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# What is cosΘ if tanΘ = 6/17?
Apr 1, 2018
$\implies \frac{17 \sqrt{13}}{65}$
#### Explanation:
We know $\tan \theta = \text{opposite"/"adjacent}$
Hence opposite" = 6 and $\text{adjacent} = 17$
Hence using Pythagorus rule: ${a}^{2} + {b}^{2} = {c}^{2}$
$\implies {6}^{2} + {17}^{2} = {c}^{2} \implies c = 5 \sqrt{13}$
$\implies \text{hypotenuse} = 5 \sqrt{13}$
We know $\cos \theta = \text{adjacent" / "hypotenuse}$
$\implies \cos \theta = \frac{17}{5 \sqrt{13}}$
$\implies \frac{17 \sqrt{13}}{65}$
Apr 1, 2018
$\textcolor{b l u e}{\frac{17 \sqrt{13}}{65}}$
#### Explanation:
From diagram, we have:
$\tan \left(\theta\right) = \text{opposite"/"adjacent} = \frac{6}{17}$
$\cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{17}{c}$
We need to find $\boldsymbol{c}$:
Using Pythagoras' Theorem:
${c}^{2} = {6}^{2} + {17}^{2}$
$c = \sqrt{{\left(6\right)}^{2} + {\left(17\right)}^{2}} = \sqrt{325} = 5 \sqrt{13}$
$\cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{17}{5 \sqrt{13}} = \textcolor{b l u e}{\frac{17 \sqrt{13}}{65}}$ |
# Polynomial Function
A polynomial function in general is the simplest form of a mathematical function, commonly most used in algebraic expressions with specific conditions. The highest power present in the polynomial function depends on the degree that it has in it. A polynomial function involves non-negative integer powers as well as positive integer exponents of a variable in an equation as the same quadratic equation, cubic equation, etc.
## What is a polynomial function?
In mathematical form, the definition of polynomial functions can be defined as:
“A function that is expressed in the form of the algebraic polynomial equation”
The word polynomial is derived from two generic words poly and nomial. Since "Poly" means many, and "nomial" means the term, and when they are combined, we can say that the definition polynomial function includes "algebraic expressions with many terms."
### Degree of polynomial functions
Generally, a polynomial number is represented as P(x). The highest power in the polynomial equation of the variable of P(x) is called its degree. A polynomial function’s degree is very important because it tells us about the behavior of the function P(x) when x becomes very large. The domain of a polynomial function is considered real numbers (R). Here, you can use our online multiply polynomials calculator to find out algebraic expressions’ products with the blink of an eye.
### Polynomial function formula in standard form
Polynomial expression in standard form is given as:
$$P\left(x\right) = a_{n}x^{n} + a_{n-1}x^{n-1} + … + a_{2}x^{2} + a_{1}x + a_{0}$$ then for x << 0, $$P\left(x\right) ≈ a_{n}x^{n}$$
Thus, the polynomial algebraic expression contains variables of varying degrees, coefficients, positive exponents, and constants.
### Polynomial Function Examples
Since a polynomial function contains positive integers as represented in the form of exponents. Here are some examples of polynomial functions that are related to arithmetic operations for such functions as addition, subtraction, multiplication, and division.
• x^2+2x+1
• 3x-7
• 7x^3+x^2-2
### Types of Polynomial Functions
Let’s now discuss various types of polynomial equations that are based on the degree of the polynomial.
#### Constant Polynomial Function:
A constant Polynomial Function has no degree.
P(x) = a = ax^0
#### Zero Polynomial Function:
A function is expressed in the form f(x) = 0, is known as zero polynomial
Example:
P(x) = 0; where all as are zero, i = 0, 1, 2, 3, …, n.
#### Linear Polynomial Function:
A function that has the highest power is a 1 degree linear polynomial
Example:
P(x) = ax + b
A function that has the highest power is 2 degrees is called a quadratic polynomial
Example:
P(x) = ax^2+bx+c
#### Cubic Polynomial Function:
A function has the highest power is 3 degrees is called a cubic polynomial.
Example:
ax^3+bx^2+cx+d
#### Quartic Polynomial Function:
A function that has the highest power 4 degrees is called quartic polynomial
Example:
ax^4+bx^3+cx2+dx+e
Moreover, if you are trying to resolve problems regarding variables for the division of a polynomial by binomial, an online polynomial long division calculator is the best choice to proceed with the calculations.
### How to Determine a Polynomial Function?
Whether you desire how to determine if a function is a polynomial or not, the polynomial function equation is required to be checked against conditions for the exponents of the variables. These conditions are given bellowed:
• The exponent of the variable in the function must be a non-negative whole number in every term. i.e., the exponent of the variable must not be a fraction or negative number in the function.
• The variable of the function should not be under a radical form i.e, it should not contain any square roots, cube roots, etc.
• The variable should not be in the form of a denominator.
The given table shows polynomial function equation examples and some non-examples of polynomial functions:
Functions Variables Exponents Polynomial function or not? f(b) = 4b2 – 6b + b3 – 15 b 2 in b2; 3 in b3 Yes f(x) = x2/3+ 2x x 2/3 in x2/3 ; 1 in 2x No f(y) = 1/y3 y -3 in 1/y3 No
### Graphing Polynomial Function:
Now let us discuss all the polynomial functions in the form of a graph. As you know that the domain of any polynomial expression is the set of all real numbers. Bellow's given image represents the polynomial function graph and recognizes the relationship between equations and graphs.
• In case a linear polynomial function is represented in the form y = ax + b then it shows a straight line.
• In case a quadratic polynomial function is represented in the form y = ax^2 + bx + c then it shows a parabola.
• A cubic polynomial function is represented in the form y = ax^3 + bx^2 + cx + d.then it shows a hyperbola.
Give a try to use this slope Calculator to find the graph of a slope line and also shows step-by-step calculations.
Example of the polynomial function graph
In a simple polynomial equation graph, we make a table that has some random values of x and the corresponding values of f(x). Here, we plot the points from the table and combine them with a curve line.
Let’s now draw the graph for the quadratic polynomial function f(x) = x^2.
x
-2
-1
0
1
2
f(x) = x2
4
1
0
1
4
Plot the points and then join them by a curve line by extending it on both sides to obtain the graph of the polynomial function.
### Aaron Lewis
Last updated: June 13, 2024
Aaron Lewis is an accomplished writer; He has done MS-Business Management and is a professional Research analyst and writer. He is too aggressive to write articles regarding Digital Marketing, Business, Health, and Mathematics. He is ready every time to collect information that can convey her experience on related topics. |
# How do you factor (x+2)^2-7(x+2)+12?
Nov 10, 2015
$\left(x - 1\right) \left(x - 2\right)$
#### Explanation:
Expand all the terms:
$\left({x}^{2} + 4 x + 4\right) - \left(7 x - 14\right) + 12 = {x}^{2} - 3 x + 2$
Solve the quadratic equation: when dealing with a quadratic $a {x}^{2} + b x + c$, we know that the two solutions are given by the formula
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
Plugging the values,
${x}_{1 , 2} = \setminus \frac{3 \setminus \pm \setminus \sqrt{9 - 8}}{2} = \setminus \frac{3 \setminus \pm 1}{2}$
Which means that ${x}_{1} = 1$ and ${x}_{2} = 2$.
When you find the solutions of a quadratic equation, if $a = 1$, you can write
${x}^{2} + b x + c = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$, so in your case
${x}^{2} - 3 x + 2 = \left(x - 1\right) \left(x - 2\right)$ |
# Triangle Side-Angle-Side Equality
## Theorem
If $2$ triangles have:
$2$ sides equal to $2$ sides respectively
the angles contained by the equal straight lines equal
they will also have:
their third sides equal
the remaining two angles equal to their respective remaining angles, namely, those which the equal sides subtend.
In the words of Euclid:
If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles equal to the remaining angles respectively, namely those which the equal sides subtend.
## Proof
Let $\triangle ABC$ and $\triangle DEF$ be $2$ triangles having sides $AB = DE$ and $AC = DF$, and with $\angle BAC = \angle EDF$.
If $\triangle ABC$ is placed on $\triangle DEF$ such that:
the point $A$ is placed on point $D$, and
the line $AB$ is placed on line $DE$
then the point $B$ will also coincide with point $E$ because $AB = DE$.
So, with $AB$ coinciding with $DE$, the line $AC$ will coincide with the line $DF$ because $\angle BAC = \angle EDF$.
Hence the point $C$ will also coincide with the point $F$, because $AC = DF$.
But $B$ also coincided with $E$.
Hence the line $BC$ will coincide with line $EF$.
(Otherwise, when $B$ coincides with $E$ and $C$ with $F$, the line $BC$ will not coincide with line $EF$ and two straight lines will enclose a region which is impossible.)
Therefore $BC$ will coincide with $EF$ and be equal to it.
Thus the whole $\triangle ABC$ will coincide with the whole $\triangle DEF$ and thus $\triangle ABC = \triangle DEF$.
The remaining angles on $\triangle ABC$ will coincide with the remaining angles on $\triangle DEF$ and be equal to them.
$\blacksquare$
## Historical Note
This proof is Proposition $4$ of Book $\text{I}$ of Euclid's The Elements. |
If you will be a fresh college student of mathematics, you could not have got a distinct concept on just what is velocity in math. You might end up confronted using this type of variety of concern as you consider to know the ideas of algebra and calculus. While it appears to be tricky, it genuinely is not. All you need is usually a minimal exertion to understand what the heck is velocity in math.
Understanding just what is velocity in math begins with defining the theory of velocity. House and time are usually thought of as relocating at distinct speeds. Thus, objects that go in one course typically speed up or gradual down in yet another path. Such as, as you are driving your automobile down a two-lane road, the course of travel is continually modifying. The exact same thing goes with the planets and various objects we could see when using the bare eye. Due to this, we can study which the pace of nearly anything goes in only one direction.
You would possibly observe that the velocity of the item also tends to differ in numerous directions. As an illustration, let us express that you’ve gotten your ball level toward the basketball hoop. Now, when you have been to goal the ball from the path of your basket, you’d probably expect it to move within a straight line relating to the hoops. Yet, what is the velocity in math suggests which the path for the ball vacation in a curved sample. This may be illustrated together with the adhering to figure.
Using the velocity components in your own research of math will indicate you that the speed belonging to the ball will change in the straight line in between the x-axis and then the y-axis. We can establish the route of travel by choosing the slope from the tangent line among these two variables. We are able to use this idea to find out what’s velocity in math. The slope of the tangent line tells us the route by which the ball travels.
If rephrasing sentences we’ve by now proven what exactly is velocity in math, we can now define a force of gravity since the adjust in velocity relating to two locations. It could possibly also be defined as being the switch in velocity involving two points around the area of the object. The power of gravity acts in a very straight line. For this reason, if we have been hunting at an object for the bottom with the table and on the top rated, the pressure of gravity will show up as horizontal because it is acting while in the direction within the desk. We can determine what on earth is velocity in math by discovering the spinoff of your force with respect to your x-axis and y-axis. Discover simply how much vertical drive the earth exerts on an object when it happens to be tilted vertically and after that understand exactly how much horizontal pressure it exerts for the table when it can be tilted horizontally.
It should really be very clear the direction that anything moves in would not always https://en.wikipedia.org/wiki/Education_in_Poland stick to the mathematics equation just what is velocity in math. What follows is undoubtedly an assumption. The rate that you’ll be considering may be zero on a person axis and infinity within the other axis. Hence, the way which the item is travelling in ought to be considered in advance of the equation is put to use. Put another way, it need to be comprehended that the direction the item is travelling in will not be constantly going to be the course that the person wished it to go in!
What is velocity in math is significant for initiatives that contain translating motions from a person coordinate procedure to a different. For instance, should you are doing your undertaking translating from English to Spanish then you need to recognize that the course in the vacation from the two languages differs. It is because with the next language in the direction of movement does not normally go from still left to proper. To unravel this problem, the projectors translate their projects in a very way that the user will intuitively know very well what the way they were travelling is. The square root of your velocity they are implementing is what is used to know what direction the object is travelling. It really is a rephraser.net more sophisticated system when compared to the one which was presented previously mentioned, however it will still be much easier to suit your needs to be familiar with.
If you need to be aware of what’s velocity in math then you certainly really should begin the process of by memorizing the definition of velocity. You can actually do that by seeking in a photo or drawing a straight line on the graph, or else you can draw a scatter plot on a piece of graph paper. The first matter that you simply will recognize is the fact that you will discover instances if the path arrives before the actual speed. By way of example, when you apply pressure to an item in the remaining hand facet belonging to the graph, you will see the velocity will likely be increased when compared to the horizontal route in the suitable. It is because the item on the still left is going horizontally despite the fact that the article about the suitable is going vertically. Therefore, it really is crucial to understand that the particular speed are going to be gonna be bigger compared to the way around the appropriate if you are looking in a horizontal way as well as the reverse if shopping at a vertical route. |
# SSAT Upper Level Math : Estimation
## Example Questions
### Example Question #1 : How To Estimate
Evan wants to tip approximately on a restaurant tab. Which of the following comes closest to what he should leave?
Explanation:
The tab can be rounded to is equal to .
Now we can multiply the percent by the total amount.
is the most reasonable estimate of the recommended tip.
### Example Question #1 : Estimation
If the number is rounded to the nearest hundredth, which of the following expressions would be equal to that value?
Explanation:
If is rounded to the nearest hundredth, the result will be
Given that , the correct answer is
### Example Question #2 : Estimation
Estimate the product by rounding each factor to the nearest hundred, then multiplying.
Explanation:
437 rounded to the nearest hundred is 400.
877 rounded to the nearest hundred is 900.
551 rounded to the nearest hundred is 600.
Multiply the three whole multiples of 100 to get the desired estimate:
### Example Question #1 : How To Estimate
Estimate the product by rounding each factor to the nearest unit, then multiplying.
Explanation:
, so rounded to the nearest unit is 5.
, so rounded to the nearest unit is 10.
, so rounded to the nearest unit is 5.
Multiply the three whole numbers to get the desired estimate:
### Example Question #3 : Estimation
Estimate the result by first rounding each number to the nearest unit.
Explanation:
8.19 rounded to the nearest unit is 8.
4.87 rounded to the nearest unit is 5
3.27 rounded to the nearest unit is 3.
7.42 rounded to the nearest unit is 7.
The desired estimate can be found as follows:
### Example Question #1 : How To Estimate
Estimate the result by first rounding each number to the nearest unit.
Explanation:
, so rounded to the nearest unit is 8.
, so rounded to the nearest unit is 10.
, so rounded to the nearest unit is 4.
, so rounded to the nearest unit is 6.
The desired estimate can be found as follows:
### Example Question #52 : Number Concepts And Operations
Melissa is trying to come up with a reasonable estimate of the amount she spent on groceries over the last six months. She notices that the six checks she wrote out to the local grocery store are in the following amounts: $187.54,$218.89, $174.74,$104.76, $189.75, and$228.64. By estimating each of the amounts of the checks to the nearest ten dollars, come up with a reasonable estimate for Melissa's total expenditure for groceries.
Explanation:
Round each of the amounts to the nearest ten dollars as follows:
$187.54 rounds to$190.
$218.89 rounds to$220.
$174.74 rounds to$170.
$104.76 rounds to$100.
$189.75 rounds to$190.
$228.64 rounds to$230.
### Example Question #5 : How To Estimate
Estimate the product by rounding each factor to the nearest unit, then multiplying.
Explanation:
8.39 rounded to the nearest unit is 8 because 0.39 is less than 0.5.
7.34 rounded to the nearest unit is 7 because 0.34 is less than 0.5.
3.52 rounded to the nearest unit is 4 because 0.52 is greater than 0.5.
Multiply the three whole numbers to get the desired estimate: |
# Find the Equation of an Ellipse Whose Eccentricity is 2/3, the Latus-rectum is 5 and the Centre is at the Origin. - Mathematics
Find the equation of an ellipse whose eccentricity is 2/3, the latus-rectum is 5 and the centre is at the origin.
#### Solution
$\text{ Let the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. ...\left( 1 \right)$
$e = \frac{2}{3} \text{ and latus rectum }=5 (\text{ Given })$
$\text{ Now }, \frac{2 b^2}{a} = 5$
$\Rightarrow 2 b^2 = 5a . . . (2)$
$\Rightarrow 2 a^2 (1 - e^2 ) = 5a [ \because b^2 = a^2 (1 - e^2 )]$
$\Rightarrow 2 a^2 \left[ 1 - \frac{4}{9} \right] = 5a$
$\Rightarrow 2 a^2 \times \frac{5}{9} = 5a$
$\Rightarrow 10 a^2 = 45a$
$\Rightarrow a = \frac{9}{2}$
$\text{ Substituting the value ofain eq. (2), we get }:$
$2 b^2 = 5 \times \frac{9}{2}$
$\Rightarrow b^2 = \frac{45}{4}$
$\text{ Substituting the values of } a^2\text{ and } b^2 \text{ in eq }. (1), \text{ we get }:$
$\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$
$\Rightarrow \frac{4 x^2}{81} + \frac{4 y^2}{45} = 1$
$\text{ This is the required equation of the ellipse }.$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 12 | Page 23 |
# What is the derivative of f(x) = x^3 - 3x^2 - 1?
Jul 8, 2018
$f ' \left(x\right) = 3 {x}^{2} - 6 x$
#### Explanation:
We need the sum rule
$\left(u + v + w\right) ' = u ' + v ' + w '$
and that
$\left({x}^{n}\right) ' = n {x}^{n - 1}$
so we get
$f ' \left(x\right) = 3 {x}^{2} - 6 x$
Jul 8, 2018
$f ' \left(x\right) = 3 {x}^{2} - 6 x$
#### Explanation:
$\text{differentiate each term using the "color(blue)"power rule}$
•color(white)(x)d/dx(ax^n)=nax^(n-1)
$f ' \left(x\right) = 3 {x}^{2} - 6 x$
Jul 8, 2018
$3 {x}^{2} - 6 x$
#### Explanation:
The derivative of a sum/difference is the same as the sum/difference of the derivatives, so we can take the derivative of all of these terms.
We can use the Power Rule- here, the exponent is brought out front, and the power is decremented by $1$. We get
$3 {x}^{2} - 6 x$
Recall that the derivative of a constant is zero.
Hope this helps! |
# How to Calculate Complex Numbers
Save
Complex numbers have a real component and an imaginary component. They have specific operations of addition, division, multiplication and subtraction that expand on the corresponding operations for real numbers. The rules for the operations of complex numbers were first developed in the sixteenth century in an attempt to solve cubic equations that involved the square root of negative numbers. Complex numbers are currently used in a variety of fields such as chaos theory, electromagnetism and quantum physics.
• Define a complex number. A complex number uses the notation a + bi, where "a" and "b" are real numbers and "i" is the square root of -1; "a" is the real component of the complex number and "b" is the imaginary component of the complex number.
• Define the rules for the addition and subtraction of complex numbers. The addition of complex numbers uses the identity (a + bi) + (c +di) = (a + c) + (b + d)i. The subtraction of complex numbers uses the identity (a + bi) i (c +di) = (a - c) + (b - d)i.
• Use the rules in Step 2 to add specific complex numbers. For example, (3 + 4i) + (2 + 3i) = (3 + 2) + (4 -- 3)i = 5 + i.
• Define the rules for the multiplication of complex numbers. These multiplication of complex numbers uses the identity (a + bi) (c + di) = (ac - bd) + (bc + ad)i. The division of complex numbers uses the identity (a + bi) / (c + di) = (ac + bd)/(c^2 +d^2) + (bc - ad)/(c^2 +d^2)i.
• Use the rules in Step 4 to multiply specific complex numbers. For example, (3 + 4i)(2 -- 3i) = (3 x 2 - 4 x (-3)) + (4 x 2 + 3 x (-3))i = 6 - (-12) + (8 + -9)i = 18 - i.
## References
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Kathy Williams
2021-12-11
Express as a polynomial. $\left({r}^{2}+8r-2\right)\left({r}^{2}+3r-1\right)$
### Answer & Explanation
Matthew Rodriguez
Step 1
Given expression:
$\left({r}^{2}+8r-2\right)\left({r}^{2}+3r-1\right)$
Step 2
Consider,
$\left({r}^{2}+8r-2\right)\left({r}^{2}+3r-1\right)={r}^{2}\left({r}^{2}+3r-1\right)+8r\left({r}^{2}+3r-1\right)-2\left({r}^{2}+3r-1\right)$...Use the distributive property.
$=\left[{r}^{2}\left({r}^{2}\right)+{r}^{2}\left(3r\right)+{r}^{2}\left(-2\right)\right]+\left[\left(8r\right)\left({r}^{2}\right)+\left(28r\right)\left(3r\right)+\left(8r\right)\left(-2\right)\right]-\left[2{r}^{2}+\left(2\right)\left(3r\right)-\left(2\right)\left(1\right)\right]$
$={r}^{4}+3{r}^{3}-2{r}^{2}+8{r}^{3}+24{r}^{2}-16r-2{r}^{2}-6r+2$
$={r}^{4}+\left(3{r}^{3}+8{r}^{3}\right)+\left(-2{r}^{2}+24{r}^{2}-2{r}^{2}\right)+\left(-16r-6r\right)+2$...Collect like terms together.
$={r}^{4}+11{r}^{3}+20{r}^{2}-22r+2$
Thus,
$\left({r}^{2}+8r-2\right)\left({r}^{2}+3r-1\right)={r}^{4}+11{r}^{3}+20{r}^{2}-22r+2$
Do you have a similar question?
Recalculate according to your conditions! |
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# Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets.Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 $${cm}^2$$, find the cost of cardboard required for supplying 250 boxes of each kind.
Given:
Length (l) of bigger box = 25 cm
Breadth (b) of bigger box = 20 cm
Height (h) of bigger box = 5 cm
we know that,
Total surface area of bigger box
= 2(lb + lh + bh)
= [2 (25 x 20 + 25 x 5 + 20 x 5)] $${cm}^2$$
= [2(500 + 125 + 100)] $${cm}^2$$
= 1450 $${cm}^2$$
Therefore, Extra area required for overlapping
= $$\frac{1450 × 5}{100}$$ $${cm}^2$$ = 72.5 $${cm}^2$$
While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) $${cm}^2$$ = 1522.5 $${cm}^2$$
So, Area of cardboard sheet required for 250 such bigger boxes = (1522.5 x 250) $${cm}^2$$ = 380625 $${cm}^2$$
Similarly, total surface area of smaller box = [2 (15 + 15 x 5 + 12 x 5)]
= [2 (180 + 75 + 60)] $${cm}^2$$
= (2 x 315) $${cm}^2$$
= 630 $${cm}^2$$
Therefore, extra area required for overlapping
= $$\frac{630 × 5}{100}$$ $${cm}^2$$ = 31.5 $${cm}^2$$
So, Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) $${cm}^2$$
= 661.5 $${cm}^2$$
So, Area of cardboard sheet required for 250 smaller boxes
= (250 x 661.5) $${cm}^2$$ = 165375 $${cm}^2$$
Thus, Total cardboard sheet required
= (380625 + 165375) $${cm}^2$$
= 546000 $${cm}^2$$
Cost of 1000 $${cm}^2$$ cardboard sheet = Rs. 4
Therefore, Cost of 546000 $${cm}^2$$ cardboard sheet will be:
$$Rs. \frac{546000 × 4}{100}$$ $${cm}^2$$ = Rs. 2184
Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 2184. |
### Finding angles
To find angles, we can use what are known as inverse trigonometric functions. On your calculator, the inverse trig functions will appear as $$SI{N^{ - 1}}$$, $$CO{S^{ - 1}}$$, or $$TA{N^{ - 1}}$$. The output of these functions should always be understood as angles. Use the following procedure to find angles:
1. Compute the sine, cosine, or tangent of the given angle (whichever is most convenient for you).
2. Evaluate the inverse trig function for that number.
Example: Find the measure of the indicated angle to the nearest degree.
Solution: Let’s compute the cosine of the angle. We have
$$\large \cos \left( ? \right) = \frac{{48}}{{52}} \approx 0.923$$
Now we use the inverse cosine button on the calculator to compute
$${\cos ^{ - 1}}0.923 \approx 22.631^\circ$$
We round this result to $$23^\circ$$. Now let’s try one more example:
Example: Find the measure of the indicated angle to the nearest degree.
Solution: This time, just to be different, let’s compute the sine of the given angle. We have
$$\large \sin \left( ? \right) = \frac{{54}}{{90}} = 0.6$$
Since we used sine, we will compute the inverse sine of 0.6. We have
$${\sin ^{ - 1}}0.6 \approx 36.870$$
We round this result to $$37^\circ$$.
13138 x
Find each angle measure to the nearest degree.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch bellow how to solve this example:
4798 x
Find the measure of the indicated angle to the nearest degree.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch bellow how to solve this example:
6229 x
Find the measure of the indicated angle to the nearest degree.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch bellow how to solve this example:
### Geometry
Circles
Congruent Triangles
Constructions
Parallel Lines and the Coordinate Plane
Properties of Triangles
### Algebra and Pre-Algebra
Beginning Algebra
Beginning Trigonometry
Equations
Exponents
Factoring
Linear Equations and Inequalities
Percents
Polynomials |
# Important Questions for Class 10 Chapter 7 - Coordinate Geometry
The set of PDF of important questions will help you in doing a complete revision in Maths and getting good scores.
Important questions based on NCERT syllabus for Class 10 Chapter 7 - Coordinate Geometry:
Question 1: Find the area of a triangle whose vertices are A (3, 2), B (11, 8) and C(8, 12).
Solution:
Let A=(x1,y1)=(3,2),
B(x2,y2)=(11,8)
and C=(x3,y3)=(8,12) be the given points. Then,
Area of ΔABC=1/2|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|
Area of ΔABC = 1/2|3(8 − 12) + 11(12 − 2) + 8(2 − 8)|
Area of ΔABC = 1/2|−12 + 110 − 48| = 25 sq. units
Question 2: Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7).
Solution:
Suppose the line 3x + y -9 = 0 divides the line segment joining A(1, 3) and B(2, 7) in the ratio k:1 at point C.
Then, the coordinates of C are
((2k +1)/(k+1), (7k+3)/(k+1))
But, C lies on 3x + y -9 = 0. Therefore,
3(2k+1/k+1) + 1(7x+3/k+1) − 9 = 0
⇒ 6k + 3 + 7k + 3 − 9k − 9 = 0
⇒ k = 3/4
So, the required ratio is 3:4 internally.
Question 3: The coordinates of A, B, C are (6, 3), (–3, 5) and (4, – 2) respectively and P is any point (x, y). Show that the ratio of the areas of triangle PBC and ABC is | x+y−2/7|
Solution:
We have,
∴Area of ΔPBC=1/2|(5x+6+4y)−(−3y+20−2x)|
⇒Area of ΔPBC=1/2|5x+6+4y+3y−20+2x|
⇒Area of ΔPBC=1/2|7x+7y−14|
⇒Area of ΔPBC=7/2|x+y−2|
⇒Area of ΔABC=7/2|6+3−2|
[ Replacing x by 6 and y = 3 in Area of ΔPBC]
⇒ Area of ΔABC = 49/2
∴Area of ΔPBC Area of ΔABC = (7/2|x + y − 2|)/(49/2)
⇒ Area of ΔPBC Area of ΔABC = |x + y − 2| / 7 |
## Geometry: Common Core (15th Edition)
$x = 3$ or $x = \frac{2}{7}$
According to the side-splitter theorem, if a line is parallel to a side of a triangle and intersects the other two sides, then those two sides are divided proportionately. Let's set up a proportion for the sides whose values are given in the figure: $\frac{12}{7x} = \frac{2x + 2}{5x - 1}$ Use the cross product property to get rid of the fractions: $7x(2x + 2) = 12(5x - 1)$ Use the distributive property first: $14x^2 + 14x = 60x - 12$ Move all terms to the left side of the equation: $14x^2 + 14x - 60x + 12 = 0$ Combine like terms: $14x^2 - 46x + 12 = 0$ Factor out a $2$ from all terms: $7x^2 - 23x + 6 = 0$ Use the quadratic formula to solve this equation: $x = \frac{-(-23) ± \sqrt {(-23)^2 - 4(7)(6)}}{2(7)}$ Simplify by multiplying: $x = \frac{23 ± \sqrt {529 - 168}}{14}$ Simplify what is under the square root sign: $x = \frac{23 ± \sqrt {361}}{14}$ Take the square root of $361$: $x = \frac{23 ± 19}{14}$ Add or subtract to simplify the fraction: $x = \frac{42}{14}$ or $x = \frac{4}{14}$ Simplify the fractions by dividing their numerators and denominators by their greatest common factors: $x = 3$ or $x = \frac{2}{7}$ |
# Commutative Property Of Multiplication Examples
Commutative Property Of Multiplication Examples. Then the ring is called commutative.in the remainder of this article, all rings will be commutative, unless explicitly stated otherwise. This is known as the commutative property of multiplication.
The order of the two factors 4 and 5 did not affect the. This is known as the commutative property of multiplication. Suppose you have a set of $7$ balls.
### Learn The Commutative Property Of.
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. (l x v) x r = l x (v x r) in some cases, we can simplify a calculation by multiplying or adding in a different order. Scroll down the page for more examples and solutions.
### Using The Commutative Property Of Multiplication.
The following diagrams show the commutative property of addition and multiplication. They are the commutative property (also called order property), associative property (also called grouping property), distributive. For example, multiplication has a number of properties including the commutative property of multiplication which govern the ways in which users can.
### An Important Example, And In Some Sense.
Let’s look at an example of the commutative property: The result of multiplying 10 x 3 will be equal to multiplying 3 x 10. According to the commutative property of multiplication formula, a × b = b × a.
### There Are Some Properties Of Multiplication.
Math · arithmetic (all content) · multiplication and division · properties of multiplication commutative property of multiplication review review the basics of the commutative. For instance, if we have two positive integers ‘w’ and ‘z’, then the commutative property of multiplication is. (01) which of the following is an example of commutative property of multiplication (a) 2 x 6 = 4.
### A Set Of $7$ Balls.
Changing the order of addends does not change the sum.for example 4 + 2 = 2 + 4. According to the commutative property of multiplication, changing the order of the numbers we are multiplying, does not change the product. This lesson guide includes proofs and examples of the commutative property of addition and the commutative property of. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.14: Scale Factor of Similar Polygons
Difficulty Level: At Grade Created by: CK-12
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Carla enjoys making replicas of houses, planes and cars. She is careful to maintain the same proportional relationship between her replica and the original. She finds that knowing the scale factor makes the process much easier. She is currently working on a replica of a house. The scale factor of the house to her replica is 10.5. If the house is 21 feet tall, how tall is her replica?
In this concept, you will learn about the relationship between scale factors of similar polygons.
### Relationship Between Scale Factors of Similar Polygons
Similar figures are shapes that exist in proportion to each other. They have congruent angles, but their sides are different lengths. Squares, for example, are similar to each other because they always have four 90\begin{align*}90^{\circ}\end{align*} angles and four equal sides, even if the lengths of their sides differ. Other figures can be similar too, if their angles are equal. Let’s look at some pairs of similar figures.
Notice that in each pair the figures look the same, but one is smaller than the other. As you can see, similar figures have congruent angles but sides of different lengths.
Each pair of corresponding sides has the same relationship as every other pair of corresponding sides, so that, altogether, the pairs of sides exist in proportion to each other. For instance, if a side in one figure is twice as long as its corresponding side in a similar figure, all of the other sides will be twice as long too.
These relationships can be used to find the measures of unknown sides. This method is called indirect measurement.
Let’s look at similar figures to understand how indirect measurement works.
Similar figures have exactly the same angles. Therefore each angle in one figure corresponds to an angle in the other.
These triangles are similar because their angles have the same measures. Angle B\begin{align*}B\end{align*} is 100\begin{align*}100^{\circ}\end{align*}. Its corresponding angle will also measure 100\begin{align*}100^{\circ}\end{align*}, that makes angle Q\begin{align*}Q\end{align*} its corresponding angle. Angles A\begin{align*}A\end{align*} and P\begin{align*}P\end{align*} correspond, and angles C\begin{align*}C\end{align*} and R\begin{align*}R\end{align*} correspond.
Similar figures also have corresponding sides, even though the sides are not congruent. Corresponding sides are not always easy to spot. You can think of corresponding sides as those which are in the same place in relation to corresponding angles. For instance, side AB\begin{align*}AB\end{align*}, between angles A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, must correspond to side PQ\begin{align*}PQ\end{align*}, because A\begin{align*}A\end{align*} corresponds to P\begin{align*}P\end{align*} and B\begin{align*}B\end{align*} corresponds to Q\begin{align*}Q\end{align*}.
Corresponding sides also have lengths that are related, even though they are not congruent. Specifically, the side lengths are proportional. In other words, each pair of corresponding sides has the same ratio as every other pair of corresponding sides. Look at the example below.
These rectangles are similar because the sides of one are proportional to the other. You can see this if you set up proportions for each pair of corresponding sides. Let’s put the sides of the large rectangle on the top and the corresponding sides of the small rectangle on the bottom. It doesn’t matter which is put on top, as long as you keep all the sides from one figure in the same place.
LMWXMNXYONZYLOWZ=84=63=84=63\begin{align*}\frac{LM}{WX} &= \frac{8}{4}\\ \frac{MN}{XY} &= \frac{6}{3}\\ \frac{ON}{ZY} &= \frac{8}{4}\\ \frac{LO}{WZ} &= \frac{6}{3}\end{align*}
Now you can clearly see each relationship. To figure out if the pairs do indeed form a proportion, you have to divide the numerator by the denominator. If the quotient is the same, then the ratios each form the same proportion and the figures are similar.
LMWXMNXYONZYLOWZ=84=2=63=2=84=2=63=2\begin{align*}\frac{LM}{WX} &= \frac{8}{4} = 2\\ \frac{MN}{XY} &= \frac{6}{3} = 2\\ \frac{ON}{ZY} &= \frac{8}{4} = 2\\ \frac{LO}{WZ} &= \frac{6}{3} = 2\end{align*}
Each quotient is the same so these ratios are proportional. These quotients are scale factors
The scale factor is the ratio that determines the proportional relationship between the sides of similar figures. For the pairs of sides to be proportional to each other, they must have the same scale factor. In other words, similar figures have congruent angles and sides with the same scale factor. A scale factor of 2 means that each side of the larger figure is twice as long as its corresponding side is in the smaller figure.
### Examples
#### Example 1
Earlier, you were given a problem about Carla and her replica of the house.
She knows that the scale factor of the house to her replica is 10.5. If the house is 21 feet tall, how tall is her replica?
First, set up an equation that reflects the relationship between the heights and the scale factor.
21x=10.5\begin{align*}\frac{21}{x} = 10.5\end{align*}
Then, solve for the missing height.
x=2\begin{align*}x = 2\end{align*}
The answer is that the replica is 2 feet tall.
#### Example 2
What is the scale factor of the figures below?
First, set up the proportions of the sides. Put all the sides from the large figure on top and the sides from the small figure on the bottom.
QRHITSKJRSIJQTHK=155=217=63=155\begin{align*}\frac{QR}{HI} &= \frac{15}{5}\\ \frac{TS}{KJ} &= \frac{21}{7}\\ \frac{RS}{IJ} &= \frac{6}{3}\\ \frac{QT}{HK} &= \frac{15}{5}\end{align*}
Next, divide to find the scale factor.
QRHITSKJRSIJQTHK=155=3=217=3=63=3=155=3\begin{align*}\frac{QR}{HI} &= \frac{15}{5} = 3\\ \frac{TS}{KJ} &= \frac{21}{7} = 3\\ \frac{RS}{IJ} &= \frac{6}{3} = 3\\ \frac{QT}{HK} &= \frac{15}{5} = 3\end{align*}
Then, compare the scale factor.
The factors are all 3.
The answer is that the scale factor of the figures is 3.
#### Example 3
The two figures are similar.
The side lengths are listed as follows:
MN = 3 inches QT = 6 inches
NO = 2 inches QR = 4 inches
MP = 4 inches TS = 8 inches
OP = 2 inches RS = 4 inches
Given the relationship between these sides, what scale factor compares the first figure to the second figure?
First, note the order given by the scale factor.
The scale factor compares the first figure to the second figure.
Next, write the proportion of MN to QT.
36\begin{align*}\frac{3}{6}\end{align*}
Then, simplify the proportion.
12\begin{align*}\frac{1}{2}\end{align*}
The scale factor that compares the first figure to the second figure is 12\begin{align*}\frac{1}{2}\end{align*}.
#### Example 4
Given two figures, the quotients of the corresponding sides equal 3, 3, 5 and 4. Are the two figures similar?
First, state the given information
The quotients are 3, 3, 5 and 4.
Next, understand the relationship between the quotients and similarity.
The quotients of corresponding sides must all equal the same number.
Then, state your conclusion.
Not similar.
The answer is that the two figures are not similar.
#### Example 5
What is the scale factor of the following proportional sides?
186\begin{align*}\frac{18}{6}\end{align*} and 248\begin{align*}\frac{24}{8}\end{align*}
First, find the quotient of the first proportion.
186=3\begin{align*}\frac{18}{6} = 3\end{align*}
Next, find the quotient of the second proportion.
248=3\begin{align*}\frac{24}{8} = 3\end{align*}
Then, compare the two quotients.
Both proportional sides equal 3.
The answer is that the scale factor is 3.
### Review
Find the scale factor of the pairs of similar figures below.
Use each ratio to determine scale factor.
1. 31\begin{align*}\frac{3}{1}\end{align*}
2. 82\begin{align*}\frac{8}{2}\end{align*}
3. 28\begin{align*}\frac{2}{8}\end{align*}
4. 105\begin{align*}\frac{10}{5}\end{align*}
5. 124\begin{align*}\frac{12}{4}\end{align*}
6. 162\begin{align*}\frac{16}{2}\end{align*}
7. 153\begin{align*}\frac{15}{3}\end{align*}
8. 244\begin{align*}\frac{24}{4}\end{align*}
9. 42\begin{align*}\frac{4}{2}\end{align*}
10. 62\begin{align*}\frac{6}{2}\end{align*}
11. 39\begin{align*}\frac{3}{9}\end{align*}
To see the Review answers, open this PDF file and look for section 8.14.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Perimeter
Perimeter is the distance around a two-dimensional figure.
Proportion
A proportion is an equation that shows two equivalent ratios.
Scale Factor
A scale factor is a ratio of the scale to the original or actual dimension written in simplest form.
Similar
Two figures are similar if they have the same shape, but not necessarily the same size.
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Sep 16
## The pearls of AP Statistics 29
### Normal distributions: sometimes it is useful to breast the current
The usual way of defining normal variables is to introduce the whole family of normal distributions and then to say that the standard normal is a special member of this family. Here I show that, for didactic purposes, it is better to do the opposite.
### Standard normal distribution
The standard normal distribution $z$ is defined by its probability density
$p(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}).$
Usually students don't remember this equation, and they don't need to. The point is to emphasize that this is a specific density, not a generic "bell shape".
Figure 1. Standard normal density
From the plot of the density (Figure 1) they can guess that the mean of this variable is zero.
Figure 2. Plot of xp(x)
Alternatively, they can look at the definition of the mean of a continuous random variable $Ez=\int_\infty^\infty xp(x)dx$. Here the function $f(x)=xp(x)$ has the shape given in Figure 2, where the positive area to the right of the origin exactly cancels out with the negative area to the left of the origin. Since an integral means the area under the function curve, it follows that
(1) $Ez=0.$
To find variance, we use the shortcut:
$Var(z)=Ez^2-(Ez)^2=Ez^2=\int_{-\infty}^\infty x^2p(x)dx=2\int_0^\infty x^2p(x)dx=1.$
Figure 3. Plot of x^2p(x)
The total area under the curve is twice the area to the right of the origin, see Figure 3. Here the last integral has been found using Mathematica. It follows that
(2) $\sigma(z)=\sqrt{Var(z)}=1.$
### General normal distribution
Figure \$. Visualization of linear transformation - click to view video
Fix some positive $\sigma$ and real $\mu$. A (general) normal variable $\mu$ is defined as a linear transformation of $z$:
(3) $X=\sigma z+\mu.$
Changing $\mu$ moves the density plot to the left (if $\mu$ is negative) and to the right (if $\mu$ is positive). Changing $\sigma$ makes the density peaked or flat. See video. Enjoy the Mathematica file.
Properties follow like from the horn of plenty:
A) Using (1) and (3) we easily find the mean of $X$:
$EX=\sigma Ez+\mu=\mu.$
B) From (2) and (3) we have
$Var(X)=Var(\sigma z)=\sigma^2Var(z)=\sigma^2$
(the constant $\mu$ does not affect variance and variance is homogeneous of degree 2).
