3. Back-substitute the solutions of (2) to find the eliminated varable in (1) ------------ First, write each equation in standard form: Ax+By+Cz=D. $\left[\begin{array}{l} 6x+3y+5z=-1\\ 2x-6y+8z=-9\\ 4+4x=-3z+9y \end{array}\right.$ $\left[\begin{array}{l} 6x+3y+5z=-1\\ 2x-6y+8z=-9\\ 4x-9y+3z==4 \end{array}\right.$ (1.) Perform elimination by addition method (eliminate x) by $\left[\begin{array}{lll} Eq.1.-3\times Eq.2\rightarrow & 21y-19z=26 & I\\ -2\times Eq.2.+Eq.3 & 3y-13z=14 & II \end{array}\right.$ , (2.) ... $(Eq.I - 7\times Eq.II)$ eliminates y, $72z=-72$ $z=-1$ Back-substitute into $21y-19z=26$ $21y-19(-1)=26$ $21y=26-19$ $y=\displaystyle \frac{7}{21}=\frac{1}{3}$ (3.) Back-substitute into one of the initial equations: $6x+3(\displaystyle \frac{1}{3})+5(-1)=-1$ $6x-4=-1$ $6x=3$ $x=\displaystyle \frac{1}{2}$ Solution set: : $\{ (\displaystyle \frac{1}{2},\frac{1}{3},-1) \}$ |
# ELEMENTARY NUMBER THEORY - LINKS; CONGRUENCES, EULER'S, FERMAT'S, CHINESE THEOREMS
Please study Number Systems before Elementary Number Theory.
We introduced Natural Numbers there.
The branch of mathematics which deals with the study of
natural numbers, is called number theory.
Here we deal with a part of it.
Here Natural Numbers are studied without use
of techniques from other mathematical fields.
The various topics include
Divisibility Rules,
Greatest Common Factor by Euclidean Algorithm,
Least Common Multiple,
Prime Factorization,
Fundamental theorem of arithmetic,
Prime Numbers,
List of Prime Numbers,
Investigation of Perfect Numbers.
Other topics include, Modular Arithmetic (Congruences),
which is simply the arithmetic of remainders.
Several important discoveries of Elementary Number Theory are
Fermat's little theorem, Euler's theorem,
the Chinese remainder theorem.
We cover these theorems of Elementary Number Theory
with examples to make
even a high school student can understand them.
For students, who study Elementary Number Theory,
## Modular Arithmetic (Congruences) of Elementary Number Theory
We know, from the knowledge of Division,
Dividend = Remainder + Quotient x Divisor.
If we denote dividend by a, Remainder by b,
Quotient by k and Divisor by m, we get
a = b + km
a = b + some multiple of m
a and b differ by some multiples of m
or if you take away some multiples of m from a, it becomes b.
Taking away some (it does n't matter, how many) multiples
of a number from another number to get a new number
has some practical significance.
Example 1 of Elementary Number Theory : Congruences
For example, look at the question
Today is Sunday. What day will it be 200 days from now ?
How do we solve the above problem ?
We take away multiples of 7 from 200.
We are interested in what remains
after taking away the mutiples of 7.
Let us show the process of dividing 200 by 7,
using our knowledge of Long Division.
```
Dividend
Divisor 7 ) 200 ( 28 Quotient
14
---
60
56
---
4 Remainder
---
```
We are not interested in how many multiples are taken away.
i. e. We are not interested in the quotient.
We only want the remainder.
We get 4 when some (28) multiples of 7 are taken away from 200.
∴ The question, "What day will it be 200 days from now ?"
now, becomes, "What day will it be 4 days from now ?"
Because, today is sunday, 4 days from now will be Thursday. Ans.
The point is, when, we are interested in taking away multiples of 7,
200 and 4 are the same for us.
Mathematically, we write this as
200 ≡ 4 (mod 7)
and read as 200 is congruent to 4 modulo 7.
The equation 200 ≡ 4 (mod 7) is called Congruence.
Here 7 is called Modulus and the process is called
Modular Arithmetic in Elementary Number Theory.
Let us see one more example.
Example 2 of Elementary Number Theory : Congruences
It is 7 O' clock in the morning.
What time will it be 80 hours from now ?
We have to take away multiples of 24 from 80.
80 ÷ 24 gives a remainder of 8.
or 80 ≡ 8 (mod 24).
∴ The time 80 hours from now is
the same as the time 8 hours from now.
7 O' clock in the morning + 8 hours = 15 O' clock
= 3 O' clock in the evening [ since 15 ≡ 3 (mod 12) ].
Let us see one last example
before we formally define Congruence.
Example 3 of Elementary Number Theory : Congruences
A person is facing East. He rotates 1260 degree anti-clockwise.
In what direction, he is facing ?
We know, rotation of 360 degrees
will bring him to the same position.
So, we have to remove multiples of 360 from 1260.
The remainder, when 1260 is divided by 360, is 180.
i. e. 1260 ≡ 180 (mod 360).
∴ rotating 1260 degrees is same as rotating 180 degrees.
∴ when he rotates 180 degrees anti-clockwise from east,
he will face west direction. Ans.
## Definition of Congruence
Let a, b and m be any integers with m not zero,
then we say a is congruent to b modulo m,
if m divides (a - b) exactly without remainder.
We write this as ab (mod m).
Other ways of defining Congruence include :
(i) a is congruent to b modulo m,
if a leaves a remainder of b when divided by m.
(ii) a is congruent to b modulo m,
if a and b leave the same remainder when divided by m.
(iii) a is congruent to b modulo m,
if a = b + km for some integer k.
In the three examples above, we have
200 ≡ 4 (mod 7); in example 1.
80 ≡ 8 (mod 24); 15 ≡ 3 (mod 12) in example 2.
1260 ≡ 180 (mod 360). in example 3.
We started our discussion with the process of division.
In division, we dealt with whole numbers only and
also, the remainder, is always less than the divisor.
In Modular Arithmetic, we deal with integers
( i.e. whole numbers + negative integers).
Also, when we write ab (mod m),
b need not necessarily be less than a.
The three most important properties of congruences modulo m are:
The reflexive property :
If a is any integer, aa (mod m).
The symmetric property :
If ab (mod m), then ba (mod m).
The transitive property :
If ab (mod m) and bc (mod m),
then a ≡ c (mod m).
Other properties :
If a, b, c and d, m, n are any integers with
ab (mod m) and cd (mod m), then
a + cb + d (mod m)
a - cb - d (mod m)
acbd (mod m)
(a)nbn (mod m)
If gcd(c,m) = 1 and acbc (mod m),
then ab (mod m)
Example 4 of Elementary Number Theory : Congruences
Find the last decimal digit of 13100.
Finding the last decimal digit of 13100 is same as
finding the remainder when 13100 is divided by 10.
We know 13 ≡ 3 (mod 10)
∴ 13100 ≡ 3100 (mod 10)
.....(i)
We know 32 ≡ -1 (mod 10)
∴ (32)50 ≡ (-1)50 (mod 10)
∴ 3100 ≡ 1 (mod 10)
.....(ii)
From (i) and (ii), we can say
last decimal digit of 13100 is 1. Ans.
The Theorems of Elementary Number Theory, that follow
are based on modular Arithmetic (congruences).
## Fermat's little theorem of Elementary Number Theory
If p is a prime number and
x is any integer that does not have p as a factor,
then x(p - 1) ÷ p gives remainder 1.
In the notation of modular arithmetic,
we can write this statement as
x(p - 1) ≡ 1 (mod p).
Examples :
Example 5 of Elementary Number Theory : Fermat's little theorem
Find the remainder when 796 is divided by 97.
Solution :
Since 97 is a prime number and is not a factor of 7
and the exponent is 96 = 97 - 1, we can apply
Fermat's little theorem.
∴ 796 ≡ 1 (mod 97).
i.e. When 796 is divided by 97, remainder = 1.Ans.
Example 6 of Elementary Number Theory : Fermat's little theorem
Find the remainder when 6330 is divided by 31.
Solution :
Since 31 is a prime number and is not a factor of 63
and the exponent = 31 - 1 = 30, we can apply
Fermat's little theorem.
6330 ≡ 1(mod31)
or 6330 ÷ 31 gives remainder 1.Ans.
Great Deals on School & Homeschool Curriculum Books
## Euler's theorem of Elementary Number Theory
If x and y are co-primes, then
xtotient of y÷y gives remainder 1,
where totient of y
= number of positive integers that are co-prime to y and less than y.
In the notation of modular arithmetic,
we can write this statement as
xtotient of y ≡ 1 (mod y).
Finding totient of y :
If y is expressed as a product of prime factors, say
y = ap x bq x cr x ......
( i.e. a, b, c,.... are distinct prime factors of y and
p, q, r,....... are their multiplicitys (exponents) respectively),
then, totient of y = yx(1- 1⁄a)x(1- 1⁄b)x(1- 1⁄c)x.....
For example, 8 = 23.
totient of 8 = 8(1- 1⁄2) = 8(1⁄2) = 4.
We can also verify that
the four numbers 1, 3, 5, 7 are co-prime to 8.
Note :
(i) totient of y is denoted by φ(y) and is called
Euler's totient function or Euler's phi function.This is an important function in Elementary Number Theory.
(ii) if y is a prime number, prime factor of y = y
then, totient of y = y((1 - 1⁄y) = y - 1.
Then Euler's theorem becomes
x(y - 1) ÷ y gives remainder 1.
Which is nothing but Fermat's Little Theorem.
Note that, here, x and y are co-primes and y ia a prime number.
Thus, Fermat's Little Theorem is a special case of Euler's theorem.
Example 7 of Elementary Number Theory : Euler's Theorem
Find the remainder when 5326 is divided by 42.
Prime Factorization of 42 gives
42 = 2 x 3 x 7.
totient of 42 = 42(1- 1⁄2)(1- 1⁄3)(1- 1⁄7)
= 42 x 1⁄2 x 2⁄3 x 6⁄7 = 12
Since 5 and 42 are co-prime, by Euler's theorem,
512 ≡ 1 (mod 42).
Here, we make use of properties of congruencesof Elementary Number Theory given above.
⇒ (512)27 ≡ 127 (mod 42).
⇒ 5324 ≡ 1 (mod 42)......(i)
We know 52 ≡ 25 (mod 42).....(ii)
(i) x (ii) gives 5324 x 52 ≡ 1 x 25 (mod 42)
⇒ 5326 ≡ 25 (mod 42)⇒ when 5326 is divided by 42, remainder = 25. Ans.
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## The Chinese remainder theorem of Elementary Number Theory
The following idea of Greatest Common Factor (G.C.F.)
is used here.
If G.C.F. of two numbers m1 and m2
= G.C.F.(m1, m2) = 1, then
we can find two integers i1, i2
such that m1i1 + m2i2 = 1.
For example, G.C.F.(3, 7) = 1.
We can find i1 = -2 and i2 = 1
by trial and error
such that 3i1 + 7i2 = 3(-2) + 7(1) = -6 + 7 = 1.
Statement of The Chinese remainder theorem
If a number a gives remainder b1, when divided by m1
and remainder b2, when divided by m2,
then it (a) gives remainder (m1i1b2 + m2i2b1)
when divided by m1m2,
where i1, i2 are two integers
such that m1i1 + m2i2 = 1.
In the notation of modular arithmetic,
we can write this statement as
if ab1(mod m1)
and ab2(mod m2),
then a ≡ (m1i1b2 + m2i2b1) (mod m1m2)
Corollary of Chinese Remainder Theorem :
In the above theorem, if b1 = b2 = b,
then ab (mod m1m2).
Example 8 of Elementary Number Theory : Chinese Remainder Theorem
Let us first take up the original problem solved by
Chinese Remainder Theorem in Elementary Number Theory.
How many soldiers are there in Han Xing's army? -
If you let them parade in rows of 3 soldiers, two soldiers will be left.
If you let them parade in rows of 5, three will be left,
and in rows of 7, two will be left.
Solution :
Let a be the number of soldiers.
Then by data,
a ≡ 2 (mod 3); ...........(i)
a ≡ 3 (mod 5); ...........(ii)
a ≡ 2 (mod 7); ...........(iii)
We have to take two of the above equations, first
and apply the Chinese Remainder Theorem.
If we take (i) and (iii), we can apply the corollary, which is simpler.
we have G.C.F.(3, 7) = 1. Applying the corollary, we get
a ≡ 2 (mod 3 x 7) or a ≡ 2 (mod 21).......(iv)
Now let us apply the theorem to (ii) and (iv).
a ≡ 3 (mod 5); ...........(ii)
a ≡ 2 (mod 21)...........(iv)
We know G.C.F.(5, 21) = 1. So we can apply the theorem.
Let us find i1 and i2 such that5i1 + 21i2 = 1.
By trial and error, i1 = -4 and i2 = 1
By the theorem, we get,
a ≡ {5(-4)2 + 21(1)(3)} (mod 5 x 21) or a ≡ 23 (mod 105)
a is such that when it is divided by 105 gives remainder 23.
or a = 23 + k (105) where k is an integer.
If k = 0 is taken, then a = 23 + 0 = 23 is a possible answer.
23 satisfies the three equations, as we can see.
23 ≡ 2 (mod 3); ...........(i)
23 ≡ 3 (mod 5); ...........(ii)
23 ≡ 2 (mod 7); ...........(iii)
Thus 23 is the lowest answer.
The general answer is 23 + k (105) where k is a whole number.
To study more about these theorems and other topics of
Elementary Number Theory, you may refer other sources.
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## Solving calculus problem
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NCERT Solutions For Class 7 Maths Chapter 3 Data Handling
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling are formulated by our professional tutors to help students with their exam preparation to attain good marks in Maths. The NCERT book is one of the top materials when it comes to providing a question bank to practice. Our main objective is to help students understand and crack these problems. We at BYJU’S, have prepared the NCERT Solutions for Class 7 Maths Chapter 3, wherein problems are solved step by step with complete descriptions.
Download Most Important Questions for Class 7 Maths Chapter – 3 Data Handling
Chapter 3 of NCERT Solutions Class 7 Maths consists of 4 exercises, and provided here are the solutions to the questions present in various exercises of the chapter. Listed below are the different concepts covered in this chapter.
• Collection of Data
• Organisation of Data
• Arithmetic Mean
• Mode
• Mode of Large Data
• Median
• Use of Bar Graphs with a Different Purpose
• Chance and Probability
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling
Access exercises of NCERT Solutions for Class 7 Maths Chapter 3 Data Handling
Exercise 3.1 Solutions
Exercise 3.2 Solutions
Exercise 3.3 Solutions
Exercise 3.4 Solutions
Access answers to Maths NCERT Solutions For Class 7 Chapter 3 – Data Handling
Exercise 3.1 Page: 62
1. Find the range of heights of any ten students in your class.
Solution:-
Let us assume the heights (in cm) of 10 students in our class be
= 130, 132, 135, 137, 139, 140, 142, 143, 145, 148
By observing the above-mentioned values, the highest value is = 148 cm
By observing the above-mentioned values, the lowest value is = 130 cm
Then,
Range of Heights = Highest value – Lowest value
= 148 – 130
= 18 cm
2. Organise the following marks in a class assessment in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
(i) Which number is the highest? (ii) Which number is the lowest?
(iii) What is the range of the data? (iv) Find the arithmetic mean.
Solution:-
First, we have to arrange the given marks in ascending order.
= 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9
Now, we will draw the frequency table of the given data.
Marks Tally Marks Frequency 1 1 2 2 3 1 4 3 5 5 6 4 7 2 8 1 9 1
(i) By observing the table clearly, the highest number among the given data is 9.
(ii) By observing the table clearly, the lowest number among the given data is 1.
(iii) We know that Range = Highest value – Lowest value
= 9 – 1
= 8
(iv) Now, we have to calculate Arithmetic Mean,
Arithmetic mean = (Sum of all observations)/ (Total number of observations)
Then,
Sum of all observation = 1 + 2 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 7 + 7
+ 8 + 9
= 100
Total Number of Observations = 20
Arithmetic mean = (100/20)
= 5
3. Find the mean of the first five whole numbers.
Solutions:-
The first five Whole numbers are 0, 1, 2, 3, and 4.
Mean = (Sum of first five whole numbers)/ (Total number of whole numbers)
Then,
Sum of five whole numbers = 0 + 1 + 2 + 3 +4
= 10
Total Number of whole numbers = 5
Mean = (10/5)
= 2
∴ The mean of the first five whole numbers is 2.
4. A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
Solution:-
Mean score = (Total runs scored by the cricketer in all innings)/ (Total number of innings
played by the cricketer)
Total runs scored by the cricketer in all innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100
= 400
Total number of innings = 8
Then,
Mean = (400/8)
= 50
∴ The mean score of the cricketer is 50.
5. Following table shows the points each player scored in four games:
Player Game 1 Game 2 Game 3 Game 4 A 14 16 10 10 B 0 8 6 4 C 8 11 Did not Play 13
Now, answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?
Solution:-
(i) A’s average number of points scored per game = Total points scored by A in 4 games/
Total number of games
= (14 + 16 + 10 + 10)/ 4
= 50/4
= 12.5 points
(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played only 3 games.
(iii) B played in all four games, so we will divide the total points by 4 to find out the mean.
Then,
Mean of B’s score = Total points scored by B in 4 games/ Total number of games
= (0 + 8 + 6 + 4)/ 4
= 18/4
= 4.5 points
(vi) Now, we have to find the best performer among the 3 players.
So, we have to find the average points of C = (8 + 11 + 13)/3
= 32/3
= 10.67 points
By observing, the average points scored A is 12.5, which is more than B and C.
Clearly, we can say that A is the best performer among the three.
6. The marks (out of 100) obtained by a group of students in a science test are 85, 76,
90, 85, 39, 48, 56, 95, 81 and 75. Find the:
(i) Highest and lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution:-
First, we have to arrange the marks obtained by a group of students in a science test in ascending order.
= 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) The highest marks obtained by the student = 95
The lowest marks obtained by the student = 39
(ii) We know that Range = Highest marks – Lowest marks
= 95 – 39
= 56
(iii) Mean of Marks = (Sum of all marks obtained by the group of students)/
(Total number of marks)
= (39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95)/ 10
= 730/10
= 73
7. The enrolment in a school for six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820.
Find the mean enrolment of the school for this period.
Solution:-
Mean enrolment = Sum of all observations / Number of observations
= (1555 + 1670 + 1750 + 2013 + 2540 + 2820)/ 6
= (12348/6)
= 2058
∴ The mean enrolment of the school for this given period is 2058.
8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
Day Mon Tue Wed Thurs Fri Sat Sun Rainfall (in mm) 0.0 12.2 2.1 0.0 20.5 5.5 1.0
(i) Find the range of rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution:-
(i) Range of rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0
= 20.5 mm
(ii) Mean of rainfall = Sum of all observations / Number of observations
= (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)/ 7
= 41.3/7
= 5.9 mm
(iii) We may observe that for 5 days, i.e. Monday, Wednesday, Thursday, Saturday and Sunday, the rainfall was less than the average rainfall.
9. The heights of 10 girls were measured in cm, and the results are as follows:
135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
(i) What is the height of the tallest girl? (ii) What is the height of the shortest girl?
(iii) What is the range of the data? (iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?
Solution:-
First, we have to arrange the given data in ascending order.
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) The height of the tallest girl is 151 cm.
(ii) The height of the shortest girl is 128 cm.
(iii) Range of given data = Tallest height – Shortest height
= 151 – 128
= 23 cm
(iv) Mean height of the girls = Sum of the height of all the girls / Number of girls
= (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150
+ 151)/ 10
= 1414/10
= 141.4 cm
(v) 5 girls have heights more than the mean height (i.e. 141.4 cm).
Exercise 3.2 Page: 68
1. The scores on the Mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they the same?
Solution:-
Arranging the given scores in ascending order, we get
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode
Mode is the value of the variable which occurs most frequently.
Clearly, 20 occurs a maximum number of times.
Hence, the mode of the given sores is 20.
Median
The value of the middle-most observation is called the median of the data.
Here, n = 15, which is odd.
Where n is the number of students.
∴ median = value of ½ (n + 1)th observation
= ½ (15 + 1)
= ½ (16)
= 16/2
= 8
Then, the value of the 8th term = 20
Hence, the median is 20.
Yes, both values are the same.
2. The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
Find the mean, mode and median of this data. Are the three same?
Solution:-
Arranging the runs scored in a cricket match by 11 players in ascending order, we get
6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Mean
Mean of the given data = Sum of all observations / Total number of observations
= (6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120)/ 11
= 429/11
= 39
Mode,
Mode is the value of the variable which occurs most frequently.
Clearly, 15 occurs a maximum number of times.
Hence, the mode of the given sores is 15.
Median,
The value of the middle-most observation is called the median of the data.
Here n = 11, which is odd.
Where n is the number of players.
∴ median = value of ½ (n + 1)th observation.
= ½ (11 + 1)
= ½ (12)
= 12/2
= 6
Then, the value of the 6th term = 15
Hence, the median is 15.
No, these three are not the same.
3. The weights (in kg.) of 15 students of a class are:
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Solution:-
Arranging the given weights of 15 students of a class in ascending order, we get
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) Mode and Median
Mode
Mode is the value of the variable which occurs most frequently.
Clearly, 38 and 43 both occur 3 times.
Hence, the modes of the given weights are 38 and 43.
Median
The value of the middle-most observation is called the median of the data.
Here, n = 15, which is odd.
Where n is the number of students.
∴ median = value of ½ (n + 1)th observation
= ½ (15 + 1)
= ½ (16)
= 16/2
= 8
Then, the value of the 8th term = 40
Hence, the median is 40.
(ii) Yes, there are 2 modes for the given weights of the students.
4. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
Solution:-
Arranging the given data in ascending order, we get
= 12, 12, 13, 13, 14, 14, 14, 16, 19
Mode
Mode is the value of the variable which occurs most frequently.
Clearly, 14 occurs the maximum number of times.
Hence, the mode of the given data is 14.
Median
The value of the middle-most observation is called the median of the data.
Here, n = 9, which is odd.
Where n is the number of students.
∴ median = value of ½ (9 + 1)th observation
= ½ (9 + 1)
= ½ (10)
= 10/2
= 5
Then, the value of the 5th term = 14
Hence, the median is 14.
5. Tell whether the statement is true or false.
(i) The mode is always one of the numbers in a data.
Solution:-
The statement given above is true.
Because Mode is the value of the variable which occurs most frequently in the given data.
Hence, a mode is always one of the numbers in the data.
(ii) The mean is one of the numbers in the data.
Solution:-
The statement given above is false.
Because mean may or may not be one of the numbers in the data.
(iii) The median is always one of the numbers in a data.
Solution:-
The statement given above is true.
Because the median is the value of the middle-most observation in the given data while arranged in ascending or descending order.
Hence, the median is always one of the numbers in a data
(iv) The data 6, 4, 3, 8, 9, 12, 13, and 9 have the mean 9.
Solution:-
Mean = Sum of all given observations / Number of observations
= (6 + 4 + 3 + 8 + 9 + 12 + 13 + 9)/8
= (64/8)
= 8
Hence, the given statement is false.
Exercise 3.3 Page: 72
1. Use the bar graph (Fig 3.3) to answer the following questions.
(a) Which is the most popular pet? (b) How many students have dogs as a pet?
Solution:-
The bar graph represents the pets owned by the students.
(a) From the bar graph, the most popular pet is cat. It is owned by 10 students out of 12 students.
(b) From the bar graph, 8 students have dogs as a pet out of 12 students.
2. Read the bar graph (Fig 3.4), which shows the number of books sold by a bookstore during five consecutive years, and answer the following questions.
(i) About how many books were sold in 1989, 1990 and 1992?
(ii) In which year was about 475 books and 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
Solution:-
(i) By observing the bar graph,
175 books were sold in the year 1989.
475 books were sold in the year 1990.
225 books were sold in the year 1992.
(ii) By observing the bar graph,
475 books were sold in the year 1990.
225 books were sold in the year 1992.
(iii) By observing the bar graph,
In the years 1989 and 1992, the number of books sold was less than 250.
(iv) By observing the bar graph, we can conclude that
The number of books sold in the year 1989 is about 1 and ¾th part of 1 cm.
WKT, Scale is taken as 1 cm = 100 books
= 100 + (¾ × 100)
= 100 + (3 × 25)
= 100 + 75
= 175
3. Number of children in six different classes is given below. Represent the data on a bar graph.
Class Fifth Sixth Seventh Eighth Ninth Tenth Number of Children 135 120 95 100 90 80
(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of Class six to the students of Class eight.
Solution:-
(a) We will take the scale as 1 unit = 10 children because we can represent a bigger and clear difference between the number of students in Class 7th and Class 9th.
(b) (i) Class 5th has the maximum number of children, i.e. 135, and Class 7th has the minimum number of children, i.e. 95.
(ii) The total number of students in Class 6th is 120, and the total number of students in Class 8th is 100.
Then,
The ratio between the number of students in Classes 6th and 8th,
= (120/100)
= 6/5
= 6: 5
4. The performance of a student in the 1st Term and 2nd Term is given. Draw a double bar graph by choosing an appropriate scale and answer the following:
Subject English Hindi Maths Science S. Science 1st Term (M.M. 100) 67 72 88 81 73 2nd Term (M.M. 100) 70 65 95 85 75
(i) In which subject has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Solution:-
(i) By observing the double bar graph, there was a maximum mark increase in the Maths subject. So, the child has improved his performance in Maths.
(ii) By observing the double bar graph, the improvement was the least in S. Science.
(iii) By observing the double bar graph, the performance in Hindi has gone down.
5. Consider this data collected from a survey of a colony.
Favourite Sport Cricket Basket Ball swimming Hockey Athletics Watching 1240 470 510 430 250 Participating 620 320 320 250 105
(i) Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Solution:-
(i) The figure above is the double bar graph, which represents the people who prefer either watching or participating in different sports. By observing the double bar graph, we came to conclude that most people like watching and participating in cricket, while the least number of people like watching and participating in athletics.
(ii) By observing the double bar graph, we came to conclude that the people who like watching and participating in cricket are the tallest among all the bars. So, cricket is the most popular sport.
(iii) By observing the double bar graph, we came to conclude that watching sports has more preference, rather than participating in sports.
6. Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this Chapter (Table 3.1). Plot a double bar graph using the data and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where the maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
Solution:-
(i) By observing the double bar graph, we came to conclude that Jammu has the largest difference in the minimum and maximum temperature on 20.6.2006.
(ii) By observing the double bar graph, we came to conclude that Jammu is the hottest city and Bangalore is the coldest city.
(iii) By observing the double bar graph, Bangalore and Jaipur, and Bangalore and Ahmedabad, we can see that for Bangalore, the maximum temperature was 28oC, while the minimum temperature of both Ahmadabad and Jaipur was 29oC.
(iv) By observing the double bar graph, Mumbai has the least difference between its minimum and maximum temperature.
Exercise 3.4 Page: 76
1. Tell whether the following is certain to happen, impossible, or can happen but not certain.
(i) You are older today than yesterday.
Solution:-
It is certain to happen.
(ii) A tossed coin will land heads up.
Solution:-
It can happen but not certain.
(iii) A die, when tossed, shall land up with 8 on top.
Solution:-
It is impossible because there are only six faces on a die marked as 1, 2, 3, 4, 5, and 6 on it.
(iv) The next traffic light seen will be green.
Solution:-
It can happen but not certain.
(v) Tomorrow will be a cloudy day.
Solution:-
It can happen but not certain.
2. There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with the number 2?
Solution:-
From the question, it is given that
There are 6 marbles in the box with numbers from 1 to 6 marked.
Probability of drawing a marble with number 2 = Number of favourable outcomes /
Number of possible outcomes
= (1/6)
(ii) What is the probability of drawing a marble with the number 5?
Solution:-
From the question, it is given that
There are 6 marbles in the box with numbers from 1 to 6 marked.
Probability of drawing a marble with number 5 = Number of favourable outcomes /
Number of possible outcomes
= (1/6)
3. A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:-
A coin has two faces: one is the Head, and the other one is the Tail.
Now, one team can choose either Head or Tail.
The probability of our team starting first= Number of favourable outcomes /
Number of possible outcomes
= ½
Disclaimer:
Dropped Topics – 3.1 Introduction, 3.2 Collecting data, 3.3 Organisation of data and 3.9 Chance and probability.
Frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 3
Q1
Why should we learn all the concepts present in NCERT Solutions for Class 7 Maths Chapter 3?
The concepts covered in Chapter 3 of NCERT Solution Maths provide questions based on the exam pattern and model question paper. So, it is necessary to learn all the concepts present in NCERT Solutions for Class 7 Maths Chapter 3.
Q2
List out the important topics present in NCERT Solutions for Class 7 Maths Chapter 3.
The topics covered in the NCERT Solutions for Class 7 Maths Chapter 3 are
1. Collection of Data
2. Organisation of Data
3. Arithmetic Mean
4. Mode
5. Mode of Large Data
6. Median
7. Use of Bar Graphs with a Different Purpose
8. Chance and Probability
These concepts are important from an exam perspective. It is strictly based on the latest syllabus of the CBSE board and also depends on the CBSE question papers and marking scheme.
Q3
How many exercises are there in NCERT Solutions for Class 7 Maths Chapter 3?
There are 4 exercises are there in NCERT Solutions for Class 7 Maths Chapter 3. The first exercise contains 9 questions, the second exercise contains 5 questions with sub-questions, the third exercise has 5 questions with sub-questions, and the fourth exercise has 9 questions based on the concepts of data handling. |
# Difference between revisions of "1981 AHSME Problems/Problem 26"
## Problem
Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\frac{1}{6}$, independent of the outcome of any other toss.) $\\ \textbf{(A) } \frac{25}{91}$
## Solution
The probability that Carol wins during the first cycle through is $\frac{5}{6}*\frac{5}{6}*\frac{1}{6}$, and the probability that Carol wins on the second cycle through is $\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}$. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: $\frac{\frac{25}{216}}{1-\frac{125}{216}}$, or $\frac{\frac{25}{216}}{\frac{91}{216}}$, which simplifies into $\boxed{\textbf{(A) } \frac{25}{91}}$ |
# How do you simplify (z^2-z-6)/(z-6) *( z^2-6z)/(z^2+2z-15)?
Jul 13, 2015
$\frac{{z}^{2} - z - 6}{z - 6} \cdot \frac{{z}^{2} - 6 z}{{z}^{2} + 2 z - 15} = \frac{{z}^{2} + 2 z}{z + 5}$
#### Explanation:
$\frac{{z}^{2} - z - 6}{z - 6} \cdot \frac{{z}^{2} - 6 z}{{z}^{2} + 2 z - 15}$
$\textcolor{w h i t e}{\text{XXXX}}$Combining the numerator factors and the denominator factors:
$= \frac{\left({z}^{2} - z - 6\right) \left({z}^{2} - 6 z\right)}{\left(z - 6\right) \left({z}^{2} + 2 z - 15\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$Factoring
$= \frac{\left(z - 3\right) \left(z + 2\right) \left(z\right) \left(z - 6\right)}{\left(z - 6\right) \left(z + 5\right) \left(z - 3\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$Cancelling out matching terms in the numerator and denominator
$= \frac{\cancel{\left(z - 3\right)} \left(z + 2\right) \left(z\right) \cancel{\left(z - 6\right)}}{\cancel{\left(z - 6\right)} \left(z + 5\right) \cancel{\left(z - 3\right)}}$
$\textcolor{w h i t e}{\text{XXXX}}$Rewrite simplified:
$= \frac{{z}^{2} + 2 z}{z + 5}$ |
# How to do a math equation
There are also many YouTube videos that can show you How to do a math equation. We will give you answers to homework.
## How can we do a math equation
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Then, work through the equation step-by-step, using the order of operations to simplify each term. Be sure to keep track of any negative signs, as they will change the direction of the operation. Finally, check your work by plugging the value of the variable back into the equation. If everything checks out, you have successfully solved the equation!
## Math checker you can trust
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## RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A
NCERT Maths Solutions for Ex 4.1 class 10 Quadratic Equations is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.
## RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A.
Other Exercises
Question 1.
Solution:
We know that a second degree of equation is called a quadratic equation. Therefore,
It is not a quadratic equation. It is of degree 5.
So, (i), (ii), (iii), (iv), (vi) and (ix) are quadratic equations.
Question 2.
Solution:
3x² + 2x – 1
= 3x² + 3x – x – 1
= 3x (x + 1) – 1 (x + 1)
= (x + 1) (3x – 1)
Either, x + 1 = 0 ⇒ x = -1
or 3x – 1 =0
⇒ 3x = 1
⇒ x = $$\frac { 1 }{ 3 }$$
Hence, (-1) and $$\frac { 1 }{ 3 }$$ are its roots.
Question 3.
Solution:
Question 4.
Solution:
Solve each of the following quadratic equations.
Question 5.
Solution:
Given : (2x – 3)(3x + 1) = 0
Either 2x – 3 = 0, then 2x = 3 ⇒ x = $$\frac { 3 }{ 2 }$$
or 3x + 1 = 0, then 3x = -1 ⇒ x = $$\frac { -1 }{ 3 }$$
x = $$\frac { 3 }{ 2 }$$ , $$\frac { -1 }{ 3 }$$
Question 6.
Solution:
4×2 + 5x = 0 ⇒ x (4x + 5) = 0
Either x = 0
or 4x + 5 = 0, then 4x = -5 ⇒ x = $$\frac { -5 }{ 4 }$$
x = $$\frac { -5 }{ 4 }$$ or 0
Question 7.
Solution:
3x² – 243 = 0
x² – 81 =0 (Dividing by 3)
⇒ (x)² – (9)² = 0
⇒ (x + 9) (x – 9) = 0
Either, x + 9 = 0, then x = -9
or x – 9 = 0, then x = 9
Hence, x = 9 or -9
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
Question 14.
Solution:
Question 15.
Solution:
Question 16.
Solution:
4x² – 9x = 100
4x² – 9x – 100 = 0
Question 17.
Solution:
Question 18.
Solution:
Question 19.
Solution:
Question 20.
Solution:
Question 21.
Solution:
√3 x² + 10x + 7√3 = 0
Question 22.
Solution:
Question 23.
Solution:
3√7 x² + 4x + √7 = 0
Question 24.
Solution:
Question 25.
Solution:
Question 26.
Solution:
3x² – 2√6x + 2 = 0
Question 27.
Solution:
Question 28.
Solution:
Question 29.
Solution:
Question 30.
Solution:
Question 31.
Solution:
Question 32.
Solution:
Question 33.
Solution:
Question 34.
Solution:
Question 35.
Solution:
Question 36.
Solution:
Question 37.
Solution:
Question 38.
Solution:
Question 39.
Solution:
Question 40.
Solution:
Question 41.
Solution:
Question 42.
Solution:
Question 43.
Solution:
Question 44.
Solution:
Question 45.
Solution:
Question 46.
Solution:
Question 47.
Solution:
Question 48.
Solution:
Question 49.
Solution:
Question 50.
Solution:
Question 51.
Solution:
Question 52.
Solution:
Question 53.
Solution:
Question 54.
Solution:
Question 55.
Solution:
Question 56.
Solution:
Question 57.
Solution:
Question 58.
Solution:
Question 59.
Solution:
⇒ x = -2
Roots, x = -2
Question 60.
Solution:
Question 61.
Solution:
Question 62.
Solution:
Question 63.
Solution:
Question 64.
Solution:
Question 65.
Solution:
Question 66.
Solution:
Question 67.
Solution:
Question 68.
Solution:
Question 69.
Solution:
Question 70.
Solution:
Question 71.
Solution:
Question 72.
Solution:
Question 73.
Solution:
Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D
NCERT Maths Solutions for Chapter 3 Ex 3.4 Class 10 acts as the best resource during your learning and helps you score well in your board exams.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.
### RS Aggarwal Solutions Class 10 Chapter 3
Show that each of the following systems of equations has a unique solution and solve it:
Question 1.
Solution:
Question 2.
Solution:
Question 3.
Solution:
This system has a unique solution.
From (ii), x = 2 + 2y
Substituting the value of x in (i),
2(2 + 2y) + 3y = 18
=> 4 + 4y + 3y = 18
=> 7y = 18 – 4 = 14
=> y = 2
and x = 2 + 2 x 2 = 2 + 4 = 6
x = 6, y = 2
Find the value of k for which each of the following systems of equations has a unique solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
Question 14.
Solution:
Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:
Question 15.
Solution:
Question 16.
Solution:
Question 17.
Solution:
Question 18.
Solution:
Question 19.
Solution:
Question 20.
Solution:
=> k (k – 6) = 0
Either k = 0, which is not true, or k – 6 = 0, then k = 6
k = 6
Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
Question 21.
Solution:
Question 22.
Solution:
Question 23.
Solution:
Question 24.
Solution:
Question 25.
Solution:
Question 26.
Solution:
Find the value of k for which each of the following systems of equations has no solution:
Question 27.
Solution:
Question 28.
Solution:
kx + 3y = 3
12x + ky = 6
Question 29.
Solution:
Question 30.
Solution:
Question 31.
Solution:
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A
NCERT Maths Solutions for Ex 2.1 class 10 Polynomials is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.
## RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A
Other Exercises
Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question 1.
Solution:
Question 2.
Solution:
x² – 2x – 8
Let f(x) = x² – 2x – 8
Question 3.
Solution:
Question 4.
Solution:
4x² – 4x – 3
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
Zeros of a quadratic polynomial are 2, -6
Question 14.
Solution:
Question 15.
Solution:
Sum of zeros = 8
Product of zeros = 12
Quadratic equation will be x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 8x + 12 = 0
=> x² – 6x – 2x + 12 = 0
=> x (x – 6) – 2 (x – 6) = 0
=> (x – 6) (x – 2) = 0
Either x – 6 = 0, then x = 6
or x – 2 = 0, then x = 2
Zeros are 6, 2
and quadratic polynomial is x² – 8x + 12
Question 16.
Solution:
Sum of zeros = 0
and product of zeros = -1
Quadratic equation will be
x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 0x – 1 = 0
=> x² – 1= 0
(x + 1)(x – 1) = 0
Either x + 1 = 0, then x = -1 or x – 1 =0, then x = 1
Zeros are 1, -1
and quadratic polynomial is x² – 1
Question 17.
Solution:
Sum of zeros = $$\frac { 5 }{ 2 }$$
Product of zeros = 1
Quadratic equation will be
x² – (Sum of zeros) x + Product of zeros = 0
and quadratic polynomial is 2x² – 5x + 2
Question 18.
Solution:
Question 19.
Solution:
Question 20.
Solution:
Question 21.
Solution:
One zero of the given polynomial is $$\frac { 2 }{ 3 }$$
=> (x + 3) (x + 3) = 0
x = -3, -3
Hence, other zeros are -3, -3
Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B
Are you looking for the best Maths NCERT Solutions Chapter 3 Ex 3.2 Class 10? Then, grab them from our page and ace up your preparation for CBSE Class 10 Exams.
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B
Solve for x and y:
Question 1.
Solution:
x + y = 3 …..(i)
4x – 3y = 26 …(ii)
From (i), x = 3 – y
Substituting the value of x in (ii),
4(3 – y) – 3y = 26
=> 12 – 4y – 3y = 26
-7y = 26 – 12 = 14
y = -2
x = 3 – y = 3 – (-2) = 3 + 2 = 5
Hence, x = 5, y = -2
Question 2.
Solution:
Question 3.
Solution:
2x + 3y= 0 ……..(i)
3x + 4y = 5 …….(ii)
Question 4.
Solution:
2x – 3y = 13 ……(i)
7x – 2y = 20 ……….(ii)
Question 5.
Solution:
=> x = -2, y = -5
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Hence, x = $$\frac { 3 }{ 2 }$$ , y = $$\frac { -2 }{ 3 }$$
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Multiplying each term by 10
4x + 3y = 17
0.7x – 2y = 8
Multiply (i) by 2 and (ii) by 3,
8x + 6y = 34
21x – 6y = 24
29x = 58
x = 2
From (i) 4 x 2 + 3y = 17
=> 8 + 3y = 17
=> 3y = 17 – 8 = 9
y = 3
x = 2, y = 3
Question 14.
Solution:
Question 15.
Solution:
Question 16.
Solution:
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
6x + 5y = 7x + 3y + 1
=> 6x + 5y – 7x – 3y = 1
=> -x + 2y = 1
=> 2y – x = 1 …(i)
7x + 3y + 1 = 2(x + 6y – 1)
7x + 3y + 1 = 2x + 12y – 2
=> 7x + 3y – 2x – 12y = -2 – 1
=> 5x – 9y = -3 …..(ii)
From (i), x = 2y – 1
Substituting the value of x in (ii),
5(2y – 1) – 9y = -3
=> 10y – 5 – 9y = -3
=> y = -3 + 5
=> y = +2
x = 2y – 1 = 2 x 2 – 1 = 4 – 1 = 3
x = 3, y = 2
Question 17.
Solution:
and x = y – 4 = 6 – 4 = 2
x = 2, y = 6
Question 18.
Solution:
Question 19.
Solution:
Question 20.
Solution:
x = 3, y = -1
Question 21.
Solution:
Question 22.
Solution:
Question 23.
Solution:
Question 24.
Solution:
Question 25.
Solution:
Question 26.
Solution:
Question 27.
Solution:
x = 3, y = 2
Question 28.
Solution:
Question 29.
Solution:
Question 30.
Solution:
Question 31.
Solution:
Question 32.
Solution:
71x + 37y = 253
37x + 71y = 287
108x + 108y = 540
x + y = 5 ………(i) (Dividing by 108)
and subtracting,
34x – 34y = -34
x – y = -1 ……..(ii) (Dividing by 34)
Adding, (i) and (ii)
2x = 4 => x = 2
and subtracting,
2y = 6 => y = 3
Hence, x = 2, y = 3
Question 33.
Solution:
217x + 131y = 913 …(i)
131x + 217y = 827 …..(ii)
348x + 348y = 1740
x + 7 = 5 …..(iii) (Dividing by 348)
and subtracting,
86x – 86y = 86
x – y = 1 …(iv) (Dividing by 86)
Now, adding (iii) and (iv)
2x = 6 => x = 3
and subtracting,
2y = 4 => y = 2
x = 3, y = 2
Question 34.
Solution:
23x – 29y = 98 ……(i)
29x – 23y = 110 ……(ii)
52x – 52y = 208
x – y = 4 ……(iii) (Dividing by 52)
and subtracting
-6x – 6y = -12
x + y = 2 …..(iv)
2x = 6 => x = 3
Subtracting (iii) from (iv)
2y = -2 => y = -1
x = 3, y = -1
Question 35.
Solution:
x = 1 and y = 2
Question 36.
Solution:
Question 37.
Solution:
Question 38.
Solution:
Question 39.
Solution:
Question 40.
Solution:
Question 41.
Solution:
Question 42.
Solution:
Question 43.
Solution:
Question 44.
Solution:
a = $$\frac { 1 }{ 2 }$$ , b = $$\frac { 1 }{ 3 }$$
Question 45.
Solution:
Question 46.
Solution:
Question 47.
Solution:
Question 48.
Solution:
Question 49.
Solution:
a²x + b²y = c² ……(i)
b²x + a²y = d² …….(ii)
Multiply (i) by a² and (ii) by b²,
Question 50.
Solution:
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E
NCERT Maths Solutions for Ex 3.5 class 10 Linear equations is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.
Question 1.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
According to the conditions,
5x + 4y = ₹ 5600 …(i)
4x + 3y = ₹ 4340 …(ii)
x = -560
and from (i)
5 x 560 + 4y = 5600
2800 + 4y = 5600
⇒ 4y = 5600 – 2800
⇒ 4y = 2800
⇒ y = 700
Cost of one chair = ₹ 560
and cost of one table = ₹ 700
Question 2.
Solution:
Let the cost of one spoon = ₹ x and cost of one fork = ₹ y
According to the conditions,
23x + 17y = 1770 …(i)
17x + 23y = 1830 …(ii)
40x + 40y = 3600
Dividing by 40,
x + y = 90 …(iii)
and subtracting,
6x – 6y = -60
Dividing by 6,
x – y = -10 …(iv)
Adding (iii) and (iv)
2x = 80 ⇒ x = 40
and subtracting,
2y = 100 ⇒ y = 50
Cost of one spoon = ₹ 40
and cost of one fork = ₹ 50
Question 3.
Solution:
Let number of 25-paisa coins = x
and number 50-paisa coins = y
Total number of coins = 50
and total amount = ₹ 19.50 = 1950 paisa
x + y = 50 …(i)
25x + 50y = 1950
⇒ x + 2y = 78 …(ii)
Subtracting (i) from (ii), y = 28
x = 50 – y = 50 – 28 = 22
Number of 25-paisa coins = 22
and 50-paisa coins = 28
Question 4.
Solution:
Sum of two numbers = 137
and difference = 43
Let first number = x
and second number = y
x + y = 137 …..(i)
x – y = 43 ……(ii)
2x = 180 ⇒ x = 90
and subtracting,
2y = 94
y = 47
First number = 90
and second number = 47
Question 5.
Solution:
Let first number = x
and second number = y
According to the conditions,
2x + 3y = 92 …(i)
4x – 7y = 2 …(ii)
Multiply (i) by 2 and (ii) by 1
4x + 6y = 184 …..(iii)
4x – 7y = 2 …….(iv)
Subtracting (iii) from (iv),
13y = 182
y = 14
From (i), 2x + 3y = 92
2x + 3 x 14 = 92
⇒ 2x + 42 = 92
⇒ 2x = 92 – 42 = 50
⇒ x = 25
First number = 25
Second number = 14
Question 6.
Solution:
Let first number = x
and second number = y
According to the conditions,
3x + y=142 …(i)
4x – y = 138 …(ii)
7x = 280
⇒ x = 40
and from (i)
3 x 40 + y = 142
⇒ 120 + y = 142
⇒ y = 142 – 120 = 22
First number = 40,
second number = 22
Question 7.
Solution:
Let first greater number = x
and second smaller number = y
According to the conditions,
2x – 45 = y …(i)
2y – 21 = x …(ii)
Substituting the value of y in (ii),
2 (2x – 45) – 21 = x
⇒ 4x – 90 – 21 = x
⇒ 4x – x = 111
⇒ 3x = 111
⇒ x = 37
From (i),
y = 2 x 37 – 45 = 74 – 45 = 29
The numbers are 37, 29
Question 8.
Solution:
Let larger number = x
and smaller number = y
According to the conditions,
3x = 4 x y + 8 ⇒ 3x = 4y + 8 …….(i)
5y = x x 3 + 5 ⇒ 5y = 3x + 5 …(ii)
Substitute the value of 3x in (ii),
5y = 4y + 8 + 5
⇒ 5y – 4y = 13
⇒ y = 13
and 3x = 4 x 13 + 8 = 60
⇒ x = 20
Larger number = 20
and smaller number = 13
Question 9.
Solution:
Let first number = x and
second number = y
According to the conditions,
⇒ 11x – 44 = 5(2x + 2) – 20
⇒ 11x – 44 = 10x + 10 – 20
⇒ 11x – 10x = 10 – 20 + 44
⇒ x = 34
and y = 2 x 34 + 2 = 68 + 2 = 70
Numbers are 34 and 70
Question 10.
Solution:
Let first number = x
and second number (smaller) = y
According to the conditions,
x – y = 14
and x² – y² = 448
⇒ (x + y) (x – y) = 448
⇒ (x + y) x 14 = 448
⇒ x + y = 32 ……(i)
and x – y = 14 ……(ii)
Adding (i) and (ii),
2x = 46 ⇒ x = 23
and subtracting (i) and (ii),
2y = 18 ⇒ y = 9
Numbers are 23, 9
Question 11.
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Number = x + 10y
By interchanging the digits,
Ones digit = y
and tens digit = x
Number = y + 10x
According to the conditions,
x + y = 12 ………. (i)
y + 10x = x + 10y + 18
⇒ y + 10x – x – 10y = 18
⇒ 9x – 9y = 18
⇒ x – y = 2 …(ii) (Dividing by 9)
Adding (i) and (ii),
2x = 14 ⇒ x = 7
and subtracting,
2y = 10 ⇒ y = 5
Number = 7 + 10 x 5 = 7 + 50 = 57
Question 12.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After reversing the digits,
Ones digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 27 = y + 10x
⇒ y + 10x – x – 10y = -27
⇒ 9x – 9y = -27
⇒ x – y = -3 …(i)
and 7 (x + y) = x + 10y
7x + 7y = x+ 10y
⇒ 7x – x = 10y – 7y
⇒ 6x = 3y
⇒ 2x = y …(ii)
Substituting the value of y in (i)
x – 2x = -3
⇒ -x = -3
⇒ x = 3
y = 2x = 2 x 3 = 6
Number = x + 10y = 3 + 10 x 6 = 3 + 60 = 63
Question 13.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(i)
and x + y= 15 …(ii)
2x = 16
x = 8
and subtracting,
2y = 14
⇒ y = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78
Question 14.
Solution:
Let one’s digit of the two digit number = x
and ten’s digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
Now, according to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and 4(x + y) + 3 = x + 10y
4x + 4y + 3 = x + 10y
⇒ 4x + 4y – x – 10y = -3
3x – 6y = -3
⇒ x – 2y = -1 ……..(ii)
Subtracting,
y = 3
and x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5
Number = x + 10y = 5 + 10 x 3 = 5 + 30 = 35
Question 15.
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 9 = y + 10x
⇒ x + 10y – y – 10x = 9
⇒ -9x + 9y = 9
⇒x – y = -1 …(i) (Dividing by -9)
Question 16.