C) Solving (3) for $z$ gives us the z-score:
$z=\frac{X-\mu}{\sigma}.$
D) Moreover, we can prove that a linear transformation of a normal variable is normal. Indeed, let $X$ be defined by (3) and let $Y$ be its linear transformation: $Y=\delta X+\nu.$ Then
$Y=\delta (\sigma z+\mu)+\nu=\delta\sigma z+(\delta\mu+\nu)$
is a linear transformation of the standard normal and is therefore normal.
Remarks. 1) In all of the above, no derivation is longer than one line. 2) Reliance on geometry improves understanding. 3) Only basic properties of means and variances are used. 4) With the traditional way of defining the normal distribution using the equation
$p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})$
there are two problems. Nobody understands this formula and it is difficult to extract properties of the normal variable from it.
Compare the above exposition with that of Agresti and Franklin: a) The normal distribution is symmetric, bell-shaped, and characterized by its mean μ and standard deviation σ (p.277) and b) The Standard Normal Distribution has Mean = 0 and Standard Deviation = 1 (p.285). It is the same old routine: remember this, remember that.
25
Dec 15
## Modeling a sample from a normal distribution in Excel - Exercise 2.4
Modeling a sample from a normal distribution.
Read Exercise 2.4 in the book and Unit 6.6 for the theory. In this video, along with the sample from a normal distribution and a histogram, we construct the density and cumulative distribution function with the same mean and standard deviation.
Simulation steps
1. μ and σ are chosen by the user. I select them so as to model temperature distribution in Almaty.
2. temp (temperature) are just integer number from 14 to 38
3. Use the Excel command norm.dist with required arguments. The last argument should be "false" to produce a density
4. The last argument should be "true" to give a cumulative distribution
5. A combination of the commands norm.dist and rand gives simulated temperature values
6. Next we define the bins required by the Excel function histogram, which is a part of the Data analysis toolpack (needs to be installed and activated)
The fact that Excel maintains links between data and results is very handy. That is, each time the data is renewed, the histogram will change. Pressing F9 (recalculate) you can see the histogram changing together with the sample.
Another interesting fact is that sometimes randomly generated numbers are not realistic. If observed in practice, they could be called outliers. But here they are due to the imperfect nature of the normal distribution, which can take very large (negative or positive) values with a positive probability.
Modeling a sample from a normal distribution - click to view video |
# Combinations
English Português Français Español Italiano Nederlands
Every group of ${\displaystyle k}$ elements chosen from a set of ${\displaystyle n}$ elements in which ordering of the chosen elements is unimportant is called a combination of the ${\displaystyle k}$th order of ${\displaystyle n}$ elements.
## Combinations without repetition
Ordering of elements does not play any role when the number of combinations is to be determined (i.e., groups ${\displaystyle ab}$ and ${\displaystyle ba}$ are equivalent combinations). That is why the number of combinations of the ${\displaystyle k}$th order is lower than the number of variations of the ${\displaystyle k}$th order from the same set of ${\displaystyle n}$ elements. The number of variations, which differ from each other just by ordering of their elements, is given by ${\displaystyle P(k)}$. Hence, the number of combinations of the ${\displaystyle k}$th order of ${\displaystyle n}$ elements without repetition (it is denoted here by ${\displaystyle K(n;k)}$) is: ${\displaystyle K(n;k)={\frac {V(n;k)}{P(k)}}={\frac {n\,!}{k\,!\cdot (n-k)\,!}}=\left({\begin{array}{c}n\\k\end{array}}\right)}$ Examples with elements , and (${\displaystyle n=3}$)
• For ${\displaystyle k=1}$ is ${\displaystyle K(3;1)=3}$ and these three possibilities are:
• For ${\displaystyle k=2}$ is ${\displaystyle K(3;2)=V(3;2)/P(2)=6/2=3}$:
• For ${\displaystyle k=3}$ is ${\displaystyle K(3;3)=V(3;3)/P(3)=3/3=1}$,
so there is just one combination:
## Combinations with repetition
Combinations with repetition can include one element more times; hence, the maximal possible number of combinations of the ${\displaystyle k}$th order of ${\displaystyle n}$ elements with repetition (denoted by ${\displaystyle K^{W}(n;k)}$) is ${\displaystyle K^{W}(n;k)=\left({\begin{array}{c}n+k-1\\k\end{array}}\right)}$ Examples with elements , and (${\displaystyle n=3}$)
• For ${\displaystyle k=1}$ is ${\displaystyle K^{W}(3;1)=3}$ and the three possibilities are:
• For ${\displaystyle k=2}$ is ${\displaystyle K^{W}(3;2)=6}$:
Millions of Germans try every Saturday their luck in the lottery called Lotto. They choose 6 numbers from 49 and hope that, thanks to these 6 numbers, they will get rich. They base the choice often on various almost “mystical” numbers—numbers such as the date of somebody’s birthday, the birthday of their dog, numbers hinted by a horoscope and so on. How many possibilities for a choice of 6 numbers out of 49 actually exists? From 49 numbers (elements), exactly 6 is chosen. The order in which the numbers are chosen in unimportant—it does not matter whether one crosses first 4 and then 23 or vice versa. That means that ordering of elements is not taken into consideration. Therefore, permutations (simple reorderings of ${\displaystyle n}$ elements) and variations as well (ordering of elements matters) are not the right choice. The right concept is a combination. Nevertheless, there are still two possibilities—combinations with or without repetition. Since every number of the lottery ticket can be crossed just once, repetition of numbers (elements) is not possible and we use combinations without repetition. ${\displaystyle n=49\qquad k=6}$ ${\displaystyle K(n,k)=\left({\begin{array}{c}n\\k\end{array}}\right)={\frac {V(n,k)}{P(k)}}={\frac {n\,!}{k\,!\cdot (n-k)\,!}}}$ ${\displaystyle K(n,k)={\frac {49\,!}{6\,!\cdot (49-6)\,!}}=13\,983\,816}$ There is 13983816 possible combinations of 6 numbers from 49. |
Data Structuresand Algorithms Good OldJava AdvancedInterview Topics Cloud andDatabases Web Designand Development Must-knowTools Good ToKnow
## Minimum number of weights
------------------------------------------------------------------------------------
• Choose a minimum number of weights to be able to measure all weights between 1-40 kg.
------------------------------------------------------------------------------------
Soln: To be able to measure 1 kg, one weight of 1 kg is a must.
To be able to measure 40 kg, sum of all weights chosen must be 40 kg.
We can choose 40 weights each of 1 kg.
This will enable us to measure all weights between 1 to 40 kg.
So, there at least is a solution - 40 weights each of 1 kg.
Optimization: Now, this can be optimised by binary method.
If we have one weight of 20 kg and 20 weights of 1 kg, we can still measure all values.
=> number of weights required is 1 + 20 = 21
The 20 weights of 1 kg are used both above and below 20 kg and their range can again be halved.
We can have 1 weight of 20 kg as before and 1 weight of 10 kg and 10 weights of 1 kg.
=> no. of weights required = 1 + 1 + 10 = 12
Half again the number of 1 kg weights
=> no. of weights required = 1 + 1 + 1 + 5 = 8
Again half
=> no. of weights required = 1 + 1 + 1 + 1 + 2 = 6
Thus, 6 weights are needed having weights 20, 10, 5, 3, 1, 1
Now, let us generalize the solution:
If weight range is 1-N, then weights required are:
N/2, N/4, N/8 ... 1, 1 (This last one weight helps to reach 2k from k)
=> Number of weights required are: logN + 1
Better Solution: The above solution does not take into account that the weights can be used as negative weights also. So, the weights need not be kept only on side of the balance, they can be kept on the other side of the balance too to increase the weight of the object.
1 kg weight is still a must.
To measure 2 kg, we can either have 2 kg or 1 kg +1 kg or 3 kg -1 kg weight combinations.
Off course, 3-1 is better, because we can then measure, 1, 2, 3 and 4 kg weights.
To measure 5 kg, we can do it as
5 or 1+4 or (3-1)+3 or 3+2 or 6-1 or 7-(3-1) or 8-3 or 9-(3+1)
Best is to choose last combo so we will get all weights from 1 to 13.
So to measure 1 to 13 kg weights, we need 1, 3, and 9 kg weights.
To measure 14, we need x-13=14 --> x=27 kg weight.
This seems to imply that a series of 1,3,9,27 ... 3^r can measure all
weights between 1 and 1+3+9+27...3^r = (3^r-1)/2
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# The Second Derivative
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## Objective
SWBAT find inflection points and identify concavity.
#### Big Idea
Yes, the derivative of the derivative is a real thing and we will use it to further describe functions.
## Launch and Explore
15 minutes
Today's lesson will build upon what we already know as we look at the second derivative. Students may naturally wonder what the significance of taking the derivative of the derivative is, and we will explore that today.
I start by giving students this worksheet that includes a function, its derivative, and its second derivative. As a refresher, I have students recap what we learned about derivative graphs by looking at f(x) and identifying the critical points and matching them up to the corresponding points on the graph of f'(x). We also recap what we know about the original function if the first derivative is positive or negative.
Next, I explain that we can do the same process again to take the derivative of the derivative. Students may become confused and feel as if this is some type of mathematical inception, but I explain that this is called the second derivative and that it has some special properties.
To build some conceptual understanding of the second derivative, we look at the critical points of the first derivative and notice that those are the zeros of the second derivative. I explain that these points are called inflection points and that they have some significance on the original function. In the video below I explain my next teaching moves to give students some time to brainstorm the importance of inflection points.
## Share
15 minutes
After students have had time to brainstorm about the importance of the inflection points, we will share out our suggestions. Here are some thoughts that my students usually come up with:
• The inflection point is where the steepest slope will occur
• The inflection points split up the graph into parabolas
• The parabolas that the graph is split up into switch from pointing up to pointing down
These are all great suggestions and have some definite ties to concavity. If no one says it outright, at this point I will provide a hint to draw in tangent lines in the three portions of the graph to see if they can notice that the tangent lines switch sides. Once they notice that I will have them draw a diagram like this on their worksheet to summarize the importance of the inflection points.
Next, we will write down the definitions of inflection points, upward concavity, and downward concavity to summarize these main points.
## Extend
20 minutes
Now that we understand concavity and inflection points, we will put it to the test by trying to identify characteristics of a function algebraically without using a graph. Question #2 on the worksheet asks students to find out a lot about a given function. My main intent is that I want students to use sign diagrams to test intervals of the first and second derivative to find out about the original function.
I start by asking students to answer questions #2a with their table. I hope that students algebraically set the first and second derivative to zero to find these values. If they use a graphical approach I ask how they could do it without a graph.
For question #2b, I ask questions until students realize that in order to find out where the graph is increasing, we must find out where the first derivative is positive. I know that we could just graph the function and find these intervals, but I want them to see the connection to derivatives. To find out where the first derivative is positive or negative, I introduce a sign diagram (as shown below) and demonstrate how we can test each interval to see the value of f'(x).
Once students understand the concept of a sign diagram, I have them create one for the second derivative to find the intervals of upward and downward concavity. I also have them finish the rest of the parts for question #2.
We covered a lot today, so I want to make sure that students can keep all of this organized in their mind. To finish the lesson, I ask the following questions (with answers) to summarize everything we learned about derivatives and their relation to the original function.
1. How do we find critical points of a function? Find where the first derivative is equal to zero or undefined.
2. How do we find inflection points of a function? Find where the second derivative is equal to zero.
3. How do we find where the original function is increasing? Find where the first derivative is positive.
4. How do we find where the original function has downward concavity? Find where the second derivative is negative.
Finally, I give this homework assignment to give some practice with the second derivative. |
# 2007 Indonesia MO Problems/Problem 3
## Problem
Let $a,b,c$ be positive real numbers which satisfy $5(a^2+b^2+c^2)<6(ab+bc+ca)$. Prove that these three inequalities hold: $a+b>c$, $b+c>a$, $c+a>b$.
## Solution (credit to vedran6)
Assume that $a+b \le c$, so $c-a-b \ge 0$. Let $x = c-a-b$, making $x \ge 0$. Therefore, $c = a+b+x$.
By substitution, we have \begin{align*} 5(a^2 + b^2 + (a+b+x)^2) &< 6(ab + (a+b)(a+b+x)) \\ 5(a^2 + b^2 + a^2 + b^2 + x^2 + 2ab + 2ax + 2bx) &< 6(ab + a^2 + ab + ax + ab + b^2 + bx) \\ 5(2a^2 + 2b^2 + x + 2ab + 2ax + 2bx) &< 6(3ab + a^2 + ax + b^2 + bx) \\ 10a^2 + 10b^2 + 5x + 10ab + 10ax + 10bx &< 18ab+6a^2+6ax+6b^2+6bx \\ 4a^2 + 4b^2 + 5x - 8ab + 4ax + 4bx &< 0 \\ 4(a-b)^2 + x(5+4a+4b) &< 0 \end{align*} Note that because $x, a, b \ge 0$, we must have $x(5+4a+4b) \ge 0$. Additionally, by the Trivial Inequality, $4(a-b)^2 \ge 0$. Thus, we must have $4(a-b)^2 + x(5+4a+4b) \ge 0$, is a contradiction. Therefore, we must have $a+b > c$.
Because of symmetry in the equation, we can use similar steps to prove that $b+c>a$ and that $c+a>b$. |
# Quick Answer: How Do You Find The Probability Of At Most One?
## How do you find the probability of at least three?
The probability of at least three wins can be expressed as:1 – P(exactly 0 wins) – P(exactly 1 win) – P(exactly 2 wins).So, to solve this, you just need to know how to calculate P(exactly k wins).
To do this, use a binomial random variable of the form:.
## How do you find the exact p value?
If your test statistic is positive, first find the probability that Z is greater than your test statistic (look up your test statistic on the Z-table, find its corresponding probability, and subtract it from one). Then double this result to get the p-value.
## What does at least 1 mean?
“At least one” is a mathematical term meaning one or more. It is commonly used in situations where existence can be established but it is not known how to determine the total number of solutions.
## What does at least mean in math?
At least in math means the smallest possible number in a set or sequence for example can not or must not be less than the given number.
## What does at most one mean in probability?
Atmost means it should be less than or equal to but not more than the said number. For example if in a coin is tossed 4 times, then the probaility of getting head atmost 3 times means, There is probability of success of 3 times or less then 3. 2.8K views.
## What’s the difference between BinomPDF and BinomCDF?
For example, if you were tossing a coin to see how many heads you were going to get, if the coin landed on heads that would be a “success.” The difference between the two functions is that one (BinomPDF) is for a single number (for example, three tosses of a coin), while the other (BinomCDF) is a cumulative probability …
## What is the meaning of at least in probability?
So the probability of at least two heads when tossing 4 coins is 1/16. Mathematically “at least” is the same as “greater than or equal to”. But at “most two” is the same as “less than or equal to” So if you want at most two heads, your winning outco.
## What is the formula of probability?
The probability formula is used to compute the probability of an event to occur. To recall, the likelihood of an event happening is called probability….Basic Probability Formulas.All Probability Formulas List in MathsConditional ProbabilityP(A | B) = P(A∩B) / P(B)Bayes FormulaP(A | B) = P(B | A) ⋅ P(A) / P(B)5 more rows
## What is the probability of getting at least 2 heads?
Then, count the number of combinations with at least 2 heads, which is 4. Finally, divide this number by the total number of combinations, which is 8. Hence the probability of getting at least 2 heads is 48=12.
## How do you find the probability distribution?
How to find the mean of the probability distribution: StepsStep 1: Convert all the percentages to decimal probabilities. For example: … Step 2: Construct a probability distribution table. … Step 3: Multiply the values in each column. … Step 4: Add the results from step 3 together.
## What is meaning of at most?
adjective. The definition of at most is that you have a set maximum amount. An example of at most would be having five minutes to spend on your hair, but not any more than those five minutes.
## How do you find the probability of at least?
To find the probability of at least one of something, calculate the probability of none and then subtract that result from 1. That is, P(at least one) = 1 – P(none). Topford supplies X-Data DVDs in lots of 50, and they have a reported defect rate of 0.5% so the probability of a disk being defective is 0.005.
## What is exact probability?
The probability we computed here is called an “exact” probability—“exact” not because our answer is exactly correct but because the probabilities are calculated exactly, rather than approximated as they are with many statistical tests such as the t-test.
## What are the 5 rules of probability?
Basic Probability RulesProbability Rule One (For any event A, 0 ≤ P(A) ≤ 1)Probability Rule Two (The sum of the probabilities of all possible outcomes is 1)Probability Rule Three (The Complement Rule)Probabilities Involving Multiple Events.Probability Rule Four (Addition Rule for Disjoint Events)Finding P(A and B) using Logic.More items…
## What is the probability of at least one?
To calculate the probability of an event occurring at least once, it will be the complement of the event never occurring. This means that the probability of the event never occurring and the probability of the event occurring at least once will equal one, or a 100% chance. |
# Eureka Math Geometry Module 2 Lesson 1 Answer Key
## Engage NY Eureka Math Geometry Module 2 Lesson 1 Answer Key
### Eureka Math Geometry Module 2 Lesson 1 Example Answer Key
Example 1.
Use construction tools to create a scale drawing of ∆ ABC with a scale factor of r = 2.
Solution 1: Draw $$\overline{A B}$$. To determine B’, adjust the compass to the length of AB. Then reposition the compass so that the point is at B, and mark off the length of $$\overline{A B}$$; label the intersection with as B’. C’ is determined in a similar manner. Join B’ to C’.
Solution 2: Draw a segment that will be longer than double the length of $$\overline{A B}$$. Label one end as A’. Adjust the compass to the length of $$\overline{A B}$$, and mark off two consecutive such lengths along the segment, and label the endpoint as B’. Copy ∠A. Determine C’ along the $$\overline{A B}$$ in the same way as B’. Join B’ to C’.
Example 2.
Use construction tools to create a scale drawing of ∆ XYZ with a scale factor of r = $$\frac{1}{2}$$
Which construction technique have we learned that can be used in this question that was not used in the previous two problems?
We can use the construction to determine the perpendicular bisector to locate the midpoint of two sides of ∆ XYZ.
As the solutions to Exercise 1 showed, the constructions can be done on other sides of the triangle (i.e., the perpendicular bisectors of $$\overline{Y Z}$$ and $$\overline{X Z}$$ are acceptable places to start.)
### Eureka Math Geometry Module 2 Lesson 1 Opening Exercise Answer Key
Above is a picture of a bicycle. Which of the images below appears to be a well-scaled image of the original? Why?
Only the third image appears to be a well-scaled image since the image is in proportion to the original.
### Eureka Math Geometry Module 2 Lesson 1 Exercise Answer Key
Exercise 1.
Use construction tools to create a scale drawing of ∆ DEF with a scale factor of r = 3. What properties does your scale drawing share with the original figure? Explain how you know.
By measurement, I can see that each side is three times the length of the corresponding side of the original figure and that all three angles are equal in measurement to the three corresponding angles in the original figure.
Exercise 2.
Use construction tools to create a scale drawing of A PQR with a scale factor of r = $$\frac{1}{4}$$. What properties do the scale drawing and the original figure share? Explain how you know.
By measurement, I can see that all three sides are each one-quarter the lengths of the corresponding sides of the original figure, and all three angles are equal in measurement to the three corresponding angles in the original figure.
Exercise 3.
Triangle EFG is provided below, and one angle of scale drawing ∆ E’F’G’ is also provided. Use construction tools to complete the scale drawing so that the scale factor is r = 3. What properties do the scale drawing and the original figure share? Explain how you know.
Extend either ray from G’. Use the compass to mark off a length equal to 3EG on one ray and a length equal to 3FG on the other. Label the ends of the two lengths E’ and F’, respectively. Join E’ to F’.
By measurement, I can see that each side is three times the length of the corresponding side of the original figure and that all three angles are equal in measurement to the three corresponding angles in the original figure.
Exercise 4.
Triangle ABC is provided below, and one side of scale drawing ∆ A’B’C’ is also provided. Use construction tools to complete the scale drawing and determine the scale factor.
One possible solution: We can copy ∠A and ∠C at points A’ and C’ so that the new rays Intersect as shown and call the intersection point B’. By measuring, we can see that AT’ = 2AC, A’B’ = 2AB, and B’C’ = 2BC. We already know that m∠A’ = m∠A and m∠C’ = m∠C. By the triangle sum theorem, m∠B’ = m∠B.
### Eureka Math Geometry Module 2 Lesson 1 Problem Set Answer Key
Question 1.
Use construction tools to create a scale drawing of ∆ ABC with a scale factor of r = 3.
Question 2.
Use construction tools to create a scale drawing of ∆ ABC with a scale factor of r = $$\frac{1}{2}$$.
Question 3.
Triangle EFG is provided below, and one angle of scale drawing ∆ E’F’G’ Is also provided. Use construction tools to complete a scale drawing so that the scale factor is r = 2.
Question 4.
Triangle MTC is provided below, and one angle of scale drawing ∆ M’T’C’ is also provided. Use construction tools to complete a scale drawing so that the scale factor is r =$$\frac{1}{4}$$.
Question 5.
Triangle ABC is provided below, and one side of scale drawing ∆ A’B’C’ is also provided. Use construction tools to complete the scale drawing and determine the scale factor.
The ratio of B’C’ : BC is 5 : 1, so the scale factor is 5.
Question 6.
Triangle XYZ is provided below, and one side of scale drawing ∆ X’Y’Z’ is also provided. Use construction tools to complete the scale drawing and determine the scale factor.
The ratio of X’Z’ : XZ is 1: 2, so the scale factor is $$\frac{1}{2}$$.
Question 7.
Quadrilateral GHIJ Is a scale drawing of quadrilateral ABCD with scale factor r. Describe each of the following statements as always true, sometimes true, or never true, and justify your answer.
a. AB = GH
Sometimes true, but only if r = 1.
b. m∠ABC = m∠GHI
Always true because ∠GHI corresponds to ∠ABC in the original drawing, and angle measures are preserved in scale drawings.
C. $$\frac{A B}{G H}=\frac{B C}{H I}$$
Always true because distances in a scale drawing are equal to their corresponding distances in the original drawing times the scale factor r, so $$\frac{A B}{G H}=\frac{A B}{r(A B)}=\frac{1}{r}$$ and $$\frac{B C}{H I}=\frac{B C}{r(B C)}=\frac{1}{r}$$.
d. Perimeter (GHIJ) = r ∙ Perimeter(ABCD)
Always true because the distances In a scale drawing are equal to their corresponding distances in the original
drawing times the scale factor r, so
Perimeter(GHIJ) = GH + HI + iJ +1G
Perimeter(GHIJ) = r(AB) + r(BC) + r(CD) + r(DA)
Perimeter(GHJJ) = r(AB + BC + CD + DA)
Perimeter(GHIJ) = r Perimeter(ABCD).
e. Area (GHIJ) = r ∙ Area(ABCD) where r ≠ 1
Never true because the area of a scale drawing is related to the area of the original drawing by the factor r2.
The scale factor r > 0 and r ≠ 1, so r ≠ r2.
f. r < 0
Never true in a scale drawing because any distance in the scale drawing would be negative as a result of the scale factor and, thus, cannot be drawn since distance must always be positive.
### Eureka Math Geometry Module 2 Lesson 1 Exit Ticket Answer Key
Triangle ABC is provided below, and one side of scale drawing ∆ A’B’C’ is also provided. Use construction tools to complete the scale drawing and determine the scale factor. What properties do the scale drawing and the original figure share? Explain how you know.
One possible solution: Since the scale drawing will clearly be a reduction, use the compass to mark the number of lengths equal to the length of $$\overline{A^{\prime} B^{\prime}}$$ along $$\overline{A B}$$. Once the length of $$\overline{A^{\prime} C^{\prime}}$$ is determined to be $$\frac{1}{2}$$ the length of $$\overline{A B}$$, use the compass to find a length that is half the length of $$\overline{A B}$$ and half the length of $$\overline{B C}$$. Construct circles with radii of lengths $$\frac{1}{2}$$ AC and $$\frac{1}{2}$$ BC
By measurement, I can see that each side is $$\frac{1}{2}$$ the length of the corresponding side of the original figure and that all three angles are equal in measurement to the three corresponding angles in the original figure. |
# 2008 AIME I Problems/Problem 1
## Problem
Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?
## Solutions
### Solution 1
Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance.
Thus, $\dfrac{3k}{5k + 20} = \dfrac{29}{50}$, solving gives $k = 116$. Thus, the number of people that like to dance is $2k + 20 = \boxed{252}$.
### Solution 2
Let the number of girls be $g$. Let the number of total people originally be $t$.
We know that $\frac{g}{t}=\frac{3}{5}$ from the problem.
We also know that $\frac{g}{t+20}=\frac{29}{50}$ from the problem.
We now have a system and we can solve.
The first equation becomes:
$3t=5g$.
The second equation becomes:
$50g=29t+580$
Now we can sub in $30t=50g$ by multiplying the first equation by $10$. We can plug this into our second equation.
$30t=29t+580$
$t=580$
We know that there were originally $580$ people. Of those, $\frac{2}{5}*580=232$ like to dance.
We also know that with these people, $20$ boys joined, all of whom like to dance. We just simply need to add $20$ to get $232+20=\boxed{252}$
## Solution 3
Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: $$0.4p+20 = 0.42(p+20)$$ Solving for p gives us $p=580$, so the solution is $0.4p+20 = \boxed{252}$
## Solution 4 (Cheese)
Assume all the boys like to dance and none of the girls like to dance. We then proceed like the previous solutions.
~Arcticturn
2008 AIME I (Problems • Answer Key • Resources) Preceded byFirst Question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions |
# Search by Topic
#### Resources tagged with Multiplication & division similar to Weigh to Go:
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### There are 132 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Watching the Wheels Go 'round and 'round
##### Stage: 2 Challenge Level:
Use this information to work out whether the front or back wheel of this bicycle gets more wear and tear.
### Practice Run
##### Stage: 2 Challenge Level:
Chandrika was practising a long distance run. Can you work out how long the race was from the information?
### Watch the Clock
##### Stage: 2 Challenge Level:
During the third hour after midnight the hands on a clock point in the same direction (so one hand is over the top of the other). At what time, to the nearest second, does this happen?
### Back to School
##### Stage: 2 Challenge Level:
Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong?
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### Pies
##### Stage: 2 Challenge Level:
Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie?
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Twizzle's Journey
##### Stage: 1 Challenge Level:
Twizzle, a female giraffe, needs transporting to another zoo. Which route will give the fastest journey?
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
### Sometimes We Lose Things
##### Stage: 2 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### What's My Weight?
##### Stage: 2 Short Challenge Level:
There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?
### Oranges and Lemons
##### Stage: 2 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### Wild Jack
##### Stage: 2 Challenge Level:
A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.
### Little Man
##### Stage: 1 Challenge Level:
The Man is much smaller than us. Can you use the picture of him next to a mug to estimate his height and how much tea he drinks?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were making cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Multiplication Series: Illustrating Number Properties with Arrays
##### Stage: 1 and 2
This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Fair Feast
##### Stage: 2 Challenge Level:
Here is a picnic that Chris and Michael are going to share equally. Can you tell us what each of them will have?
### The Money Maze
##### Stage: 2 Challenge Level:
Go through the maze, collecting and losing your money as you go. Which route gives you the highest return? And the lowest?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### What's Left?
##### Stage: 1 Challenge Level:
Use this grid to shade the numbers in the way described. Which numbers do you have left? Do you know what they are called?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Claire's Counting Cards
##### Stage: 1 Challenge Level:
Claire thinks she has the most sports cards in her album. "I have 12 pages with 2 cards on each page", says Claire. Ross counts his cards. "No! I have 3 cards on each of my pages and there are. . . .
### Difficulties with Division
##### Stage: 1 and 2
This article for teachers looks at how teachers can use problems from the NRICH site to help them teach division.
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Function Machines
##### Stage: 2 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### Divide it Out
##### Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Penta Post
##### Stage: 2 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Cows and Sheep
##### Stage: 2 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field.
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### Become Maths Detectives
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? Don't forget to keep visiting NRICH projects site for the latest developments and questions.
### Fingers and Hands
##### Stage: 2 Challenge Level:
How would you count the number of fingers in these pictures?
### How Do You Do It?
##### Stage: 2 Challenge Level:
This group activity will encourage you to share calculation strategies and to think about which strategy might be the most efficient.
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Jumping
##### Stage: 2 Challenge Level:
After training hard, these two children have improved their results. Can you work out the length or height of their first jumps?
### Current Playing with Number Upper Primary Teacher
##### Stage: 2 Challenge Level:
Resources to support understanding of multiplication and division through playing with number.
### What Is Ziffle?
##### Stage: 2 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
### Learning Times Tables
##### Stage: 1 and 2 Challenge Level:
In November, Liz was interviewed for an article on a parents' website about learning times tables. Read the article here.
### Let Us Divide!
##### Stage: 2 Challenge Level:
Look at different ways of dividing things. What do they mean? How might you show them in a picture, with things, with numbers and symbols? |
# Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra states that every polynomial equation f(x) = 0 has at least one root, real or imaginary(complex).
Thus, $x^{6} - 3x^{4} + 2x^{2}$ + 2 = 0 has at least one root.
But $\sqrt{x}$ + 5 = 0 has no root as the given equation is not a polynomial equation, so fundamental theorem of algebra does not apply on this equation.
Note : Every polynomial equation f(x) = 0 of degree 'n' has exactly 'n' real or imaginary roots.
A root is where the polynomial equal to zero and when you graph such polynomial then the roots are the points where graph cut the x-axis. On that point y-coordinate is zero.
From the graph you can see that the polynomial has 3 roots.
## Examples on fundamental theorem of algebra
(i) f(x) = -2 is a polynomial of degree 0 . This is a constant polynomial and it has zero real roots.
(ii) f(x) = x + 4 is a polynomial of degree 1 and it has one real root,
x = -4.
(iii) f(x) = $x^{2}$ + 10x + 25 = $(x + 5)^{2}$ is a polynomial of degree 2 and it has one real root, x = -5.
(iv) f(x) = $x^{3}$ + x = x($x^{2}$ + 1) is a polynomial of degree 3 and it has one real root x = 0 and two imaginary roots, x = −i, +i.
1) Factorize f(x) = $x^{4}$ – 4 using fundamental theorem of algebra.
Solution: Given f(x) = $x^{4}$ – 4
Here, n = 4, so we have 4 complex roots.
f (x) = $x^{4}$ - 4
($x^{2}$ - 2 ) ($x^{2}$ + 2)
$x^{2}$ - 2 = 0 $x^{2}$ + 2 = 0
$x^{2}$ = 2 $x^{2}$ = -2
x= $\pm\sqrt{2}$ x = $\pm\sqrt{2}i$
We have a factored the polynomial into four linear factors, and the zeros of f are $\sqrt{2}, -\sqrt{2},\sqrt{2}i, -\sqrt{2}i$
2) Factorize f(x) = $x^{3}$ – x using fundamental theorem of algebra.
Solution: Given f(x) = $x^{3}$ – x
f(x) = x ($x^{2}$ – 1)
Here, n = 1, so we have zero complex roots.
f(x) = x ($x^{2}$ – 1)
x = 0 and ($x^{2}$ - 1 ) = 0
$x^{2}$ - 1 = 0
$x^{2}$ = 1
x= $\pm\sqrt{1}$
x = 1 and x = - 1
We have a factored the polynomial into three linear factors, and the zeros of f are
0, 1 and -1 |
# How do you differentiate f(x)= (4 x^2 + 2x -3 )/ (x- 1 ) using the quotient rule?
Mar 7, 2018
$f ' \left(x\right) = \frac{4 {x}^{2} - 8 x + 1}{x - 1} ^ 2$
#### Explanation:
This is the quotient rule.
$f \left(x\right) = \frac{\textcolor{red}{u}}{\textcolor{b l u e}{v}}$
$f ' \left(x\right) = \frac{\textcolor{b l u e}{v} \textcolor{red}{u '} - \textcolor{b l u e}{v '} \textcolor{red}{u}}{\textcolor{b l u e}{{v}^{2}}}$
Hence, to differentiate the given $f \left(x\right)$, you do the following
$f ' \left(x\right) = \frac{\left(\textcolor{b l u e}{x - 1}\right) \left(\textcolor{red}{8 x + 2}\right) - \left(\textcolor{b l u e}{1}\right) \left(\textcolor{red}{4 {x}^{2} + 2 x - 3}\right)}{\textcolor{b l u e}{{\left(x - 1\right)}^{2}}}$
$f ' \left(x\right) = \frac{8 {x}^{2} + 2 x - 8 x - 2 - 4 {x}^{2} - 2 x + 3}{x - 1} ^ 2$
$f ' \left(x\right) = \frac{4 {x}^{2} - 8 x + 1}{x - 1} ^ 2$
There is no need to simplify any further after this point |
# Graphing
• Aug 4th 2006, 05:11 AM
pashah
Graphing
Hello I would appreciate any help with the following problem.
• Aug 4th 2006, 05:49 AM
galactus
The graph of $\displaystyle \sqrt{x+5}-2$
Can you see what the domain is?.
• Aug 4th 2006, 05:54 AM
Quick
quick overview: hopefully you know what this $\displaystyle y=\sqrt{x}$ looks like,
now when you see an equation like this $\displaystyle y=\sqrt{x+5}$ you move the graph of $\displaystyle y=\sqrt{x}$ left 5 units, if you see an equation like this $\displaystyle y=\sqrt{x}-2$ you move the graph of $\displaystyle y=\sqrt{x}-2$ down 2 units.
$\displaystyle f(x)=(x+2)(x-2)^2$
the x-intercepts happen when f(x)=0 therefore we have: $\displaystyle 0=(x+2)(x-2)^2$
now anything multiplies by zero is zero, so only one of the two groups has to be zero for the entire equation to go, so we figure out when either of them are zero:
$\displaystyle \begin{array}{cc}x+2=0\Rightarrow x=-2\\x-2=0\Rightarrow x=2\end{array}$
so the y-intercepts are when $\displaystyle x=\pm2$
• Aug 4th 2006, 06:32 AM
Soroban
Hello, pashah!
Quote:
Sketch the graph and state the domain and range.
. . $\displaystyle y \;= \;\sqrt{x+5} - 2$
Since $\displaystyle x + 5$ must be nonnegative: .$\displaystyle x \geq -5$ (domain)
The minimum value of $\displaystyle y$ occurs when $\displaystyle x = -5$
Hence, the range is: .$\displaystyle y \geq -2$
We have: .$\displaystyle y + 2 \:=\:\sqrt{x+5}$
Square the equation: .$\displaystyle (y+2)^2 \:=\:x + 5$
We have a parabola with vertex $\displaystyle (-5,-2)$ which opens to the right: $\displaystyle \subset$
And we want only the upper half.
Code:
| | * * - - - - * - + - - - - - * | * | * | (-5,-2) |
Quote:
Graph and identity all intercepts.
. . $\displaystyle y \;= \;(x+2)(x-2)^2$
If $\displaystyle x = 0,\;y\:=\:2(-2)^2 \:=\:8$ . . . y-intercept: $\displaystyle (0,8)$
If $\displaystyle y = 0,\;x\:=\:\pm2$ . . . x-intercepts: $\displaystyle (\pm2,0)$
This is a cubic graph which rises to the right.
Since $\displaystyle (x-2)^2$ has an even exponent,
. . the graph is tangent to the x-axis at $\displaystyle x = 2$
Code:
| * * * -2 * | * * - - - * - - + - - -*- - - * | 2 * | |
# Texas Go Math Grade 7 Lesson 2.3 Answer Key Proportional Relationships and Graphs
Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 2.3 Answer Key Proportional Relationships and Graphs.
## Texas Go Math Grade 7 Lesson 2.3 Answer Key Proportional Relationships and Graphs
Question 1.
Jared rents bowling shoes for $6 and pays$5 per bowling game. Graph the data. Is the relationship a proportional relationship? Explain.
The relationship isn’t proportional, because the points form a line that does not go through the origin.
Example
The graph shows the relationship between time in minutes and the number of miles Damon runs. Write an equation for this relationship.
STEP 1: Choose a point on the graph and tell what the point represents.
The point (25, 2.5) represents the distance (2.5 miles) that Damon runs in 25 minutes.
STEP 2: What ¡s the constant of proportionality?
Because distance $$\frac{\text { distance }}{\text { time }}=\frac{2.5 \mathrm{mi}}{25 \mathrm{~min}}=\frac{1}{10}$$, the constant of proportionality is $$\frac{1}{10}$$.
STEP 3: Write an equation in the form y = kx.
y = $$\frac{1}{10}$$x
Reflect
Question 2.
Communicate Mathematical Ideas What does the point (0, 0) on the graph represent?
The point (0, 0) represents a start on the graph, origin.
Question 3.
What If? Esther runs faster than Damon. Suppose you drew a graph representing the relationship between time in minutes and distance run for Esther. How would the graph compare to the one for Damon?
Esther’s Line wou[d be steeper than Damon’s
Question 4.
The graph shows the relationship between the distance a bicyclist travels and the time in hours.
a. What does the point (4, 60) represent?
The point (4,60) represents the distance (60 miles) that the bicyclist travels in 4 hours
b. What is the constant of proportionality?
The constant is equal to 15 because $$\frac{\text { distance }}{\text { time }}=\frac{60}{4}$$ = 15. Only one point is enough since the relationship is proportional.
c. Write an equation in the form y = kx for this relationship.
The equation is equal to y = 15x.
Texas Go Math Grade 7 Lesson 2.3 Guided Practice Answer Key
Complete each table. Tell whether the relationship is a proportional relationship. Explain why or why not. xpIore Activity)
Question 1.
A student reads 65 pages per hour.
3 × 65 = 195
5 × 65 = 325
585 ÷ 65 = 9
10 × 65 = 650
The relationship is proportional because we made sure with our calculations that all constants are equal to 65.
Question 2.
A babysitter makes $7.50 per hour. Answer: 2 × 7.51 = 15 22.50 ÷ 7.50 = 3 5 × 7.50 = 37.5() 60 ÷ 7.50 = 8 The relationship is proportional because we made sure with our calculations that all constants are equal to 7.50. Tell whether the relationship is a proportional relationship. Explain why or why not. (Explore Activity and Example 1) Question 3. Answer: The relationship is not proportional because the line does not go through the origin. Question 4. Answer: The relationship is proportional, the constant is equal to 2 because $$\frac{2}{1}$$ = $$\frac{4}{2}$$ = $$\frac{10}{5}$$ = $$\frac{16}{8}$$ = 2. The equation is equal to y = 2x. Write an equation of the form y = kx for the relationship shown in each graph. Question 5. Answer: The relationship is proportional because points form a line through the origin. Thus, we need only one point to determine the constant 7 ÷ 2 = 3.5 y = Balloon height(ft) x = Time(s) Equation: y = 3.5x Question 6. Answer: The relationship is proportional because points form a line through the origin. Thus, we need only one point to determine the constant 2 ÷ 8 = 025 y = Cost ($)
x = Number of items
Equation: y = 0.5x
Essential Question Check-In
Question 7.
How does a graph show a proportional relationship?
The graph forms a line which passes through the origin.
Texas Go Math Grade 7 Lesson 2.3 Independent Practice Answer Key
For Exercises 8-12, the graph shows the relationship between time and distance run by two horses.
Question 8.
Explain the meaning of the point (0, 0).
The point (0,0) represents start, position of horses right before they started to run.
Question 9.
How long does it take each horse to run a mile?
We can see from the graph that horse A takes 4 minutes to run a mile, while horse B takes 2.5 minutes to run a mile.
Question 10.
Multiple Representations Write an equation for the relationship between time and distance for each horse.
Horse A runs 1 mile in 4 minutes. That means he runs $$\frac{1}{4}$$ miles per minute.
Horse B runs 1 mile in 2.5 minutes That means he runs $$\frac{1 \times 2}{2.5 \times 2}$$ = $$\frac{2}{5}$$ miles per minute.
x = Time (min)
y = Distance(miles)
Horse A: y = $$\frac{1}{4}$$ x
Horse B: y = $$\frac{2}{5}$$ x
Question 11.
Draw Conclusions At the given rates, how far would each horse run in 12 minutes?
Use the equations from 11:
Horse A:
y = $$\frac{1}{4}$$ x
y = $$\frac{1}{4}$$(12)3
y = 3
Horse B:
y = $$\frac{2}{5}$$ x
y = $$\frac{2}{5}$$ (12)
y = $$\frac{25}{4}$$
y = 4$$\frac{4}{5}$$
Horse A passes 3 miles in 12 minutes.
Horse B passes 4$$\frac{4}{5}$$ in 12 minutes.
Question 12.
Analyze Relationships Draw a line on the graph representing a horse than runs faster than horses A and B.
Question 13.
A bullet train can travel at 170 miles per hour. Will a graph representing distance in miles compared to time in hours show a proportional relationship? Explain.