Solution:
Let the one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and xy = 35 …(ii)
Now, (x + y)² = (x – y)² + 4xy = (2)² + 4 x 35 = 4 + 140 = 144 = (12)²
⇒ (x + y) = 12 …(iii)
Subtracting (i) from (iii), we get
Question 17.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits One’s digit = y
Ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y – 63 = y + 10x
Question 18.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + 10y + y + 10x = 121
⇒ 11x + 11y = 121
⇒ x + y = 11 …(i)
x – y = 3 …(ii)
2x = 14 ⇒ x = 7
Subtracting,
2y = 8 ⇒ y = 4
Number = 7 + 10 x 4 = 7 + 40 = 47
or 4 + 10 x 7 = 4 + 70 = 74
Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
x + y = 8 …(i)
Question 20.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
$$\frac { x + 2 }{ y }$$ = $$\frac { 1 }{ 2 }$$
$$\frac { x }{ y – 1 }$$ = $$\frac { 1 }{ 3 }$$
⇒ 2x + 4 = y …(i)
3x = y – 1 …(ii)
⇒ 3x = 2x + 4 – 1
⇒ 3x = 2x + 3
⇒ 3x – 2x = 3
⇒ x = 3
and y = 2x + 4 = 2 x 3 + 4 = 6 + 4 = 10
Fraction = $$\frac { 3 }{ 10 }$$
Question 21.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
y – x = 11
y = 11 + x …(i)
Question 22.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
Question 23.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
x + y = 4 + 2x
⇒ y = 4 + x …(i)
Fraction = $$\frac { x }{ y }$$ = $$\frac { 5 }{ 9 }$$
Question 24.
Solution:
Let first number = x
and second number = y
According to the conditions,
x + y = 16
Question 25.
Solution:
Let in classroom A, the number of students = x
and in classroom B = y
According to the conditions,
x – 10 = y + 10
⇒ x – y = 10 + 10 = 20
⇒ x – y = 20 …(i)
and x + 20 = 2 (y – 20)
⇒ x + 20 = 2y – 40
⇒ x – 2y = -(40 + 20) = -60
x – 2y = -60 …(ii)
Subtracting, y = 80
and x – y = 20
⇒ x – 80 = 20
⇒ x = 20 + 80 = 100
Number of students in classroom A = 100 and in B = 80
Question 26.
Solution:
Let fixed charges = ₹ x
and other charges = ₹ y per km
According to the conditions,
For 80 km,
x + 80y = ₹ 1330 …(i)
and x + 90y = ₹ 1490 …(ii)
Subtracting (i) from (ii),
10y = 160 ⇒ y = 16
and from (i)
x + 80 x 16 = 1330
⇒ x + 1280 = 1330
⇒ x = 1330 – 1280 = 50
Fixed charges = ₹ 50
and rate per km = ₹ 16
Question 27.
Solution:
Let fixed charges of the hostel = ₹ x
and other charges per day = ₹ y
According to the conditions,
x + 25y = 4500 ……..(i)
x + 30y = 5200 ……(ii)
Subtracting (i) from (ii),
5y = 700
y = 140
and from (i),
x + 25 x 140 = 4500
⇒ x + 3500 = 4500
⇒ x = 4500 – 3500 = 1000
Fixed charges = ₹ 1000
and per day charges = ₹ 140
Question 28.
Solution:
Let first investment = ₹ x
and second investment = ₹ y
Rate of interest = 10% p.a. for first kind and 8% per second
Interest is for the first investment = ₹ 1350
and for the second = ₹ 1350 – ₹45 = ₹ 1305
According to the conditions,
Question 29.
Solution:
Ratio in the income of A and B = 5 : 4
Let A’s income = ₹ 5x and
B’s income = ₹ 4x
and ratio in their expenditures = 7 : 5
Let A’s expenditure = 7y
and B’s expenditure = 5y
According to the conditions,
5x – 7y = 9000 …(i)
and 4x – 5y = 9000 …(ii)
Multiply (i) by 5 and (ii) by 7,
25x – 35y = 45000
28x – 37y = 63000
Subtracting, we get
3x = 18000
⇒ x = 6000
A’s income = 5x = 5 x 6000 = ₹ 30000
and B’s income = 4x = 4 x 6000 = ₹ 24000
Question 30.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
In first case,
Profit on chair = 25%
and on table = 10%
and selling price = ₹ 1520
According to the conditions,
Question 31.
Solution:
Distance between two stations A and B = 70 km
Let speed of first car (starting from A) = x km/hr
and speed of second car = y km/hr
According to the conditions,
7x – 7y = 70
⇒ x – y = 10 …(i)
and x + y = 70 …(ii)
Adding (i) and (ii),
2x = 80 ⇒ x = 40
Subtracting (i) and (ii),
2y = 60 ⇒ y = 30
Speed of car A = 40 km/hr
and speed of car B = 30 km/hr
Question 32.
Solution:
Let uniform speed of the train = x km/hr
and time taken = y hours
Distance = x x y = xy km
Case I:
Speed = (x + 5) km/hr
and Time = (y – 3) hours
Distance = (x + 5) (y – 3)
(x + 5) (y – 3) = xy
⇒ xy – 3x + 5y – 15 = xy
-3x + 5y = 15 …(i)
Case II:
Speed = (x – 4) km/hr
and Time = (y + 3) hours
Distance = (x – 4) (y + 3)
(x – 4) (y + 3) = xy
⇒ xy + 3x – 4y – 12 = xy
3x – 4y = 12 …(ii)
Adding (i) and (ii),
y = 27
and from (i),
-3x + 5 x 27 = 15
⇒ -3x + 135 = 15
⇒ -3x = 15 – 135 = -120
⇒ x = 40
Speed of the train = 40 km/hr
and distance = 27 x 40 = 1080 km
Question 33.
Solution:
Let the speed of the train = x km/hr
and speed of taxi = y km/hr
According to the conditions,
Question 34.
Solution:
Distance between stations A and B = 160 km
Let the speed of the car starts from A = x km/hr
and speed of car starts from B = y km/hr
8x – 8y = 160
⇒ x – y = 20 …(i)
and 2x + 2y = 160
⇒ x + y = 80 …(ii)
Adding (i) and (ii)
2x = 100 ⇒ x = 50
and subtracting,
2y = 60 ⇒ y = 30
Speed of car starting from A = 50 km/hr
and from B = 30 km/hr
Question 35.
Solution:
Distance = 8 km
Let speed of sailor in still water = x km/hr
and speed of water = y km/hr
According to the conditions,
Question 36.
Solution:
Let speed of a boat = x km/hr
and speed of stream = y km/hr
According to the conditions,
Question 37.
Solution:
Let a man can do a work in x days
His 1 day’s work = $$\frac { 1 }{ x }$$
and a boy can do a work in y days
His 1 day’s work = $$\frac { 1 }{ y }$$
According to the conditions,
Question 38.
Solution:
Let length of a room = x m
and breadth = y m
and area = xy m²
According to the conditions,
x = y + 3 …(i)
(x + 3) (y – 2) = xy
xy – 2x + 3y – 6 = xy
-2x + 3y = 6 …(ii)
-2 (y + 3) + 3y = 6 [From (i)]
-2y – 6 + 3y = 6
⇒ y = 6 + 6 = 12
x = y + 3 = 12 + 3 = 15 …(ii)
Length of room = 15 m
and breadth = 12 m
Question 39.
Solution:
Let length of a rectangle = x m
and breadth = y m
Then area = x x y = xy m²
According to the conditions,
(x – 5) (y + 3) = xy – 8
⇒ xy + 3x – 5y – 15 = xy – 8
⇒ 3x – 5y = -8 + 15 = 7 …..(i)
and (x + 3) (y + 2) = xy + 74
⇒ xy + 2x + 3y + 6 = xy + 74
⇒ 2x + 3y = 74 – 6 = 68 …(ii)
Multiply (i) by 3 and (ii) by 5
Question 40.
Solution:
Let length of a rectangle = x m
and breadth = y m
Then area = xy m²
According to the conditions,
(x + 3) (y – 4) = xy – 67
⇒ xy – 4x + 3y – 12 = xy – 67
⇒ -4x + 3y = -67 + 12 = -55
⇒ 4x – 3y = 55 …(i)
and (x – 1) (y + 4) = xy + 89
⇒ xy + 4x – y – 4 = xy + 89
⇒ 4x – y = 89 + 4 = 93 ….(ii)
⇒ y = 4x – 93
Substituting the value of y in (i),
4x – 3(4x – 93) = 55
⇒ 4x – 12x + 279 = 55
⇒ -8x = 55 – 279 = -224
⇒ x = 28
and y = 4x – 93 = 4 x 28 – 93 = 112 – 93 = 19
Length of rectangle = 28 m
and breadth = 19 m
Question 41.
Solution:
Let reservation charges = ₹ x
and cost of full ticket from Mumbai to Delhi
According to the conditions,
x + y = 4150 …(i)
2x + $$\frac { 3 }{ 2 }$$ y = 6255
⇒ 4x + 3y = 12510 …(ii)
From (i), x = 4150 – y
Substituting the value of x in (ii),
4 (4150 – y) + 3y = 12510
⇒ 16600 – 4y + 3y = 12510
-y = 12510 – 16600
-y = -4090
⇒ y = 4090
and x = 4150 – y = 4150 – 4090 = 60
Reservation charges = ₹ 60
and cost of 1 ticket = ₹ 4090
Question 42.
Solution:
Let present age of a man = x years
and age of a son = y years
5 year’s hence,
Man’s age = x + 5 years
and son’s age = y + 5 years
x + 5 = 3 (y + 5) = 3y + 15
⇒ x – 3y = 15 – 5 = 10
x = 10 + 3y …(i)
and 5 years ago,
Man’s age = x – 5 years
and son’s age = y – 5 years
x – 5 = 7 (y – 5) = 7y – 35
x = 7y – 35 + 5 = 7y – 30 …(ii)
From (i) and (ii),
10 + 3y = 7y – 30
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
⇒ y = 10
and x = 10 + 3y = 10 + 3 x 10 = 10 + 30 = 40
Present age of a man = 40 years
and of son’s age = 10 years
Question 43.
Solution:
Let present age of a man = x years
and age of his son = y years
2 years ago,
Man’s age = x – 2 years
Son’s age = y – 2 years
x – 2 = 5 (y – 2)
⇒ x – 2 = 5y – 10
x = 5y – 10 + 2 = 5y – 8 …(i)
2 years later,
Man’s age = x + 2 years
and son’s age = y + 2 years
x + 2 = 3(y + 2) + 8
x + 2 = 3y + 6 + 8
⇒ x = 3y + 6 + 8 – 2 = 3y + 12 …(ii)
From (i) and (ii),
5y – 8 = 3y + 12
⇒ 5y – 3y = 12 + 8
⇒ 2y = 20
⇒ y = 10
and x = 5y – 8 = 5 x 10 – 8 = 50 – 8 = 42
Present age of man = 42 years
and age of son = 10 years
Question 44.
Solution:
Let age of father = x years
and age of his son = y years
According to the conditions,
2y + x = 10 …(i)
2x + y = 95 …(ii)
From (i),
x = 70 – 2y
Substituting the value of x in (ii),
2 (70 – 2y) + y = 95
⇒ 140 – 4y + y = 95
⇒ -3y = 95 – 140 = -45
⇒ -3y = -45
⇒ y = 15
and x = 70 – 2y = 70 – 2 x 15 = 70 – 30 = 40
Age of father = 40 years
and age of his son = 15 years
Question 45.
Solution:
Let present age of a woman = x years
and age of her daughter = y years
According to the conditions,
x = 3y + 3 …(i)
3 years hence,
Age of woman = x + 3 years
and age of her daughter = y + 3 years
x + 3 = 2 (y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒x = 2y + 16 – 3 = 2y + 13 …(ii)
From (i),
3y + 3 = 2y + 13
⇒ 3y – 2y = 13 – 3
⇒ y = 10
and x = 3y + 3 = 3 x 10 + 3 = 30 + 3 = 33
Present age of woman = 33 years
and age of her daughter = 10 years
Question 46.
Solution:
Let cost price of tea set = ₹ x
and of lemon set = ₹ y
According to the conditions,
Question 47.
Solution:
Let fixed charges = ₹ x (for first three days)
and then additional charges for each day = ₹ y
According to the conditions,
Mona paid ₹ 27 for 7 dyas
x + (7 – 3) x y = 27
⇒ x + 4y = 27
and Tanvy paid ₹ 21 for 5 days
x + (5 – 3) y = 21
⇒ x + 2y = 21 …(ii)
Subtracting,
2y = 6 ⇒ y = 3
But x + 2y = 21
⇒ x + 2 x 3 = 21
⇒ x + 6 = 21
⇒ x = 21 – 6 = 15
Fixed charges = ₹ 15
and additional charges per day = ₹ 3
Question 48.
Solution:
Let x litres of 50% solution be mixed with y litres of 25% solution, then
x + y = 10 …(i)
Subtracting (i) from (ii),
x = 6
and x + y = 10
⇒ 6 + y = 10
⇒ y = 10 – 6 = 4
50% solution = 6 litres
and 25% solution = 4 litres
Question 49.
Solution:
Let x g of 18 carat be mixed with y g of 12 carat gold to get 120 g of 16 carat gold, then
x + y = 120 …(i)
Now, gold % in 18-carat gold = $$\frac { 18 }{ 24 }$$ x 100 = 75%
⇒ 3x + 2y = 320 …(ii)
From (i),
x = 120 – y
Substituting the value of x in (ii),
3 (120 – y) + 2y = 320
⇒ 360 – 3y + 2y = 320
⇒ -y = 320 – 360
⇒ -y = -40
⇒ y = 40
and 40 + x = 120
⇒ x = 120 – 40 = 80
Hence, 18 carat gold = 80 g
and 12-carat gold = 40 g
Question 50.
Solution:
Let x litres of 90% pure solution be mixed withy litres of 97% pure solution to get 21 litres of 95% pure solution. Then,
x + y = 21 …(i)
⇒ 90x + 97y = 1995
From (i), x = 21 – y
Substituting the value of x in (ii),
90 (21 – y) + 97y = 1995
⇒ 1890 – 90y + 97y = 1995
⇒ 7y = 1995 – 1890 = 105
⇒ y =15
and x = 21 – y = 21 – 15 = 6
90% pure solution = 6 litres
and 97% pure solution = 15 litres
Question 51.
Solution:
Let larger supplementary angle = x°
and smaller angle = y°
According to the conditions,
x + y = 180° …(i)
x = y + 18° …(ii)
From (i),
y + 18° + y = 180°
⇒ 2y = 180° – 18° = 162°
⇒ 2y = 162°
⇒ y = 81°
and x= 180°- 81° = 99°
Hence, angles are 99° and 81°
Question 52.
Solution:
In ∆ABC,
∠A = x, ∠B = (3x – 2)°, ∠C = y°, ∠C – ∠B = 9°
Question 53.
Solution:
In a cyclic quadrilateral ABCD,
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C.
### RS Aggarwal Solutions Class 10 Chapter 3
Solve each of the following systems of equations by using the method of cross multiplication:
Question 1.
Solution:
Question 2.
Solution:
Question 3.
Solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
x = 15, y= 5
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself.
### RS Aggarwal Solutions Class 10 Chapter 3
MCQ
Question 1.
Solution:
(a)
Question 2.
Solution:
(d)
Question 3.
Solution:
(a)
Question 4.
Solution:
(d)
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Let first number = x
and second number = y
According to the conditions, x – y = 26 …(i)
and x = 3y …..(ii)
From (i),
3y – y = 26
⇒ 2y = 26
⇒ y = 13
and x = 3 x 13 = 39
Numbers are 39 and 13
Short-Answer Questions (3 marks)
Question 11.
Solution:
23x + 29y = 98 …..(i)
29x + 23y = 110 …..(ii)
Adding, we get 52x + 52y = 208
x + y = 4 …..(iii) (Dividing by 52)
and subtracting,
-6x + 6y = -12
x – y = 2. …..(iv) (Dividing by -6)
Adding (iii) and (iv),
2x = 6 ⇒ x = 3
Subtracting,
2x = 2 ⇒ y = 1
Hence, x = 3, y = 1
Question 12.
Solution:
x = 1, y = $$\frac { 3 }{ 2 }$$
Question 13.
Solution:
Question 14.
Solution:
Question 15.
Solution:
Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the condition,
5x + 7y = 195 …(i)
7x + 5y= 153 …(ii)
Adding, (i) and (ii)
12x + 12y = 348
x + y = 29 ….(iii) (Dividing by 12)
and subtracting,
-2x + 2y = 42
-x + y = 21 …..(iv) (Dividing by -2)
Now, Adding (iii) and (iv),
2y = 50 ⇒ y = 25
and from (iv),
-x + 25 = 21 ⇒ -x = 21 – 25 = -4
x = 4
Cost of one pencil = ₹ 4
and cost of one pen = ₹ 25
Question 16.
Solution:
2x – 3y = 1, 4x – 3y + 1 = 0
2x – 3y = 1
2x = 1 + 3y
x = $$\frac { 1 + 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below
Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.
Similarly,
4x – 3y + 1 = 0
⇒ 4x = 3y – 1
⇒ x = $$\frac { 3y – 1 }{ 4 }$$
Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).
Hence, x = -1, y = -1
Question 17.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°
But ∠A + ∠C = 180°
4x + 20° + 4y° = 180°
⇒ 4x + 4y = 180° – 20 = 160°
x + y = 40° …(i) (Dividing by 4)
and ∠B + ∠D = 180°
⇒ 3x – 5 + 7y + 5 = 180°
⇒ 3x + 7y = 180° …(ii)
From (i), x = 40° – y
Substituting the value of x in (ii),
3(40° – y) + 7y = 180°
⇒ 120° – 3y + 7y = 180°
⇒ 4y = 180°- 120° = 60°
y = 15°
and x = 40° – y = 40° – 15° = 25°
∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°
∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°
∠C = 4y = 4 x 15 = 60°
∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°
Question 18.
Solution:
Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Fraction = $$\frac { x }{ y }$$
According to the conditions,
Question 20.
Solution:
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself are helpful to complete your math homework.
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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS.
### RS Aggarwal Solutions Class 10 Chapter 3
Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c)
Question 2.
Solution:
(c)
Question 3.
Solution:
(a)
Question 4.
Solution:
(d)
Question 5.
Solution:
(a)
Question 6.
Solution:
(b)
Question 7.
Solution:
(c) 4x + 6y = 3xy, 8x + 9y = 5xy
Dividing each term by xy,
Question 8.
Solution:
(a)
Question 9.
Solution:
(c)
Question 10.
Solution:
(b)
Question 11.
Solution:
(d)
Question 12.
Solution:
(b)
Question 13.
Solution:
(a)
Question 14.
Solution:
(d)
Question 15.
Solution:
(d)
Question 16.
Solution:
(d)
Question 17.
Solution:
(d)
Question 18.
Solution:
(d) The system of equations is consistent then their graph lines will be either intersecting or coincident.
Question 19.
Solution:
(a) The pair of lines of equation is inconsistent then the system will not have no solution i.e., their lines will be parallel.
Question 20.
Solution:
(b)
Question 21.
Solution:
(b) ABCD is a cyclic quadrilateral
∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°
∠A + ∠C = 180°
Now, x + y + 10°+ x + y – 30° = 180°
⇒ 2x + 2y – 20 = 180°
⇒ 2x + 2y = 180° + 20° = 200°
⇒ x + y = 100° …(i)
and ∠B + ∠D = 180°
⇒ y + 20° + x + y = 180°
⇒ x + 2y = 180° – 20° = 160° …(ii)
Subtracting,
-y = -60° ⇒ y = 60°
and x + 60° = 100°
⇒ x = 100° – 60° = 40°
Now, ∠B = y + 20° = 60° + 20° = 80°
Question 22.
Solution:
(d) Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + y = 15 …(i)
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(ii)
Adding (i) and (ii),
2x = 16 ⇒ x = 8
and x + y = 15
⇒ 8 + y = 15
⇒ y = 15 – 8 = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78
Question 23.
Solution:
(b) Let the numerator of a fractions = x
and denominator = y
Question 24.
Solution:
(d) Let present age of man = x years
and age of his son = y years
5 years hence,
Age of man = (x + 5) years
and age of son = y + 5 years
(x + 5) = 3 (y + 5)
⇒ x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……(i)
and 5 years earlier
Age of man = x – 5 years
and age of son = y – 5 years
x – 5 = 7 (y – 5)
x – 5 = 7y – 35
⇒ x = 7y – 35 + 5
x = 7y – 30 ……….(ii)
From (i) and (ii),
7y – 30 = 3y + 10
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
y = 10
x = 3y + 10 = 3 x 10 + 10 = 30 + 10 = 40
Present age of father = 40 years
Question 25.
Solution:
(b)
Question 26.
Solution:
(c)
Question 27.
Solution:
(a)
The system has infinitely many solutions.
The lines are coincident.
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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F.
### RS Aggarwal Solutions Class 10 Chapter 3
Very-Short and Short-Answer Questions
Question 1.
Solution:
Question 2.
Solution:
Question 3.
Solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Let first, number = x
and second number = y
x – y = 5
Question 8.
Solution:
Let cost of one pen = ₹ x
and cost of one pencil = ₹ y
According to the conditions,
5x + 8y = 120 …(i)
8x + 5y = 153 …(ii)
13x + 13y = 273
x + y = 21 …(iii) (Dividing by 13)
and subtracting (i) from (ii),
3x – 3y = 33
⇒ x – y = 11 …….(iv) (Dividing by 3)
Again adding (iii) and (iv),
2x = 32 ⇒ x = 16
Subtracting,
2y = 10 ⇒ y = 5
Cost of 1 pen = ₹ 16
and cost of 1 pencil = ₹ 5
Question 9.
Solution:
Let first number = x
and second number = y
According to the conditions,
and x + y = 80
⇒ x + 15 = 80
x = 80 – 15 = 65
Numbers are : 65, 15
Question 10.
Solution:
Let one’s digit of a two digits number = x
and ten’s digit = y
Number = x + 10y
By reversing its digits One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + y = 10 …(i)
x + 10y – 18 = y + 10x
x+ 10y – y – 10x = 18
⇒ -9x + 9y = 18
⇒ x – y = -2 (Dividing by -9) …..(ii)
Adding (i) and (ii),
2x = 8 ⇒ x = 4
and by subtracting,
2y = 12 ⇒ y = 6
Number = x + 10y = 4 + 10 x 6 = 4 + 60 = 64
Question 11.
Solution:
Let number of stamps of 20p = x
and stamps of 25 p = y
According to the conditions,
x + y = 47 …..(i)
20x + 25y = 1000
4x + 5y = 200 …(ii)
From (i), x = 47 – y
Substituting the value of x in (ii),
4 (47 – y) + 5y = 200
188 – 4y + 5y = 200
⇒ y = 200 – 188 = 12
and x + y = 47
⇒ x + 12 = 47
⇒ x = 47 – 12 = 35
Hence, number of stamps of 20 p = 35
and number of stamps of 25 p = 12
Question 12.
Solution:
Let number of hens = x
and number of cows = y
According to the conditions,
x + y = 48 …..(i)
x x 2 + y x 4 = 140
⇒ 2x + 4y = 140
⇒ x + 2y = 70 ……(ii)
Subtracting (i) from (ii),
y = 22
and x + y = 48
⇒ x + 22 = 48
⇒ x = 48 – 22 = 26
Number of hens = 26
and number of cows = 22
Question 13.
Solution:
Question 14.
Solution:
Question 15.
Solution:
12x + 17y = 53 …(i)
17x + 12y = 63 …(ii)
Adding, 29x + 29y = 116
Dividing by 29,
x + y = 4 …(iii)
Subtracting,
-5x + 5y = -10
⇒ x – y = 2 …(iv) (Dividing by -5)
Adding (iii) and (iv)
2x = 6 ⇒ x = 3
Subtracting,
2y = 2 ⇒ y = 1
x = 3, y = 1
x + y = 3 + 1 = 4
Question 16.
Solution:
Question 17.
Solution:
kx – y = 2
6x – 2y = 3
Question 18.
Solution:
Question 19.
Solution:
Question 20.
Solution:
Question 21.
Solution:
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F are helpful to complete your math homework.
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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A.
### RS Aggarwal Solutions Class 10 Chapter 3
Solve each of the following systems of equations graphically.
Question 1.
Solution:
2x + 3y = 2 …..(i)
x – 2y = 8 …(ii)
From Eq. (i),
⇒ 2x = 2 – 3y
⇒x = $$\frac { 2 – 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Now, plot the points (1, 0), (-2, 2) and (4, -2) on the graph and join them to get a line.
Similarly x – 2y = 8 ⇒ x = 8 + 2y
Now plot the points (6, -1), (4, -2) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at point(4, -2).
x = 4, y = -2
Question 2.
Solution:
3x + 2y = 4
⇒ 3x = 4 – 2y
⇒ x = $$\frac { 4 – 2y }{ 3 }$$
Giving some different values to y, we get corresponding values of x as given below:
Now plot the points (0, 2), (2, -1) and (4, -4) on the graph and join them to get a line. Similarly,
2x – 3y = 7
2x = 3y+ 7
x = $$\frac { 3y+ 7 }{ 2 }$$
Now plot the points on the graph and join them to get another line.
We see that these two lines intersect each other at point (2, -1).
x = 2, y = -1
Question 3.
Solution:
2x + 3y = 8
⇒ 2x = 8 – 3y
x = $$\frac { 8 – 3y }{ 2 }$$
Now, giving some different values to y, we get corresponding values of x as given below
Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
x – 2y + 3 = 0
x = 2y – 3
Now, plot the points (-3, 0), (-1, 1) and (1, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (1, 2).
x = 1, y = 2
Question 4.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = $$\frac { 5y – 4 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Now, plot the points (-2, 0) (3, 2) and (8,4) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
⇒ y = 8 – 2x
Plot the points (1, 6), (2, 1) and (3, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (3, 2).
x = 3, y = 2
Question 5.
Solution:
3x + 2y = 12
⇒ 3x = 12 – 2y
x = $$\frac { 12 – 2y }{ 2 }$$
Giving some different values to y, we get corresponding the values of x as given below:
Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
5x – 2y = 4
⇒ 5x = 4 + 2 y
⇒ x = $$\frac { 4 + 2 y }{ 5 }$$
Plot the points (0, -2), (2, 3) and (4, 8) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, 3).
x = 2, y = 3
Question 6.
Solution:
3x + y + 1 = 0 ⇒y = -3x – 1
Giving some different values to x, we get corresponding values of y as given below:
Now, plot the points (0, -1), (-1, 2) and (-2, 5) on the graph and join them to get a line.
Similarly,
2x – 3y + 8 = 0
2x = 3y – 8
x = $$\frac { 3y – 8 }{ 2 }$$
Now, plot the points (-4, 0), (-1, 2) and (2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the points (-1, 2).
x = -1, y = 2
Question 7.
Solution:
2x + 3y + 5 = 0
2x = -3y – 5
x = $$\frac { -3y – 5 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Now, plot the points (-4, 1), (-1, -1) and (2, -3) on the graph and join them to get a line. Similarly,
3x – 2y – 12 = 0
⇒ 3x = 2y + 12
x = $$\frac { 2y + 12 }{ 3 }$$
Plot the points (4, 0), (0, -6) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -3).
x = 2, y = -3
Question 8.
Solution:
2x – 3y + 13 = 0
⇒ 2x = 3y – 13
⇒ x = $$\frac { 3y – 13 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Now, plot the points (-5, 1), (-2, 3) and (1, 5) on the graph and join them to get a line.
Similarly,
3x – 2y + 12 = 0
3x = 2y – 12
x = $$\frac { 2y – 12 }{ 3 }$$
Now, plot the points (-4, 0), (-2, 3) and (0, 6) on the graph and join them to get another line.
We see that these lines intersect each other at the point (-2, 3).
x = -2, y = 3
Question 9.
Solution:
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
⇒ x = $$\frac { 4 – 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (2, 0) (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
3x – y + 5 = 0
⇒ -y = -5 – 3x
⇒ y = 5 + 3x
Plot the points (0, 5), (-1, 2) and (-2, -1) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (-1, 2).
x = -1, y = 2
Question 10.
Solution:
x + 2y + 2 = 0
⇒ x = – (2y + 2)
Giving some different values to y, we get corresponding the values of x as given below:
Plot the points (-2,0), (0, -1) and (2, -2) on the graph and join them to get a line.
Similarly,
3x + 2y – 2 = 0
⇒ 3x = 2 – 2y
x = $$\frac { 2 – 2y }{ 3 }$$
Plot the points (0, 1), (2, -2) and (-2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -2).
x = 2, y = -2
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:
Question 11.
Solution:
x – y + 3 = 0
⇒ x = y – 3
Giving some different value to y, we get corresponding values of x as given below:
Plot the points (-3, 0), (-1, 2) and (0, 3) on the graph and join them to get a line.
Similarly,
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
x = $$\frac { 4 – 3y }{ 2 }$$
Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get another line.
We see that these line intersect each other at (-1, 2) and x-axis at A (-3, 0) and D (2, 0).
Area of ∆BAD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AD x BL
= $$\frac { 1 }{ 2 }$$ x 5 x 2 = 5 sq.units
Question 12.
Solution:
2x – 3y + 4 = 0
⇒ 2x = 3y – 4
x = $$\frac { 3y – 4 }{ 2 }$$
Giving some different values to y, we get corresponding value of x as given below:
Plot the points (-2, 0), (1, 2) and (4, 4) on the graph and join them to get a line.
Similarly,
x + 2y – 5 = 0
x = 5 – 2y
Now, plot the points (5, 0), (3, 1) and (1, 2) on the graph and join them to get another line.
We see that there two lines intersect each other at the point B (1, 2) and intersect x- axis at A (-2, 0) and D (5, 0) respectively.
Now, area of ∆BAD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AD x BL
= $$\frac { 1 }{ 2 }$$ x 7 x 2
= 7 sq. units
Question 13.
Solution:
4x – 3y + 4 = 0
⇒ 4x = 3y – 4
x = $$\frac { 3y – 4 }{ 4 }$$
Giving some different values to y, we get corresponding value of x as given below:
Plot the points (-1, 0), (2, 4) and (-4, -4) on the graph and join them, to get a line.
Similarly,
4x + 3y – 20 = 0
4x = 20 – 3y
x = $$\frac { 20 – 3y }{ 4 }$$
Plot the points (5, 0), (2, 4) and (-1, 8) on the graph and join them to get a line.
We see that these two lines intersect each other at the point B (2, 4) and intersect x-axis at A (-1, 0) and D (5, 0).
Area ∆BAD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AD x BL
= $$\frac { 1 }{ 2 }$$ x 6 x 4 = 12 sq. units
Question 14.
Solution:
x – y + 1 = 0 ⇒ x = y – 1
Giving some different values toy, we get the values of x as given below:
Plot the points (-1, 0), (0, 1) and (1, 2) on the graph and join them to get a line.
Similarly,
3x + 2y – 12 = 0
⇒ 3x = 12 – 2y
x = $$\frac { 12 – 2y }{ 3 }$$
Plot the points (4, 0), (2, 3) and (0, 6) on the graph and join them to get another line.
We see that these two lines intersect each ohter at the point E (2, 3) and intersect x- axis at A (-1, 0) and D (4, 0).
Area of ∆EAD = $$\frac { 1 }{ 2 }$$ x base x altitude 1
= $$\frac { 1 }{ 2 }$$ x AD x EL
= $$\frac { 1 }{ 2 }$$ x 5 x 3
= $$\frac { 15 }{ 2 }$$
= 7.5 sq. units
Question 15.
Solution:
x – 2y + 2 = 0 ⇒ x = 2y – 2
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (-2, 0), (0, 1), (2, 2) on the graph and join them to get a line.
Similarly,
2x + y – 6 = 0
⇒ y = 6 – 2x
Plot the points on the graph and join them to get a line.
We see that these two lines intersect each other at the point C (2, 2) and intersect the x-axis at A (-2, 0) and F (3, 0).
Now, area of ∆CAF = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AF x CL
= $$\frac { 1 }{ 2 }$$ x 5 x 2 = 5 sq. units
Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
Question 16.
Solution:
2x – 3y + 6 = 0
⇒ 2x = 3y – 6
x = $$\frac { 3y – 6 }{ 2 }$$
Giving some different values to y, we get corresponding values of x, as given below:
Now, plot the points (-3, 0), (0, 2), (3, 4) on the graph and join them to get a line.
Similarly,
2x + 3y – 18 = 0
2x = 18 – 3y
x = $$\frac { 18 – 3y }{ 2 }$$
Plot the points (0, 6), (3, 4) and (6, 2) on the graph and join them to get a line.
We see that these two lines intersect each other at C (3, 4) and intersect y-axis at B (0, 2) and D (0, 6).
Now, area of ∆CBD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x BD x CL
= $$\frac { 1 }{ 2 }$$ x 4 x 3 sq. units = 6 sq. units
Question 17.
Solution:
4x – y – 4 = 0
⇒ 4x = y + 4
x = $$\frac { y + 4 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (1, 0), (0, -4) and (2, 4) on the graph and join them to get a line.
Similarly,
3x + 2y – 14 = 0
2y = 14 – 3x
y = $$\frac { 14 – 3x }{ 2 }$$
Now, plot the points (0, 7), (2, 4) and (4, 1) on the graph and join them to get another line.
We see that these two lines intersect each other at C (2, 4) and y-axis at B (0, -4) and D (0, 7) respectively.
Area of ∆CBD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x BD x CL
= $$\frac { 1 }{ 2 }$$ x 11 x 2 sq. units = 11 sq. units
Question 18.
Solution:
x – y – 5 = 0 ⇒ x = y + 5
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (5, 0), (0, -5) and (1, -4) on the graph and join these to get a line.
Similarly,
3x + 5y – 15 = 0
3x = 15 – 5y
x = $$\frac { 15 – 5y }{ 3 }$$
Plot the points (5, 0), (0, 3) and (-5, 6) on the graph and join them to get a line.
We see that these two lines intersect each other at A (5, 0) and y-axis at B (0, -5) and E (0, 3) respectively.
Now area of ∆ABE = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x BE x AL
= $$\frac { 1 }{ 2 }$$ x 8 x 5 sq. units = 20 sq. units
Question 19.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = $$\frac { 5y – 4 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (-2, 0), (3, 2) and (-7, -2) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
2x = 8 – y
x = $$\frac { 8 – y }{ 2 }$$
Plot the points (4, 0), (3, 2) and (2, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at B (3, 2) and y-axis at G (0, 1) and H (0, 8) respectively.
Now area of ∆EGH = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x GH x EL
= $$\frac { 1 }{ 2 }$$ x 7 x 3 sq. units
= $$\frac { 21 }{ 2 }$$ = 10.5 sq. units
Question 20.
Solution:
5x – y – 7 = 0 ⇒ y = 5x – 7
Giving some different values to x, we get corresponding values of y as given below:
Plot the points (0, -7), (1, -2), (2, 3) on the graph and join them to get a line.
Similarly,
x – y + 1 = 0 ⇒ x = y – 1
Plot the points (-1, 0), (1, 2) and (2, 3) on the graph and join them to get a line.
We see that these two lines intersect each other at C (2, 3) and intersect y-axis at A (0, -7) and F (0, 1) respectively.
Now, area of ∆CAF = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AF x CL
= $$\frac { 1 }{ 2 }$$ x 8 x 2 = 8 sq.units
Question 21.
Solution:
2x – 3y – 12 ⇒ 2x = 12 + 3y
x = $$\frac { 12 + 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (6, 0), (3, -2) and (0, -4) on the graph and join them to get a line.
Similarly,
x + 3y = 6 ⇒ x = 6 – 3y
Plot the points (6, 0), (0, 2) and (-6, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at the points A (6, 0) and intersect the y-axis at C (0, -4) and E (0, 2) respectively.
Now area of ∆ACE = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x CE x AO
= $$\frac { 1 }{ 2 }$$ x 6 x 6 sq. units = 18 sq.units
Show graphically that each of the following given systems of equations has infinitely many solutions:
Question 22.
Solution:
2x + 3y = 6
2x = 6 – 3y
x = $$\frac { 6 – 3y }{ 2 }$$
Giving some different values to y, we get the corresponding val ues of x as given below:
Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get a line.
Similarly,
4x + 6y = 12
4x = 12 – 6y
x = $$\frac { 12 – 6y }{ 2 }$$
on the graph and join them to get a line.
We see that all the points lie on the same straight line.
This system has infinite many solutions.
Question 23.
Solution:
3x – y = 5
⇒ y = 3x – 5
Giving some different values to x, we get corresponding values of y as shown below:
Now, plot the points (0, -5), (1, -2), (2, 1) on the graph and join them to get a line:
Similarly,
6x – 2y = 10
⇒ 6x = 10 + 2y
x = $$\frac { 10 + 2y }{ 6 }$$
Plot the points (2, 1), (4, 7), (3, 4) on the graph and join them to get another line.
We see that these lines coincide each other.
This system has infinite many solutions.
Question 24.
Solution:
2x + y = 6
y = 6 – 2x
Giving some different values to x, we get corresponding values of y as given below:
Plot the points (0, 6), (2, 2), (4, -2) on the graph and join them to get a line.
Similarly,
6x + 3y = 18
⇒ 6x = 18 – 3y
⇒ x = $$\frac { 18 – 3y }{ 2 }$$
Now, plot the points (3, 0), (1, 4) and (5, -4) on the graph and join them to get another line.
We see that these two lines coincide each other.
This system has infinitely many solutions
Question 25.
Solution:
x – 2y = 5 ⇒ x = 5 + 2y
Giving some different values to y, we get corresponding values of x as shown below:
Plot the points (5, 0), (3, -1), (1, -2) on the graph and join them to get a line.
similarly,
3x – 6y = 15
⇒ 3x = 15 + 6y
x = $$\frac { 15 + 6y }{ 3 }$$
Plot the points (7, 1), (-1, -3) and (-3, -4) on the graph and join them to get another line.
We see that all the points lie on the same line.
Lines coincide each other.
Hence, the system has infinite many solutions.
Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:
Question 26.
Solution:
x – 2y = 6 ⇒ x = 6 + 2y
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (6, 0), (4, -1) and (0, -3) on the graph and join them to get a line.
Similarly,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y
Plot the points (0, 0), (2, 1), (4, 2) on the graph and join them to get a line.
We see that these two lines are parallel i.e., do not intersect each other.
This system has no solution.
Question 27.
Solution:
2x + 3y = 4
⇒ 2x = 4 – 3y
⇒ x = $$\frac { 4 – 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:
Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
4x + 6y = 12 ⇒ 4x = 12 – 6y
x = $$\frac { 12 – 6y }{ 4 }$$
Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get another line.
We see that two lines are parallel i.e., these do not intersect each other at any point.
Therefore the system has no solution.
Question 28.
Solution:
2x + y = 6, y = 6 – 2x
Giving some different values to x, we get corresponding values ofy as given below:
Plot the points (0, 6), (1, 4) and (3, 0) on the graph and join them to get a line.
Similarly,
6x + 3y = 20 ⇒ 6x = 20 – 3y
We see that these two lines are parallel and do not intersect each other.
Therefore this system has no solution.
Question 29.
Solution:
2x + y = 2 ⇒ y = 2 – 2x
Giving some different values to x, we get corresponding values of y as given below:
Plot the points (0, 2), (1, 0) and (-2, 6) on the graph and join them to get a line.
Similarly,
2x + y = 6 ⇒ y = 6 – 2x
Plot the points (0, 6), (2, 2) and (3, 0) on the graph and join them to get another line.
ABFD is the trapezium whose vertices are A (0, 2), B (1, 0), F (3, 0), D (0, 6).
Area of trapezium ABFD = Area ∆DOF – Area ∆AOB
= $$\frac { 1 }{ 2 }$$ (DO x OF) – $$\frac { 1 }{ 2 }$$ (AO x OB)
= $$\frac { 1 }{ 2 }$$ (6 x 3) – $$\frac { 1 }{ 2 }$$ (2 x 1) sq.units
= 9 – 1 = 8 sq.units
Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself.
Other Exercises
MCQ
Question 1.
Solution:
In the given figure,
PT is the tangent and PQ is the chord of the circle with centre O.
∠OPT = 50°
OP is radius and PT is the tangent.
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ + ∠QPT = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 40°
In ∆OPQ,
∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°) = 180° – 80°
= 100° (b)
Question 2.
Solution:
Angle between two radii of a circle = 130°
Then ∠APB = 180° – ∠AOB
= 180°- 130° = 50° (c)
Question 3.
Solution:
In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O
∠APB = 80°
OA is radius of the circle and AP is the tangent
OA ⊥ AP ⇒ ∠OAP = 90°
OP bisects ∠APB,
∠APO = $$\frac { 1 }{ 2 }$$ x 80 = 40°
∠POA = 180° – (90° + 40°)
= 180° – 130° = 50° (b)
Question 4.
Solution:
In the given figure, AD and AE are tangents to the circle with centre O.
BC is the tangent at F which meets AD at C and AE at B
AE = 5 cm
AE and AD are the tangents to the circle
AE = AD = 5 cm
Tangents from an external point drawn to the circle are equal
CD = CF and BE = BF
Now, perimeter of ∆ABC = AB + AC + BC
= AB + AC + BF + CF (BE = BF and CF = CD)
= AB + AC + BE + CD
= AB + BE + AC + CD
= AE + AD
= 5 + 5 = 10 cm (b)
Question 5.
Solution:
In the given figure, a quadrilateral ABCD is circumscribed a circle touching its sides at P, Q, R and S respectively.
AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm
A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD …(i)
Now, AP and AS are tangents to the circle
AP = AS = 5 cm …(ii)
Similarly, CQ = CR = 3 cm
BP = BQ = x – 5 = 4
BQ = BC – CQ = 7 – 3 = 4 cm
x – 5 = 4
⇒ x = 4 + 5 = 9cm
Question 6.
Solution:
In the given figure, PA and PB are the tangents drawn from P to the circle.
OA and OB are the radii of the circle and AP and BP are the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
∠A + ∠B = 90° + 90° = 180°
But these are opposite angles of a quadrilateral
AOBP is a cyclic quadrilateral
A, O, B, P are concyclic
Question 7.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O from an external point P.
∠PBA = 65°,
To find : ∠OAB and ∠APB
In ∆APB
AP = BP (Tangents from P to the circle)
∠PAB = ∠PBA = 65°
∠APB = 180° – (∠PAB + ∠PBA)
= 180° – (65° + 65°) = 180° – 130° = 50°
OA is radius and AP is tangent
OA ⊥ AP
∠OAP = 90°
∠OAB = ∠OAP – ∠PAB = 90° – 65° = 25°
Hence, ∠OAB = 25° and ∠APB = 50°
Question 8.
Solution:
Given : In the figure,
BC and BD are the tangents drawn from B
to the circle with centre O.
∠CBD = 120°
To prove : OB = 2BC
Construction : Join OB.
Proof: OB bisects ∠CBD (OC = OD and BC = BD)
Question 9.
Solution:
(i) A line intersecting a circle in two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinitely many tangents.
Question 10.
Solution:
Given : In a circle, from an external point P, PA and PB are the tangents drawn to the circle with centre O.
To prove : PA = PB
Construction : Join OA, OB and OP.
Proof : OA and OB are the radii of the circle and AP and BP are tangents.
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
PA = PB (c.p.c.t.)
Hence proved.
Question 11.
Solution:
Given : In a circle with centre O and AB is its diameter.
From A and B, PQ and RS are the tangents drawn to the circle
To prove : PQ || RS
Proof : OA is radius and PAQ is the tangent
OA ⊥ PQ
∠PAO = 90° …(i)
Similarly, OB is the radius and RBS is tangent
∠OBS = 90° …(ii)
From (i) and (ii)
∠PAO = ∠OBS
But there are alternate angles
PQ || RS
Question 12.
Solution:
Given : In the given figure,
In ∆ABC,
AB = AC.
A circle is inscribed the triangle which touches it at D, E and F
To prove : BE = CE
Proof: AD and AF are the tangents drawn from A to the circle
But, AB = AC
AB – AD = AC -AF
⇒ BD = CF
But BD = BE and CF = CE (tangent drawn to the circle)
But BD = CF
BE = CE
Hence proved.
Question 13.
Solution:
Given : In a circle from an external point P, PA and PB are the tangents to the circle
OP, OA and OB are joined.
To prove: ∠POA = ∠POB
Proof: OA and OB are the radii of the circle and PA and PB are the tangents to the circle
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
∠POA = ∠POB (c.p.c.t.)
Hence proved.
Question 14.
Solution:
Given : A circle with centre O, PA and PB are the tangents drawn from A and B which meets at P.
AB is chord of the circle
To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP
Proof: OA is radius and AP is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
Similarly, OB ⊥ BP ⇒ ∠OBP = 90°
In ∆OAB, OA = OB (radii of the circle)
∠OAB = ∠OBA
⇒ ∠OAP – ∠OAB = ∠OBP – ∠OBA
⇒ ∠PAB = ∠PBA
Hence proved.
Question 15.
Solution:
Given : A parallelogram ABCD is circumscribed a circle.
To prove : ABCD is a rhombus.
Proof: In a parallelogram ABCD.
Opposite sides are equal and parallel.
AB = CD and AD = BC
Tangents drawn from an external point of a circle to the circle are equal.
AP = AS BP = BQ
CQ = CR and DR = DS
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC [AB = CD and AD = BC]
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
But AB = CD and BC = AD
AB = BC = CD = AD
Hence || gm ABCD is a rhombus.
Question 16.
Solution:
Given : O is the centre of two concentric circles with radii 5 cm and 3 cm respectively.
AB is the chord of the larger circle which touches the smaller circle at P.
OP and OA are joined.
To find : Length of AB
Proof: OP is the radius of the smaller circle and touches the smaller circle at P
OP ⊥ AB and also bisects AB at P
AP = PB = $$\frac { 1 }{ 2 }$$ AB
Now, in right ∆OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
Hence AB = 2 x AP = 2 x 4 = 8 cm
Question 17.
Solution:
In the figure, quad. ABCD is circumscribed about a circle which touches its sides at P, Q, R and S respectively
To prove : AB + CD = AD + BC
Proof: Tangents drawn from an external point to a circle are equal
AP = AS
BP = BQ
CR = CQ
DR = DS
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence AB + CD = AD + BC
Question 18.
Solution:
Given : A quad. ABCD circumscribe a circle with centre O and touches at P, Q, R and S respectively
OA, OB, OC and OD are joined forming angles AOB, BOC, COD and DOA
To prove : ∠AOB + ∠COD = 180°
and ∠BOC + ∠AOD = 180°
Construction : Join OP, OQ, OR and OS
Proof: In right ∆AOP and ∆AOS,
Side OP = OS (radii of the same circle)
Hyp. OA = OA (common)
∆AOP = ∆AOS (RHS axiom)
∠1 = ∠2 (c.p.c.t.)
Similarly, we can prove that
∠4 = ∠3
∠5 = ∠6
∠8 = ∠7
∠1 + ∠4 + ∠5 + ∠8 = ∠2 + ∠3 + ∠6 + ∠7
⇒ (∠1 + ∠8) + (∠4 + ∠5) = (∠2 + ∠3) + (∠6 + ∠7)
⇒ ∠AOB + ∠COD = ∠AOD + ∠BOC
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (angles at a point)
∠AOB + ∠COD = ∠AOD + ∠BOC = 180°
Hence proved
Question 19.
Solution:
Given : From an external point P, PA and PB are the tangents drawn to the circle,
OA and OB are joined.
To prove : ∠APB + ∠AOB = 180°
Construction : Join OP.
Proof : Now, in ∆POA and ∆PBO,
OP = OP (common)
PA = PB (Tangents from P to the circle)
OA = OB (Radii of the same circle)
∆POA = ∆PBO (SSS axiom)
∠APO = ∠BPO (c.p.c.t.)
and ∠AOP = ∠BOP (c.p.c.t.)
OA and OB are the radii and PA and PB are the tangents
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
In ∆POA,
∠OAP = 90°
∠APO + ∠AOP = 90°
Similarly, ∠BPO + ∠BOP = 90°
(∠APO + ∠BPO) + (∠AOP + ∠BOP) = 90° + 90°
⇒ ∠APB + ∠AOB = 180°.
Hence proved.
Question 20.
Solution:
Given : PQ is chord of a circle with centre O.
TP and TQ are tangents to the circle
Radius of the circle = 10 cm
i.e. OP = OQ = 10 cm and PQ = 16 cm
To find : The length of TP.
OT bisects the chord PQ at M at right angle.
PM = MQ = $$\frac { 16 }{ 2 }$$ = 8 cm
In right ∆PMO,
OP² = PM² + MO² (Pythagoras Theorem)
⇒ (10)² = (8)² + MO²
⇒ 100 = 64 + MO²
⇒ MO² = 100 – 64 = 36 = (6)²
MO = 6 cm
Let TP = x and TM = y
In right ∆TPM,
TP² = TM² + PM²
⇒ x² = y² + 8²
⇒ x² = y² + 64 …(i)
and in right ∆TPM
OT² = TP² + OP²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y + 36 = y2 + 64 + 100 {From (i)}
⇒ 12y = 64 + 100 – 36 = 128
Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
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# Expressions with One or More Variables
## Evaluate expressions given values for variables.
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Expressions with One or More Variables
Joshua loves people and so he is working at the ticket counter at the zoo. His job is to count the people entering the zoo each day. He does this twice. He counts them in the morning and in the afternoon. Sometimes he has more people come in the morning and sometimes the counts are higher in the afternoon. Joshua loves his job. He loves figuring out how much money the zoo has made from the ticket sales. Joshua has a thing for mental math. While many of his friends think it is too difficult, Joshua enjoys figuring it out in his head.