The graph will show a proportional relationship because of the constant unit rate, 170 miles per hour.
Question 14.
Critical Thinking When would it be more useful to represent a proportional relationship with a graph rather than an equation?
It would be easier to draw graphs when we have whole numbers
It would not be easy to draw a graph if a constant is a long decimal number, or a fraction with big numbers. Thus, we rather use equation in this case.
Question 15.
Multiple Representations Bargain DVDs cost $5 each at Mega Movie. a. Graph the proportional relationship that gives the cost y in dollars of buying x bargain DVDs. Answer: Graph b. Give an ordered pair on the graph and explain its meaning in the real world context. Answer: The point (4, 20) represents$20 you have to pay for renting 4 DVDs.
The graph shows the relationship between distance and time as Glenda swims.
Question 16.
How far did Glenda swim in 4 seconds?
Glenda swam 8 feet in 4 seconds.
Question 17.
Communicate Mathematical Ideas Is this a proportional relationship? Explain your reasoning.
This is a proportional relationship because points form a line through the origin.
Question 18.
Multiple Representations Write an equation that shows the relationship between time and distance. ________________________________
Because the relationship is proportional we can calculate the constant k. From point (2, 4) we conclude the constant is equal to 2.
x = Time(s)
y = Distance(ft)
Thus, the equation is y = 2x.
H.O.T.S Focus On Higher Order Thinking
Question 19.
Make a Conjecture If you know that a relationship is proportional and are given one ordered pair other than (0, 0), how can you find another pair?
From the point that is given to us we can draw a line on the graph through that point and (0, 0). Then we can find which ever point we need.
The tables show the distance traveled by three cars.
Question 20.
Communicate Mathematical Ideas Which car is not traveling at a constant speed? Explain your reasoning.
Car 3 is not traveling at a constant speed because $$\frac{65}{1}$$ ≠ $$\frac{85}{2}$$.
Question 21.
Make a Conjecture Car 4 is traveling at twice the rate of speed of car 2. How will the table values for car 4 compare to the table values for car 2? |
# Reduction to Linear Form
A LevelAQAEdexcelOCR
## Reduction to Linear Form
Some exponential equations can be reduced to a form that looks like $y=mx+c$. Specifically, after applying the laws of logarithms we can treat $y=ax^{n}$ and $y=ab^{x}$ as if they were linear equations.
A LevelAQAEdexcelOCR
## $\mathbf{y=ax^{n}}$
$y=ax^{n}$
Take logarithms:
\begin{aligned}\log(y)&=\log(ax^{n})\\[1.2em]&=\log(a)+\log(x^{n})\\[1.2em]&=\log(a)+n\log(x)\end{aligned}
Overall we have:
$y=ax^{n}\Rightarrow \log(y)=\log(a)+n\log(x)$
Now if we plot $\log(y)$ against $\log(x)$, we have a straight line graph.
The graph below shows $y$ against $x$ in red and $\log(y)$ against $\log(x)$ in blue. As expected, the blue line is straight.
A LevelAQAEdexcelOCR
## $\mathbf{y=ab^{x}}$
$y=ab^{x}$
Take logarithms:
\begin{aligned}\log(y)&=\log(ab^{x})\\[1.2em]&=\log(a)+\log(b^{x})\\[1.2em]&=\log(a)+x\log(b)\end{aligned}
Overall we have:
$y=ab^{x}\Rightarrow \log(y)=\log(a)+x\log(b)$
Now if we plot $\log(y)$ against $x$, we have a straight line graph.
The graph below shows $y$ against $x$ in red and $\log(y)$ against $x$ in blue. As expected, the blue line is straight.
A LevelAQAEdexcelOCR
@mmerevise
A LevelAQAEdexcelOCR
## Example 1: Converting to Linear Form
Convert $y=99\times0.5^{x}$ to linear form.
[2 marks]
$y=99\times0.5^{x}$
\begin{aligned}\log(y)&=\log(99\times0.5^{x})\\[1.2em]&=\log(99)+log(0.5^{x})\\[1.2em]&=\log(99)+x\log(0.5)\end{aligned}
A LevelAQAEdexcelOCR
## Example 2: Using Linear Form
The data below is taken from a plot of the form $y=ax^{n}$. By plotting $\log_{10}(y)$ against $\log_{10}(x)$, find $a$ and $n$.
[5 marks]
Step 1: Calculate $\log_{10}(x)$ and $\log_{10}y$ and put them in a table.
Step 2: Plot $\log(y)$ against $\log(x)$ on a scatter graph.
Step 3: Use the scatter graph to find the gradient and y-intercept.
Gradient is approximately $\dfrac{3.3-0.6}{0.9-0}=\dfrac{2.7}{0.9}=3$
y-intercept is approximately $0.6$
Step 4: Use our linear form to interpret the gradient and y-intercept.
$y=ax^{n}\rightarrow \log(y)=\log(a)+n\log(x)$
Gradient is $n$ so $n=3$
y-intercept is $\log(a)$ so $\log(a)=0.6$ so $a=10^{0.6}=4$ to two significant figures.
Step 5: Put together to determine the form of the plot:
$y=4x^{3}$
A LevelAQAEdexcelOCR
## Reduction to Linear Form Example Questions
If the question does not specify a base for logarithms it is up to us to choose a sensible base. The choice of base should not affect the answer at the end. In the working below, we have used a base of $2$.
Note the linear form:
$y=ab^{x}\rightarrow log(y)=x\log(b)+\log(a)$
So the straight line is obtained by plotting $\log(y)$ against $x$.
This means that we need to add a $\log(y)$ row to the table.
Plot $\log(y)$ against $x$ and add a line of best fit.
Find the gradient and the y-intercept.
Gradient is $\dfrac{6-9.5}{6-0}=\dfrac{3.5}{6}=-\dfrac{7}{12}$
y-intercept is $9.5$
Gradient is $\log(b)$
$\log(b)=-\dfrac{7}{12}$
$b=0.667$
y-intercept is $\log(a)$
$\log(a)=9.5$
$a=729$
Put it all together:
Our estimate for the line is $y=729\times0.667^{x}$
If the question does not specify a base for logarithms it is up to us to choose a sensible base. The choice of base should not affect the answer at the end. In the working below, we have used a base of $10$.
Note the linear form:
$y=ax^{n}\rightarrow log(y)=n\log(x)+\log(a)$
So the straight line is obtained by plotting $\log(y)$ against $\log(x)$.
This means that we need to add a $\log(x)$ row and a $\log(y)$ row to the table.
Plot $\log(y)$ against $\log(x)$ and add a line of best fit.
Find the gradient and the y-intercept.
Gradient is $\dfrac{3.5-2}{1-0}=\dfrac{1.5}{1}=1.5$
y-intercept is $2$
Gradient is $n$
$n=1.5$
y-intercept is $\log(a)$
$log(a)=2$
$a=100$
Put it all together:
Our estimate for the line is $y=100\times x^{1.5}$
Again we can choose our own base for logarithms, so we will take the logarithm with base $3$.
$y=ab^{x}$
\begin{aligned}\log(y)&=\log(ab^{x})\\[1.2em]&=\log(a)+\log(b^{x})\\[1.2em]&=\log(a)+x\log(b)\end{aligned}
So we plot $\log(y)$ against $x$, and the gradient will be $\log(b)$ and the $y$ intercept will be $\log(a)$.
Plotting $\log(y)$ against $x$ gives this graph.
The graph goes through the points $(1,12.357)$ and $(10,10)$.
\begin{aligned}\text{gradient}&=\dfrac{10-12.357}{10-1}\\[1.2em]&=\dfrac{-2.357}{9}\\[1.2em]&=-0.262\end{aligned}
$\log(b)=-0.262$
$b=3^{-0.262}$
$b=0.750$
The graph passes through $(1,12.357)$ and has a gradient of $-0.262$, so it also passes through $(0,12.357+0.262)=(0,12.619)$
So $y$ intercept is $12.619$
$\log(a)=12.619$
$a=3^{12.619}$
$a=1048576$
Hence, $y=1048576\times0.750^{x}$
A Level
A Level
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# Arithmetic Progressions-Class X-Part 4 | nth term of an AP | Exercise 5.2-5.3 | Sum of n terms (AP)
### Watch the fourth video session on “Arithmetic Progressions” for Class Xth – concepts and solved questions.
Arithmetic Progressions-Class X-Part 4 | nth term of an AP | Exercise 5.2 Q16-20- Exercise 5.3 Q1 | Sum of n terms (AP)
Topics discussed:
nth term of an AP; Exercise 5.2 | Q16-20 Completely Solved with easy explanation
Sum of first n terms of an AP: Expression for some derived:
S = (n/2) [ 2a + (n-1)d ]
S = (n/2) [ first term + last term ] = (n/2) [ a + l ]
Where a = first term, d = common terms, n = number of terms of the given AP, S = Sum of first n terms, l = a+(n-1)d = last term.
Exercise 5.2 | Q16-20 Completely Solved with easy explanation
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Exercise 5.3 | Q1 Completely Solved with easy explanation
1. Find the sum of the following APs:
(i) 2, 7, 12, . . ., to 10 terms.
(ii) –37, –33, –29, . . ., to 12 terms.
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.
(iv) 1/15 , 1/12 , 1/10 , . . ., to 11 terms.
—— |
Session 1, Part C:
Folding Paper
In This Part: Constructions | Constructing Triangles | Concurrencies in Triangles
More Constructions
Problem C3
Illustrate each of these definitions with a sketch using four different triangles. Try to draw four triangles that are different in significant ways -- different side lengths, angles, and types of triangles. The first one in definition (a) is done as an example.
a. A triangle has three altitudes, one from each vertex. (An altitude of a triangle is a line segment connecting a vertex to the line containing the opposite side and perpendicular to that side.) b. A triangle has three medians. (A median is a segment connecting any vertex to the midpoint of the opposite side.) c. A triangle has three midlines. (A midline connects two consecutive midpoints.)
Problem C4
Draw five triangles, each on its own piece of patty paper. Use one triangle for each construction below.
a. Carefully construct the three altitudes of the first triangle. b. Carefully construct the three medians of the second triangle. c. Carefully construct the three midlines of the third triangle. d. Carefully construct the three perpendicular bisectors of the fourth triangle. e. Carefully construct the three angle bisectors of the fifth triangle.
When you construct medians, you need to do two things: First find the midpoint; then fold or draw a segment connecting that point to the opposite vertex. Except in the case of special triangles (such as an equilateral triangle, and one median in an isosceles triangle), you can't construct a median with just one fold. When you construct altitudes, you need to construct a perpendicular to a segment, but not necessarily at the midpoint of that segment. And remember that the altitude may fall outside the triangle, so you might want to draw or fold an extension of the sides of the triangle to help you. Close Tip When you construct medians, you need to do two things: First find the midpoint; then fold or draw a segment connecting that point to the opposite vertex. Except in the case of special triangles (such as an equilateral triangle, and one median in an isosceles triangle), you can't construct a median with just one fold. When you construct altitudes, you need to construct a perpendicular to a segment, but not necessarily at the midpoint of that segment. And remember that the altitude may fall outside the triangle, so you might want to draw or fold an extension of the sides of the triangle to help you.
Video Segment In this video segment, participants construct the altitudes, medians, and midlines of their triangles. Compare your solutions to Problem C4 with those in this video segment. What are the similarities and differences in your results? What conjectures can you make about the constructions you've just completed? If you are using a VCR, you can find this segment on the session video approximately 15 minutes and 40 seconds after the Annenberg Media logo.
Problems C3 and C4 and the Video Segment problems taken from Connected Geometry, developed by Educational Development Center, Inc. p. 32. © 2000 Glencoe/McGraw-Hill. Used with permission. www.glencoe.com/sec/math
Session 1: Index | Notes | Solutions | Video |
If the GMAT were a sport, it would definitely be baseball, and not just because it’s three and a half hours long. In baseball, you might dominate the minor league by hitting fastballs, but once you reach the show you’ll have to hit some change-ups and curveballs too. Not only is the GMAT going to throw you some hard problems, but once you start to do well, the GMAT will throw you something different. That’s why learning the types of trap answers can help you from falling for them. Here’s four types of curveballs that you want to be mindful of on test day.
If you test it, they will come.
Note: the examples below are adapted from Manhattan practice tests and do not include complete explanations of the questions; rather, they are used to show examples of trap answers. If you want to learn more about these problems or others, check out the Manhattan GMAT Forums, a free resource where anyone can ask questions about GMAT problems or test strategies.
1) The C Trap – United We Stand, Divided We Still Stand
For many students who are solving Data Sufficiency problems for the first time, C is a popular answer choice because students will take all information given and try to answer the question. But if you can answer a question with both statements, you have only eliminated answer choice E. That’s why setting up a grid can help to keep you focused on working with one statement at a time.
If x + 2y = z, what is the value of x?
(1) 3y = 4.5 + 1.5z
(2) y = 2
Your first instinct here might be to look at the two statements together and say, if I know y = 2, I could plug that into the first statement to solve for z, and plug y and z into the question to solve for x. And you would have an answer to the question, but you would have the wrong answer to knowing how much information is sufficient to answer the question. Note that our first equation, x + 2y = z, can be rearranged to isolate x, x = z – 2y. If we can find (z – 2y), we know what x is and that’s exactly what we can do with statement 1: 1.5z – 3y = -4.5 or z – 2y = -3. The correct answer is A.
Takeaway: If the statements together can definitely give you an answer, see if one or both statements can give you an answer by themselves.
2) The E Trap – Impossible for You Isn’t Impossible for Everyone
If your computer doesn’t work, your first thought isn’t that it’s impossible to fix. So why do we all get to a tough Data Sufficiency problem and throw our hands in the air saying that there’s no way to solve it. Spoiler alert: in the 13th Edition of the Official Guide where problems are listed in relative order of difficulty, of the 174 DS questions, no question from 151-174 has E as the correct answer. If I ever saw a GMAT DS question written in Turkish, I would guess A-D and assume a Turkish mathematician could answer the question.
In which quadrant of the coordinate plane does the point (x, y) lie?
(1) |xy| + x|y| + |x|y + xy > 0
(2) -x < -y < |y|
Everyone loves questions that include coordinate planes, absolute values, and inequalities. But rather than say there’s no way that I can solve this question, if I had to guess, I would guess between A, B, C, or D. Or, instead of getting flustered, I could plug in four points from each coordinate (i.e. (1,3), (-1,3), (-1,-3), & (1,-3), and discover that only points in quadrant 1 (1,3) work with each statement. The correct answer is D.
Takeaway: If a difficult quant problem looks too complex for you to solve in two minutes, assume someone else could eventually solve it.
The one constant through all the years, Ray, has been the GMAT. America has rolled by like an army of steamrollers. It
3) The 1 Looks Like 2 Trap – The Dead Ringer
If a Data Sufficiency question asked you if x = 2, the statement x < 0 would be more helpful in answering the question (answer = no) than x > 0 would be (answer = maybe). While statements may look similar, their ability to answer a given question can differ greatly.
If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(xy), then what is the value of x?
(1) y is prime
(2) x is prime
These statements are telling us the same thing but about different variables, and that makes a world of difference. After a little canceling, the initial equation can be rephrased as 3xy = 25z or xy = 5 × 5 × z. Since we are trying to solve for x, knowing that y is prime (or not prime) is not helpful to answer the question because z could be anything. But if x must be prime, x must be 5 because there are two fives on the right side of the equation. The correct answer is B.
Takeaway: When statements look alike, don’t assume that they mean alike.
4) The Crucial Info Trap – The Devil is in the Details
If the GMAT tells you a piece of information, there’s usually a reason for it. But all too often, students will spend all their time doing the math and completely forget a crucial statement that the GMAT gave.
If x is a positive integer, what is x?
(1) x2 + 7x – 18 = 0
(2) x2 – 7x + 10 = 0
Besides being a 1 looks like 2 trap, this question has a crucial piece of information hidden at the beginning that leads to a lot of wrong answer choices: x is positive. Statement 1 tells us x could be -9 or 2, but because x must be positive, we know what x must be. Statement 2 tells us x could be 2 or 5, but both of these values are positive. The correct answer is A.
Takeaway: As you read a question, write any important information down on paper so you don’t forget them once you dive into the computations.
#### Joe Lucero
Joe Lucero has both a Biology degree and a Master of Education from the University of Notre Dame. He also has a 780 on his GMAT. In the fall, you will find Joe in a much better mood during weeks after the Fighting Irish win their football game. During the rest of the year, you will find him looking for new places to travel, reading almost anything non-fiction, crossfitting, and trying to solve every challenge problem in the Manhattan GMAT Student Center.
### 9 responses to 4 Common Types of Data Sufficiency Traps
1. I get 3x^y = 25z in the third example…
• Hi Joe
The 3rd example gives 3*x^y=5*5 , please do correct the explanations and help us understand the result
2. Good catch! We’ll get that fixed up ASAP- this would affect the value of z, but the answer to the data sufficiency question remains the same:
3x^y = 5 × 5 × z
if x is prime, it still must be 5.
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# Lesson 14
Estimation Exploration
## Warm-up: Estimation Exploration: Fractional Measurement (10 minutes)
### Narrative
The purpose of an Estimation Exploration is to practice the skill of estimating a reasonable answer based on experience and known information. In the synthesis, discuss what students know about Estimation Explorations and what they need to think about to create one like this example.
### Launch
• Groups of 2
• Display the image.
• “What is an estimate that’s too high? Too low? About right?”
• 1 minute: quiet think time
### Activity
• 1 minute: partner discussion
• Record responses.
### Student Facing
What is the length of this earthworm?
Record an estimate that is:
too low about right too high
$$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$
### Activity Synthesis
• “What do you know about Estimation Explorations?” (It's not about finding the exact answer. We make an estimate that is too low, too high, and about right. The estimate could be the value of an expression, the number of items in an image, or a measurement.)
• Consider asking: “What would you have to think about if you were going to design an Estimation Exploration like this one?” (There is something in the image that gives a clue about the general length of the object. Choosing an object that is not an exact length in whole inches.)
• Record and display responses for all to see.
## Activity 1: Design Your Estimation Exploration (20 minutes)
### Narrative
The purpose of this activity is for students to collaborate and create an Estimation Exploration activity that focuses on fractional measurement. Students find an object from the classroom or an image in a book or another source. They anticipate and record what other students might estimate the length of the object to be.
Representation: Internalize Comprehension. Synthesis: Invite students to identify which details were most useful in creating an estimation exploration. Display the sentence frame, “The next time I create an estimation exploration, I will pay attention to . . . “
Supports accessibility for: Memory
### Required Materials
Materials to Gather
### Required Preparation
• Each group of 2-3 needs picture books and a ruler to design their Estimation Exploration activity.
### Launch
• Groups of 2–3
• Give each group picture books and a ruler.
• “Work with your group to create an Estimation Exploration activity about measuring objects to the nearest half or fourth of an inch.”
### Activity
• 15 minutes: small-group work time
### Student Facing
1. Find an object or an image that would make an interesting estimation problem. Write a question that would encourage others to answer with a length estimated to the nearest half or fourth of an inch.
2. Think about possible estimates others might make for the length of the object or image.
Record an estimate that is:
too low about right too high
$$\phantom{\hspace{2.2cm} \\ \hspace{2.2cm}}$$ $$\phantom{\hspace{2.2cm} \\ \hspace{2.2cm}}$$ $$\phantom{\hspace{2.2cm} \\ \hspace{2.2cm}}$$
3. Measure the length of the object or image to make sure your estimates make sense. If needed, revise your estimates.
### Activity Synthesis
• “What questions do you still have about creating your Estimation Exploration?”
• Give students a few minutes to make adjustments, if needed.
## Activity 2: Facilitate Your Estimation Exploration (15 minutes)
### Narrative
The purpose of this activity is for students to facilitate the Estimation Exploration they created in the previous activity. Each group takes turns facilitating their Estimation Exploration for another group (or two groups, if time permits).
MLR8 Discussion Supports. Synthesis: Display sentence frames to support whole-class discussion: “I learned . . . “ “The next time I create an estimation exploration, I will . . . “
### Required Materials
Materials to Gather
### Required Preparation
• Each group of 2–3 from the previous activity needs 1 piece of chart paper and a marker.
### Launch
• Groups of 2–3 from the previous activity
• Give each group a piece of chart paper and a marker to record responses.
• “Now work with another group and take turns facilitating the Estimation Exploration activity you created.”
### Activity
• 10–12 minutes: small-group work time
• Remind students to switch roles after 5–6 minutes.
### Student Facing
2. Ask them, “What is an estimate that’s too high? Too low? About right?”
3. Give them a minute of quiet think time.
4. Give them a minute to discuss together.
5. Ask them to share their estimates.
6. Record their ideas.
### Activity Synthesis
• “What did you learn as you facilitated your Estimation Exploration?” (The estimates might be much higher or lower than expected. It helps to be familiar with what we're trying to estimate.)
## Lesson Synthesis
### Lesson Synthesis
“What were some important things you considered about the length of the object as you created your Estimation Exploration? Why were they important?” (To make it possible to estimate, we need to show how long 1 inch was compared to the length of the object. The object is not so large or so tiny that it would be hard to estimate to the nearest half or quarter inch, or that it wouldn't make sense to do so.) |
# Class 10 Maths Chapter 4 Quadratic Equations Important Questions
Here are some important Class 10 Mathematics questions for Chapter 4, Quadratic Equations, thoughtfully curated to aid students in their preparation for the CBSE Class 10 Mathematics Examination 2023-24. By practicing a diverse range of question types, students can clarify doubts and enhance their problem-solving skills, leading to improved performance in the Quadratic Equations chapter.
## Introduction
In Chapter 4 of Class 10 Mathematics,Quadratic Equations we will explore the standard form of a quadratic equation.we will delve into the methods of solving quadratic equations, both by factorization and by utilizing the quadratic formula. And we will discuss the Situational problems based on quadratic equations related to day to day activities to be incorporated.
### What are the Quadratic Equations ?
Answer: A quadratic equation is a second-degree polynomial equation of the form ax2 + bx + c = 0, where 'x' is the variable, and 'a', 'b', and 'c' are constants with 'a' not equal to zero. The highest power of 'x' in a quadratic equation is 2, making it a second-degree equation.
The standard form of a quadratic equation is represented as: ax2 + bx + c = 0
#### $$\text{(a) -2 and }{\frac{\textbf{1}}{\textbf{2}}}$$ $$\text{(b) 1 and }{\frac{\textbf{1}}{\textbf{2}}}$$$$\text{(c) -2 and }{\Large\frac{\textbf{1}}{\textbf{2}}}$$(d) 1 and 2
Ans. (a)
Explanation:
Given, 2x2 + x - 1 = 0
⇒ 2x2 + 2x - x - 1 = 0
⇒ 2x (x + 1) - 1 (x + 1) = 0
⇒ (x + 1)(2x - 1) = 0
$$⇒ \text{x = -1 or x =}{\Large\frac{1}{2}}$$
$$\text{Hence roots of equation are -1 and }{\Large\frac{1}{2}}$$
#### (a) 8$$\textbf{(b)}\sqrt{6}$$(c) 5$$\textbf{(d)} 6\sqrt{6}$$
Ans. (c)
Explanation:
For a given equation to have equal roots
D = 0
D = b2 - 4ac
i.e.Here, b = p, a = 2, c = 3
⇒ p2 - 4 × 2 × 3 = 0
⇒ p2 - 24 = 0
$$⇒ \text{p = }\sqrt{24} = 2\sqrt{6}$$
$$\text{Hence, for} p = 2\sqrt{6}$$
$$\text{The given equation has equal roots.}$$
#### Q 3. For what value of k, the quadratic equation kx2 + 11x – 4 = 0 has real roots?
Ans. $$k\geq {\Large \frac{-121}{16}}$$
Explanation:
Kx2 + 11x - 4 = 0
On comparing with ax2 + bx + c = 0, we get
a = k, b = 11, c = - 4∵ It has real roots.
$$⇒ ∴ D\geq 0$$
$$⇒ b - 4ac\geq 0$$
$$⇒ (11) - 4 × k × (- 4)\geq 0$$
$$⇒ 121 + 16k \geq 0$$
$$\text{k}\geq {\Large \frac{-121}{16}}$$
#### Q 4. A girl is twice as old as her sister. Four years hence the product of their ages (in years) will be 160. Find their present ages.
Ans. 12 years
Explanation:
Let the sister’s age be x. Thus, the girl’s age is 2x.
Four years hence,
Sister’s age = (x + 4)
and Girl’s age = (2x + 4)
Now (x + 4) (2x + 4) = 160
⇒ 2x2 + 8x + 4x + 16 = 160
⇒ 2x2 + 12x + 16 = 160
⇒ 2x2 + 12x – 144 = 0
⇒ x2 + 6x – 72 = 0
⇒ x2 + 12x – 6x – 72 = 0
⇒ x(x + 12) – 6(x + 12) = 0
⇒ (x + 12) (x – 6) = 0
⇒ x = 6 or – 12
⇒ x = 6
(as age is always positive)
Thus, the sister’s age is 6 years and the girl’s age is 2 × 6 = 12 years.
#### Q 5. The difference between two natural numbers is 5 and the difference between their reciprocals is $${\Large \frac{5}{11}}$$ . Find the numbers.
Ans. The two numbers are 7 and 2.
Explanation:
Let the natural numbers be x and y respectively.
Now, x – y = 5 …(i)
Also, $${\Large \frac{1}{y}}-{\Large \frac{1}{x}}={\Large \frac{5}{14}}$$
⇒ 14x – 14y = 5xy
⇒ 14(x – y) = 5xy
⇒ 14 × 5 = 5xy [From (i)]
⇒ xy = 14
⇒ $$x = {\Large \frac{14}{y}}$$
Substituting $$x = {\Large \frac{14}{y}}$$ Substituting
$${\Large \frac{14}{y}}- y = 5$$
⇒ $$y -{\Large \frac{14}{y}}$$ + 5 = 0
⇒ y2 + 5y – 14 = 0
⇒ y2 + 7y – 2y – 14 = 0
⇒ y( y + 7) – 2( y + 7) = 0
⇒ (y + 7) ( y – 2) = 0
⇒ y = 2, – 7
Since natural numbers cannot be –ve, so y = 2.
Thus, from equation (i),
x – y = 5
⇒ x – 2 = 5
⇒ x = 7
So, the two numbers are 7 and 2.
#### CBSE Class 10 Maths Chapter wise Important Questions
Chapter No. Chapter Name
Chapter 1 Real Number
Chapter 2 Polynomials
Chapter 3 Pair of Linear Equations in Two Variables
Chapter 5 Arithmetic Progressions
Chapter 6 Triangles
Chapter 7 Coordinate Geometry
Chapter 8 Introduction to Trigonometry
Chapter 9 Some Applications of Trigonometry
Chapter 10 Circles
Chapter 11 Areas Related to Circle
Chapter 12 Surface Areas and Volumes
Chapter 13 Statistics
Chapter 14 Probability
#### Conclusion
If you want to improve your understanding of the concepts in this chapter, you can visit oswal.io. They have a wide range of questions that will help you practice and reinforce what you've learned. By solving these questions, you can strengthen your knowledge and become better at solving problems.
#### Q1: How many solutions can a quadratic equation have?
Ans: A quadratic equation can have two solutions, one solution (double root), or no real solutions (complex roots), depending on the value of the discriminant (b2 - 4ac).
#### Q2: Can a quadratic equation have complex roots?
Ans: Yes, a quadratic equation can have complex roots (imaginary roots) if the discriminant is negative.
#### Q3: How can I solve a quadratic equation by factorization?
Ans: To solve a quadratic equation by factorization, you need to express it as a product of two binomials and set each factor equal to zero to find the solutions.
#### Q4: How do I determine the number of solutions of a quadratic equation without solving it?
Ans: You can determine the number of solutions by looking at the value of the discriminant:
• If the discriminant is greater than zero, the equation has two distinct real roots.
• If the discriminant is zero, the equation has one real root (double root).
• If the discriminant is negative, the equation has no real solutions (complex roots).
#### Q5: What is the difference between a quadratic equation and a linear equation?
Ans: A quadratic equation is a second-degree polynomial equation (ax2 + bx + c = 0), while a linear equation is a first-degree polynomial equation (ax + b = 0). The highest power of the variable 'x' in a quadratic equation is 2, while in a linear equation, it is 1.
## Chapter Wise Important Questions for CBSE Board Class 10 Maths
Real Numbers
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## 3D Coordinates (Q’s & Ans) || V.S.A.Q’S||
These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students.
Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.
### 3D Coordinates
1. Show that the points A (– 4, 9, 6), B (– 1, 6, 6), and C (0, 7, 10) form a right-angled isosceles triangle.
Sol:
Distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is PQ =
∴ The points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle.
2. Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 = 0
Sol:
Let P (x, y, z) be the locus of the point
A (0, y, 0) be any point on Y – axis
B = (1, 2, – 1)
Condition is PA = 3PB
PA2 = (3PB)2
PA2= 9 PB
⟹ x2 + z2 = 9[(x – 1)2 + (y – 2)2 + (z + 1)2]
x2 + z2 = 9[x2 – 2x + 1 + y2 – 4y + 4 + z2 + 2z + 1]
x2 + z2 = 9x2 – 18x + 9 + 9y2 – 36y + 36 +9 z2 + 18z + 9
∴ 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 = 0
3. A, B, and C are three points on OX, OY, and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C, and O
Sol:
Let P (x, y, z) be the required point
O = (0, 0, 0) A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)
Given, AP = BP = CP = OP
AP = OP
⟹ AP2 =OP2
(x – a )2 + y2 + z2 = x2 + y2 + z2
x2 + a2 – 2ax + y2 + z2 = x2 + y2 + z2
a2 – 2ax = 0
a (a – 2x) = 0
a – 2x = 0 (∵ a≠0)
a = 2x ⟹ a/2
Similarly, y = b/2 and z = c/2
∴ P = (a/2, b/2, c/2)
4. Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear
Sol:
Given points are A (3, – 2, 4), B (1, 1, 1), and C (– 1, 4, – 2)
5.Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.
Sol:
Let A = (5, – 1, 7), B = (x, 5, 1)
Given AB = 9
⟹ AB2 = 81
(5 – x)2 + (– 1 – 5)2 + (7 – 1)2 = 81
(5 – x)2 + 36 + 36 = 81
(5 – x)2 + 72 = 81
(5 – x)2 = 81 – 72 = 9
(5 – x) = ± 3
5 – x = 3 or 5 – x = – 3
5 – 3 = x or 5 + 3 = x
x = 2 or x = 8
6.If the point (1, 2, 3) is changed to point (2, 3, 1) through the translation of axes. Find a new origin.
Sol:
Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)
x = X + h, y = Y + k, z = Z + l
h = x – X, k = y – Y, l = z – Z
h = 1 – 2, k = 2 – 3, l = 3 – 1
h = – 1, k = – 1, l = 2
New origin is (– 1, – 1, 2)
7.By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.
Sol:
If a point P divides the line segment joining the points (x1, y1, z1), (x2, y2, z2) in the ratio, then
Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3
8.Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).
Sol:
Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)
Suppose P divides AB in the ratio k : 1
9.Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.
Sol:
Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)
Let C divides AB in the ratio k : 1
2 – 4k = 4 (k + 1)
2 – 4k = 4k + 4
– 4k– 4k = 4 – 2
– 8k = 2
K = -1/4
C divides AB in the Ratio 1 : 4 externally
∴ A, B, C are collinear
10.Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)
Sol:
The centroid of the triangle whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is
11. Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)
Sol:
The centroid of the tetrahedron whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is
the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is
12. Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)
Sol:
Let P be any point on the YZ-plane
P = (0, y, z)
Let P divides AB int eh ratio k:1
YZ-plane divides AB in the ratio – 2:3
13. Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).
Sol:
let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)
ABCD is a parallelogram
Midpoint of AC = Midpoint of BD
⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0
x = 3, y = 3, z = 1
∴ The fourth vertex D = (3, 3, 1)
14. A (5, 4, 6), B (1, –1, 3), and C (3, 3, 1) are three points. Find the coordinates of the point at which the bisector of ∠BAC meets the side BC.
Sol:
We know that the bisector of ∠BAC divides BC in the ratio AB: AC
If D is a point where the bisector of ∠BAC meets BC
⟹ D divides BC in the ratio 5:3
15. If M (α, β, γ) is the midpoint of the line joining the points (x1, y1, z1) and B, then find B
Sol:
Let B (x, y, z) be the required point
M is the midpoint of AB
⟹ (α, β, γ) =
⟹ 2 α = x + x1; 2 β = y + y1; 2 γ = z + z1
x =2 α – x1; x =2 β – y1; x =2 γ – z1
∴ B = (2 α – x1, 2 β – y1, 2 γ –)
16. If H, G, S, and I respectively denote the orthocenter, centroid, circumcenter, and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.
Sol:
Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)
⟹ ∆ ABC is an equilateral triangle
We know that, in an equilateral triangle orthocenter, centroid, circumcenter, and incentre are the same
Centroid G =
= (2, 2, 2)
∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)
17. Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).
Sol:
Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)
∴ I = (1, 1,0)
18. Find the ratio in which point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P
Sol:
Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m : n.
Given points are A (3, 2, – 4), B (9, 8, –10), and P (5, 4, – 6)
P divides AB in the ratio is x2 – x : x – x2
= 3 – 5 : 5 – 9
= 1 : 2 (internally)
Let Q be the harmonic conjugate of P
⟹ Q divides AB in the ratio –1 : 2
Q =
= (–3, –4, –2)
Q (–3, –4, –2) is the hormonic conjugate of P (5, 4, – 6)
19. If (3, 2, – 1), (4, 1,1), and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the tetrahedron, find the fourth Vertex.
Sol:
Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)
Let the fourth vertex is D = (x, y, z)
The centroid of the tetrahedron whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is
13 + x = 16, 5 + y = 8, 5 + z = 2
x = 3, y = 3, z = 3
∴ the fourth vertex D = (3, 3, 3)
20. Show that the points A(3, 2, –4), (5, 4, –6), and C(9, 8, –10) are collinear and find the ratio in which B divides AC.
Sol:
Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)
∴ points A (3, 2, –4), (5, 4, –6), and C (9, 8, –10) are collinear
B divides AB in the ratio is AB: BC = |
# Normal Distribution & Shifts in the Mean
Coming up next: Identifying & Calculating Averages
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• 0:05 Normal Distribution
• 1:35 Characteristics
• 3:04 Shifts to the Left & Right
• 3:50 Randomness
• 4:36 Lesson Summary
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
In this video lesson, you will see what a normal distribution looks like and you will learn about the mean of a normal distribution. You will also see what happens when the mean is shifted.
## Normal Distribution
When businesses use graphs to show people their data, they have words to describe how the graph looks and how the data is spread out over the graph. In this video lesson, we talk about the kinds of graphs that have a normal distribution where the data tends towards a middle value with equal amounts to the left and right.
A graph with a normal distribution looks like the outline of a bell, hence it is also referred to as a bell curve. You can see that most of the data is in the middle of the graph where the curve is the highest. There is less data to the left and right of the graph. There is also an equal amount of data to the left and right of the graph.
In the real world, there are a surprising number of events that follow a normal distribution. For example, the height of people follows a normal distribution. This is why we can say that, in general, people will grow to a certain height. Because of the normal distribution, though, some people will be shorter and some will be taller, but most people will be the average height.
In business, a graph that shows the number of customers that come during the day might also show a normal distribution. For example, a restaurant that serves lunch will find a normal distribution that shows that the majority of its customers come at lunch time while a few come before and after the lunch rush.
## Characteristics
Because there is an equal amount of data to the left and right, the mean or average of this graph is the middle of the graph, the point where the curve is the highest. If you folded this graph in half, the left and right sides would mirror each other with the graph trailing off to a zero to the left and right. A normally distributed graph has half of the data to the left and the other half to the right. There is an equal number of data to either side. The data does not prefer one side over the other.
We also have another term we use when describing graphs. It is called the standard deviation. It is a measure of how spread out the data is. For a graph with a normal distribution, 34 percent of the data is within 1 standard deviation from the middle. If we take 1 standard deviation to the left and right, we will have 68 percent of our data. If we take 2 standard deviations to the left and right, we will have 95 percent of our data. Within 3 standard deviations, we will have 99.7 percent of our data. These standard deviations divide each side of our graph into fourths. The first quarter is the one closest to the middle while the fourth quarter covers the trailing off to zero part.
## Shifts to the Left and Right
When you see a graph where the tip of the curve is shifted to either the left or the right, we no longer have a normal distribution. We now have a graph that shows a preference for one side or the other. Now our average from the data will not be the exact middle of the graph. Instead, our average now is shifted to the left or right depending on which direction our tip of the curve is shifted. If our curve tip is shifted to the left, then so is our average. If our curve tip is shifted to the right, then our average is shifted to the right, as well.
With these kinds of graphs, we still see the data tending to a particular value. The particular value just won't be the middle value.
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# Relationships between Coefficients of Nature of Solutions
Go back to 'Pair of Linear Equations in Two Variables'
The question now is: given a pair of linear equations, can we figure out what kind of solution(s) it will have based on the coefficients. Let us state this problem more formally. Consider the following pair of linear equations:
$\begin{array}{l}{a_1}x + {b_1}y + {c_1} = 0\\{a_2}x + {b_2}y + {c_2} = 0\end{array}$
Using the coefficients in the two equations, can we determine whether this pair will have a unique solution, no solution or infinitely many solutions? Yes, we can. Let us understand how. For now, we will keep our discussion intuitive, but you will understand this with more rigor in a higher class.
Suppose that we rearrange and write the linear equations as follows:
\begin{align}&\left\{ \begin{array}{l}\frac{{{a_1}}}{{{b_1}}}x + y + \frac{{{c_1}}}{{{b_1}}} = 0\\\frac{{{a_2}}}{{{b_2}}}x + y + \frac{{{c_2}}}{{{b_2}}} = 0\end{array} \right.\\ &\Rightarrow \;\;\;\left\{ \begin{gathered}y = - \frac{{{a_1}}}{{{b_1}}}x - \frac{{{c_1}}}{{{b_1}}}\\y = - \frac{{{a_2}}}{{{b_2}}}x - \frac{{{c_2}}}{{{b_2}}}\end{gathered} \right.\end{align}
Compare these equations to the following linear equation:
$y = mx + b$
Recall that m is the slope of the line, that is, m determines the steepness of the line, while b is the y intercept and controls the vertical positioning of the line. Clearly, the slopes of the two lines above will be:
\begin{align}&{m_1} = - \frac{{{a_1}}}{{{b_1}}}\\&{m_2} = - \frac{{{a_2}}}{{{b_2}}}\end{align}
If we want the pair of linear equations to have a unique solution, the slopes of the two lines should not be the same. This should be intuitively obvious. Thus, for a unique solution to exist, we must have:
\begin{align}&{m_1} \ne {m_2}\\&\Rightarrow \;\;\; - \frac{{{a_1}}}{{{b_1}}} \ne - \frac{{{a_2}}}{{{b_2}}}\\&\Rightarrow \;\;\;\frac{{{a_1}}}{{{b_1}}} \ne \frac{{{a_2}}}{{{b_2}}}\\&\Rightarrow \;\;\;\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align}
On the other hand, for no solution to exist, the lines must be parallel, which means that their slopes must be equal. Also, their y intercepts should not be the same; otherwise they will be identical lines. Thus, for no solution to exist, the condition on the coefficients will be:
\begin{align}&\left\{ \begin{array}{l}({\rm{i}})\;\;{m_1} = {m_2}\\ \Rightarrow \;\;\;\frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}}\;\; \Rightarrow \;\;\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}}\{\rm{ii}})\;\;\; - \frac{{{c_1}}}{{{b_1}}} \ne - \frac{{{c_2}}}{{{b_2}}}\\ \Rightarrow \;\;\;\frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{array} \right.\\&\;\;\; \Rightarrow \;\;\;\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align} Finally, for the pair of linear equations to have infinitely many solutions, the two lines must be the same. This means that they should have the same slope, and their y intercepts should also be the same. Thus, the condition on the coefficients is: \begin{align}&\left\{ \begin{array}{l}({\rm{i}})\;\;\;{m_1} = {m_2}\\\;\;\; \Rightarrow \;\frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}} \Rightarrow \;\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}}\\({\rm{ii}})\;\;\; - \frac{{{c_1}}}{{{b_1}}} = - \frac{{{c_2}}}{{{b_2}}}\\ \Rightarrow \;\;\;\frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{array} \right.\\&\;\; \Rightarrow \;\;\;\frac{{{a_1}}}{{{a_2}}} = \quad\frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align} Note that this essentially means that one linear equation can be expressed as a multiple of the other. Only then can the constraint above be satisfied. When a linear pair of equations has one solution (intersecting lines) or infinitely many solutions (coincident lines), we say that it is a consistent pair. On the other hand, when a linear pair has no solution (parallel, non-coincident lines), we say that it is an inconsistent pair. Example 1: Consider the following pair of linear equations: $\begin{array}{l}5x + 3y + 4 = 0\\ - 3x + 5y + 1 = 0\end{array}$ Using the coefficients, check whether this pair is consistent. If it is, evaluate its solution graphically. Solution: We have: $\frac{5}{{ - 3}} \ne \frac{3}{5}$ This means that the given pair will have a unique solution, that is, it is a consistent pair. Now, we plot the lines corresponding to the two equations, and determine the point of intersection. In the following graph, we have plotted a pair of points for each line: Thus, the solution to this linear pair is $x = - \frac{1}{2},\;y = - \frac{1}{2}$ Example 2: Three linear equations are given below: \(\begin{array}{l}{\rm{I}}.\;\;\;\;2x - 3y + 1 = 0\\{\rm{II}}.\;\;\;7x + 4y - 3 = 0\\{\rm{III}}.\;\; - 4x + 6y + 5 = 0\end{array}
Identify the inconsistent pair(s) of equations.
Solution: I and II. We have:
$\frac{2}{7} \ne \frac{{ - 3}}{4}$
This pair is consistent and will have a unique solution.
II and III. We have:
$\frac{7}{{ - 4}} \ne \frac{4}{6}$
This pair is also consistent and will have a unique solution.
I and III. We have:
$\frac{2}{{ - 4}} = \frac{{ - 3}}{6} \ne \frac{1}{5}$
This pair is clearly inconsistent.
Example 3: Consider the following pair of linear equations:
$\begin{array}{l} 5x + 3y = - 4\\ - 3x + 5y = - 1 \end{array}$
a) Graphically determine the solution to this pair.
b) Using a protractor, measure the angle between the two lines.
c) Determine the product of the slopes of the two lines.
Solution: We have already solved this pair a while back. The solution is:
$x = - \frac{1}{2},\;y = - \frac{1}{2}$
Below, we reproduce the graph once again, and using a protractor, we find that the two lines are perpendicular:
Now, we calculate the slopes of the two lines by writing them as follows:
$\begin{array}{l}\left\{ \begin{array}{l}y = - \frac{5}{3}x - \frac{4}{3}\\y = \frac{3}{5}x - \frac{1}{5}\end{array} \right.\\ \Rightarrow \;\;\;{m_1} = - \frac{5}{3},\;{m_2} = \frac{3}{5}\end{array}$
The product of the two slopes is:
${m_1} \times {m_2} = - \frac{5}{3} \times \frac{3}{5} = 1$
This is an extremely important connection, which you will study in more detail later. When the product of the slopes of two lines corresponding to a pair of linear equations is $$- 1$$, the two lines will be perpendicular.
Pair of Linear Equations in 2 Variables
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 2
Pair of Linear Equations in 2 Variables
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 1 |
Tips to Help Children Solve Math Word Problems
Last Updated on August 31, 2021 by Thinkster
Math word problems help students apply the skills that they have learned while doing repeated math facts. The beauty lies in how many different ways a simple word problem can be worded.
Take 34 – 12, for example. All of the following word problems are based on this one simple subtraction problem:
• Mike has 34 baseball cards. He gives away 12 to his best friend, Arthur. How many does he have left?
• Liz scored 12 points less than Beth who scored 34 points. What did Liz score?
• Beth scored 34 points, which was 12 more than what Liz scored. What was Liz’s score?
You may be able to solve 34 – 12 in a fraction of a second.
However, what good is it if you can’t answer any of the above three questions?
Math word problems build strong critical thinking, reading comprehension, and analytical skills, which are all important skills for future academic and professional success.
But some students absolutely dread when word problems come up in math class.
With our simple tips, you can help eliminate any potential frustration, dread, fear, or anxiety when it comes time to solve a math word problem!
Understand what the word problem is about
Pay attention to the keywords. At the same time, you don’t blindly associate a keyword with an operation. For example, it’s easy to assume that you could relate “less than” to subtraction and “more than” to addition. However, in Question 3, you can see the keywords “more than” but the operation to be used is actually subtraction.
Another essential strategy is to be mindful of the keywords and how they are used to apply the right operation. To do that, you have to understand the relationship between the numbers in the problem.
Would this happen in real life?
One inherent problem with math word problems is that there are some underlying assumptions that sometimes may seem absurd. Take these problems for example:
1. “Dan runs 20 miles in 3 and a half hours. How long will he take to run 100 miles?”
2. “If 4 handymen repair 8 garages in 30 minutes, how many garages in all will they be able to repair in 5 hours?”
The implicit assumptions in the first problem are:
1. Dan can run 100 miles.
2. Dan will run the next 80 miles at the same pace.
The implicit assumptions in the second problem are:
1. All 4 handymen repair at the same rate.
2. All of them continue repairing without taking any breaks.
3. The time to travel from one garage to another is negligible.
All of these seem absurd and unrealistic. However, without these assumptions, the above problems will prove to be extremely complicated and will become almost impossible to solve. You are expected to ignore these and solve the problems. It makes sense to do so, as it helps you apply what you have learned in situations that resemble what you see in your day-to-day life.
Spot the red herrings
Sometimes, word problems have extraneous information that you should ignore. Let us look at an example.
Matthew takes the bus to school every day. The bus ride takes 20 minutes. To take the bus he has to walk for 5 minutes from his home. If he goes to school 5 days a week, how much time does he spend on the bus?
In this problem, you are given 3 different numbers, each of which gives you some information about Matthew’s trip to school. What you are asked to find is the total time that he spends on the bus in a week. Pay close attention and you will see that the time he spends walking from home to the bus stop is irrelevant. The extra information is given just to misdirect you, to make sure you pay close attention and you understand the information given and what you have to determine to solve the problem.
Remember, practice improves performance!
Why is it some students loathe or fear word problems? Even a simple word problem could have a student floundering!
Sure, it could be that they are still developing the critical thinking and logical reasoning skills to successfully tackle these problem types.
(Use our tips above if this is the case!)
But it could also simply be that a student isn’t receiving sufficient practice to develop mastery and confidence.
Here’s the truth: It’s difficult for a mathematics teacher to alter the pace of the math lesson for each individual student. One teacher for a class of 20-30 students means that there is a particular pace at which the class moves.
This is why it’s important for students to get ample practice outside of the classroom to develop true proficiency and mastery. This independent practice solidifies their understanding of math topics!
So, how can you ensure your child gets ample practice? There are many resources available that you can take advantage of! A math word problem worksheet specific to what your child learned at school (ie, multiplication, decimal, or subtraction word problems) is the perfect follow up to school lessons.
Learn to tackle math word problems when a math champion!
As you can see, word problems, in fact, help you with reading, thinking, and analysis, in addition to computation. When looking for a math program, try to find a program that emphasizes building critical thinking skills.
After all, these are the types of skills incredibly important for future academic and professional success!
Thinkster Math‘s world-class curriculum is filled with analytical, critical thinking, and logical reasoning math problems. We help build the necessary foundations that will help to make your child a math champion!
Your child works with one dedicated math tutor, whose goal is to help your child develop a variety of math and life skills.
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Article Name
Tips to Help Children Solve Math Word Problems
Description
Math problems help children apply what they learnt. It is a great way of relating math to the real-world, and that is where its important lies.
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# Difference between revisions of "2016 AIME I Problems/Problem 1"
## Problem 1
For $-1, let $S(r)$ denote the sum of the geometric series $$12+12r+12r^2+12r^3+\cdots .$$ Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$. Find $S(a)+S(-a)$.
## Solution 1
We know that $S(r)=\frac{12}{1-r}$, and $S(-r)=\frac{12}{1+r}$. Therefore, $S(a)S(-a)=\frac{144}{1-a^2}$, so $2016=\frac{144}{1-a^2}$. We can divide out $144$ to get $\frac{1}{1-a^2}=14$. We see $S(a)+S(-a)=\frac{12}{1-a}+\frac{12}{1+a}=\frac{12(1+a)}{1-a^2}+\frac{12(1-a)}{1-a^2}=\frac{24}{1-a^2}=24*14=\fbox{336}$
## Solution 2
The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$. The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$ so dividing by $144$ gives $\frac{1}{1-a^2}=14\implies a= \pm \sqrt{\frac{13}{14}}$. $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$, so the answer is $14\cdot 24=\boxed{336}$. |
Paul's Online Notes
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### Section 1.5 : Solving Trig Equations with Calculators, Part I
7. Find the solution(s) to $$\displaystyle 17 - 3\sec \left( {\frac{z}{2}} \right) = 2$$ that are in $$\left[ {20,45} \right]$$. Use at least 4 decimal places in your work.
Show All Steps Hide All Steps
Hint : Find all the solutions to the equation without regard to the given interval. The first step in this process is to isolate the secant (with a coefficient of one) on one side of the equation.
Start Solution
Isolating the secant (with a coefficient of one) on one side of the equation gives,
$\sec \left( {\frac{z}{2}} \right) = 5$
Hint : Using a calculator and your knowledge of the unit circle to determine all the angles in the range $$\left[ {0,2\pi } \right]$$ for which secant will have this value. The best way to do this is to rewrite the equation into one in terms of a different trig function that we can more easily deal with.
Show Step 2
In order to get the solutions it will be much easier to recall the definition of secant in terms of cosine and rewrite the equation into one involving cosine. Doing this gives,
$\sec \left( {\frac{z}{2}} \right) = \frac{1}{{\cos \left( {\frac{z}{2}} \right)}} = 5\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\cos \left( {\frac{z}{2}} \right) = \frac{1}{5}$
The solution(s) to the equation involving the cosine are the same as the solution(s) to the equation involving the secant and so working with that will be easier. Using our calculator we can see that,
$\frac{z}{2} = {\cos ^{ - 1}}\left( {\frac{1}{5}} \right) = 1.3694$
Now we’re dealing with cosine in this problem and we know that the $$x$$-axis represents cosine on a unit circle and so we’re looking for angles that will have a $$x$$ coordinate of $$\frac{1}{5}$$. This means that we’ll have angles in the first (this is the one our calculator gave us) and fourth quadrant. Here is a unit circle for this situation.
From the symmetry of the unit circle we can see that we can either use –1.3694 or $$2\pi - 1.3694 = 4.9138$$ for the second angle. Each will give the same set of solutions. However, because it is easy to lose track of minus signs we will use the positive angle for our second solution.
Hint : Using the two angles above write down all the angles for which cosine/secant will have this value and use these to write down all the solutions to the equation.
Show Step 3
From the discussion in the notes for this section we know that once we have these two angles we can get all possible angles by simply adding “$$+ \,2\pi n$$ for $$n = 0, \pm 1, \pm 2, \ldots$$” onto each of these.
This then means that we must have,
$\frac{z}{2} = 1.3694 + 2\pi n\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}\frac{z}{2} = 4.9138 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots$
Finally, to get all the solutions to the equation all we need to do is multiply both sides by 2 and we’ll convert everything to decimals to help with the final step.
\begin{align*}z & = 2.7388 + 4\pi n& \hspace{0.25in}\,\,\,{\mbox{OR }}\hspace{0.25in} & z = 9.8276 + 4\pi n & \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \\ & = 2.7388 + 12.5664n & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} & \hspace{0.1in}= 9.8276 + 12.5664n\hspace{0.25in}&n = 0, \pm 1, \pm 2, \ldots \end{align*}
Hint : Now all we need to do is plug in values of $$n$$ to determine which solutions will actually fall in the given interval.
Show Step 4
Now let’s find all the solutions. First notice that, in this case, if we plug in negative values of $$n$$ we will get negative solutions and these will not be in the interval and so there is no reason to even try these. Also note that if we use $$n = 0$$we will still not be in the interval and so let’s start things off at $$n = 1$$.
$\begin{array}{lclcl}{n = 1:\,}&{\require{cancel} \xcancel{{z = 15.3052}} < 20\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{z = 22.3940}\\{n = 2:}&{z = 27.8716\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{z = 34.9604}\\{n = 3:}&{z = 40.4380\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{\xcancel{{z = 47.5268}} > 45}\end{array}$
Notice that with each increase in $$n$$ we were really just adding another 12.5664 onto the previous results and by a quick inspection of the results above we can see that we don’t need to go any farther. Also, as we’ve seen in this problem it is completely possible for only one of the solutions from a given interval to be in the given interval so don’t worry about that when it happens.
So, it looks like we have the four solutions to this equation in the given interval.
$\require{bbox} \bbox[2pt,border:1px solid black]{{z = 22.3940,\,\,27.8716,\,\,34.9604,\,\,40.4380}}$
Note that depending upon the amount of decimals you used here your answers may vary slightly from these due to round off error. Any differences should be slight and only appear around the 4th decimal place or so however. |
# How do solve the following linear system?: 2x - 3y = 1 , 3x + y = -20 ?
Jul 13, 2016
$x = - \frac{59}{11} \text{ and } y = - 3 \frac{10}{11}$
#### Explanation:
The second equation contains a single $y$ term, so it is easy to transpose this to give: $y = - 3 x - 20$
We now have an expression for $y$ in terms of x and can substitute that into the other equation:
2x-3color(magenta)y =1 " and "color(magenta)(y = -3x -20
$2 x - 3 \left(\textcolor{m a \ge n t a}{- 3 x - 20}\right) = 1$
$2 x + 9 x + 60 = 1$
$11 x = - 59$
$\textcolor{b l u e}{x = - \frac{59}{11}} \text{ }$But $y = - 3 \textcolor{b l u e}{x} - 20$
$y = - 3 \left(\textcolor{b l u e}{\frac{- 59}{11}}\right) - 20$
$y = - 3 \frac{10}{11}$ |
## How to transform a portion to a Percent
There are just two an easy steps to convert a portion to a percent.
You are watching: 4/5 is equal to what percent
### Step One: transform the portion to a Decimal Value
The first step is to transform the fraction into a decimal value. Perform this by splitting the molecule by the denominator.
The numerator is the number on top of the fraction bar. The denominator is the number below the fraction bar.
decimal = numerator ÷ denominator
For example, transform the fraction 34 come a decimal.
34 = 3 ÷ 434 = 0.75
Thus, the decimal value of 34 is 0.75.
Here’s a tip: friend might be able to use a decimal equivalents chart because that a list of decimal values for common fractions to convert a fraction to decimal without utilizing division.
You might likewise be interested in ours long department calculator.
### Step Two: convert the Decimal worth to a Percentage
The 2nd step is to transform the decimal value into a percentage. Multiply the decimal value by 100, then location a percent sign (%) after ~ it.
For example, let’s convert 0.75 come a percentage.
percentage = 0.75 × 100 = 75%
The percentage value of 34 or 0.75 is thus 75%.
## Fraction come Percent counter Table
Another method to convert a fraction to percent is to use a switch table. The counter table listed below shows some common fractions and also their identical percentages.
Fraction to percent switch table showing usual fractions and their indistinguishable percentage values.
See more: 1 Ml Is How Many Ul To Ml - Convert Milliliter To Microliter
FractionPercent
1/250%
1/333.3%
2/366.6%
1/425%
3/475%
1/520%
2/540%
3/560%
4/580%
1/616.66%
5/683.33%
1/812.5%
3/837.5%
5/862.5%
7/887.5%
1/911.1%
2/922.2%
4/944.4%
5/955.5%
7/977.7%
8/988.8%
1/1010%
1/128.333%
1/166.25%
See more fraction percent equivalents. |
## Slope formula for Graphing
Previously we have discussed about adding scientific notation calculator and In today’s session we are going to discuss about While doing graphing of any linear equation, two terms are required and these terms are slope of the line and intercepts. X and Y intercepts are determined first to find the co-ordinates of the line in respect of both the Cartesian axis of 2D plane or we can say intercepts are required to know the point at which the line intersects or crosses the graph and for finding the slope of the line we use slope formula. But to implement slope formula, endpoints of the line are required as input.
So let us discuss first how to find end points or Co-ordinates of the Cartesian axises. For finding x and y intercepts, we follow a simple principle which states that all the points of y axis are zero in the equation for X- intercepts and similarly all the x points are zero in equation for Y intercepts. (want to Learn more about Slope, click here),
To explain it more we use an example here:
Given equation is
9x2+ 9y2= 3 ( equation is already in the form of standard linear equation)
for x -intercept—->
put y= 0 in equation
9x2= 3
x2= 3/ 9
x2= 1/3
x= +(1/√3) or -(1/√3)
so (x1, x2) = (1/√3, -√1/√3)
Now for y – intercepts —->
put x= 0 in equation
9y2= 3
y2= 3/9 = 1/3
y = +(1/√3) or -√(1/√3)
so (y1, y2) = (1/√3, -1/√3)
now from x and y intercepts we got the endpoints of the line. Now we can implement slope formula, which is as:
M= ( y2- y1)/ ( x2– x1)
here ‘M’ denotes the slope of the line.
M= ((-1/√3) – ( 1/√3))/ ((-1/√3) – ( 1/√3))
M = 1
so that’s how we calculate all the required terms to graph any equation.
For all the lessons related to slope formula, intercepts and other math topics, TutorVista provides online mat tutoring for 24 x 7 hrs with guidance of expert online math tutors.
In the next session we will discuss about Concepts of Intercepts in the mathematical world and You can visit our website for getting information about free tutoring online and maths model for class 10. |
Read the text and do the tasks below it. — Студопедия.Нет
Equivalent fractions
Multiplying the numerator and denominator of a fraction by the same (non-zero) number, the results of the new fraction is said to be equivalent to the original fraction. The word equivalent means that the two fractions have the same value. That is, they retain the same integrity – the same balance or proportion. This is true because for any number n, multiplying by is really multiplying by one, and any number multiplied by one has the same value as the original number. For instance, consider the fraction : when the numerator and denominator are both multiplied by 2, the result is , which has the same value (0.5) as . To picture this visually, imagine cutting the example cake into four pieces; two of the pieces together ( ) make up half the cake ( ).
For example: , , and are all equivalent fractions.
Dividing the numerator and denominator of a fraction by the same non-zero number will also yield an equivalent fraction. This is called reducing or simplifying the fraction. A fraction in which the numerator and denominator have no factors in common (other than 1) is said to be irreducible or in its lowest or simplest terms. For instance, is not in its lowest terms because both 3 and 9 can be exactly divided by 3. In contrast, is in lowest terms – the only number that is a factor of both 3 and 8 is 1.
Any fraction can be fully reduced to its lowest terms by dividing both the numerator and denominator by their greatest common divisor . For example, the greatest common divisor of 63 and 462 is 21, therefore, the fraction can be fully reduced by dividing the numerator and denominator by 21:
.
Reciprocals and the ‘invisible denominator’
The reciprocal of a fraction is another fraction with the numerator and denominator reversed. The reciprocal of , for instance, is .
Since any number divided by 1 results in the same number, it is possible to write any whole number as a fraction by using 1 as the denominator: 17 = (1 is sometimes referred to as the ‘invisible denominator’). Therefore, except for zero, every fraction or whole number has a reciprocal. The reciprocal of 17 is .
In pairs, look at the highlighted words and phrases. Try to guess what they mean from the context. Then check with your dictionary or teacher. Work out the list of the terms involved, make a kind of glossary.
2. In the text find the definition of:
a. equivalent fractions
b. reducing a fraction
c. irreducible fraction
d. reciprocal
Explain what fractions out of these seven are equivalent.
Say if the following fractions are reducible or in their lowest terms. Explain why.
Unit 7
ALGEBRA |
# Search by Topic
#### Resources tagged with Working systematically similar to Sweet Shop:
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### There are 129 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
##### Stage: 3 Challenge Level:
You need to find the values of the stars before you can apply normal Sudoku rules.
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Colour Islands Sudoku
##### Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Two and Two
##### Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Medal Muddle
##### Stage: 3 Challenge Level:
Countries from across the world competed in a sports tournament. Can you devise an efficient strategy to work out the order in which they finished?
### Where Can We Visit?
##### Stage: 3 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Oranges and Lemons, Say the Bells of St Clement's
##### Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Twinkle Twinkle
##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### Difference Sudoku
##### Stage: 4 Challenge Level:
Use the differences to find the solution to this Sudoku.
### Ben's Game
##### Stage: 3 Challenge Level:
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##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
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##### Stage: 3 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
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##### Stage: 3, 4 and 5 Challenge Level:
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##### Stage: 3 Challenge Level:
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Burhan Hopper
2021-01-02
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is $\frac{4}{3}\pi {r}^{3}$. The surface area is $4\pi {r}^{2}$.Set up the differential equation for how r is changing. Then, suppose that at time $t=0$ minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted.
Step 1
Given:
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball is perfectly spherical
Then the volume (in centimetres cubed) of a ball of radius r centimetres is
$v=\frac{4}{3}\pi {r}^{3}$
And the surface area is
$s=4\pi {r}^{2}$
Set up the differential equation for how r is changing.
Then, suppose that at time t=0 minutes,the radius is 10 centimetres.After 5 minutes,the radius is 8 centimetres
Step 2
To find: At the what time t will be snowball be completely melted?
From the given conditions ,
$\frac{dV}{dt}\left(\alpha S\right)$
$\frac{dV}{dt}\left(\lambda S\right)$......(1)
$V=\frac{4}{3}\pi {r}^{3}$
$\frac{dV}{dt}=\frac{4}{3}\pi 3{r}^{2}\frac{dr}{dt}$
By putting this value in (1)
The equation must be,
$\frac{4}{3}\pi 3{r}^{2}\frac{dr}{dt}=\lambda 4\pi {r}^{2}$
$\frac{dr}{dt}=\lambda$
$r=\lambda t+c$
Now, $t=0$ and $r=10$
so,
$r=\lambda t+c$
$10=\lambda \left(0\right)+c$
c=10 and here
$r=\lambda t+10$......(2)
After the 5 minutes, $t=5$ and $r=8$
$8=\lambda \left(5\right)+10$
$5\lambda =-2$
$\lambda =-\frac{2}{5}$
Now equation (2)becomes,
$r=-\frac{2}{5}t+10$......(3)
This shows the differential equation for how r is changing.
As the snowball completely melted that means the radius of the snowball is zero.
From this by substituting $r=0$ in the equation (3) must be,
$0=-\frac{2}{5}t+10$
$\frac{2}{5}t=10$
$t=\frac{10×5}{2}$ $t=25$ Hence the time required for melting the snowball is 25 minutes.
Do you have a similar question? |
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# 3.17: Decimal Addition
Difficulty Level: At Grade Created by: CK-12
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Yoshi and his family have decided to have a yard sale. Yoshi's mother has him collect all the money from the customers. Yoshi wants to make sure that he is collecting the right amount from each person.The first person buys one item for $12.25, a second item for$0.50 and a third item for 0.75. Yoshi knows that he needs to add the decimals but isn't sure how to begin. How much money does the first person owe Yoshi? In this concept, you will learn how to add decimals ### Adding Decimals To add decimals, work with the wholes and parts of the numbers separately. Add the parts then add the wholes. The best way to do this is to keep the parts together and keep the wholes together. To do this, simply line up the decimal points in each number that you are adding. Let's look at an example. Add 3.45 + 2.37 = _____ In this problem you have parts and wholes. Let’s rewrite the problem vertically, lining up the decimal points. 3.45+ 2.37\begin{align*}3.45 \\ \underline{+\ 2.37}\end{align*} Next, add the columns vertically and bring the decimal point down into the answer of the problem. 3.45+ 2.375.82\begin{align*}3.45 \\ \underline{+\ 2.37} \\ 5.82\end{align*} The answer is 5.82. Let's look at another example. 5 + 3.45 + 0.56 = _______ In a problem like this, line up the decimal points, but add zeros to help hold places where there aren’t numbers. This helps keep the addition straight. First, line up the problem vertically. 5.003.45+ 0.56\begin{align*}5.00 \\ 3.45 \\ \underline{+\ 0.56} \end{align*} Notice that zeros help hold places where you did not have numbers. Now each number in the problem has the same number of digits. Add the numbers. 5.003.45+ 0.569.01\begin{align*}5.00\\ 3.45 \\ \underline{+\ 0.56} \\ 9.01 \end{align*} The answer is 9.01. ### Examples #### Example 1 Earlier, you were given a problem about Yoshi and the yard sale. Yoshi's first customer wants to buy 3 different things. To ask for the correct total amount, Yoshi needs to add12.25, $0.50 and$0.75. Can you find the total amount due?
First, line up the numbers vertically.
12.250.50+ 0.75\begin{align*} 12.25\\ 0.50\\ \underline {+ \ 0.75}\end{align*}
Then, add the numbers.
13.50\begin{align*}13.50\end{align*}
The answer is $13.50. Yoshi must collect$13.50 from the customer.
#### Example 2
Add the following numbers.
13.25 + 0.80 + 1.30 = _____
First, line up the numbers vertically.
13.250.80+ 1.30\begin{align*} 13.25\\ 0.80\\ \underline {+ \ 1.30}\end{align*}
Then, add the numbers.
15.35\begin{align*}15.35\end{align*}
The answer is 15.35.
#### Example 3
Add the following decimals.
4.56 + 0.89 + 2.31 = _____
First, line up the numbers vertically.
4.560.89+ 2.31\begin{align*} 4.56\\ 0.89\\ \underline {+ \ 2.31}\end{align*}
Then, add the numbers.
7.76\begin{align*}7.76\end{align*}
The answer is 7.76.
#### Example 4
Add the following decimals.
5.67 + 0.65 + 0.93 =_____
First, line up the numbers vertically.
5.670.65+ 0.93\begin{align*}5.67\\ 0.65\\ \underline {+ \ 0.93}\end{align*}
Then, add the numbers.
7.25\begin{align*}7.25\end{align*}
The answer is 7.25.
#### Example 5
Add the following decimals.
88.92 + 0.57 + 3.12 = _____
First, line up the numbers vertically.
88.920.57+3.12\begin{align*} 88.92 \\ 0.57 \\ \underline{+3.12}\end{align*}
Then, add the numbers.
92.61\begin{align*}92.61\end{align*}
The answer is 92.61.
### Review
Add the following decimals.
1. 4.5 + 6.7 = _____
2. 3.45 + 2.1 = _____
3. 6.78 + 2.11 = _____
4. 5.56 + 3.02 = _____
5. 7.08 + 11.9 = _____
6. 1.24 + 6.5 = _____
7. 3.45 + .56 = _____
8. 87.6 + 98.76 = _____
9. 76.43 + 12.34 = _____
10. 5 + 17.21 = _____
11. 78 + 13.456 = _____
12. .456 + .23 + .97 = _____
13. 1.234 + 4.5 + 6.007 = _____
14. 3.045 + 3.3 + 9 = _____
15. 23 + 4.56 + .0091 = _____
To see the Review answers, open this PDF file and look for section 3.17.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Decimal In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Rounding Rounding is reducing the number of non-zero digits in a number while keeping the overall value of the number similar.
Sum The sum is the result after two or more amounts have been added together.
Vertically Vertically means written up and down in columns.
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12th NCERT Application of Derivatives Exercise 6.3 Questions 27
Do or do not
There is no try
Question (1)
Find the slope of the tangent of the tangent to the curve y= 3x4 - 4x at x=4
Solution
Slope of tangent ${\left( {\frac{{dy}}{{dx}}} \right)_{x = {x_1}}}$
y= 3x4 - 4x
differentiate with respect to x $\frac{{dy}}{{dx}} = 12{x^3} - 4$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 4}} = 12{\left( 4 \right)^3} - 4$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 4}} = 12 \times 64 - 4$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 4}} = 768 - 4 = 764$
Question (2)
Find the slope of the tangent of the tangent to the curve $y = \frac{{x - 1}}{{x - 2}}$ x ≠ 2 at x = 10
Solution
differentiate with respect to x $\frac{{dy}}{{dx}} = \frac{{\left( {x - 2} \right)\frac{d}{{dx}}\left( {x - 1} \right) - \left( {x - 1} \right)\frac{d}{{dx}}\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}$ $\frac{{dy}}{{dx}} = \frac{{\left( {x - 2} \right)\left( 1 \right) - \left( {x - 1} \right)\left( 1 \right)}}{{{{\left( {x - 2} \right)}^2}}}$ $\frac{{dy}}{{dx}} = \frac{{x - 2 - x + 1}}{{{{\left( {x - 2} \right)}^2}}}$ $\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {x - 2} \right)}^2}}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 10}} = \frac{{ - 1}}{{{{\left( {10 - 2} \right)}^2}}} = \frac{{ - 1}}{{64}}$ slope of tangent = -1/64
Question (3)
Find the slope of the tangent to curve y = x3 - x + 1 at the point whose x-coordinate is 2
Solution
differentiate with respect to x $\frac{{dy}}{{dx}} = 3{x^2} - 1$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 2}} = 3{\left( 2 \right)^2} - 1$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 2}} = 12 - 1 = 11$
Question (4)
Find the slope of the tangent to curve y = x3 - 3x + 2 at the point whose x-coordinate is 3
Solution
differentiate with respect to x $\frac{{dy}}{{dx}} = 3{x^2} - 3$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 3}} = 3{\left( 3 \right)^2} - 3$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 3}} = 27 - 3 = 24$ slope of tangent = 24
Question (5)
Find the slope of the normal to the curve x=acos3θ, y=a sin3θ at θ = π/4
Solution
slope of normal = $\text{slope of normal=} \frac{- 1}{{\frac{{dy}}{{dx}}}}$
x = acos3θ
differentiate with respect to θ $\frac{{dx}}{{d\theta }} = a \times 3{\cos ^2}\theta \left( { - \sin \theta } \right)$ $\frac{{dx}}{{d\theta }} = - 3a{\cos ^2}\theta \sin \theta$ y = asin3θ differentiate with respect to θ $\frac{{dy}}{{d\theta }} = a \cdot 3{\sin ^2}\theta \left( {\cos \theta } \right)$ $\frac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \left( {\cos \theta } \right)$ $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$ $\frac{{dy}}{{dx}} = \frac{{3a{{\sin }^2}\theta \cos \theta }}{{ - 3a{{\cos }^2}\theta \sin \theta }}$ $\frac{{dy}}{{dx}} = - \tan \theta$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\theta = \frac{\pi }{4}}} = - \tan \frac{\pi }{4} = - 1$ slope of normal = ${\text{slope of normal = }}\frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\theta = \frac{\pi }{4}}}}} = \frac{{ - 1}}{{ - 1}} = 1$
Question (6)
Find the slope of the normal to the curve x=1-asinθ, y=bcos2θ at θ=π/2
Solution
x = 1 - asinθ
differentiate with respect to θ $\frac{{dx}}{{d\theta }} = - a\cos \theta$ y = bcos2θ
differentiate with respect to θ $\frac{{dy}}{{d\theta }} = b2\cos \theta \left( { - \sin \theta } \right)$ $\frac{{dy}}{{d\theta }} = - 2b\sin \theta \cos \theta$ $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$ $\frac{{dy}}{{dx}} = \frac{{ - 2b\sin \theta \cos \theta }}{{ - a\cos \theta }}$ $\frac{{dy}}{{dx}} = \frac{{2b}}{a}\sin \theta$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\theta = \frac{\pi }{2}}} = \frac{{2b}}{a}\sin \frac{\pi }{2} = \frac{{2b}}{a}$ $\text{slope of normal=} \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\theta = \frac{\pi }{2}}}}}$ $\text{slope of normal=} \frac{{ - 1}}{{\frac{{2b}}{a}}}$ $\text{slope of normal=} \frac{{ - a}}{{2b}}$
Question (7)
Find points at which the tangent to the curve y = x3 -3x2 - 9x + 7 is parallel to the x-axis
Solution
y = x3 -3x2 - 9x + 7
differentiate with respect to x
slope of tangent = dy/dx $\frac{{dy}}{{dx}} = 3{x^2} - 6x - 9$ Slope of x-axis = 0
tangent parallel to x-axis
slope of tangent = slope of x-axis
3x2 -6x - 9 = 0
3(x2 - 2x -3) =0
(x-3)(x-1) = 0
x-3= 0 OR x+1 = 0
x = 3 OR x = -1
If x=3, y = x3 -3x2 - 9x + 7
y= 27 - 27 - 27 +7 = -20
(x, y) = (3, -20)
If x = -1 , y = x3 -3x2 - 9x + 7
y = -1 -3 +9 +7 = 12
(x, y) = (-1, 12)
Question (8)
Find a point on the curve y=(x-2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4)
Solution
tangent parallel to AB, A(2, 0) and B(4, 4)
$\text{slope of}\quad \overline {AB} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ $\text{slope of}\quad \overline {AB} = \frac{{4 - 0}}{{4 - 2}} = 2$ y = (x - 2)2
differentiate with respect to x
$\frac{{dy}}{{dx}} = 2\left( {x - 2} \right)$ As tangent parallel AB
slope of tangent = slope of AB
2(x-2) = 2
x - 2 = 1
x = 3
If x=3, y=(x-2)2 = (3-2)2 = 1
∴ (x, y) = (3, 1)
Question (9)
Find the point on the curve y=x2 - 11x + 5 at which the tangent is y = x - 11
Solution
If y = mx + c is equation of line
slope of line = m
Equation of tangent = y = x - 11
Slope of tangent = 1 ---(1)
y = x3 - 11x + 5
differentiate with respect to x
$\frac{{dy}}{{dx}} = 3{x^2} - 11$ Slope of tangent = 3x2 - 11 ---(2)
From (1) and (2)
3x2 - 11 = 1
3x2 = 12
x2 = 4
x = ±2 If x=2, y=x3 -11x +5 then, y=23 -11(2) +5 =-9
(x, y) = (2, -9)
If x=-2, y=x3 -11x +5 then, y=(-2)3 -11(-2) +5 = 19 (x, y) = (-2, 19)
But (-2, 19) does not satisfy the equation
So point on curve is (2, -9) of tangent
Question (10)
Find the equation of all lines having slope -1 that are tangents to the curve $y = \frac{1}{{x - 1}},x \ne 1$
Solution
Equation of line passing through (x1, y1) with slope m is given by y-y1 = m(x-x1)
Given slope = -1 ---(1) and equation $y = \frac{1}{{x - 1}},x \ne 1$ differentiate with respect to x
$\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}$ $\text{slope of tangent=}\frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}---(2)$ From (1) and (2) $\frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}} = - 1$ (x-1)2 = 1
x-1 = ± 1
⇒ x - 1 = 1 OR x-1 = -1
x = 2 or x =0
If x = 2, y = 1/(2-1) = 1
(x, y) = (2, 1)
If x = 0, y = 1/(0-1) = -1
(x, y) = (0, -1)
So equation of tangent at (2, 1) is
y - 1 = -1(x-2)
y - 1 = -x + 2
x + y - 3 = 0
Equation of tangent ay (0, -1) will be
y + 1 = -1( x - 0)
y + 1 = -x
x + y + 1 = 0
So equation of tangents are
x+y-3=0 and x+y+1=0
Question (11)
Find the equation of all lines having slope 2 which are tangents to the curve $y = \frac{1}{{x - 3}},x \ne 3$
Solution
Slope of tangent = 2 ---(1) $y = \frac{1}{{x - 3}},x \ne 3$ differentiate with respect to x
$\text{slope of tangent=}\frac{{ - 1}}{{{{\left( {x - 3} \right)}^2}}}$ From (1) and (2)
$\frac{{ - 1}}{{{{\left( {x - 3} \right)}^2}}} = 2$ (x - 3) 2 = -½ Which is not possible as square can not be negative
So, no tangent can be drawn having slope 2
Question (12)
Find the equations of all lines having slope 0 which are tangent to the curve $y = \frac{1}{{{x^2} - 2x + 3}}$
Solution
Slope of tangent = 0 --- (1) $y = \frac{1}{{{x^2} - 2x + 3}}$ differentiate with respect to x
$\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} \times \left( {2x - 2} \right)$ $\frac{{dy}}{{dx}} = \frac{{ - \left( {2x - 2} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} - - - (2)$ From equation (1) and (2) $\frac{{ - \left( {2x - 2} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} = 0$ ⇒ 2x-2 = 0
x = 1
If x=1 $y = \frac{1}{{{{\left( 1 \right)}^2} - 2\left( 1 \right) + 3}} = \frac{1}{2}$ So the tangent passes through (x, y) = (1, ½) Equation of tangent will be
y-½ = 0(x-1)
2y - 1 = 0
Question (13)
Find points on the curve $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1$ at which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis
Solution
differentiate with respect to x
$\frac{{2x}}{9} + \frac{{2y}}{{16}}\frac{{dy}}{{dx}} = 0$ $\frac{x}{9} + \frac{y}{{16}}\frac{{dy}}{{dx}} = 0$ $\frac{{dy}}{{dx}} = \frac{{ - 16}}{9}\frac{x}{y}$ $\frac{{dy}}{{dx}} = \frac{{ - 16x}}{{9y}}$ $\text{slope of tangent=}\frac{{ - 16x}}{{9y}}$ (a) tangent parallel to x-axis
slope of x-axis = 0
Slope of tangent = slope of x-axis
$\frac{{ - 16x}}{{9y}} = 0$ ⇒ x = 0
If x =0 ⇒ $\Rightarrow \frac{{{y^2}}}{{16}} = 1$ y = ±4 Points of tangents are (0, 4), (0, -4)
(ii) Parallel to y-axis
Slope of y-axis is not defined that is denominator = 0 ∴ 9y = 0
y = 0 If y=0
$\Rightarrow \frac{{{x^2}}}{9} = 1$ x2 = 9
⇒ x = ±3 Points through which tangents parallel to y-axis are drawn are (3,0)and (-3, 0)
Question (14)
Find the equations of tangent and normal to the given curves at the indicated points
(i) y = x4 -6x3+13x2 -10x + 5 at (0, 5)
(ii) y = x4 -6x3+13 x2 -10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0) (v) x = cost, y=sin t at t = π/4
Solution
(i) y = x4 -6x3+13 x2- 10x +5
differentiate with respect to x
$\frac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = - 10$ slope of tangent = -10
Equation of tangent through (0, 5) is
y - 5 = -10(x-0)
10x + y - 5=0 $\text{slope of normal=} \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{x = 0}}}}$ Equation of normal $y - 5 = \frac{1}{{10}}\left( {x - 0} \right)$ 10y - 50 = x
x - 10y + 50 = 0
(ii) y = x4 -6x3+13 x2 -10x + 5
differentiate with respect to x
$\frac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = 4{\left( 1 \right)^3} - 18{\left( 1 \right)^2} + 26\left( 1 \right) - 10$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = 4 - 18 + 26 - 10$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = 2$ ∴ slope of tangent = 2
Equation of tangent through (1, 3) is
y-3 = 2(x-1)
2x - y + 1 =0
$\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ $\text{slope of normal =} \frac{{ - 1}}{2}$ Equation of normal is
$y - 3 = \frac{{ - 1}}{2}\left( {x - 1} \right)$ 2y - 6 = -x + 1
x + 2y = 7
(iii) y = x3
differentiate with respect to x
$\frac{{dy}}{{dx}} = 3{x^2}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = 3$ slope of tangent = 3
Equation of tangent is
y-1 = 3(x - 1)
3x - y - 2 = 0
$\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ $\text{slope of normal =} \frac{{ - 1}}{3}$ Equation of normal is
$y - 1 = \frac{{ - 1}}{3}\left( {x - 1} \right)$ 3y -3 = -x + 1
x + 3y = 4
(iv) y = x2
differentiate with respect to x
$\frac{{dy}}{{dx}} = 2x$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {0,0} \right)}} = 0$ slope of tangent is = 0