To enter the zoo for the day, it costs an adult \begin{align*}\7.00\end{align*} and a child \begin{align*}\5.00\end{align*}. Joshua has written the following expression to help him to figure out the amount of money that the zoo makes in half a day. He divides his arithmetic up between the morning and the afternoon.
Here are his counts for Monday.
AM - 65 adults and 75 children
PM - 35 adults and 50 children
Here are his counts for Tuesday.
AM - 70 adults and 85 children
PM - 50 adults and 35 children
Given these counts, how much revenue (money) was collected at the zoo for the entire day on Monday? How much money was collected at the zoo for the entire day on Tuesday? How much money was collected in the two days combined? Joshua can figure this out using his expression.
Can you?
In this Concept, you will learn how to use a variable expression to solve a real-world problem. Pay close attention. You will need these skills to figure out the zoo revenue for Monday and Tuesday.
### Guidance
In this Concept, you are going to learn how to evaluate expressions that have multiple variables and multiple operations.
Let’s see what this looks like.
Evaluate \begin{align*} 6a+b \end{align*} when \begin{align*}a\end{align*} is 4 and \begin{align*}b\end{align*} is 6. First, of all, you can see that there are two variables in this expression. There are also two operations here. The first one is multiplication: \begin{align*}6a\end{align*} lets us know that we are going to multiply 6 times the value of \begin{align*}a\end{align*}. The second one is addition: the \begin{align*}+ \ b\end{align*} lets us know that we are going to add the value of \begin{align*}b\end{align*}. We have also been given the values of \begin{align*}a\end{align*} and \begin{align*}b\end{align*}. We substitute the given values for each variable into the expression and evaluate it.
Our answer is 29. Notice that we used the order of operations when working through this problem.
Order of Operations
P - parentheses
E - exponents
MD - multiplication and division in order from left to right
AS - addition and subtraction in order from left to right
Whenever we are evaluating expressions with more than one operation in them, always refer back and use the order of operations.
Let’s look at another example with multiple variables and expressions.
Evaluate \begin{align*} 7b-d \end{align*} when \begin{align*}b\end{align*} is 7 and \begin{align*}d\end{align*} is 11. First, we substitute the given values in for the variables.
What about when we have a dilemma that is all variables?
Evaluate \begin{align*} ab + cd \end{align*} when \begin{align*}a\end{align*} is 4, \begin{align*}b\end{align*} is 3, \begin{align*}c\end{align*} is 10 and \begin{align*}d\end{align*} is 6.
We work on this one in the same way as the other examples. Begin by substituting the given values in for the variables.
We have two multiplication problems here and one addition. Next, we follow the order of operations to evaluate the expression.
Now it is time for you to try a few on your own.
#### Example A
Evaluate \begin{align*}12x - y \end{align*} when \begin{align*}x\end{align*} is 4 and \begin{align*}y\end{align*} is 9.
Solution: 39
#### Example B
Evaluate \begin{align*}\frac{12}{a} + 4\end{align*} when \begin{align*}a\end{align*} is 3.
Solution: 8
#### Example C
Evaluate \begin{align*}5x + 3y \end{align*} when \begin{align*}x\end{align*} is 4 and \begin{align*}y\end{align*} is 8.
Solution: 52
Now back to Joshua and the tickets. Here is the problem once again.
Joshua loves people and so he is working at the ticket counter. His job is to count the people entering the zoo each day. He does this twice. He counts them in the morning and in the afternoon. Sometimes he has more people come in the morning and sometimes the counts are higher in the afternoon. Joshua loves his job. He loves figuring out how much money the zoo has made from the ticket sales. Joshua has a thing for mental math. While many of his friends think it is too difficult, Joshua enjoys figuring it out in his head.
To enter the zoo for the day, it costs an adult \begin{align*}\7.00\end{align*} and a child \begin{align*}\5.00\end{align*}. Joshua has written the following expression to help him to figure out the amount of money that the zoo makes in half a day. He divides his arithmetic up between the morning and the afternoon.
Here are his counts for Monday.
AM - 65 adults and 75 children
PM - 35 adults and 50 children
Here are his counts for Tuesday.
AM - 70 adults and 85 children
PM - 50 adults and 35 children
Begin by underlining all of the important information in the problem, done here already.
First, we can start with Monday. Our expression remains the same. We can use \begin{align*} 7x+5y \end{align*}.
For Monday morning, the zoo had 65 adults and 75 children visit. Those are the given values that we can substitute into our expression for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
For Monday afternoon, the zoo had 35 adults and 50 children visit. Those are the given values that we can substitute into our expression for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
The total amount of money made on Monday is \begin{align*}830 + 495 = \1325.\end{align*} Next, we can figure out Tuesday. For Tuesday morning, the zoo had 70 adults and 85 children visit. Those are the given values for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
For Tuesday afternoon, the zoo had 50 adults and 35 children visit. Those are the given values for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
The total amount of money made on Tuesday is \begin{align*}915 + 525 = \1440\end{align*}.
If we wanted to figure out the total amount of revenue for both days combined, we simply add the two totals together.
### Vocabulary
Evaluate
to simplify an expression that does not have an equals sign.
Variable
a letter, usually lowercase, that is used to represent an unknown quantity.
Expression
a number sentence that uses operations but does not have an equals sign
Variable Expression
a number sentence that has variables or unknown quantities in it with one or more operations and no equals sign.
Revenue
means money
### Guided Practice
Here is one for you to try on your own.
Evaluate \begin{align*} a + ab + cd \end{align*} when \begin{align*}a\end{align*} is 4, \begin{align*}b\end{align*} is 9, \begin{align*}c\end{align*} is 6 and \begin{align*}d\end{align*} is 4.
First, we have to substitute the given values into the expression.
\begin{align*}4 + 4(9) + 6(4)\end{align*}
Now we can evaluate using the order of operations.
\begin{align*}4 + 36 + 24\end{align*}
\begin{align*}64\end{align*}
### Practice
Directions: Evaluate each multi-variable expression when \begin{align*}x = 2 \end{align*} and \begin{align*}y = 3\end{align*}.
1. \begin{align*}2x + y\end{align*}
2. \begin{align*}9x - y \end{align*}
3. \begin{align*}x + y \end{align*}
4. \begin{align*}xy \end{align*}
5. \begin{align*}xy + 3 \end{align*}
6. \begin{align*}9y - 5 \end{align*}
7. \begin{align*}10x - 2y \end{align*}
8. \begin{align*}3x + 6y \end{align*}
9. \begin{align*}2x + 2y \end{align*}
10. \begin{align*}7x - 3y \end{align*}
11. \begin{align*}3y - 2 \end{align*}
12. \begin{align*}10x - 8 \end{align*}
13. \begin{align*}12x - 3y \end{align*}
14. \begin{align*}9x + 7y \end{align*}
15. \begin{align*}11x - 7y \end{align*}
### Vocabulary Language: English
algebraic
algebraic
The word algebraic indicates that a given expression or equation includes variables.
Algebraic Expression
Algebraic Expression
An expression that has numbers, operations and variables, but no equals sign.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Order of Operations
Order of Operations
The order of operations specifies the order in which to perform each of multiple operations in an expression or equation. The order of operations is: P - parentheses, E - exponents, M/D - multiplication and division in order from left to right, A/S - addition and subtraction in order from left to right.
Parentheses
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
revenue
revenue
Revenue is money that is earned.
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
Variable Expression
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity. |
# How would you find the slope of the line parallel to the line y=1/3x+2?
Apr 25, 2017
See below.
#### Explanation:
When two lines are parallel, that means they do not intersect each other at all. For this to be true, the two lines must have the same slope. If you want a line parallel to one line with the slope of $\frac{1}{3}$ , then the slope of that line must also, by definition, be $\frac{1}{3}$.
Apr 25, 2017
$\frac{1}{3}$
#### Explanation:
By definition, two lines that are parallel to each other have the same slope. Otherwise, the two lines will cross one another not being parallel.
The given equation is given in slope-intercept form:
y = mx + b
Note: m = slope; b = y-intercept
Therefore, the slope of the equation,
y = $\frac{1}{3}$x + 2
is $\frac{1}{3}$ |
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# Let f(x) = $\alpha {{x}^{2}}-2+\dfrac{1}{x}$ where $\alpha$ is a real constant. The smallest $\alpha$ for which f(x)$\ge$0 for all x>0 is ?(a) $\dfrac{{{2}^{2}}}{{{3}^{3}}}$(b) $\dfrac{{{2}^{3}}}{{{3}^{3}}}$(c) $\dfrac{{{2}^{4}}}{{{3}^{3}}}$(d) $\dfrac{{{2}^{5}}}{{{3}^{3}}}$
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Hint: To solve the above problem, we need to be aware about the basic concepts of the extremum (that is, minimum and maximum) of a function. We will use the principle of derivatives to solve this function. We will use the property that for minimum, we have,
f’(x) = 0 and f’’(x) > 0 [Here, f’(x) is same as $\dfrac{d}{dx}\left( f(x) \right)$ and f’’(x) is same as $\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f(x) \right)$.
Now, before solving we try to understand the basics about extremum points. Basically, an extremum is a point of a function at which it has the highest (maximum) or lowest (minimum) value. In general, these points occur where there is a change in sign of the slope of the graph at that point. For example, we take a simple case of f(x) = ${{x}^{2}}$. Here, we see that the sign of the slope changes when we cross x=0. Thus, x = 0 is an extremum point. Now, we try to solve the problem in hand keeping above points in mind.
We have, f(x) = $\alpha {{x}^{2}}-2+\dfrac{1}{x}$. Now, to find extremum point, in general, we perform f’(x) = 0 [f’(x) is same as $\dfrac{d}{dx}\left( f(x) \right)$]
We know that, $\dfrac{d({{x}^{2}})}{dx}=2x,\text{ }\dfrac{d(\text{constant})}{dx}=0,\text{ }\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$, thus, we get,
f’(x) = $2\alpha x-\dfrac{1}{{{x}^{2}}}$ -- (1)
We equate this to zero, we get,
$2\alpha x-\dfrac{1}{{{x}^{2}}}$=0
\begin{align} & 2\alpha x=\dfrac{1}{{{x}^{2}}} \\ & {{x}^{3}}=\dfrac{1}{2\alpha } \\ & x={{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}} \\ \end{align}
Now, we want this point to be minimum since, we have to find the smallest $\alpha$, we have another condition,
f’’(x) > 0
Using $\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)=-\dfrac{2}{{{x}^{3}}}$, thus doing derivative of (1), we have,
f’’(x) = 2$\alpha$- ($-\dfrac{2}{{{x}^{3}}}$)
f’’(x) = 2$\alpha$+$\dfrac{2}{{{x}^{3}}}$
Now, in the question, for f’’(x) to be greater than zero, we have,
2$\alpha$+$\dfrac{2}{{{x}^{3}}}$> 0
$\alpha$> $-\dfrac{1}{{{x}^{3}}}$ - (2)
Now, we try to satisfy, f(x)$\ge$0 for $x={{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}}$,
thus, we have,
f${{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}}$$\ge$0
\begin{align} & \alpha {{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{2}{3}}}-2+{{\left( \dfrac{1}{2\alpha } \right)}^{-\dfrac{1}{3}}}\ge 0 \\ & {{\alpha }^{\dfrac{1}{3}}}{{2}^{-\dfrac{2}{3}}}-2+{{2}^{\dfrac{1}{3}}}{{\alpha }^{\dfrac{1}{3}}}\ge 0 \\ & {{\alpha }^{\dfrac{1}{3}}}\left( \dfrac{1+2}{{{2}^{\dfrac{2}{3}}}} \right)-2\ge 0 \\ \end{align}
\begin{align} & {{\alpha }^{\dfrac{1}{3}}}\left( \dfrac{3}{{{2}^{\dfrac{2}{3}}}} \right)\ge 2 \\ & {{\alpha }^{\dfrac{1}{3}}}\ge \dfrac{{{2}^{\dfrac{5}{3}}}}{3} \\ & \alpha \ge \dfrac{{{2}^{5}}}{{{3}^{3}}} \\ \end{align}
Thus, the minimum value of $\alpha =\dfrac{{{2}^{5}}}{{{3}^{3}}}$.
Note: We found the minimum of the function, since we had to find the minimum value of $\alpha$ for f(x) to be just greater than or equal to zero. Further, it is important to keep in mind the range of x while arriving at the answer. For example, in the question, it is given that x>0, thus, if we get an extremum point which is negative, we reject that point. |
# Significant part math courses
It’s crucial to understand exactly what the word signifies, so students can create their incline calculator in addition to examine their understanding about z abilities.
The incline is that the angle made. It is utilised in engineering and forestry to fix the incline of a surface. The incline of a street, for example, may be determined by choosing a road’s https://efinor.no/blog/2020/01/31/how-does-one-do-gcf-in-q/ parallel or perpendicular advantages together with some u shaped piece of timber or the assistance of the plank.
For example, the slope of an street could possibly be described by first using the term”continuous”. This means the horizontal space between the center of the point and the center of the lineup is equivalent to the length of the line divided from the point’s height. Subsequently a angle between both perpendicular and parallel lines can also be calculated. The incline of a road is in fact the angle between the plane of the slope along with the her response perpendicular edge of the trail.
The word”intermediate incline” describes the angle that’s between 2 angles. The intermediate slope is named 90 degrees if the two angles are both positive. The notion is it is the middle of the lineup, perhaps not the line, that is being considered.
The grade level may also be described as a means to describe a curve’s slope. Degree grades’ essential feature , however, is they are used to rate students’ grades. Levels are a tool employed in scoring for students with achievements.
The word”axis” is traditionally utilised to describe the line to which the slope of a curve goes. In the event the lineup is really a straight line, then the incline goes from left to right and from top to bottom. Even a”triangle” is among the many intersecting lines that intersect the point.
Actual pitch is a sort of incline that’s exhibited in statistics for mathematical functions. It may be utilised to represent variations within the curve. The incline of the cubic www.paramountessays.com or quadratic equation, for example, is popularly known as the derivative. Derivatives can be utilised generally in elementary school classes in trigonometry.
The word”reversed slope” is utilized to describe an inverted-U contour which moves through the center of the place. The shape is called a Mercator’s projection, After a plane is lying around the surface of the sphere. The regeneration of the square or square area is popularly called the hyperbola.
The term”acceleration slope” is just a special case of incline. In this case, means of a curve that’s going at the same speed because the rate of lighting describes the incline. The speed could be measured by using the rate of light (do ) along with also a management (from which the lighting is currently shifting ).
Because it’s two sides, the medial side that meets the side which meet precisely the parallel and the vertical A curve can be clarified by using slope for its equation. The incline might be plotted at a chart, along with some graphs which reveal it incorporate the tangent of the chart.
A curve’s incline is also essential in physics, so since it joins the rate of a projectile in accordance. The term”incline” was perhaps not a part of math, also it was used only in physics. In the 19th century, even the incline of this face was known to by”smeared” surfaces, and today,”incline” is still a portion of most math equations.
# Inside Math is definitely an outstanding class.
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Inside Mathematics is usually a study of chemistry, physics, and physical science fiction. These actually are fascinating fields we’ve got under no circumstances learned about in faculty. What I’d find I can use for my life and education and though my personal know-how, at the same time as reading this novel, could be the principles involved.
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## A Dog Shop is a Great Place to Purchase a Pet
If you own a dog and/or thinking about buying one, it’s https://wonder-pets.net/ probably a smart idea to check out your dog shop initial. In most cases you can save a lot of money on your new dog and have a happy animal companion that will grow up with you for many years to come.
While there are plenty of numerous places to acquire a pet, your dog shop supplies the convenience of buying in your own home and even at the office. Not only is this easy, but dog owners are much more likely to get their dogs from a pet store than anywhere else.
Naturally , some pet owners choose to shop by a pet shop, since they understand the items they want are precisely what they want. Most people like to shop at a pet store because the pets are so much cheaper. When buying a puppy, for example , the cost of the puppy plus the vaccinations could be substantially lower than buying a huge, dangerous doggie from a breeder.
When viewing a dog store, you can also want to view the various breeds. There are several prevalent breeds, including the Golden Retriever, Labrador retriever Retriever, Dachshund, Poodle, and Pekinese, and lastly, the Faustkämpfer.
Once you’ve chosen a breed, you’ll find that not only can you find a family pet store that provides everything you need, nevertheless also many more options to choose from. These include gadgets, collars, leashes, etc .
One of many nice reasons for having purchasing a family pet from an animal store is they usually take care of a pet as long as you’re away. Some family pet stores present free delivery to your home, which can be nice as the last thing you want is for the animal to get still left alone in the cold, or perhaps wet. You can also be able to produce phone calls might questions, along with see pictures of your pet in its new home. House animals can be very pricey, but in most cases you’ll find that you can get a great deal on a great family pet from an area pet retail outlet. You’ll find all the info you need by a dog store, so look around and see whatever you think.
## Selecting the best Kind Of Canine beds
Choosing the right kind of dog beds for your pet can be a real challenge. A lot of use cheap bed furniture and do not find the right a person. There are several choices, yet which ones will continue to work well to your dog?
Some folk like to maintain their canines on soft and luxurious beds. These kinds of could be vinyl, faux soft or even organic cotton. Plastic headboards can be a bit expensive because of the materials utilised in making them. Nevertheless , if you want to give your dog a lot of firm and nice comfort this is definitely an option.
Bed linens can be used in a lot of methods and can make a massive difference when it comes to your dog’s bed. You should use it with regards to tummy time, for walking around the property, even just for brushing its teeth. A couple of cute and colorful styles can be used to brighten up the look of your pet’s the sack.
Dog beds can be found in so many sizes and shapes and out of so many different brands. But keep in mind that the large amount of cash can be spent on a fantastic bed. There are a few other options you are able to consider.
The first thing you should do should be to determine how large your dog. You can aquire several sizes in the market, from portable to enormous sized. Also you can ask the help of a pet retail outlet clerk to determine if it is really necessary to buy extra bed linens.
It would be best to purchase a larger dog crib and the simply thing you need to worry about as if it will be cozy for your dog. When you buy bed linen for your puppy, you will also ought to provide right feeding and care. Nourishing is very important and you should not leave it until after when you decide to switch the dog’s bedding.
Should you really love your pup and really want him to feel good about being inside best dog beds your house, providing him with a dog bed can certainly help. Check out different choices available for you plus your dog.
## A glass Dog Doorways Is Great Options for Your Residence
Glass puppy doors are becoming very popular, nevertheless there are a lot of concerns that people currently have about this a glass type of door. First, a handful of definitions. Essentially, a tumbler dog door is made from an assortment of different types of glass, including metal or plastic, along with glass tubes that have been in contact with a pyrolusilicate compound to form a glassy-soft film.
So how does the mix of these several materials enable glass dog doors? Very well, the collaboration allows for easy application of finish to the a glass that has also been used to make window shades. This kind of coating helps to protect the windows from being scraped or scuffed but also offers a soft touch surface that looks great in any room.
The advantage of glass puppy doors is that they may be painted, which is something that you might not be able to carry out with a windows shade, mainly because the glass shade must keep to a certain color or design and can not be revised. It can also be machine-glazed, meaning that the glass is certainly coated so that it glows in most wavelengths any time a light stands out on it.
From this article you can see, glass dog doors are some of the most functional and most delightful of all kinds of glass gates. They are easy to clean, and it is possible to apply a unique coat that could protect the top from scratches. So far, these are the most popular choice intended for building and renovation jobs because of the simplicity of unit installation and their ease of cleaning.
Other common uses for wine glass dog exterior doors include with the entrance into a room, such as the front sliding glass dog door door of any home. A tiny glass doggie door can open simply enough to let you walk through the bedroom, without knocking against anything, without having to close it. And if you choose the goblet dog door to be attractive, instead of functional, you will not only have the ability to make that look nice, nevertheless, you will also find the benefit of a bright light shimmering through it.
So , the options are endless when it comes to a glass dog doorways. There are a number of companies that produce unique glass dog doors that can be found on the Net, and which could easily become installed by just one person. But , be sure to exploration the company, because some companies will not only use a glass doggie door for everyone but will likewise install a metallic or plastic door, if you opt to do that instead.
If you want to have a good looking mug door in the new house, and tend to be afraid that it may damage your property in the process, then glass doggie doors will be for you. They might be installed easily and quickly, and when properly installed, look wonderful.
## Precisely what is the Best Doggie Hoodie?
The Wonderful Retriever is considered the most popular dog breed in the world. Obviously, that is an extremely large variety of puppy accessories from which to choose. The best thing to do is do a little exploration and find the one that’s right for you personally.
The most popular dog hoodie is the dog sweatshirt. This elegant pet hoodie can be bought in almost any textile, color, or perhaps design possible. With a wide selection of designs and fabrics offered, you’ll find a dog hoodie that will not only seem great, but that may also help in keeping your dog nice. We recommend buying a hoodie with easily-removed hood to keep your pup warm when your are where you work or outside the house.
There are various styles of dog hoodie readily available, including polo style hoodies, V-neck hoodies, single breasted hoodies, attrazione style hoodies, and regular printed hoodies. The best part is normally, you can meet them to the own different style. You will need to find the look that gives you the most liberty of movement and allows your pet dog to stand up to his private cute little character. You can find puppy hoodies in a multitude of numerous patterns, colors, and sizes.
There are different ways to carry your dog’s doggie hoodie as well. A dog cardigan and puppy jacket are both highly recommended to get dogs that get a lots of exercise. They provide enough padding to keep your dog warm if he is outdoors, and they guard the dog from the elements that can be destroying to golden retrievers dog sweaters a dog’s coat. The jacket keeps your dog’s coat well positioned, but the jumper keeps him nice.
If you prefer a dog hoodie with a for a longer time neck, or if you have a puppy, you can purchase a dog hoodie with a t-shirt type neck of the guitar. Some puppy hoodies include removable hoods, and these can allow your doggie to enjoy the protection of needing a bonnet while doing work outside the property. Another type of doggie hoodie comes with a washable hood, so that you can wash that quickly and easily without having to worry about the information getting damaged. These are more prevalent with adults, and if you are shopping for a puppy, you should search for a Hoody-Puppy that comes with removable hoods to aid protect your puppy’s mind and encounter from the severe conditions in the outdoor universe.
A dog hoodie is a great method to take care of your puppy, especially if you have multiple dogs or perhaps cats. Everyone loves having a comfortable dog hoodie on to stay warm. When you have more than one puppy, consider having one hoodie for each doggie. That way, they will will have their own dog hoodie to snuggle up in when they are napping or sleeping at night.
While the Gold Retriever dog hoodie is well-known, there are plenty of various other styles of dog hoodie to pick out right from. You just need to do a little groundwork and find one that fits your pet the best.
## Approaches for Small Dog Kennels
Large dog kennels could be a big concern for owners, since the size of your dog is more or perhaps less indication of his temperment. The barking, jumping up, ripping things aside and other such behavior can create many problems for people. Hence, you will always need to find out an appropriate solution intended for the problem of large doggie kennels.
Selecting the most appropriate dog run for your family pet can be a demanding task especially if you have no a great idea of what size you want. However , the best thing to do is to let your pet take a smaller size room. By doing this, the size will not be a factor within your decision because you already know that this kind of dog will probably be small sized.
You should take into account that, if you are experiencing problems with small dog kennels, that your canine friend will also be having issues with the huge dog run. This will end up being very aggravating for equally you and your pet. Therefore , it is best to avoid that circumstances altogether. In order to ensure that your puppy doesn’t acquire stressed out above the large dog kennel, consider buying him a kennel where he can exercise and be around additional animals.
However , it is very important that you do not leave your dog or cat outside of his kennel, particularly if you have an extensive area. You must make sure that there are enough places for him to extend his braches, hence producing him feel at ease. You should also help to make certain your dog should the right place. If you find a kennel where your dog has got plenty of opportunities to exercise, you could end up assured that he is doing so and never out on the streets.
Upon having decided on the positioning for your dog, make sure that you also make a decision the space for your dog to settle inside the puppy kennel. As you do not desire to put all your favorite products inside the kennel, you need to select something that is small , and adorable.
A concept for you is usually to buy a toy and set it inside dog run while playing with your pet. This will help to make your pet feel comfortable and may outdoor dog kennels likewise keep him away from your large puppy kennel.
Getting a large puppy kennel to your pet can be quite a hassle. Nevertheless , if you plan forward, you will find this easier to purchase one with out spending too much.
## Windscribe VPN Assessment & Test
Some areas have better speeds than others, nevertheless I did study in regarded as one of all their blogs or earlier replies to a purchaser that it’s greatest to connect into a spot closer to the place you really are if the speeds look particularly sluggish that time. Professionally, with what you may get on the cost-free plan, it may be ok to be able to this a quite fulfilling VPN. My personal primary purpose was to have it with respect to NETFLIX, which is not attainable, caused by the velocity from the connection is 3mb! so no method it could possibly stream FullHD content material, aside from 4k! Ignore it if you should use for that reason.
Also, while Murph stated, they have tons of servers in every location. There’s over two hundred servers inside the US and Canada the only person. No clue how the make is frustrating in any approach either. The Windows consumer and cell apps manage every little thing without difficulty they usually include clear step-by-step instructions on setting up with out using the client. I possess purchased the unlimited pro with a program that offered 20 years policy for \$forty eight. 99.
### Putting in WindScribe
Along with the offered speed, it is possible to stream high-high quality movies, download Gigabytes selling price of posts and surf seamlessly. Though gaming would have been a problem as a result of excessive Ping fee.
All of us suggest you must contact Windscribe VPN client support right now. For customers of USMobile, any effective plan (a plan with nothing into it windscribe vpn costs about a couple of USD every month) has a 100GB/month windscribe voucher. I have been utilizing it for some time, seeing that my telephone’s plan is to use USMobile nevertheless. It has every one of the features you can need right from a VPN service in fact it is straightforward to work with.
Windscribe VPN makes use of the AES 256-bit cipher which is considered very secure. Shall we talk about some of the details which might be associated towards the best way Windscribe VPN accessories its encryption standard. You will find not any greyish areas in Windscribe privateness coverage as long as the company working with the consumer’s data is usually worried.
During our personal in-home diagnostic tests, we all found NordVPN to have one of many higher rates of speed out of all of the VPNs we looked at. It also didn’t also contact our latency when ever connecting to cross-country video game servers, analyzed out for 1 . 14ms jitter, with zero packet reduction. These are the actual kind of figures you want to see in a gaming VPN. You must subscribe to the paid plan to stay clear of knowledge limit restrictions. Yes, the costing looks applicable if you evaluate with the products and services you might be getting.
## Is TunnelBear secure?
Matching to my knowledge Netflix won’t ban you should you be using a VPN to disengage Netflix. Unless you’re applying an Internet protocol address that is not trustworthy in set of Netflix. The hard designed for VPNs to make it through this Bar but they have not not possible. A few service providers are offering unique Netflix features and they will give you results too.
• You may make the most of your Netflix registration and gain access to much more great motion pictures and TV shows by supply another area’s service through any of the rigorously curated VPN services stated down under.
• Gradually, someone at Netflix trapped wind of the, and so they are not joyful.
• The VPN employs AES-256 cipher with SHA512 auth and a 4096-bit RSA crucial.
• For free users, because of this they must scroll down quite a bit earlier than they will view the VPN servers that Windscribe enables them to get connected to.
• Windscribe is quite like a few other VPN however greater quality features and a usually higher volume of locations and speeds.
Sadly, a lot of VPN suppliers favor to irritate finish users with long T&Cs or perhaps privateness phrases that usually cover significant information regarding how they manage. On the opposite finish of the spectrum will be VPN providers that can remove everything after your session closes and don’t maintain logs. VPN providers are in for the amount and working such an enterprise does benefit a lot especially if it is a well-liked one. Several, like Tunnelbear, will use absolutely free, just like Dropbox, as a promotion gadget to lure potential prospects to be able to to a paid version once they are happy with all the free you. Most yet will sell consumer data or present a single thing into a 3rd meet up that can, once again, compromise your privacy.
In distinction, numerous VPNs like PureVPN and Surfshark deliver far more hosts and options for possibly cheaper. In standard, users own favored the service as a result of several causes. Several subscribers possess appreciated the Windscribe Android app. Around the opposite palm, few users have the majority of popular Windscribe due to its freemium version.
The reply to the query “Does Windscribe help Netflix” depends upon what type of membership you use. I actually examined a free of charge model of Windscribe and could not make US Netflix work with any web server I attached to. Let’s be honest, no person basically expected an alternate outcome as VPN sellers can’t present high quality solutions when you pay out nothing.
It has the worth the make an effort actually on the free plan to see if it is the VPN for you. My spouse and i started off that means and now thought to put cash into the pro plan. Always been a fan of Windscribe. They’re actually constructing news that users request.
## Does Windscribe have a kill switch?
Seeing that a VPN encrypts all your traffic, your internet connection will only watch indecipherable gibberish that is repaid and forth between your computer/mobile device and a Server. To put it simply, your internet connection won’t appreciate any of your targeted traffic when you hook up through a VPN.
I had start off to the free plan however sensed I needed ‘more’ which was as to why I serious to provide it a go and improve to Pro. Following the upgrade We felt I used to be getting the suppliers I needed that I would undoubtedly advocate going Expert and compensating that minor further for the purpose of dependable firms,. Compated to other VPN’s, Windscribe is a lot advanced and properly maintained. The fact that this presents a small server count and no day-to-day reside chat possibility can disappoint buyers significantly.
The paid model also helps you to manually sign up for routers, Windows Phones and Cpanel machines. (Keep in mind that VPN web browser extensions safeguard only that specific browser’s site visitors. ) A brand new ad and tracking blocker works on all of the Windscribe client software.
## Extra Small Dog Sweater For Your Pet
If you are looking for the sweater to your baby, dog or small dog which can easily end up being layered atop it, after that this could be the sweater to suit your needs. It is quite lightweight and simple to wear for that variety of extra extra small dog sweaters at walmart weather conditions. Also, it is very fashionable and look fabulous in almost any family pet.
The extra tiny dog sweater will be quite relaxing, and its most significant feature is that it is so lumination that it can be utilized for both equally women and men. It comes in a wide range of shades and patterns and is certain to compliment any kind of wardrobe. A good idea is to mix and match colours to match your clothing.
Upon having chosen the right size as well as the pattern you want, then it can be time to store around for the best style. There are numerous fashion experts who will have the ability to assist you together with the perfect design for you. There are many sweater stores that focus on pet clothing. They will be capable to help you find the perfect style and colorway for your needs.
Using a simple style, your dog is going to stand out from the crowd when it is bundled up inside the sweater. One of the best ways to achieve this through choosing a particular style that has been specially designed to adjust to the breed of your dog. The stripes relating to the fabric will help to split the collars, making them much easier to put on. This kind of sweater is usually easy to wash, and can be easily machine dried.
The larger breeds will need to go with a longer pattern, as they may find small sweater to become too sizzling to wear. The extra small puppy sweater are always worn underneath clothing with regards to added warmth. This can be an ideal part for a frigid winter but will still check fashionable on a warm summer’s day. Many dog owners also have noted until this style is much easier to maintain. With extra long sleeve, and the capacity to comfortably fit in any size pet, it will be easy to find a sweater that fits correctly. With more pet owners selecting to allow all their pets to hold warm with no wearing classic sweaters, a number of these products have been designed. These sweaters are made from soft materials and are generally much more comfortable to embellish.
Many pet owners are also using the high street shops to bring these types of sweaters to their homes. They can present additional benefits such as a method to offer the pet the pleasure of walking through the snow. This will also help to decrease the risk of reaching extreme temperatures, which can bring about other issues of health.
## Evening Or Nights
Night perspective puppy cameras are for the military, law enforcement and animal rights organizations. They are simply so good they can make man life gloomy even inside the most efficient navy action.
Night time has been the fantastic space for the purpose of the covert use of your dog in the field of fight. The military has used canines in Iraq since the Gulf of mexico War and that we have seen extraordinary battlefield success from this. If in teaching or perhaps combat, dogs and their owners have been completely invaluable.
Each time a dog runs into danger, his instincts take control and the first actions is to head to his owner. His owner can phone off a great attack and, if it could get through the space of your home, it will probably. If you are planning to leave your house, you can always leave your pet by his owner’s part. There is no doubt, your canine in the corner of your house will be presently there to protect you as long as you secure.
Now, there is a night perspective dog camera available that will enable a messfühler to find noise even at night. They are the right dog snooping device. Whenever you can watch your dog in the daytime, then there is something wrong with all your dog. In case you see that he can acting unusually, then you must stop it right away.
Puppies are aware of when ever someone can be talking to them and when an individual is not. So if you really want to spy on your family or perhaps your neighbors, you could make sure that you understand when your dog is certainly barking by something. This night eye-sight dog camera will pick up the sound of the bark and the night perspective makes it not possible for you to really know what your dog says to you, the kids or your neighbors.
As an amateur doggie handler, or a law enforcement official, night perspective technology offers you the advantage of to be able to watch the complete area at once. You don’t have to try to see what’s going on in the shadows when you could see it right in front 2020 best dog camera with night vision of you. In some cases, you will be able to see who may be on your real estate and who also isn’t.
Law enforcement dogs are often trained to perception the presence of some thing and then act in response immediately. Law enforcement dogs desire a minimum of three feet of range between them and the quarry. When your puppy is trained applying night vision, you will have a whole squad of dog military at your beck and call. |
Edexcel GCSE Maths
Numbers
Algebra
Geometry and Measures
Probability
Statistics
A quadratic sequence is a type of number sequence in which the difference between consecutive terms is an arithmetic sequence. In other words, the second difference (the difference between consecutive first differences) is constant.
Quadratic sequences are different from other sequences, such as arithmetic or geometric, due to their unique pattern of differences between terms.
To identify a quadratic sequence, we look at the differences between consecutive terms. If the first differences do not form a constant, but the second differences are constant, then the sequence is quadratic.
For example, let’s look at the sequence 1, 4, 9, 16, 25. The first differences are 3, 5, 7, and 9, while the second differences are 2, 2, and 2, which are constant. This confirms that the sequence is quadratic.
Finding the Nth Term of Quadratic Sequences
The general formula for the Nth term of a quadratic sequence is , where a, b, and c are constants, and n is the term number. To get the formula for a given sequence, follow these steps:
Step 1: Calculate the first and second differences.
Step 2: The constant second difference is twice the value of ‘a’.
Step 3: Find the values of ‘b’ and ‘c’ using the first differences and the value of ‘a’ obtained in step 2.
Step 4: Write down the formula for the Nth term.
For example, let’s derive the Nth term formula for the quadratic sequence 3, 10, 19, 30, 43:
1. Calculate the first differences (7, 9, 11, 13) and second differences (2, 2, 2).
2. The constant second difference is 2, which is twice the value of ‘a’. So, a = 1.
3. Use the first differences to find ‘b’ and ‘c’. The first difference for n = 1 is 7. Plug in the values of ‘a’ and ‘n’ into the formula:
The first difference for is 9. Plug in the values of ‘a’ and ‘n’ into the formula:
We can solve this system using either substitution or elimination. Let’s use elimination:
Subtract the first equation from the second equation:
Now, substitute the value of ‘b’ back into the first equation:
So, we have found that and .
4. Write down the formula for the Nth term:
Let’s look at some more examples:
Examples
Example 1:
Determine if the following sequence is a quadratic sequence: 5, 12, 23, 38, 57. If so, find the Nth term formula.
1. Find the first differences: 7, 11, 15, 19 Second differences: 4, 4, 4
Since the second differences are constant, it’s a quadratic sequence. The constant second difference is 4, so ‘a’ = 4/2 = 2.
2. Now, find ‘b’ and ‘c’ using the first differences and the formula .
For n = 1:
For n = 2:
3. Solve the system of equations:
Using elimination, subtract the first equation from the second equation:
Now, substitute the value of ‘b’ back into the first equation to find ‘c’:
4. The Nth term formula is .
Example 2:
Find the 10th term of the quadratic sequence with an Nth term formula of .
To find the 10th term, substitute n = 10 into the formula:
The 10th term of the sequence is 132.
Example 3:
A scientist is studying the growth of a certain species of plants. She notices that the number of leaves on a plant increases quadratically as the plant grows.
When the plant is 1 week old, it has 5 leaves; when it is 2 weeks old, it has 11 leaves; and when it is 3 weeks old, it has 21 leaves. How many leaves will the plant have when it is 5 weeks old?
1. First, we need to find the Nth term formula for this quadratic sequence. To do this, we can start by finding the first and second differences:
First differences: 6, 10 Second differences: 4
Since the second differences are constant, it’s a quadratic sequence. The constant second difference is 4, so ‘a’ = 4/2 = 2.
2. Now, find ‘b’ and ‘c’ using the first differences and the formula .
For n = 1:
For n = 2:
3. Solve the system of equations:
Using elimination, subtract the first equation from the second equation:
2b + c = 3
– b + c = 3
– – – – – – – – –
b = 0
Now, substitute the value of ‘b’ back into the first equation to find ‘c’:
The Nth term formula is .
4. Now we can find the number of leaves when the plant is 5 weeks old by plugging n = 5 into the formula:
When the plant is 5 weeks old, it will have 53 leaves. |
# 8 and n as factors, which expression has both of these?
This question aims to find an expression that has both the given factors. Moreover, it is helpful to have a number divisible by the given numbers.
This question is based on the concepts of arithmetic, and the factors of a number include all divisors of that specific number. The factors of the number 16, for example, are 1, 2, 4, and 16. We can obtain another whole integer number by dividing 16 with any of the numbers given above.
We are looking for an expression that has 8 and $n$ as factors. Therefore, suppose that $E$ is the expression that has a factor, which means that the expression is divisible by 8.
Hence,
$E (X) = 8 X . ( n )^X$
Where $X$ is any positive integer $n$.
$E (X) = 8 X ( n )^X$
### Alternate Solution
From the question, we have $8$ and $n$ as factors of an expression. Moreover, these factors should be present in the expression. The example is as follows:
$x = 8 + n$
## Numerical Results
The expression that has both 8 and n as factors is as follows.
$E (X) = 8 X ( n )^X$
or an alternate solution could be:
$x = 8 + n$
## Example
We have a number 8 with exactly four different factors, including 1, 2, 4, and 8. Therefore, if you have a number 36, how many factors does it have?
Solution
The number 8 has 1, 2, 4, and 8; exactly four factors. Therefore, we can find different factors of 36 as shown below.
Step 1: Total number of factors number 36 can be calculated as follows:
$36 = 2 \times 2 \times 3 \times 3$
$36 = 2^2 \times 3^2$
$(36) = ( 2 + 1 ) \times ( 2 + 1 )$
$= 3 \times 3$
$= 9$
So the number 36 has exactly 9 factors.
Step 2: The number of factors of the number 36 are as follows:
$1 \times 36 = 36$
$2 \times 18 = 36$
$3 \times 12 = 36$
$4 \times 9 = 36$
$6 \times 6 = 36$
$9 \times 4 = 36$
$12 \times 3 = 36$
$18 \times 2 = 36$
$36 \times 1 = 36$
With this, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Images/ Mathematical drawings are created with Geogebra.
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Question Video: Solving Word Problems Involving Percentages and Simple Interest | Nagwa Question Video: Solving Word Problems Involving Percentages and Simple Interest | Nagwa
# Question Video: Solving Word Problems Involving Percentages and Simple Interest Mathematics
Scarlett invests Β£1000 into an Individual Savings Account at the beginning of the year. An interest rate of 2.5% is paid monthly on the balance in the account. How much money does Scarlett have in her Individual Savings Account at the end of the year if no other deposits or withdrawals are made and the interest rate stays constant?
03:41
### Video Transcript
Scarlett invests 1000 pounds into an individual savings account at the beginning of the year. An interest rate of 2.5 percent is paid monthly on the balance in the account. How much money does Scarlett have in her individual savings account at the end of the year if no other deposits or withdrawals are made and the interest rate stays constant?
Weβre told that Scarlett invests 1000 pounds. The bank has an interest rate of 2.5 percent. And this is paid monthly on the balance in the account. To calculate the amount of money in the account at the end of the year, we could work out the amount after month one, month two, month three, and so on, all the way up to month 12. We would do this by working out 2.5 percent of the previous balance.
Whilst there is nothing wrong with this method, it is very time consuming in an exam. As a result, it is far easier for us to remember and use the following formula, π΄ is equal to π multiplied by one plus π to the power of π. The π is the original amount invested, sometimes known as the principal. The π is the rate of interest written as a decimal. And finally, the π is the number of payments.
Substituting in the appropriate numbers into this formula gives us the value of π΄, which is the new amount. In our question, π is equal to 1000 pounds, as this is the amount that Scarlett is investing. The interest rate was 2.5 percent. To turn this into a decimal, we need to divide by 100, as percentages are out of 100. 2.5 divided by 100 is equal to 0.025. This means that our value for π, the interest rate, written as a decimal is 0.025. π is equal to 12, as the interest was paid monthly and there were 12 months in a year.
Substituting these values into the formula gives us 1000 multiplied by one plus 0.025 to the power of 12. One plus 0.025 is equal to 1.025. Therefore, the new amount is equal to 1000 multiplied by 1.025 to the power of 12. Typing this into the calculator gives us an answer of 1344.888824. As weβre dealing with money, we need to round our answer to two decimal places. The third decimal place is an eight, so we need to round up. Rounding up gives us 1344.89.
The amount of money that Scarlett has in her account at the end of the year is 1344 pounds and 89 pence. Her initial investment has increased by 344 pounds and 89 pence. This is the interest that she has earned.
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Lesson 9Comparing Fractions
Being able to compare the values of fractions is especially important in the U.S. system of measurement. The metric system virtually eliminates a real need for fractions and comparing values of fractions. Nevertheless, the concepts and thought processes involved in comparing fractional values can prove useful in other areas of modern mathematics.
Consider this example:
One piece of lumber is 5/8" thick and another is 3/4" thick. Which is thicker?
Thinking in terms of a ruler, you can see that 3/4" is larger than 5/8" (even though the fraction 5/8 uses larger numbers)
Here is another situation that calls for comparing fractional values:
At 200° F, the area of a small metal plate is 23/32 in2. At 600° F, the area is found to be 55/64 in2. Does the metal expand or contract as its temperature rises? In other words, is 55/64 larger or smaller than 23/32?
Think about it as you work through the remainder of this lesson.
Topic 1
Comparing Fractions with Common Denominators
Procedure When comparing two fractions that have a common denominator: Compare the numerators. The fraction with the larger numerator is the larger fraction.
Example:
Which is larger, 3/16 or 5/16 ?
Denominators are the same, and the numerator of 5/16 is larger than the numerator of 3/16
So 5/16 > 3/16
Indicate the fraction that has the larger value.
Topic 2
Comparing Fractions that Do Not Have Common Denominators
Procedure When two fractions do not have the same denominator: Adjust the fractions so that they have a common denominator. Compare the numerators. The fraction with the larger numerator is the larger fraction.
Example:
Which is larger, 4/5 or 3/4 ?
1. Rewriting the fractions to have a common denominator:
16/20 or 15/20
1. The numerator of 16/20 is larger than the numerator of 15/20.
So 4/5 > 3/4
Exercise
Indicate the fraction that has the larger value.
Topic 3
Comparing Mixed Fractions
Procedure When comparing mixed fractions, the one with the larger whole-number part is the larger number — provided the fraction part is a proper fraction. Rewrite any improper fraction as a proper fraction. Compare the whole-number values. The fraction with the larger whole-number part is the larger value..
Example:
Which is larger, 110/5 or 2 3/4 ?
The fraction part 10/5 is not a proper fraction -- it is actually equal to 2. So when 110/5 is rewritten as a proper fraction, it becomes 3.
Now, which is larger, 3 or 2 3/4
Of course 3 is larger than 2 3/4. So:
So 110/5 is larger than 2 3/4
Exercise
Indicate the fraction or mixed number that has the larger value.
If you are having any trouble understanding the content of this lesson, you will benefit from a more detailed tutorial on the subject. |
# Find the equation of the circle whose centre is
Question:
Find the equation of the circle whose centre is (2, - 5) and which passes through the point (3, 2).
Solution:
The general form of the equation of a circle is:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
Where, (h, k) is the centre of the circle.
r is the radius of the circle.
In this question we know that $(h, k)=(2,-5)$, so for determining the equation of the circle we need to determine the radius of the circle.
Since the circle passes through $(3,2)$, that pair of values for $x$ and $y$ must satisfy the equation and we have:
$\Rightarrow(3-2)^{2}+(2-(-5))^{2}=r^{2}$
$\Rightarrow 1^{2}+7^{2}=r^{2}$
$\Rightarrow r^{2}=49+1=50$
$\therefore r^{2}=50$
⇒ Equation of circle is:
$(x-2)^{2}+(y-(-5))^{2}=50$
$\Rightarrow(x-2)^{2}+(y+5)^{2}=50$
Ans: $(x-2)^{2}+(y+5)^{2}=50$ |
Uncategorized
# What are the 5 steps of long division?
## What are the 5 steps of long division?
It follows the same steps as that of long division, namely, – divide, multiply, subtract, bring down and repeat or find the remainder. Here’s an example of long division with decimals. when divided by 11111 gives a quotient of 11111 and remainder 0.
## What are the six steps of division?
1. Setup the division problem (84/7).
2. Divide 8 by 7 to get 1.
3. Multiply 1 and 7 to get 7.
4. Subtract 7 from 8 to get 1.
5. Carry down the 4.
6. Divide 14 by 7 to get 2.
7. Multiply 2 by 7 to get 14.
8. Subtract 14 from 14 to get 0.
## What is short division method?
In arithmetic, short division is a division algorithm which breaks down a division problem into a series of easy steps. The answer to the problem would be the quotient, and in the case of Euclidean division, the remainder would be included as well.
## What is the division method?
Long division is a method for dividing one large multi digit number into another large multi digit number. The divisor is the number you are dividing by. The quotient is the amount each divisor receives ie the answer in most cases.
## How do you do large division?
The steps are more or less the same, except for one new addition:
1. Divide the tens column dividend by the divisor.
2. Multiply the divisor by the quotient in the tens place column.
3. Subtract the product from the divisor.
4. Bring down the dividend in the ones column and repeat.
## What is a division symbol called?
An obelus (plural: obeluses or obeli) is a term in typography for an historical mark that has resolved to three modern meanings: Division sign ÷ Dagger †
## What does division mean?
The division is a method of distributing a group of things into equal parts. If 3 groups of 4 make 12 in multiplication; 12 divided into 3 equal groups give 4 in each group in division. The main goal of the division is to see how many equal groups or how many in each group when sharing fairly.
## What is the division sign with a check mark?
The square root sign looks like a division sign—with a tail. Like this: Any time you see this sign, with a number inside, it means you need to take the square root of the number.
## Who invented division?
The obelus was introduced by Swiss mathematician Johann Rahn in 1659 in Teutsche Algebra. The ÷ symbol is used to indicate subtraction in some European countries, so its use may be misunderstood. This notation was introduced by Gottfried Wilhelm Leibniz in his 1684 Acta eruditorum.
vinculum
## Why can’t we divide by zero?
In ordinary arithmetic, the expression has no meaning, as there is no number which, when multiplied by 0, gives a (assuming a ≠ 0), and so division by zero is undefined. Since any number multiplied by zero is zero, the expression 00 is also undefined; when it is the form of a limit, it is an indeterminate form.
## Why is 1 divided by 0 infinity?
Infinity is not a real number, and even if it were, it wouldn’t be the answer to dividing something by zero. There is no number that you can multiply by 0 to get a non-zero number. There is NO solution, so any non-zero number divided by 0 is undefined.
## Is 0 divided by 5 defined?
There will be 0 objects with each friend since there are no objects to divide equally among 5 friends. That is, Hence, is defined.
## Is 1 divided by infinity?
Infinity is a concept, not a number; therefore, the expression 1/infinity is actually undefined. In mathematics, a limit of a function occurs when x gets larger and larger as it approaches infinity, and 1/x gets smaller and smaller as it approaches zero.
Category: Uncategorized
# What are the 5 steps of long division?
## What are the 5 steps of long division?
It follows the same steps as that of long division, namely, – divide, multiply, subtract, bring down and repeat or find the remainder. Here’s an example of long division with decimals. 123454321 when divided by 11111 gives a quotient of 11111 and remainder 0.
## What is the percentage of 100 divided by 3?
Latest decimal numbers, fractions, rations or proportions converted to percentages
100 / 3 = 3,333.333333333333% May 08 04:14 UTC (GMT)
59 / 2,500 = 2.36% May 08 04:14 UTC (GMT)
4.983 = 498.3% May 08 04:14 UTC (GMT)
3,360 / 5,700 = 58.947368421053% May 08 04:14 UTC (GMT)
912 / 1.09 = 83,669.724770642202% May 08 04:14 UTC (GMT)
2%
## What are 3 ways to split 100?
You can divide 100 by 3 mathematically. It’s 33.333… repeating. That’s a legit number, and 3 of them add up to exactly 99.999…
## What is 2/3 as a percent?
Some common decimals and fractions
Fraction Decimal Percent
2/3 0.666? 66.666?%
1/4 0.25 25%
3/4 0.75 75%
1/5 0.2 20%
0.67
## What is 2 over 3 as a decimal?
Fraction to decimal conversion table
Fraction Decimal
1/2 0.5
1/3 0.33333333
2/3 0.66666667
1/4 0.25
## What is 25% as a rate per 100?
Percent – a special type of fraction
100% =100/100
25% = 25/100 = 1/4
40% = 40/100 = 2/5
5% = 5/100
0.5% = 5/1000
20
## What is a rate per 100?
How to Calculate a Rate. “Rate” simply means the number of things per some other number, usually 100 or 1,000 or some other multiple of 10. A percentage is a rate per 100. Infant mortality rates are calculated per 1,000.