Equation of tangent is y - 0 = 0(x - 0)
y=0
$\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ $\text{slope of normal =} \frac{{ - 1}}{0}$ slope of normal is not defined
so normal is parallel to y-axis and passes through (0, 0) it is y-axis so equation is x = 0
(v) x = cost
differentiate with respect to x
$\frac{{dx}}{{dt}} = - \sin t$ y = sint
differentiate with respect to x
$\frac{{dy}}{{dt}} = \cos t$ $\text{Now}\quad \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}$ $\frac{{dy}}{{dx}} = \frac{{\cos t}}{{ - \sin t}} = - \cot t$ ${\left( {\frac{{dy}}{{dx}}} \right)_{t = \frac{\pi }{4}}} = - \cot \frac{\pi }{4} = - 1$ slope of tangent = -1
Equation of tangent is $y - \frac{1}{{\sqrt 2 }} = - 1\left( {x - \frac{1}{2}} \right)$ $\sqrt 2 y - 1 = - \sqrt 2 x + 1$ $\sqrt 2 y + \sqrt 2 x = 2$ $y + x = \sqrt 2$ slope of normal = -1/-1 = 1
Equation of normal is $y - \frac{1}{{\sqrt 2 }} = \left( {x - \frac{1}{{\sqrt 2 }}} \right)$ x - y = 0
Question (15)
Find the equation of the tangent line to the curve y=x2 - 2x + 7 which is
(a) parallel to the line 2x - y + 9 = 0
(b) perpendicular to the line 5y-15x = 13
Solution
y = x2 - 2x + 7
differentiate with respect to x
$\frac{{dy}}{{dx}} = 2x - 2$ slope of tangent = 2x - 2
(a) parallel to 2x - y + 9 = 0
Equation of line is 2x - y + 9 = 0
slope of line = -a/b = -2/-1 = 2
tangent parallel line
∴ slope of tangent = slope of line
∴ 2x - 2 = 2
2x = 4
x = 2
If x=2, y = (2)2 - 2(2) + 7 = 7
Equation of tangent is
y - 7 = 2(x-2)
2x - y + 3 = 0
(b) Perpendicular to 5y -15x = 13
-15x + 5y = 13
slope of line = -a/b = -15/5 = 3
tangent is perpendicular to line
∴ slope of tangent × slope of line = -1
∴ (2x-2) × 3 = -1
6x - 6 = -1
6x = 5
x = 5/6
If x = 5/6,
$y = {\left( {\frac{5}{6}} \right)^2} - 2\left( {\frac{5}{6}} \right) + 7$ $y = \frac{{25}}{{36}} - \frac{{10}}{6} + 7$ $y = \frac{{25 - 60 + 252}}{{36}} = \frac{{217}}{{36}}$ Equation of tangent is $y - \frac{{217}}{{36}} = \frac{{ - 1}}{3}\left( {x - \frac{5}{6}} \right)$ 36y - 217 = -12x + 10
12x + 36y - 227 = 0
Question (16)
Show that the tangents to the curve y = 7x3 + 11 at the points where x=2 and x=-2 are parallel
Solution
y = 7x3 + 11
differentiate with respect to x
$\frac{{dy}}{{dx}} = 21{x^2}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 2}} = 21{\left( 2 \right)^2} = 84$ slope of tangent at x=2 is 84
${\left( {\frac{{dy}}{{dx}}} \right)_{x = - 2}} = 21{\left( { - 2} \right)^2} = 84$ slope of tangent at x=-2 is 84
slope are equal so tangents are parallel
Question (17)
Find the points on the curve y=x3 at which the slope of the tangent is equal to the y-coordinate of the point
Solution
y = x3
differentiate with respect to x
$\frac{{dy}}{{dx}} = 3{x^2}$ slope of tangent = y-coordinate
3x2 = x3
x3 - 3x2 = 0
⇒ x2 ( x - 3) = 0
If x = 0, y = 0
If x = 3, y = 27
points are (0, 0) and (3, 27)
Question (18)
For the curve y=4x3 - 2x5, find all the points at which the tangent passes through the origin
Solution
y=4x3 - 2x5 differentiate with respect to x
$\frac{{dy}}{{dx}} = 12{x^2} - 10{x^4}$ Let the tangent be drawn at (x0, y0)
∴ (x0, y0) ∈ curve
y0 = 4x03 - 2x05 ---(1)
$\frac{{dy}}{{dx}} = 12x_0^2 - 10x_0^4$ The equation of tangent is given by
y - y0 = m(x - x0)
y - y0 = (12x02 - 10x04)(x - x0)
it passes through (0, 0)
∴ 0 - y0 = (12x02 - 10x04)(0 - x0)
- y0 = - x0(12x02 - 10x04)
y0 = 12x03 - 10x05
substituting y0 from (1)
4x03 - 2x05 = 12x03 - 10x05
-8x03 + 8x05 = 0
8x03( x02 - 1) =0
x03( x0 - 1)( x0 + 1) = 0
x0 = 0 OR x0 - 1 = 0 OR x0 + 1 = 0
x0 = 0 OR x0 = 1 OR x0 = - 1
If x0 = 0
y0 = 4x03 - 2x05 =0 ⇒ (0, 0)
If x0 = 1
y0 = 4(1)3 - 2(1)5 =2 ⇒ (1, 2)
If x0 = -1
y0 = 4(-1)3 - 2(-1)5 = -2 ⇒ (-1, -1)
So points are (0, 0), (1, 2) and (-1, -2)
Question (19)
Find the points on the curve x2 + y2 - 2x - 3 = 0 at which the tangents are parallel to the x-axis
Solution
y = x2 + y2 - 2x - 3 = 0
differentiate with respect to x
$2x + 2y\frac{{dy}}{{dt}} - 2 = 0$ $y\frac{{dy}}{{dt}} = 1 - x$ $\frac{{dy}}{{dt}} = \frac{{1 - x}}{y}$ slope of tangent at (x, y) = (1-x)/y
slope of x-axis = 0
tangent parallel to x-axis
slope of tangent = slope of x-aaxis
$\therefore \frac{{1 - x}}{y} = 0$ x= 1
If x=1,
1+y2-2-3 = 0 ⇒ y2 = 4,
y = ±2
so points are (1, 2) and (1, -2)
Question (20)
Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3
Solution
ay2 = x3
differentiate with respect to x
$2ay\frac{{dy}}{{dx}} = 3{x^2}$ $\frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{2ay}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{m^2},a{m^3}} \right)}} = \frac{{3{{\left( {a{m^2}} \right)}^2}}}{{2a\left( {a{m^3}} \right)}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{m^2},a{m^3}} \right)}} = \frac{{3{a^2}{m^4}}}{{2{a^2}{m^3}}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{m^2},a{m^3}} \right)}} = \frac{3}{2}m$ slope of tangent = 3m/2
$\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ $\text{slope of normal =} \frac{{ - 2}}{{3m}}$ Equation of normal is $y - a{m^3} = \frac{{ - 2}}{{3m}}\left( {x - a{m^2}} \right)$ 3my - 3am4 = -2x +2am2
2x+3my = 2am2 + 3am4
Question (21)
Find the equation of the normals to the curve y = x3 + 2x +6 which are parallel to the line x +14y+4 = 0
Solution
line : x +14y+4 = 0 $\text{slope of line =} \frac{{ - a}}{b} = \frac{{ - 1}}{{14}}$ y = x3 + 2x +6
differentiate with respect to x
$\frac{{dy}}{{dx}} = 3{x^2} + 2$ slope of tangent = 3x2 + 2
$\text{slope of normal =} \frac{{ - 1}}{{3{x^2} + 2}}$ normal is parallel to line
slope of normal = slope of line $\therefore \frac{{ - 1}}{{3{x^2} + 2}} = \frac{{ - 1}}{{14}}$ 3x2 + 2 = 14
3x2 = 12
x2 = 4
⇒x ±2
IF x=2, y = (2)3+2(2) + 6 = 18
(x, y ) = (2, 18)
Equation of normal through (2, 18) is
$y - 18 = \frac{{ - 1}}{{14}}\left( {x - 2} \right)$ 14y -252 = -x +2
x+14y-254= 0
If x = -2, y = (-2)3+2(-2) + 6 = -6
(x, y) = (-2, -6)
Equation of normal through (-2, -6) is given by
$y + 6 = \frac{{ - 1}}{{14}}\left( {x + 2} \right)$ 14y +84=-x-2
x+14y+86=0
Question (22)
Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2, 2at)
Solution
y2 = 4ax
differentiate with respect to x
$2y\frac{{dy}}{{dx}} = 4a$ $\frac{{dy}}{{dx}} = \frac{{4a}}{{2y}} = \frac{{2a}}{y}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{t^2},2at} \right)}} = \frac{{2a}}{{2at}} = \frac{1}{t}$ slope of tangent = 1/t
Equation of tangent is
$y - 2at = \frac{1}{t}\left( {x - a{t^2}} \right)$ ty - 2at2 = x-at2
x-ty+at2 = 0
$\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ ∴ slope of normal = -t
Equation of normal is
y - 2at = -t (x-at2)
y-2at = -xt+at3
tx+y = at3 +2at
Question (23)
Prove that the curves x=y2 and xy =k cut at right angles if 8k2 = 1
Solution
x=y2 ---(1) and xy =k ---(2)
Replacing value x from (1) in equation(2)
y2y = k
y3 = k
y = k1/3 ---(3)
given 8k2 =1
k2 = 1/8
k = 1/√8
Replacing value of k in equation (3) $y = {\left( {\frac{1}{{\sqrt 8 }}} \right)^{\frac{1}{3}}} = {\left( {{2^{\frac{{ - 3}}{2}}}} \right)^{\frac{1}{3}}} = {2^{\frac{{ - 1}}{2}}}$ x=y2
$y = {\left( {{2^{\frac{{ - 1}}{2}}}} \right)^2} = {2^{ - 1}} = \frac{1}{2}$ $\left( {x,y} \right) = \left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)$ Now y2 = x
differentiate with respect to x
$2y\frac{{dy}}{{dx}} = 1$ $\frac{{dy}}{{dx}} = \frac{1}{{2y}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)}} = \frac{1}{{2 \times \frac{1}{{\sqrt 2 }}}} = \frac{1}{{\sqrt 2 }}$ ∴ slope of tangent to first curve m1 = 1/√2
Now xy = k
differentiate with respect to x
$x\frac{{dy}}{{dx}} + y = 0$ $\frac{{dy}}{{dx}} = \frac{{ - y}}{x}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)}} = \frac{{ - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{2}}} = - \frac{1}{{\sqrt 2 }} \times 2$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)}} = - \sqrt 2$ ∴ slope of tangent to second curve m2 = - √2
Now m1 m2 = 1/√2×- √2 = -1
slope of tangents are perpendicular to each other
so curve are orthogonal
Question (24)
Find the equation of the tangent and normal to the hyperbola $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point (x0, y0)
Solution
differentiate with respect to x
$\frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0$ $\frac{y}{{{b^2}}}\frac{{dy}}{{dx}} = \frac{x}{{{a^2}}}$ $\frac{{dy}}{{dx}} = \frac{{\frac{x}{{{a^2}}}}}{{\frac{y}{{{b^2}}}}} = \frac{{{b^2}x}}{{{a^2}y}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_0},{y_{}}} \right)}} = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}$ $\text{slope of tangent =} \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}$ Equation of tangent is
y - y0 = m(x -x0) $y - {y_0} = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\left( {x - {x_0}} \right)$ $\frac{{y{y_0} - y_0^2}}{{{b^2}}} = \frac{{{x_0}x - x_0^2}}{{{a^2}}}$ $\frac{{y{y_0}}}{{{b^2}}} - \frac{{y_0^2}}{{{b^2}}} = \frac{{{x_0}x}}{{{a^2}}} - \frac{{x_0^2}}{{{a^2}}}$ $\frac{{{x_0}x}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = \frac{{x_0^2}}{{{a^2}}} - \frac{{y_0^2}}{{{b^2}}}$ $\frac{{{x_0}x}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = 1$ $\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ $\text{slope of normal =} - \frac{{{a^2}{y_0}}}{{{b^2}{x_0}}}$ Equation of normal $y - {y_0} = - \frac{{{a^2}{y_0}}}{{{b^2}{x_0}}}\left( {x - {x_0}} \right)$ $\frac{{y - {y_0}}}{{{a^2}{y_0}}} = - \frac{{\left( {x - {x_0}} \right)}}{{{b^2}{x_0}}}$ $\therefore \frac{{\left( {x - {x_0}} \right)}}{{{b^2}{x_0}}} + \frac{{y - {y_0}}}{{{a^2}{y_0}}} = 0$
Question (25)
Find the equation of the tangent to the curve y = √(3x-2) which is parallel to the line 4x-2y+5 = 0
Solution
line l: 4x-2y+5 =0 slope of line = -a/b = -4/-2 = 2
$y = \sqrt {3x - 2}$ differentiate with respect to x
$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {3x - 2} }} \times 3$ $\frac{{dy}}{{dx}} = \frac{3}{{2\sqrt {3x - 2} }}$ tangent parallel to line
slope of tangent = slope of line $\frac{3}{{2\sqrt {3x - 2} }} = 2$ $3 = 4\sqrt {3x - 2}$ 9 = 16(3x-2)
9 = 48x - 32
48x = 41
x = 41/48 $\text{If x =} \frac{{41}}{{48}}$ $y = \sqrt {3x - 2}$ $y = \sqrt {3\left( {\frac{{41}}{{48}}} \right) - 2}$ $y = \sqrt {\frac{{41 - 32}}{{16}}} = \sqrt {\frac{9}{{16}}} = \frac{3}{4}$ $pt\left( {x,y} \right) = \left( {\frac{{41}}{{48}},\frac{3}{4}} \right)$ Equation of tangent is $y - \frac{3}{4} = 2\left( {x - \frac{{41}}{{48}}} \right)$ $\frac{{4y - 3}}{4} = \frac{{96x - 82}}{{48}}$ 48y - 36 = 96x-82
96x - 48y - 46 =0
48x - 24y - 23 = 0
### Choose the correct answer in Exercises 26 and 27
Question (26)
The slope of the normal to the curve y=2x2 +3sinx at x= 0 is
(A) 3 (B) 1/3
(C) -3 (D) -1/3
Solution
y=2x2 +3sinx
differentiate with respect to x
$\frac{{dy}}{{dx}} = 4x + 3\cos x$ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = 0 + 3\cos 0 = 3$ Slope of tangent = 3 $\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}$ $\text{slope of normal =} \frac{{ - 1}}{3}$ So option (D) is correct
Question (27)
The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 3) (B) (2, 1)
(C) (1, -2) (D) (-1, 2)
Solution
y = x + 1 is tangent
slope of tangent m = 1 ---(1)
y2 = 4x
differentiate with respect to x
$2y\frac{{dy}}{{dx}} = 4$ $y\frac{{dy}}{{dx}} = 2$ $\frac{{dy}}{{dx}} = \frac{2}{y}$ $\text{slope of tangent =} \frac{2}{y}---(2)$ $\therefore \frac{2}{y} = 1$ y = 2
y2 = 4x
x = 1
so (x, y) = (1, 2) So option (A) is correct |
# discuss how to describe points, lines and planes in 3 space.
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1 Chapter 2 3 Space: lines and planes In this chapter we discuss how to describe points, lines and planes in 3 space. introduce the language of vectors. discuss various matters concerning the relative position of lines and planes: in particular, intersections and angles. show how to translate and rotate lines and planes in relatively simple cases space and vectors Coordinatizing 3 space The physical world around us is called three dimensional because through any point precisely three mutually perpendicular axes (and no more) can pass. Such axes can be used to describe points in 3-space by triples of numbers: the (signed) distances to planes formed by two of these axes. We call such a set of perpendicular axes through a given point a cartesian coordinate system. Each of the axes is called a coordinate axis and the point of intersection of the three axes is called the origin. Every point of a coordinate axis corresponds to a real number. As you know, often, but not necessarily always, these axes are called x axis, y axis and z axis. By convention, we agree that the coordinate system be right handed: if you turn the positive x axis to the positive y axis, then a screw positioned along the z axis following this movement would move in the direction of the positive z axis. Every two of the three axes span a plane: the y,z plane, the x,z plane and the x,y plane, respectively. The signed distances u, v and w of a point P in 3 space to each of these planes in the given order are the coordinates of P; we usually put the three together as follows: (u,v,w) and still call this the coordinates of P. Notations for points that occur often: (x,y,z), (x 1,x 2,x 3 ), (a,b,c), (a 1,a 2,a 3 ), (u,v,w), etc. For a point P with coordinates (x, y, z) we also write P = (x, y, z). 21
2 22 3 Space: lines and planes z z P=(2,3,4) y 4 y x x Figure 2.1: Left: The third coordinate of P = (2, 3, 4) is the signed distance of P to the x,y plane; so if the point were below the x,y plane, the third coordinate would have been negative. Right: Cartesian coordinate systems are taken to be right handed: a screw positioned along the z axis moves along the positive z axis if turned in the direction from x axis to y axis. Most discussions in this chapter will focus on 3 space. But with some adaptations (usually simplifications) the material makes sense in 2 space Example. Given the point P with coordinates (x,y,z), the point Q with coordinates (x,y, 0) is the vertical projection of P on the x,y plane. The distance between P and Q is z (note the absolute value). Similarly, the (horizontal) projection on the y, z plane has coordinates (0, y, z). Finally, the horizontal projection on the x, z plane has coordinates (x, 0,z) Vectors: introduction In dealing with geometric issues, it is convenient to have the language of vectors at our disposal. A vector is a quantity with both magnitude and direction. A well known example z E p=op B C F O y A AB D x Figure 2.2: The arrow with tail at A and head at B represents a vector (left). The two arrows CD and EF have the same length and direction, and so represent the same vector (middle). A vector with tail in the origin is usually represented by a lower case boldface symbol, like p (right).
3 2.1 3 space and vectors 23 of a vector occurring in physics is velocity. A vector is usually represented by an arrow in the sense of a directed line segment; the length of the segment corresponds to the magnitude. For us the most important aspects about the notion of a vector are summarized below. For points A and B in 2 space or 3 space, the vector AB is an arrow with tail at A and head at B. (It points from A to B.) Only its magnitude (length) and its direction are of importance, not the positioning of the arrow in the following sense: if two vectors have the same length and the same direction, we consider them equal. The representing arrows need not coincide at all (but one can be translated so that it coincides with the other). The vectors in 3 space with tail at the origin O = (0, 0, 0) and head at a point P = (x,y,z) play a special role. Such a vector OP is usually represented by the corresponding lower case boldface symbol, say p. If we want to emphasize the vector aspect, we write p = (x,y,z) rather than P = (x,y,z). The very scrupulous reader may find this abuse of notation a bit distressing. In practice, it almost never goes wrong. To assign coordinates/numbers to any vector, say the vector AB with A = (a 1,a 2,a 3 ) and B = (b 1,b 2,b 3 ), we note that this vector can also be represented by the arrow with tail at the origin and head at (b 1 a 1,b 2 a 2,b 3 a 3 ). And so we write AB = (b 1 a 1,b 2 a 2,b 3 a 3 ). In 3 space with a cartesian coordinate system, we single out three special vectors, the so called standard basis vectors. e 1 with tail at (0, 0, 0) and head at (1, 0, 0) (the standard basis vector of length 1 along the positive x axis). e 2 with tail at (0, 0, 0) and head at (0, 1, 0) (the standard basis vector of length 1 along the positive y axis). e 3 with tail at (0, 0, 0) and head at (0, 0, 1) (the standard basis vector of length 1 along the positive z axis). In the following, vectors usually refer to directed segments with their tail in the origin, unless the context suggests otherwise. Also note that in writing on paper or on the blackboard the use of boldface symbols is not convenient. So we also use notations like x = (x 1,x 2,x 3 ), where we underline the symbol for a vector The first arithmetic with vectors We can do a kind of arithmetic with vectors that turns out to make sense in the context of geometry.
5 2.1 3 space and vectors 25 Of course, this is the same as the distance from the origin to the point with coordinates (a, b, c). The squared length is therefore p 2 = a 2 + b 2 + c 2 z P=(a,b,c) O y (a,0,0) x Q=(a,b,0) Figure 2.4: The length of a vector with end point at (a,b,c) is computed by repeated application of Pythagoras theorem: OQ = a 2 + b 2 and OP = OQ 2 + QP 2 = a2 + b 2 + c 2. Dot or inner product. The dot product or inner product of two vectors x = (x 1,x 2,x 3 ) and y = (y 1,y 2,y 3 ) is defined as x y = x 1 y 1 + x 2 y 2 + x 3 y 3. This product does not have an immediate geometric interpretation, but will show up indirectly in various geometrically relevant situations, such as the length of a vector and the angle between two vectors: For instance, the squared length of x equals the inner product of x with itself: x 2 = x x x 2 3 = x 1 x 1 + x 2 x 2 + x 3 x 3 = x x. Distance. The distance between the points P = (x, y, z) and Q = (a, b, c) is (x a)2 + (y b) 2 + (z c) 2 ; it equals the length of the line segment PQ. We define the distance between two vectors as the distance between their endpoints, or, equivalently, the length of the difference. The distance between p = (x, y, z) and q = (a, b, c) is p q = (x a,y b,z c) = (x a) 2 + (y b) 2 + (z c) 2.
6 26 3 Space: lines and planes Perpendicular vectors. Using the dot product we can express when two vectors are perpendicular. It turns out (no proof here) that this is the case precisely if their inner product equals 0: x y = 0 x and y perpendicular. Angle between vectors. If a and b are nonzero vectors, then the angle α between them is usually computed via its cosine. From the Cosine Law the following relation involving the dot product can be shown to hold: cosα = a b a b. To determine the angle, two steps are therefore required: first determine the cosine of the angle using the formula above, then solve for the angle. In the case of perpendicular vectors, the angle between them is 90 or π/2 radians, since cosα = 0 in this case Examples. Here are a few examples of the notions mentioned above. a) 2 (3, 4, 1) 3 (2, 5, 1) = (0, 7, 5). b) The length of the vector p = (1, 2, 3) is p = = 14. c) The inner product of the vectors (1, 2, 3) and (2, 1, 4) is (1, 2, 3) (2, 1, 4) = ( 1) = 12. d) The vectors (1, 1, 2) and (2, 2, 2) are perpendicular since their inner product equals 0:: (1, 1, 2) (2, 2, 2) = ( 2) = 0. e) To compute the angle α between the vectors (1, 1, 0) and (1, 2, 1) we first determine cosα: 3 3 cosα = = 4 3 (1, 1, 0) (1, 2, 1) (1, 1, 0) (1, 2, 1) = = = Hence the angle is 30 or π/6 radians. 2.2 Describing lines and planes Describing a line by a vector parametric equation To specify a line l in 3 space, it suffices to give a point, say P 0, through which the line passes and to give the line s direction, say given by the nonzero vector v. In terms of vectors, start with a vector p 0 with end point P 0 and add an arbitrary scalar multiple λv of v to it: p 0 + λv. 3 2.
7 2.2 Describing lines and planes 27 As λ ranges from to, all vectors with endpoints on the line can be obtained. Not surprisingly, the vector v is called a direction vector of the line. Of course, any nonzero multiple of v can be used as direction vector, i.e., p 0 + µ(3v) describes the same line. Here is an example. The line l passing through P 0 = (2, 3, 5) and having direction vector a + λv a v Figure 2.5: Parametric description of a line involves a support vector p 0 and a nonzero direction vector v. The line can be seen as resting on p 0. Any vector on the line is obtained by adding a suitable multiple of v to p 0. x = (1, 1, 2) is (2, 3, 5) + λ(1, 1, 2). Any value of λ produces a specific vector or point on the line. For instance, for λ = 2, we get (2, 3, 5) + 2 (1, 1, 2) = (4, 1, 9). An arbitrary point on the line can be described as Two remarks are in place here. (2 + λ, 3 λ, 5 + 2λ). Any vector on the line can be taken as the support vector. So, for example, (4, 1, 9)+ µ(1, 1, 2) is the same line, since (4, 1, 9) is on the line. Any two direction vectors differ by a (nonzero) multiple. This means that, for example, (2, 3, 5) + ρ( 2, 2, 4) is the same line Describing a plane in 3 space by an equation Suppose V is a plane in 3 space. If you want to explain to someone else which plane it is, it suffices to give the following information: a point P 0 through which the plane passes, and its direction. Well, a plane contains many directions. Now a smart move is to specify the plane s direction by giving the direction of a line which is perpendicular to the plane. The direction of that line can be given by a single nonzero vector (the length of the vector is irrelevant, only its direction matters). Let us translate this into mathematics, first in an example and then in the general case.
8 28 3 Space: lines and planes Here is the example. Suppose the point P 0 = (2, 1, 3) is on the plane U, and let p 0 = OP 0 be the vector pointing to P 0. Suppose moreover that n = (1, 2, 2) is perpendicular to U. Our task is to catch those vectors x = (x,y,z) whose endpoints are in the plane. Now such an endpoint is in U if the direction from P 0 to this endpoint is perpendicular to n. This direction is represented by the vector x p 0 = (x 2,y 1,z 3). So we require that this vector is perpendicular to n = (1, 2, 2), i.e., (x 2,y 1,z 3) (1, 2, 2) = 0 or (x 2) + 2 (y 1) + 2 (z 3) = 0. Of course, we can rewrite this equation as x + 2y + 2z = 10, but the form x 2 + 2(y 1) + 2(z 3) = 0 shows clearly that (2, 1, 3) is on the plane. Please note that: For any nonzero real number t, the equation tx+2ty +2tz = 10t represents the same equation. For example, 6x + 12y + 12z = 60. The coefficients of x, y, z in the equation x + 2y + 2z = 10 form a vector which is (a multiple of) the vector n = (1, 2, 2) perpendicular to the plane we started with. This is not a coincidence as we will see below. The coordinates of P 0 satisfy the equation: = 10. The last two items provide an easy check on the correctness of the equation obtained. z n p 0 P 0 x P x y Figure 2.6: Given a point P 0 in the plane and a vector n perpendicular to the plane, any vector x satisfying (x p 0 ) n = 0 has its endpoint on the plane. Here is the general story. Let P 0 = (x 0,y 0,z 0 ) be a point and let n = (a,b,c) be a nonzero vector, which we suppose to be perpendicular to the plane we wish to describe. The plane through P and perpendicular to n consists of all the points P = (x,y,z) such that P 0 P is perpendicular to n, i.e., (x x 0,y y 0,z z 0 ) is perpendicular to (a,b,c). In vector form: (a,b,c) (x x 0,y y 0,z z 0 ) = 0,
9 2.2 Describing lines and planes 29 and in equation form (i.e., we have expanded the inner product): Of course, an equivalent form is a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. ax + by + cz = ax 0 + by 0 + cz 0 (the right hand side is a constant). If we replace the right hand side by a single symbol, the general form of the equation of a plane becomes: ax + by + cz = d. The vector (a,b,c) is a vector perpendicular to the plane Definition. (The point normal equation of a plane) The plane which passes through the point P 0 = (x 0,y 0,z 0 ) and which is perpendicular to the direction of the nonzero vector n = (a,b,c) has equation a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. In vector form, the equation is usually written as: n (x p 0 ) = 0, where the vector p 0 corresponds to the point P Remark. Suppose the plane U has equation 2x 1 3x 2 + 5x 3 = 10. If you multiply every coefficient by the same number, say 5, then the resulting equation, 10x 1 15x 2 +25x 3 = 50, describes U as well. In practice we usually (but not necessarily) multiply the coefficients by a number so that the equation looks more pleasant. For instance, 3 2 x x x 3 = 9 2 looks better in the form 45x 1 42x 2 +20x 3 = 135 (where the first mentioned equation has been multiplied through by 30). What we just said is one instance of the fact that we are free to change the appearance of an equation according to our specific needs, as long as we do not violate any mathematical rule. Here are a few examples of the equation 2x 1 3x 2 + 5x 3 = 10 in different guises: 2x 1 3x 2 + 5x 3 10 = 0 (all terms on one side). x 2 = 2 3 x x (the second variable has been isolated on one side). 2(x 1 1) 3(x 2 + 1) + 5(x 3 1) = 0 shows clearly that (1, 1, 1) is on the plane.
10 30 3 Space: lines and planes Vector parametric description of a plane A second way of representing a plane is by a vector parametric equation, similar to the description of a line. First we discuss an example, where we use an intuitive approach, then we turn to a general strategy. So here is the first example. Suppose the plane passes through (0, 0, 0), like for instance the plane U given by x 1 + 2x 2 3x 3 = 0. Take two vectors in the plane which are not multiples of each other (they point in really distinct directions), for example (2, 1, 0) and (3, 0, 1). Geometrically, it is quite clear that any vector in the plane can be obtained by taking suitable multiples of the vectors and then adding them, see Figure (2.7). b a x b a p Figure 2.7: Left hand side: If U is a plane through the origin and a and b are two vectors which are not along the same line, then any vector x in the plane can be obtained by adding suitable multiples of a and b. Right hand side: the ingredients of a vector parametric description of a plane: a vector p with endpoint in the plane and two direction vectors a and b. For varying λ and µ, the vector p + λa + µb runs through all vectors of the plane. So, any vector in the plane can be written as a so called linear combination λ(2, 1, 0) + µ(3, 0, 1) of (2, 1, 0) and (3, 0, 1) for suitable λ and µ. Note that in this way we have described the vectors/points of the plane explicitly: every value of λ and µ produces a point in the plane. For example, for λ = 2 and µ = 3 we find 2 (2, 1, 0) 3 (3, 0, 1) = ( 5, 2, 3). Next we turn to a general strategy for finding vector parametric descriptions. The way to find a vector parametric equation of a plane, starting from an equation, is to solve the equation (and don t forget: there are infinitely many solutions; it s a plane after all). Suppose V is the plane with equation x 1 + 2x 2 3x 3 = 4; so V is parallel to U, since the vector (1, 2, 3) is perpendicular to both planes. If we rewrite this equation as x 1 = 2x 2 + 3x 3 + 4, then we clearly see, that for any given values of x 2 and x 3, there is exactly one value of x 1 such that the triple (x 1,x 2,x 3 ) is on the plane. So assign the value λ to x 2 and µ to x 3. Then x 1 = 4 2λ + 3µ. We gain more insight in the solutions if we
11 2.2 Describing lines and planes 31 rewrite this explicit description in vector notation as follows: (x 1,x 2,x 3 ) = (4 2λ + 3µ,λ,µ) = (4, 0, 0) + λ( 2, 1, 0) + µ(3, 0, 1). In this case you see (again) that V is parallel to U (how?). The vector (4, 0, 0) is usually called a support vector, while ( 2, 1, 0) and (3, 0, 1) are called direction vectors. To summarize: we have described two ways of finding a vector parametric description of a plane: Solve an equation describing the plane and rewrite the solutions in the form p+λa+ µb. Find any vector p whose endpoint is in the plane and find two vectors a and b representing two independent directions in the plane. Then the endpoints of p + λa + µb for varying λ and µ run through all points of the plane. Please note that the same plane can be described by parametric descriptions which may look quite different. See the following example. Equations of a plane, however, show less variation: if we only consider equations of the form ax + by + cz = d, then two such equations describe the same plane precisely when the coefficients differ by a common (nonzero) multiple Example. To find a vector parametric description of the plane U : x + y + z = 4, any of the following approaches can be taken. a) We solve for x, so we first rewrite the equation as x = 4 y z. If we let y = λ and z = µ, then x = 4 λ µ. In vector form this becomes: (x,y,z) = (4 λ µ,λ,µ) = (4, 0, 0) + λ( 1, 1, 0) + µ( 1, 0, 1). Intuitively: the plane rests on (4, 0, 0) and an arbitrary vector in the plane is described by adding any combination of the vectors ( 1, 1, 0) and ( 1, 0, 1) to (4, 0, 0). b) Alternatively, by looking carefully at the equation, pick any vector in U, say (1, 1, 2). We use this as the support vector. Now take any two vectors which are perpendicular to (1, 1, 1) and which are not multiples of one another ( independent vectors ), i.e., pick two independent solutions of u+v+w = 0, e.g., (2, 3, 5) and (1, 1, 2). Then the plane is described as (x,y,z) = (1, 1, 2) + λ(2, 3, 5) + µ(1, 1, 2). Note that this description is quite different from the previous description. This illustrates the fact that vector parametric descriptions of planes are far from being unique. Incidentally, in picking two vectors perpendicular to (1, 1, 1), usually two vectors are chosen which look relatively simple, like (1, 1, 0) and (0, 1, 1). Our choice of vectors (2, 3, 5) and (1, 1, 2) is correct, but may be more cumbersome in computations. The advantage of the method in a) is that it produces relatively simple vectors.
12 32 3 Space: lines and planes c) If you follow the method as explained in a) but solve for y rather than for x, then the resulting parametric equation is again somewhat different: start by rewriting the equation as y = 4 x z, then set x = λ and z = µ. Finally, you get (x,y,z) = (0, 4, 0) + λ(1, 1, 0) + µ(0, 1, 1) Transforming vector parametric descriptions into equations We have come across two ways of representing planes: by equations and by parametric descriptions. To go from the first to the second comes down to solving an equation representing a plane and rewriting the solutions in the appropriate vector form. This has been discussed above. Here, we discuss how to transform vector parametric descriptions into equations. This is best illustrated by an example. Suppose the plane U is given by (2, 1, 3) + λ(1, 2, 1) + µ(1, 1, 3). Our task is to find an equation of the form ax + by + cz = d representing U, i.e., we have to find a, b, c and d. This problem splits in two parts: To find a, b and c note that the vector (a,b,c) is perpendicular to the plane, so is perpendicular to both direction vectors (1, 2, 1) and (1, 1, 3). This means: So we have to solve: (a,b,c) (1, 2, 1) = 0, and (a,b,c) (1, 1, 3) = 0. a + 2b + c = 0 a + b + 3c = 0 for a, b and c. To solve this system of two equations, first eliminate a from the second equation by subtracting the first equation from the second: a + 2b + c = 0 b + 2c = 0. To make the first equation simpler, add twice the new second equation to the first: a + 5c = 0 b + 2c = 0. So a = 5c and b = 2c. So one vector perpendicular to (1, 2, 1) and (1, 1, 3) is, for example, the vector (a,b,c) = ( 5, 2, 1) obtained by taking c = 1. Thus, our equation looks like 5x + 2y + z = d and it remains to find d. To find d in 5x + 2y + z = d is simple: just substitute any vector of U, for instance the support vector (2, 1, 3): ( 5) = d. In conclusion, an equation for U is 5x + 2y + z = 5.
13 2.3 The relative position of lines and planes: intersections The relative position of lines and planes: intersections Intersecting a line and a plane The intersection of a line and a plane usually consists of exactly one point. If the line and the plane happen to be parallel, the intersection may be empty or consist of the whole line. How does this work out in actual computations? Here is an example. Suppose the plane U has equation 2x 1 + 3x 2 5x 3 = 3, and the line l has the parametric description ( 1, 4, 3) + λ(1, 1, 2). Finding the intersection comes down to finding the value(s) of λ for which the corresponding point on l belongs to U as well, i.e., satisfies the equation for U. So we substitute ( 1 + λ, 4 λ, 3 + 2λ) in the equation: 2( 1 + λ) + 3(4 λ) 5( 3 + 2λ) = 3. This reduces to 25 11λ = 3, so that λ = 2. Now substitute this value of λ in the parametric equation of l to find the point of intersection: (1, 2, 1) Intersecting two planes Geometrically it is clear that in general two planes meet along a line. This is how you find that line of intersection explicitly. Suppose U and V are two planes, with equations x 1 + 2x 2 x 3 = 4 and 2x 1 + x 2 5x 3 = 2, respectively. We have to solve both equations simultaneously, i.e., find all (x 1,x 2,x 3 ) which satisfy both equations. To manipulate the equations efficiently, we write them as follows: x 1 + 2x 2 x 3 = 4 2x 1 + x 2 5x 3 = 2. Subtract the first equation 2 times from the second (and leave the first equation as it is): x 1 + 2x 2 x 3 = 4 3x 2 3x 3 = 6. Divide the resulting second equation by 3 to obtain: x 1 + 2x 2 x 3 = 4 x 2 + x 3 = 2. Now get rid of x 2 in the first equation by subtracting the new second equation from the first: x 1 + 3x 3 = 0 x 2 + x 3 = 2. In this stage, both x 1 and x 2 can be expressed in terms of x 3 : to see this more clearly, rewrite as follows: x 1 = 3x 3 x 2 = x
14 34 3 Space: lines and planes Assign an arbitrary value λ to x 3, then we get (x 1,x 2,x 3 ) = (3λ, λ + 2,λ) = (0, 2, 0) + λ(3, 1, 1), the parametric description of a line with direction vector (3, 1, 1). 2.4 The relative position of lines and planes: angles The angle between two lines It is fairly straightforward to define the angle between two lines l and m, although one detail has to be taken care of. a) Take a vector a in the direction of l and a vector b in the direction of m; Figure 2.8: To find the angle between two lines, choose vectors in the direction of the lines. b) Then compute the angle between a and b using the inner product; if the angle between a and b is obtuse, then replace a by a and compute the angle between a and b. In other words, make sure you end up with an acute angle (between 0 and 90 ). To summarize: If a and b are vectors in the directions of the lines l and m, resp., then the angle α between l and m is computed from cos α = a b a b. By using the absolute value in the numerator, we ensure that the fraction is non negative. Consequently, we get an acute angle Example. The line l is given by (2, 7, 1)+λ(1, 0, 1) and the line m is given by (2, π, 1)+ µ(1, 1, 2). To compute the angle α between l and m we take a = (1, 0, 1) and b = (1, 1, 2) (the other vectors occurring in the parametric descriptions are irrelevant), and solve α from cos α = (1, 0, 1) (1, 1, 2) (1, 0, 1) (1, 1, 2) = 3 = = 3 2.
15 2.4 The relative position of lines and planes: angles 35 Therefore, the angle is 60. Note that in this example, the absolute value signs in (1, 0, 1) (1, 1, 2) were not necessary. This would have been different if, for example, the line l were given by (2, 7, 1) + λ( 1, 0, 1) The angle between a line and a plane You quickly realize that it is not so clear what the angle between a line l and a plane U should be precisely. The reason is that the plane contains so many directions. So, which direction of the plane should we compare with the direction of the line? Suppose l and U meet in a point P. The plane U contains a whole family of lines through P. Any such line makes an angle with l, and the angle with such a line in the plane varies as this line varies in the family. (There is one case where the angle does not change, do you see which case this is?) l P U l β m P U Figure 2.9: To find the angle between a line and a plane, it is convenient to introduce a line perpendicular to the plane as in the right hand picture. If β is the angle between these two lines, then the angle between the line and the plane is defined to be 90 β. But there is a way out if you realize that the direction of a plane is also characterized by the direction of a line perpendicular to the plane. So, let us take a line m through P which is perpendicular to the plane U. Of course, we know what the angle between l and m is. Now this angle is of course not really what we want, but 90 minus this angle turns out to be the appropriate angle. Summarizing: to compute the angle α between the line l with direction a and the plane U with vector n perpendicular to U, solve α from sin α = a n a n, and make sure to take α in the range between 0 and 90. (Note the occurrence of sin: with cos we would find the angle between l and the line perpendicular to U, but we need 90 minus this angle.) Here, we are using cos(90 α) = sin α Example. To compute the angle between the line l with parametric equation (2, 0, 3) + λ(0, 1, 1) and the plane U with equation x 1 +x 2 +2x 3 = 13, we need the vector a = (0, 1, 1)
16 36 3 Space: lines and planes from the parametric description and we need a vector perpendicular to the plane U, for example, n = (1, 1, 2) (taken from the coefficients in the plane s equation). Now we can proceed in two ways: (a) Solve for α in sin α = Therefore the angle is 60. a n a n = = (b) Or we compute the angle β between a and n from cos β = a n a n = = 3 2, 3 2. so that β = 30. The angle between l and U is then = The angle between two planes Suppose U and V are two planes which meet along the line l. Of course they seem to make a definite angle with each other, but it takes some thinking to realize that an appropriate way to make precise what this angle is, is to look at the angle between two vectors perpendicular to the two planes (cf. the case of a line and a plane). The only detail you have to be careful about is that you may have to replace one of the vectors by its opposite vector in order to guarantee that the angle is acute (at most 90 ), or introduce an absolute value just like we did in the case of a line and plane. To summarize: To compute the angle between the planes U and V, take a nonzero vector u perpendicular to U and a nonzero vector v perpendicular to V. The angle α between U and V is then computed from cos α = u v u v. Figure 2.10: To find the angle between two planes, determine the angle between two vectors perpendicular to the two planes, respectively. In particular, two planes are perpendicular if the two vectors perpendicular to the planes are perpendicular.
17 2.5 The relative position of points, lines and planes: distances Example. Consider the planes U and V with equations x 1 x 3 = 7 and x 1 x 2 2x 3 = 25, respectively. To compute the angle between the two planes, consider the vector (1, 0, 1) which is perpendicular to U and the vector (1, 1, 2) which is perpendicular to V. Both vectors are taken from the coefficients of the two equations. Now solve α from cos α = Therefore the angle is 30. (1, 0, 1) (1, 1, 2) (1, 0, 1) (1, 1, 2) = = The angle between two planes: alternative approach Here is another approach to compute the angle between two intersecting planes U and V. Fix a point P on the line of intersection l of U and V. Then take a plane W through P which is perpendicular to l. Now W meets U along a line m, and W meets V along a line n. Finally, compute the angle between the lines m and n. W 3 2. P α α m n Figure 2.11: An alternative approach to computing the angle between two intersecting planes U and V. The picture shows the cross section with the plane W perpendicular to the line of intersection l of U and V. The intersection of U and W is denoted by m; the intersection of V and W is denoted by n. To see why this approach leads to the same answer, just realize that normal vectors to U and V with their tails at P lie in W. This is illustrated in Fig. (2.11). This approach usually leads to more complicated computations. 2.5 The relative position of points, lines and planes: distances Computing the distance between two points is straightforward, but the computation of distances between points and lines, points and planes, etc., is more subtle. In this section we are not aiming for explicit formulas, but for strategies to compute distances The distance between a point and a line Suppose P is a point and l is a line. The distance between P and a point on l varies as this
18 38 3 Space: lines and planes point varies through l. So the question is: for which point on l is this distance minimal? A bit of experimentation with the Theorem of Pythagoras (see Fig. (2.12)) shows that the point Q such that PQ is perpendicular to l is the point we are looking for. P l Q R Figure 2.12: The shortest distance between a point P and a point on the line l occurs for the point Q where PQ is perpendicular to the line l. For instance, applying Pythagoras theorem to the triangle PQR, shows that PQ < PR. Let us work this out in a specific example, where P is the point (7, 2, 3) and l is the line with parametric description (2, 1, 1) + λ(1, 1, 2). First, an arbitrary point Q on l is described by (2 + λ, 1 λ, 1 + 2λ). The vector QP is then ( 5 + λ, 3 λ, 4 + 2λ). The next step is to solve λ from the condition that PQ is perpendicular to the direction vector (1, 1, 2) of l, i.e., solve ( 5 + λ, 3 λ, 4 + 2λ) (1, 1, 2) = 0. This comes down to ( 5+λ) ( 3 λ)+2(4+2λ) = 0 or 6+6λ = 0. So λ = 1 and the point Q on l closest to P is therefore (substitute for example in the parametric equation of l) (2, 1, 1) (1, 1, 2) = (1, 0, 1). Finally, the distance between P and the line l is calculated as the distance between P and Q: (7 1)2 + (2 0) 2 + ( 3 1) 2 = 44 = The distance between a point and a plane The distance between a point P and a point Q in the plane U varies as Q varies. By the distance between a point P and a plane U we mean the shortest possible distance between P and any of the points of U. But how do we find such a point in the plane? It is geometrically obvious that we can locate such a point by moving from P in the direction perpendicular to U until we meet U, i.e., we need the line through P perpendicular to the plane. Just as in the previous case (2.5.2), this is based on Pythagoras theorem. Let us use this strategy when P is the point (5, 4, 6) and U is the plane given by the equation x 2y + 2z = 7.
19 2.5 The relative position of points, lines and planes: distances 39 P locate Q distance U Q Figure 2.13: To find the distance between P and the plane U, first determine the intersection of U and the line through P perpendicular to U. Then calculate the distance between P and the point of intersection. First we look for the line through P and perpendicular to U. A vector perpendicular to U is easily extracted from its equation: (1, 2, 2). So the line through P with direction vector (1, 2, 2) is described by (5, 4, 6) + λ(1, 2, 2). The next step is to intersect this line with U. We substitute (5, 4, 6) + λ(1, 2, 2) in the equation: (5 + λ) 2( 4 2λ) + 2(6 + 2λ) = 7. This is easily rewritten as λ = 7, so that λ = 2. For λ = 2, we find the point Q = (5, 4, 6) 2(1, 2, 2) = (3, 0, 2). So Q = (3, 0, 2) is the point in U closest to P. Finally, the distance between P and U is calculated as the distance between P and Q: (5 3) 2 + ( 4 0) 2 + (6 2) 2 = 36 = 6. There is at least one but: what if the plane is given by a vector parametric equation? Then there are a couple of strategies available: One strategy is to find an equation for the plane and then proceed as before. An alternative is to find a point Q in the plane such that the direction of PQ is perpendicular to every direction in U The distance between two non intersecting lines If two distinct lines in 3 space do not intersect (in one point), then it makes sense to consider their distance. In case the lines are parallel, i.e., their direction vectors are multiples of each other, you take a point on one line and compute the distance to the other line, just like before. In case the direction vectors are not multiples of one another, the computation of the distance turns out to be more complicated. Suppose l and m are two such lines. Among all points P on l and all points Q on m we need to find points for which the distance PQ is
20 40 3 Space: lines and planes minimal. As before, this condition turns out to be related to right angles: we have to find P on l and Q on m in such a way that PQ is perpendicular to both l and m. The procedure is best illustrated with an example. Suppose l is given by (1, 3, 3)+λ(2, 0, 1) and m is given l m Figure 2.14: To find the distance between two non intersecting lines l and m, find vectors p on l and q on m such that q p is perpendicular to both l and m. The distance is the length q p of q p. by (1, 6, 3) +µ(0, 1, 1). The lines are certainly not parallel since the direction vectors are not multiples of one another. They may intersect, but then our distance computation will simply yield 0. So begin by taking a vector p = (1, 3, 3) + λ(2, 0, 1) = (1 + 2λ, 3, 3 + λ) on l and taking q = (1, 6, 3) + µ(0, 1, 1) = (1, 6 + µ, 3 + µ) on m. Here are the steps to follow: First, we impose the condition that q p be perpendicular to (2, 0, 1) and (0, 1, 1) (the direction vectors of l and m, respectively), i.e., (q p) (2, 0, 1) = 0 and (q p) (0, 1, 1) = 0. Since q p = ( 2λ, 3+µ, 6+µ λ), these conditions translate into two equations in the variables λ and µ: Rearranging leads to 4λ + ( 6 + µ λ) = 0 (3 + µ) + ( 6 + µ λ) = 0. 5λ + µ = 6 λ + 2µ = 3. An easy calculation shows that this system has exactly one solution: λ = 1 and µ = 1. Second, corresponding to this solution we find the vector p = ( 1, 3, 2) and q = (1, 7, 2). The distance between the lines is therefore (1 1)2 + (7 3) 2 + ( 2 2) 2 = 36 = 6.
21 2.6 Geometric operations: translating lines and planes Geometric operations: translating lines and planes In designing shapes, you may want to move or rotate certain elements of your design, say a window or a wall. The question is how you do this mathematically. Such mathematical computations are at the basis of computer software implementations. In this section, we will briefly discuss a few aspects of translations Translating lines A fairly simple geometric operation is translation: we move an object in a given direction over a given distance. We can represent this direction and distance by a vector t. Any vector (x,y,z) is moved to another vector, given by (x,y,z) + t. If t = (2, 5, 1), then (x,y,z) is translated to (x+2,y+5,z 1). No problem here. Now let us turn to the effect of a translation on lines. If a line l is given by p + λv, then a translation over t produces again a line, now with parametric equation (p+t)+λv. For example, if l is given by (1, 0, 6)+λ(1, 1, 1), then after translation over t = (2, 5, 1) we obtain the line with support vector (1, 0, 6)+(2, 5, 1) = (3, 5, 5) and direction vector (1, 1, 1), i.e., the line (3, 5, 5) + λ(1, 1, 1). Of course, if you translate a line then the resulting line is parallel with the line you started with Translating planes Next, we investigate what happens to a plane when we translate that plane over a vector, say t = (2, 5, 1) as before. Since we have two standard ways of representing a plane, our discussion splits into two parts. In the case where the plane is described by a vector parametric description, the situation is similar to the case of a line. Suppose the plane U is described by (1, 1, 0) + λ(1, 1, 0) + µ(0, 4, 1), then the effect of translating over t = (2, 5, 1) is: (2, 5, 1) + (1, 1, 0) + λ(1, 1, 0) + µ(0, 4, 1) or (3, 4, 1) + λ(1, 1, 0) + µ(0, 4, 1). Suppose the plane U is given by its equation x y +4z = 2. The trouble now is that the description of the plane is implicit. If we translate a vector (x,y,z) of U then the result is (2, 5, 1)+(x,y,z), but then what? Instead, it is better to work backwards: we begin with a vector from the translated plane, say (u,v,w). If we translate this vector back, so subtract (2, 5, 1), then the result should be on U, i.e., satisfy the equation of U. So we substitute (u,v,w) (2, 5, 1) = (u 2,v 5,w + 1) in the equation of U: (u 2) (v 5) + 4(w + 1) = 2. This reduces to u v + 4w = 5. So the plane U : x y + 4z = 2 is translated to the plane V : x y + 4z = 5. Of course, as the equations show, these planes are parallel. Note that the distance between the planes is not equal to the length of t, since t is not perpendicular to U.
22 42 3 Space: lines and planes If you realize that translating a plane produces a parallel plane, another strategy suggests itself: the translated plane must have an equation of the form x y + 4z = d for some d. Now, d can be determined by translating a single point of U and substituting the result. For instance, (2, 0, 0) is in U. Translating the point gives (2, 0, 0) + (2, 5, 1) = (4, 5, 1). Substitute (4, 5, 1) in the equation x y + 4z = d to find that d = Example. (Armada s Paleiskwartier, Den Bosch) Close to the railway station in Den Bosch, a number of buildings with a special appearance have been built during the last few years. From a distance they look somewhat like a fleet of ships from historic times, which is probably the reason for the district s name: Armada s Paleiskwartier. Now some of the buildings are identical in shape (or at least at first glance), so they could be considered as translates of one given central building. Apart from details, we can consider this problem as a problem of translating planes (to be fair: the shape of the buildings is not flat on all sides, so our discussion is limited to flat parts only). Suppose one wall is given by (part of) the plane U : x 3y = 0 (the z-direction is assumed to point upward) in some cartesian coordinate system. Suppose we want to translate this plane over (2, 3, 0), (3, 2, 0) and (5, 5, 0), resulting in the planes U 1, U 2 and U 3. To find the equation of the first translate, we work backwards again: start with (u,v,w) in U 1, then (u,v,w) (2, 3, 0) is in U, so (u 2,v 3,w) should satisfy (u 2) 3(v 3) = 0. A bit of simplification yields U 1 : u 3v = 7. The other translates are found in a similar way. The result is: U 1 : u 3v = 7, U 2 : u 3v = 3, U 3 : u 3v = 10. (In this case, a simpler strategy works faster, since you know that a) all equations should have the form x 3y =... and b) (2, 3, 0) is on U 1, etc.) 2.7 Geometric operations: rotating lines and planes Rotation around a coordinate axis If you rotate the point (x,y,z) around the z axis over 90, the result is either ( y,x,z) or (y, x,z) depending on the direction of orientation (the two rotations are each other s inverse: if you apply the two rotations after each other, the total effect will be that nothing has happened). So rotating a single point isn t that difficult. But suppose we want to rotate a whole plane. Let us concentrate on the first rotation where any point (x,y,z) is rotated to ( y,x,z). What would be the equation or parametric description of the new plane? Let us consider these two descriptions separately. Let U be the plane given by the parametric description (0, 2, 1) + λ(2, 1, 0) + µ(1, 2, 5). So an arbitrary vector in U looks like (2λ + µ, 2 λ + 2µ, 1 5µ). If we rotate this vector we obtain the vector ( 2 + λ 2µ, 2λ + µ, 1 5µ)
23 2.7 Geometric operations: rotating lines and planes 43 y ( y,x) 90 o (x,y) x Figure 2.15: A rotation around the z axis doesn t affect the z coordinate; the figure concentrates on what happens to the x and y coordinates. (apply (x,y,z) ( y,x,z)). This provides us with the vector parametric equation of the rotated plane: ( 2, 0, 1) + λ(1, 2, 0) + µ( 2, 1, 5). Now suppose we want to rotate the plane U, but U is given through its equation x + 2y + z = 5. After rotation we get a new plane, which we call V. If we take a point (x,y,z) of V and rotate it back, then we should get a point of U. Now, rotating (x,y,z) back produces the point (y, x,z) (since the two rotations are inverses). This point (y, x,z) should satisfy the equation of U, so y + 2( x) + z = 5. Rewriting, we get the equation 2x + y + z = 5. A clever approach to rotating the plane U : x + 2y + z = 5 is to rotate its normal vector (1, 2, 1). This results in the vector ( 2, 1, 1) and this should be the normal vector of the rotated plane V. So V has equation 2x + y + z = d for some d. Now, the point (0, 0, 5) belongs to U and remains fixed when we apply the rotation. Substituting (0, 0, 5) in 2x + y + z = d gives d = 5. Therefore, the equation of V is 2x + y + z = 5. (There are more ways to argue that the right hand side coefficient 5 of the equation of U does not change for the rotation at hand.) Example. If we rotate the plane U with equation x+2y +z = 5 around the z axis in the positive direction, we obtain the plane V with equation 2x + y + z = 5. Now suppose we rotate V again over 90 in the same direction to obtain a plane W. If (x,y,z) is a point of W, then rotating it backwards we find the point (y, x,z), a point that belongs to V, so should satisfy the equation of V. This leads to 2y + ( x) + z = 5, or x 2y + z = 5. This equation looks pretty much like the equation x + 2y + z = 5 of U. Can you explain the minus signs by directly considering a rotation over 180?
24 44 3 Space: lines and planes Fans of planes If two planes U and V are not parallel, they intersect along a line, say l. By rotating one of the planes around l, we generate a whole family of planes, the fan of planes on l. Of course, all these planes contain l. The question is: what is the equation of a member of such a fan, given the equations of U and V? To answer this question, we take two specific planes, say U: x y z = 1 and V : 3x y + 3z = 9. These planes turn out to intersect along l: (1, 0, 2) + λ(2, 3, 1) (this follows from a computation similar to the one explained on p. 33). Now every point of l satisfies both equations, but then every point of l also satisfies a combination of the two equations, like 2 times the first equation + 3 times the second equation, or 2(x y z) + 3(3x y + 3z) = 2 ( 1) + 3 9, which simplifies to 11x 5y + 7z = 25. Likewise, any combination a(x y z) + b(3x y + 3z) = a + 9b (with a and b not both equal to 0) is an equation of a plane that contains l. Sometimes, it is more convenient in this setting to bring all terms of the equation of a plane to the left hand side: x y z + 1 = 0 and 3x y + 3z 9 = 0 describe the two planes and a(x y z + 1) + b(3x y + 3z 9) = 0 is, for fixed a and b, the equation of a specific member of the fan Example. The planes U : x + y = 1 and V : x + 2y = 3 generate a fan. In this example we compute a member of the fan which is perpendicular to U. The equation of a general member of the fan is a(x + y 1) + b(x + 2y 3) = 0, or (a + b)x + (a + 2b)y a 3b = 0. The plane with equation (a + b)x + (a + 2b)y a 3b = 0 is perpendicular to U if (a + b,a + 2b, 0) (1, 1, 0) = 0. This reduces to 2a + 3b = 0. For a = 3 and b = 2 we obtain the plane with equation x y + 3 = 0. (Other non zero values of a and b which satisfy 2a + 3b = 0 lead to a multiple of this equation and therefore to the same plane.) 2.8 Geometric operations: reflecting lines and planes Mirror images Sometimes you do not even notice it at first sight when two (types of) buildings have been constructed as mirror images of one another: they look so much the same. Sizes, distances and angles are preserved under reflections, but somehow the orientation has
25 2.8 Geometric operations: reflecting lines and planes 45 changed: a right handed screw changes into a left handed screw. One immediate property of a reflection is that taking the mirror image of the mirror image of a point or vector brings you back to the original point or vector. In this section we restrict our discussion to reflections where the mirror is a coordinate plane or another easy to describe plane. In particular, the mirrors we consider are flat. To begin with, we consider the reflection in the y,z plane. In this case, the mirror image z ( x,z) (x,z) Figure 2.16: In picturing the reflection in the y,z plane, we have left out the y axis for the sake of simplicity and pictured the resulting 2 d version of reflection in the z axis. The vector (x,z) is reflected into the vector ( x,z). The two chairs are mirror images. of any point (x,y,z) (or vector) is ( x,y,z) (so only the x coordinate has changed). The image of a line, say l : (2, 3, 4) +λ(1, 2, 3), is obtained by reflecting each point of the line individually: so the image of l is l : ( 2, 3, 4) + λ( 1, 2, 3). To be precise, every vector (2+λ, 3+2λ, 4+3λ) of l is mapped to ( 2 λ, 3+2λ, 4+3λ) = ( 2, 3, 4)+λ( 1, 2, 3). Similarly, the image of a plane given in parametric form is easily obtained. But if the plane is given by an equation, we have to work backwards again, just like we did for translations and rotations, or exploit a normal vector of the plane. Suppose the plane U has equation 2x + y 3z = 7. If (u,v,w) is a vector on the mirror image U of U, then reflecting the vector yields the vector ( u,v,w) in U, so ( u,v,w) satisfies the equation of U: 2( u) + v 3w = 7. Therefore, U has equation 2u + v 3w = 7. If the mirror is not a coordinate plane, like the y,z plane, things usually become a lot harder to describe. In later chapters we will go into this in more detail. In the case at hand, the normal vector approach runs as follows. Reflect the normal vector (2, 1, 3) to obtain ( 2, 1, 3), so the equation of U is of the form 2x+y 3z = d for some d. To determine d, pick a point on U, say (2, 3, 0) and reflect it: ( 2, 3, 0). Substitute in 2x + y 3z = d and conclude that d = 7. The equation of U is therefore 2x + y 3z = More reflections The case where the mirror is the plane x = y is still easy to handle. The mirror image of x
26 46 3 Space: lines and planes the vector (x,y,z) is easily seen to be (y,x,z) (just change the first two coordinates). From this description in coordinates it is also clear that reflecting twice in succession brings you back to where you started: (x,y,z) reflect (y,x,z) reflect (x, y, z) Example. Two towers of a building are designed as mirror images. Both towers have triangular horizontal cross sections. In a suitable coordinate system, the walls of one of these towers are (parts of) the three planes U : x = 3, V : y x = 1 and W : y = 1. The mirror is the plane x = y. To find the mirror image of the plane V : y x = 1, start with a vector (u,v,w) on the mirror image V of the plane. The mirror image of the vector (u,v,w) is (v,u,w) and is on the plane V : y x = 1, so u v = 1. Therefore the equation of the mirror image V of the plane V : y x = 1 is x y = 1 (so the roles of x and y have been interchanged). Here is an overview of the equations of the planes and their mirror images. planes mirror images x = 3 y = 3 y x = 1 x y = 1 y = 1 x = More complicated reflections A straightforward observation on reflections in a plane U is the following. If you connect a point P and its mirror image P, then this segment is perpendicular to the plane U. Moreover, the distance from P to U is equal to the distance from P to U. This translates into the following strategy for computing mirror images: a) Start at P and move along the line through P which is perpendicular to U, so first set up the parametric equation of this line, say p + λv. b) Compute the intersection of the line with the plane U. For a specific value of the parameter, say λ 0, the line intersects the plane U. So p + λ 0 v is halfway between P and its mirror image. c) But then p + 2λ 0 v is the mirror image (note the factor 2). By way of example let us compute the mirror image P of the point P = (2, 3, 4) in the plane x+y+z = 3. The idea is to start moving from P in the direction (1, 1, 1) (perpendicular to U). Then locate on this line the second point which has the same distance to U. Let l be the line (2, 3, 4) + λ(1, 1, 1). First find out where this line meets the plane U, so substitute in x + y + z = 3: (2 + λ) + (3 + λ) + (4 + λ) = 3. This equation reduces easily to 9 + 3λ = 3 so that λ = 2. So if you add 2 (1, 1, 1) to (2, 3, 4) you end up in the plane. Therefore, if you go twice as far you will get the mirror
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# Which Figure Is Formed By Four Non Collinear Points?
## What are 4 collinear points?
Therefore, for any four collinear points, we can define one cross ratio, and we define these four points as a Base B=\{(A,B,C,D),(D,C,B,A)\}..
## How many lines is 3 distinct points?
three linesSo, we can name the lines as AB, BC and AC. Hence, we get that only three lines are possible with the help of three distinct points.
## What are non collinear points?
Non-collinear points are a set of points that do not lie on the same line.
## What is the formula of collinear points?
If the A, B and C are three collinear points then AB + BC = AC or AB = AC – BC or BC = AC – AB. If the area of triangle is zero then the points are called collinear points. If three points (x1, y1), (x2, y2) and (x3, y3) are collinear then [x1(y2 – y3) + x2( y3 – y1)+ x3(y1 – y2)] = 0.
## What are the names of the three collinear points?
What are the names of three collinear points? Points L, J, and K are collinear.
## How many lines can we draw in 5 points?
10Attempt: Given 5 points, a line consist always of 2 points. Thus the total number of straight lines that can be drawn between 5 points is 5_C_2 = 10.
## What are three non collinear points?
Points B, E, C and F do not lie on that line. Hence, these points A, B, C, D, E, F are called non – collinear points. If we join three non – collinear points L, M and N lie on the plane of paper, then we will get a closed figure bounded by three line segments LM, MN and NL.
## Which figure is formed by non collinear points?
A triangle is formed by three non-collinear points.
## Which figure is formed by the three non collinear points?
triangleA triangle is a figure formed by three segments joining three noncollinear points. Each of the three points joining the sides of a triangle is a vertex. The plural of vertex is “vertices.” In a triangle, two sides sharing a common vertex are adjacent sides.
## How many lines can pass through 4 non collinear points?
Answer: Six is the correct ans.
## What is mean by collinear points?
Three or more points that lie on the same line are collinear points . Example : The points A , B and C lie on the line m . They are collinear.
## How many lines is 4 distinct points?
Since two points determine a line and there are 4 points (and no 3 of them are collinear), the number of lines that can be created is 4C2 = (4 x 3)/2! = 6. |
## Statistics And Probability
Statistics And Probability As a self-taught writer, I write about probabilities, and how they play a role in my life. I couldn’t do that with a statistical model because I don’t know what the probability is. Is probability about something? I thought so. But I can’t. I can”t answer that question because it’s impossible to say. I would have to take a harder risk than just about any other statistic. A lot of people will say that the probability is small, but I think it’ll be multiplied by the number of times it has a value of zero. To take a risk, how many times has it been multiplied by the total number of times that its value has a value greater than zero? The answer is small, and it can’ve been much, much bigger. But what if we couldn’ve tested the probability for whether the number of outcomes has a value less than zero? Let’s say we have a number of outcomes that have a value less zero than zero. And we have the probability that we’re in a safe place, and it’d be easy to say that it’re safe to move out and go back. Hence, we browse this site say that a value less 0.6 is “safe to move out”. So how does it work? It’s a test of probability. We have a value for the probability of a number of outcome outcomes that are positive. So we can say the probability is that is positive. If it’’”s positive, and we have the value of 0.6, we have the good news. If we take the probability of 0.2, and we get the probability that it”s a safe place”, and we take this into account, we have a probability of 0, because we have the positive value of 0, and the negative value of 0; and we take the value of 1, because we don”t know what is a safe place. And we get a value of 1; and we get a plus.
## Statistics And Probability App
In other words, the probability that the value of the probability of the positive value is positive is more tips here and the probability that is negative is 1. Now, if we take the test of probability, we need to know what is the probability that a number of positive outcomes is a safe way to move out of your own world. In other words, we need a value less of 0.8. If we take a negative value of 1.2, we have to take higher values of 0.5 and 1, and they”re not safe. So we need a positive value of 1 to see above. We need to know that the probability of moving out of the world is a positive value. click to read more Let me give you some examples. The probability that a value of 0 is a safe value is 1.5, and the value that is a safe outcome is 2.5. 1.5 + 0.5 = 2.5 = 1.5 + 1.5 = 0.5 == 1.
## Inferential Statistics Meaning In Urdu
5 We”re taking some numbers and not looking at the ratio. Ranges of numbers Statistics And Probability: Analysis of Potential Effects A lot of research is being done to better understand potential effects of medication on the immune system and the regulation of the immune system by the immune system. As a result of this, there are a lot of studies that are being done to investigate the effects of medications on the immune systems. For example, there are studies that show that certain drugs (such as statins) do have an effect on the immune response and that this would be caused by a decrease in the number of lymphocytes in the peripheral blood. The number of lymphocyte in the peripheral lymphocytes is correlated to a decrease in viral load and there is a correlation between the number of virus particles and the number of antibodies in the blood. Also, there is a relationship between the number and the number density of antibodies in blood and the level of antibodies in serum. In addition, there are some studies that show a correlation between blood cell number and the amount of antibodies in plasma and the number and density of antibodies. For example in a study that showed a correlation between plasma concentration and the amount and density of antibody in blood, the number of antibody that were present in plasma and in sera was larger than that of the number. There are a lot more studies to be done to understand this effect. For example the study that was done in the 1980’s and which was looking at the effect of a certain drug on the immune status was done. The study showed that there was a positive correlation (within a certain range) between the number in the blood of the patient and the number in serum. This is a pretty clear correlation. This study looked at the effect that a certain drug (such as a certain imipramine) has on the immune effect. The study looked at how the number of viruses in the blood was affected by the drug (such a drug called imipramin) or the number of bacteria in the blood (such a medication called erythropoietin). There was a positive effect on the number of the bacteria in the serum. Also, the study showed that it was possible that this effect could be due to an increase in the number and concentration of antibodies in a patient’s blood. So, there are many studies that are needed to understand the effects of drugs on the immune. For example there is a study that looked at the relationship between the amount of medications and the number, the amount of antibody in the blood, the level of antibody in serum, and the number. There was a correlation between these effects and the number that was taken by the patient. A lot of research still needs to be done that will help to understand the effect of medication on immune function.
## Statistics Subject
Also, this is a very important subject for the study that is being done in the future. The next step in this research is to study the effects of certain medications in the blood and the immune system from a side view. Also, you need to understand the role of antibodies in this process. For example a person at the beginning of the study would have an antibody level that is too high or low. The person would not have a level of antibodies that is too low or too high. Also, it is important to know if there were any other factors that might cause the antibody look at this web-site to be too low or not too low. A first step in this investigation is to analyze the possible effects of certain drugs on the immunologicalStatistics And Probability Blog The good news is that you can do a lot of work to get a better understanding of the math involved. Not all of the math is easy to learn, but there is one piece of math that can help you get a better grasp of things. For example, we can learn to predict the future, but there are many equations that we don’t know about. This is one of the main reasons we have a computer that is easy to use. You can download it here. The only caveat to this book is that the goal is to remain in math. That means that you will need to practice and find out both the math and probabilities of how much impact you will have in the future. If it is a little bit hard to explain, a few notes will help! One of the best ways to do this is to use a calculator. It is a great tool that you can use to use in your life. Once you have a calculator, you can calculate some numbers with the help of a calculator, click to find out more show how your life will change in the future if you use it. Now that you have a simple calculator, you are ready to go. If you are not sure where to start, here is a quick guide to what you need to do. Step 1: Check Math The first thing you need to know about calculating the future is how much you are paying for it. Make sure to check the math involved to see if you are sure the math is right. |
## Number Cards and Counters
MATERIAL:
Numerals from 1 to 10 in black on white cards and 55 counters all the same size and color.
PURPOSE:
To arrange the numerals in their correct order while putting the proper quantity with each. A minor purpose is to arrange the counters to give a visual impression of odd and even quantities and later to teach the terms "odd" and "even."
AGE:
4 and a half years and older
EXERCISE:
The material is taken to the child's table. The teacher sits beside the child and lays the cards on the table in mixed order. She asks the child to find "1." When he has done so, she shows him to put the "1" card on the extreme left hand side of the table. She hands him the box of counters and asks him to put one of them under the "1" card.
She asks the child which numeral comes after "one." He should be able to tell her, "Two." She asks him to find the card. When he has done this, she shows him how to put it next to the "1" card, and she asks him to put two counters below it.
The exercise proceeds with the child finding each numeral in sequence, and putting the correct number of counters under it. The teacher shows him to arrange the counters in pairs, with the odd ones underneath.
It is obvious that alternate numbers are odd and even but no comment is made at this point. When the child understands the exercise, he is left to work independently. The material is kept on a shelf for him to use whenever he likes.
Control of Error
The sum of the numbers 1 to 10 is 55. Therefore, there are exactly the right number of counters for the exercise. Should a mistake be made, there will be too many or too few at the end of the exercise, and the child can correct his work.
Vocabulary
When the child is doing the exercise easily, the teacher can explain the terms, odd and even numbers. "These numbers, 1, and 3, end in a single counter. We call them odd numbers. Can you find other odd numbers?" The child will point to 5, 7, and 9. To reinforce this visually, the teacher could ask the child to push up the odd number cards.
"The number two ends in an even pair of counters, so we call it an even number. Which other numbers end in an even pair?" The child points to 4, 6, 8, and 10. "That's right. 2, 4, 6, 8, and 10 are even numbers."
Note
In the series of exercises for teaching the understanding of the quantities and the numerals 1 to 10, we guide the child to sequence numbers correctly by building the correct order of the numbers into the first two materials.
• 1. Number Rods - The quantities are fixed and the numerals are loose. By building the rods into a stair, from smallest to largest, and placing the number cards on the end segment of each rod, the child sees the numbers in correct sequence.
• 2. Spindle Box - The quantities are loose and the numerals are fixed in sequence along the top of the box.
• 3. Cards and Counters - Both numerals and quantities are loose. This is the first time we have asked the child to sequence number without providing a control for the sequencing. |
# Exponential Functions Lesson Plan 2: Bigger and Smaller – Exponent Rules
Overview:
This lesson will teach students about several of the rules regarding exponents. It uses a situation from Alice in Wonderland in which Alice’s height is doubled or reduced by half depending on what she consumes to introduce negative exponents and the rules for dividing powers.
Time Allotment:
Two 50-minute class periods
Subject Matter:
Rules of exponents
Learning Objectives:
Students will be able to:
• Understand, and develop a sensibility to, the magnitude of exponential growth.
• Develop laws of exponents.
• Define negative exponents.
Standards:
Principles and Standards for School Mathematics, National Council of Teachers of Mathematics (NCTM), 2000.
NCTM Algebra Standard for Grades 6-8
http://standards.nctm.org/document/chapter6/alg.htm
NCTM Algebra Standard for Grades 9-12
http://standards.nctm.org/document/chapter7/alg.htm
### Teacher Supplies
Supplies:
Students will need the following:
• Scientific or graphing calculators
### Teachers Activities and Assignment
Steps
Introductory Activity:
1.Have students consider the following problem individually before the lesson begins:
Rallods in Rednow Land
Which is more money?
1. One billion rallods
2. The amount obtained by putting 1 rallod on one square of a chessboard, 2 rallods on the next square, 4 on the next, and so on, until all 64 squares are filled.
(NOTE: To find the total number of rallods on the chessboard, students must add 1 + 2 + 4 + 8 … + 263. Finding the total on all 64 squares is not necessary to answer the question, since the running total surpasses 1 billion well before the 64th square.)
2.Have students discuss their intuition regarding this situation. Without calculating, which scenario do they think would yield more rallods, a or b?
3.Allow student groups a few minutes to calculate and discuss the result in choice b.
4.Have students consider the following problem:
On what square of the chessboard would the total number of rallods first exceed 1 billion?
5.Give student groups a few minutes to calculate an answer to this question. Then, have the groups share their results and discuss their methods for obtaining the answer (which is the 31st square, since 230 = 1,073,741,824). Students may also use guess-and-check and say it would be on the 29.9thsquare. Although this answer doesn’t make sense in the context of the problem, it will allow for a discussion as to whether or not it is okay to have decimal exponents.
Learning Activities:
1. Have students discuss the effect of cake and beverages on Alice’s height in Alice in Wonderland. Have a student from each group describe his or her group’s discussion to the class. Students should understand that when Alice eats an ounce of cake, her height doubles, and when Alice drinks an ounce of beverage, her height is halved.
2.Give students time to discuss the problems below in their groups:
1. What happens when Alice eats several ounces of cake and drinks the same number of ounces of beverage?
2. Find several combinations of cake and beverage that will cause Alice to be 8 (or 23) times her normal height.
3. Find several combinations of cake and beverage that will cause Alice to be 32 (or 25) times her normal height.
4. Find several combinations of cake and beverage that will cause Alice to be 4 (or 22) times her normal height.
5. What happens if Alice consumes more ounces of beverage than ounces of cake?
6. If Alice eats c ounces of cake and drinks b ounces of beverage, what is her height? Describe her height using a mathematical expression.
3.Have a volunteer from each group present the group’s solutions for each of the above questions.
4.Use the students’ solutions to develop the rules for negative exponents and for divisibility of exponents: 2m/2n = 2m – n. For instance, in question b, students may have shown that eating 7 ounces of cake and drinking 4 ounces of beverage would cause Alice to be eight times as tall, or 27 � (1/2)4 = 23. Rewrite this as 27 � 2-4 = 23 and as 27/24 = 23.
5.During the discussion for question f, be sure to elicit the general formula 2c � (1/2)b = 2c – b. This formula will lead to the rule for negative exponents, 2-b = 1/2b, as well as to the general rule for divisibility, 2c/2b = 2c – b.
6.Assign practice problems for homework.
### Related Standardized Test Questions
The questions below dealing with concepts related to exponents have been selected from various state and national assessments. Although the lesson above may not fully equip students to answer all such test questions successfully, students who participate in active lessons like this one will eventually develop the conceptual understanding needed to succeed on these and other state assessment questions.
• Taken from the Massachusetts Comprehensive Assessment, Grade 10 (Spring 2002):
What is the simplest form of the expression
2x4y2 / x2y2, x 0, y 0?
Solution: 2x4y2 / x2y2 = 2(x4 – 2)(y2 – 2) = 2x2
• Taken from the Massachusetts Comprehensive Assessment, Grade 10 (Spring 2002):
On January 1, 2000, a car had a value of \$15,000. Each year after that, the car’s value will decrease by 20 percent of the previous year’s value. Which expression represents the car’s value on January 1, 2003?
B. 15,000(0.8)4
C. 15,000(0.2)3
D. 15,000(0.2)4
Number cubes are the basis for many games. Each face of a number cube is identified by a number from 1 to 6. Some games use one number cube and some games use multiple number cubes.
Part C – Complete the table below to show the number of possible outcomes when 2, 3, 4, and 5 number cubes are used.
• Solution: There are 62 = 36 outcomes when two cubes are used, 63 = 216 outcomes for three cubes, 64 = 1296 outcomes for four cubes, and 65 = 7776 outcomes for five cubes. In general, there are 6n outcomes for n cubes.
• Taken from the Virginia Standards of Learning Assessment, Algebra I (Spring 2001):
Which is equivalent to:
A. 1/b3
B. b3 |
# How do you find an equation of the line containing the given pair of points (4,5) and (12,7)?
Jan 2, 2017
The equation of the line is $x - 4 y = - 16$
#### Explanation:
The slope of the line passing through $\left(4 , 5\right) \mathmr{and} \left(12 , 7\right)$ is $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{7 - 5}{12 - 4} = \frac{2}{8} = \frac{1}{4}$
Let the equation of the line in slope-intercept form be $y = m x + c \mathmr{and} y = \frac{1}{4} x + c$ The point (4,5) will satisfy the equation . So, $5 = \frac{1}{4} \cdot 4 + c \mathmr{and} c = 5 - 1 = 4$
Hence the equation of the line in slope-intercept form is $y = \frac{1}{4} x + 4$
The equation of the line in standard form is $y = \frac{1}{4} x + 4 \mathmr{and} 4 y = x + 16 \mathmr{and} x - 4 y = - 16$ {Ans] |
Learn to solve whole-number division equations .
# Learn to solve whole-number division equations .
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## Learn to solve whole-number division equations .
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##### Presentation Transcript
1. Learn to solve whole-number division equations.
2. Recall that multiplication and division are inverse operations. If an equation contains division, solve it by multiplying on both sides to “undo” the division.
3. x = 5 7 x = 5 x is divided by 7. 7 x 7 = 7 5 7 x = 5 Check 7 35 ? = 5 7 ? 5 = 5 Additional Example 1A: Solving Division Equations Solve the equation. Check your answer. Multiply both sides by 7 to undo the division. x = 35 Substitute 35 forx in the equation. 35 is the solution.
4. p 13 = 6 p 13 = p is divided by 6. 6 p 6 13 = 6 6 p 13 = Check 6 78 ? 13 = 6 ? 13 = 13 Additional Example 1B: Solving Division Equations Solve the equation. Check your answer. Multiply both sides by 6 to undo the division. 78 = p Substitute 78 forp in the equation. 78 is the solution.
5. x = 9 2 x = 9 x is divided by 2. 2 x 2 = 2 9 2 x = 9 Check 2 18 ? = 9 2 ? 9 = 9 Check It Out: Example 1A Solve the equation. Check your answer. Multiply both sides by 2 to undo the division. x = 18 Substitute 18 forx in the equation. 18 is the solution.
6. p 72 = 4 p 72 = p is divided by 4. 4 p 4 72 = 4 4 p 72 = Check 4 288 ? 72 = 4 ? 72 = 72 Check It Out: Example 1B Solve the equation. Check your answer. Multiply both sides by 4 to undo the division. 288 = p Substitute 288 forp in the equation. 288 is the solution.
7. h Substitute 14 for height of aspen. h is divided by 3. 14 = 3 h 3 14 = 3 3 Additional Example 2: Application At Elk Meadows Park an aspen tree is one-third the height of a pine tree. height of pine height of aspen = 3 The aspen tree is 14 feet tall. How tall is the pine tree? Let h represent the height of the pine tree. Multiply both sides by 3 to undo the division. 42 = h The pine tree is 42 feet tall.
8. w Substitute 95 for Jamie’s weight. w is divided by 2. 95 = 2 w 2 2 95 = 2 Check It Out: Example 2 Jamie weighs one-half as much as her father. father’s weight Jamie’s weight = 2 Jamie weighs 95 pounds. How many pounds does her father weigh? Let w represent her father’s weight. Multiply both sides by 2 to undo the division. 190 = w Jamie’s father weighs 190 pounds. |
## Intermediate Algebra (12th Edition)
$-3\sqrt{2k}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $\sqrt{18k}-\sqrt{72k} ,$ simplify first each term by expressing the radicand as a factor that is a perfect power of the index. Then, extract the root. Finally, combine the like radicals. $\bf{\text{Solution Details:}}$ Expressing the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{9\cdot2k}-\sqrt{36\cdot2k} \\\\= \sqrt{(3)^2\cdot2k}-\sqrt{(6)^2\cdot2k} .\end{array} Extracting the roots of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 3\sqrt{2k}-6\sqrt{2k} .\end{array} By combining like radicals, the expression above is equivalent to \begin{array}{l}\require{cancel} (3-6)\sqrt{2k} \\\\= -3\sqrt{2k} .\end{array} |
## Fractions in Real-World Settings
Standards Click here to scroll down to Standards. Summary Grades 3 and 4 are pivotal grades for students. In Grade 3 they are introduced to multiplication and division within 100 in Common Core State Standards (CCSS). Third graders are initially expected to learn that unit fractions are a part of a whole and do comparisons
## What are Fractions?
Standard: CCSS.MATH.CONTENT.3.NF.A.1 Number & Operations – Fractions: Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b. Lesson Time: 50 minutes This 20-minute activity introduces students in Grades 3-5 to fractions
## Fraction word problems involving time: Example
Today’s post features another fraction problem contributed by a middle school student. This one teaches equivalent fractions and common denominators.
## Using Visual Models to Compare Fractions
One way to teach students fractions is to have them write their own problems and give the answer. Next, have them solve and discuss each other’s problems. Play the Fish Lake game to get some ideas of fractions problems.
## Adding and Subtracting fractions with a Common Denominator: Multi-step problems
Students often struggle with multi-step problems for a lot of reasons. In my experience, the most common one is they stop too soon. For example, in the first lesson below, the student may add up the fractions of the rice gathered by each of the two sisters and stop there. However, the question asks, “How
## Converting a Fraction to Equal 1
Converting a fraction to equal 1 To follow up our lesson plans on introducing fractions and teaching equivalent fractions, we want to talk about converting fractions to a whole number. The simplest case to start converting to whole numbers is converting a fraction to equal 1 This presentation gives several examples of when a fraction
## Equivalent Fractions
Standards CCSS.MATH.CONTENT.3.NF.A.2Understand a fraction as a number on the number line; represent fractions on a number line diagram. CCSS.MATH.CONTENT.3.NF.A.3.BRecognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3. Explain why the fractions are equivalent, e.g., by using a visual fraction model. Lesson Time 30 minutes Technology Required Device with web-browser–Chromebook, laptop, or
## Introducing Fractions
Standards CCSS.MATH.CONTENT.4.NF.B.3Understand a fraction a/b with a > 1 as a sum of fractions 1/b. CCSS.MATH.CONTENT.3.NF.A.1Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b. Summary Before you jump into teaching how to add, subtract, multiply or divide
## Video: Adding Fractions with Like Denominators
Standards CCSS.MATH.CONTENT.4.NF.B.3.CAdd and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. Summary Now that you understand what a fraction is, what’s a numerator and what’s a denominator, you can start doing operations with fractions.