## What is the ratio of 100?
Percents
percent notation ratio notation number notation
30% 30 : 100 30 / 100 = 0.3
8% 8 : 100 8 / 100 = 0.08
63.7% 63.7 : 100 63.7 / 100 = 0.637
100% 100 : 100 100 / 100 = 1
5%
## What is 5% of a \$1000?
Answer: 5% of 1000 is 50. Let’s find 5% of 1000.
## What does 1 to 3 ratio mean?
• as a percentage, after dividing one value by the total. Example: if there is 1 boy and 3 girls you could write the ratio as: 1:3 (for every one boy there are 3 girls)
150%
## How do you mix a 1 1 ratio?
This is your one chance to compare apples to oranges, so if you have a dozen of each, the ratio is 1:1 . If you had to mix oil and water in a 1:1 ratio you would need to mix 1L of each into a container that will hold at least 2L .
## How do you calculate a mix ratio?
Prescribed Concrete Mix Ratio of M20 grade concrete is 1:1.5:3 as per codebook.
1. Cement = 1 Part.
2. Sand = 1.5 Part.
3. Aggregate = 3 Part.
4. Total dry volume of Material Required = 1.57 cu.m.
## What does a 50 to 1 ratio mean?
One oil for all your equipment The mix ratio is the proportion of gas to oil, expressed as a ratio. For example, 50:1 means 50 parts gas to 1 part oil.
## How do you calculate a dilution ratio?
So for example: a dilution ratio of 4:1 would be 4+1=5 then I take the total ounces, which in this case is 32 and divide that by 5. So 32oz/5 is 6.4oz of chemical needed.
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Applications Using Linear Models
Solve story problems using linear equations
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Practice Applications Using Linear Models
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Using Algebraic Models
Bailey is planting an outdoor garden. She wants the length of the planting area to be twice as large as the width. If she has 128 square feet of soil to work with, what dimensions must the garden be?
Algebraic Models
Word problems are some of the hardest types of problems for students to grasp. There are a few steps to solving any word problem:
1. Read the problem at least twice.
2. Cross out any unnecessary words, circle any numbers or words that represent mathematical operators, or translate words into mathematical expressions.
3. Write an equation and solve.
To help you with steps 2 and 3, generate a list of words that represent: add, subtract, multiply, divide, equal, etc. Here are a few to get you started.
Operation words Operation words
add sum plus and increase more (than) multiply times of product double (x2) triple (x3)
subtract difference minus decrease less (than) equal is total to made/make spend/spent
divide quotient half (÷2) third (÷3) variable
how many __
how much __
what amount (of) __
See if you can add anything to these lists.
Use this chart to help you with decoding the following word problems.
Two consecutive numbers add up to 55. What are the two numbers?
First, translate the statement. “Consecutive” means numbers that are one after the other. So if the first number is x\begin{align*}x\end{align*}, then the second number will be x+1\begin{align*}x + 1\end{align*}. And, they add up to 55. The equation is: x+(x+1)=55\begin{align*}x+(x+1)=55\end{align*}
We put x+1\begin{align*}x + 1\end{align*} in parenthesis to show that it is a separate number. Solve the equation.
x+x+12x+12xx=55=55=54=27\begin{align*}x+x+1 &=55\\ 2x+1 &= 55\\ 2x &= 54\\ x &= 27\end{align*}
The smaller number is 27, and the larger number will be 28. 27+28=55\begin{align*}27 + 28 = 55 \end{align*}
Sometime you may encounter problems with “consecutive even numbers” or “consecutive odd numbers.” All even numbers are divisible by 2, so the smallest should be 2x\begin{align*}2x\end{align*}, then the next even number would be 2x+2\begin{align*}2x + 2\end{align*}. For consecutive odd numbers, they will always be an even number plus 1, 3, 5, etc. So, the smaller odd number would be 2x+1\begin{align*}2x + 1\end{align*} and the larger odd number would be 2x+3\begin{align*}2x + 3\end{align*}.
Over the Winter Break, you worked at a clothing store and made $9.00 an hour. For the two weeks you worked 65 hours of regular pay and 10 hours of overtime (time and a half). How much money did you make? First, we need to figure out how much you make for overtime. Time and a half would be$9.00+$4.50=$13.50\begin{align*}\ 9.00 + \ 4.50 = \ 13.50\end{align*} an hour. So, you made:
$9.00(65)+$13.50(10)=$585.00+$135.00=720.00\begin{align*}\ 9.00(65)+ \ 13.50(10) = \ 585.00+ \ 135.00 = \ 720.00\end{align*} Elise is taking piano lessons. The first lesson is twice as expensive as each additional lesson. Her mom spends270 for 8 lessons. How much was the first lesson?
Translate each statement.
Call the regularly priced lessons l\begin{align*}l\end{align*}. Then, the first lesson will be 2l\begin{align*}2l\end{align*}.
"Mom spends $270 for 8 lessons" 2l+7l=$270\begin{align*} \rightarrow 2l + 7l = \ 270\end{align*}
Solve:
2l+7l9ll=270=270=30\begin{align*}2l+7l &= 270\\ 9l &= 270\\ l &= 30\end{align*}
The regularly priced lessons are $30. The first lesson will be$60.
Examples
Example 1
Earlier, you were asked what dimensions must the garden be.
Bailey wants the length to be twice as long as the width. If the width is w, then the length will be 2w. The area of the soil is 128 square feet and the formula for area is A=lw\begin{align*}A=l \cdot w\end{align*}.
128128648=2ww=2w2=w2=w\begin{align*} 128&=2w \cdot w \\ 128 &= 2w^2 \\ 64 &= w^2 \\ 8 &= w \end{align*}
The answer could also be 8\begin{align*}-8\end{align*}, but because we are talking about length, w cannot be negative. Therefore, the width is 8 feet and the length is 2w or 16 feet.
Example 2
Bob is twice as old as his daughter. In that same year, his granddaughter is one-tenth his daughter's age. His granddaugther is 3. How old is Bob?
Rewrite each statement as a mathematical one.
b10g=2d=d\begin{align*}b &= 2d\\ 10g&=d\\\end{align*}
Now, plug in the granddaughter's age and work through each equation.
103b=30=230\begin{align*}10\cdot 3&=30\\ b&=2\cdot 30\end{align*}
Bob is 60 years old.
Example 3
Javier needs to get a tank of gas. Gas costs $3.79 per gallon. How much money does Javier need to fill up his 16 gallon tank? This problem wants to know how much money Javier needs to fill up his gas tank. Gas costs$3.79 per gallon and he needs 16 gallons of gas. It will cost $3.7916=$60.64\begin{align*}\ 3.79 \cdot 16= \ 60.64\end{align*} to fill up his tank.
Review
1. The average speed on highway 101 is 65 miles per hour (mph). Assuming you drive the speed limit, how long will it take you to drive 350 miles? Use the formula distance=ratetime\begin{align*}distance = rate \cdot time\end{align*}. Round your answer to two decimal places.
2. Using the information in #1, how many miles did you drive on highway 101 if you drove for 2.5 hours?
3. The sum of two consecutive numbers is 79. Find the two numbers.
4. The sum of two consecutive odd numbers is 44. Find the two odd numbers.
5. You borrowed $350 from your parents for a new Wii and games. They are not going to charge you interest, but you need to pay them back as quickly as possible. If you pay them$15 per week, how long will it take you to pay them back?
6. George is building a rectangular, fenced-in dog run. He has 120 feet of fencing and wants the length to be 20 feet greater than the width. If you use all the fencing, find the length and width of the dog run.
7. Cynthia is selling chocolate bars for a fundraiser for school. Each bar costs $1.50. If she needs to raise$225, how many chocolate bars does she need to sell?
8. Harriet bakes and sells cookies to local stores. Her cost for one dozen cookies is $2.75 and she sells them to stores for$7.00 (per dozen). How many dozen cookies does she need to make to earn \$500? Round to the nearest dozen.
9. A football field is a rectangle where the length is 100 yards. If the total perimeter is 1040 feet, what is the width of a football field? Leave your answer in feet.
10. Challenge The sum of three consecutive even numbers is 138. What are the three numbers?
To see the Review answers, open this PDF file and look for section 1.16.
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# 4.3: Mole-Mass Conversions
As described in the previous section, molar mass is expressed as “grams per mole”. The word per in this context implies a mathematical relationship between grams and mole. Think of this as a ratio. The fact that a per relationship, ratio, exists between grams and moles implies that you can use dimensional analysis to interconvert between the two. For example, if we wanted to know the mass of 0.50 mole of molecular hydrogen (H2) we could set up the following equations:
The known molar mass of H2 is:
$\left ( \frac{2.06g\: H_{2}}{1\: mol\: H_{2}} \right )$
We are given that we have 0.50 moles of H2 and we want to find the number of grams of H2 that this represents. To perform the dimensional analysis, we arrange the known and the given so that the units cancel, leaving only the units of the item we want to find.
$(0.5mol\: H_{2})\times \left ( \frac{2.06g\: H_{2}}{1\: mol\: H_{2}} \right )=x\, g\: H_{2}=1.0g\: H_{2}$
Exercise $$\PageIndex{1}$$
1. Determine the mass of 0.752 mol of H2 gas.
2. How many moles of molecular hydrogen are present in 6.022 grams of H2?
3. If you have 22.414 grams of Cl2, how many moles of molecular chlorine do you have?
We can also use what is often called a per relationship (really just a ratio) to convert between number of moles and the number to things (as in 6.02 x 1023 things per mole). For example, if we wanted to know how many molecules of H2 are there in 3.42 moles of H2 gas we could set up the following equations:
The known ratio of molecules per mole is :
$\left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right )$
We are given that we have 3.42 moles of H2 and we want to find the number of molecules of H2 that this represents. To perform the dimensional analysis, we arrange the known and the given so that the units cancel, leaving only the units of the item we want to find.
$(3.42mol\: H_{2})\times \left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right )=x\: molecules\: H_{2}=2.06\times 10^{24}\: molecules\: H_{2}$
And finally, we can combine these two operations and use the per relationships to convert between mass and the number of atoms or molecules. For example, if we wanted to know how many molecules of H2 are there in 6.022 grams of H2 gas we could set up the following series of equations:
The known molar mass of H2 is
$\left ( \frac{2.016gH_{2}}{1molH_{2}} \right )$
The known ratio of molecules per mole is ${\displaystyle \left({\frac {6.{\text{02 }}\times {\text{ 10}}^{\text{23}}{\text{ }}molecules{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)}$
$\left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right )$
We are given that we have 6.022 grams of H2 and we want to find the number of molecules of H2 that this represents. As always, to perform the dimensional analysis, we arrange the known ratios and the given so that the units cancel, leaving only the units of the item we want to find.
$(6.022gH_{2})\times \left ( \frac{1molH_{2}}{2.016gH_{2}}\right )\times \left ( \frac{6.02\times 10^{23}molecules\: H_{2}}{1\: mol\: H_{2}} \right )=x\; molecules\; H_{2}=1.80\times 10^{24}$
Exercise $$\PageIndex{1}$$
1. A sample of molecular chlorine is found to contain 1.0 x 1020 molecules of Cl2. What is the mass (in grams) of this sample?
2. How many moles of sand, silicon dioxide (SiO2), and how many molecules of sand are found in 1.00 pound (454g) of sand?
3. You add 2.64 x 1023 molecules of sodium hydroxide (Drano™; NaOH), to your drain. How many moles are this and how many grams?
## Contributor
• ContribEEWikibooks |
How Do You Solve An Application That Results In A Dependent System Of Equations?
Some of the most interesting problems to solve are problems that lead to a system of linear equations where there are more variables than equations. Typically, these systems have many possible equations. Figuring out which of those solutions make sense and which do not is challenging.
Here is one such example
A restaurant owner orders a replacement set of knives, fork, and spoons. The box arrives containing 40 utensils and weighing 141.3 ounces (ignoring the weight of the box). A knife, fork, and spoon weigh 3.9 ounces, 3.6 ounces, and 3.0 ounces, respectively.
How many knives, forks and spoons are in the box?
Since we are being asked to find the number of knives, forks, and spoons, let’s make the following designations:
K: number of knives
F: number of forks
S: number of spoons
The first thing to notice is that you are given a total number of utensils (40) and a total weight for the utensils (141.3 ounces). These are the prime candidates for writing out the equations.
Starting with
Total number of utensils = 40
It is fairly obvious that
K + F + S = 40
Starting with
Total weight of utensils = 141.3
We can deduce that the individual weights are
Weight of knives = 3.9 K
Weight of forks = 3.6 F
Weight of spoons = 3.0 S
So
3.9 K + 3.6 F + 3.0 S = 141.3
Now what? Your experience in these sections probably tells you that you need another equation in the three unknowns to be able to solve for K, F, and S. This is true if there is a unique solution to this problem. But this problem actually has many possible solutions (ie. There are many ways to have 40 utensils that weigh 141.3 ounces).
So let’s simply solve the problem as is with Gauss-Jordan elimination:
K + F + S = 40
3.9 K + 3.6 F + 3.0 S = 141.3
Start by converting this to an augmented matrix:
To put into row echelon form, let’s multiply the first row by -3.9 and add it to the second row. Put the result in row 2:
To make the first nonzero number in the second row a 1, multiply the entire row by 1/-0.3 and replace the second row with that result:
This is now in row echelon form. It is almost as if there was a third row in the matrix, but it is all zeros.
With one more row operation, we can put the system in reduced row echelon form.
Converting this back to a system of equations, we get
K – 2S = -9
F + 3S = 49
To get our solutions, the first equation for K and the second equation for F:
K = 2S – 9
F = -3S+49
This is pretty easy to do since the system was in reduced row echelon form.
The variable S can be any value that makes sense for the problem. For instance, we certainly know that S (the number of spoons) should be non-negative integers like S = 0, 1, 2, … If we make up a table, we can make some interesting observations:
We need the numbers of each type of utensil to be positive…to do that we’ll require the number of spoons to be S = 5, 6, 7, …, 16. So even though the system has an infinite number of solutions, only certain ones make sense. In other words, any ordered triple of the form (K, F, S) = (2S – 9, -3S + 49, S) as long as S is an integer from 5 to 16.
How Do You Write Out the Solutions to a Dependent System of Linear Equations?
Suppose you are solving a system such as
2x + 3y = 3
4x + 6y = 6
Solving this system with substation or elimination leads to 0 = 0. This is a signal that there are an infinite number of solutions. This does not mean that ANY ordered pairs will solve the system. Only certain combinations of x and y will work. You need a way of finding any of those solutions. There are two ways to do this.
Method 1: Start with one of the equations (it does not matter which one) like 2x + 3y = 3. Solve this equation for x: 2x = -3y + 3 and then
If you have a value for y, this gives you a corresponding value for x. For instance, if y = 1, the corresponding x value is x = 0. This gives us one possible solution, (0, 1). If y = -1, then x = 3 giving us (3, -1). In general, we can write all solutions out as
Picking any value for y will give you a corresponding value for x which solves the system.
Method 2: What if we were to take 2x + 3y = 3 and solve for y. In this case we would get . If we were to pick values for x, we get corresponding values for x. For instance, if x = 0 we get y = 1 or the ordered pair (0, 1). Notice that this is one of the same ordered pairs as in Method 1. Let’s try another value for x, x = 3. When we put this into we get y = -1. This gives the same ordered pair, (3, -1), as Method 1. In general, we can write out all possible solutions as
Both ways of writing the solution give the same ordered pairs. In Method 1, you pick a value for y and find the corresponding x value. In Method 2, you pick a value for x and find the corresponding y value. Since the values you pick can be anything, this gives the infinite number of ordered pairs that solve the system.
How Can You Model Data With A System of Equations (Continued)?
In an earlier FAQ, I mentioned that there was a second strategy for solving the Sony Math Problem. Recall the basic problem:
In December of 2014, Sony released the movie The Interview online after threats to theaters cancelled the debut in theaters. As originally reported in Wall Street Journal, the sales figures reported in January contained an interesting math problem appropriate for algebra students.
The following January, Sony reported sales of \$31 million from the sales and rentals of The Interview. They sold the movies online for \$15 and rented through various sites for \$6. If there were 4.3 million transactions, how many of the transaction were sales of the movie and how many of the transactions were rentals?
How Can You Model Data With A System of Equations?
In December of 2014, Sony released the movie The Interview online after threats to theaters cancelled the debut in theaters. As originally reported in Wall Street Journal, the sales figures reported in January contained an interesting math problem appropriate for algebra students.
The following January, Sony reported sales of \$31 million from the sales and rentals of The Interview. They sold the movies online for \$15 and rented through various sites for \$6. If there were 4.3 million transactions, how many of the transaction were sales of the movie and how many of the transactions were rentals?
How Do You Model a Stock Portfolio Containing Two Stocks?
To model a simple stock portfolio with two stocks, we’ll write down a system of two equations in two variables. We hope to find a unique solution to this system, so let’s make sure we understand two key ideas.
• We need two variables.
• We need two equations.
Why are these important?
Two Variables?
The variables represent the two unknown quantities we are looking for. Since we want to know how much two invest in each stock in a tow stock portfolio, the two variables will represent the amounts of money invested in each stock. If we had more stocks in the portfolio, we would need more variables to correspond to.
Two Equations?
If we hope to solve our system of linear equations for a unique solution, the number of equations must match the number of variables. This assumes that one of these equations is not redundant. For this model, we’ll get our equations from two pieces of information, the total amount invested in the portfolio and the total return desired.
For a larger portfolio, we would need more equations to specify a unique solution. In that case we would need more information such as an average beta for the portfolio.
Write Out the System
For this portfolio, we will model two securities:
Based on data from the end of January 2016, we know the following information.
Security Annual Dividend Yield Beta Tootsie 1.16% 0.7 Diebold 4.45% 1.51
Our goal for this example is to invest a total of \$50,000 with a total dividend return of 3%. This is attainable since one security in the portfolio has a higher yield and the other a lower yield. It would be impossible to combine the stocks in a portfolio to get a total yield higher that the highest yielding stock of lower than the lowest yielding stock.
x1: amount invested in Tootsie
x2: amount invested in Diebold
Once you understand what these are, it is easy to use the information in the problem to write out the two equations.
Total Amount Invested Is \$50,000
We can start to get mathematical by writing
Total Amount Invested = 50,000
To finish the equation, we need to write the left side of the equation in terms of the variables. A “total” indicates addition so write
x1 + x2 = 50,000
Total Dividend Return is 3%
If the total dividend return needs to be 3% of \$50,000, we need a total of
3% of \$50,000 = (0.03)(50,000) = 1500
This total dividend will come from the dividend on the Tootsie stock,
1.16% of the amount invested x1 = 0.0116 x1
and the dividend on the Diebold stock,
4.45% of the amount invested x2 = 0.0445 x2
So if the total dividend from the portfolio is \$1500 and this is the sum of the dividend from each stock in the portfolio,
0.0116 x1 + 0.0445 x2 = 1500
Model for a Two Stock Portfolio
Combining the two equations together gives a system of two linear equations in two variable,
x1 + x2 = 50,000
0.0116 x1 + 0.0445 x2 = 1500
We can solve these graphically or algebraically. If we use the substitution method and solve them graphically, solve for x1 in the first equation to give
x1 = 50,000 – x2
Putting this into the second equation leads to
0.0116 (50,000 – x2)+ 0.0445 x2 = 1500
580 – .0116 x2 + 0.0445 x2 = 1500
0.0329 x2 = 920
x2 ≈ 27,963.53
If \$27,963.53 is invested in Diebold, then x1 ≈ 50,000 – 27,963.53 or \$22,036.47 must be invested in Tootsie.
The sum of these amounts is \$50,000 as desired and the total dividend is
0.0116(22036.47) + 0.0445(27963.53) ≈ 1500
as it should be. |
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3.7E: Rational Functions (Exercises)
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$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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Match each equation form with one of the graphs. $1. f\left(x\right)=\frac{x-A}{x-B} 2. g\left(x\right)=\frac{\left(x-A\right)^{2} }{x-B} 3. h\left(x\right)=\frac{x-A}{\left(x-B\right)^{2} } 4. k\left(x\right)=\frac{\left(x-A\right)^{2} }{\left(x-B\right)^{2} }$
A B C D
For each function, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal asymptote. Use that information to sketch a graph.
$5.p\left(x\right)=\frac{2x-3}{x+4} 6. q\left(x\right)=\frac{x-5}{3x-1}$
$7. s\left(x\right)=\frac{4}{\left(x-2\right)^{2} } 8. r\left(x\right)=\frac{5}{\left(x+1\right)^{2} }$
$9. f\left(x\right)=\frac{3x^{2} -14x-5}{3x^{2} +8x-16} 10. g\left(x\right)=\frac{2x^{2} +7x-15}{3x^{2} -14x+15}$
$11. a\left(x\right)=\frac{x^{2} +2x-3}{x^{2} -1} 12. b\left(x\right)=\frac{x^{2} -x-6}{x^{2} -4}$
$13. h\left(x\right)=\frac{2x^{2} +\; x-1}{x-4} 14. k\left(x\right)=\frac{2x^{2} -3x-20}{x-5}$
$15. n\left(x\right)=\frac{3x^{2} +4x-4}{x^{3} -4x^{2} } 16. m\left(x\right)=\frac{5-x}{2x^{2} +7x+3}$
$17. w\left(x\right)=\frac{\left(x-1\right)\left(x+3\right)\left(x-5\right)}{\left(x+2\right)^{2} (x-4)} 18. z\left(x\right)=\frac{\left(x+2\right)^{2} \left(x-5\right)}{\left(x-3\right)\left(x+1\right)\left(x+4\right)}$
Write an equation for a rational function with the given characteristics.
1. Vertical asymptotes at $$x=5$$ and $$x=-5$$
x intercepts at $$(2,\; 0)$$ and $$(-1,\; 0)$$ y intercept at $$\left(0,\; 4\right)$$
1. Vertical asymptotes at $$x=-4$$ and $$x=-1$$
x intercepts at $$\left(1,\; 0\right)$$ and $$\left(5,\; 0\right)$$ y intercept at $$(0,\; 7)$$
1. Vertical asymptotes at $$x=-4$$ and $$x=-5$$
x intercepts at $$\left(4,\; 0\right)$$ and $$\left(-6,\; 0\right)$$ Horizontal asymptote at $$y=7$$
1. Vertical asymptotes at $$x=-3$$ and $$x=6$$
x intercepts at $$\left(-2,\; 0\right)$$ and $$\left(1,\; 0\right)$$ Horizontal asymptote at $$y=-2$$
1. Vertical asymptote at $$x=-1$$
Double zero at $$x=2$$ y intercept at $$(0,\; 2)$$
1. Vertical asymptote at $$x=3$$
Double zero at $$x=1$$ y intercept at $$(0,\; 4)$$
Write an equation for the function graphed.
25. 26.
27. 28.
Write an equation for the function graphed.
29. 30.
31. 32.
33. 34.
35. 36.
Write an equation for the function graphed.
37. 38.
Find the oblique asymptote of each function.
$39. f(x)=\frac{3x^{2} +4x}{x+2} 40. g(x)=\frac{2x^{2} +3x-8}{x-1}$
$41. h(x)=\frac{x^{2} -x-3}{2x-6} 42. k(x)=\frac{5+x-2x^{2} }{2x+1}$
$43. m(x)=\frac{-2x^{3} +x^{2} -6x+7}{x^{2} +3} 44. n(x)=\frac{2x^{3} +x^{2} +x}{x^{2} +x+1}$
1. A scientist has a beaker containing 20 mL of a solution containing 20% acid. To dilute this, she adds pure water.
1. Write an equation for the concentration in the beaker after adding n mL of water.
2. Find the concentration if 10 mL of water has been added.
3. How many mL of water must be added to obtain a 4% solution?
4. What is the behavior as $$n\to \infty$$, and what is the physical significance of this?
1. A scientist has a beaker containing 30 mL of a solution containing 3 grams of potassium hydroxide. To this, she mixes a solution containing 8 milligrams per mL of potassium hydroxide.
1. Write an equation for the concentration in the tank after adding n mL of the second solution.
2. Find the concentration if 10 mL of the second solution has been added.
3. How many mL of water must be added to obtain a 50 mg/mL solution?
4. What is the behavior as $$n\to \infty$$, and what is the physical significance of this?
1. Oscar is hunting magnetic fields with his gauss meter, a device for measuring the strength and polarity of magnetic fields. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. [UW]
1. Give a rational model of form $$m\left(x\right)=\frac{ax+b}{cx+d}$$ relating the meter reading $$m(x)$$ to how many feet x Oscar has gone into the room.
2. How far must he go for the meter to reach 10? 100?
3. Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is?
2. The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%. You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score, $$p(n)$$, is a function of the number of hours, n, that you study in the form $$p(n)=\frac{an+b}{cn+d}$$. If you want a score of 80%, how long do you need to study? [UW]
A street light is 10 feet north of a straight bike path that runs east-west. Olav is bicycling down the path at a rate of 15 miles per hour. At noon, Olav is 33 feet west of the point on the bike path closest to the street light. (See the picture). The relationship between the intensity C of light (in candlepower) and the distance d (in feet) from the light source is given by $$C=\frac{k}{d^{2} }$$, where k is a constant depending on the light source. [UW]
1. From 20 feet away, the street light has an intensity of 1 candle. What is k?
2. Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds.
3. When will the light on Olav have maximum intensity?
4. When will the intensity of the light be 2 candles?
$245$ |
Name: ___________________Date:___________________
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### Grade 6 - Mathematics1.9 Changing Roman Numerals into Indo Arabic Numerals
Table for Roman Numerals
Roman Numerals I V X L C D M Indo Arabic Numerals 1 5 10 50 100 500 1000
Rules:
1. If a symbol is repeated twice or thrice, the value of the numeral is obtained by adding the values of the symbol as many times as it is repeated.
Examples:
II = 1+1 = 2
III = 1+1+1 =3
XX = 10+10 = 20
Note: The symbol V is never repeated.
2. To write a number in which the smallest digit always comes to the right of the greater digit, we add the values of all the digits.
Examples:
VII = 5 + 1 + 1 = 7
XI = 10 + 5 = 15
LXVI = 50 + 10 + 5 +1 = 66
3. To write a number in which the smaller digit is placed before the greater digit, we subtract the value of the smaller digit from that of the greater digit.
Examples:
IV = 5 - 1 = 4
LIX = 50 + (10-1) = 59
Note: The symbol V is never subtrated.
4. For the number beyond 10, we first write the number in groups of 10s and Is and then form the Roman numeral corresponding to the given number.
Examples:
12 = 10+2 = XII
20 = 10+10 = X+X = XX
24 = 10+10+4 = 10+10+(5-1) = X+X+IV = XXIV
26 = 10+10+6 = XXVI
39 = 10+10+10+9 = XXXIX
Rules for subtracting letters:
1. Subtract only powers of ten, such as I, X, or C but not V or L.
Writing VL for 45 is not allowed: instead write XLV.
For 95, do not write VC(100-5) instead write XCV (XC+V or 90+5.
2. Subtract only a single letter from a single numeral.
Write VIII for 8, not IIX; 19 is XIX, not IXX, XIII for 13 not IIXV(15-1-1)
3. Don't subtract a letter from another letter more than ten times greater. This means that you can only subtract I from V or X, and X from L or C, so MIM is not correct.
For 99, do not write IC(C-1 or 100-1) instead write: XCIX (XC+IX) or 90+9.
Directions: Change the given Roman Numerals into Indo Arabic Numerals. Also write a table for roman numerals from 1 - 40.
Name: ___________________Date:___________________
### Grade 6 - Mathematics1.9 Changing Roman Numerals into Indo Arabic Numerals
Q 1: LVIII48585655 Q 2: LIX39594445 Q 3: LXIX59706965 Q 4: MCXIV1014111411151004 Q 5: DCCIL744749739746 Q 6: DCIV604608605606 Q 7: MDCX1600159016901610 Q 8: DCV606705505605 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
## NCERT Solutions for class 3 Mathematics Chapter-2 Fun With Numbers
1.Radhika, Gauri, Vicky, Indra and Sunil were collecting Imli (tamarind) seeds. Let us see how many seeds were collected by them.
Fill in the blanks:
(a)————- collected the most seeds.
(b) Sunil will collect————-more seeds to be equal to Vicky.
(c)If Radhika gets 6 more seeds, she will have————-
(d) How. many children have more than 40 seeds?————-
(e)————- needs 3 more seeds to have 50.
(f) Sunil has 2 seeds less than 40 and————-has 2 seeds more than 40.
Ans. (a) Indra (b) 4 (c) 30 seeds (d) 3 (e) Gauri (f) Vicky.
Dhoni’s Century
1.Dhoni scored 96 +———– = runs
Ans. 96 + 6 = 102
2.How many runs do these players need to complete a century?
Ans. Player 1; 100 – 93 = 7 Player 2; 100 – 97 = 3
Player 3; 100 – 89 = 11 Player 4; 100 – 99 = 1
3.Fill in the blanks:
Ans.
4.Oh! 206! Guess how many more to make a triple century?
Ans. Triple century = 300
So, 300 – 206 = 94
Top Tea Scores in the Cricket World Cup
1.Fill in the blanks:
1.C. K just missed his century. How many runs did he need to make a century?————
2.————and———— scored almost equal runs.
3.————scored a complete century, no less, no more.
4. Most runs scored by any batsman are ————
5.————and———— have a difference of just 1 run between them.
6. ————scored 2 more runs than one and a half century.
Ans. 1. 100 -9 = 1 Hence C. K. missed his century by 1 run.
2. A. S. and C. K. or A. P. J. and V. V. S.
3. A. S.
4. 178 scored by M. D.
5. A. S. and C. K. or A. P. J. and V. V. S.
6. K S. P.
2.Fill in the blahks (Counting in 10’s)
Ans.
3.Fill in the blanks (Counting in 50’s)
Ans.
4.How For can you go like this?
Ans. We can go on and on, because there is no end to these numbers.
5.What is the biggest number you can call out?
Ans. There is no biggest number and hence we cannot call that number.
Colour the Numbers
1.Find these numbers in the above chart and colour them.
Ans.
Jumping Animals
Find out:
(а)Gabru and Bunny both jump on squares 104,————— and—————
(b)Tarru and Bunny both jump on squares,—————,————— ,
—————and —————
(c)Is there any square where all three of them jump?—————
(d)Guess who will finish in the least jumps?————— In how many jumps?
Ans. (a) 139 and 174 (b) 114, 134, 154 and 174 (c) 174 (d) Gabru, 12
Hint: The last square is numbered 175. Multiply number of square in one jump by number of jumps for an animal. Add the result with the number on starting ; ‘square. You can also take Ludo coins of different colours to count the number of squares for an animal.
Class, Jump!
(a) Jump 2 steps forward:
104,106, 108,————- ,————- ,————- ,————-
(b)Jump 2 steps backward:
262, 260, 258,————- ,————- ,————- ,————-
(c)Jump 10 steps forward:
110,120,130,————- ,————- ,————- ,————-
(d) Jump 10 steps backward:
200,190,180,————- ,————- ,————- ,————-
Continue the pattern:
(e)550, 560, 570,————- ,————- ,————- ,————-
(f) 910,920,930,940, ————- ,————- ,————- ,————-
(g)209, 207, 205,————- ,————- ,————- ,————-
(h)401, 402, 403,————- ,————- ,————- ,————-
Ans. (a)110, 112, 114, 116
(b) 256, 254, 252, 250
(c)140, 150,160, 170
(d) 170,160, 150, 140
(e)580,590,600,610
(f) 950, 960, 970, 980
(g) 203, 201,199, 197
(h)404,405,406,407
Lazy Crazy Shop
1.Find out how many packets of tens, hundreds and loose items each animal will take. Fill in the blanks:
Ans.
2.Lazy Crazy also has a crazy way of taking money. He takes only in ?.100 notes, ? 10 notes and coins. Now find out how they will play him for what they have taken.
1.Rs 420 2.Rs 143 3.Rs 242 4. Rs 55
Ans.1.Rs 420 4 Rs 100 notes and 2 Rs 10 notes.
2.Rs 143 1 Rs 100 notes, 4 Rs 10 notes and 3 Re 1 coins.
3.Rs 242 2 Rs 100 notes, 4 Rs 10 notes and 2 Re 1 coins.
4.Rs 55 5 Rs 10 notes and 5 Re 1 coins.
3.Who am I? Match with the number.
Ans. 1. 45 2. 89 3. 96 4. 100 5. 150 6. 87
4.How many are these?
Ans. (a) 240 rupees (b) 319 sticks (c) 212 blocks (d) 138 beads (e) 320 rupees.
Moon Mama Counts His Starry Friends
1.Which cards will I have in my pocket if I have counted up to
(а)19 (b)21 (c) 95 (d)201 (e) 260
(f) 300 (g) 306 (h) 344 (i) 350 (j) 400
Ans. (a) T9 [10] [1] [1] [1] [1] [1] [1] [1] [1] [1]
(b)21 [10] [10] [1]
(c)95 [10] [10] [10] [10] [10] [10] [10] [10] [10] [1] [1] [1] [1] [1]
(d) 201 [100] [100] [1]
(e) 260 [100] [100] [10] [10] [10] [10] [10] [10]
(f) 300 [100] [100] [100]
(g)306 [100] [100] [100] [1] [1] [1] [1] [1] [1]
(h)344 [100] [100] [100] [10] [10] [10] [10] [1] [1] [1] [1]
(i)350 [100] [100] [100] [10] [10] [10] [10] [10]
(j) 400 [100] [100] [100] [100]
2.When I had [10] [10] cards in my pocket, I knew I had counted 20 stars. Now you tell me the number of stars counted in each case. Write the answer in the blank space.
Ans. (a) [10] [10]. —————–20
(b)[10]'[10] 1 11111 . —————–26
(c)[100] [100] [10] [100] —————–310
(d)[100] [10] [10] [10] .—————–130
(e)[100] [100] [100] [100].—————–400
(f)[100] 1 1 [100] 1 1[10] 1 1 [100].—————–316
(g)[100] 1 [100] [100] 1 [100] [100] 1 [100].—————–603
(h)[100] [10] [100] 1 [100] [10] [100] [10].—————–431
(i)[100] [100] [100] [100] [100] [100] [100] [100] [100] [100].—————–1000
3.Guess how many starry friends I have in all……..!!!
Ans. These are 38 in numbers. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.2: Matrices and linear systems
#### 3.2.1 Matrices and vectors
Before we can start talking about linear systems of ODEs, we will need to talk about matrices, so let us review these briefly. A matrix is an $$m \times n$$ array of numbers ($$m$$ rows and $$n$$ columns). For example, we denote a $$3 \times 5$$ matrix as follows
$A = \begin {bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35} \end {bmatrix}$
By a vector we will usually mean a column vector, that is an $$m \times 1$$ matrix. If we mean a row vector we will explicitly say so (a row vector is a $$1 \times n$$ matrix). We will usually denote matrices by upper case letters and vectors by lower case letters with an arrow such as $$\vec {\text {x}}$$ or $$\vec {b}$$. By $$\vec {0}$$ we will mean the vector of all zeros.
It is easy to define some operations on matrices. Note that we will want $$1 \times 1$$ matrices to really act like numbers, so our operations will have to be compatible with this viewpoint.
First, we can multiply by a scalar (a number). This means just multiplying each entry by the same number. For example,
$2 {\begin {bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end {bmatrix}} = \begin {bmatrix} 2 & 4 & 6 \\ 8 & 10 & 12 \end {bmatrix}$
Matrix addition is also easy. We add matrices element by element. For example,
$\begin {bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end {bmatrix} + \begin {bmatrix} 1 & 1 & -1 \\ 0 & 2 & 4 \end {bmatrix} = \begin {bmatrix} 2 & 3 & 2 \\ 4 & 7 & 10 \end {bmatrix}$
If the sizes do not match, then addition is not defined.
If we denote by 0 the matrix of with all zero entries, by $$c, d$$ scalars, and by $$A, B, C$$ matrices, we have the following familiar rules.
$A + 0 = A = 0 + A$
$A + B = B + A$
$(A + B) + C = A + ( B + C)$
$c( A + B) = cA + cB$
$( c + d) A = cA + dA$
Another useful operation for matrices is the so-called transpose. This operation just swaps rows and columns of a matrix. The transpose of $$A$$ is denoted by $$A^T$$. Example:
${ \begin {bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end {bmatrix}}^T = \begin {bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end {bmatrix}$
#### 3.2.2 Matrix Multiplication
Let us now define matrix multiplication. First we define the so-called dot product (or inner product) of two vectors. Usually this will be a row vector multiplied with a column vector of the same size. For the dot product we multiply each pair of entries from the first and the second vector and we sum these products. The result is a single number. For example,
$\begin {bmatrix} a_1 & a_2 & a_3 \end {bmatrix} \cdot \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix} = \begin {bmatrix} a_1b_1 + a_2b_2 + a_3b_3 \end {bmatrix}$
And similarly for larger (or smaller) vectors.
Armed with the dot product we can define the product of matrices. First let us denote by $$\text {row}_i (A)$$ the $$i^{th}$$ row of $$A$$ and by $$\text {column}_j (A)$$ the $$j^{th}$$ column of $$A$$. For an $$m \times n$$ matrix $$A$$ and an $$n \times p$$ matrix $$B$$ we can define the product $$AB$$. We let $$AB$$ be an $$m \times p$$ matrix whose $$ij^{th}$$ entry is
$\text {row}_i (A) \cdot \text {column}_j (B)$
Do note how the sizes match up. Example:
$\begin {bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end {bmatrix} \begin {bmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end {bmatrix} = \begin {bmatrix} 1. 1 + 2.1 + 3.1 & 1. 0 + 2.1 + 3.0 & 1. (-1) + 2.1 + 3.0 \\ 4.1 + 5.1 + 6.1 & 4.0 + 5.1 + 6.0 & 4.(-1) + 5.1 + 6.0 \end {bmatrix} = \begin {bmatrix} 6 & 2 & 1 \\ 15 & 5 & 1 \end {bmatrix}$
For multiplication we want an analog of a 1. This analog is the so-called identity matrix. The identity matrix is a square matrix with 1s on the main diagonal and zeros everywhere else. It is usually denoted by $$I$$. For each size we have a different identity matrix and so sometimes we may denote the size as a subscript. For example, the $$I_3$$ would be the $$3 \times 3$$ identity matrix
$I = I_3 = \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix}$
We have the following rules for matrix multiplication. Suppose that $$A, B, C$$ are matrices of the correct sizes so that the following make sense. Let $$\alpha$$ denote a scalar (number).
$A (BC) = (AB) C$
$A (B + C) = AB + AC$
$(B + C) A = BA + CA$
$\alpha (AB) = ( \alpha A )B = A ( \alpha B)$
$IA = A = AI$
A few warnings are in order.
• $$AB \ne BA$$ in general (it may be true by fluke sometimes). That is, matrices do not commute. For example take $$A = \begin {bmatrix} 1 & 1 \\ 1 & 1 \end {bmatrix}$$ and $$B = \begin {bmatrix} 1 & 0 \\ 0 & 2 \end {bmatrix}$$.
• $$AB = AC$$ does not necessarily imply $$B = C$$, even if $$A$$ is not 0.
• $$AB = 0$$ does not necessarily mean that $$A = 0$$ or $$B = 0$$. For example take $$A = B = \begin {bmatrix} 0 & 1 \\ 0 & 0 \end {bmatrix}$$.
For the last two items to hold we would need to “divide” by a matrix. This is where the matrix inverse comes in. Suppose that $$A$$ and $$B$$ are $$n \times n$$ matrices such that
$AB = I = BA$
Then we call $$B$$ the inverse of $$A$$ and we denote $$B$$ by $$A^{-1}$$. If the inverse of $$A$$ exists, then we call $$A$$ invertible. If $$A$$ is not invertible we sometimes say $$A$$ is singular.
If $$A$$ is invertible, then $$AB = AC$$ does imply that $$B = C$$ (in particular the inverse of $$A$$ is unique). We just multiply both sides by $$A^{-1}$$ to get $$A^{-1} AB = A^{-1} AC$$ or $$IB = IC$$ or $$B = C$$. It is also not hard to see that $${(A^{-1})}^{-1} = A$$.
#### 3.2.3 The determinant
We can now talk about determinants of square matrices. We define the determinant of a $$1 \times 1$$ matrix as the value of its only entry. For a $$2 \times 2$$ matrix we define
$\text {det} \left ( \begin {bmatrix} a & b \\ c & d \end {bmatrix} \right ) \overset {\text {def}}{=} ad - bc$
Before trying to compute the determinant for larger matrices, let us first note the meaning of the determinant. Consider an $$n \times n$$ matrix as a mapping of the $$n$$ dimensional euclidean space $$\mathbb {R}^n$$ to $$\mathbb {R}^n$$. In particular, a $$2 \times 2$$ matrix $$A$$ is a mapping of the plane to itself, where $$\vec {x}$$ gets sent to $$A \vec {x}$$. Then the determinant of $$A$$ is the factor by which the area of objects gets changed. If we take the unit square (square of side 1) in the plane, then $$A$$ takes the square to a parallelogram of area $$\mid \text {det} (A) \mid$$. The sign of $$\text {det} (A)$$ denotes changing of orientation (negative if the axes got flipped). For example, let
$A = \begin {bmatrix} 1 & 1 \\ -1 & 1 \end {bmatrix}$
Then $$\text {det} (A) = 1 + 1 = 2$$. Let us see where the square with vertices $$(0, 0), (1, 0), (0, 1)$$ and $$(1, 1)$$ gets sent. Clearly $$(0, 0 )$$ gets sent to $$(0, 0)$$.
$\begin {bmatrix} 1 & 1 \\ -1 & 1 \end {bmatrix} \begin {bmatrix} 1 \\ 0 \end {bmatrix} = \begin {bmatrix} 1 \\ -1 \end {bmatrix}, \begin {bmatrix} 1 & 1 \\ -1 & 1 \end {bmatrix} \begin {bmatrix} 0 \\ 1 \end {bmatrix} = \begin {bmatrix} 1 \\ 1 \end {bmatrix}, \begin {bmatrix} 1 & 1 \\ -1 & 1 \end {bmatrix} \begin {bmatrix} 1 \\ 1 \end {bmatrix} = \begin {bmatrix} 2 \\ 0 \end {bmatrix}$
So the image of the square is another square. The image square has a side of length $$\sqrt {2}$$ and is therefore of area 2.
If you think back to high school geometry, you may have seen a formula for computing the area of a parallelogram with vertices $$(0, 0), (a, c), (b, d)$$ and $$(a + b, c + d )$$. And it is precisely
$\mid \text {det} \left ( \begin {bmatrix} a & b \\ c & d \end {bmatrix} \right ) \mid$
The vertical lines above mean absolute value. The matrix $$\begin {bmatrix} a & b \\ c & d \end {bmatrix}$$ carries the unit square to the given parallelogram.
Now we can define the determinant for larger matrices. We define $$A_{ij}$$ as the matrix $$A$$ with the $$i^{th}$$ row and the $$j^{th}$$ column deleted. To compute the determinant of a matrix, pick one row, say the $$i^{th}$$ row and compute.
$\text {det} (A) = \sum _ {j=1}^n (-1)^{i+j} a_{ij} \text {det} (A_{ij})$
For the first row we get
$\text {det} (A) = a_{11} \text {det} (A_{11}) - a_{12} \text {det} (A_{12}) + a_{13} \text {det} (A_{13}) - \dots \begin {cases} +a_{1n} \text {det} (A_{1n} & \text {if n is odd} \\ -a_{1n} \text {det} (A_{1n} & \text {if n even} \end {cases}$
We alternately add and subtract the determinants of the submatrices $$A_{ij}$$ for a fixed $$i$$ and all $$j$$. For a $$3 \times 3$$ matrix, picking the first row, we would get $$\text {det} (A) = a_{11} \text {det} (A_{11}) - a_{12} \text {det} (A_{12}) + a_{13} \text {det} (A_{13})$$. For example,
$\text {det} \left ( \begin {bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end {bmatrix} \right ) = 1 \cdot \text {det} \left ( \begin {bmatrix} 5 & 6 \\ 8 & 9 \end {bmatrix} \right ) - 2 \cdot \text {det} \left ( \begin {bmatrix} 4 & 6 \\ 7 & 9 \end {bmatrix} \right ) + 3 \cdot \text {det} \left ( \begin {bmatrix} 4 & 5 \\ 7 & 8 \end {bmatrix} \right )$
$= 1(5 \cdot 9 - 6 \cdot 8) - 2 ( 4 \cdot 9 - 6 \cdot 7) + 3 ( 4 \cdot 8 - 5 \cdot 7 ) = 0$
The numbers $$(-1)^{i+j} \text {det} (A_{ij})$$ are called cofactors of the matrix and this way of computing the determinant is called the cofactor expansion. It is also possible to compute the determinant by expanding along columns (picking a column instead of a row above).
Note that a common notation for the determinant is a pair of vertical lines:
$\begin {bmatrix} a & b \\ c & d \end {bmatrix} = \text {det} \left ( \begin {bmatrix} a & b \\ c & d \end {bmatrix} \right )$
I personally find this notation confusing as vertical lines usually mean a positive quantity, while determinants can be negative. I will not use this notation in this book. One of the most important properties of determinants (in the context of this course) is the following theorem.
Theorem 3.2.1. An $$n \times n$$ matrix $$A$$ is invertible if and only if $$\text {det} (A) \ne 0$$.
In fact, there is a formula for the inverse of a $$2 \times 2$$ matrix
${\begin {bmatrix} a & b \\ c & d \end {bmatrix}}^ {-1} = \frac {1}{ad - bc} \begin {bmatrix} d & -b \\ -c & a \end {bmatrix}$
Notice the determinant of the matrix in the denominator of the fraction. The formula only works if the determinant is nonzero, otherwise we are dividing by zero.
#### 3.2.4 Solving linear systems
One application of matrices we will need is to solve systems of linear equations. This is best shown by example. Suppose that we have the following system of linear equations
$2x_1 + 2x_2 + 2x_3 = 2$
$x_1 + x_2 + 3x_3 = 5$
$x_1 + 4x_2 + x_3 = 10$
Without changing the solution, we could swap equations in this system, we could multiply any of the equations by a nonzero number, and we could add a multiple of one equation to another equation. It turns out these operations always suffice to find a solution.
It is easier to write the system as a matrix equation. Note that the system can be written as
$\begin {bmatrix} 2 & 2 & 2 \\ 1 & 1 & 3 \\ 1 & 4 & 1 \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \\ x_3 \end {bmatrix} = \begin {bmatrix} 2 \\ 5 \\ 10 \end {bmatrix}$
To solve the system we put the coefficient matrix (the matrix on the left hand side of the equation) together with the vector on the right and side and get the so-called augmented matrix
$\left [ \begin {array}{ccc|c} 2 & 2 & 2 & 2 \\ 1 & 1 & 3 & 5 \\ 1 & 4 & 1&10 \end {array} \right ]$
We apply the following three elementary operations.
• Swap two rows.
• Add a multiple of one row to another row.
• Multiply a row by a nonzero number.
We will keep doing these operations until we get into a state where it is easy to read off the answer, or until we get into a contradiction indicating no solution, for example if we come up with an equation such as $$0 = 1$$.
Let us work through the example. First multiply the first row by $$\frac {1}{2}$$ to obtain
$\left [ \begin {array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 1 & 3 & 5 \\ 1 & 4 & 1&10 \end {array} \right ]$
Now subtract the first row from the second and third row.
$\left [ \begin {array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 0 & 2 & 4 \\ 0 & 3 & 0 & 9 \end {array} \right ]$
Multiply the last row by $$\frac {1}{3}$$ and the second row by $$\frac {1}{2}$$.
$\left [ \begin {array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & 0 & 3 \end {array} \right ]$
Swap rows 2 and 3.
$\left [ \begin {array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end {array} \right ]$
Subtract the last row from the first, then subtract the second row from the first.
$\left [ \begin {array}{ccc|c} 1 & 0 & 0 & -4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end {array} \right ]$
If we think about what equations this augmented matrix represents, we see that $$x_1 = -4, x_2 = 3$$ and $$x_3 = 2$$. We try this solution in the original system and, voilà, it works!
Exercise $$\PageIndex{1}$$:
Check that the solution above really solves the given equations.
If we write this equation in matrix notation as
$A \vec {x} = \vec {b}$
where $$A$$ is the matrix $$\begin {bmatrix} 2 & 2 & 2 \\ 1 & 1 & 3 \\ 1 & 4 & 1 \end {bmatrix}$$ and $$\vec {b}$$ is the vector $$\begin {bmatrix} 2 \\ 5 \\ 10 \end {bmatrix}$$. The solution can be also computed via the inverse,
$\vec {x} = A^{-1} A \vec {x} = A^{-1} \vec {b}$
One last note to make about linear systems of equations is that it is possible that the solution is not unique (or that no solution exists). It is easy to tell if a solution does not exist. If during the row reduction you come up with a row where all the entries except the last one are zero (the last entry in a row corresponds to the right hand side of the equation) the system is inconsistent and has no solution. For example if for a system of 3 equations and 3 unknowns you find a row such as $$\begin {bmatrix} 0 & 0 & 0 \mid & 1 \end {bmatrix}$$ in the augmented matrix, you know the system is inconsistent.
You generally try to use row operations until the following conditions are satisfied. The first nonzero entry in each row is called the leading entry.
• There is only one leading entry in each column.
• All the entries above and below a leading entry are zero.
• All leading entries are 1.
Such a matrix is said to be in reduced row echelon form. The variables corresponding to columns with no leading entries are said to be free variables. Free variables mean that we can pick those variables to be anything we want and then solve for the rest of the unknowns.
Example $$\PageIndex{1}$$:
The following augmented matrix is in reduced row echelon form.
$\left [ \begin {array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end {array} \right ]$
Suppose the variables are $$x_1, x_2$$ and $$x_3$$. Then $$x_2$$ is the free variable, $$x_1 = 3 - 2x_2$$, and $$x_3 = 1$$.
On the other hand if during the row reduction process you come up with the matrix
$\left [ \begin {array}{ccc|c} 1 & 2 & 13 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 3 \end {array} \right ]$
there is no need to go further. The last row corresponds to the equation $$0x_1 + 0x_2 + 0x_3 = 3$$, which is preposterous. Hence, no solution exists.
### 3.2.5 Computing the inverse
If the coefficient matrix is square and there exists a unique solution $$\vec {x}$$ to $$A \vec {x} = \vec {b}$$ for any $$\vec {b}$$, then $$A$$ is invertible. In fact by multiplying both sides by $$A^{-1}$$ you can see that $$\vec {x} = A^{-1} \vec {b}$$. So it is useful to compute the inverse if you want to solve the equation for many different right hand sides $$\vec {b}$$.
The $$2 \times 2$$ inverse can be given by a formula, but it is also not hard to compute inverses of larger matrices. While we will not have too much occasion to compute inverses for larger matrices than $$2 \times 2$$ by hand, let us touch on how to do it. Finding the inverse of $$A$$ is actually just solving a bunch of linear equations. If we can solve $$A \vec {x}_k = \vec {e}_k$$ where $$\vec {e}_k$$ is the vector with all zeros except a 1 at the $$k^{th}$$ position, then the inverse is the matrix with the columns $$\vec {x}_k$$ for $$k = 1, \dots , n$$ (exercise: why?). Therefore, to find the inverse we can write a larger $$n \times 2n$$ augmented matrix $$[ A \mid I ]$$, where $$I$$ is the identity. We then perform row reduction. The reduced row echelon form of $$[ A \mid I ]$$ will be of the form $$[ I \mid A^{-1} ]$$ if and only if $$A$$ is invertible. We can then just read off the inverse $$A^{-1}$$. |
Class 7 CBSE maths Integers
This page contains main concepts and results for Class 7, Chapter-1 INTEGERS. This chapter is mainly about representation of integers on the number line and their addition and subtraction.
Numbers
We use numbers to count anything. So, what are various types of numbers?
1. Natural numbers
Natural numbers are counting numbers but these set of numbers do not include zero. This is because you cannot count zero. So numbers,
1,2,3,4,5,6……etc are all natural numbers
2. Whole numbers
All natural numbers along with zero are called whole numbers. For example
0, 1, 2, 3, 4, 5, 6………etc are all whole numbers.
NOTE:- Thes types of numbers do not include fractions.
From the definition of natural numbers we can conclude that every natural or counting number is a whole number.
3. Integers
Integers include all natural numbers, zero and negative numbers for example,
-4, -3, -2, -1, 0, 1, 2, 3, ………. etc are all integers.
So now we have,
1. Positive integers:- 1, 2, 3, ……
2. Negative integers:- -1, -2, -3, …….
3. 0 (zero):- which is an integer that is neither negative nor positive.
Note:- Integers like whole numbers do not include fractions for example 3.5 , ½ etc.
Number line for integers is
Now from this number line you can see that numbers to the left of zero are all negative. Again from this number line you can observe that numbers to the right of zero are all positive.
Important note:- If the number has no sign attached to it as prefix then it means that it is a positive number.
For example number 3 is really number +3
Important Points
On the number line when we
1. Add a positive integer, we move to the right.
2. Add a negative integer we move to the left.
3. Subtract a positive integer we move to the left.
4. Subtract a negative integer, we move to the right.
Properties of integers
• Integers are closed under addition, subtraction and multiplication. Which means that sum of integers will also give integers.
• Addition and multiplication are commutative for integers, i.e.,
1. a + b = b + a
2. a × b = b × a
For any two integers a and b.
• Addition and multiplication are associative for integers, i.e.,
1. (a + b) + c = a + (b + c)
2. (a × b) × c = a × (b × c)
for any three integers a, b and c.
• Existence of identity
1. Zero (0) is an additive identity for integers, i.e.,
a + 0 = 0 + a = a
for any integer a.
2. 1 is multiplicative identity for integers, i.e.,
a × 1 = 1 × a = a
for any integer a.
• Integers show distributive property of multiplication over addition, i.e.,
a × (b + c) = a × b + a × c
for any three integers a, b and c.
• Product of a positive integer and a negative integer is a negative integer, i.e,
a × (–b) = – ab
where a and b are positive integers.
• Product of two negative integers is a positive integer, i.e.,
• (–a) × (–b) = ab
where a and b are positive integers.
• Product of even number of negative integers is positive, whereas the product of odd number of negative integers is negative, i.e.,
• When a positive integer is divided by a negative integer or vice-versa and the quotient obtained is an integer then it is a negative integer, i.e.,
$a\div \left( b \right)=\left( a \right)\div b\text{ =}\frac{a}{b}$
where a and b are positive integers and $-\frac{a}{b}$is an integer
• When a negative integer is divided by another negative integer to give an integer then it gives a positive integer, i.e.,
$\left( a \right)\div \left( b \right)=\frac{a}{b}$ , where a and b are positive integers and $\frac{a}{b}$ is also an integer.
• For any integer a,
$a\div 1=a~$
and
a ÷ 0 is not defined.
Problem solving strategy
Now let us consider solving a problem and apply the above mentioned problem solving strategy. For this purpose we will solve a NCERT book problem.
Question:
Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan's account after the withdrawal.
Solution:
Step 1:- Understand the problem
• What do you know from the problem?
Here find ‘what is the amount Mohan deposited in his bank account and how much he withdraws’.
• What are you trying to find?
Balance in Mohan’s bank account after the withdrawal.
Amount deposited = Rs 2000
Amount withdrawn = Rs 1642
Here we have to find the amount he removed from his account. So,
Balance in Mohan's account = Money deposited - Money withdrawn
Step 3:- Solve the problem
Now that you have all the known and unknown quantities and you also have a strategy to solve your problem, you can now carefully carry out your calculations.
Balance in Mohan's account = 2000 + (-1642) = 2000 - 1642 = 358
Therefore, balance in Mohan's account after withdrawal is Rs 358.
Step 4:- Revise |
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### High School Mathematics - 27.3 Binomial Theorem
Coefficient of a particular power of x Let the particular power occur in the (r+1)th term. Write the (r+1)th term of the given binomial. Equate the power of x in the (r+1)th term and the given power. Evaluate r and substitute in in step 2. Eample: Find the co-efficient of x152)10. Solution: Let x15 occur in the (r+1)th term. Tr+1 = 10Crx10-r .(-x2)r. 10Cr.x10-r.x2r.(-1)r. = (-1)r.10Crx10+r x10+r = x15 r = 5 Terms Independent of x Procedure Let (r+1)th term be the term independent of x. Put the power of x in this term equal to zero and evaluate it. Example: Find the term independent of x in the expansion of (3/2x2 - 1/3x)9. Solution: Let (r+1)th term be independent of x. Tr+1 = 9Cr.(3/2x2)9-r.(-1/3x)r Putting 18-3r = 0, we get r = 6. The required term is (-1)6.9C6. 3-3/23 = 9C3.(1/33.23 = 7/18 Answer: 7/18 Example: Find the greatest coefficient in the expansion of (1+x)10 Solution: Let Tr+1 have the greatest coefficient. Then coefficient of Tr+1 >= the coefficient of T. 10Cr >= 10Cr-1 (10-r+1)/r >= 1 or 11 >= 2r, or r <= 51/2 Hence upto r = 5 the coefficients increase, the greatest of them being r = 5. The greatest coefficient = 10C5 = 252 Answer : 252 Directions: Answer the following
Q 1: In the expansion (x2+1/x)n, the coefficient of the 4th term is equal to the coefficient of the ninth term Find 6th term.Answer: Q 2: Find the term containing x2 if any in (3x-1/2x)8.Answer: Q 3: Find the greatest term in the expansion of (2-3x)7 when x = 1.Answer: Q 4: Find the coefficient of x7 in the expansion of (x2+1/x)11Answer: Q 5: Find numerically the greatest term in the expansion of (5x+4)7 when x = 1(write 1st, 2nd etc)Answer: Q 6: Find the coefficient of a6b3 in the expansion of (2a-b/3)9Answer: Q 7: Find numerically the greatest term in the expansion of (2+3x)7 when x = 4/5Answer: Q 8: Find the coefficient of a6b3 in the expansion of (2a-b/3)9Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
# How to integrate by partial fractions? – 6 Steps
Integration functions can be a little confusing especially when it involves polynomials in the denominator. In such cases use of integration fractions can be easy and can solve the problem in minutes. Students of calculus will find it handy to learn how to decompose functions into partial fractions for other parts of calculus.
• Check to make sure that the fraction you are trying to integrate is proper: A proper fraction has a smaller in the denominator than in the numerator. If the power of the numerator is larger than or equal to the power of denominator, it is improper and must be divided using long division.
• After dividing your improper fraction with long division, your new fraction can be put in the form of quotient +(remainder / divisor)
• Factor the polynomials in the denominator
• Separate the fraction that you wish to decompose in to multiple fractions: The number of fractions in decomposition should be equal to the number of factors of x. The numerators of these decomposed fractions should be represented with constants.
• Multiply both sides by the denominator of the original fraction in order to get rid of all denominators.
• Set the coefficients equal on both the sides.
These are few steps that can be used to integrate by partial fractions. People who are into calculus this function is very beneficial to them. It makes it very easy for them to calculate their problems in less time and without making it complex. Get the answers right at the very first time, once you know how to integrate by partial fraction. This principle is very useful and once we learn it then it is easy to solve problems using this technique.
Be open minded and learn how to use integrated fractions to solve your problems and you will know how easy it is once learned well. Now colleges teach this subject as well when it is necessary for the students in their said subjects. It is an interesting subject and many calculus students learn this quick and fast |
# 12.2 Circles
Page 1 / 2
This module introduces the definition and formula of a circle, including example problems.
## The definition of a circle
You’ve known all your life what a circle looks like. You probably know how to find the area and the circumference of a circle, given its radius. But what is the exact mathematical definition of a circle? Before you read the answer, you may want to think about the question for a minute. Try to think of a precise, specific definition of exactly what a circle is.
Below is the definition mathematicians use.
## Definition of a circle
The set of all points in a plane that are the same distance from a given point forms a circle . The point is known as the center of the circle , and the distance is known as the radius .
Mathematicians often seem to be deliberately obscuring things by creating complicated definitions for things you already understood anyway. But if you try to find a simpler definition of exactly what a circle is, you will be surprised at how difficult it is. Most people start with something like “a shape that is round all the way around.” That does describe a circle, but it also describes many other shapes, such as this pretzel:
So you start adding caveats like “it can’t cross itself” and “it can’t have any loose ends.” And then somebody draws an egg shape that fits all your criteria, and yet is still not a circle:
So you try to modify your definition further to exclude that ... and by that time, the mathematician’s definition is starting to look beautifully simple.
But does that original definition actually produce a circle? The following experiment is one of the best ways to convince yourself that it does.
## Experiment: drawing the perfect circle
1. Lay a piece of cardboard on the floor.
2. Thumbtack one end of a string to the cardboard.
3. Tie the other end of the string to your pen.
4. Pull the string as tight as you can, and then put the pen on the cardboard.
5. Pull the pen all the way around the thumbtack, keeping the string taut at all times.
The pen will touch every point on the cardboard that is exactly one string-length away from the thumbtack. And the resulting shape will be a circle. The cardboard is the plane in our definition, the thumbtack is the center , and the string length is the radius .
The purpose of this experiment is to convince yourself that if you take all the points in a plane that are a given distance from a given point, the result is a circle. We’ll come back to this definition shortly, to clarify it and to show how it connects to the mathematical formula for a circle.
## The mathematical formula for a circle
You already know the formula for a line: $y=mx+b$ . You know that $m$ is the slope, and $b$ is the y-intercept. Knowing all this, you can easily answer questions such as: “Draw the graph of $y=2x–3$ ” or “Find the equation of a line that contains the points (3,5) and (4,4).” If you are given the equation $3x+2y=6$ , you know how to graph it in two steps: first put it in the standard $y=mx+b$ form, and then graph it.
All the conic sections are graphed in a similar way. There is a standard form which is very easy to graph, once you understand what all the parts mean. If you are given an equation that is not in standard form, you put it into the standard form, and then graph it.
#### Questions & Answers
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
Ramkumar Reply
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#### Get the best Algebra and trigonometry course in your pocket!
Source: OpenStax, Advanced algebra ii: conceptual explanations. OpenStax CNX. May 04, 2010 Download for free at http://cnx.org/content/col10624/1.15
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# How to Solve Two-Step Equations
1. Math Lessons >
2. Two-step Equations
What is a two-step equation?
As the name suggests, a two-step equation requires two steps to solve a problem. A two-step equation has more than one operation.
An example of two-step equation is
9k + 3 = 21
How do you solve a two-step equation?
It doesn't take much to solve a two-step equation. Isolate the variable to solve the equation. To isolate the variable, use inverse operations wherever required.
Let's consider the below instance:
5x + 7 = 17
Step 1: In the above-given equation, undo the addition using inverse operations (subtraction) on both sides of the equation.
5x = 10
Step 2: To remove the coefficient of the variable, perform the inverse of multiplication (division). Divide by 5 on both sides of the equation.
x = 2
Plug-in the value of the variable in the equation to check your answer.
For the equation 5x + 7 = 17, substitute x = 2,
(5 × 2) + 7 = 17
10 + 7 = 17
17 = 17
The lesson in a nutshell:
To obtain the value of a variable, we have to isolate the variable.
Perform inverse operations on both sides of the equation to isolate the variable.
Simplify the equation and find the value of the variable.
Check your answer by substituting the variable value in the equation.
Boost your skills using our free printable worksheets on Two-step Equations.
Solved Examples:
Example 1
1. Solve 4x – 7 = 15
4x – 7 = 15
4x – 7 + 7 = 15 + 7
[ Adding 7 on both sides ]
4x = 22
4x4 = 224
[ Dividing by 4 on both sides ]
x = 112
Example 2
2. Solve 8y = 40 – 2y
8y = 40 – 2y
8y + 2y = 40 – 2y + 2y
[ Adding 2y on both sides ]
10y = 40
10y10 = 4010
[ Dividing by 10 on both sides ]
y = 4
Example 3
3. On a clearance sale, a jumpsuit was up for grabs at half the price. Sara grabs the deal, and she further uses a coupon for \$7. If she pays \$21 to the cashier, what was the original cost of the jumpsuit?
The above scenario forms an equation: , where k refers to the original cost of the jumpsuit.
k2 – 7 = 21
k2 – 7 + 7 = 21 + 7
[ Adding 7 on both sides ]
k2 = 28
k2 × 2 = 28 × 2
[ Multiplying 2 on both sides ]
k = 56
The original cost of the jumpsuit is \$56. |
# Lecture 4 - PowerPoint PPT Presentation
1 / 23
Lecture 4. Recurrences. 4.1 The substitution method The substitution method : (i) 猜一個答案 (ii) 用歸納法證明 (for both upper and lower bounds). 範例 : 找到 T ( n ) = 2 T ( n /2 ) + n 的 upper bound ( T (1) = 1) 猜測 T ( n ) = O ( n lg n )
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Lecture 4
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## Lecture 4
Recurrences
4.1 The substitution method
The substitution method:
(i)猜一個答案
(ii)用歸納法證明
(for both upper and lower bounds)
Recurrences
(T(1) = 1)
Recurrences
Basis: (n=n0)
n0 2 且 c T(n0)/(n0 lg n0) ………(1)
Recurrences
Induction: (n>n0)
T(n) 2(cn/2 lg n/2) + n(Substitution)
cn lg (n/2) + n
= cn lg n – cn lg 2 + n
= cn lg n – cn + n
• cn lg n，
其中最後一步在 c 1 的時候成立………(2)
根據(1)(2)，我們可以選擇 n0 = 2 以及 c = max{1， T(2)/(2 lg 2)} = 2 讓 basis 及 induction 都成立。
Recurrences
Substitution Method
證明存在 c和 n0使得T(n) cg(n) 對於全部的 nn0都成立……(1)
Recurrences
如果我們已知 c和 n0，那麼我們就可以用歸納法證明(1)
(a) Basis step: (1) 在 n = n0時成立
(b) Induction step: (1) 在 n > n0時成立
如何找到 c和 n0符合歸納法中的證明？
(i) 找到符合 basis step 的 c和 n0的條件
(ii) 找到符合 induction step 的 c和 n0的條件
(iii) 綜合 (i) 和 (ii) 的條件
Recurrences
Subtleties:
(修改假設：減一個低次方項)
Recurrences
Basis: ok!
Induction: T(n)= T(n/2) + T(n/2) + 1
c(n/2 + n/2) + 1
= cn + 1
Induction: T(n) (cn/2-b) + (cn/2-b) + 1
= cn – 2b + 1
cn – b，
Recurrences
Avoiding pitfalls:
T(n) = 2T(n/2) + n
Induction: T(n) 2cn/2 + n
cn + n
=O(n) 錯 !!
Recurrences
Changing variable:
T(n) = 2T( ) + lg n
T(2m) = 2T(2m/2) +m
S(m) = 2S(m/2) + m (把 m換成 lg n)
Recurrences
4.2 The iteration(recursion-tree) method
T(n) = n + 3(n/4 + 3T(n/16))
= n + 3n/4 + 9(n/16 + 3T(n/64))
.
.(note that n/( 1)
n + 3n/4 + 9n/16 + 27n/64 + ... + (1)
= 4n+o(n) = O(n)
Recurrences
Recursion trees: (for visualizing the iteration)
• T(n)=2T(n/2) + n2
T(n)
n2
Recurrences
n2
n2
lg n
Total：
Recurrences
n
n
n
n
Total：
Recurrences
4.3 The master method
Theorem 4.1 (Master theorem)
1. 若存在常數 > 0 使得 f(n) = O(nlogb a)，則 T(n) = (nlogb a)
2. 若 f(n) = (nlogb a)，則 T(n) = (nlogb a lg n)
3. 若存在常數 > 0 使得 f(n) = (nlogb a+)，且存在常數 c < 1 使得對所有夠大的 n，af(n/b) cf(n)，則 T(n) = (f(n))。
Recurrences
Recurrences
Recurrences
Problem 1:
Recurrences
Recurrences
Recurrences
Problem 2:
Recurrences
Recurrences |
Example
Example 1:
The average of five numbers is 29, if one number is exclude the average becomes 27. what is the exclude number ?
let the exclude number is
= ( 29 x 5 ) – ( 27 x 4 )
= 145 – 108
= 37.
The exclude number 37.
Example 2:
What would be the average of first 20 natural numbers ?
Sum of first n natural numbers = n ( n + 1 ) /2
So, we can find easily average of first 20 natural numbers 20 x 21 / 2 = 210
So, then Required average is = 210 / 20 = 10.5.
Example 3:
What would be the average of first 20 multiplies of 5.
Required average = 5 ( 1 + 2 + 3 +……………….. + 20) /20
= ( 5 x 20 x 21 / 20 x 2) = 2100 / 40 = 52.5.
So the Required average is 52.5.
Example 4 :
The average marks of 13 students is 40, the first six students is 30 and that of the last six students is 32. Find the value of the 7th number.
Shortcut tricks :
7th student marks number = Total of 13 students result – ( Total of first six students + Total of last six students results )
= 13 x 40 – ( 6 x 30 + 6 x 32 )
= 520 – 180 + 192
= 148.
Example 5
In a college, 16 girls has the average age is 18 years and 14 boys has the average age 17 years. What would be the average age of entire college ?
Answer: 16 girls has the average age is 18 years ( 16 x 18 ) = 288
4 boys has the average age 17 years ( 14 x 17 ) = 238
Average age is = 288 + 238 / 30 = 526 / 30 = 17.54 years |
### Home > CCA2 > Chapter 9 > Lesson 9.3.1 > Problem9-80
9-80.
Denae and Gina were trying to solve $\sqrt { x + 5 }+\sqrt { x }= 5$. They knew they would have to square the equation to get rid of the radicals, but when they tried it, they got $x + 5 +2\sqrt { x ( x + 5 ) }+ x = 25$ , which did not look much easier to deal with. Gina had an idea.
Let’s start by putting the radicals on opposite sides of the equation and then squaring,” she said. Her work is shown at right.
But there’s still a radical!” she exclaimed, disappointed.
$\sqrt{x+5}+\sqrt{x}=5$
$\sqrt{x+5}=5-\sqrt{x}$
$x+5=25-10\sqrt{x}+x$
That’s okay,” said Denae, “there’s only one now, so we can get the term with the radical by itself and then square everything again.
$x+5=25-10\sqrt{x}+x$
$-20=-10\sqrt{x}$
$2=\sqrt{x}$
$?=x$
1. Use Denae and Gina’s method to solve $\sqrt { 2 x - 2 } - \sqrt { x } = 1$.
$\sqrt{2x-2}=1+\sqrt{x}$
$\left(\sqrt{2x-2}\right)^{2}=\left(1+\sqrt{x}\right)^2$
$2x-2=1+1\sqrt{x}+1\sqrt{x}+x$
$x-3=2\sqrt{x}$
$\left(x-3\right)^2=\left(2\sqrt{x}\right)^2$
$x^2-3x-3x+9=4x$
Continue solving. Note that you will need to factor and use the Zero Product Property.
Be sure to check both of your answers in the original equation. |
# Solve the following differential equation : $(1+x^2)\large\frac{dy}{dx}$$-2xy=(x^2+2)(x^2+1) ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• (i) Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • (ii) Find the integrating factor (I.F) = e^{\int Pdx}. • (iii) Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: (1+x^2)\large\frac{dy}{dx}$$-2xy=(x^2+2)(x^2+1)$
Clearly this is a linear differential equation .
Divide throughout by $(1+x^2)$
$\large\frac{dy}{dx}-\frac{2x}{1+x^2}$$y=x^2+2 This is of the form \large\frac{dy}{dx}$$+Py=Q$
Hence $P=\large\frac{-2x}{1+x^2}$ and $Q=x^2+2$
The general solution for this equation is
$ye^{\int Pdx}=\int Qe^{\int Pdx}dx+c$
Step 2:
Let us find the integral factor I.F
$\int Pdx=\int \large\frac{2x}{1+x^2}$$dx Put 1+x^2=t,then 2xdx=dt \therefore \int\large\frac{dt}{t}$$=\log t$
$\qquad\;\;=\log (1+x^2)$
$\therefore e^{\int Pdx}=e^{\log (1+x^2)}$
$\Rightarrow 1+x^2$
$\therefore y(1+x^2)=\int (x^2+2).(1+x^2)dx$
Step 3:
$y(1+x^2)=\int [x^4+2x^2+x^2+2]$
$y(1+x^2)=\int( x^4+3x^2+2)dx$
On integrating we get,
$y(1+x^2)=\large\frac{x^5}{5}+\frac{3x^3}{3}$$+2x+c y(1+x^2)=\large\frac{x^5}{5}$$+x^3+2x+c$
$5y(1+x^2)=x^5+5x^3+10x+c$ |
## How do you find the percentage of variation?
To calculate the percentage increase:First: work out the difference (increase) between the two numbers you are comparing.Increase = New Number – Original Number.Then: divide the increase by the original number and multiply the answer by 100.% increase = Increase Original Number 100.
Is variance a percentage?
The variance formula is used to calculate the difference between a forecast and the actual result. The variance can be expressed as a percentage or as an integer (dollar value or the number of units).
How do you find the delta between two numbers?
If you have a random pair of numbers and you want to know the delta – or difference – between them, just subtract the smaller one from the larger one. For example, the delta between 3 and 6 is (6 – 3) = 3. If one of the numbers is negative, add the two numbers together.
### What is variance in simple terms?
Variance describes how much a random variable differs from its expected value. The variance is defined as the average of the squares of the differences between the individual (observed) and the expected value. This means that it is always positive.
What does variance tell us in statistics?
The term variance refers to a statistical measurement of the spread between numbers in a data set. More specifically, variance measures how far each number in the set is from the mean and thus from every other number in the set.
What’s the difference between standard deviation and variance?
Standard deviation looks at how spread out a group of numbers is from the mean, by looking at the square root of the variance. The variance measures the average degree to which each point differs from the mean—the average of all data points.
## How do you interpret a sample variance?
A small variance indicates that the data points tend to be very close to the mean, and to each other. A high variance indicates that the data points are very spread out from the mean, and from one another.
How do you find a sample mean?
How to calculate the sample meanAdd up the sample items.Divide sum by the number of samples.The result is the mean.Use the mean to find the variance.Use the variance to find the standard deviation.
What is variance in probability?
In probability theory and statistics, variance is the expectation of the squared deviation of a random variable from its mean. Informally, it measures how far a set of numbers is spread out from their average value. |
Working with Equations
# 9 Transposing Equations
Click play on the following audio player to listen along as you read this section.
Have you ever come across a situation during your math studies where you’re required to solve for a variable which doesn’t seem to be in the right place? Take a look at the following example to see what I mean.
$\Large \text{A} = {\text{B}}^{2} \times 0.7854 \times \text{H}$
In a perfect world, you would like to solve for “A” and at the same time be given the values of both “B” and “H.”
But what if you were given “A” and you had to solve for “B”? How would you go about doing this?
The idea here would be to move the variables around and isolate “B.” What this means is that “B” is on one side of the equation by itself, and everything else is on the other side. Take a look at the equation again when this has been done.
$\Large \text{B} = \sqrt{\dfrac{\text{A}}{.7854 \times \text{H}}}$
Changing the formula around is referred to as “transposing” an equation.
It’s not as simple as just moving stuff around though. There are rules to get to this point, and those rules and their application are what we are going to deal with in this part of the chapter.
The most important thing to remember when transposing equations is that whatever is done to one side of the equation must also be done to the other side of the equation.
If you look at it mathematically, this makes sense. We already determined that an equation is two mathematical expressions that are separated by an equal sign.
What this means is the addition, subtraction, division, and multiplication variables and constants on one side of the equation are equal to all the addition, subtraction, division, and multiplication variables on the other side of the equation.
So if you decide to add 5 to one side, you must add 5 to the other side. What this does is keep the equation equal.
Take a look at the following example.
$\Large \begin{array}{c}10+7=9+8 \\ \text{This works out to be:} \\ 17=17\end{array}$
Here we have an equation that is true. Now add 5 to the left hand side of the equation, and you’ll see that, in order to keep the equation true, you’ll have to add 5 to the right hand side of the equation.
$\Large \begin{array}{c} 10 + 7 + \mathbf{5} = 9 + 8 + \textbf{?} \\ 22 = 9 + 8 + ? \\ 22 = 9 + 8 + \mathbf{5} \\ 22 = 22 \end{array}$
Keep in mind that this is an example where we added to one side. If we had subtracted, divided, or multiplied, things would be different. We would have to do the same thing to the other side.
# Transposing Equations Using Addition and Subtraction
$\Large \begin{array}{c}7+2=8+1 \\ \text{This works out to be:}\\ 9=9\end{array}$
Then add 4 to one side of the equation and solve.
$\Large 7+2+4=8+1+ \text{ ?}$
As the left hand side of the equation has had 4 added to it, the right hand side of the equation also has to have 4 added to it. We get:
$\Large \begin{array}{c}7+2+ \mathbf{4} = 8 + 1 + \textbf{?} \\ 13 = 9 + \text{ ?} \\ 13 = 9 + \mathbf{4} \\ 13 =13 \end{array}$
As stated before, the rule is that whatever you do to one side you must do to the other. In this case, if we add 4 to one side then we have to add 4 to the other side to keep it equal.
Subtraction would work the same way. If we were to subtract 4 from one side we would have had to subtract 4 from the other side in order to keep it equal.
Let’s try out this concept with an equation that has an unknown variable in it. Take a look at the following equation and solve for “J.”
$\Large 7+ \text{J}=5+8$
In order to solve for “J” we must isolate “J” on one side of the equation, and it actually doesn’t matter which side we isolate “J” on. If we follow our rule, in order to isolate “J” we would have to get rid of the 7 on the left side. The question becomes, how is that done? What we essentially have to do is move the 7 from the left side to the right side.
Once again, we always have to keep in mind that whatever we do to one side, we must do to the other.
Start with this. Would you agree that 7 − 7 = 0?
What if we subtracted 7 from the left hand side of the equation? What we would be left with is just “J” on the left side, which solves the problem of isolating “J.”
$\Large \begin{array}{c} \mathbf{(7-7)} +\text{J} = 5+8 \\ 0 + \text{J} = 5+8 \\ \text{J}= 5+8 \end{array}$
Mathematically this is not correct yet as we only dealt with the left side of the equation. What we need to do now is to do the same thing to the right side and then solve the equation. So we end up subtracting 7 from the right side of the equation to make everything equal once again.
$\Large \begin{array}{c}(7-7)+\text{J} = 5+8-7 \\ 0 + \text{J} = 13-7 \\ \text{J} = 13 - 7 \\ \text{J}=6 \end{array}$
You can always check that your answer is correct by taking the answer and putting it back in the equation to replace the variable.
$\Large \begin{array}{rl}\text{Replace J with 6} & \\ \downarrow & \\ 7 + \text{J} & = 5+8 \\ 7+6 & = 5+8 \\ 13 & = 13 \end{array}$
Example
Solve for G.
$\Large 27+ \text{G} = 43+49$
Step 1: Isolate the variable you are trying to find. In this case “G.” In order to do this, we must remove the 27 from the left side of the equation. This is done by subtracting 27 from the left side and then also from the right side.
$\Large \begin{array}{c}27+\text{G} = 43+49 \\ \mathbf{(27-27)}+\text{G} = 43+49 - \mathbf{27}\end{array}$
Step 2: Work through the equation.
$\Large \begin{array}{c}(27-27) +\text{G} = 43+49-27 \\ 0 + \text{G} = 92-27 \\ \text{G}=65\end{array}$
$\Large \begin{array}{rl}\text{Replace G with 65} &\\ \downarrow & \\ 27 + \text{G} & = 43 + 49 \\ 27 + 65 & = 43 + 49 \\ 92 & =92\end{array}$
Example
Solve for H.
$\Large \text{H}-16=13+19$
Step 1: Isolate the variable you are trying to find. In this case “H.” In order to do this we must remove the 16 from the left side of
the equation. This is done by adding 16 to the left hand side of the equation and then also adding 16 to the right hand side of
the equation.
$\Large \begin{array}{c}\text{H}-16 = 13+19 \\ \text{H} + \mathbf{(-16+16)}=13 + 19 \mathbf{ +16}\end{array}$
Step 2: Work through the equation.
$\Large \begin{array}{c}\text{H}+(-16+16)=13+19+16 \\ \text{H}+0=32+16 \\ \text{H}=48 \end{array}$
$\Large \begin{array}{l}\text{Replace H with 48} \\ \downarrow \\ \text{H}-16 = 13+19 \end{array}$
$\Large \begin{array}{c} 48-16=13+19 \\32=32 \end{array}$
# Practice Questions
Now it’s time to do a couple of practice questions for yourself. Make sure to check the video answers to see how you did.
Question 1
Solve for S.
$\Large 142+\text{S}=198+257$
Question 2
Solve for Y.
$\Large \text{Y} - 22 = 51+53$
# Transposing Equations Using Multiplication and Division
Transposing equations using multiplication and division uses the same basic principle as addition and subtraction in that whatever you do to one side you must do to the other.
Before we dive into this part of the chapter, we should refresh our memories a little and talk about reciprocals as it relates to math. Take a look at the following two numbers with one being a whole number and the other a fraction.
$\Large 4 \text{ and } \dfrac{1}{4}$
Don’t forget that when we write the number 4, we could also write it as:
$\Large\dfrac{4}{1}$
Do you remember what will happen if we multiplied those two together?
$\Large 4 \times \dfrac{1}{4} = \dfrac{4}{4} = 1$
Reciprocals are numbers that when multiplied together equal 1. This is an important concept when transposing using multiplication and division. If we divide a number by its reciprocal and end up with one, we have essentially removed that number from the equation.
Here is an example. Solve for K.
$\Large 8 \times \text{K} = 14 \times 17$
Similar to transposing equations with addition and subtraction, the first thing to do is to isolate the variable we are trying to find. In this case, “K.” This means that the 8 needs to be removed from the left hand side of the equation. Multiplying 8 by its reciprocal will give us a value of 1. Perfect!
$\Large 8 \times \dfrac{1}{8} = \dfrac{8}{8} = 1$
Now adding this information to the equation, we would end up with this:
$\Large \begin{array}{c} 8 \times \dfrac{1}{8} = \dfrac{8}{8} = 1 \\ \mathbf{\dfrac{1}{8}} \times 8 \times \text{K}= 14 \times 17 \\ \dfrac{8}{8} \times \text{K} = 14 \times 17\\ 1\times \text{K}=14\times 17 \\ \text{K}=14\times 17 \end{array}$
This would leave us with 1 × K on the left hand side of the equation which would end up being just “K” when multiplied together. This is exactly what we are looking for. We are not quite finished yet though. Now back to the golden rule.
Whatever you do to one side of the equation you must do to the other. Therefore, as we multiplied the left hand side of the equation by we need to multiply the right hand side of the equation by .
$\Large \dfrac{1}{8} \times 8 \times \text{K} = 14 \times 17 \times \dfrac{1}{8}$
So now if we followed this through we would end up with:
$\Large \begin{array}\\ \mathbf{\dfrac{1}{8}} \times 8 \times \text{K} = 14 \times 17 \times \mathbf{ \dfrac{1}{8}} \\ \dfrac{8}{8} \times \text{K} = \dfrac{238}{8} \\ \text{K} = 29.75 \end{array}$
We should do a check of the answer like we did previously to see if we are correct.
$\Large \begin{array}{rl}\text{Replace K with 29.75} & \\ \downarrow & \\ 8 \times \text{K} & = 14 \times 17 \\ 8 \times 29.75 & = 14 \times 17 \\ 238 & = 238\end{array}$
Example
We’ll try another example, but this time we’ll deal with fractions that will have to be moved around. We’ll also start to do this in steps as we have done with previous questions.
Solve for L.
$\Large \dfrac{4}{9} \times \text{L} = 12 \times 12$
Step 1: Isolate L. In this case, we have to move the 4/9 from one side to the other. In order to do this we have to multiply both sides by the reciprocal of 4/9. This will essentially remove 4/9 from the left hand side of the equation. The reciprocal is of 4/9 is 9/4.
$\Large \dfrac{4}{9} \times \dfrac{9}{4} = \dfrac{36}{36} = 1$
Therefore we get:
$\Large \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4}$
Step 2: Solve the equation
$\Large \begin{array}{c} \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4} \\ 1 \times \text{L} = 144 \times \dfrac{9}{4} \\ \text{L} = \dfrac{1296}{4} \\ \text{L}=324 \end{array}$
$\Large \begin{array}{rl}\text{Replace L with 324} & \\ \downarrow & \\ \dfrac{4}{9} \times \text{L} & = 12 \times 12 \\ \dfrac{4}{9} \times 324 & = 12 \times 12 \\ \dfrac{1296}{9} & = 144 \\ 144 & = 144 \end{array}$
Example
This one is a bit more challenging. You’ll note in the question that there are no numbers, only letters. Take a look at the question, and see if you can come up with any ideas on how you would solve this equation.
Solve for “D” in the following equation.
$\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}$
Step 1: Determine which variable you must isolate. In this case it’s given for us and it’s “D.”
$\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\textbf{D}}$
The challenge here is that “D” is in the denominator (bottom) of the fraction. If we were to isolate it but it was still in the denominator of a fraction we would not have something that we could work with.
When “D” is isolated it must not only be by itself on one side of the equation but also appear as a whole number and not as the denominator of a fraction.
Step 2: To make this process easier break the equation down to look like the following.
$\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}}$
Remember that:
$\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{A}}{\text{B}}$
We haven’t really changed anything mathematically we have just made the equation into something that is easier to work with.
Step 3: Isolate “D.” In order to do this we must remove “C” from the right hand side of the equation. Do this by multiplying both the right and left hand side by 1/C.
$\Large \mathbf{\dfrac{1}{\text{C}}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}} \times \mathbf{\dfrac{1}{\text{C}}}$
Now this might start looking a little bit complicated but if you follow it through you’ll start to see the answer form. Look at the right hand side of the equation. It now has both and C/1 and a 1/C. Multiplying those reciprocals together you get 1 and effectively take out the “C” on the right hand side.
$\Large \dfrac{\text{C}}{1} \times \dfrac{1}{\text{C}}= \dfrac{\text{C}}{\text{C}}=1$
So we end up with:
$\Large \dfrac{1}{\text{C}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= 1 \times \dfrac{1}{\text{D}}$
If we wanted to simplify this we could just do the following:
$\Large \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}}$
Step 4: Get D into the numerator of the equation. What we need to do will take a little bit of math and some patience. We need to follow the same rules that we have been following.
The idea here is to multiply each side by D/1. This will make the right hand side of the equation equal to 1 but now the left hand side of the equation will have D in the numerator.
$\Large \begin{array}{c}\dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}} \times \dfrac{\text{D}}{1} \\ \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C}\times \text{B}} = 1 \end{array}$
What you’ll notice is that it actually creates more work but we have managed to get “D” into the numerator of the equation. The only problem is that we also have a bunch of other stuff on the same side as “D” and now our job becomes getting rid of all that stuff.
Now all we have to do is follow the rules and get the A, B and C over to the right hand side and isolate D. I’ll do this simply and in one quick calculation.
$\Large \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} \times \dfrac{\text{C} \times \text{B}}{\text{A}} = \dfrac{\text{C} \times \text{B}}{\text{A}}$
What we end up with is:
$\Large \text{D}= \dfrac{\text{C} \times \text{B}}{\text{A}}$
There is a lot of math involved there but if you follow the rules and go one step at a time you’ll eventually get there. My suggestion here is that you go over what we just went through a couple times before moving on. Understanding the math involved here is quite important when it comes to transposing formulas and equations.
At this point you might be wondering if there was a shortcut and as luck would have it there is for this procedure which I’ll go through now.. We’ll start with an example.
Take the following equation:
$\Large \dfrac{3}{4}=\dfrac{3}{4}$
This equation is true. Now take each of the fractions and flip them around.
$\Large \dfrac{4}{3} = \dfrac{4}{3}$
It seems that if you flip both fractions around then the equation is still true. Mathematically you are doing the same thing to one side as you are doing to the other. There is a lot of math involved here but the main point is that if you were to do all the math you would end up with the same answer.
How we can use this to help us simplify our question is to just do the following. Remember we started with:
$\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}$
We needed to isolate D but the main problem (and the thing that causes us to do a lot of work) was that D was in the denominator and we needed it in the numerator.
Through our math wizardry we can get D into the numerator by just flipping around the left hand fraction. Whatever we do to that side we then must do to the right side. What we end up with is:
$\Large \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}}$
All we need to do now is get C out of the left hand side by multiplying it by C/1 and then doing the same thing to the left hand side.
$\Large \dfrac{\text{C}}{1} \times \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}} \times \dfrac{\text{D}}{1}$
We end up with…
$\Large \dfrac{\text{C} \times \text{B}}{\text{A}} = \text{D}$
Now to take all that in in one shot might be a little much so you may want to go back and reread that whole explanation. But always keep in mind the math that goes along with that.
Example
In this example, we are going to use the formula for the area of a circle. Calculating the area of a circle is actually something we are going to do further into this volume, but for now we are just going to deal with the formula itself.
$\Large \begin{array}{c} \text{Area of a circle} \\ \text{A} = {\text{D}}^{2} \times 0.7854 \end{array}$
$\Large \begin{array}{cl} \text{Where:} & \text{A = area of the circle} \\ & \text{D = diameter of the circle} \end{array}$
Solving for the area of the circle would be pretty straight forward but as we are dealing with transposing equations in this section what we’ll do is solve for the diameter (D).
Step 1: Identify the variable you are trying to solve for.
$\Large \text{A} = \mathbf{{\text{D}}^{2}} \times 0.7854$
From this you can see we have a couple problems we have to deal with. The first one is that we have to isolate D. The second one is that D has the exponent 2 attached to it so we have to somehow eliminate that.
Step 2: Isolate D.
For this we have to move the 0.7854 from the right hand side of the equation to the left hand side. We simply follow the rules we have used up to this point.
$\Large \begin{array}{c} \mathbf{\dfrac{1}{0.7854}} \times \text{A} = {\text{D}}^{2} \times 0.7854 \times \mathbf{\dfrac{1}{0.7854}} \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times \dfrac{0.7854}{0.7854} \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times 1 \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \end{array}$
That takes care of the first problem. Now we have to tackle the D2 issue.
Step 3: Remove the exponent from D.
Once again we go back to our original rule. Whatever you do to one side you must to the same thing to the other side.
We have to refer to the video that you watched earlier in this section regarding exponents and square roots. We’ll do a quick refresher before we solve for D.
Remember that:
$\Large \begin{array}{c} {\text{D}}^{2} = \text{D} \times \text{D} \\ \text{and... } \\ \sqrt{\text{D} \times \text{D}} = \text{D} \end{array}$
So what we see is that if you square root a number that is squared (has an exponent of 2) you end up with the number itself. I’ll quickly show you with numbers.
$\Large \begin{array}{c} {4}^{2} = 4 \times 4 \\ {4}^{2} = 16 \\ \sqrt{16} = \sqrt{4 \times 4} \\ \sqrt{16} = 4 \end{array}$
Having gone through all that we can now solve the problem.
$\Large \begin{array}{c} \sqrt{\dfrac{\text{A}}{0.7854}}= \sqrt{{\text{D}}^{2}} \\ \sqrt{\dfrac{\text{A}}{0.7854}} = \text{D} \end{array}$
# Practice Questions
Now it’s time to do a couple of practice questions for yourself. Make sure to check the videos answer to see how you did.
Question 1
Solve for B.
$\Large \dfrac{\text{A} \times \text{B}}{\text{C}}=\dfrac{\text{D} \times \text{E}}{\text{F}}$
Question 2
Solve for C.
$\Large \dfrac{\text{A} \times \text{B}}{\text{C}}=\dfrac{\text{D} \times \text{E}}{\text{F}}$ |
In geometry, a quadrant is one of the four sections of a rectangular coordinate plane. The four quadrants make up the area contained by the x- and y-axes and are labeled I through IV, starting in the upper right quadrant and going counterclockwise, as shown in the figure below.
A quadrant is one of the four infinite sections defined by the x- and y-axes. Together, the 4 quadrants make up the coordinate plane.
• Quadrant I - the first quadrant is in the upper right hand corner of the coordinate plane. The x- and y-values in quadrant I are all positive (+).
• Quadrant II - the second quadrant is in the upper left hand corner of the coordinate plane. The x-values in quadrant II are negative (-) and the y-values are positive (+).
• Quadrant III - the third quadrant is in the bottom left hand corner of the coordinate plane. The x- and y-values in quadrant III are all negative (-).
• Quadrant IV - the fourth quadrant is in the bottom right hand corner of the coordinate plane. The x-values in quadrant IV are positive (+) and the y-values are negative (-).
## Signs of x and y values in a quadrant
The signs of coordinates of any point plotted in a particular quadrant are the same. A point cannot lie in both a quadrant and on an axis. Points that lie on either axis have one coordinate that is 0, which is neither positive nor negative. The diagram below shows the signs of coordinates in different quadrants.
Example:
Determine which quadrant each of the following points are located in. A: (2, 2); B: (-3, 1); C: (-1, -2); D: (3, -4).
Point A is in quadrant I since its coordinates are (+, +).
Point B is in quadrant II since its coordinates are (-, +).
Points C is in quadrant III since its coordinates are (-, -).
Point D is in quadrant IV since its coordinates are (+, -).
In the Cartesian coordinate system, points are defined by the x- and y-axis and are represented by an ordered pair, (x, y). The sign of x and y determine which quadrant the point will lie in. The value of x and y determines the number of units along the x- or y-axes from which the given point lies. Also, note that the origin does not lie in any quadrant since it is located at (0, 0). Refer to the table below, which summarizes the sign of the x- and y-coordinates in each quadrant:
I + +
II - +
III - -
IV + -
As an example, given the point (-2, 1), we know that the point lies in quadrant II, so we count 2 units left on the x-axis and 1 unit up to locate the point in the coordinate plane.
## Sign of trigonometric functions
The signs of the three basic trigonometric functions, sin, cos, and tan, vary based on which quadrant they are in. The sign of each trigonometric function in each quadrant can be determined using the signs of the coordinates along with basic trigonometric relationships. The diagram below shows the signs of these functions in different quadrants.
### Did you know?
The first part of the word "quadrant" is from a Latin root meaning four.
A popular name for an all-terrain vehicle (ATV) is "quad," named for its four large tires. |
## Simple Interest Calculator
Simple Interest <a href="https://studysaga.in/calculator-studysaga/">Calculator</a>
## Simple Interest Calculator
When you borrow money from someone or a bank, you pay interest on the principal amount. Similarly, when you lend money to someone or a bank, you earn interest on the principal amount. Simple interest is the type of interest that is charged on the principal amount borrowed or lent, and it is calculated based on the principal amount, interest rate, and the time period of borrowing or lending.
Formula: The formula for simple interest is straightforward, and it is given as:
Simple Interest = (P * R * T) / 100
where, P = Principal amount R = Rate of interest T = Time period
Examples:
1. Suppose you borrow \$10,000 at an interest rate of 5% per annum for a period of 3 years. The simple interest on this loan would be:
Simple Interest = (10,000 * 5 * 3) / 100 = \$1,500
Therefore, you would have to pay \$1,500 as interest on this loan over a period of 3 years.
1. Let’s say you lend \$15,000 to someone at an interest rate of 8% per annum for a period of 2 years. The simple interest earned on this loan would be:
Simple Interest = (15,000 * 8 * 2) / 100 = \$2,400
Therefore, you would earn \$2,400 as interest on this loan over a period of 2 years.
1. Simple interest is easy to calculate, and the formula is simple.
2. It is useful in calculating interest on short-term loans.
3. It is less complicated than compound interest.
1. Simple interest does not take into account the effect of inflation.
2. It does not consider the compounding effect, which means you do not earn interest on interest.
3. It is not suitable for long-term investments as the interest earned may not be substantial.
Here are some examples of simple interest problems with solutions:
Example 1: Calculate the simple interest on a principal of \$5,000 at a rate of 4% per annum for 3 years.
Solution: Given, Principal (P) = \$5,000, Rate (R) = 4% per annum, Time (T) = 3 years Simple Interest (SI) = (P x R x T) / 100 = (5000 x 4 x 3) / 100 = \$600 Therefore, the simple interest on \$5,000 at a rate of 4% per annum for 3 years is \$600.
Example 2: Find the amount obtained on a principal of \$10,000 at a rate of 6% per annum for 2 years, when the interest is compounded annually.
Solution: Given, Principal (P) = \$10,000, Rate (R) = 6% per annum, Time (T) = 2 years, Compounding Period (n) = 1 Amount (A) = P(1 + R/n)^(nT) = 10000(1 + 0.06/1)^(1 x 2) = \$11,236 Therefore, the amount obtained on a principal of \$10,000 at a rate of 6% per annum for 2 years, when the interest is compounded annually, is \$11,236.
Example 3: If \$2,000 is invested for 5 years at a simple interest rate of 8% per annum, what will be the total amount at the end of the period?
Solution: Given, Principal (P) = \$2,000, Rate (R) = 8% per annum, Time (T) = 5 years Simple Interest (SI) = (P x R x T) / 100 = (2000 x 8 x 5) / 100 = \$800 Amount (A) = P + SI = \$2,000 + \$800 = \$2,800 Therefore, the total amount at the end of 5 years on an investment of \$2,000 at a simple interest rate of 8% per annum is \$2,800.
Example 4: If \$5,000 is borrowed at a simple interest rate of 12% per annum for 3 years, what will be the total amount to be paid back at the end of the period?
Solution: Given, Principal (P) = \$5,000, Rate (R) = 12% per annum, Time (T) = 3 years Simple Interest (SI) = (P x R x T) / 100 = (5000 x 12 x 3) / 100 = \$1,800 Total amount to be paid back (A) = P + SI = \$5,000 + \$1,800 = \$6,800 Therefore, the total amount to be paid back at the end of 3 years on a loan of \$5,000 at a simple interest rate of 12% per annum is \$6,800.
Example 5: Suppose you borrow \$2,000 from a friend and agree to pay back the loan with 5% simple interest after 3 years. How much interest will you have to pay, and what will be the total amount you owe at the end of the loan term?
Solution: The formula for calculating simple interest is:
I = P * r * t
where I is the interest, P is the principal (or amount borrowed), r is the interest rate, and t is the time period in years.
Plugging in the given values, we get:
I = 2000 * 0.05 * 3 = \$300
Therefore, the interest you will have to pay is \$300.
To calculate the total amount owed at the end of the loan term, we add the interest to the principal:
Total amount owed = Principal + Interest = \$2000 + \$300 = \$2300
Therefore, at the end of the 3-year loan term, you will owe your friend a total of \$2300.
Simple interest is a useful tool for calculating interest on short-term loans, and it is easy to calculate. However, it may not be the best option for long-term investments or loans as it does not take into account the effect of inflation or compounding. It is essential to consider all the factors before choosing the type of interest for your investment or loan. |
# SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE BECAUSE GRAPHING IS SOMETIMES INACCURATE, ALGEBRA CAN BE USED TO FIND EXACT SOLUTIONS. ONE OF THOSE.