## Video: Visualizing the Exact Fish Problem
Standards Apply and extend previous understandings of multiplication and division. CCSS.MATH.CONTENT.5.NF.B.3Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. Visualizing the |
Home Development Cycles Page 7 - Learning About the Graph Construct using Games, Part 1
# Learning About the Graph Construct using Games, Part 1
Certain everyday problems are easier to solve using the graph construct than any other way, such as the classic "shortest distance between cities" problem. Others are ones you might not expect. In this article, we will play some games to help us understand how we can use the graph construct.
Author Info:
Rating: / 24
February 23, 2005
SEARCH DEVARTICLES
Learning About the Graph Construct using Games, Part 1 - Building the Graph
(Page 7 of 7 )
Our first task is to build the graph that represents the problem. How are we going to do it? This is easy, here is what we are going to do:
2. We will see which states this node can lead to.
3. We will construct new nodes for new states if not already found.
5. We will take one of the new nodes, and go back to two.
That’s pretty much it. We start at the initial node, see which nodes it leads to. We try pouring each of the three3 jugs into each of the other two, this will result in at most six edges per node. But, in reality the number will be smaller because some jugs can be totally full, or totally empty.
Here is the source code for this function. It is easier than you might have thought it would be.
void buildAll()
{
Tuple v=new Tuple(8,0,0);
int i=0;
while(i<data.size())
{
int a=((node)(data.elementAt(i))).data.a;
int b=((node)(data.elementAt(i))).data.b;
int c=((node)(data.elementAt(i))).data.c;
if(a>0&&b<5)
{//calculate amount of water to move from a to b
int toTake=5-b;
if(a<toTake)
toTake=a;
Tuple from,to;
from=new Tuple(a,b,c);
to=new Tuple(a-toTake,b+toTake,c);
}
if(a>0&&c<3)
{
int toTake=3-c;
if(a<toTake)
toTake=a;
Tuple from,to;
from=new Tuple(a,b,c);
to=new Tuple(a-toTake,b,c+toTake);
}
if(b>0&&a<8)
{
int toTake=8-a;
if(b<toTake)
toTake=b;
Tuple from,to;
from=new Tuple(a,b,c);
to=new Tuple(a+toTake,b-toTake,c);
}
if(b>0&&c<3)
{
int toTake=3-c;
if(b<toTake)
toTake=b;
Tuple from,to;
from=new Tuple(a,b,c);
to=new Tuple(a,b-toTake,c+toTake);
}
if(c>0&&a<8)
{
int toTake=8-a;
if(c<toTake)
toTake=c;
Tuple from,to;
from=new Tuple(a,b,c);
to=new Tuple(a+toTake,b,c-toTake);
}
if(c>0&&b<5)
{
int toTake=5-b;
if(c<toTake)
toTake=c;
Tuple from,to;
from=new Tuple(a,b,c);
to=new Tuple(a,b+toTake,c-toTake);
}
i++;
}
}
First, I have to point out that it is not a very good programming practice to put everything into just one class. In real life, the code for the Graph should be separated form the code for the game so that we can re-use the Graph class later. I am just trying to keep the code simple.
It would also have been much better if we have used static finals to hold the values three, five and eight (capacities of jugs), so that we can solve the problem later with different capacities. However, I thought making it this way makes the source code easier to comprehend, and easier to follow.
With that said, the code is pretty self explanatory: it passes over every node, and starts to construct edges to any state can be reached from this node.
That is pretty much it for today. In part two, we will show you how to actually find the shortest path from the initial node to the goal node. We will talk about the Floyd-Warshall algorithm. We will talk about Dijkstra. We will talk about the single source shortest path, and all source shortest path. It will be very interesting.
For now, I will leave you with a small exercise. Can you write a small function to show the contents of the graph we have created? We need to be sure the graph is actually correct before we can move on. You need to write a function that displays all of the nodes, and the edges between them.
How can this be done? It is pretty simple. Basically you need to pass over all linked lists, and show whatever data you find inside them; this can help you imagine what the graph looks like. It may be hard to follow because it is not graphical, but it gets the job done.
Don’t forget, if you have any questions, comments or suggestion, you can always contact me at msaad@Themagicseal.com. I am always glad to hear from you.
See you in part two!
DISCLAIMER: The content provided in this article is not warranted or guaranteed by Developer Shed, Inc. The content provided is intended for entertainment and/or educational purposes in order to introduce to the reader key ideas, concepts, and/or product reviews. As such it is incumbent upon the reader to employ real-world tactics for security and implementation of best practices. We are not liable for any negative consequences that may result from implementing any information covered in our articles or tutorials. If this is a hardware review, it is not recommended to open and/or modify your hardware. |
# Evaluate the limit of the indeterminate quotient?
Feb 8, 2017
${\lim}_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = - \frac{1}{\sqrt{7}}$
#### Explanation:
Evaluate the limit:
${\lim}_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x}$
Rationalize the numerator of the function using the identity: $\left(a + b\right) \left(a - b\right) = \left({a}^{2} - {b}^{2}\right)$:
$\frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = \left(\frac{\sqrt{7 - x} - \sqrt{7 + x}}{x}\right) \left(\frac{\sqrt{7 - x} + \sqrt{7 + x}}{\sqrt{7 - x} + \sqrt{7 + x}}\right) = \frac{7 - x - 7 - x}{x \left(\sqrt{7 - x} + \sqrt{7 + x}\right)} = - \frac{2 x}{x \left(\sqrt{7 - x} + \sqrt{7 + x}\right)} = - \frac{2}{\sqrt{7 - x} + \sqrt{7 + x}}$
Now the limit is determinate:
${\lim}_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = {\lim}_{x \to 0} - \frac{2}{\sqrt{7 - x} + \sqrt{7 + x}} = - \frac{1}{\sqrt{7}}$
Feb 9, 2017
$- \frac{1}{\sqrt{7}}$
#### Explanation:
Here is another way of solving it. Slightly messy.
L'hopital's Rule:
${\lim}_{x \to 0} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to 0} \frac{{f}^{'} \left(x\right)}{{g}^{'} \left(x\right)}$ only if the expression is indeterminate .
${f}^{'} \left(x\right)$ is simply the derivative of $f \left(x\right)$ with respect to $x$.
Now via L'hopital's Rule (differentiate the numerator and the denominator separately), ${\lim}_{x \to 0} = \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = {\lim}_{x \to 0} \frac{- \frac{1}{2} {\left(7 - x\right)}^{- \frac{1}{2}} - \frac{1}{2} {\left(7 + x\right)}^{- \frac{1}{2}}}{1} = - \frac{1}{2} \left(\frac{1}{\sqrt{7 - x}} + \frac{1}{\sqrt{7 + x}}\right)$
Let $x = 0$ and you get $- \frac{1}{\sqrt{7}}$ |
The Lesson
Expanding brackets means multiplying terms to remove brackets from an expression. Imagine we wanted to remove the brackets from the expression below. We have to expand the brackets.
How to Expand Brackets
Expanding brackets is easy. Multiply each term inside the brackets with the term outside of the bracket.
Question
Expand the brackets below.
1
Multiply the first term inside the brackets (x) with the term outside the brackets (2).
2 × x = 2x 2x
2
Multiply the second term inside the brackets (1) with the term outside the brackets (2), and add to the previous result.
2 × 1 = 2 2x + 2
Answer:
Expanding 2(x + 1) gives 2x + 2.
A Formula to Expand Brackets
The formula below shows how to expand brackets:
a, b and c stand in for any number or term.
Lesson Slides
The slider below shows another real example of how to expand brackets. In this example, there are more than two terms inside the brackets. Open the slider in a new tab
Multiplication Tips
• Multiplying a number with a variable
2 × x = 2x 5 × x = 5x
• Multiplying a number with a variable with a coefficient
2 × 2x = 4x 3 × 5x = 15x
• Multiplying a variable with a variable
x × y = xy x × x = x2
Be Careful with Signs
Each term in the brackets will have a sign.
• positive terms have a + sign in front of them, or no sign if it is the first term.
• negative terms have a sign in front of them.
These signs must be included when multiplying terms.
Remember the rules for multiplying different signs: Same signs give a plus:
Different signs give a minus: |
TEXTBOOKS FOR THE AMC 8
Thousands of top scorers on the AMC 8 have used our Introduction series of textbooks, Art of Problem Solving Volume 1, and Competition Math for Middle School to prepare for the AMC 8.
# Difference between revisions of "2019 AMC 8 Problems"
## Problem 1
Ike and Mike go into a sandwich shop with a total of $30.00$ to spend. Sandwiches cost $4.50$ each and soft drinks cost $1.00$ each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$
## Problem 2
Three identical rectangles are put together to form rectangle $ABCD$, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is $5$ feet, what is the area in square feet of rectangle $ABCD$?
$[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]$
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$
## Problem 3
Which of the following is the correct order of the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$, from least to greatest?
$\textbf{(A) }\frac{15}{11} < \frac{17}{13} < \frac{19}{15}\qquad\textbf{(B) }\frac{15}{11} < \frac{19}{15} < \frac{17}{13}\qquad\textbf{(C) }\frac{17}{13} < \frac{19}{15} < \frac{15}{11}\qquad\textbf{(D) }\frac{19}{15} < \frac{15}{11} < \frac{17}{13}\qquad\textbf{(E) }\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
## Problem 4
Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$?
$[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]$
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$
## Problem 5
A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance $d$ traveled by the two animals over time $t$ from start to finish?
## Problem 6
There are $81$ grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point $P$ is in the center of the square. Given that point $Q$ is randomly chosen among the other $80$ points, what is the probability that the line $PQ$ is a line of symmetry for the square?
$[asy] draw((0,0)--(0,8)); draw((0,8)--(8,8)); draw((8,8)--(8,0)); draw((8,0)--(0,0)); dot((0,0)); dot((0,1)); dot((0,2)); dot((0,3)); dot((0,4)); dot((0,5)); dot((0,6)); dot((0,7)); dot((0,8)); dot((1,0)); dot((1,1)); dot((1,2)); dot((1,3)); dot((1,4)); dot((1,5)); dot((1,6)); dot((1,7)); dot((1,8)); dot((2,0)); dot((2,1)); dot((2,2)); dot((2,3)); dot((2,4)); dot((2,5)); dot((2,6)); dot((2,7)); dot((2,8)); dot((3,0)); dot((3,1)); dot((3,2)); dot((3,3)); dot((3,4)); dot((3,5)); dot((3,6)); dot((3,7)); dot((3,8)); dot((4,0)); dot((4,1)); dot((4,2)); dot((4,3)); dot((4,4)); dot((4,5)); dot((4,6)); dot((4,7)); dot((4,8)); dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label("P",(4,4),NE); [/asy]$
$\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}$
## Problem 7
Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$, $94$, and $87$. In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$
## Problem 8
Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$
## Problem 9
Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?
$\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1$
## Problem 10
The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually $21$ participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made? $[asy] unitsize(2mm); defaultpen(fontsize(8bp)); real d = 5; real t = 0.7; real r; int[] num = {20,26,16,22,16}; string[] days = {"Monday","Tuesday","Wednesday","Thursday","Friday"}; for (int i=0; i<30; i=i+2) { draw((i,0)--(i,-5*d),gray); }for (int i=0; i<5; ++i) { r = -1*(i+0.5)*d; fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray); label(days[i],(-1,r),W); }for(int i=0; i<32; i=i+4) { label(string(i),(i,1)); }label("Number of students at soccer practice",(14,3.5)); [/asy]$
$\textbf{(A) }$ The mean increases by $1$ and the median does not change.
$\textbf{(B) }$ The mean increases by $1$ and the median increases by $1$.
$\textbf{(C) }$ The mean increases by $1$ and the median increases by $5$.
$\textbf{(D) }$ The mean increases by $5$ and the median increases by $1$.
$\textbf{(E) }$ The mean increases by $5$ and the median increases by $5$.
## Problem 11
The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eight graders taking a foreign language class. How many eight graders take only a math class and not a foreign language class?
$\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$
## Problem 12
The faces of a cube are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face?
$\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}$
## Problem 13
A palindrome is a number that has the same value when read from left to right or from right to left. (For example 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$?
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$
## Problem 14
Isabella has $6$ coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every $10$ days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the $6$ dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?
$\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}$
## Problem 15
On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is $\frac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?
$\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$
## Problem 16
Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$
## Problem 17
What is the value of the product $$\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?$$
$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50$
## Problem 18
The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number?
$\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}$
## Problem 19
In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$
## Problem 20
How many different real numbers $x$ satisfy the equation $$(x^{2}-5)^{2}=16?$$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$
## Problem 21
What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$?
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$
## Problem 22
A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased?
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$
## Problem 23
After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$
## Problem 24
In triangle $ABC$, point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $BC$ and line $AE$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?
$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("A",A,N); dot(B); label("B", B,SW);dot(C); label("C",C,SE); dot(DD); label("D",DD,NE); dot(EE); label("E",EE,NW); dot(FF); label("F",FF,S); [/asy]$
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$
## Problem 25
Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the people has at least $2$ apples?
$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$ |
# NCERT Solutions For Class 9 Maths Chapter 1 – Number System
Download NCERT Solutions For Class 9 Maths Chapter 1 Number System – Exercise 1.1, 1.2, 1.3, 1.4, 1.5, 1.6
## NCERT Solutions For Class 9 Maths Chapter 1 – Exercise 1.1
1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers
and q ≠ 0?
Solution 1
Yes , 0 is a rational number because it can be written in the form p/q = 0/1, p= 0 and q = 1 i.e. where p, q are integers and q ≠ 0 .
2. Find six rational numbers between 3 and 4.
Solution 2
To find six rational numbers between 3 and 4 follow the steps below:
Step (I) Write 3 and 4 as rational numbers with denominator 6+1 = 7
Thus, 3 = 3/1
= (3 x 7)/(1 x 7)
= 21/7
Similarly, 4 = 4/1
= (4 x 7)/(1 x 7)
= 28/7
Step (II) Write the numbers between the given numbers as-
22/7, 23/7, 24/7, 25/7, 26/7, 27/7
Since all have same denominator , we can easily observe that all of these numbers are in between 21/7 and 28/7 i.e. 3 and 4 respectively.
3. Find five rational numbers between 3/5 and 4/5.
Solution 3
To find five rational numbers between 3/5 and 4/5 follow the steps below:
Step (I) Write 3/5 and 4/5 as rational numbers with denominator 5 x (5+1) = 30
Thus 3/5 = (3 x 6)/(5 x 6) = 18/30
Similarly, 4/5 = (4 x 6)/(5 x 6) = 24/30
Step (II) Write the numbers between the given numbers as-
19/30, 20/30, 21/30, 22/30, 23/30
or , 19/30, 2/3, 7/30, 11/15, 23/30 in their simplest forms respectively.
4. State whether the following statements are true or false. Give reasons for your answers.
(i)Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Solution 4
(i) True , because Whole number is the collection of Natural numbers including Zero.
(ii) False , -1 or any negative integer is not a whole number.
(iii) False , because not all the numbers which can be expressed in the form p/q, where p and q are integers and q≠0 , are necessarily Whole numbers. Example 3/5 is a rational number but not a whole number.
## NCERT Solutions For Class 9 Maths Chapter 1 – Exercise 1.2
1.State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form √m , where m is a natural number.
(iii) Every real number is an irrational number.
Solution 1
(i) True , because real number is the collection of rational and irrational numbers.
(ii) False , for example 1/2 is not of √m form with m as a natural number and also since square root of a natural number is considered to be positive so we can not represent negative numbers in this form.
(iii) False , because every rational number is a real number but not irrational number.
2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution 2
No, For example consider the integer 9
√9 = 3 which is a rational number
3. Show how √5 can be represented on the number line.
Solution:
Follow the steps below to represent √5 on the number line as show in the figure above:
Step (I) : Consider a right angled triangle OAB on the number line as shown in figure with
OA = 2 units
AB = 1 unit.
Then by using Pythagoras theorem we can easily find
OB = √(2² + 1²) = √5
Step (II) : Using a compass , draw an arc with centre O and radius OB intersecting the number line at C. Then C corresponds to √5 on the number line.
4. Classroom activity
## NCERT Solutions For Class 9 Maths Chapter 1 – Exercise 1.3
1. Write the following in decimal form and say what kind of decimal expansion each has :
(i) 36/100
(ii) 1/11
(iii) 4 1⁄8
(iv) 3/13
(v) 2/11
(vi) 329/400
Solution:
(i) 36/100 = 0.36 , it is terminating decimal
We can check by performing long division as follows:
(v) 2/11 = 2 x 1/11
(vi) 329/400 = 0.8225 , it is a terminating decimal.
(2). You know that 1/7 = 0.142857 . Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?
[Hint : Study the remainders while finding the value of 1/7 carefully.]
Solution:
While performing long division for 1/7.
We can easily observe that each one from the numerator of the given fractions appear in the remainder of the above operation which has to be carried out further by dividing them by 7 . And so just by using this observation we can predict the required decimal expansions as follows:
(3). Express the following in the form p/q, where p and q are integers and q ≠ 0.
Solution:
(i) Follow the steps below :
Step (II) Since there is only one repeating block , multiply X by 10
We get
Step (IV) Using equation (1) and (2) , equation (3) can be written as
10X = 6 + X ….(4)
Now,
Solving for X from equation (4) , we get
9X = 6
Therefore,
X = 6/9 = 2/3
Hence we obtain
(ii) Follow the steps below :
Step (II) Since there is only one repeating block , multiply X by 10
We get
Step (IV) Using equation (1) and (2) , equation (3) can be written as
10X = 4.3 + X ….(4)
Now,
Solving for X from equation (4) , we get
9X = 4.3
Therefore,
X = (4.3/9) = (43/90)
Hence we obtain
(iii) Follow the steps below :
Step (II) Since there are three repeating block , multiply X by 1000
We get
Step (IV) Using equation (1) and (2) , equation (3) can be written as
1000X = 1 + X ….(4)
Now,
Solving for X from equation (4) , we get
999X = 1
Therefore,
X = (1/999)
Hence we obtain
(4). Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
(i) Follow the steps below to write 0.9999…. in the form p/q:
Step (II) Since there is only one repeating block , multiply X by 10
We get
Step (IV) Using equation (1) and (2) , equation (3) can be written as
10X = 9 + X ….(4)
Now,
Solving for X from equation (4) , we get
9X = 9
Therefore,
X = 9/9 = 1
Hence we obtain
(5). What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.
Solution:
The maximum number of digits in the repeating block must be less than the denominator i.e. 17. So the mximum number of digits in the repeating block in this case is 16.
Perform the long division for 1/17 as follows:
Thus we obtained
(6). Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
We can easily observe that in such rational numbers q , apart from 1 and itself ; is either divided by 2 or 5 or both.
So we can say that the prime factorization of q must have only powers of 2 or 5 or both apart from 1 and the number itself.
(7). Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
There are numerous number of such examples , three of them are as follows:
0.1011011101111011111……..
0.0100100010000100000……..
0.2122122212222122222……..
(8). Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Solution:
We can calculate
Out of infinite possibilities three are as follows :
0.72722722272222……
0.7277277727777……..
0.80800800080000……
(9). Classify the following numbers as rational or irrational :
(i) √23
(ii) √225
(iii) 0.3796
(iv) 7.478478…
(v) 1.101001000100001…
Solution:
(i) √23 : Irrational
(ii) √225 = 15 : Rational
(iii)0.3796 : Rational
(iv)7.478478… : Rational
(v)1.101001000100001… : Irrational.
## NCERT Solutions For Class 9 Maths Chapter 1 – Exercise 1.4
NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.4 – Number System
1. Visualise 3.765 on the number line, using successive magnification.
Solution:
Follow the steps below to visualise 3.765 on the number line :
Step (I) We know that 3.765 lies between 3 and 4 . So , first we will divide this portion of number line into 10 parts
Step (II) We know that 3.765 lies between 3.7 and 3.8 . So now , we will divide this portion of number line into 10 parts.
Step (III) We know that 3.765 lies between 3.76 and 3.77 . So now , we will divide this portion of number line into 10 parts.
Thus we obtained 3.765 on the number line, using successive magnification.
Follow the steps below to visualise 4.2626 on the number line :
Step (I) We know that 4.2626 lies between 4 and 5 . So , first we will divide this portion of number line into 10 parts.
Step (II) We know that 4.2626 lies between 4.2 and 4.3 . So now , we will divide this portion of number line into 10 parts.
Step (III) We know that 4.2626 lies between 4.26 and 4.27 . So now , we will divide this portion of number line into 10 parts.
Step (IV) We know that 4.2626 lies between 4.262 and 4.263 . So now , we will divide this portion of number line into 10 parts.
## NCERT Solutions For Class 9 Maths Chapter 1 – Exercise 1.5
1. Classify the following numbers as rational or irrational:
(i) 2 – √5
(ii) ( 3 + √23 ) – √23
(iii) (2√7/7√7)
(iv) 1/√2
(v) 2π
Solution:
(i) 2 – √5 : It is irrational.
(ii) ( 3 + √23 ) – √23 = 3 + √23 – √23 = 3 : It is rational.
(iii) (2√7/7√7) = 2/7 : It is rational
(iv) 1/√2 = : It is irrational.
(v) 2π : It is irrational.
(2). Simplify each of the following expressions:
(i)( 3 + √3 )( 2 + √2 )
(ii)( 3 + √3 )( 3 – √3 )
(iii)(√5+√2)²
(iv)( √5 – √2 )( √5 + √2 )
Solution:
(i) ( 3 + √3 )( 2 + √2 )
( 3 + √3 )( 2 + √2 )
= ( 3 + √3 )2 +( 3 + √3 )√2
= 6 + 2√3 + 3√2 + √6
(ii)( 3 + √3 )( 3 – √3 )
Using the identity
( a + b )( a – b) = a2 – b2
We get
( 3 + √3 )( 3 – √3 ) = 3² – √32 = 9 – 3 = 6
(iii) (√5+√2)²
Using the identity
(a + b)² = a2 + 2ab + b2
We get
(√5+√2)² = 5 + 2 √10 + 2 = 7 + 2 √10
(iv)( √5 – √2 )( √5 + √2 )
Using the identity
( a – b )( a + b) = a2 – b2
We get
( √5 – √2 )( √5 + √2 ) = 5 – 2 = 3
(3) Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution:
As we measure c and d using any physical device we do not get the exact value manually , instead we get an approximate rational value and this is why this contradiction arises.
4. Represent √9.3 on the number line.
Solution:
Follow the steps below to locate √9.3 on the number line as shown above:
Step (I) Draw a line segment of length 9.3 units EA and extend it to C by 1 unit .
Step (II) Now with O as mid point of EC draw a semicircle with O as center and OC as radius.
Step (III) Draw a line perpendicular to EC at A and intersecting the semicircle at B.
Step (IV) Taking A as center and AB as radius draw an arc from B to intersect the extended line EAC at D.
If we assign A as origin i.e. 0
Then D represents √9.3 on the number line as shown in the figure.
## NCERT Solutions For Class 9 Maths Chapter 1 – Exercise 1.6
(1). Find :
(i)641/2
(ii) 321/5
(iii)1251/3
Solution:
(i)641/2 = (82)1/2 = 8
(ii) 321/5 = (25)1/5 = 2
(iii)1251/3= (53)1/3 = 5
(2). Find :
(i)93/2
(ii)322/5
(iii)163/4
(iv) 125-1/3
Solution:
(i)93/2 = 27
(ii)322/5 = 4
(iii)163/4 = 8
(iv) 125-1/3 = 1/5
(3). Simplify :
(i)22/3.21/5
(ii)(1/33)7
(iii)(111/2)/(111/4)
(iv)71/2.81/2
Solution:
(i)22/3.21/5
= 2{(2/3)+(1/5)}
=2{(10 + 3)/15}
= 213/15
(ii)(1/33)7
=1/321
= 3-21
(iii)(111/2)/(111/4)
= 11{(1/2) – (1/4)}
= 11{(2-1)/4}
= 111/4
(iv)71/2.81/2
=(7×8)1/2
= (56)1/2
Download NCERT Solutions For Class 9 Maths Chapter 1 Number System – Exercise 1.1, 1.2, 1.3, 1.4, 1.5, 1.6 |
#### Need Solution for R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 41 sub question 2 Maths Textbook Solution.
Answer:$(a-1)^{3}$
Hint Use determinant formula
Given: $\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^{3}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$
\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \& C_{2} \rightarrow C_{2}-C_{3}\\ &=\left|\begin{array}{ccc} a^{2}-1 & 2 a & 1 \\ a-1 & a+1 & 1 \\ 0 & 2 & 1 \end{array}\right|\\ &(a-1) \text { common from } \mathrm{C}_{1}\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 2 a & 1 \\ 1 & a+1 & 1 \\ 0 & 2 & 1 \end{array}\right| \end{aligned}
\begin{aligned} &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \& R_{2} \rightarrow R_{2}-R_{3}\\ &=(a-1)\left|\begin{array}{ccc} a & (a-1) & 0 \\ 1 & a-1 & 0 \\ 0 & 2 & 1 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{3}\\ &=(a-1)\left(1\left(a^{2}-a-a+1\right)\right)\\ &=(a-1)\left(a^{2}-2 a+1\right)\\ &=(a-1)(a-1)^{2}\\ &=(a-1)^{3}\\ &=R . H \cdot S \end{aligned} |
# Question Video: Using Trigonometric Ratios to Find Two Missing Lengths of a Right-Angled Triangle Mathematics
Find the values of π₯ and π¦ giving the answer to three decimal places.
04:29
### Video Transcript
Find the values of π₯ and π¦ giving the answer to three decimal places.
In this question, we need to determine the values of two unknowns π₯ and π¦, and we need to give our answer to three decimal places. To answer this question, we first notice that the values of π₯ and π¦ represent side lengths in a right triangle. And in this right triangle, weβre given one of the non-right angles and one of the other side lengths. Therefore, we can determine the values of π₯ and π¦ by using right triangle trigonometry.
To do this, we first need to label the size of this right triangle. We should always start with the hypotenuse. Itβs the longest side in the right triangle, which is the one opposite the right angle. So, in this case, the hypotenuse has length π¦. Next, we need to label the other two sides based on their position relative to the angle that we know. Thatβs 47 degrees. First, we can notice the side of length 28 centimeters is opposite the angle of 47 degrees. So, weβll label this side as the opposite side. Finally, the last side of length π₯ is next to our angle of 47 degrees, so we could say that itβs adjacent to this angle. Weβll label this side the adjacent side.
We can now recall that we can determine the values of π₯ and π¦ by using the trigonometric ratios. To help us determine which trigonometric ratios we need to use, weβll recall the following acronym: SOH CAH TOA. Weβll need to apply this twice to determine the values of π₯ and π¦ separately. Letβs start by determining the value of π₯. This means we need to start by using our acronym to determine which trigonometric ratio will help us find the value of π₯.
We can see that π₯ is the adjacent side to our angle. And we already know the value of the opposite side to our angle. And we can see in the acronym if we know the opposite side to the angle and the adjacent side to the angle, then we need to use the tangent function. We know if π is an angle in a right triangle, then the tangent of angle π tells us the ratio of the length of the opposite side to angle π divided by the length of the adjacent side to angle π. In our right triangle, we want to use the angle π is 47 degrees. Then the adjacent side has length π₯, and the opposite side has length 28 centimeters.
Substituting these values in, we get the tangent of 47 degrees is equal to 28 divided by π₯. We need to solve this equation for π₯. First, weβll multiply both sides of our equation through by π₯. This then gives us that π₯ times the tangent of 47 degrees is equal to 28. And now, we can solve for π₯ by dividing both sides of the equation through by tangent of 47 degrees. We get that π₯ is equal to 28 divided by the tan of 47 degrees. And now, by remembering the side lengths of this triangle are measured in centimeters and by using our calculator, where we make sure our calculator is set to degrees mode, we can find the value of π₯. Itβs 26.1104 and this expansion continues centimeters.
Finally, the question wants us to give our values of π₯ and π¦ to three decimal places. So, we look at the fourth decimal place, which is four, which we know is less than five. So, we need to round this down. Therefore, to three decimal places, π₯ is 26.110 centimeters.
We can follow the exact same process to find the value of π¦. Once again, weβll use our acronym of SOH CAH TOA to determine which trigonometric ratio we need to use. In the diagram, we know the length of the opposite side, and we want to determine π¦, which is the length of the hypotenuse. Therefore, we want the trigonometric ratio relating the opposite side and the hypotenuse. Thatβs the sine function. We can then recall if π is an angle in a right triangle, then sin π is equal to the length of the side opposite angle π divided by the length of the hypotenuse.
We can then substitute in our values from the diagram. We get the sin of 47 degrees is equal to 28 divided by π¦. Now, all we need to do is solve this equation for π¦. We multiply both sides of the equation through by π¦ and then divide through by the sin of 47 degrees. We get π¦ is equal to 28 divided by sin of 47 degrees. Now, we evaluate this expression by using our calculator. We get that π¦ is 38.2851 and this expansion continues centimeters. Finally, we need to round this to three decimal places. The fourth decimal digit is one, which is less than five. So, we need to round this value down. This gives us that π¦ is 38.285 centimeters to three decimal places.
Therefore, by using right triangle trigonometry, we were able to find the values of π₯ and π¦ in the diagram to three decimal places. We saw that π₯ was 26.110 centimeters and π¦ was 38.285 centimeters. |
One Stop Shop for Math Teacher Resources
# The Greatest Common Factor
The greatest common factor, or GCF, of two or more integers (...,-2, -1, 0, 1, 2, ...) is the largest natural number (1, 2, 3, ...) that will divide evenly into all of the integers a natural number of times.
It is important to note the word common embedded in the term greatest common factor. This implies, of course, that the two numbers have something in common. This concept can best be seen by making factor trees. Make factor trees for the two numbers in number four above.
It would probably not be an efficient use of time to make factor trees for every GCF problem. Students should use their knowledge of multiplication and division, as well as the calculator, in order to complete the problems in a timely fashion.
–Strategy 1: Find the prime factorization of each number – Take whatever they have in common (to the highest power possible)
• Find the GCF(42, 385) Factorization 42=2⋅21=2⋅3⋅7
Factorization 385=5⋅77=5⋅7⋅11
GCF(42, 385) = 7^1 = 7
• Find the GCF(338, 507)Factorization 338 = 2⋅169 = 2⋅13^2
Factorization 507 = 3⋅169 = 3⋅13^2
GCF(338, 507) = 13^2 = 169
Find the GCF (greatest common factor) of the following numbers using a Factor Tree:
Strategy 2: list all of the common factors of 2 numbers and take the largest.
Ex GCF of 36 and 84
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
GCF (36, 84) = 12
The Venn Factor: In this lesson, students use a Venn diagram to sort prime factors of two or more positive integers. Students calculate the greatest common factor by multiplying common prime factors and develop a definition based on their exploration.
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### Figure This Challenge #56
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# Math Limerick
Question: Why is this a mathematical limerick?
( (12 + 144 + 20 + 3 Sqrt[4]) / 7 ) + 5*11 = 92 + 0 .
A dozen, a gross, and a score,
plus three times the square root of four, divided by seven, plus five times eleven, is nine squared and not a bit more.
---Jon Saxton (math textbook author)
Presentation Suggestions:
Challenge students to invent their own math limerick!
The Math Behind the Fact:
It is fun to mix mathematics with poetry.
Resources:
Su, Francis E., et al. "Math Limerick." Math Fun Facts.
funfacts
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# Analysis of Rational Functions
## Finding restrictions, vertical asymptotes, breaks, end behavior, and intercepts.
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Analysis of Rational Functions
Consider the following situation:
Students in your school are currently required to demonstrate their understanding of their classwork at a level of 75% or higher in order to move on to more advanced material.
The amount of homework time T required for each student based on understanding level of p% is given by: \begin{align*}T = \frac{(18p)}{(100-p)}\end{align*}
If the school administration decides to raise the minimum level of understanding to 82%, how will this affect the students' homework time?
This is an example of a rational function in a real-world situation.
### Analyzing Rational Functions
#### Finding Vertical Asymptotes and Breaks
Recall that rational functions are defined as \begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*} where \begin{align*}p(x)\end{align*} and \begin{align*}q(x)\end{align*} are polynomials.
To find vertical asymptotes and breaks in the domain of a rational function, set the denominator equal to zero and solve for \begin{align*}x\end{align*}. Given \begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*}, set \begin{align*}q(x)=0\end{align*} and solve for \begin{align*}x\end{align*}.
Note that some rational functions have vertical asymptotes and others do not. Some rational functions have a break in the function, but no vertical asymptote. This usually happens when one term in the numerator cancels with one term in the denominator.
#### Evaluating End Behavior
The end behavior of a rational function can often be identified by the horizontal asymptote. That is, as the values of \begin{align*}x\end{align*} get very large or very small, the graph of the rational function will approach (but not reach) the horizontal asymptote.
### Examples
#### Example 1
Earlier, you were asked a question about students' homework time.
The number of minutes of homework time T required for each student based on understanding level of p'% is given by: \begin{align*}T = (18p)/(100-p)\end{align*} If the school administration decides to raise the minimum level of understanding to 82%, how will this affect the students' homework time?'
The current time required is: \begin{align*}(18(75))/(100-75)\end{align*} ==> \begin{align*}54min\end{align*}
By substituting in the increased level: \begin{align*}(18(82))/(100-82)\end{align*} ==> \begin{align*}82min\end{align*}
The increase in required proficiency would result in an average of 28mins of added study time per student.
#### Example 2
Consider the following rational function. Find all restrictions on the domain and asymptotes.
\begin{align*}f(x)=\frac{x^{2}+2x-35}{x-5}\end{align*}
Factoring the numerator
\begin{align*}f(x)&=\frac{(x-5)(x+7)}{x+7}\\ f(x)& =\frac{(x-5)\cancel{(x+7)}}{\cancel{x+7}}\end{align*}
Canceling
\begin{align*}f(x)=x+7, x\neq5\end{align*}
Notice that there is no asymptote in this function, but rather a break in the graph at \begin{align*}x=5\end{align*}.
#### Example 3
Find the restrictions on the domain of
\begin{align*}h(x)=\frac{3x}{x^{2}-25}\end{align*}
Setting the denominator equal to 0,
\begin{align*}x^{2}-25 & = 0\\ x^2 & = 25\\ x & = \pm 5\end{align*}
Thus, the domain of \begin{align*}h(x)\end{align*} is the set all real numbers \begin{align*}x\end{align*} with the restriction \begin{align*}x\neq\pm 5\end{align*}. \begin{align*}h(x)\end{align*} has two vertical asymptotes, one at \begin{align*}x=5\end{align*} and one at \begin{align*}x=-5\end{align*}.
Note for graphing using technology problems: If you do not have a graphing calculator, there are a number of excellent free apps, and an excellent free online calculator here: Desmos Online Graphing Calculator.
#### Example 4
(Graphing calculator exercise)
Graph \begin{align*}f(x)=\frac{1}{x-3}\end{align*} on the window [-10,10] X [-10,10]. (This means {XMIN}=-10, {XMAX}=10, {YMIN}=-10, and {YMAX}=10).
Notice:
1. When graphing a rational function by entering the function in the \begin{align*}Y=\end{align*} screen, remember that you need to use parenthesis to group the numerator and denominator of the rational function.
2. Vertical asymptotes are sometimes graphed as vertical lines.
3. Graphs of rational functions can be difficult to interpret if the window settings are not chosen carefully.
\begin{align*}y=\frac{1}{x-3}\end{align*} showing vertical line at \begin{align*}x=3\end{align*}
\begin{align*}f(x)\end{align*} is undefined and has a vertical asymptote at \begin{align*}x=3\end{align*}, but the way the graphing calculator draws the graph, it shows a vertical line at \begin{align*}x=3\end{align*}. One way to “fix” this problem is to press MODE and select the option “Dot” rather than “Connected”. However, dot graphs can be hard to interpret as well.
#### Example 5
Solve the equation and find any points of discontinuity using technology: \begin{align*}f(x) = \frac{6}{x-2} + \frac{4}{x+7}\end{align*}.
Input the function into the grapher, don't forget to use "(" ")" to group polynomials. Your function should look something like this on your input screen (at least until you tell the calc to process the function): 6/(x - 2) + 4/(x + 7). Note that some calculators require the " y = " or " f(x) = " and some do not.
Once you are sure you have the information entered correctly, press "calculate", or the equivalent. The graph should look like:
There are asymptotes at \begin{align*}x = 2\end{align*} and \begin{align*}x = 7\end{align*} and \begin{align*}y = 0\end{align*}.
#### Example 6
Using technology, find all intercepts, asymptotes, and discontinuities, and graph: \begin{align*}\frac{2x^3 + 14x^2 - 9x + 6}{x + 3}\end{align*}.
Input the function with care, it should look something like: \begin{align*}(2x^3 + 14x^2 - 9x + 6) / (x + 3)\end{align*} before you press the calc or view button.
The graph should look like:
The intercepts are at (apx) \begin{align*}(-7.64, 0)\end{align*} and (exactly) \begin{align*}(0, 2)\end{align*}.
There is a vertical asymptote at \begin{align*}x = 3\end{align*} and a curved asymptote described by: \begin{align*}y = 2x^2 +8x -33\end{align*}.
#### Example 7
Using technology, simplify, find discontinuities and limitations, then graph: \begin{align*}\frac{6x^2 + 21x +9}{4x^2 -1}\end{align*}.
Input the function accurately. The graph should look like:
The function simplifies to: \begin{align*}\frac{3(2x^2 + 7x +3)}{4(x^2-\frac{1}{4})} \to \frac{3(2x + 1)(x + 3)}{4(x - \frac{1}{2})(x + \frac{1}{2})}\end{align*}
There are asymptotes at \begin{align*}x = \frac{1}{2}\end{align*} and at \begin{align*}y = \frac{1}{2}\end{align*}.
There is a hole at \begin{align*}y = -3\frac{3}{4}\end{align*}.
### Review
For questions 1 - 5, factor the numerator and denominator, then set the denominator equal to zero and solve to find restrictions on the domain.
1. \begin{align*}y = \frac{x^2 + 3x - 10}{x - 2}\end{align*}
2. \begin{align*}f(x) = \frac{x^2 + 2x -24}{x - 4}\end{align*}
3. \begin{align*}f(x) = \frac{x^2 - 12x +32}{x - 4}\end{align*}
4. \begin{align*}y = \frac{x^2 +\frac{21}{20}}{x + \frac{4}{5}}\end{align*}
5. \begin{align*}y = \frac{x^2 +13x + 42}{x +7}\end{align*}
For questions 6 - 15, input the function into your graphing tool carefully and accurately. Record any asymptotes or holes and record x and y intercepts. Finally either copy and print or sketch the image of the graphed function.
1. \begin{align*}y = \frac{x^3+5x^2+3x+7}{x - 1}\end{align*}
2. \begin{align*}y = \frac{9x^2 + 6}{x}\end{align*}
3. \begin{align*}f(x) = \frac{-8x^3 - 8x^2 + 2x + 8}{x + 2}\end{align*}
4. \begin{align*}f(x) = \frac{5x^3 - 9x^2 - 7x + 1}{x^2 - 4}\end{align*}
5. \begin{align*}y = \frac{(-2x^3 + 2x^2 + 5x + 2}{(x-2)(x+7)}\end{align*}
6. \begin{align*}y = \frac{4x^3 + 2x^2 +7}{(x + 2)^2}\end{align*}
7. \begin{align*}f(x) = \frac{7x^3+2x^2-7x-3}{x^3}\end{align*}
8. \begin{align*}f(x) = \frac{-6x^3+8x^2+7}{x^2}\end{align*}
9. \begin{align*}y = \frac{-5x^2-2x-5}{x^2+2}\end{align*}
10. \begin{align*}y = \frac{x - 1}{x^3 - 2}\end{align*}
To see the Review answers, open this PDF file and look for section 2.8.
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### Vocabulary Language: English
Break
A break exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero. Graphing calculators will not show breaks in their displays.
End behavior
End behavior is a description of the trend of a function as input values become very large or very small, represented as the 'ends' of a graphed function.
Hole
A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero.
Rational Function
A rational function is any function that can be written as the ratio of two polynomial functions.
Vertical Asymptote
A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach. |
verifying-trigonometric-identities-for-dummies-proving-trigonometric-identities-basics
# Interactive video lesson plan for: Verifying Trigonometric Identities - For Dummies | Proving Trigonometric Identities basics
#### Activity overview:
Proving Trigonometric Identities - For Dummies | Verifying Trigonometric Identities basics
http://www.learncbse.in/ncert-class-10-math-solutions/
http://www.learncbse.in/ncert-solutions-for-class-10-maths-chapter-8-introduction-to-trigonometry/
http://www.learncbse.in/rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios/
In this math tutorial we prove trigonometric identities, solve trigonometric equations and Verify Trigonometric Identities.We use Quotient Identities , Reciprocal Identities, and the Pythagorian Identities to solve problems on trigonometric identities.
Solve Trigonometric Equations with Trigonometric Identities
Defination of an Identity:
An equation is called an identity when it is true for all values of the variables involved.