## Presentation on theme: "SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE BECAUSE GRAPHING IS SOMETIMES INACCURATE, ALGEBRA CAN BE USED TO FIND EXACT SOLUTIONS. ONE OF THOSE."— Presentation transcript:
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE BECAUSE GRAPHING IS SOMETIMES INACCURATE, ALGEBRA CAN BE USED TO FIND EXACT SOLUTIONS. ONE OF THOSE ALGEBRAIC METHODS IS “COMPLETING THE SQUARE”
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE Let’s solve x 2 – 10x + 18 = 0 Step 1: Get rid of constant on the left side x 2 – 10x + 18 = 0 -18 x 2 – 10x = -18
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE Let’s solve x 2 – 10x + 18 = 0 Step 2: Add constant to left side to create PST Half of middle term, then square it. x 2 – 10x = -18 + 25 Must add it to BOTH sides. + 25 (x – 5) 2 = 7
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE Let’s solve x 2 – 10x + 18 = 0 Step 3: Square root of both sides. (x – 5) 2 = 7
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE Let’s solve x 2 – 10x + 18 = 0 Step 4: Solve left side for x (x – 5) 2 = 7 +5
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE Try this one x 2 + 6x – 3 = 0 (x + 3) 2 = 12 -3 x 2 + 6x = 3 +3 Half of 6, squared +9 x 2 + 6x + 9 = 12
COMPLETE THE SQUARE x 2 + 12x + _____ = 3 + _____ 36
COMPLETE THE SQUARE x 2 – 8x + _____ = 10 + _____ 16
COMPLETE THE SQUARE x 2 – 20x + _____ = 1 + _____ 100
SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA Standard form for quadratic equations is ax 2 + bx + c = 0 and can be solved using the Quadratic Formula:
SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA Example:3x 2 + 7x – 2 = 0
THE DISCRIMINANT In a quadratic formula, the discriminant is the expression under the racical sign. What is the discriminant for 4x 2 + 2x – 7 = 0 ? b 2 – 4ac = 2 2 – 4(4)(-7)= 4 + 112 = 116
THE DISCRIMINANT The discriminant tells you something about the roots of the equation. If the discriminant is negative, (b 2 – 4ac < 0), then there are no real roots (no solutions). If the discriminant is zero, (b 2 – 4ac = 0), then there is a double root (one solution). If the discriminant is positive, (b 2 – 4ac > 0), then there are two real roots.
FLASH CARDS In the equation, x 2 + 5x – 6 = 0 a = 1
FLASH CARDS In the equation, x 2 + 5x – 6 = 0 b = 5
FLASH CARDS In the equation, x 2 + 5x – 6 = 0 c = -6
FLASH CARDS In the equation, x 2 + 5x – 6 = 0 the discriminant = 49
FLASH CARDS In the equation, 3x 2 – 6 = 0 the discriminant = 72
FLASH CARDS How many roots if the discriminant is equal to 120 Two real roots
FLASH CARDS How many roots if the discriminant is equal to 0 A double root
FLASH CARDS How many roots if the discriminant is equal to 13 Two real roots
FLASH CARDS How many roots if the discriminant is equal to -15 No real roots
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# Operations with negative and positive numbers
Absolute value (modulus) of a number. Addition.
Subtraction. Multiplication. Division.
Rules of signs at multiplication and division.
Absolute value (modulus): for a negative number this is a positive number, received by changing the sign " – " by " + "; for a positive number and zero this is the number itself. The designation of an absolute value (modulus) of a number is the two straight brackets insideof which the number is written.
E x a m p l e s :
| – 5 | = 5, | 7 | = 7, | 0 | = 0.
Addition: 1) at addition of two numbers of the same sign their absolute values are added and before the sum their common sign is written. E x a m p l e s : ( + 6 ) + ( + 5 ) = 11 ; ( – 6 ) + ( – 5 ) = – 11 ; 2) at addition of two numbers with different signs their absolute values are subtracted(the smaller from the greater) and a sign of a number, having a greater absolute value is chosen. E x a m p l e s : ( – 6 ) + ( + 9 ) = 3 ; ( – 6 ) + ( + 3 ) = – 3 .
Subtraction: it is possible to change subtraction of two numbers by addition, thereat a minuend saves its sign, and a subtrahend is taken with the back sign.
E x a m p l e s :
( + 8 ) – ( + 5 ) = ( + 8 ) + ( – 5 ) = 3;
( + 8 ) – ( – 5 ) = ( + 8 ) + ( + 5 ) = 13;
( – 8 ) – ( – 5 ) = ( – 8 ) + ( + 5 ) = – 3;
( – 8 ) – ( + 5 ) = ( – 8 ) + ( – 5 ) = – 13.
Multiplication: at multiplication of two numbers their absolute values are multiplied, and a product has the sign " + ", if signs of factors are the same, and " – ", if the signs are different. The next scheme ( a rule of signs at multiplication ) is useful:
+ · + = +
+ · – = –
– · + = –
– · – = +
At multiplication of some factors (two and more ) a product has the sign " + ", if a number of negative factors is even, and the sign " – ", if this number is odd.
E x a m p l e :
Division: at division of two numbers the first absolutevalue is divided by the second and a quotient has the sign " + ", if signs of dividend and divisor are the same, and " – ", if they are different. The same rule of signs as at multiplication acts:
+ : + = +
+ : – = –
– : + = –
– : – = +
E x a m p l e :
( – 12 ) : ( + 4 ) = – 3 . |
# 2019 AIME I Problems/Problem 6
## Problem 6
In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$.
## Solution 1 (Trig)
Let $\angle MKN=\alpha$ and $\angle LNK=\beta$. Note $\angle KLP=\beta$.
Then, $KP=28\sin\beta=8\cos\alpha$. Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$.
Dividing the equations gives $$\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}$$
Thus, $MK=\frac{MN}{\tan\alpha}=98$, so $MO=MK-KO=\boxed{090}$.
## Solution 2 (Similar triangles)
$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$
First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similarities we see that $$\frac{KP}{KL} = \frac{KL}{KN}$$ $$KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}$$ and $$\frac{KP}{KO} = \frac{KM}{KN}$$ $$KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}$$ Combining the two equations, we get $$\frac{8\cdot KM}{KN} = \frac{784}{KN}$$ $$8 \cdot KM = 28^2$$ $$KM = 98$$ Since $KM = KO + MO$, we get $MO = 98 -8 = \boxed{090}$.
## Solution 3 (Similar triangles, orthocenters)
Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$. Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\triangle KNP$).
As $\triangle KOL \sim \triangle KHP$ (as $LO \parallel PH$, using the fact that $H$ is the orthocenter), we may let $OH = 8k$ and $LP = 28k$.
Then using similarity with triangles $\triangle KLH$ and $\triangle KMP$ we have
$$\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}$$
Cross-multiplying and dividing by $4+4k$ gives $2(8+8k+HM) = 28 \cdot 7 = 196$ so $MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}$. (Solution by scrabbler94)
## Solution 4 (Algebraic Bashing)
First, let $P$ be the intersection of $LO$ and $KN$. We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are $$4225+d^2=c^2,$$ $$4225+d^2+16d+64=a^2+2ab+b^2,$$ $$a^2+e^2=c^2,$$ $$b^2+e^2=64,$$ $$b^2+e^2+2ef+f^2=784,$$ $$a^2+e^2+2ef+f^2=g^2,$$ and $$g^2+784=a^2+2ab+b^2.$$ We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$. Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator)
## Solution 5 (5-second PoP)
$[asy] size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("K", K, SW); dot("L", L, unit(L)); dot("M", M, unit(M)); dot("N", NN, SE); dot("X", X, S); [/asy]$ Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, $$KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,$$and $MO=KM-KO=\boxed{090}$.
(Solution by TheUltimate123)
## Solution 6 (Alternative PoP)
$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$
Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KO=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have
$$(KO)(OM)=(LO)(OQ)$$ ($Q$ is the intersection of $OL$ with the circle $\neq L$)
$$8(MO)=(\sqrt{720+x^2}+x)(\sqrt{720+x^2}-x)$$
$$8(MO)=720$$
$$MO=\boxed{090}$$.
(Solution by RootThreeOverTwo) |
## Is rate of change x or y
Mar 30, 2016 If f(x) is a function defined on an interval [a,a+h], then the amount of change of f(x) over the interval is the change in the y values of the function Feb 23, 2012 Demonstrate an understanding of the instantaneous rate of change. A car speeding down \begin{align*}y - y_0 = m_{tan}(x - x_0).\end{align*}.
Mar 30, 2016 If f(x) is a function defined on an interval [a,a+h], then the amount of change of f(x) over the interval is the change in the y values of the function Feb 23, 2012 Demonstrate an understanding of the instantaneous rate of change. A car speeding down \begin{align*}y - y_0 = m_{tan}(x - x_0).\end{align*}. Sep 23, 2007 x y . Again, this is the slope of a secant, except this time it has negative slope. Here's the formal definition: the average rate of change of f(x) on rate of change = change in y change in x = change in distance change in time = 160 − 80 4 − 2 = 80 2 = 40 1 The rate of change is 40 1 or 40 . This means a vehicle is traveling at a rate of 40 miles per hour.
## The Rate of Change of a function is how much the y-value changes as the x- value changes by some amount. For a linear function, we would call that "slope".
Solution for Suppose the constant rate of change y with respect to x is -2.5 and we know that y=-6.75 when x=1.5 what is the value of y when x=-1 ? In the following picture, we marked two points to help you better understand how to find the average rate of change. The average rate of change formula is: A = [f(x is a function with 2 inputs and 1 output, what does "rate of change" mean? The graph of $z = f(x, y)$ is a surface in 3 dimensions. Imagine standing somewhere on Mar 21, 2016 What if you pick two other points? Did you notice in a linear function the slope or rate of change of y was always the same when the x-values Mar 30, 2016 If f(x) is a function defined on an interval [a,a+h], then the amount of change of f(x) over the interval is the change in the y values of the function Feb 23, 2012 Demonstrate an understanding of the instantaneous rate of change. A car speeding down \begin{align*}y - y_0 = m_{tan}(x - x_0).\end{align*}.
### rate of change = change in y change in x = change in distance change in time = 160 − 80 4 − 2 = 80 2 = 40 1 The rate of change is 40 1 or 40 . This means a vehicle is traveling at a rate of 40 miles per hour.
The rate of change of y with respect to x, if one has the original function, can be found by taking the derivative of that function. This will measure the rate of change at a specific point. However, if one wishes to find the average rate of change over an interval, Q. A climber is on a hike. After 2 hours he is at an altitude of 400 feet. After 6 hours, he is at an altitude of 700 feet. What is the average rate of change? You are correct, the expression "the rate of change of y with respect to x" does mean how fast y is changing in comparison to x. In your example of velocity, if y is the distance travelled in miles, and x is the time taken in hours then y/x is the average velocity in miles per hour. Velocity is the rate of change of distance with respect to time.
### Definition 1 is a special case of the following general definition of average rate of change. Definition 2 The average rate of change of y = f(x) with respect to x
Rate of Change. In the examples above the slope of line corresponds to the rate of change. e.g. in an x-y graph, a slope of 2 means that y increases by 2 for every increase of 1 in x. The examples below show how the slope shows the rate of change using real-life examples in place of just numbers. So in linear functions, the slope or rate of change of y with respect to x is always the same. Since we are used to writing straight lines in slope intercept form ( y=mx+b ), the slope will always be the m value in the form. This is because x is always multiplied by the m . When a function is not linear, This is called the rate of change per month. By finding the slope of the line, we would be calculating the rate of change. We can't count the rise over the run like we did in the calculating slope lesson because our units on the x and y axis are not the same. The rate of change of y with respect to x, if one has the original function, can be found by taking the derivative of that function. This will measure the rate of change at a specific point. However, if one wishes to find the average rate of change over an interval,
## Let us Find a Derivative! To find the derivative of a function y = f(x) we use the slope formula: Slope = Change in Y Change in X = ΔyΔx. slope delta x and delta y.
The rate of change in y from period t to t + 1 is given by yt"$yt yt. (1) &xy φ 2x + 2y. 2. From the definition of growth rate, we have. 1 + ' x y! φ xt"$ yt"$!/ xt. Solution for Suppose the constant rate of change y with respect to x is -2.5 and we know that y=-6.75 when x=1.5 what is the value of y when x=-1 ? In the following picture, we marked two points to help you better understand how to find the average rate of change. The average rate of change formula is: A = [f(x is a function with 2 inputs and 1 output, what does "rate of change" mean? The graph of$z = f(x, y)\$ is a surface in 3 dimensions. Imagine standing somewhere on Mar 21, 2016 What if you pick two other points? Did you notice in a linear function the slope or rate of change of y was always the same when the x-values
So in linear functions, the slope or rate of change of y with respect to x is always the same. Since we are used to writing straight lines in slope intercept form ( y=mx+b ), the slope will always be the m value in the form. This is because x is always multiplied by the m . When a function is not linear, This is called the rate of change per month. By finding the slope of the line, we would be calculating the rate of change. We can't count the rise over the run like we did in the calculating slope lesson because our units on the x and y axis are not the same. The rate of change of y with respect to x, if one has the original function, can be found by taking the derivative of that function. This will measure the rate of change at a specific point. However, if one wishes to find the average rate of change over an interval, Q. A climber is on a hike. After 2 hours he is at an altitude of 400 feet. After 6 hours, he is at an altitude of 700 feet. What is the average rate of change? You are correct, the expression "the rate of change of y with respect to x" does mean how fast y is changing in comparison to x. In your example of velocity, if y is the distance travelled in miles, and x is the time taken in hours then y/x is the average velocity in miles per hour. Velocity is the rate of change of distance with respect to time. |
# Solving equations with square roots
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We will also provide some tips for Solving equations with square roots quickly and efficiently Solving natural log equations requires algebraic skills as well as a strong understanding of exponential growth and decay. The key is to remember that the natural log function is the inverse of the exponential function. This means that if you have an equation that can be written in exponential form, you can solve it by taking the natural log of both sides. For example, suppose you want to solve for x in the equation 3^x = 9. Taking the natural log of both sides gives us: ln(3^x) = ln(9). Since ln(a^b) = b*ln(a), this reduces to x*ln(3) = ln(9). Solving for x, we get x = ln(9)/ln(3), or about 1.62. Natural log equations can be tricky, but with a little practice, you'll be able to solve them like a pro!
Solving for exponents can be a tricky business, but there are a few basic rules that can help to make the process a bit easier. First, it is important to remember that any number raised to the power of zero is equal to one. This means that when solving for an exponent, you can simply ignore anyterms that have a zero exponent. For example, if you are solving for x in the equation x^5 = 25, you can rewrite the equation as x^5 = 5^3. Next, remember that any number raised to the power of one is equal to itself. So, in the same equation, you could also rewrite it as x^5 = 5^5. Finally, when solving for an exponent, it is often helpful to use logs. For instance, if you are trying to find x in the equation 2^x = 8, you can take the log of both sides to get Log2(8) = x. By using these simple rules, solving for exponents can be a breeze.
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How to solve logarithmic functions has been a mystery for many students. The concept seems difficult, but it is not as hard as it looks. There are three steps in solving logarithmic functions. First, identify the base of the logarithm. Second, use properties of logs to rewrite the equation. Third, solve for the unknown using basic algebra. These steps may seem confusing at first, but with practice they will become easy. With a little effort, anyone can learn how to solve logarithmic functions.
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# RD Sharma Solutions for Class 10 Maths Chapter 13 Probability
### RD Sharma Class 10 Chapter 13 Exercise 13.1 Page No: 13.20
1. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
Solution:
Given: Probability that it will rain tomorrow P(E) = 0.85
Required to find: Probability that it will not rain tomorrow P(E)
We know that sum of the probability of occurrence of an event and the probability of non-occurrence of an event is 1.
Therefore, the probability that it will not rain tomorrow is = 0.15
2. A die is thrown. Find the probability of getting:
(i) a prime number (ii) 2 or 4
(iii) a multiple of 2 or 3 (iv) an even prime number
(v) a number greater than 5 (vi) a number lying between 2 and 6
Solution:
Given: A dice is thrown once
Required to find:
(i) Probability of getting a prime number
(ii) Probability of getting 2 or 4
(iii) Probability of getting a multiple of 2 or 3.
(iv) Probability of getting an even number
(v) Probability of getting a number greater than five.
(vi) Probability of lying between 2 and 6
Total number on a dice is 6 i.e., 1, 2, 3, 4, 5 and 6.
(i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, probability of getting a prime number = 3/6 = 1/2
(ii) For getting 2 and 4, clearly the number of favourable outcomes is 2.
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting 2 or 4 = 2/6 = 1/3
(iii) Multiple of 2 are 3 are 2, 3, 4 and 6.
So, the number of favourable outcomes is 4
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting an multiple of 2 or 3 = 4/6 = 2/3
(iv) An even prime number is 2 only.
So, the number of favourable outcomes is 1.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting an even prime number = 1/6
(v) A number greater than 5 is 6 only.
So, the number of favourable outcomes is 1.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a number greater than 5 = 1/6
(vi) Total number on a dice is 6.
Numbers lying between 2 and 6 are 3, 4 and 5
So, the total number of numbers lying between 2 and 6 is 3.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a number lying between 2 and 6 = 3/6 =1/2
3. Three coins are tossed together. Find the probability of getting:
(iii) at least one head and one tail (iv) no tails
Solution:
Given: Three coins are tossed simultaneously.
When three coins are tossed then the outcome will be anyone of these combinations.
TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.
So, the total number of outcomes is 8.
(i) For exactly two heads, the favourable outcome are THH, HHT, HTH
So, the total number of favourable outcomes is 3.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting exactly two heads is 3/8
(ii) For getting at least two heads the favourable outcomes are HHT, HTH, HHH, and THH
So, the total number of favourable outcomes is 4.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting at least two heads when three coins are tossed simultaneously = 4/8 = 1/2
(iii) For getting at least one head and one tail the cases are THT, TTH, THH, HTT, HHT, and HTH.
So, the total number of favourable outcomes i.e. at least one tail and one head is 6
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting at least one head and one tail = 6/8 = 3/4
(iv) For getting an outcome of no tail, the only possibility is HHH.
So, the total number of favourable outcomes is 1.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting no tails is 1/8.
4. A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number
Solution:
Given A pair of dice is thrown
Required to find: Probability that the total of numbers on the dice is greater than 9
First let’s write the all possible events that can occur
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),
It’s seen that the total number of events is 62 = 36
Favourable events i.e. getting the total of numbers on the dice greater than 9 are
(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6).
So, the total number of favourable events i.e. getting the total of numbers on the dice greater than 9 is 6
We know that, Probability = = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting the total of numbers on the dice greater than 9 = 6/36 = 1/6
5. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.
Solution:
Given A pair of dice is thrown
Required to find: Probability that the total of numbers on the dice is greater than 10
First let’s write the all possible events that can occur
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),
It’s seen that the total number of events is 62 = 36
Favourable events i.e. getting the total of numbers on the dice greater than 10 are (5, 6), (6, 5) and (6, 6).
So, the total number of favourable events i.e. getting the total of numbers on the dice greater than 10 is 3.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting the total of numbers on the dice greater than 10 = 3/36 = 1/12
6. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vii) neither an ace nor a king
(viii) neither a red card nor a queen
(ix) the seven of clubs
(x) a ten
(xii) a black card
(xiii) a seven of clubs
(xiv) jack
(xvi) a queen
(xvii) a heart
(xviii) a red card
Solution:
Given: A card is drawn at random from a pack of 52 cards
Required to Find: Probability of the following
Total number of cards in a pack = 52
(i) Number of cards which are black king = 2
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a black king = 2/52 = 1/26
(ii) Total number of black cards is (13 + 13) 26
Total number of kings are 4 in which 2 black kings are also included.
So, the total number of black cards or king will be 26+2 = 28
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a black cards or a king = 28/52 = 7/13
(iii) Total number of cards which are black and a king cards is 2
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a black cards and a king is 2/52 = 1/26
(iv) A jack, queen or a king are 3 from each 4 suits.
So, the total number of a jack, queen and king are 12.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a jack, queen or a king is 12/52 = 3/13
(v) Total number of heart cards are 13 and king are 4 in which king of heart is also included.
So, the total number of cards that are a heart and a king = 13 + 3 = 16
Hence, the total number of cards that are neither a heart nor a king = 52 – 16 = 36
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting cards neither a heart nor a king = 36/52 = 9/13
(vi) Total number of spade cards is 13
Total number of aces are 4 in which ace of spade is included in the number of spade cards.
Hence, the total number of card which are spade or ace = 13 + 3 = 16
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting cards that is spade or an ace = 16/52 = 4/13
(vii) Total number of ace cards are 4 and king are 4
Total number of cards that are an ace or a king = 4 + 4 = 8
So, the total number of cards that are neither an ace nor a king is 52 – 8 = 44
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting cards which are neither an ace nor a king = 44/52 = 11/13
(viii) It’s know that the total number of red cards is 26.
Total number of queens are 4 in which 2 red queens are also included
Hence, total number of red cards or queen will be 26 + 2 = 28
So, the total number of cards that are neither a red nor a queen= 52 -28 = 24
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting neither a red card nor a queen = 24/52 = 6/13
(ix) Total number of card other than ace is 52 – 4 = 48
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting other than ace = 48/52 = 12/13
(x) Total number of tens in the pack of cards is 4.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a ten = 4/52 = 1/13
(xi) Total number of spade is 13
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a spade = 13/52 = 1/4
(xii) Total number of black cards in the pack is 26
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting black cards is 26/52 = 1/2
(xiii) Total number of 7 of club is 1 only.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a 7 of club = 1/52
(xiv) Total number of jacks are 4
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a jack = 4/52 = 1/13
(xv) Total number of ace of spade is 1
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting an ace of spade = 1/52
(xvi) Total number of queens is 4
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a queen = 4/52 = 1/13
(xvii) Total number of heart cards is 13
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a heart card = 13/52 = 1/4
(xviii) Total number of red cards is 26
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a red card = 26/52 = 1/2
(xix) Total number of kings and queen is 4 + 4 = 8
So, the total number of cards that are neither a king nor a queen is 52 – 8 = 44
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting card which is neither an queen nor a king = 44/52 = 11/13
7. In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.
Solution:
Given: Tickets are marked numbers from 1 to 50. And, one ticket is drawn at random.
Required to find: Probability of getting a prime number on the drawn ticket
Total number of tickets is 50.
Tickets which are number as prime number are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
Total number of tickets marked as prime is 15.
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a prime number on the ticket = 15/50 = 3/10
8. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:
Given: A bag contains 10 red and 8 white balls
Required to find: Probability that one ball is drawn at random and getting a white ball
Total number of balls 10 + 8 = 18
Total number of white balls is 8
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a white ball from the urn is 8/18 = 4/9
9. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) white? (ii) red?
(iii) black? (iv) not red
Solution:
Given: A bag contains 3 red, 5 black and 4 white balls
Required to find: Probability of getting a
(i) White ball
(ii) Red ball
(iii) Black ball
(iv) Not red ball
Total number of balls 3 + 5 + 4 =12
(i) Total number of white balls is 4
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a white ball = 4/12 = 1/3
(ii) Total number red balls are 3
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a red ball = 3/12 = 1/4
(iii) Total number of black balls is 5
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a black ball = 5/12
(iv) Total number of balls which are not red are 4 white balls and 5 black balls i.e. 4 + 5 = 9
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting no red ball = 9/12 = 3/4
10. What is the probability that a number selected from the numbers 1, 2, 3, …, 15 is a multiple of 4?
Solution:
Given: Numbers are from 1 to 15. One number is selected
Required to find: Probability that the selected number is a multiple of 4
Total number between from 1 to 15 to 15
Numbers that are multiple of 4 are 4, 8 and 12.
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of selecting a number which a multiple of 4 is 3/15 = 1/5
11. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that the ball drawn is not black?
Solution:
Given: A bag contains 6 red, 8 black and 4 white balls and a ball is drawn at random
Required to find: Probability that the ball drawn is not black
Total number of balls 6 + 8 + 4 = 18
Total number of black balls is 8
So, the total number of balls which are not black is 18 – 8 = 10
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a ball which is not black = 10/18 = 5/9
12. A bag contains 5 white balls and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?
Solution:
Given: A bag contains 7 red and 5 white balls and a ball is drawn at random
Required to find: Probability that the ball drawn is white
Total number of balls 7 + 5 = 12
Total number of white balls is 5
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a white ball = 5/12
13. Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?
Solution:
Given: Tickets are marked from 1 to 20 are mixed up. One ticket is picked at random.
Required to find: Probability that the ticket bears a multiple of 3 or 7
Total number of cards is 20.
And, the cards marked which is multiple of 3 or 7 are 3, 6, 7, 9, 12, 14, 15 and 18.
So, the total number of cards marked multiple of 3 or 7 is 8.
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a card that is a multiple of 3 or 7 is 8/20 = 2/5
14. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
Solution:
Given: In a lottery there are 10 prizes and 25 blanks.
Required to find: Probability of winning a prize
Total number of tickets is 10 + 25 = 35
Total number of prize carrying tickets is 10
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of winning a prize = 10/35 = 2/7
15. If the probability of winning a game is 0.3, what is the probability of losing it?
Solution:
Given: probability of winning a game P(E) = 0.3
To Find: Probability of losing the game
We know that the sum of probability of occurrence of an event and probability of non-occurrence of an event is 1.
So,
Thus, the probability of losing the game is
16. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) red (ii) black or white (iii) not black
Solution:
Given: A bag contains 7 red, 5 black and 3 white balls and a ball is drawn at random
Required to find: Probability of getting a
(i) Red ball
(ii) Black or white ball
(iii) Not black ball
Total number of balls 7 + 5 + 3 = 15
(i) Total number red balls is 7
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a red ball = 7/15
(ii) Total number of black or white balls is 5 + 3 = 8
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing white or black ball = 8/15
(iii) Total number of black balls is 5
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing black ball P(E) = 5/15 = 1/3
But, this is not what is required.
And, we know that sum of probability of occurrence of an event and probability of non-occurrence of an event is 1
Thus, the probability of drawing a card that is not black is 2/3
17. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) White (ii) Red
(iii) Not black (iv) Red or White
Solution:
Given: A bag contains 4 red, 5 black and 6white balls and a ball is drawn at random
Required to Find: Probability of getting a
(i) white ball
(ii) red ball
(iii) not black ball
(iv) red or white
Total number of balls 4 + 5 + 6 = 15
(i) Total number of white balls in the bag is 6
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing white a ball = 6/15 = 2/5
(ii) Total number of red balls in the bag is 4
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing red a ball = 4/15
(iii) Total number of black balls are 5
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing black ball P(E) = 5/15 = 1/3
We know that sum of probability of occurrence of an event and probability of non-occurrence of an event is 1.
Thus, the probability of drawing a ball that is not black is 2/3
(iv) Total number of red or white balls 4 + 6 = 10
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a white or red ball = 10/15 = 2/3
18. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) A king of red suit (ii) A face card (iii) A red face card
(iv) A queen of black suit (v) A jack of hearts (vi) A spade
Solution:
Given: One card is drawn from a well shuffled deck of 52 playing cards
Required to find: Probability of following
Total number of cards is 52
(i) Total number of cards which are king of red suit is 2
Number of favourable outcomes i.e. Total number of cards which are king of red suit is 2
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting cards which is a king of red suit = 2/52 = 1/26
(ii) Total number of face cards are 12
Number of favourable outcomes i.e. total number of face cards is 12
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a face cards = 12/52 = 3/13
(iii) Total number of red face cards are 6
Number of favourable outcomes i.e. total number of red face cards is 6
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a red face cards = 6/52 = 3/26
(iv) Total number of queen of black suit cards is 2
Total number of favourable outcomes i.e. total number of queen of black suit cards is 2
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting cards which is a queen of black suit = 2/52 = 1/26
(v) Total number of jack of hearts is 1
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a card which is a jack of hearts = 1/52
(vi) Total number of spade cards are 13
Total numbers of favourable outcomes i.e. total number spade cards is 13
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a spade card = 13/52 = 1/4
19. Five cards – ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.
(i) What is the probability that the card is a queen?
(ii) If a king is drawn first and put aside, what is the probability that the second card picked up is the (a) ace? (b) king?
Solution:
Given: Five cards-ten, jack, queen, king and Ace of diamond are shuffled face downwards.
Required to find: Probability of following
Total number of cards is 5
(i) Total number of cards which is a queen is 1
Number of favourable outcomes i.e. Total number of cards which is queen = 1
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting cards which is a queen = 1/5
(ii) If a king is drawn first and put aside then
Total number of cards becomes 4
(a) Number of favourable outcomes i.e. Total number of ace card is 1
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting an ace card = 1/4
(b) Number of favourable outcomes i.e. Total number of king cards is 0
We know that Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a king = 0
20. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) Red (ii) Back
Solution:
Given: A bag contains 3 red, and 5 black balls. A ball is drawn at random
Required to find: Probability of getting a
(i) red ball
(ii) white ball
Total number of balls 3 + 5 = 8
(i) Total number red balls are 3
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a red ball = 3/8
(ii) Total number of black ball are 5
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of drawing a black ball = 5/8
21. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, …., 12 as shown in figure. What is the probability that it will point to:
(i) 10? (ii) an odd number?
(iii) a number which is multiple of 3? (iv) an even number?
Solution:
Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing number 1, 2, 3 ….12
Required to find: Probability of following
Total numbers on the spin is 12
(i) Favourable outcomes i.e. to get 10 is 1
So, total number of favourable outcomes i.e. to get 10 is 1
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a 10 = 1/12
(ii) Favourable outcomes i.e. to get an odd number are 1, 3, 5, 7, 9, and 11
So, total number of favourable outcomes i.e. to get a prime number is 6
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a prime number = 6/12 = 1/2
(iii) Favourable outcomes i.e. to get a multiple of 3 are 3, 6, 9, and 12
So, total number of favourable outcomes i.e. to get a multiple of 3 is 4
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting multiple of 3 = 4/12 = 1/3
(iv) Favourable outcomes i.e. to get an even number are 2, 4, 6, 8, 10, and 12
So, total number of favourable outcomes i.e. to get an even number is 6
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting an even number = 6/12 = 1/2
22. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is:
(i) The name of a girl (ii) The name of a boy?
Solution:
Given: In a class there are 18 girls and 16 boys, the class teacher wants to choose one name. The class teacher writes all pupils’ name on a card and puts them in basket and mixes well thoroughly. A child picks one card
Required to find: The probability that the name written on the card is
(i) The name of a girl
(ii) The name of a boy
Total number of students in the class = 18 + 16 = 34
(i) The names of a girl are 18, so the number of favourable cases is 18
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a name of girl on the card = 18/34 = 9/17
(ii) The names of a boy are 16, so the number of favourable cases is 16
We know that, Probability = Number of favourable outcomes/ Total number of outcomes
Thus, the probability of getting a name of boy on the card = 16/34 = 8/17
23. Why is tossing a coin considered to be a fair way of deciding which team should choose ends in a game of cricket?
Solution:
No. of possible outcomes while tossing a coin = 2 i.e., 1 head or 1 tail
Probability = Number of favourable outcomes/ Total number of outcomes
P(getting tail) = 1/2
As we can see that the probability of both the events are equal, these are called equally like events.
Thus, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.
24. What is the probability that a number selected at random from the number 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 will be their average?
Solution:
Given numbers are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4
Total number of possible outcomes = 10
= 30/10
= 3
Now, let E be the event of getting 3.
Number of favourable outcomes = 3 {3, 3, 3}
P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 3/10
Therefore, the probability that a number selected at random will be the average is 3/10.
25. There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.
Solution:
Given: 30 cards of same size in a bag on which numbers 1 to 30 are written. And, one card is taken out of the bag at random.
Required to find: Probability that the number on the selected card is not divisible by 3.
Total number of possible outcomes are 30 {1, 2, 3, … 30}
Let E = event of getting a number that is divisible by 3
So, the number of favourable outcomes = 10{3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 10/30
= 1/3
Then, = Event of getting number not divisible by 3
= 2/3
Thus, the probability that the number on the selected card is not divisible by 3 = 2/3
26. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is (i) red or white (ii) not black (iii) neither white nor black.
Solution:
Total number of possible outcomes = 20 (5 red, 8 white & 7 black}
(i) Let E = event of drawing a red or white ball
No. of favourable outcomes = 13 (5 red + 8 white)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 13/20
(ii) Let E = event of getting a black ball
No. of favourable outcomes =7 (7 black balls)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 7/20
= Event of not getting black ball
(iii) Let E = Event of getting neither a white nor a black ball
No. of favourable outcomes = 20 – 8 – 7 = 5(total balls – no. of white balls – no. of black balls)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 5/20 = 1/4
27. Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.
Solution:
Total no. of possible outcomes = 25 {1, 2, 3…. 25}
Let E = Event of getting a prime no.
So, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19, 23
No. of favourable outcomes = 9
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 9/25
The,
= Event of not getting a prime
Therefore, the probability of selecting a number which is not prime is 16/25.
28. A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is
(i) Red or white (ii) Not black (iii) Neither white nor black
Solution:
Total number of balls = 8 + 6 + 4 = 18
Total no. of possible outcomes =18
(i) Let E = Event of getting red or white ball
No. of favourable outcomes = 14 (8 red balls + 6 white balls)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 14/18
P(E) = 7/9
(ii) Let E = Event of getting a black ball
Number of favourable outcomes = 4 (4 black balls)
P(E) = 4/18
P(E) = 2/9
Then,
= Event of not getting a black ball
(iii) Let E = event of getting neither a white nor a black ball
No. of favourable outcomes = 18 – 6 – 4
= 8(Total balls – no. of white balls – no. of black balls)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 8/18 = 4/9
29. Find the probability that a number selected at random from the numbers 1, 2, 3…. 35 is a:
(i) Prime number (ii) Multiple of 7 (iii) Multiple of 3 or 5
Solution:
Numbers from 1, 2, 3….. 35 are a total of 35.
Total no. of possible outcomes = 35
(i) Let E = event of getting a prime number
No. of favourable outcomes =11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 11/35
(ii) Let E = event of getting a number which is a multiple of 7
No. of favourable outcomes = 5 {7, 14, 21, 28, 35}
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 5/35 = 1/7
(iii) Let E = Event of getting no which is multiple of 3 or 5
No. of favourable outcomes = 16 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35}
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 16/35
30. From a pack of 52 playing cards Jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is
(i) a black queen (ii) a red card
(iii) a black jack (iv) a picture card (Jacks. queens and kings are picture cards)
Solution:
We know that,
Total no. of cards = 52
All jacks, queens & kings, aces of red colour are removed.
Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 (remaining cards)
(i) Let E = event of getting a black queen
No. of favourable outcomes = 2 (queen of spade & club)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 1/22
(ii) Let E = event of getting a red card
No. of favourable outcomes = 26 – 8
= 18 (total red cards jacks – queens, kings, aces of red colour)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 18/44 = 9/22
(iii) Let E = event of getting a black jack
No. of favourable outcomes = 2 (jack of club & spade)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 2/44 = 1/22
(iv) Let E = event of getting a picture card
No. of favourable outcomes = 6 (2 jacks, 2 kings & 2 queens of black colour)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 6/44 = 3/22
31. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:
(i) an orange flavoured candy
(ii) a lemon flavoured candy
Solution:
(i) We know that the bag contains lemon flavoured candies only. So, the event that Malini will take out an orange flavoured candy is an impossible event.
Thus, the probability of impossible event is 0
P(an orange flavoured candy) = 0
(ii) As the bag contains lemon flavoured candies only. Then, the event that Malini will take out a lemon flavoured candy is sure event. Thus, the probability of sure event is 1.
P(a lemon flavoured candy) = 1
32. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
If E = event of 2 students not having same birthday
Given, P(E) = 0.992
Let,
= event of 2 students having the same birthday.
We know that,
Therefore, the probability that the 2 students have the same birthday is 0.008
33. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red (ii) not red
Solution:
Given,
A bag contains 3 red and 5 black balls.
So, the total no. of possible outcomes = 8 (3 red + 5 black)
(i) Let E = event of getting red ball.
No. of favourable outcomes = 3 (as there are 3 red)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 3/8
(ii) Let
= event of getting no red ball.
From the previous question we already have P(E) = 3/8
34. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red (ii) not green
Solution:
Given,
A box containing 5 red, 8 white and 4 green marbles.
So, the total no. of possible outcomes = 17 (5 red + 8 white + 4 green)
(i) Let E = Event of getting a red marble
Number of favourable outcomes = 5 (as 5 red marbles)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 5/17
(ii) Let E= event of getting a green marble
Number of favourable outcomes = 4 (as 4 green marbles)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 4/17
So,
= Event of getting not a green marble
Then we know that,
Therefore, the probability that the marble taken out is not green is 13/17.
35. A lot consists of 144 ball pens of which 20 are defective and others good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
Solution:
No. of good pens = 144 – 20 = 124
No. of detective pens = 20
Total no. of possible outcomes =144 (total no. of pens)
(i) For her to buy it the pen should be a good one.
So, let E = event of buying a pen which is good.
No. of favourable outcomes = 124 (124 good pens)
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 124/144 = 31/36
(ii) Now, Let
= Event of she not buying a pen as it was a defective one.
Therefore, the probability that she will not buy = 5/36
36. 12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is good one.
Solution:
We have,
No. of good pens = 132
No. of defective pens = 12
So, the total no. of pens = 132 + 12 = 144
Then, the total no. of possible outcomes = 144
Now, let E = event of getting a good pen.
No. of favourable out comes = 132 {132 good pens}
Probability, P(E) = Number of favourable outcomes/ Total number of outcomes
P(E) = 132/144 = 11/12
### RD Sharma Class 10 Chapter 13 Exercise 13.2 Page No: 13.32
1. Suppose you drop a tie at random on the rectangular region shown in fig. below. What is the probability that it will land inside the circle with diameter 1 m?
Solution:
Area of a circle with radius 0.5 m A circle = (0.5)2 = 0.25 πm2
Area of rectangle = 3 x 2 = 6m2
The probability that tie will land inside the circle with diameter 1m
Therefore, the probability that the tie will land inside the circle = π/24
2. In the accompanying diagram, a fair spinner is placed at the centre O of the circle. Diameter AOB and radius OC divide the circle into three regions labelled X, Y and Z.? If ∠BOC = 45°. What is the probability that the spinner will land in the region X?
Solution:
Given,
∠BOC = 45°
∠AOC = 180 – 45 = 135° [Linear Pair]
Area of circle = πr2
Area of region x = θ/360 × πr2
= 135/360 × πr2
= 3/8 × πr2
The probability that the spinner will land in the region
x = 3/8
Therefore, the probability that the spinner will land in region X is 3/8.
3. A target is shown in fig. below consists of three concentric circles of radii, 3, 7 and 9 cm respectively. A dart is thrown and lands on the target. What is the probability that the dart will land on the shaded region?
Solution:
We have,
1st circle – with radius 3
2nd circle – with radius 7
3rd circle – with radius 9
So, their areas would be
Area of 1st circle = π(3)2 = 9π
Area of 2nd circle = π(7)2 = 49π
Area of 3rd circle = π(9)2 = 81π
Area of shaded region = Area of 2nd circle – Area of 1st circle
= 49π − 9π
= 40π
Probability that it will land on the shaded region
Therefore, the probability that the dart will land on the shaded region is 40/81.
Courtesy : CBSE |
# How do I find log 10 526
## 3.3 logarithms
### From online math bridging course 1
theory Exercises
Content:
• Logarithms
• The laws of logarithms
Learning goals:
After this section you should be able to:
• Calculate with bases and exponents
• Know the meaning of the expressions \ displaystyle \ ln, \ displaystyle \ lg, \ displaystyle \ log and \ displaystyle \ log_ {a}.
• Calculate simple logarithms with the definition of the logarithm.
• Know that logarithms are only defined for positive numbers.
• Know the meaning of the number \ displaystyle e.
• Use the laws of logarithms to simplify logarithmic expressions.
• Know when the logarithmic laws are valid.
• Change the base of logarithms
Are you still very familiar with the learning objectives from school and do you know exactly how to do the calculations? Then you can do the same with the exams begin (you can find the link in the Student Lounge).
### A - base 10 logarithm
Often one uses powers with the base \ displaystyle 10 to write large numbers, for example:
\ displaystyle \ begin {align *} 10 ^ 3 & = 10 \ times 10 \ times 10 = 1000 \ ,, \ 10 ^ {- 2} & = \ frac {1} {10 \ times 10} = \ frac {1} {100} = 0 \ textrm {.} 01 \, \ mbox {.} \ end {align *}
If you look at the exponent, you can see that:
"the exponent of 1000 is 3", or that
"the exponent of 0.01 is -2".
The logarithm is defined in the same way. We wrote more formally:
"The logarithm of 1000 is 3 ". This is written \ displaystyle \ lg 1000 = 3,
"The logarithm from 0.01 is -2 ". This is written \ displaystyle \ lg 0 \ textrm {.} 01 = -2.
The following applies more generally:
The logarithm of a number \ displaystyle y is called \ displaystyle \ lg y and is the exponent that makes up the equation
\ displaystyle 10 ^ {\ \ bbox [# AAEEFF, 2pt] {\, \ phantom {a} \,}} = y \, \ mbox {.}
Fulfills. \ displaystyle y must be a positive number so that the logarithm \ displaystyle \ lg y should be defined after a power with a positive base (like 10) is always positive.
example 1
1. \ displaystyle \ lg 100000 = 5 \ quad because \ displaystyle 10 ^ {\, \ bbox [# AAEEFF, 1pt] {\ scriptstyle \, 5 \ vphantom {,} \,}} = 100 \, 000
2. \ displaystyle \ lg 0 \ textrm {.} 0001 = -4 \ quad because \ displaystyle 10 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, - 4 \ vphantom {,} \,}} = 0 \ textrm {.} 0001
3. \ displaystyle \ lg \ sqrt {10} = \ frac {1} {2} \ quad because \ displaystyle 10 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 1/2 \,}} = \ sqrt {10}
4. \ displaystyle \ lg 1 = 0 \ quad because \ displaystyle 10 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 0 \ vphantom {,} \,}} = 1
5. \ displaystyle \ lg 10 ^ {78} = 78 \ quad because \ displaystyle 10 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 78 \ vphantom {,} \,}} = 10 ^ {78}
6. \ displaystyle \ lg 50 \ approx 1 \ textrm {.} 699 \ quad because \ displaystyle 10 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 1 \ textrm {.} 699 \,}} \ approx 50
7. \ displaystyle \ lg (-10) does not exist because \ displaystyle 10 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, a \ vphantom {b,} \,}} can never become -10, no matter how to choose \ displaystyle a
In the example above you can easily see that \ displaystyle \ lg 50 must be between 1 and 2 after \ displaystyle 10 ^ 1 < 50="">< 10^2.="" um="" einen="" genaueren="" wert="" von="" \displaystyle="" \lg="" 50="1\textrm{.}69897\ldots" zu="" erhalten,="" braucht="" man="" aber="" einen="" taschenrechner.="">
Example 2
1. \ displaystyle 10 ^ {\ textstyle \, \ lg 100} = 100
2. \ displaystyle 10 ^ {\ textstyle \, \ lg a} = a
3. \ displaystyle 10 ^ {\ textstyle \, \ lg 50} = 50
### B - Different bases
You can also define logarithms for bases other than 10 (except for base 1). In this case, however, one must clearly show which number is the basis. For example, if you use base 2, you write \ displaystyle \ log _ {\, 2} and this means "the logarithm of base 2".
Example 3
1. \ displaystyle \ log _ {\, 2} 8 = 3 \ quad because \ displaystyle 2 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 3 \ vphantom {,} \,}} = 8.
2. \ displaystyle \ log _ {\, 2} 2 = 1 \ quad because \ displaystyle 2 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 1 \ vphantom {,} \,}} = 2.
3. \ displaystyle \ log _ {\, 2} 1024 = 10 \ quad because \ displaystyle 2 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 10 \ vphantom {,} \,}} = 1024.
4. \ displaystyle \ log _ {\, 2} \ frac {1} {4} = -2 \ quad because \ displaystyle 2 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, - 2 \ vphantom {,} \,}} = \ frac {1} {2 ^ 2} = \ frac {1} {4}.
The calculations with bases other than 2 are quite similar.
Example 4
1. \ displaystyle \ log _ {\, 3} 9 = 2 \ quad because \ displaystyle 3 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 2 \ vphantom {,} \,}} = 9.
2. \ displaystyle \ log _ {\, 5} 125 = 3 \ quad because \ displaystyle 5 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 3 \ vphantom {,} \,}} = 125.
3. \ displaystyle \ log _ {\, 4} \ frac {1} {16} = -2 \ quad because \ displaystyle 4 ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, - 2 \ vphantom {,} \,}} = \ frac {1} {4 ^ 2} = \ frac {1} {16}.
4. \ displaystyle \ log _ {\, b} \ frac {1} {\ sqrt {b}} = - \ frac {1} {2} \ quad because \ displaystyle b ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt ] {\, - 1/2 \,}} = \ frac {1} {b ^ {1/2}} = \ frac {1} {\ sqrt {b}} (if \ displaystyle b> 0 and \ displaystyle b \ not = 1).
When calculating to base 10, one rarely writes \ displaystyle \ log _ {\, 10}, but one simply writes lg or log.
### C - The natural logarithm
The two most commonly used logarithms are those with a base of 10, and the number \ displaystyle e \ displaystyle ({} \ approx 2 \ textrm {.} 71828 \ ldots \,). The logarithms to the base e become natural logarithms called. Instead of \ displaystyle \ log _ {\, e}, write \ displaystyle \ ln when calculating natural logarithms.
Example 5
1. \ displaystyle \ ln 10 \ approx 2 {,} 3 \ quad because \ displaystyle e ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 2 {,} 3 \,}} \ approx 10.
2. \ displaystyle \ ln e = 1 \ quad because \ displaystyle e ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 1 \ vphantom {,} \,}} = e.
3. \ displaystyle \ ln \ frac {1} {e ^ 3} = -3 \ quad because \ displaystyle e ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, - 3 \ vphantom {,} \,} } = \ frac {1} {e ^ 3}.
4. \ displaystyle \ ln 1 = 0 \ quad because \ displaystyle e ^ {\ scriptstyle \, \ bbox [# AAEEFF, 1pt] {\, 0 \ vphantom {,} \,}} = 1.
5. If \ displaystyle y = e ^ {\, a} then \ displaystyle a = \ ln y.
6. \ displaystyle e ^ {\, \ bbox [# AAEEFF, 1pt] {\, \ ln 5 \ vphantom {,} \,}} = 5
7. \ displaystyle e ^ {\, \ bbox [# AAEEFF, 1pt] {\, \ ln x \ vphantom {,} \,}} = x
Most good calculators can calculate 10 logarithms and natural logarithms.
### D - laws of logarithms
In 1617-1624 Henry Biggs published a table with all the logarithms of the numbers up to 20,000, and in 1628 Adriaan Vlacq expanded the table with numbers up to 100,000 Can add logarithms of the two numbers and then calculate the number from the logarithm (this is much more effective than multiplying the numbers directly).
Example 6
Calculate \ displaystyle \ .35 \ cdot 54.
If we know that \ displaystyle 35 \ approx 10 ^ {\, 1 \ textrm {.} 5441} and \ displaystyle 54 \ approx 10 ^ {\, 1 \ textrm {.} 7324} (i.e. \ displaystyle \ lg 35 \ approx 1 \ textrm {.} 5441 and \ displaystyle \ lg 54 \ approx 1 \ textrm {.} 7324), we can easily calculate the product:
\ displaystyle 35 \ cdot 54 \ approx 10 ^ {\, 1 \ textrm {.} 5441} \ cdot 10 ^ {\, 1 \ textrm {.} 7324} = 10 ^ {\, 1 \ textrm {.} 5441 + 1 \ textrm {.} 7324} = 10 ^ {\, 3 \ textrm {.} 2765} \ ,.
If we also know that \ displaystyle 10 ^ {\, 3 \ textrm {.} 2765} \ approx 1890 (i.e. \ displaystyle \ lg 1890 \ approx 3 \ textrm {.} 2765) we have made it, the product
\ displaystyle 35 \ cdot 54 = 1890
can only be calculated with the addition of the exponents \ displaystyle 1 \ textrm {.} 5441 and \ displaystyle 1 \ textrm {.} 7324.
This is an example of the logarithmic laws, viz
\ displaystyle \ log (ab) = \ log a + \ log b
This comes from the rules of arithmetic for powers. On the one hand we have
\ displaystyle a \ cdot b = 10 ^ {\ textstyle \ log a} \ cdot 10 ^ {\ textstyle \ log b} = 10 ^ {\, \ bbox [# AAEEFF, 1pt] {\, \ log a + \ log b \, }}
but on the other hand we also have
\ displaystyle a \ cdot b = 10 ^ {\, \ bbox [# AAEEFF, 1pt] {\, \ log (ab) \,}} \, \ mbox {.}
With the calculation rules for powers one can similarly derive the following logarithmic laws:
\ displaystyle \ begin {align *} \ log (ab) & = \ log a + \ log b, \ [4pt] \ log \ frac {a} {b} & = \ log a - \ log b, \ [4pt] \ log a ^ b & = b \ cdot \ log a \, \ mbox {.} \ \ end {align *}
The logarithmic laws apply regardless of the base.
Example 7
1. \ displaystyle \ lg 4 + \ lg 7 = \ lg (4 \ cdot 7) = \ lg 28
2. \ displaystyle \ lg 6 - \ lg 3 = \ lg \ frac {6} {3} = \ lg 2
3. \ displaystyle 2 \ cdot \ lg 5 = \ lg 5 ^ 2 = \ lg 25
4. \ displaystyle \ lg 200 = \ lg (2 \ cdot 100) = \ lg 2 + \ lg 100 = \ lg 2 + 2
Example 8
1. \ displaystyle \ lg 9 + \ lg 1000 - \ lg 3 + \ lg 0 {,} 001 = \ lg 9 + 3 - \ lg 3 - 3 = \ lg 9- \ lg 3 = \ lg \ displaystyle \ frac {9 } {3} = \ lg 3
2. \ displaystyle \ ln \ frac {1} {e} + \ ln \ sqrt {e} = \ ln \ left (\ frac {1} {e} \ cdot \ sqrt {e} \, \ right) = \ ln \ left (\ frac {1} {(\ sqrt {e} \,) ^ 2} \ cdot \ sqrt {e} \, \ right) = \ ln \ frac {1} {\ sqrt {e}}
\ displaystyle \ phantom {\ ln \ frac {1} {e} + \ ln \ sqrt {e}} {} = \ ln e ^ {- 1/2} = - \ frac {1} {2} \ cdot \ ln e = - \ frac {1} {2} \ cdot 1 = - \ frac {1} {2} \ vphantom {\ biggl (}
3. \ displaystyle \ log_2 36 - \ frac {1} {2} \ log_2 81 = \ log_2 (6 \ cdot 6) - \ frac {1} {2} \ log_2 (9 \ cdot 9)
\ displaystyle \ phantom {\ log_2 36 - \ frac {1} {2} \ log_2 81} {} = \ log_2 (2 \ cdot 2 \ cdot 3 \ cdot 3) - \ frac {1} {2} \ log_2 ( 3 \ times 3 \ times 3 \ times 3)
\ displaystyle \ phantom {\ log_2 36 - \ frac {1} {2} \ log_2 81} {} = \ log_2 (2 ^ 2 \ cdot 3 ^ 2) - \ frac {1} {2} \ log_2 (3 ^ 4) \ vphantom {\ Bigl (}
\ displaystyle \ phantom {\ log_2 36 - \ frac {1} {2} \ log_2 81} {} = \ log_2 2 ^ 2 + \ log_2 3 ^ 2 - \ frac {1} {2} \ log_2 3 ^ 4
\ displaystyle \ phantom {\ log_2 36 - \ frac {1} {2} \ log_2 81} {} = 2 \ log_2 2 + 2 \ log_2 3 - \ frac {1} {2} \ cdot 4 \ log_2 3
\ displaystyle \ phantom {\ log_2 36 - \ frac {1} {2} \ log_2 81} {} = 2 \ cdot 1 + 2 \ log_2 3 - 2 \ log_2 3 = 2 \ vphantom {\ Bigl (}
4. \ displaystyle \ lg a ^ 3 - 2 \ lg a + \ lg \ frac {1} {a} = 3 \ lg a - 2 \ lg a + \ lg a ^ {- 1}
\ displaystyle \ phantom {\ lg a ^ 3 - 2 \ lg a + \ lg \ frac {1} {a}} {} = (3-2) \ lg a + (-1) \ lg a = \ lg a - \ lg a = 0
### E - change base
Sometimes you want to write logarithms in one base than logarithms in another base.
Example 9
1. Write \ displaystyle \ lg 5 as a natural logarithm.
By definition, \ displaystyle \ lg 5 is the number that makes up the equation
\ displaystyle 10 ^ {\ lg 5} = 5 \, \ mbox {.}
Fulfills. By calculating the natural logarithm of both sides, we get
\ displaystyle \ ln 10 ^ {\ lg 5} = \ ln 5 \, \ mbox {.}
With the logarithmic law \ displaystyle \ ln a ^ b = b \ ln a we write the left side as \ displaystyle \ lg 5 \ cdot \ ln 10 and get the equation
\ displaystyle \ lg 5 \ cdot \ ln 10 = \ ln 5 \, \ mbox {.}
Division by \ displaystyle \ ln 10 gives the answer
\ displaystyle \ lg 5 = \ frac {\ ln 5} {\ ln 10} \ qquad (\ approx 0 \ textrm {.} 699 \ ,, \ quad \ text {so} \ 10 ^ {0 \ textrm {.} 699 } \ approx 5) \, \ mbox {.}
2. Write the 2-logarithm of 100 as a 10-logarithm, lg.
According to the definition of the logarithm, it is clear that \ displaystyle \ log_2 100 is the equation
\ displaystyle 2 ^ {\ log _ {\ scriptstyle 2} 100} = 100
Fulfills. We log both sides (using the 10 logarithm) and get
\ displaystyle \ lg 2 ^ {\ log _ {\ scriptstyle 2} 100} = \ lg 100 \, \ mbox {.}
After \ displaystyle \ lg a ^ b = b \ lg a, we get \ displaystyle \ lg 2 ^ {\ log_2 100} = \ log _ {\ scriptstyle 2} 100 \ cdot \ lg 2 and the right side is just \ displaystyle \ lg 100 = 2. This gives the equation
\ displaystyle \ log _ {\ scriptstyle 2} 100 \ cdot \ lg 2 = 2 \, \ mbox {.}
Division by \ displaystyle \ lg 2 shows that
\ displaystyle \ log _ {\ scriptstyle 2} 100 = \ frac {2} {\ lg 2} \ qquad ({} \ approx 6 \ textrm {.} 64 \ ,, \ quad \ text {so is} \ 2 ^ {6 \ textrm {.} 64} \ approx 100) \, \ mbox {.}
The general formula for changing the base from \ displaystyle a to \ displaystyle b in logarithms is
\ displaystyle \ log _ {\ scriptstyle \, a} x = \ frac {\ log _ {\ scriptstyle \, b} x} {\ log _ {\ scriptstyle \, b} a} \, \ mbox {.}
For example, if we want to write \ displaystyle 2 ^ 5 to base 10, we first write 2 to base 10
\ displaystyle 2 = 10 ^ {\ lg 2}
and use the calculation rules for powers
\ displaystyle 2 ^ 5 = (10 ^ {\ lg 2}) ^ 5 = 10 ^ {5 \ cdot \ lg 2} \ quad ({} \ approx 10 ^ {1 \ textrm {.} 505} \,) \, \ mbox {.}
Example 10
1. Write \ displaystyle 10 ^ x to the natural base e.
First we write 10 to the base e,
\ displaystyle 10 = e ^ {\ ln 10}
and use the calculation rules for powers
\ displaystyle 10 ^ x = (e ^ {\ ln 10}) ^ x = e ^ {\, x \ cdot \ ln 10} \ approx e ^ {2 \ textrm {.} 3 x} \, \ mbox {.}
2. Write \ displaystyle e ^ {\, a} to base 10
The number \ displaystyle e can be written like \ displaystyle e = 10 ^ {\ lg e}, and therefore is
\ displaystyle e ^ a = (10 ^ {\ lg e}) ^ a = 10 ^ {\, a \ cdot \ lg e} \ approx 10 ^ {\, 0 \ textrm {.} 434a} \, \ mbox {.} |
# Abelian Group Example
Last Updated : 15 Mar, 2021
Problem-: Prove that ( I, + ) is an abelian group. i.e. The set of all integers I form an abelian group with respect to binary operation ‘+’.
Solution-:
Set= I ={ ……………..-3, -2 , -1 , 0, 1, 2 , 3……………… }.
Binary Operation= ‘+’
Algebraic Structure= (I ,+)
We have to prove that (I,+) is an abelian group.
To prove that set of integers I is an abelian group we must satisfy the following five properties that is Closure Property, Associative Property, Identity Property, Inverse Property, and Commutative Property.
1) Closure Property
∀ a , b ∈ I ⇒ a + b ∈ I
2,-3 ∈ I ⇒ -1 ∈ I
Hence Closure Property is satisfied.
2) Associative Property
( a+ b ) + c = a+( b +c) ∀ a , b , c ∈ I
2 ∈ I, -6 ∈ I , 8 ∈ I
So, LHS= ( a + b )+c
= (2+ ( -6 ) ) + 8 = 4
RHS= a + ( b + c )
=2 + ( – 6 + 8 ) = 4
Hence RHS = LHS
Associative Property is also Satisfied
3) Identity Property
a + 0 = a ∀ a ∈ I , 0 ∈ I
5 ∈ I
5+0 = 5
-17 ∈ I
-17 + 0 = – 17
Identity property is also satisfied.
4) Inverse Property
a + ( -a ) = 0 ∀ a ∈ I , -a ∈ I ,0 ∈ I
a=18 ∈ I then ∋ a number -18 such that 18 + ( -18 ) = 0
So, Inverse property is also satisfied.
5) Commutative Property
a + b = b + a ∀ a , b ∈ I
Let a=19, b=20
LHS = a + b
= 19+( -20 ) = -1
RHS = b + a
= -20 +19 = -1
LSH=RHS
Commutative Property is also satisfied.
We can see that all five property is satisfied. Hence (I,+) is an Abelian Group.
Note-: (I,+) is also Groupoid, Monoid and Semigroup. |
# Tarski's High School Algebra Problem
In high school, we all learn algebraic rules for simplifying expressions. The following eleven identities over the positive integers cover most of the cases we encounter in school:
• $$x + y = y + x$$
• $$(x + y) + z = x + (y + z)$$
• $$x \cdot 1 = x$$
• $$x \cdot y = y \cdot x$$
• $$(x \cdot y) \cdot z = x \cdot (y \cdot z)$$
• $$1^x = 1$$
• $$x^1 = x$$
• $$x^{y+z} = x^y \cdot x^z$$
• $$(x \cdot y)^z = x^z \cdot y^z$$
• $$(x^y)^z = x^{y \cdot z}$$
One question that arises is whether these rules are complete. That is, can we prove every valid identity over the positive integers holds just using these eleven rules?
This problem is known as Tarski's high school algebra problem. The Wikipedia page gives a good summary of the problem, its solution, and related work. In short, the answer is no. Andrew Wilkie proved this by giving an equation:
$$$\begin{split} {\left( (1 + x)^y + (1 + x + x^2)^y\right)^x \cdot \left((1 + x^3)^x + (1 + x^2 + x^4)^x\right)^y} \\ = {\left((1 + x)^x + (1 + x + x^2)^x\right)^y \cdot \left((1 + x^3)^y + (1 + x^2 + x^4)^y\right)^x} \end{split}$$$
that is true for all positive integers $$x$$ and $$y$$. He then showed that it is not possible to give a proof of this equation just using the eleven rewrite rules above.
In this post, I will give a mechanized proof of Wilkie's theorem in the Coq proof assistant. However, I will formalize a proof due to Burris and Yeats instead of Wilkie's original proof.
We begin by defining the algebraic expressions we are talking about as a Coq data type:
Instead of representing variables by letters like "x" and "y", we will index them using natural numbers. Var 0 will be the first variable, Var 1 will be the second, and so on. We also define some notation to make it easier to write and read these expressions.
Infix "+" := Eplus : expression_scope.
Infix "*" := Etimes : expression_scope.
Infix "^" := Epow : expression_scope.
Next, we would like to give a denotation to these algebraic expressions. What we mean here is, when we have an expression like:
$x^2 + x + 4$
We can think of this as the description of a function from positive integers (which will be bound to the variable $$x$$) to positive integers. We would like to capture this idea formally.
Unfortunately, we first need to define exponentiation on the positive integers. Although the Coq standard library has definitions for exponentiation over general integers and natural numbers, I did not find one for the positive integer type. This is easy enough to do ourselves. This definition is just like Zpower from the standard library, but uses iter_pos instead of iter_nat
Definition Ppower (x y: positive) :=
iter_pos y positive (fun q: positive => x * q) 1.
Infix "^" := Ppower : positive_scope.
We also prove some lemmas about this operation. The proofs are omitted here, but they can be found in the full source file
Lemma unit_power: forall x, 1^x = 1.
Lemma exp_one: forall x, x^1 = x.
Lemma exp_plus_aux: forall y z x,
Ppower_nat x (y + z) = Ppower_nat x y * Ppower_nat x z.
Lemma exp_plus: forall x y z, x ^ (y + z) = x^y * x^z.
Lemma exp_mult: forall z x y, (x * y) ^ z = x^z * y^z.
Lemma exp_exp: forall z x y, (x ^ y) ^ z = x ^ (y * z).
Since all of the other operators we need are in the standard library, we are ready to define the denotation of algebraic expressions:
Fixpoint denote (e: expression) (f: nat -> positive) : positive :=
match e with
| Var n => f n
| Const n => n
| Eplus e1 e2 => denote e1 f + denote e2 f
| Etimes e1 e2 => denote e1 f * denote e2 f
| Epow e1 e2 => denote e1 f ^ denote e2 f
end.
Given an expression e and a function f from nat -> positive, denote uses the function f as map to look up the values of the variables in the expression. That is, Var n gets mapped to f n. Constant terms are mapped to the corresponding positive, and Eplus, Etimes, and Epow are mapped to the corresponding operator for the positive type.
Now we give an inductive data type corresponding to proofs using the eleven high school identities:
Local Open Scope expression_scope.
Inductive td : expression -> expression -> Prop :=
| plus: forall x x' y y', td x x' -> td y y' -> td (x + y) (x' + y')
| times: forall x x' y y', td x x' -> td y y' -> td (x * y) (x' * y')
| pow: forall x x' y y', td x x' -> td y y' -> td (x ^ y) (x' ^ y')
| cplus: forall p1 p2, td (Const p1 + Const p2) (Const (p1 + p2))
| ctimes: forall p1 p2, td (Const p1 * Const p2) (Const (p1 * p2))
| cpow: forall p1 p2, td (Const p1 ^ Const p2) (Const (p1 ^ p2))
| refl: forall x, td x x
| trans: forall x y z, td x y -> td y z -> td x z
| sym: forall x y, td x y -> td y x
| t1: forall x y, td (x + y) (y + x)
| t2: forall x y z, td ((x + y) + z) (x + (y + z))
| t3: forall x, td (x * (Const 1)) x
| t4: forall x y, td (x * y) (y * x)
| t5: forall x y z, td ((x * y) * z) (x * (y * z))
| t6: forall x y z, td (x * (y + z)) (x * y + x * z)
| t7: forall x, td ((Const 1) ^ x) (Const 1)
| t8: forall x, td (x^(Const 1)) x
| t9: forall x y z, td (x^(y + z)) (x^y * x^z)
| t10: forall x y z, td ((x * y)^z) (x^z * y^z)
| t11: forall x y z, td ((x^y) ^ z) (x^(y * z)).
Given two expression e1 and e2, if we can build a term of type td e1 e2 then there is a proof that e1 = e2 using one of the eleven identities. The constructors t1 through t11 correspond to these identities, but we also have some extras. plus, times, and pow state that if we have a proof that e1 and e1' are equal, and e2 and e2' are equal then e1 + e2 is equal to e1' + e2', and e1 * e2 is equal to e1' * e2', etc. cplus, ctimes, and cpow let us fold constants so that we can conclude td (Const 1 + Const 1) (Const 2)
Then we have refl, sym and trans which correspond to the idea that e1 is always equal to itself, if e1 equals e2 then e2 equals e1, and that if we have a proof that e1 is equal to e2, and e2 is equal to e3, then we also have a proof that e1 is equal to e3.
We also establish notation. We use [=] instead of "=" because this equality is distinct from the built-in equality in Coq
Infix "[=]" := td (at level 70) : expression_scope.
Now we can prove that these rules are sound. That is, if we can construct a proof that e1 [=] e2, then for each possible denotation of e1 and e2, the resulting values are equal.
Lemma td_sound: forall x y, x [=] y -> forall f, denote x f = denote y f.
Proof.
intros.
induction H; simpl; auto.
rewrite IHtd1; rewrite IHtd2; auto with *.
rewrite IHtd1; rewrite IHtd2; auto with *.
rewrite IHtd1; rewrite IHtd2; auto with *.
rewrite IHtd1; rewrite IHtd2; auto with *.
apply Pplus_comm.
rewrite Pplus_assoc; auto.
apply Pmult_1_r.
apply Pmult_comm.
rewrite Pmult_assoc; auto.
apply Pmult_plus_distr_l.
apply unit_power.
apply exp_one.
apply exp_plus.
apply exp_mult.
apply exp_exp.
Qed.
Now we can prove identities in Coq about positive integers by constructing corresponding expressions, say e1 and e2, building a term of the type td e1 e2, and then using the previous lemma.
In practice, we don't really want to be doing this. Proving the following fact "directly" using the usual Coq tactics is a lot more pleasant and efficient, this is just merely to demonstrate that it can be done.
Remark example:
(forall x y, (x + y) * (x + y) = x ^ 2 + 2 * (x * y) + y ^2)%positive.
Proof.
intros.
pose (e1 := (Var 0 + Var 1) * (Var 0 + Var 1)).
pose (e2 := (Var 0) ^ (Const 2) + (Const 2) * (Var 0 * Var 1) +
(Var 1) ^ (Const 2)).
pose (f := fun (n:nat) => match n with O => x | _ => y end).
specialize (td_sound e1 e2). intros.
pose (e1' := (Var 0 + Var 1) * (Var 0) + (Var 0 + Var 1) * Var 1).
pose (e1'' := (Var 0 * Var 0) + (Var 1 * Var 0) + (Var 0 + Var 1)
* Var 1).
assert (e1 [=] e2).
eapply trans.
eapply t6.
eapply trans.
eapply plus.
I will omit the rest of this proof since it's quite tedious
specialize (H H0 f).
simpl in H. auto.
Qed.
Now, what we'd like to prove is that there is some pair of expressions e1 and e2 such that forall f, denote e1 f = denote e2 f (which implies that they correspond to a valid identity over the positive integers) but for which there does not exist a term of type td e1 e2 (ie we cannot prove the identity using the high school identities
We begin by constructing expression terms w1 and w2 corresponding to Wilkie's counterexample.
We'll now want to prove that forall f, denote w1 f = denote w2 f. Let's first talk about how this proof would proceed on paper. In Wilkie's original paper he lets $$A = 1 + x$$, $$B = 1 + x + x^2$$, $$C = 1 + x^3$$ and $$D = 1 + x^2 + x^4$$, so that the left hand side of his counterexample becomes $$(A^x + B^x)^y(C^y + d^y)^x$$ and the right hand side becomes $$(A^y + B^y)^x (C^x + D^x)^y$$.
Then by observing that $$C = A \cdot (1 + x + x^2)$$ and $$D = B \cdot (1 + x + x^2)$$, we can factor out $$(1 + x + x^2)^{xy}$$ from both sides and then the equality of the two expressions is clear.
To do this in Coq we first construct a function denote_z that instead of mapping our expressions onto positives, will instead map them onto integers
Fixpoint denote_z (e: expression) (f: nat -> positive) : Z :=
(match e with
| Var n => Zpos (f n)
| Const n => Zpos n
| Eplus e1 e2 => denote_z e1 f + denote_z e2 f
| Etimes e1 e2 => denote_z e1 f * denote_z e2 f
| Epow e1 e2 => denote_z e1 f ^ denote_z e2 f
end)%Z.
The reason for doing this is that our proof will involve factoring out terms of the form $$[1 - x + x^2$$. This term involves subtraction. However, the positives are not closed under subtraction. The definition for subtraction in Coq's standard library deals with this by defining x - y to be 1 if y >= x.
Obviously, such a definition of subtraction does not have the properties we usually expect. By mapping things onto Z instead, we will work in an expanded domain where intermediary values can be negative, even though the final expression will still evaluate to a positive number.
We first quickly prove an intermediate result we need
Require Import Zpow_facts.
Lemma zpow_pos: forall p1 p2, (Zpos p1 ^ Zpos p2 = Zpos (p1 ^ p2))%Z.
The next two lemmas capture the idea that if we prove an identity is valid using our expanded domain of Z, it will still hold under our normal definition of denote
Lemma denote_to_denotez: forall e f, denote_z e f = Zpos (denote e f).
Proof.
induction e; simpl; auto;
intros;
try (specialize (IHe1 f); specialize (IHe2 f);
rewrite IHe1; rewrite IHe2; auto).
apply zpow_pos.
Qed.
Lemma denote_z_implies_denote:
forall e1 e2, (forall f, denote_z e1 f = denote_z e2 f)
-> (forall f, denote e1 f = denote e2 f).
Proof.
intros.
specialize (denote_to_denotez e1 f).
specialize (denote_to_denotez e2 f).
intros.
specialize (H f).
rewrite H0 in H.
rewrite H1 in H.
inversion H.
auto.
Qed.
Now we are ready to prove that w1 = w2 is a valid identity using our usual denote function. This proof is a little long, and it's probably not good Coq style, but I have tried to structure it according to the factorization suggested by Wilkie in his initial paper.
Local Open Scope Z_scope.
Lemma Zexp_exp:
forall p2 x p1, (x ^ (Zpos p1)) ^ (Zpos p2) = x ^ (Zpos p1 * Zpos p2).
Lemma Zpower_pos_mult:
forall x y p, (x * y) ^ (Zpos p) = x^(Zpos p) * y^(Zpos p).
Lemma w1_eq_w2: forall f, denote w1 f = denote w2 f.
Proof.
apply denote_z_implies_denote.
intros. unfold w1. unfold w2.
unfold denote_z.
pose (x := Zpos (f 0%nat)).
pose (y := Zpos (f 1%nat)).
fold x. fold y.
pose (A := 1 + x).
pose (B := 1 + x + x^2).
pose (C := 1 + x^3).
pose (D := 1 + x^2 + x^4).
fold B. fold A.
fold C. fold D.
assert (C = A * (1 - x + x^2)).
unfold C. unfold A.
ring.
assert (D = B * (1 - x + x^2)).
unfold D. unfold B.
ring.
assert ((C ^ x + D ^ x) ^ y = (1 - x + x^2)^(x * y) * (A ^ x + B ^x)^y).
rewrite H. rewrite H0.
unfold x, y.
rewrite Zpower_pos_mult.
rewrite Zpower_pos_mult.
fold x. fold y.
rewrite <-Zmult_plus_distr_l.
unfold y.
rewrite Zpower_pos_mult.
fold y.
unfold x, y.
rewrite Zexp_exp.
fold x. fold y.
ring.
rewrite H1.
assert ((C ^ y + D ^ y) ^ x = (1 - x + x^2)^(x * y) * (A ^ y + B ^y)^x).
rewrite H. rewrite H0.
unfold x, y.
rewrite Zpower_pos_mult.
rewrite Zpower_pos_mult.
fold x. fold y.
rewrite <-Zmult_plus_distr_l.
unfold x.
rewrite Zpower_pos_mult.
fold x.
unfold x, y.
rewrite Zexp_exp.
fold x. fold y.
replace (y * x) with (x * y) by (auto with * ).
ring.
rewrite H2.
ring.
Qed.
We must now prove that ~ td w1 w2. This is the critical step of the proof. How can we possibly prove that there does not exist a proof using these rules? Following the proof from Burris and Yeats, we define an algebra of 12 elements along with definitions of +, *, and ^ on this algebra.
Then we'll show that this algebra satisfies all of the eleven high school identities. We'll also show that w1 = w2 is not a valid identity over this 12 element algebra, by explicitly picking values for the variables at which the two expressions differ.
This will show that there cannot be a proof of w1 = w2 using the high school identities, because if there were that would mean that the equation w1 = w2 would also hold over the 12 element algebra.
We start by defining the algebra. We omit the definitions of addition, multiplication, and exponentiation on this algebra since they are just long match statements that elucidate little. This algebra was discovered by Burris and Yeats through the use of a computer program, so it is not very "natural".
Inductive byalg :=
| n1 | n2 | n3 | n4
| a | b | c | d
| e | f | g | h.
We now prove that these operators have the desired properties. These "proofs" are just computation. We do proof by cases over all possible values for the variables and have Coq check each case holds.
Lemma byplus_comm: forall x y, byplus x y = byplus y x.
Proof.
destruct x; destruct y; auto.
Qed.
Lemma byplus_assoc:
forall x y z, byplus x (byplus y z) = byplus (byplus x y) z.
Lemma bymult_comm: forall x y, bymult x y = bymult y x.
Lemma bymult_assoc:
forall x y z, bymult x (bymult y z) = bymult (bymult x y) z.
Lemma bymult_ident:
forall x, bymult x n1 = x.
Lemma bymult_distr:
forall x y z, bymult x (byplus y z) = byplus (bymult x y) (bymult x z).
Lemma byexp_one:
forall x, byexp x n1 = x.
Lemma byexp_base_one:
forall x, byexp n1 x = n1.
Lemma byexp_byplus:
forall x y z, byexp x (byplus y z) = bymult (byexp x y) (byexp x z).
Lemma byexp_bymult:
forall x y z, byexp (bymult x y) z = bymult (byexp x z) (byexp y z).
Lemma byexp_exp:
forall x y z, byexp (byexp x y) z = byexp x (bymult y z).
Now we can define a denotation for expressions to byalg terms, just as we did with denote and denotez
Fixpoint denote_by (e: expression) (f: nat -> byalg) : byalg :=
(match e with
| Var n => f n
| Const n =>
match n with
| 1 => n1
| 2 => n2
| 3 => n3
| _ => n4
end
| Eplus e1 e2 => byplus (denote_by e1 f) (denote_by e2 f)
| Etimes e1 e2 => bymult (denote_by e1 f) (denote_by e2 f)
| Epow e1 e2 => byexp (denote_by e1 f) (denote_by e2 f)
end)%positive.
And we can also prove that the td proofs are sound under this denotation. The properties of byplus, bymult, and byexp that we've shown above cover the eleven cases corresponding to the high school identities.
However, recall that we also had rules showing that constants could be "folded", e.g. that Const p1 + Const p2 [=] Const (p1 + p2) and so on. These turn out to be the most annoying part of the proof. Luckily, the proof works because of the way the operators treat the elements n1, n2, n3, and n4. For example:
• byplus n1 n2 = n3
• bymult n2 n2 = n4
• bymult n3 n2 = n4
What's happening is that in the case of byplus n1 n2, we treat n1 and n2 as though they are 1 and 2 respectively, add them to get 3 and then we return n3. The same thing happens for the other operators. Now what happens if the resulting number is greater than 4, as in the case with bymult n3 n2, (since 3*2 = 6)? In that case we just return n4.
Once we realize this pattern, we can prove that the constant folding rules are sound. Addition and multiplication are handled just by brute force case work. However, this didn't work for exponentiation, so we need a few helper lemmas first.
Lemma denote_by_const_gt_3:
(forall p1, p1 > 3 -> (forall f, denote_by (Const p1) f = n4))%positive.
Lemma Ppow_increase: (forall p2 p1, p1 ^ p2 > p1 \/ p1 ^ p2 = p1)%positive.
Lemma Ppow_gt_3: (forall p2 p1, p1 > 1 -> p2 > 1 -> p1 ^ p2 > 3)%positive.
Local Open Scope positive_scope.
Local Open Scope expression_scope.
Lemma positive_range: forall p: positive, p > 1 \/ p = 1.
Lemma constants_pow_gt_1: forall p1 p2 f, p1 > 1 -> p2 > 1 ->
denote_by (Const p1 ^ Const p2) f =
denote_by (Const (p1 ^ p2)) f.
Lemma constant_pow:
forall p2 p1 f, denote_by (Const p1 ^ Const p2) f =
denote_by (Const (p1 ^ p2)) f.
With those out of the way, we proceed with the soundness proof.
Lemma td_sound_by:
forall x y, x [=] y -> forall f, denote_by x f = denote_by y f.
Proof.
intros.
induction H; try (apply constant_pow); simpl.
simpl; rewrite IHtd1; rewrite IHtd2; auto with *.
simpl; rewrite IHtd1; rewrite IHtd2; auto with *.
simpl; rewrite IHtd1; rewrite IHtd2; auto with *.
intros; simpl; repeat (induction p1; induction p2; auto with *);
repeat (induction p1; auto with *);
repeat (induction p2; auto with *).
intros; simpl; repeat (induction p1; induction p2; auto with *);
repeat (induction p1; auto with *);
repeat (induction p2; auto with *).
auto.
rewrite IHtd1; rewrite IHtd2; auto.
auto.
apply byplus_comm.
rewrite byplus_assoc; auto.
apply bymult_ident.
apply bymult_comm.
rewrite bymult_assoc; auto.
apply bymult_distr.
apply byexp_base_one.
apply byexp_one.
apply byexp_byplus.
apply byexp_bymult.
apply byexp_exp.
Qed.
Next we show that under the Burris and Yeats algebra denotation, w1 is not equal to w2. Specifically, when x = a and y = e, the two expressions are not equal.
Lemma w1_neq_w2_by: ~ (forall f, denote_by w1 f = denote_by w2 f).
Proof.
intros. unfold not.
intros.
specialize (H (fun x => match x with | O => a | _ => e end)).
compute in H.
inversion H.
Qed.
Which implies that we cannot have w1 [=] w2, as discussed earlier.
Theorem not_td_w1_w2: ~ (w1 [=] w2).
Proof.
unfold not; intro.
eapply w1_neq_w2_by.
apply td_sound_by; auto.
Qed.
One might ask what needs to be added to the set of eleven identities to make them complete. Unfortunately, R. Gurevic showed that a finite list of identities cannot be complete. However, that proof is considerably harder, so this seems like a good place to wrap up!
Thanks to Edward Gan for his feedback.
References:
R. Gurevic. Equational theory of positive numbers with exponentiation is not finitely axiomatizable. Ann. Pure Appl. Logic, 49(1):1-30, 1990.
Stanley N. Burris and Karen A. Yeats. The saga of the high school identities. Algebra Universalis, 52(2-3):325-342, 2004.
A. J. Wilkie. On exponentiation - a solution to Tarski's high school algebra problem. |
# Sketching Linear Graphs
Lesson
## How do I graph a linear relationship?
To graph any liner relationship you only need two points that are on the line. You can use any two points from a table of values, or substitute in any two values of $x$x into the equation and solve for corresponding $y$y-value to create your own two points. Often, using the intercepts is one of the easiest ways to sketch the line.
#### Example - sketch from table of values
x 1 2 3 4 y 3 5 7 9
To sketch from a table of values, we need just any two points from the table. From this table we have 4 coordinates, $\left(1,3\right)$(1,3), $\left(2,5\right)$(2,5), $\left(3,7\right)$(3,7), $\left(4,9\right)$(4,9).
Drag the $2$2 of the points on this interactive to the correct positions and graph this linear relationship.
#### Example - sketch from any two points
If we are given the equation of a linear relationship, like $y=3x+5$y=3x+5, then to sketch it we need two points. We can pick any two points we like.
Start by picking any two $x$x-values you like, often the $x$x-value of $0$0 is a good one to pick because the calculation for y can be quite simple. For our example, $y=3x+5$y=3x+5 becomes $y=0+5$y=0+5, $y=5$y=5. This gives us the point $\left(0,5\right)$(0,5)
Similarly look for other easy values to calculate such as $1$1, $10$10, $2$2. I'll pick $x=1$x=1. Then for $y=3x+5$y=3x+5, we have $y=3\times1+5$y=3×1+5, $y=8$y=8.This gives us the point $\left(1,8\right)$(1,8)
Now we plot the two points and create a line.
#### Example - sketch from the intercepts
The general form of a line is great for identifying both the x and y intercepts easily.
For example, the line $3y+2x-6=0$3y+2x6=0
The x intercept happens when the $y$y value is $0$0. $3y+2x-6=0$3y+2x−6=0 $0+2x-6=0$0+2x−6=0 $2x=6$2x=6 $x=3$x=3 The y intercept happens when the $x$x value is $0$0. $3y+2x-6=0$3y+2x−6=0 $3y+0-6=0$3y+0−6=0 $3y=6$3y=6 $y=2$y=2
From here it is pretty easy to sketch, we find the $x$x intercept $3$3, and the $y$y intercept $2$2, and draw the line through both.
#### Example - sketch from the gradient and a point
Start by plotting the single point that you are given.
Remembering that gradient is a measure of change in the rise per change in run, we can step out one measure of the gradient from the original point given.
For a gradient of $4$4 $1$1 unit across and $4$4 units up. For a gradient of $-3$−3 $1$1 unit across and $3$3 units down. For a gradient of $\frac{1}{2}$12 $1$1 unit across and $\frac{1}{2}$12 unit up.
The point can be any point $\left(x,y\right)$(x,y), or it could be an intercept. Either way, plot the point, step out the gradient and draw your line!
For example, plot the line with gradient $-2$2 and has $y$y intercept of $4$4.
Start with the point, ($y$y intercept of $4$4) Step out the gradient, (-$2$2 means $2$2 units down) Draw the line
To sketch linear graphs, it's easiest to substitute in values to find coordinates to put it in gradient-intercept form.
$y=mx+b$y=mx+b
where $m$m is the gradient and $b$b is the $y$y-intercept
Our graphs may not always be in this form so we may need to rearrange the equation to make $y$y the subject (that means $y$y is on one side of the equation and everything else is on the other side).
## What happens if we're not given the equation of a line?
Sometimes, it doesn't matter. We can sketch a straight line on a graph just by knowing a couple of its features such as a point that lies on the line and it's gradient. At other times, we may need to generate an equation before we sketch it. So other than the gradient-intercept form, we can use:
• Gradient-point formula: $y-y_1=m\left(x-x_1\right)$yy1=m(xx1)
• Two point formula: $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$yy1xx1=y2y1x2x1
Ok let's look at this in action with some examples.
#### Examples
##### Question 1
Plot the graph of the line whose gradient is $-3$3 and passes through the point $\left(-2,4\right)$(2,4).
##### Question 2
Graph the linear equation $y=3x-1$y=3x1 using the point $Y$Y as the $y$y-intercept.
##### Question 3
Graph the linear equation $-6x+3y+24=0$6x+3y+24=0 by finding any two points on the line.
## Horizontal or Vertical Graphs
On horizontal lines, the $y$y value is always the same for every point on the line.
On vertical lines, the $x$x value is always the same for every point on the line.
#### Example
##### Question 4
Graph the line $y=-3$y=3. |
STATISTICS (UPDATED 7/30/2009)
NOT EVERYTHING THAT LOOKS IMPRESSIVE IS IMPRESSIVE
*ROFFMAN SKIP FORMULA.
The number of ways that a term (either forwards, backwards, or diagonal) can fit into a matrix is determined as follows:
(1) Let the number of skips possible in a forward direction on a row of length (r), where r = the number of columns in the matrix, be equal to "Sr."
(2) Likewise, let the number of skips possible in a vertical direction on a column of length (c), where c = the number of rows in the matrix, be equal to "Sc."
(3) The Roffman Skip Formula for total skips is as follows:
Skips = 2(Sr + Sc + 2[Sr][Sc]) = 2Sr + 2Sc + 4SrSc.
An example of skip value determined through use of the above tables and formula follows: Find the number of skips possible for a 4-letter word in a Matrix 28 columns by 11 rows.
SolutionSkip Tables for words ranging between 3 and 8 letters are posted on this site. For a 4-letter word use Table 1B. On it find that 28 columns = 9 possible skips forward. Thus Sr = 9. Now note that 11 rows = 3 possible skips vertically. Thus Sc = 3. Now apply the formula which is Skips = 2(Sr + Sc + 2[Sr][Sc]) 2(9 + 3 + 2(9)[3]) = 2(12 + 54) = 2(66) = 132 SKIPS. Now, let us suppose that the 4-letter term occurred at skip 100. To get an idea of how likely such a term is to be found at an ELS, search a range of 132 skips, such as from skip 101 to skip 232. The number of "hits" for this term is then divided by letters in the Control (if this is scrambled Torah the number of letters in the Control is the same as in Torah, i.e., 304,805). The quotient is the Word Frequency Per Letter. This is multiplied by the number of letters on each matrix to reveal Word Expectancy Per Matrix. It is inherent in this procedure that the larger the number of letters in the matrix, the larger the number of placements possible for any given key word at any ELS. After determining Word Frequency Per Plot we apply the Poisson Equation to see the probability that they are present at least once. This is necessary to determine a true probability for each word. Just because a word is likely to appear once per plot does not imply it will always be there. Words may average out to many times per plot area without actually being in a given plot of that area. Of course, if the expected frequency is sufficiently high we eventually reach a probability like .9999999 which we simply round off as 1.0.
HOW TO FIND THE CHANCE OF A TERM APPEARING AT LEAST ONCE*
1. FIND PROBABILITY IT DOES NOT OCCUR BY POISSON EQUATION.
x (-lambda)
f(x) = Lambda e x = 0; lambda = expected frequency per matrix
x!
2. 1 ‑ f(0) = THE PROBABILITY OF OCCURRING AT LEAST ONCE.
(where f(0) = the probability it will not occur)
3. On an Excel or Works spreadsheet, head columns as follows: A: Whatever identifies the calculation; B: Skips Used on the Matrix, C: Number of hits (on CodeFinder or similar software) in Skip Range; D: Divide by 304,805 Letters in Torah or Control; E: The Quotient Equals Frequency Per letter; F: E Quotient Multiplied by Letters on Matrix = Word Expectancy; G: Poisson Equation = 1-EXP(-F#) where # equals the row number of the item in Column F in question on the spreadsheet. If you want to know the chance for the item to be on the matrix, head Column H accordingly. The value of Column H will be the reciprocal of the value found in Column G by Poisson Equation.
* Note: While this author (Barry S. Roffman) discovered the Roffman Skip Formula, my son (an MIT geophysics graduate), Rabbi Robert Roffman, is the author of the spreadsheets and the man who first introduced use of the Poisson Equation into my research.
ROW SPLIT AND WRAPPED MATRICES
There is some indication that when a row skip or row split function for the axis term is employed, that the true value of an open text match must be the value computed by standard means divided by the row spit. The lowest ELS of Ark of the Covenant at skip -306 (cylinder circumference 306 letters) had about one chance in 2,931 of being in a 104-letter matrix with Egyptians were burying. At skip -306 there was no row split function enabled on CodeFinder. Had it been enabled and a row split of 2 were used (with a cylinder circumference of 153 letters), if the matrix size (area) were the same, I would have divided 2,931 by 2 to arrive at a value of 1 chance in 1,465. In this case however, the matrix with circumference 153 would have been larger because the match on the matrix with circumference 306 was already about as tight as it could be with the row skip function disabled. There is also a discussion about dividing the value of a matrix by the number of passes through the Torah made by CodeFinder on a wrapped (rounded torus) search before acquiring an axis term. See the permutation experiment
SPECIAL CASE SKIPS
Finally, when computing the value of a-priori open text terms on a matrix, it is my practice to only employ the frequency of this term at Skip +1 (in unwrapped Torah) on my spreadsheet in column C. However, if the a priori term appears at skip -1, N (parallel to, in the same direction, and at the skip of the axis term), or -N (parallel to, in the opposite direction, and at the skip of the term) in column C, I list the frequency (number of hits) as the total hits at skips +1, -1, N, and -N (with wrapped Torah allowed if that was required to find the axis term. These skips are considered special because they seem to leap out at the eye of the researcher and make the case for deliberate encoding seem more plausible. |
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# Determine the minimum value of current gain $\beta$ (up to nearest integer) required to put the transistor in saturation when ${V_{BB}} = + 5\,V$ . Assume that in saturation state, ${V_{BE}} = 0.8\,V$ and ${V_{CE}} = 0.12\,V$.
Last updated date: 21st Jun 2024
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Hint: In this problem we need to find the value of current gain. Current gain is the ratio of the collector current to the base current. In order to find the collector current and the base current, apply Kirchhoff’s voltage law in the base-emitter loop and in the collector loop.
Complete step by step solution:
Let us understand current gain and voltage gain first. The current gain in the common-base configuration is defined as the change in collector current divided by the change in emitter current, when base-to-collector voltage is constant. Similarly, the voltage gain for the common base amplifier is the ratio of output voltage to the input voltage.
Let ${I_B}$ and ${I_C}$ be base current and collector current respectively, taking $+ 5V$ as input voltage ${V_{in}}$ , the summation of voltages in the base-emitter loop gives
$8 \times {10^3} \times {I_B} + {V_{BE}} - 5V = 0$
We assume that the transistor is in saturation, this implies
${V_{BE}} = {V_{BE(sat)}} = 0.8V$
$\Rightarrow 8 \times {10^3} \times {I_B} + 0.8V - 5V = 0$
$$\Rightarrow {I_B} = \dfrac{{4.2V}}{{80 \times {{10}^3}}} = 0.0525mA$$ --equation $$1$$
The base current is $$0.0525mA$$
Applying Kirchhoff’s voltage law in the collector loop, we have
$$\Rightarrow 5 \times {10^3} \times {I_C} + {V_{CE}} - 12\,V = 0$$
We assume that the transistor is in saturation, this implies
${V_{CE}} = {V_{CE(sat)}} = 0.12V$
Substituting this value in the above equation, we get
$$\Rightarrow 5 \times {10^3} \times {I_C} + 0.12V - 12\,V = 0$$
$$\Rightarrow {I_C} = \dfrac{{12 - 0.12}}{{5 \times {{10}^3}}} = 2.376mA$$
The emitter current is $$2.376mA$$
The current gain is given as
$\beta = \dfrac{{{I_c}}}{{{I_B}}}$
Substituting the values, we get
$$\beta = \dfrac{{2.376}}{{0.0525}} \approx 45.25$$
The minimum value of current gain $\beta$ (up to nearest integer) is $$45$$.
Note: In saturation mode, the transistor acts like a short circuit between the collector and the emitter. In saturation mode both the diodes in the transistor are forward biased. Be careful of the sign conventions when applying Kirchhoff’s voltage law. Remember that the current gain is the ratio of the collector current to the base current. |
# Class 8 RD Sharma Solutions – Chapter 17 Understanding Shapes Special Types Of Quadrilaterals – Exercise 17.1 | Set 2
• Last Updated : 16 Apr, 2021
### Question 16. The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:
Given that,
Perimeter of the parallelogram = 150 cm
Let us assume that one of the sides as = ‘x’ cm
and other side as = (x + 25) cm
As we know that opposite sides of a parallelogram are parallel and equal.
Therefore, Perimeter = Sum of all sides
x + x + 25 + x + x + 25 = 150
4x + 50 = 150
4x = 150 – 50
x = 100/4 = 25
Hence, Sides of the parallelogram are (x) = 25 cm and (x+25) = 50 cm.
### Question 17. The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.
Solution:
Given that,
Shorter side of the parallelogram = 4.8 cm
and longer side of the parallelogram = 4.8 + 4.8/2 = 4.8 + 2.4 = 7.2cm
As we know that opposite sides of a parallelogram are parallel and equal.
Therefore, Perimeter = Sum of all sides
Perimeter of the parallelogram = 4.8 + 7.2 + 4.8 + 7.2 = 24cm
Hence, Perimeter of the parallelogram is 24 cm.
### Question 18. Two adjacent angles of a parallelogram are (3x-4)o and (3x+10)°. Find the angles of the parallelogram.
Solution:
As we know that adjacent angles of a parallelogram are equal.
Therefore, (3x – 4)° + (3x + 10)° = 180°
3x° + 3xo – 4 + 10 = 180°
6x = 180° – 6°
x = 174°/6 = 29°
The adjacent angles are,
(3x – 4)° = 3×29 – 4 = 83°
(3x + 10)° = 3×29 + 10 = 97°
As we know that Sum of adjacent angles = 180°
Hence, each angle is 83°, 97°, 83°, 97°.
### Question 19. In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC =30°, ∠BDC= 10° and ∠CAB =70°. Find:
∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC, and ∠DBA.
Solution:
Given that,
∠ABC = 30°,
∠ABC = ∠ADC = 30° (We know that measure of opposite angles are equal in a parallelogram)
∠BDC = 10°
∠CAB =70°
∠BDA = ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20° (From figure we concluded)
∠DAB = 180° – 30° = 150°
∠ADB = ∠DBC = 20° (alternate angles)
∠BCD = ∠DAB = 150° (we know, opposite angles are equal in a parallelogram)
∠DBA = ∠BDC = 10° (we know, Alternate interior angles are equal)
In ΔABC ∠CAB + ∠ABC + ∠BCA = 180° (since, sum of all angles of a triangle is 180°)
70° + 30° + ∠BCA = 180°
∠BCA = 180° – 100° = 80°
∠DAB = ∠DAC + ∠CAB = 70° + 80° = 150°
∠BCD = 150° (opposite angle of the parallelogram)
∠DCA = ∠CAB = 70°
In ΔDOC ∠BDC + ∠ACD + ∠DOC = 180° (since, sum of all angles of a triangle is 180°)
10° + 70° + ∠DOC = 180°
∠DOC = 180°- 80°
∠DOC = 100°
Therefore, ∠DOC = ∠AOB = 100° (Vertically opposite angles are equal)
∠DOC + ∠AOD = 180° [Linear pair]
100° + ∠AOD = 180°
∠AOD = 180°- 100°
∠AOD = 80°
Therefore, ∠AOD = ∠BOC = 80° (Vertically opposite angles are equal)
∠CAB = 70o
∠ABC + ∠BCD = 180° (In a parallelogram sum of adjacent angles is 180°)
30° + ∠ACB + ∠ACD = 180°
30° + ∠ACB + 70° = 180°
∠ACB = 180° – 100°
∠ACB = 80°
Hence ∠DAB = 150o, ∠ADC = 30°, ∠BCD = 150°, ∠AOD = 80°, ∠DOC = 100°, ∠BOC = 80°, ∠AOB = 100°, ∠ACD = 70°, ∠CAB = 70°, ∠ADB = 20°, ∠ACB = 80°, ∠DBC = 20°, and ∠DBA = 10°.
### Question 20. Find the angles marked with a question mark shown in Figure.
Solution:
In ΔBEC ∠BEC + ∠ECB +∠CBE = 180° (Sum of angles of a triangle is 180°)
90° + 40° + ∠CBE = 180°
∠CBE = 180°-130°
∠CBE = 50°
∠CBE = ∠ADC = 50° (Opposite angles of a parallelogram are equal)
∠B = ∠D = 50° (opposite angles of a parallelogram are equal)
∠A + ∠B = 180° (Sum of adjacent angles of a triangle is 180°)
∠A + 50° = 180°
∠A = 180°-50°
therefore, ∠A = 130°
In ΔDFC ∠DFC + ∠FCD +∠CDF = 180° (Sum of angles of a triangle is 180°)
90° + ∠FCD + 50° = 180°
∠FCD = 180°-140°
∠FCD = 40°
∠A = ∠C = 130° (Opposite angles of a parallelogram are equal)
∠C = ∠FCE +∠BCE + ∠FCD
∠FCD + 40° + 40° = 130°
∠FCD = 130° – 80°
∠FCD = 50°
Hence ∠EBC = 50°, ∠ADC = 50° and ∠FCD = 50°.
### Question 21. The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram, is 60°. Find the angles of the parallelogram.
Solution:
Let us consider parallelogram ABCD, where DP⊥ AB and DQ ⊥ BC.
Given that ∠PDQ = 60°
In quadrilateral DPBQ ∠PDQ + ∠DPB + ∠B + ∠BQD = 360° (Sum of all the angles of a Quadrilateral is 360°)
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
∠B = 120°
∠B = ∠D = 120° (Opposite angles of parallelogram are equal)
∠B + ∠C = 180° (Sum of adjacent interior angles in a parallelogram is 180°)
120° + ∠C = 180°
∠C = 180° – 120° = 60°
∠A = ∠C = 60° (Opposite angles of parallelogram are equal)
Hence, Angles of a parallelogram are 60°, 120°, 60°, 120°
### Question 22. In Figure, ABCD and AEFG are parallelograms. If ∠C =55°, what is the measure of ∠F?
Solution:
From figure, we conclude that,
In parallelogram ABCD ∠C = ∠A = 55° (In a parallelogram opposite angles are equal in a parallelogram)
In parallelogram AEFG ∠A = ∠F = 55° (In a parallelogram opposite angles are equal in a parallelogram)
Hence ∠F = 55°
### Question 23. In Figure, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?
Solution:
From figure, we conclude that,
In parallelogram BDEF BD = EF (In a parallelogram opposite sides are equal)
In parallelogram DCEF DC = EF (In a parallelogram opposite sides are equal)
Since, BD = EF = DC
Therefore, BD = DC
### Question 24. In Figure, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not?
Solution:
From figure we conclude that,
In parallelogram BDEF BD = EF and BF = DE (opposite sides are equal in a parallelogram)
In parallelogram DCEF DC = EF and DF = CE (opposite sides are equal in a parallelogram)
In parallelogram AFDE AF = DE and DF = AE (opposite sides are equal in a parallelogram)
therefore, DE = AF = BF
similarly, DF = CE = AE
Given that, DE = DF
Since, DF = DF
AF + BF = CE + AE
AB = AC
Hence, ΔABC is an isosceles triangle.
### Question 25. Diagonals of parallelogram ABCD intersect at O as shown in Figure. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ΔBOY = ΔDOX
Now, state if XY is bisected at O.
Solution:
(i) OB = OD
OB = OD. Since diagonals bisect each other in a parallelogram.
(ii) ∠OBY =∠ODX
∠OBY =∠ODX. Since alternate interior angles are equal in a parallelogram.
(iii) ∠BOY= ∠DOX
∠BOY= ∠DOX. Since vertical opposite angles are equal in a parallelogram.
(iv) ΔBOY ≅ ΔDOX
ΔBOY and ΔDOX. Since OB = OD, where diagonals bisect each other in a parallelogram.
∠OBY =∠ODX (Alternate interior angles are equal)
∠BOY= ∠DOX (Vertically opposite angles are equal)
ΔBOY ≅ΔDOX (by ASA congruence rule)
OX = OY (Corresponding parts of congruent triangles)
Hence XY is bisected at O.
### Question 26. In Fig., ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same:
(i) ∠A = ∠C
(ii) ∠FAB = ½ ∠A
(iii) ∠DCE = ½ ∠C
(iv) ∠CEB = ∠FAB
(v) CE ∥ AF
Solution:
(i) ∠A = ∠C
True, Since ∠A =∠C = 55° [opposite angles are equal in a parallelogram]
(ii) ∠FAB = ½ ∠A
True, Since AF is the angle bisector of ∠A.
(iii) ∠DCE= ½ ∠C
True, Since CE is the angle bisector of angle ∠C.
(iv) ∠CEB= ∠FAB
True, Since ∠DCE = ∠FAB (opposite angles are equal in a parallelogram).
∠CEB = ∠DCE (alternate angles)
½ ∠C = ½ ∠A [AF and CE are angle bisectors]
(v) CE || AF
True, since one pair of opposite angles are equal, therefore quad. AEFC is a parallelogram.
### Question 27. Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM? Why or why not?
Solution:
Given that,
AL and CM are perpendiculars on diagonal BD.
In ΔAOL and ΔCOM,
∠AOL = ∠COM (vertically opposite angle) ———–(i)
∠ALO = ∠CMO = 90° (each right angle) ——–(ii)
By using angle sum property,
∠AOL + ∠ALO + ∠LAO = 180° ———(iii)
∠COM + ∠CMO + ∠OCM = 180° ———- (iv)
From (iii) and (iv)
∠AOL + ∠ALO + ∠LAO = ∠COM + ∠CMO + ∠OCM
∠LAO = ∠OCM (from (i) and (ii))
In ΔAOL and ΔCOM
∠ALO = ∠CMO (each right angle)
AO = OC (diagonals of a parallelogram bisect each other)
∠LAO = ∠OCM (proved)
therefore, ΔAOL is congruent to ΔCOM
Hence AL = CM (Corresponding parts of congruent triangles)
### Question 28. Points E and F lie on diagonals AC of a parallelogram ABCD such that AE = CF. what type of quadrilateral is BFDE?
Solution:
From figure, we conclude that,
In parallelogram ABCD,
AO = OC ———- (i) (Diagonals of a parallelogram bisect each other)
AE = CF ———-(ii) Given
On subtracting (ii) from (i)
AO – AE = OC – CF
EO = OF ——-(iii)
In ΔDOE and ΔBOF,
EO = OF (proved)
DO = OB (Diagonals of a parallelogram bisect each other)
∠DOE = ∠BOF (vertically opposite angles are equal in a parallelogram)
By SAS congruence ΔDOE ≅ ΔBOF
therefore, DE = BF (Corresponding parts of congruent triangles)
In ΔBOE and ΔDOF,
EO = OF (proved)
DO = OB (diagonals of a parallelogram bisect each other)
∠DOF = ∠BOE (vertically opposite angles are equal in a parallelogram)
By SAS congruence ΔDOE ≅ ΔBOF
Hence, DF = BE (Corresponding parts of congruent triangles).
Hence, BFDE is a parallelogram, since one pair of opposite sides are equal and parallel.
### Question 29. In a parallelogram ABCD, AB = 10cm, AD = 6 cm. The bisector of ∠A meets DC in E, AE and BC produced meet at F. Find the length CF.
Solution:
From figure we concluded that,
In a parallelogram ABCD,
Given, AB = 10 cm, AD = 6 cm
CD = AB = 10 cm and AD = BC = 6 cm (In a parallelogram opposite sides are equal)
AE is the bisector of ∠DAE = ∠BAE = x
∠BAE = ∠AED = x (alternate angles are equal)
ΔADE is an isosceles triangle. Since opposite angles in ΔADE are equal.
AD = DE = 6cm (opposite sides are equal)
CD = DE + EC
EC = CD – DE = 10 – 6 = 4cm
∠DEA = ∠CEF = x (vertically opposite angle are equal)
∠EAD = ∠EFC = x (alternate angles are equal)
ΔEFC is an isosceles triangle. Since opposite angles in ΔEFC are equal.
CF = CE = 4cm (opposite side are equal to angles)
Hence CF = 4cm.
My Personal Notes arrow_drop_up |
# Helping a Struggling Maths Student: Pascal’s Triangle, Magic Triangles & Number Bond Games
Practising Number Bonds using Pascal’s Triangle
I thought I would have a fun frivolous lesson as both L10 and C10 seem to be handling any number bond related question I threw their way. Pascal’s triangle is a triangle of numbers made up of lots of number bonds:
The whole triangle fascinates me and I knew it would feature heavily in our future maths so I thought now would be an ideal time to introduce it. This would not be an in-depth lesson looking at all the patterns. That would come later. No, this would simply be an introduction to the fun elements of maths. The one I gave them was much larger than the one above and went into much larger 3 digit numbers. My goal was for the girls to look at a partially filled in one and see the number bond pattern for themselves. The reality was they began to see all the other patterns first. However, they picked up a pattern they assumed was correct but in fact wasn’t. That was my chance to ask them to have a look for any number bonds. It didn’t take long. I then asked them to disprove the previous pattern they had been so sure of by filling in the rest of the numbers in Pascal’s triangle. As the numbers got larger, I could see L10 beginning to use her fingers to count. I hadn’t seen her do that all summer, so asked why? She was struggling to do the larger numbers in her head. I used this as an opportunity to give her some mental arithmetic strategies, some we had covered before others were new.
Mental Arithmetic Strategies
1. Count from largest to smallest amounts ie Add the hundreds first, then the tens then the units (this is completely counter to what she would normally do on paper)
2. Think in terms of placement, carrying if necessary. So 367 +596 becomes 800+150+13 which is easier to calculate: 950+13 then 963
3. Adding large numbers is linked intrinsically to number bonds of ten and another method is to break down and put back together the numbers being added. Using the same example above:
367+596=
300+500+60+40+50+7+3+3=
800+100+50+10+3=
900+60+3=
963
On paper these look cumbersome, but with instruction and practice one’s mind can do these almost imperceptibly in a matter of seconds. I am a very visual learner, so in my head the numbers split up and re-join, dancing about, with very little effort required from me. I am certain this would have taken time and effort to achieve when first learning mental arithmetic. In teaching the girls these strategies and discouraging the use of finger counting, my goal was to begin the process of training their own brains to calculate without much thought. Funnily enough C10 seems to do this naturally and doesn’t have too much of an issue with mental calculations. For L10 it is, as everything mathematical seems to be, very difficult. That said, she found calculating the smaller numbers much easier than she did five short weeks ago, so we are moving in the right direction.
Practising number bonds by using the Magic triangle
This is a fun activity whereby an equilateral triangle is cut and counters with 1-6 written are placed one in each corner and one in the middle of each triangle side. The goal is to find a combination of numbers whereby the three numbers on each side of the triangle add up to the same as the other sides:
Progressively larger triangles can be used, increasing the difficulty of the maths required to find the answer.
We started off simply, using equate tiles and wooden pencils for the numbers and triangle outlines. I was pleased by how easy the girls found it:
Number Bond Games
We had as family started to play games to reinforce all I was teaching. Their success in our family came down to one thing. If mummy or daddy were playing, it was seen as a great family games night. If they weren’t involved, whilst there were no complaints, there was definitely less excitement. As for Gary and I? Our number bonds are coming along nicely, thank you very much!!
• Totally Tut
This is a gentle game for 2-4 players and practises the four operations. For now we just used it for addition and subtraction.
• Equate
We are huge fans of Scrabble in this household, so to find what is effectively a number Scrabble was quite exciting. This is a maths facts game which can include fractions, decimals, whole numbers, addition, subtraction, multiplication and division. It is easy to limit the facts practice to just whole numbers and addition and subtraction, which is what we are doing at the moment. The rules are easy to follow and if you know how to play word Scrabble this is very similar.
• 21
This is a simple card game, which a friend taught us, and requires the individual to make up to a value of 21 with two or more cards. Great for practicing addition facts. You can find the rules here. I am sure there are lots of other card games but we are not a family to play many card games so we are learning them as we go along or someone tells us about them.
• Sudoku
This is fairly new to our home-school, and while it doesn’t require adding or subtraction, it is number logic at its most simple, ensuring the children are playing with numbers and therefore increasing in confidence using them. This website is particularly useful as it has a clear set of rules and tips as well as providing a daily printable Sudoku puzzle just for children
I intend to print one each day for all of the older children to do when they have time on their hands, if they finish an activity early:
We have already moved onto Place Value, which we are thoroughly enjoying. I’ll be sharing that next week.
## 15 comments
1. Maths looks like a lot of fun in your household these days! 🙂
2. I love things like Pascal’s triangle for teaching maths ideas. They give a great flavour of the beauty of maths.
Have you ever come across Lockhart’s Lament? http://worrydream.com/refs/Lockhart-MathematiciansLament.pdf
I found it absolutely fascinating and perspective-shifting. I wonder if even L10 might get something from reading parts of it.
1. Thanks Lucinda. I’m half way through Elephant in a classroom. Brilliant book!
3. I get frustrated by Sudoku because I don’t want to sit there and puzzle out the answers, that’s so bad of me……
1. The children are really enjoying the challenge! I know what you mean though, I’m not sure I would ever do one just for fun!
4. Beenie says:
The magic triangle was class! I think we’ll give that a go.
1. They’re great because you can alter the difficulty level so easily.
5. We’re big Sodoku fans in our house. I’d love to try that Equate game, what a good find!
1. We wouldn’t naturally be a gaming family, although the children really enjoy them. Gary and I have to constantly remind ourselves to MAKE AN EFFORT!!
6. Fun, Fun, Fun. 🙂
1. Ha! It’s certainly an improvement on sitting to do copious numbers of sums each day!
1. Ooh, thank you Denise! I shall check that out straight away. I just love your visions for maths, so thank you so much for sharing!
7. Great post! I like your new approach to math and hope to try several of your ideas with my kids this fall. |
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# Exploring Fractions
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### Exploring Fractions
1. 1. Exploring Fractions<br />Susan Muir <br />K-4 Math Coach<br />January 29th , 2010<br />
2. 2. Discuss with your elbow partner, instances in your professional life where “one half” has significance.<br />Instructional Strategies: Accessing Prior Knowledge through Connections<br /> The Meaning of One Half<br />
3. 3. <ul><li>What is a Fraction?
4. 4. Fraction Principles- Exploration and Discussion
5. 5. Fraction Quantity
6. 6. Exploring Part-Whole Relationships
7. 7. Let’s Experience a Problem
8. 8. Where Do I Go From Here?</li></ul>Agenda<br />
9. 9. Fractions are used frequently in real life situations.<br />Fractions are all Around Us<br />
10. 10. 1.A fraction has a numerator and a denominator.<br /><ul><li>The denominator tells the total number of equal parts in a whole, and the numerator tells the number of parts accounted for.</li></ul> Fraction of a Set<br />3 of the 4 balls are soccer balls.3/4 are soccer balls.<br />Fraction Principles<br />
11. 11. 2.You have to know what the whole is to say what the fraction is.<br /><ul><li>Fractions represent parts of the whole.
12. 12. As the whole changes, so does the fraction.</li></ul>Fraction Principles<br />
13. 13. 3. The equal parts into which the whole is divided are equal but do not have to be identical.<br />4. Fraction parts do not have to be adjacent.<br />Fraction Principles<br />
14. 14. 5. The parts of a set or group do not have to be identical.<br />6. If the numerator and denominator of a fraction are equal, the fraction represents one whole, or 1. Therefore the whole number can be represented as fractions.<br /><ul><li>2/2, 3/3, 4/4, and so on.</li></ul>Fraction Principles<br />
15. 15. 7. Fractions have more than one name.<br />8. Fractions with a numerator greater than their denominator are greater than 1.<br /><ul><li>7/4 is greater than 1</li></ul>Fraction Principles<br />
16. 16. Fraction Quantity<br />Area Models<br />Set Models<br />Linear Models<br />
17. 17. Exploring Part Whole Relationships<br />Counting Activities<br />Trading Activities<br />Find the Whole/Part/Fraction<br />
18. 18. Three Part Lesson Model<br /><ul><li>Before- Gains Background Knowledge
19. 19. During-Explore Activity
20. 20. After- Show and Share/ Connect and Reflect</li></ul> Let’s Experience a Problem<br />
21. 21. <ul><li>Learning Materials</li></ul>http://hoodamath.com/games/fractioneaters.php<br />fraction game<br />http://nces.ed.gov/nceskids/createagraph/<br />graph maker<br />http://themathworksheetsite.com/numline.html<br />Number line generator<br />http://www.mrnussbaum.com/tonyfraction.htm<br />Pizza Shop ( Michelle Morley’s Website)<br />http://www.curriculumsupport.education.nsw.gov.au/secondary/mathematics/years7_10/teaching/frac.htm<br />Fraction website with printable resources- using pattern blocks to teach fractions<br /><ul><li>http://www.eworkshop.on.ca/edu/core.cfm
22. 22. Ontario Online Teaching Resource
23. 23. Planning Outcome-based Lessons
24. 24. Learning More About Teaching Fractions</li></ul> Workshop Wednesdays<br /> Where Do I Go From Here?<br /> |
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# Classroom Case Studies, Grades K-2 Part D: Critiquing Student Lessons (20 minutes)
## Session 10: K-2, Part D
In the K-2 curriculum, students are frequently asked to think about patterns, but often their “pattern sniffing” skills end with simply finding the next object. Note 7
Here is a “pattern” problem.
Subtract, then look at the differences in each row.
Continue the pattern.
### Problems
Problem D1
What algebraic ideas are in this lesson?
Problem D2
How are patterns used in this lesson?
Problem D3
What mathematics do you think students would learn from this lesson?
Problem D4
Are there any misconceptions that students might develop from this lesson?
Problem D5
How would you modify the problem, or what additional questions might you ask, to incorporate the framework for analyzing patterns?
### Notes
Note 7
You may be puzzled by this series of subtraction problems, as well you should! After working on Problems D1-D5, you may notice that in certain problems — where students “see” the pattern and then mindlessly fill in numbers that satisfy the pattern — the mathematical thinking and reasoning gets lost.
### Solutions
Problem D1
Problem D2
Answers will vary. The patterns shown only relate to the answers. For the first row, the answers increase by 10, but the problems themselves are not directly related. The second row’s answers are consecutive multiples of 11, but the problems are not related. The third row’s answers decrease by one with unrelated problems, and the fourth row’s answers decrease by five with unrelated problems.
Problem D3
Answers will vary. Students should strengthen their subtraction with regrouping skills.
Problem D4
Answers will vary. Students could learn to look for patterns in answers without actually looking at the problems. The answers may not have any pattern.
Problem D5
Answers will vary. One way to modify the problem would be to make the problems relate to the pattern of the answers. For example, the first row could show 63 – 55, 63 – 45, 63 – 35, 63 – 25, 63 – 15, then ask what comes next. |
# How to Solve Quadratic Equations for Bank PO | Know Here
An equation of the form ax2+bx+c=0 is called quadratic equation where a= coefficient of x2, b= coefficient of x and c= constant. A Quadratic Equation ax2+bx+c=0 always have two solutions.
Discriminant Method to find the Value of x:
x = (-b±√(b^2-4ac))/2a
Eg- 3x2+14x+8=0
Here a=3, b=14, c=8
x = (-b±√(b^2-4ac))/2a
(-14±√(14^2-4✕3✕8))/(2✕3)
(-14±√(196-96))/6
=(-14±√(100))/6
=(-14±10)/6
Take +ve sign, x=(-14+10)/6 = -4/6 = -0.226
Take -ve sign , x=(-14-10)/6 = -24/6 = -4
Factorization Method to find the value of x:
ax2+bx+c=0
acx2 = Multiplication of factor bx which must result bx on addition
Eg- 3x2+14x+8=0
Here a=3, b=14, c=8
acx2=8✕3x2 = 24x2
Now see the factor of 14x which gives 24x2 i.e. 12x✕2x
You also need to see that addition of the factor of 14x must result 14x and multiplication mus result 24x2
Here factor of 14x are 12x and 2x
On Multiplication = 12x✕2x = 24x2
Now, 3x2+14x+8=0
3x2+12x+2x+8=0
3x(x+4)+2(x+4)=0
(3x+2)(x+4)=0
Now, Take 3x+2=0, x= -2/3 and x+4=0, x=-4
Hence, the values of x are -2/3 and -4.
Q1. Solve the Equations given below and Mark Answer given:
I. 3x2+14x+8=0
II. 4y2+16y+7=0
1. x>y
2. x<y
3. x≥y
4. x≤y
5. x=y or No relationship can be establish between x and y
Explanations:
I. 3x2+14x+8=0
3x2+12x+2x+8=0
3x(x+4)+2(x+4)=0
(3x+2)(x+4)=0
Now, Take 3x+2=0, x= -2/3 and x+4=0, x=-4
Hence, the values of x are -2/3=-0.226 and -4.
II. 4y2+16y+7=0
4y2+14y+2y+7=0
2y(2y+7)+1(2y+7)=0
(2y+1)(2y+7)=0
Now, Take 2y+1=0, y= -1/2 and 2y+7=0, y=-7/2
Hence, the values of y are -1/2=-0.5 and -7/2=-3.5
Now, x=-0.226,-4 and y=-0.5,-3.5
Compare Now,
-0.226>-0.5 it means x>y _____________(1)
-0.226>-3.5 it means x>y ______________(2)
-4<-0.5 it means x<y __________________(3)
-4<-3.5 it means x<y ___________________(4)
> and < never occur together, so we can not establish any relationship between x and y, Hence, Option(5) is correct.
Q2. Solve the Equations given below and Mark Answer given:
I. x2=√196
II. y=√196
1. x>y
2. x<y
3. x≥y
4. x≤y
5. x=y or No relationship can be establish between x and y
Explanations:
I. x2=√196
x= ±14 means x=+14, -14
II. y=√196
y=14
Compare Now,
+14=+14 it means x=y _____________(1)
-14<+14 it means x<y ______________(2)
On combining, (1) and (2), we get x≤y, Hence Option(4) is the correct answer.
[Note: We take ± sign only when we make square root if square root is already made in any specific sign then we take only the sign in which its given.]
x= x1=First Obtained Value of x, x2= Second Obtained Value of x
y= y1=First Obtained Value of y, y2= Second Obtained Value of y
#01. Comparison Condition when we need to select Answer x>y.
All the four equations must show x>y.
First Obtained Value of x>First Obtained Value of y means x>y _____________(1)
First Obtained Value of x>Second Obtained Value of y means x>y ___________(2)
Second Obtained Value of x>First Obtained Value of y means x>y ____________(3)
Second Obtained Value of x>Second Obtained Value of y means x>y___________(4)
#02. Comparison Condition when we need to select Answer x<y.
All the four equations must show x<y.
First Obtained Value of x<First Obtained Value of y means x<y _____________(1)
First Obtained Value of x<Second Obtained Value of y means x<y ___________(2)
Second Obtained Value of x<First Obtained Value of y means x<y ____________(3)
Second Obtained Value of x<Second Obtained Value of y means x<y___________(4)
#03. Comparison Condition when we need to select Answer x=y.
All the four equations must show x=y.
First Obtained Value of x=First Obtained Value of y means x=y _____________(1)
First Obtained Value of x=Second Obtained Value of y means x=y ___________(2)
Second Obtained Value of x=First Obtained Value of y means x=y ____________(3)
Second Obtained Value of x=Second Obtained Value of y means x=y___________(4)
#04. Comparison Condition when we need to select Answer x≥y.
When at least on of four equations show x=y and rest show x>y or When at least on of four equations show x>y and rest show x=y.
#05. Comparison Condition when we need to select Answer x≤y.
When at least on of four equations show x=y and rest show x<y or When at least on of four equations show x<y and rest show x=y.
#06. Comparison Condition when we need to select Answer "No relationship can be establish between x and y".
When at least on of four equations show x>y and also When at least on of four equations show x<y. |
PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for A Circular Metal Plate Expends Under Heating So that Its Radius Increases by K%. Find the Approximate Increase in the Area of the Plate, If the Radius of the Plate before Heating is 10 Cm. - PUC Karnataka Science Class 12 - Mathematics
#### Question
A circular metal plate expends under heating so that its radius increases by k%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm.
#### Solution
Let at any time, x be the radius and y be the area of the plate.
$\text { Then,}$
$y = x^2$
$\text { Let ∆ x be the change in the radius and }\bigtriangleup y \text { be the change in the area of the plate }.$
$\text { We have }$
$\frac{∆ x}{x} \times 100 = k$
$\text { When }x = 10,\text { we get }$
$∆ x = \frac{10k}{100} = \frac{k}{10}$
$\text { Now,} y = \pi x^2$
$\Rightarrow \frac{dy}{dx} = 2\pi x$
$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 10 cm} = 20\pi {cm}^2 /cm$
$\therefore ∆ y = dy = \frac{dy}{dx}dx = 20\pi \times \frac{k}{10} = 2k\pi \ {cm}^2$
Hence, the approximate change in the area of the plate is 2
$\pi$ cm2 .
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [2]
Solution A Circular Metal Plate Expends Under Heating So that Its Radius Increases by K%. Find the Approximate Increase in the Area of the Plate, If the Radius of the Plate before Heating is 10 Cm. Concept: Approximations.
S |
In compound interest, the interest for each period is added to the principle before interest is calculated for the next period. With this method the principle grows as the interest is added to it. This method is mostly used in investments such as savings account and bonds.
To understand compound interest clearly, let’s take an example.
1000 is borrowed for three years at 10% compound interest. What is the total amount after three years?
You can understand the process of compound interest by image shown below.
Year Principle Interest (10%) Amount 1st 1000 100 1100 2nd 1100 110 1210 3rd 1210 121 1331
## Difference between Simple Interest and compound interest
After three years,
In simple interest, the total amount would be 1300
And in compound interest, the total amount would be 1331.
## Basic Formulas of Compound Interest
If A = Amount
P = Principle
C.I. = Compound Interest
T = Time in years
R = Interest Rate Per Year
## Shortcut Formulas for Compound Interest
Rule 1: If rate of interest is R1% for first year, R2% for second year and R3% for third year, then
Example
Find the total amount after three years on Rs 1000 if the compound interest rate for first year is 4%, for second year is 5% and for third year is 10%
Sol:
P = 1000
R1 = 4%, R2 = 5% and R3 = 10%
(From the table given at the bottom of the page)
A = 1201.2
Rule 2:
If principle = P, Rate = R% and Time = T years then
1. If the interest is compounded annually:
2. If the interest is compounded half yearly (two times in year):
3. If the interest is compounded quarterly (four times in year):
Example
Find the total amount on 1000 after 2 years at the rate of 4% if
1. The interest is compounded annually
2. The interest is compounded half yearly
3. The interest is compounded quarterly.
Sol:
Here P = 1000
R = 4%
T = 2 years
If the interest is compounded annually
(From the table given at the bottom of the page)
A = 1081.6
If the interest is compounded half yearly
A = 1082.4
If the interest is compounded quarterly
A = 1082.9
Rule 3: If difference between Simple Interest and Compound Interest is given.
• If the difference between Simple Interest and Compound Interest on a certain sum of money for 2 years at R% rate is given then
Example
If the difference between simple interest and compound interest on a certain sum of money at 10% per annum for 2 years is Rs 2 then find the sum.
Sum:
• If the difference between Simple Interest and Compound Interest on a certain sum of money for 3 years at R% is given then
Example
If the difference between simple interest and compound interest on a certain sum of money at 10% per annum for 3 years is Rs 2 then find the sum.
Sol:
Rule 4: If sum A becomes B in T1 years at compound interest, then after T2 years
Example
Rs 1000 becomes 1100 after 4 years at certain compound interest rate. What will be the sum after 8 years?
Sum:
Here A = 1000, B = 1100
T1 = 4, T2 = 8
Look up Table
Simple Interest |
What is the first step in solving In(x – 1) = In6 – Inx for x?
Question
What is the first step in solving In(x – 1) = In6 – Inx for x?
in progress 0
2 weeks 2021-09-10T15:10:01+00:00 2 Answers 0
The first step would to be use quotient rule.
3
Step-by-step explanation:
ln(x-1)=ln(6)-ln(x)
The first step would to be use quotient rule there on the right hand side:
ln(x-1)=ln(6/x)
*Quotient rule says ln(a/b)=ln(a)-ln(b).
Now that since we have ln(c)=ln(d) then c must equal d, that is c=d.
ln(x-1)=ln(6/x)
implies
x-1=6/x
So you want to shove a 1 underneath the (x-1) and just cross multiply that might be easier.
Cross multiplying:
Multiplying/distribute[/tex]
Subtract 6 on both sides:
Now this is not too bad to factor since the coefficient of x^2 is 1. All you have to do is find two numbers that multiply to be -6 and add up to be -1.
These numbers are -3 and 2 since -3(2)=-6 and -3+2=-1.
So the factored form of our equation is
This implies that x-3=0 or x+2=0.
So solving x-3=0 gives us x=3 (just added 3 on both sides).
So solve x+2=0 gives us x=-2 (just subtracted 2 on both sides).
We need to see if these are actually the solutions by plugging them in.
Just a heads up: You can’t do log(negative number).
Checking x=3:
ln(3-1)=ln(6)-ln(3)
ln(2)=ln(6/3)
ln(2)=ln(2)
This is true.
Checking x=-2:
ln(-2-1)=ln(6)-ln(-2)
ln(-3)=ln(6)-ln(-2)
We don’t need to go further -2 makes the inside of our logarithms negative above.
The only solution is 3.
2. Step-by-step explanation:
Solution: |
The Decimal and Binary Number Systems
# The Decimal and Binary Number Systems
Before we start looking into the more involved stuff in the Numerical Analysis section, we must first establish a greater understanding of the decimal number system and the binary number system. We will first secure a foundation of the decimal number system which we are all too familiar with.
## The Decimal Number System
Definition: The Decimal Number System is a number system for which every real number $x$ can be written in terms of the ten digits $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$ as the sum of powers of $10$. A number in the decimal number system is said to be of base $10$ and to specify this we attach a subscript $10$ to $x$, written $(x)_{10}$.
We will specify the base of a number for the rest of this page and after that, we will will only specify the base of a number when the context is unclear.
For example, the number $(1539)_{10}$ is a decimal number, and we can decompose $(1539)_{10}$ into its respectively powers of $10$ as follows:
(1)
\begin{align} \quad (1539)_{10} = \left ( 1 \cdot 10^3 \right ) + \left ( 5 \cdot 10^2 \right ) + \left ( 3 \cdot 10^1\right ) + \left ( 9 \cdot 10^0 \right ) \end{align}
Similarly, we can decompose a decimal number that contains a fractional component similarly. For example, the number $(746.2901)_{10}$ can be written as the following sum:
(2)
\begin{align} \quad (746.2901)_{10} = \left ( 7 \cdot 10^2 \right ) + \left (4 \cdot 10^1 \right ) + \left (6 \cdot 10^0 \right ) + \left ( 2 \cdot 10^{-1} \right ) + \left (9 \cdot 10^{-2} \right ) + \left ( 0 \cdot 10^{-3} \right ) + \left (1 \cdot 10^{-4} \right ) \end{align}
In the case where we have a non-terminating decimal, for example, $\left ( \frac{1}{3} \right)_{10}$, then we can write this number as an infinite sum of powers of $10$:
(3)
\begin{align} \quad \left (\frac{1}{3} \right )_{10} = 0.333... = \left ( 3 \cdot 10^{-1} \right ) + \left ( 3 \cdot 10^{-2} \right ) + ... + \left ( 3 \cdot 10^{-n} \right ) + ... \end{align}
## The Binary Number System
We are most accustomed to using the decimal number system explained above, and the basic arithmetic associated with it, however, the binary number system is also extremely important - especially in computing. We will formally define the binary number system below. Note the similarities to the definition given for the decimal number system.
Definition: The Binary Number System is a number system for which every real number $x$ can be written in terms of the two digits $0$ and $1$ as the sum of powers of $2$. A number written in the binary number system is said to be of base $2$, and to specify this we attach a subscript $2$ to $x$, written $(x)_{2}$.
With the binary number system, the only possibly digits to represent a real number $x$ are $0$'s and $1$'s, and the place holders for each digit correspond to powers of $2$. For example, consider the binary number $(1101)_2$. We note that:
(4)
\begin{align} \quad (1101)_2 = \left ( 1 \cdot 2^3 \right ) + \left ( 1 \cdot 2^2 \right ) + \left (0 \cdot 2^1 \right ) + \left ( 1 \cdot 2^0 \right ) = (13)_{10} \end{align}
Of course, binary numbers can also have fractional components. For example, consider the binary number $(101.0101)_2$ :
(5)
\begin{align} \quad (101.0101)_2 = \left ( 1 \cdot 2^2 \right ) + \left ( 0 \cdot 2^1 \right ) + \left ( 1 \cdot 2^0 \right ) + \left ( 0 \cdot 2^{-1} \right ) + \left ( 1 \cdot 2^{-2} \right ) + \left ( 0 \cdot 2^{-3} \right ) + \left ( 1 \cdot 2^{-4} \right ) = (5.3125)_{10} \end{align}
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## Thursday, February 27, 2014
Simple math: Obtaining a common denominator to add fractions
1/2 + 1/4
To add these two fractions, we must get a denominator that is the same. This is called the common denominator. We can get a common denominator by multiplying the two denominators, in this problem it would be 2 x 4 = 8. But that is not the lowest common denominator, which is preferred when adding and subtracting fractions. To get the lowest common denominator, let's look at all the multiples of 2 and then the multiples of 4.
Multiples of 2 are
2, 4, 6, 8, 10, 12, …...
Multiples of 4 are
4, 8, 12, 16, 20, …..
The lowest common denominator is the smallest number that is the same in both sets of multiples. Notice the 4 is bold in each set. That is the lowest common denominator.
Now that we have the common denominator, we have to make ½ into an equivalent fraction with 4 as the denominator. We learned how to do this in the previous section. If we multiply the numerator and denominator by 2, we get 2/4.. Notice ¼ already has a denominator of 4, so we don't have to change this fraction in order to add.
2/4 + 1/4 = 3/4
Let's now consider a little more difficult problem where we'll have to change both fractions before adding.
3/5 + 1/6
We need to get a common denominator, so get all the multiples of 5.
5, 10, 15, 20, 25, 30, 35, ….
Next get all the multiples of 6.
6, 12, 18, 24, 30, 36, 42, …..
Notice that 30 is the smallest of the numbers that are common to both lists, so 30 is the common denominator.
Next we get equivalent fractions with 30 in the denominator. Multiply the first fraction by 6/6 to get 18/30. Multiply the second fraction by 5/5 to get 5/30. Now we can add the fractions to get 18/30 + 5/30 = 23/30 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# 6.2: Correlation
Difficulty Level: At Grade Created by: Bruce DeWItt
### Learning Objectives
• Understand the properties of the linear correlation coefficient
• Estimate and interpret linear correlation coefficients
• Understand the difference between correlation and causation
• Identify possible lurking variables in bivariate data
• Understand the effects outliers and influential points can have on correlation
### The Correlation Coefficient
The correlation coefficient is a statistic that measures the strength and direction of a linear relationship between two numeric variables. The symbol for correlation is r, and r can take any value from -1.0 to +1.0. The correlation coefficient (r) tells us two things about the linear relationship between the two variables, its strength and its direction. The direction of the relationship, positive or negative, is given by the sign of the r value. A positive value for r indicates that the relationship is positive (increasing to the right), and a negative r value indicates a negative relationship between the two variables (decreasing to the right). Bivariate data with a positive correlation tells us that as the explanatory variable increases, so does the response variable. And, bivariate data with a negative correlation tells us that as the explanatory variable increases, the response variable decreases. A correlation of zero indicates neither of these trends.
The second thing that the correlation coefficient tells us is the strength of the linear relationship - how close the points are to forming a perfect line. An r of exactly 1 or -1 has a perfect correlation, the relationship forms a perfect, exact line. An r value of exactly +1 means that the relationship forms a perfect line with a positive slope and a r value of exactly -1 means that the scatterplot will show a perfect line with a negative slope. The closer the correlation value is to either +1 or -1, the stronger the linear relationship is. And, as r gets closer to zero (either positive or negative), the weaker the linear relationship is. It is important to note that this is only measuring the linear relationship between the two variables. If the relationship shows a clear curved pattern for example, the correlation will tell us nothing about the strength of the relationship.
Here are some sample scatterplots with their correlation coefficients given:
We will be using either our calculator or a computer to calculate the correlation coefficient. The formula to calculate the correlation coefficient is quite tedious. It involves calculating the mean and standard deviation of all of the x-values and the mean and standard deviation of all of the y-values. It then compares the x-value of each ordered pair to the mean of x and every y-value to the mean of y (by subtracting and then dividing by the standard deviation), multiplies these newly calculated values, adds all of them, and divides by one less than the sample size. The correlation formula is shown below, but we will be using technology rather than calculating by hand. See appendix for calculator instructions.
#### Example 1
Estimate the correlation coefficient for each of the following scatterplots.
#### Solution
Nevada: The correlation will be negative and fairly strong, so my estimate is r ≈ -0.85.
Height & IQ: There seems to be no pattern to the graph, so my estimate is r ≈ 0.
### Properties of Correlation
When considering using correlation as a measure of the strength between two variables, you should construct and examine a scatterplot first. It is important to check for outliers, be sure that the relationship appears to be linear, be sure that your sample size is sufficient, and consider whether the individuals being examined were too much alike in some way to begin with. Thus, when examining correlation, there are four things that could affect our results: outliers, linearity, size of the sample and homogeneity of the group.
An outlier, or a data point that lies outside of our overall pattern, can have a great effect on correlation. How great of an affect is determined by the sample size of the data and by the magnitude by which the outlier lies outside of the pattern. The three plots below show scatterplots with their correlation coefficients (r). The first plot shows a positive and reasonably linear graph. Its correlation is r = .897, which is positive and fairly strong. The second plot shows the same data as plot one, with one outlier (upper left) added. Its correlation has dropped to r = .374, which is still positive, but much weaker. This demonstrates how outliers can bring the correlation closer to zero. However, some outliers can actually strengthen the correlation. This is demonstrated in the third plot, which shows the same data as the first with one outlier (upper right) added. With this outlier, the linear relationship becomes even stronger than the first plot, at r = .973.
If the relationship is not linear, calculating the correlation coefficient is meaningless. It is only testing the linear relationship between the two variables. Imagine a scatterplot that shows a perfect parabolic relationship. We would know that there is a strong relationship between these two variables, but if we calculated the correlation coefficient, we would arrive at a figure around zero. Therefore, the correlation coefficient is not always the best statistic to use to understand the relationship between variables.
As we discussed in experimental design, a small sample size can be misleading. It can either appear to have a stronger or weaker relationship than is really accurate. The larger the sample, the more accurate of a predictor the correlation coefficient will be on the linear relationship between the two variables.
When a group is too much alike in regard to some characteristics (homogeneous), the range of scores on either or both variables is restricted. For example, suppose we are interested in finding out the correlation between IQ and salary. If only members of the Mensa Club (a club for people with IQ's over 140) are sampled, we will most likely find a very low correlation between IQ and salary since most members will have a consistently high IQ, but their salaries will vary. This does not mean that there is not a relationship – it simply means that the restriction of the sample limited the magnitude of the correlation coefficient.
Correlation is just a number, it has no units. Also, a change in units of measurement will not affect the correlation. For example, suppose that you had measured several people's heights to the nearest inch and weight to the nearest pound and calculated the correlation coefficient. If you were to then convert the heights to centimeters or weights to kilograms, or both, and then calculate the correlation again, it would be the same value.
### Lurking Variables
It is very important to know that a high correlation does not mean causation! Often times studies that showing a high correlation between two variables will influence readers into thinking that one variable is the cause of the relationship. This is not always true! A high correlation simply does not prove that one variable is causing the other. In some situations we would agree that one variable is in fact causing the response in another. The best way to prove such a direct cause-and-effect relationship is by carrying out a well designed experiment. For example, smoking is strongly correlated with lung disease, and, based on much scientific evidence, we can now say that cigarette smoking causes lung disease. However, this topic was highly debated for many years before the surgeon general announced that it was accepted that cigarette smoking causes lung cancer and emphysema. Many people refused to accept this for many years. People who stood to lose money if smoking was proven to be unsafe, suggested every possible other explanation that they could think of. They suggested that it was simply a coincidence, or that all people who choose to smoke might have something else in common that was actually the cause of the lung disease, not the cigarettes. Because it was not ethical to experiment on humans in order to prove the direct cause-and-effect relationship, the debates went on for a long time.
Sometimes the relationship between variables is a cause-and-effect one, but many times it can be simply a coincidence that the two variables are highly correlated. It is also possible that some other outside factor, a lurking variable, is causing both variables to change. A situation where we have two variables that are both being affected by some other, outside, lurking variable is called common response. For example, we can show a high correlation between the number of TV's per household and the life expectancy per person among many countries. However, it makes no sense that TV's cause people to live longer. Some lurking variable is having an effect here. It is likely that the economic status of the countries is causing both variables to change: more money means more TV's and more money means better health care. If a country is wealthy it is much more likely to have citizens who own TV's. Also, if a country is wealthy it is much more likely to have good hospitals, roads, health education, access to clean water and food, all things that contribute to longer life.
In some situations we will have two variables that are highly correlated, but we are unsure of the exact cause of the relationship. We may be unclear as to whether or not one is causing the other, if there is a lurking variable causing a common response, or if there is some unknown lurking variable that is related in some other unknown way (lurking variables are not always obvious to the researchers). Such a situation is called confounding, because it is confusing to determine how the variables are related (if at all), and whether there may be some lurking variable and if it is related to the variables in question. The variables seem all mixed up and the relationship is unclear, even if highly correlated. An example of confounding is global warming. This is a highly debated topic in social media and web-blogs. Some people argue that human pollution is a major cause of the increase in CO2 and other green house gasses in the atmosphere. While others argue that it is a part of a natural cycle that has normally occurred in our Earth's history. Still some may think both explanations are at work. This is an example of confounding because there is confusion about the cause of global warming.
And don't forget that some relationships are occurring completely by chance, and their high correlation is then just a coincidence. For example, if you researched divorce rates and gas prices over the past 50 years you may note that both have gone up. A scatterplot comparing divorce rates and gas prices would show a strong positive relationship. The correlation would likely be a high, positive value. However, it makes no sense that divorce rates are causing high gas prices. It also is unlikely that there exists a common response or some form of confounding. So in this case, we would say that this is a relationship that is best explained by sheer coincidence.
#### Example 2
Suggest possible lurking variables to explain the high correlations between the following variables. Explain your reasoning. Consider whether common response, confounding, or coincidence may be involved.
a) It has been shown that cities with more police officers also have higher numbers of violent crimes. Does this mean that more police officers are causing more violent crimes to occur?
b) Over the past 25 years, the percent of parents using car-seats has increased significantly. During this same time period, the rate of DUI arrests has also increased significantly. These two variables, when graphed, show a very high, positive correlation. Does this mean that car-seat use is causing DUI's to increase?
c) A study published in USA Today claimed that, "Teens who text a lot [are] more likely to try sex, drugs, alcohol." Does this mean that texting causes teens to try sex, drugs and alcohol? Could we then limit teen behaviors such as these by canceling their texting plans?
#### Solutions
a) It makes no sense that the number of police officers would be causing the violent crime to occur. It is much more likely that it is the reverse, that communities with high numbers of violent crimes need higher numbers of police officers. It is also probable that both variables increase in cities with higher populations. Due to the fact that we can think of more than one possible lurking variable and it is difficult to know how all of these variables actually relate, we would say that this is an example of confounding (the variables in question and the lurking variables are all mixed up).
b) It is clearly ridiculous to think that car-seat use is causing an increase in the rate of DUI's. It also makes no sense that DUI's cause car-seats to be used. It may be simply a coincidence that these are both increasing. Or, perhaps there has been in increase in law enforcement for both over this time period. The awareness of the dangers of both have increased over the past 25 years, so maybe this is an example of common response. Or, maybe many factors contribute to the increase of both, so perhaps this is an example of confounding. But, no matter what, this is not cause-and-effect.
c) It is unlikely that texting is actually the cause of these behaviors. There is most likely some other, lurking variable(s) that are the cause(s). One probable lurking variable, when it comes to teenagers, is the parents. Perhaps this is an example of a common response to parents who are not very involved in their teens' lives. Parents who are not very involved would not be aware that their teen is texting too much and would also not be aware of what choices their teen is making during his or her free time. Perhaps teens who spend a lot of time unsupervised would be more likely to text and would also be more likely to try sex, drugs, and alcohol. All of these behaviors might be a common response to not having parents who prohibit or limit teens from doing these things. Canceling texting plans would have little to no affect on other teen behaviors.
Calculating Correlation on the Internet,
There are several websites where you can enter in data points and find their correlation one of them is below.
If this site no longer works, trying googling "finding correlation applet" and see what you get for results.
For an explanation of the correlation coefficient,
see kbower50, The Correlation Coefficient (3:59).
Another, more lighthearted example of Correlation ≠ Causation can be found at the following website, which discusses the evil of the pickle.
For a better understanding of correlation try these fun links below,
http://www.istics.net/stat/Correlations Match the graph to its correlation.
### Problem Set 6.2
#### Section 6.2 Exercises
1) What are the two things that the correlation coefficient measures?
2) The program used to create this scatterplot found the line-of-best-fit and reported the r-squared value as r2 = 0.805 for the relationship between arm-span and height for several individuals. What is the correlation coefficient? Is it positive or negative? Explain how you know.
3) During the summer Ms. Statsteacher lets her two daughters stay up later than during the school year. Their bedtimes during the summer range from 8:30 p.m. to 12:30 a.m. She has discovered that her older daughter Reily will wake up between 8:00 and 9:00 a.m. no matter what time she goes to bed. However, her younger daughter Neila tends to wake up later after she gets to stay up later, and earlier when she goes to bed earlier. Neila has been known to wake up anytime between 8:00 and 11:45 a.m.
a) Sketch a separate (approximate) scatterplot for each daughter, that compares time going to sleep and time waking up. Which will be explanatory and which will be response?
b) Which of these do you think will best approximate the correlation for Reily?
A. close to r = +1
B. close to r = +.75
C. close to r = 0
D. close to r = -.75
E. close to r = -1
c) Which of these do you think will best approximate the correlation for Neila?
A. close to r = +1
B. close to r = +.75
C. close to r = 0
D. close to r = -.75
E. close to r = -1
4) Suggest possible lurking variables to explain the high correlations between the following variables. Explain your reasoning. Consider whether common response, confounding, or coincidence may be involved.
a) As ice cream sales increase, the rate of drowning deaths increases sharply. Does this mean that ice cream causes drowning?
b) With a decrease in the number of pirates, there has been an increase in global warming over the same time period. Does this mean global warming is caused by a lack of pirates?
c) The higher the number of fire-fighters fighting a fire, the more damage done by the fire. Does this mean that we can limit damage by sending fewer fire-fighters to fires?
d) Suppose that each of the hockey players on the high school team supplies his or her own hockey stick, with varying degrees of flex. The assistant coach has been keeping a record of the degree of flex for each player's stick and their respective point totals (goals and assists). He has noted that there is a strong, negative correlation between these two variables. In other words, the players with less flex in their sticks are scoring more points and those with more flex are scoring fewer points. Does this prove that the amount of flex in a stick will cause the point totals for the players? Can we then give players less flexible sticks and expect to increase scoring?
5) In a recent study in Resource Manual, it was noted that divorced men were twice as likely to abuse alcohol as married men. The authors concluded that getting divorced caused alcohol abuse. Do you agree? Explain your reasoning.
6) A commercial for a new diet pill claims "You will lose weight while you sleep! No exercise needed!". They then show several before-and-after photos of people who have lost weight. People who were obese are now very buff. They then give the information for you to order the pills ("for three payments of just \$19.95 each, plus shipping and handling"). Is this proof that these diet pills caused these people to lose weight? Suggest possible lurking variables. Explain your reasoning.
7) Match each graph with its correlation coefficient:
8) A correlation of r = 0 indicates no linear relationship between the two given variables. But, this does not mean that there is no relationship between the two variables. Sketch a scatterplot in which there is a strong relationship between the variables, but the correlation would be near r = 0.
9) Use the "Beach Visitors" scatterplot to answer the questions that follow.
a) Identify the explanatory and response variables.
b) Estimate the correlation coefficient for the graph.
c) Describe what the scatterplot shows. (remember S.C.O.F.D)
#### Review Exercises
10) Zeke flips a coin 93 times and tails shows up 34 of those times. Based on these results, what is the experimental probability of getting tails?
11) If Stephanie's batting average is 0.258, how many hits would you expect her to get out of her next 20 times at bat?
12) You have been playing the game Yahtzee with some friends and you have been keeping track of how often someone gets a Yahtzee (5 of the same dice) when they roll all 5 dice at once. The results today have been 3 Yahtzee's, on a single roll, out of 79 trials. Based on these results, what is the experimental probability of getting a Yahtzee in one roll?
13) What is the theoretical probability of getting a Yahtzee in one roll?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
# How to tell if a Number is Divisible by 8
I’ve explained a number of divisibility rules lately, offering tricks to tell if numbers are divisible by 2, 3, 4, 5, 6 and 7.
There is also a trick for divisibility by 8, and that’s what I’d like to explain in this post.
Essentially the trick for 8 is a lot like the trick for 4. If you’d like to refresh your memory on how that trick works, just go here.
The basic idea is that you use the DPM and the DPS, the Divisibility Principle of Multiples and the Divisibility Principle of Sums. My post on the trick for 4 explains both of these principles, so I won’t explain them in depth here.
But I will say that the basic approach we use is this.
If we’re wondering if a number, call it x, divides into another number, call it y, we do the following.
1) Break y apart into two addends, call them p and q. We choose these p and q numbers in such a way that we know that x goes into p evenly.
2) Then all we have to do is check to see if x also goes into q. If it does, then x DOES go into y; if x does not go into q, then we can be certain that x does NOT go into y. That is how powerful these two DPM and DPS principles are!
Here’s a quick example of how this works with 4. Suppose we want to know if 4 divides evenly into 252. We know (somehow) that 4 does go into 240. Since 252 = 240 + 12, all we need to do is check to see if 4 also goes into the difference, which is 12. This number, 12, is like the q-number described above. If 4 does go into 12, we can be sure that 4 DOES divide evenly into 252. And if 4 does NOT go into 12, we can be equally sure that 4 does NOT go into 252.
Of course we do know that 4 goes into 12 evenly, so we can be certain that 4 DOES divide evenly into 252.
With 4, we realized that 4 divides evenly into 20 and 100, and those two facts turned out to be really helpful. Knowing that, we were able to subtract multiples of both 20 and 100 from the numbers we were checking out. Doing that, we were able to get a small manageable number (12 in the example), and checking if 4 went into that small number, we figured out if 4 went into the whole number.
With 8, we use the same approach, only now we use the facts that 8 divides evenly into 40 and 200.
The fact that 8 goes into 40 lets us know that we can subtract out multiples of 40: 40, 80, 120, 160, and 200.
The fact that 8 divides into 200 allows us to subtract out the multiples of 200: 200, 400, 600, 800, and 1,000.
And there’s also a bonus to the fact that 8 divides evenly into 1,000. Since that’s true, we know that 8 also divides evenly into all multiples of 1,000. That being the case, we never have to even bother worrying about digits in the thousands place or any place value larger than 1,000. We can confine our search to the number’s final three digits: the digits in the hundreds, tens, and ones places.
So here’s an example of how we check to see if 8 divides evenly into a number. We’ll check if 8 divides evenly into 736.
1st) Subtract the nearest (but lesser) multiple of 200. That would be 600. 736 – 600 = 136. We’ve whittled our search down to 136.
2nd) Subtract the nearest (but lesser) multiple of 40. That would be 120. 136 – 120 = 16.
3rd) Check to see if 8 goes into the difference. 8 does go into 16 evenly, so 8 DOES divide into 736 evenly. End of story.
Another example: Does 8 divide evenly into 486.
1st) Subtract the nearest (but lesser) multiple of 200. That would be 400. 486 – 400 = 86. We’ve now narrowed our search down to 86.
2nd) Subtract the nearest (but lesser) multiple of 40. That would be 80. 86 – 80 = 6.
3rd) Check to see if 8 goes into the difference, 6. Of course, 8 does not go into 6, so 8 does NOT divide evenly into 486. End of story.
Example #3: Does 8 divide evenly into 73,984.
1st) Disregard the digits greater than the hundreds digit. That means we confine our search to the 984.
2nd) Subtract the nearest (but lesser) multiple of 200. That would be 800. 984 – 800 = 184. Our search has now narrowed down to 184.
2nd) Subtract the nearest (but lesser) multiple of 40. That would be 160. 184 – 160 = 24. Our search is now down to just 24.
3rd) Check to see if 8 goes into the difference, 24. As we know, 8 does divide into 24, so 8 DOES divide evenly into 73,984. End of story, once again.
PRACTICE: Check to see if 8 divides evenly into these numbers.
a) 104
b) 178
c) 472
d) 544
e) 770
f) 853
g) 2,496
h) 5,952
i) 17,038
j) 456,618
k) 2,438,296
a) 104: Divisible by 8
b) 178: Not divisible by 8
c) 472: Divisible by 8
d) 544: Divisible by 8
e) 770: Not divisible by 8
f) 853: Not divisible by 8
g) 2,496: Divisible by 8
h) 5,952: Divisible by 8
i) 17,038: Not divisible by 8
j) 456,618: Not divisible by 8
k) 2,438,288: Divisible by 8 |
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