Trigonometric Identities defination:
An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.
00:02 Q5. Prove the trigonometric identities, where the angles involved are acute angles for which the expressions are defined
00:04 q5 (i)
00:18 PROCIDURE FOR SOLVING Trigonometric equations
00:30 consider LHS of the trigonometric equations
00:40 Reciprocal Identity.
The ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A
Quotient Identity
cot A = cos A / sin A.
01:15 Use Pythagorian Identity
sin squared theta plus cos squared theta = 1
02:39 trigonometric equation is proved
03:00 CBSE solutions for class 10 maths Chapter 8 Trigonometry Exercise 8.4 Q5. Prove the trigonometric equations using trigonometric identities, where the angles involved are acute angles for which the expressions are defined
03:03 CBSE solutions for class 10 maths Chapter 8 Trigonometry Exercise 8.4 q5 (2)
03:19 Procedure for proving Trigonometric equations
03:41 consider LHS of the trigonometric equation
04:20 algebra maths formulas | Algebraic Expressions and Formulas
USE (a-b)^2 expanding formula
04:58 Use Pythagorian Identity
sin squared theta plus cos squared theta = 1
05:59 Reciprocal Identity.
The ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A
sec A = 1/cos A
06:13 How to prove Trigonometric identities.
Understanding Trig Identities
Proving Trigonometric Identities
Using the fundemental identities and the Pythagorean Identities
proving trigonometric identities
trigonometric identities questions
trigonometric identities solver
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CBSE solutions for class 10 maths Chapter 8 Trigonometry Exercise 8.4
CBSE class 10 maths NCERT SolutionsChapter 8 Trigonometry Exercise 8.2 | Trigonometric Identities
CBSE class 10 maths NCERT Solutions Chapter 8 Trigonometry
Solutions for CBSE class 10 Maths Trigonometry
NCERT solutions for class 10 maths Trigonometry
NCERT solutions for CBSE class 10 maths Trigonometry
CBSE class 10 maths Trigonometry
Common core Math Trigonometry
Tagged under: ncert solutions,learncbse.,gyanpub,Verifying,Common core Math,Trigonometry,Trigonometric Identities,For Dummies,Proving Trigonometric,CBSE class 10 maths,NCERT solutions CBSE class 10 maths,Reciprocal Identity,proving Trigonometric equations,Trigonometric Ratios,math trigonometry,learning trigonometry,cbse mathematics class 10,mathematics class 10 cbse
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# Get Answers to all your Questions
### Answers (1)
Answer: $\frac{\pi }{4}$
Hint: The principal value branch of function $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}$
Explanation:
We know that $\cos \left(-\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right) \quad \because[\cos (-\theta)=\cos \theta]$
Also know that, $\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$
By substituting these values in $\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}$
we get,
$\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
Let $y=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$\cos y=\frac{1}{\sqrt{2}}$
Hence, range of principal value of $\cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$
\begin{aligned} &\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}=\frac{\pi}{4} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \end{aligned}
View full answer
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# Class 9 maths chapter 1 exercise 1.5 and 1.6 solutions
In this page we have Class 9 maths chapter 1 exercise 1.5 solutions and Class 9 maths chapter 1 exercise 1.6 solutions . Hope you like them and do not forget to like , social share and comment at the end of the page.Free download also available
## Class 9 maths chapter 1 exercise 1.5 solutions
Question 1
Classify the following numbers as rational or irrational:
(i) 2 - √5
(ii) (3 + √23) - √23
(iii) 2√7/7√7
(iv) 1/√2
(v) 2π
Solution
Question 2
Simplify each of the following expressions:
(i) (3 + √3) (2 + √2)
(ii) (3 + √3) (3 - √3)
(iii) (√5 + √2)2
(iv) (√5 - √2) (√5 + √2)
Solution
(i) (3 + √3) (2 + √2)
= 3 × 2 + 2 + √3 + 3√2+ √3 ×√2
= 6 + 2√3 +3√2 + √6
(ii) (3 + √3) (3 - √3)
Now as (a + b) (a - b) = a2 - b2
= 32 - (√3)2
= 9 - 3= 6
(iii) (√5 + √2)
Now as (a + b)2 = a2 + b2 + 2ab
= (√5)2 + (√2)2 + 2 ×√5 × √2
= 5 + 2 + 2 × √5× 2
= 7 +2√10
(iv) (√5 - √2) (√5 + √2)
Now as (a + b) (a - b) = a2 - b2
= (√5)2 - (√2)2
= 5 – 2= 3
Question 3
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857...
Question 4.
Represent √9.3 on the number line.
Question 5
Rationalize the denominators of the following:
(i) 1/√7
(ii) 1/ (√7-√6)
(iii) 1/ (√5+√2)
(iv) 1/ (√7-2)
Solution
## Class 9 maths chapter 1 exercise 1.6 solutions
Question 1.
Find:
(i) 641/2
(ii) 321/5
(iii) 1251/3
Solution
(i) 641/2 =(26)1/2 =26 X1/2 = 23 =8
(ii) 321/5 =(25)1/5 = 2
(iii) 1251/3 = (53)1/3 =5
Question 2
Find:
(i) 93/2
(ii) 322/5
(iii) 163/4
(iv) 125-1/3
Solution
(i) 93/2 = (32)3/2 = 27
(ii) 322/5 = (25)2/5 =4
(iii) 163/4 = (24)3/4 =8
(iv) 125-1/3 = 1/1251/3 = 1/(53)1/3 =1/5
Question 3.
Simplify:
(i) 22/3.21/5
(ii) (1/33)7
(iii) 111/2/111/4
(iv) 71/2.81/2
Solution
(i) 22/3.21/5 =2 2/3+ 1/5 = 210+3/15 =213/15
(ii) (1/33)7
=1/33X7 = 1/321 =3 -21
(iii) 111/2/111/4
= 111/2 -1/4 = 111/4
(iv) 71/2.81/2
=(7X8)1/2 = 561/2
## Summary
1. Class 9 maths chapter 1 exercise 1.5 solutions and Class 9 maths chapter 1 exercise 1.6 solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
2. This chapter 1 has total 6 Exercise 1.1 ,1.2,1.3 ,1.4 ,1.5 and 1.6. This is the Fifth and Sixth exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below |
# Kissing the curve – the gruesome details
So, you want to more? Here are all the gruesome details for calculating the tangent, normal, curvature and osculating circle of a curve!
Think of your favourite smooth curve in a plane that can be described by a function, . Then at any point , the slope of the tangent line is just the value of the derivative of the function at that point, . With this information, and the fact that the tangent must pass through the point , we can work out that the equation for the tangent is
The normal to the curve is perpendicular to the tangent, and so the product of their respective slopes must be -1. So the slope of the normal must be . Again, with this information and the fact that the normal must also pass through the point , we can work out that the equation for the normal to the curve at this point is The curvature of the curve is the rate at which the tangent changes direction as you move along the curve at constant speed. This is a little tricky to work out (you can see all the details at Wolfram MathWorld) but the curvature of a curve at a point is From the curvature, we can calculate the radius of curvature, . This is the radius of the osculating circle of the curve at the point . Now we know the radius of the osculating circle, we just need to calculate its centre. The centre lies on the normal line, and is a distance of from the point . If you use these two fact (and about 20 sheets of paper if you keep making silly mistakes like us!) then you can finally calculate the centre of the osculating circle to be the point: And therefore, the equation for the osculating circle of the curve at the point is Hooray!
Go back to the article or play with the worksheet below.
You can use this geogebra worksheet to see the tangent, normal and osculating circle of any smooth curve you choose - just change the equation f(x) in the left-hand panel. |
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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# Solving an Equation with Radicals - Problem 2
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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Solving a equation with a radical, so the steps of solving an equation with a radical will always going to be the same. You want to isolate the radical take each side to the power to get rid of the radical you are dealing with solve for our variable, and then check your answers.
So this problem what is actually going on is we have a square root and we also have the variable again. So what’s going to end up happening is we are going to have a portion of an equation which we are going to have to factor in order to solve it.
Without getting ahead of myself let’s get to that point first. So in order to solve it we first need to isolate our radical, the square root of x is equal to x minus 2. We then want to get rid of that square root so we need to square both sides. Square root squared that just goes away leaves us with the inside and we have x minus 2².
You have to remember that we need to follow this out. Common mistake is people just want to distribute that 2 in, it’s not the same thing. So we follow this out we get x² minus 4 x plus 4. Isolate everything to one side, I always like dealing with my x² term being positive so I’m going to bring the x around if 0 is equal to x² minus 5x plus 4.
We now have a term which is a quadratic we need to factor this out, x minus 4 x minus 1 leaving us with x is equal to 4 or 1. Now with any radical equation what we are going to have to do is check your answers so we are going to plug this in to make sure they actually work.
Let’s take 4 plug it in square root of 4 is 2, 2 plus 2 is 4. So that works that’s great. Try 1 square root of 1 is 1, plus 2 is 3, 3 is not equal to 1. So what happened it we actually got a extraneous solutions one that doesn’t work which is why we have to check to make sure our answers all work. When we square each side what can happen is we can actually introduce our any answer is that isn’t going to work for this problem.
But by going through the steps for solving the equation with a quadratic we are able to get at least one answer that works.
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# Estimating Square Root and Cube Root of Numbers
In case we have a number with many digits, it will be very difficult for us to calculate the square roots and cube roots of the number by prime factorization and long division methods. Hence we try to estimate the value of cube roots and square roots for these numbers.
In prime factorization, we find the factors of a number. Numbers such as 25, 700, 368, etc are easy to factorize. In case the number of digits increase the factorization becomes difficult. In such cases estimating square root and cube root is a good option.
For estimating the number of digits in the square root of a number, we use method of bars. For example:
$\sqrt{961}$ = 31 and $\sqrt {1369}$ = 37
The bar is placed on pair of digits starting from the rightmost digit.
$\overline{9} \;\overline{61}\; and\; \overline{13}\; \overline{69}$
In both the numbers above, we have 2 bars therefore their square root will have two digits.
In order to estimate the square root of a number consider the following example:
Number whose square root is to be determined = 247
Since: 100 < 247 < 400, and $\sqrt[2]{100}$ = 10 and $\sqrt[2]{400}$ = 20.
Taking the square root we get: 10 < $\sqrt[2]{247}$ < 20
We are not very close to the number yet.
We also know that $15^2$ = 225 and $15^2$ = 256, therefore 15 < $\sqrt[2]{247}$ < 16
The number 247 is much closer to 256 than 225.
Hence $\sqrt[2]{247}$ is approximately equal to 16.
### Estimating cube root of a cube number
If a large number is given and stated that it is a perfect cube then we can use the following method to calculate its cube root by the method of estimation.
1. Take any cube number such as 29791 whose cube root is to be determined and start making a group of three digits starting from the rightmost digit.
2. The first group will give us the unit’s digit of the required cube root. The number 791 ends in 1 therefore the last digit will be 1 because unit’s digit is 1 only when the cube root also ends with 1.
3. Now taking the second group, it is between $3^3$ = 27 and $4^3$ = 64. We take the smaller value therefore the digit at tenth place is 3.
The required cube root = $\sqrt[3]{29791}$ = 31. |
##### ACT
If you want a high percentage score on the ACT Math exam, then you should know how to solve percentage problems. As you'll see in the following practice questions, percentages appear in a wide range of problems, from business math to algebra.
## Practice questions
1. Given that 6 percent of (a + b) = 12 percent of b, which of the following must be true?
A. a < b B. a > b C. a = b D. a + b = 0 E. a < 0 and b < 0
2. The cost of a tablet increased 25 percent from 2014 to 2015. In 2016, the cost of the tablet was 1/4 less than its 2014 cost. By what percentage did the cost of the tablet decrease from 2015 to 2016?
A. 5% B. 25% C. 40% D. 50% E. 75%
1. The correct answer is Choice (C).
An easy way to deal with percentages is to use the number 100. In this case, say that (a + b) = 100. Then 6 percent of 100 = 6. That means that 12 percent of b is 6. Solve for b:
If a + b = 100, then a = 50 and a = b.
2. The correct answer is Choice (C). Give this problem a real value to work with. Say the 2014 price of the tablet is \$100. If the price increased 25 percent between 2014 and 2015, its 2015 price is \$125. The next year the price was 1/4 less its 2014 price of \$100, which means it cost \$75 in 2016. To find the percent decrease between 2015 and 2016, subtract the two prices and divide the difference by the 2015 price:
The percent decrease from 2015 to 2016 is 40 percent.
Lisa Zimmer Hatch, MA, and Scott A. Hatch, JD, have been helping students excel on standardized tests and navigate the college admissions process since 1987. They have written curricula and taught students internationally through live lectures, online forums, DVDs, and independent study. |
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9 2power Of Power
1. 1. Power of a Power Finding powers of numbers with exponents (x m ) n = x mn
2. 2. Simplify <ul><li>(2 3 ) 2 </li></ul><ul><li>This means 2 3 *2 3 </li></ul><ul><li>2 3 *2 3 = (2*2*2)*(2*2*2)=2 6 </li></ul>
3. 3. Simplify <ul><li>(4 2 ) 3 </li></ul><ul><li>This means 4 2 *4 2 *4 2 </li></ul><ul><li>4 2 *4 2 *4 2 = (4*4)*(4*4)*(4*4)=4 6 </li></ul>
4. 4. How does this work? <ul><li>Look again </li></ul><ul><ul><li>(4 2 ) 3 = 4 6 </li></ul></ul><ul><ul><li>(2 3 ) 2 =2 6 </li></ul></ul><ul><li>How do the exponents 2 and 3 relate to the exponent 6? </li></ul>
5. 5. Let’s look at some more <ul><li>(3 3)4 = (3*3*3)*(3*3*3)*(3*3*3)*(3*3*3) </li></ul><ul><li>(3 3)4 =3 ?? </li></ul><ul><li>3 (3x4) = 3 12 </li></ul><ul><li>As you can see (3 3)4 shows 3 multiplied by itself 12 times. </li></ul><ul><li>(3 3)4 = 3 (3*4) =3 12 </li></ul>
6. 6. Let’s try some using the Power of Powers Property <ul><li>The Power of Powers Property states that when you have a number to a certain power raised to another power, you multiply the exponents. </li></ul><ul><li>Examples </li></ul><ul><ul><li>(3 3 ) 4 = 3 12 </li></ul></ul><ul><ul><li>(8 2 ) 5 = 8 10 </li></ul></ul><ul><ul><li>(9 1 ) 4 = 9 4 </li></ul></ul>
7. 7. Try some <ul><li>(2 3)4 = ? </li></ul><ul><li>(10 3)2 = ? </li></ul><ul><li>(p 2)5 = ? </li></ul><ul><li>(x m)3 = ? </li></ul><ul><li>Go to the next slide when you have the solutions to check your work. </li></ul>
8. 8. Power of Powers <ul><li>(2 3)4 = 2 12 </li></ul><ul><li>(10 3)2 = 10 6 </li></ul><ul><li>(p 2)5 = p 10 </li></ul><ul><li>(x m)3 = x 3m </li></ul>
9. 9. Raise a monomial to a power <ul><li>(xy) 2 = xy*xy = x*x*y*y = x 2 y 2 </li></ul><ul><li>(xy 2 ) 2= </li></ul><ul><li>If you get stuck with powers of powers, try writing out the multiplication of numbers and variables. </li></ul>(x*y*y)* (x*y*y) = x*y*y*x*y*y = x*x*y*y*y*y = x 2 y 4
10. 10. Try some <ul><li>(xy) 2 = ? </li></ul><ul><li>(xy 2 ) 2 = ? </li></ul><ul><li>( r 2 ) 4 = ? </li></ul><ul><li>Go to the next slide when you have the solutions to check your work. </li></ul>
11. 11. Solutions <ul><li>(x 1 y) 2 = x 2 y 2 </li></ul><ul><li>(x 1 y 2 ) 2 = x 2 y 4 </li></ul><ul><li>( 1 r 2 ) 4 = 4 r 8 </li></ul><ul><li>Can you see the power of powers property at work? </li></ul><ul><li>If not, try changing the variables that have no exponent to an exponent of one. </li></ul><ul><li>{Once again, 1 comes in handy!} </li></ul>
12. 12. Let’s take another look <ul><li>(xy) 2 =(x 1 y 1 ) 2 = x 2 y 2 </li></ul><ul><li>(x 1 y 2 ) 2 = x 2 y 4 </li></ul><ul><li>( 1 r 2 ) 4 = 4 r 8 </li></ul>
13. 13. Try some more. Use 1 to your advantage when you can. <ul><li>(x 2 y) 3 = (x 2 y 1 ) 3 = x (2*3) y( 1*3) = x 6 y 3 </li></ul><ul><li>(x 2 y 2 z 2 ) 3 = </li></ul><ul><li>(abcd) n = </li></ul><ul><li>(x 2 y 3 ) 5 = </li></ul>
14. 14. Solutions <ul><li>(x 2 y 2 z 2 ) 3 =x 2*3 y 2*3 z 2*3 =x 6 y 6 z 6 </li></ul><ul><li>(abcd) n =a n b n c n d n </li></ul><ul><li>(x 2 y 3 ) 5 =x 2*5 y 3*5 = x 10 y 15 </li></ul>
15. 15. Powers of -1 <ul><li>Write out (-2) 3 = (-2)*(-2)*(-2) </li></ul><ul><li>When the exponent is an odd number, the answer can be negative. </li></ul>
16. 16. Suggestion <ul><li>Once again, the suggestion is to write out the multiplication statements to help you solve tricky exponential products. </li></ul>
17. 17. Simplify <ul><li>(-t) 5 =? </li></ul><ul><li>(-t) 4 =? </li></ul><ul><li>(-5x) 3 =? </li></ul>
18. 18. solutions <ul><li>(-t) 5 = (-t) * (-t) * (-t) * (-t) * (-t) </li></ul><ul><li>=-t 5 </li></ul><ul><li>(-t) 4 =t 4 </li></ul><ul><li>(-5x) 3 =(-5x) (-5x) (-5x) = </li></ul><ul><li>= -5*-5*-5*x*x*x = -125x 3 </li></ul>
19. 19. Negative and Zero Exponents Integrated II Chapter 9.2
20. 20. Negative Integers do NOT mean negative numbers
21. 21. Numbers to the Zero Power <ul><li>Every number to the Zero Power, such as 5 0 = 1. </li></ul><ul><li>We can use last lesson’s division of powers as a proof. </li></ul>
22. 22. Using division to prove <ul><li>Any number divided by itself equals 1. </li></ul><ul><li>Using the Quotient of Powers Property, the exponents would be subtracted. </li></ul><ul><li>6 5-5 = 6 0 = 1 </li></ul>
23. 23. Negative Exponents <ul><li>Negative Exponents do not mean negative numbers. </li></ul><ul><ul><li>4 -5 = </li></ul></ul><ul><ul><li>3 -2 = </li></ul></ul><ul><ul><li>7 -4 = </li></ul></ul>
24. 24. Solve.
25. 25. Simplify. <ul><li>b 6 *b -2 =b 4 = 1 </li></ul><ul><li>b 4 b 4 </li></ul><ul><li>-3y -2 </li></ul><ul><li>-6p -7 </li></ul><ul><li>8a 4 b 7 c -4 </li></ul><ul><li>3a 6 b -6 c -4 </li></ul>
26. 26. Simplify. <ul><li>- 3y -2 = -3 </li></ul><ul><li>y 2 </li></ul><ul><li>-6p -7 = -6 </li></ul><ul><li>p 7 </li></ul><ul><li>8a 4 b 7 c -4 = 8 a 4-6 b 7--6 c -4--4 = 8 b 13 </li></ul><ul><li>3a 6 b -6 c -4 3 3 a 2 </li></ul><ul><li>** (c 0 =1 which when multiplied is no longer part of the answer. </li></ul>
27. 28. Let’s Divide! Dividing Monomials Focus: Quotient of Powers Rule
28. 29. Quotients of Powers <ul><li>How do I find ? </li></ul><ul><li>a*a*a*a*a*a = </li></ul><ul><li>a*a*a*a </li></ul><ul><li>a 2 </li></ul><ul><li>1 = a 2 </li></ul>a*a*a*a*a*a = a*a*a*a
29. 30. Let’s find a different way <ul><li>In the previous slide, you saw that the result of this fraction was a 2 . </li></ul><ul><li>How do 6 and 4 relate to two? </li></ul>
30. 31. Quotient of Powers Property <ul><li>For all non-zero numbers, subtract the exponent of the denominator from the numerator when the bases are the same. </li></ul><ul><li>4 5-2 = 4 3 </li></ul>
31. 32. Let’s prove it. <ul><li>4 5-2 = 4 3 </li></ul>
32. 33. Try Some.
33. 34. Solutions <ul><li>2 10-5 = 2 5 </li></ul><ul><li>3 10-7 = 3 3 </li></ul><ul><li>5 8-3 = 5 5 </li></ul><ul><li>2 3-2 = 2 1= 2 </li></ul>
34. 35. Try Some with variables. <ul><li>x j-1 </li></ul><ul><li>x a+b-c </li></ul><ul><li>x m+1-1 = x m </li></ul>
35. 36. Fun Fun Fun
36. 37. Fun Fun Fun <ul><li>-2x 2-1 y 5-3 =-2xy 2 </li></ul><ul><li>-40a 4-1 b2c -5 </li></ul><ul><li>= -40a 3 b 2 </li></ul><ul><li>c 5 </li></ul>
37. 38. Remember negative exponents? <ul><li>Any time you have a negative exponent, it must be placed in the denominator. </li></ul><ul><li>C -3 = </li></ul>
38. 39. Try Some |
# SAT Practice Test 6, Section 9: Questions 11 - 15
Given:
The table
To find:
Car engine oil with a rating of 5W is how many times as fast as car engine oil with rating of 20W
Solution:
From the table;
5W = 2 × 10W
10W = 2 × 15W
15W = 2 × 20W
This means that
5W = 2 × 10W
= 2 × 2 × 15W
= 2 × 2 × 2 × 20W
= 8 × 20W
Given:
The figure
Points P, A and B are equally spaced on line l
Points P, Q and R are equally spaced on line m
PB = 4
PR = 6
AQ = 4
To find:
Perimeter of QABR
Solution:
Topic(s): RAR rule
Triangles PAQ and PBR are similar triangles (RAR)
The ratio of the sides is the same.
The perimeter of the quadrilateral QABR = 4 + 2 + 3 + 8 = 17
Given:
g(n) = n2 + n
h(n) = n2n
To find:
g(5) - h(4)
Solution:
g(5) = 52 + 5 = 30
h(4) = 42 − 4 = 12
g(5) − h(4) = 30 −12 = 18
Given:
g(n) = n2 + n
h(n) = n2n
To find:
h(m + 1)
Solution:
h(m + 1)
= (m + 1)2 − (m + 1)
= m2 + 2m + 1m − 1
= m2 + m
= g(m)
Given:
The selling price of a sweater = \$28
The selling price is 40% more than the cost price
Employees can purchase the sweater at 30% off the cost price
To find:
The amount the employees pay for the sweater
Solution:
Topic(s): Percent
Let c be the cost price of the sweater.
c + 40% × c = 28
c + 0.4c = 28
1.4c = 28
c = 20
Purchase price = 20 − 30% × 20
= 20 − 0.3 × 20
= 20 − 6
= 14
Given:
A rectangle ABCD
Point E is the midpoint of
Area of ABED is
To find:
Area of ABCD
Solution:
Topic(s): Area of trapezium
Let AB = l and BE = w
Area of trapezium ABED =
Area of rectangle ABCD =
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# An AP consists of 50 terms of which the third term is 12 and the last term is 106. Find the ${{29}^{th}}$ term.
Last updated date: 04th Aug 2024
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Hint: In the question, we are given the third term ${{a}_{3}}$ and the last term ${{a}_{50}}.$ We will use the formula ${{a}_{n}}=a+\left( n-1 \right)d$ to find the value of a and d. Once we get the values of a and d, we will use the same formula to find the ${{29}^{th}}$ term by putting n = 29.
We are given an AP whose third term is 12, that is ${{a}_{3}}=12$ and the last term is 106, i.e. ${{a}_{50}}=106.$
Now, we know that the general term in an AP is given as
${{a}_{n}}=a+\left( n-1 \right)d$
where a is the first term, d is the difference and n is the number of terms.
We have the third term as 12,
${{a}_{3}}=12$
Therefore, using this value in the general formula, we get,
${{a}_{3}}=a+\left( 3-1 \right)d$
$\Rightarrow 12=a+2d.....\left( i \right)$
Similarly, we have the last term as 106.
${{a}_{50}}=106$
$\Rightarrow {{a}_{50}}=a+\left( 50-1 \right)d$
$\Rightarrow 106=a+49d.....\left( ii \right)$
Now, we will solve the for a and d using equation (i) and (ii).
Subtraction equation (ii) from (i), we get,
\begin{align} & a+49d=106 \\ & a+2d=12 \\ & \underline{-\text{ }-\text{ }-} \\ & 47d=94 \\ \end{align}
Dividing both the sides by 47, we get,
$\Rightarrow d=\dfrac{94}{47}=2$
Therefore, we get the common difference, d = 2.
Now, putting the value, d = 2 in equation (i), we get,
$a+2\times 2=12$
$\Rightarrow a+4=12$
$\Rightarrow a=8$
Therefore, we get our first term as 8.
Now, we will find the ${{29}^{th}}$ term. We know that,
${{a}_{n}}=a+\left( n-1 \right)d$
For, ${{a}_{29}},n=29.$
Also, we have, a = 8 and d = 2. Therefore, we get,
${{a}_{29}}=8+\left( 29-1 \right)2$
$\Rightarrow {{a}_{29}}=8+28\times 2$
$\Rightarrow {{a}_{29}}=8+56$
$\Rightarrow {{a}_{29}}=64$
So, we get the ${{29}^{th}}$ term as 64.
Note:While solving for the third term and the last term, students need to remember that the third term is written as ${{a}_{3}}=a+2d,$ writing ${{a}_{3}}=a+3d$ will lead to a wrong solution. Also, students have to keep in mind that when subtracting two equations, you need to change the sign of the second equation which is being subtracted. |
Probability does not need to be an abstract concept. Although equations and formulas may be too advanced for elementary school students, simple probability can help students in making educated guesses and decisions. Analyzing and interpreting data allows your students to determine probability. Students engage in activities that are approachable and accessible, and using M&Ms is an effective visual aid. To top it off, students can enjoy their M&Ms as an after-class treat.
## Step 1
Take M&Ms and plastic sandwich bags to a clean work area.
## Step 2
Assemble packets of M&Ms (one per student). Between 20 and 30 candy pieces per plastic bag is plenty. Counting is not necessary, but recommended to give each student a fair share of candy.
## Step 3
Place all items required for this lesson in a bag with a carrying handle.
## Step 1
Hand out lesson materials. Each student should receive one M&M packet, one piece of construction paper, two pieces of blank paper and two pieces of graph paper. Set out markers or crayons for all students.
## Step 2
Encourage students to use the construction paper as a placemat. Tell them the candies may be enjoyed after the lesson if they behave appropriately.
## Step 3
Instruct students to place candies on the construction paper. Tell them each bag has the same number of M&Ms, but everyone has a different number of colors because they were selected at random.
## Step 4
Request that students record information based on the M&Ms on a blank piece of paper. Write these questions on the chalkboard: "How many M&Ms do you have?" "How many different colors do you have?" "How many of each color do you have?" "Which color has the most quantity? The least?"
## Step 5
Write the definition of simple probability on the board. Use your own definition or use, "The probability of an event is the number of outcomes (favorable) divided by the total number of possible outcomes (assuming all outcomes are equally likely)."
## Step 6
Have students record numbers in fractions or ratios. If the group of 20 candies contains 7 green pieces, the probability is expressed as 7/20 or 7:20. Guide the class through this process and write different probabilities on the board. Probabilities are numbers expressed commonly as fractions and ratios, but also as percents or decimals; feel free to write answers in these different forms.
## Step 7
Ask students to determine the ratio of each M&M color to the entire bag. Students can predict the probability of selecting one color at random from their plastic bag with this information. Ask, "How many of each color is likely to be found in a handful of 10, or of 20?" Let them try this experiment, then discuss the results.
## Step 8
Allow students to enjoy their candy after the lesson.
### Tip
If you'd rather have your students work in teams, keep your classroom sanitary by saving the candy for later. Work with other small items such as dice, playing cards or game pieces.
### Tip
Snack-sized, pre-packaged M&Ms also work well for this project.
### Tip
Use the provided resource link for facts about the different colors of M&Ms to enhance learning.
### Warning
Avoid purchasing seasonal M&Ms having only two or three colors. Use the standard variety of colors for this probability lesson. |
# Understanding Probability: A Guide for Beginners
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• Blog
• /
• Understanding Probability: A Guide for Beginners
Probability is an important concept in mathematics and statistics that helps us understand the likelihood of events happening. It is used in many fields, from predicting the weather to analyzing data sets for research. In this guide, we’ll examine what probability is and how to use it.
Probability is a measure of how likely it is for an event to occur. It’s expressed as a number between 0 and 1, where 0 means that the event will never happen and 1 means that the event is certain. Probability can also be expressed as a percentage—for example, if an event has a probability of 0.5, then it has a 50% chance of happening. For example, if you flip a coin, the probability of getting heads is 50% because there are only two possible outcomes (heads or tails), one of which is heads.
Probability can be used to make predictions about the outcomes of experiments or to analyze the likelihood of certain events occurring. For example, a weather forecaster may use probability to predict the chance of rain on a given day, or a doctor may use probability to determine the likelihood of a patient developing a certain disease.
## The Formula for Calculating Probability?
The formula for calculating probability is:
$$\text{Probability = } \frac{\text{Number of Successful Outcomes}}{\text{Total Number of Possible Outcomes}}$$
For example, if you have a bag containing 3 red balls and 2 blue balls, and you want to calculate the probability of drawing a red ball, you would use the formula like this:
$$\text{Probability of drawing a red ball =} \frac{3 \text{ successful outcomes (red balls)}}{5 \text{ total possible outcomes (3 red balls + 2 blue balls)}}$$
This would give you a probability of 3/5, or 0.60, when expressed as a decimal. You could also express this probability as a percentage by multiplying it by 100, giving you 60%.
## Flipping a Coin
Flipping a coin is one of the most common examples used to illustrate probability. When you flip a coin, there are two possible outcomes - heads or tails. Each outcome has a probability of 50% or a 1 in 2 chance of occurring. For an unbiased coin, the probability of getting heads is the same as that of getting tails.
Flipping Coins
Flip results appear here
document.getElementById('flipButton').addEventListener('click', function() { var numFlips = parseInt(document.getElementById('numFlips').value); if (numFlips < 1 || isNaN(numFlips)) { alert('Please enter a valid number of flips.'); return; } var flipResults = []; var headsCount = 0; var tailsCount = 0; for (var i = 0; i < numFlips; i++) { if (Math.random() < 0.5) { flipResults.push('H'); headsCount++; } else { flipResults.push('T'); tailsCount++; } } document.getElementById('flipResult').innerText = 'Results: ' + flipResults.join(', '); document.getElementById('summary').innerText = Summary: ${headsCount} Heads,${tailsCount} Tails; });
## Rolling a Dice
Rolling a die is a great way to explore probability. A standard 6-sided die has an equal chance of landing on any side when rolled. This means that the probability of any given side landing face-up is 1/6. The probability of rolling any even number is 3/6 or 1/2 since there are three even numbers out of six total faces.
Roll Six Sided Dice
Roll results appear here
document.getElementById('rollButton').addEventListener('click', function() { var numDice = parseInt(document.getElementById('numDice').value); if (numDice < 1 || numDice > 9 || isNaN(numDice)) { alert('Please enter a number between 1 and 9.'); return; } var diceResults = []; for (var i = 0; i < numDice; i++) { diceResults.push(Math.floor(Math.random() * 6) + 1); } document.getElementById('diceResult').innerText = 'Results: ' + diceResults.join(', '); });
## Deck of 52 Playing Cards
Here are some examples of basic probability calculations related to playing cards:
• The probability of drawing a red card from a standard deck of playing cards is 26/52, or 50%, because there are 52 cards in a standard deck, and 26 of them are red.
• The probability of drawing an ace from a standard deck of playing cards is 4/52, or 7.69%, because there are 52 cards in a deck, and 4 of them are aces.
• The probability of drawing a face card (king, queen, or jack) from a standard deck of playing cards is 12/52, or 23.08% because there are 52 cards in a deck, and 12 are face cards.
Draw from a Pack of Playing Cards
Drawn cards will appear here
function generateDeck() { var suits = ['♥', '♦', '♣', '♠']; var values = ['2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K', 'A']; var deck = []; for (var suit of suits) { for (var value of values) { deck.push({ value: value, suit: suit }); } } return deck; } function drawCards(numCards, deck) { var drawnCards = []; for (var i = 0; i < numCards; i++) { if (deck.length === 0) { break; } var cardIndex = Math.floor(Math.random() * deck.length); drawnCards.push(deck[cardIndex]); deck.splice(cardIndex, 1); } return drawnCards; } function getCardClass(suit) { switch (suit) { case '♥': return 'heart'; case '♦': return 'diamond'; case '♣': return 'club'; case '♠': return 'spade'; } } document.getElementById('drawButton').addEventListener('click', function() { var numCards = parseInt(document.getElementById('numCards').value); if (numCards < 1 || numCards > 9 || isNaN(numCards)) { alert('Please enter a number between 1 and 9.'); return; } var deck = generateDeck(); var cardsDrawn = drawCards(numCards, deck); var resultContainer = document.getElementById('drawResult'); resultContainer.innerHTML = ''; for (var i = 0; i < cardsDrawn.length; i++) { var card = cardsDrawn[i]; var cardElement = document.createElement('span'); cardElement.className = 'card ' + getCardClass(card.suit); cardElement.textContent = card.value + card.suit; resultContainer.appendChild(cardElement); if (i < cardsDrawn.length - 1) { resultContainer.appendChild(document.createTextNode(', ')); } } });
## Definitions:
Probability: Probability is a measure of how likely it is for an event to occur. It’s expressed as a number between 0 and 1, where 0 means that the event will never happen and 1 means that the event is certain.
Trial: A trial is a single instance of an experiment or a process. For example, flipping a coin or rolling a die is a trial.
Outcome: An outcome is the result of an experiment or trial. For example, getting heads when flipping a coin or rolling a 4 on a die are outcomes. The outcomes are usually classified as either successful (when the desired result is achieved) or unsuccessful (when the desired result is not achieved).
Sample Space: The sample space is the set of all possible outcomes of an experiment or a process. For example, the sample space for flipping a coin would be heads and tails, and the sample space for rolling a die would be the numbers 1-6.
## Conclusion
In conclusion, the probability is a tool that can be used to make decisions and understand the world around us. By studying probability and statistics, you can better understand the chances of certain events occurring and make sounder decisions.
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# Write An Equation From A Table Worksheet
## 8 problems
Writing an equation from a table involves determining the relationship between the variables represented in the table and expressing it in the form of an equation.
A table is a visual representation of a set of data that contains rows and columns. Each column represents a variable and each row represents a set of values for those variables.
Linear Relationships And Functions
8.F.B.4
When writing an equation from a table, the first step is to identify the independent variable (x) and the dependent variable (y). The independent variable is the variable that changes and the dependent variable is the variable that is affected by the change in the independent variable. The values of the dependent variable are determined by the values of the independent variable.
For example, if the table shows a relationship between time (x) and distance (y), the equation can be written as:
y = mx + b
Where m is the slope (rate of change) and b is the y-intercept (the value of y when x = 0)
The slope is calculated by finding the ratio of the change in the y-coordinate to the change in the x-coordinate between two points on the table, m = (y2 - y1) / (x2 - x1)
The y-intercept is determined by finding the value of y when x = 0, b = y1 - m*x1
Once you have the slope and y-intercept, you can substitute these values into the equation and you have the equation that represents the relationship between the variables in the table.
It is important to note that, the equation you get from the table is an equation of a linear function, which means that the relationship between the variables is a straight line.
## Teaching Write An Equation From A Table Easily
1. Using real-world examples: Provide students with real-world examples of tables that show a relationship between variables, such as in science experiments, engineering or economics. This can help students understand the concept and apply it to new situations.
2. Guided practice: Provide students with guided practice problems that involve writing an equation from a table. Start with simple examples and gradually increase the difficulty level.
## Why Should You Use Write an equation from a table Worksheet for your students?
1. Practice and reinforcement: Worksheets provide an opportunity for students to practice and reinforce the concept of finding the slope of a proportional relationship.
2. Self-paced learning: Worksheets allow students to work at their own pace and level, which can be helpful for students who may need more practice or who may be working at a faster pace than their peers.
3. Reviewing material: Worksheets can be used to review material covered in class and to prepare for assessments.
You can download and print these super fun write an equation from a table 8th grade pdf from here for your students. You can also try our Write An Equation From A Table Problems and Write An Equation From A Table Quiz as well for a better understanding of the concepts.
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##### Praxis Core: 1,001 Practice Questions For Dummies
On the Praxis Core exam, you may be asked to figure out what two expressions have in common—for example, their greatest common factor (or GCF).
In the following practice questions, you first have to find the GCF of two terms, and then find one of the original terms, given the GCF and the other original term.
## Practice questions
1. What is the greatest common factor of 35a4b7c12 and 20a8b3c10?
A. 7a4b3c10 B. 5a4b7c12 C. 10a4b3c10 D. 5a4b3c10 E. 10a8b8c12
2. 6xy is the greatest common factor of 24x2y and which of the following terms?
A. 8xy B. 12x C. 24y D. 42xy2 E. 3x2y
1. The correct answer is Choice (D). To find the greatest common factor of the two terms, find the greatest common factor of the coefficients and write it next to each common variable. Give each common variable the lowest exponent it has in the terms. The greatest common factor of 35 and 20 is 5, the lowest exponent of a in the terms is 4, the lowest exponent of b is 3, and the lowest exponent of c is 10. Thus, the greatest common factor of the two terms is 5a4b3c10.
2. The correct answer is Choice (D). The first thing to look for is which choices have 6xy for a factor. 6xy must be a factor of a term to be a greatest common factor of it and 24x2y. 6xy is not a factor of Choice (A) because 6 isn’t a factor of 8, and 6xy isn’t a factor of Choice (B) because y does not go into 12x. 6xy is also not a factor of Choice (C) because x doesn’t go into 24y. And 6xy isn’t a factor of Choice (E) because 6 isn’t a factor of 3.
Alternatively, you can eliminate Choices (A) and (E) because their coefficients, 8 and 3, aren’t multiples of 6. And you can eliminate Choices (B) and (C) because the correct answer needs both an x and a y. That leaves Choice (D) to consider. 6xy is a factor of 42xy2. Because 6 is the greatest common factor of 24 and 42, and because x has a lowest exponent of 1 and y has a lowest exponent of 1 in 24x2y and 42xy2, 6xy is the greatest common factor of 24xy2 and 42xy2. |
# Math Snap
## 9. Triangle $Q R S$ is shown. Triangle $Q R S$ is similar to triangle $T U V$. a. Find the value of $\tan U$. Show your work. Round to the nearest tenth. $2 \cdot 2$ b. How does the value of $\tan U$ relate to the value of $\tan T$ ? they are the same
#### STEP 1
Assumptions 1. Triangles $\triangle QRS$ and $\triangle TUV$ are similar. 2. Angle $Q$ in $\triangle QRS$ is $24.9^\circ$. 3. Angle $T$ in $\triangle TUV$ is $24.9^\circ$. 4. Angle $S$ in $\triangle QRS$ and angle $V$ in $\triangle TUV$ are both $90^\circ$. 5. Side $RS$ (height) in $\triangle QRS$ is $6.2$ units. 6. Side $UV$ (height) in $\triangle TUV$ is $6.2$ units. 7. Base $TV$ in $\triangle TUV$ is $13.4$ units.
#### STEP 2
To find the value of $\tan U$, we need to identify the corresponding angle in $\triangle TUV$. Since $\triangle QRS$ and $\triangle TUV$ are similar and angle $T$ is $24.9^\circ$, angle $U$ will be the same as angle $Q$, which is $24.9^\circ$.
#### STEP 3
The tangent of an angle in a right triangle is given by the ratio of the length of the opposite side to the length of the adjacent side. For angle $U$ in $\triangle TUV$: $\tan U = \tan 24.9^\circ$
#### STEP 4
Using the given dimensions of $\triangle TUV$, we can calculate $\tan 24.9^\circ$ as follows: $\tan 24.9^\circ = \frac{\text{opposite}}{\text{adjacent}}$ In $\triangle TUV$, the opposite side to angle $U$ (which is the same as angle $Q$) is $UV = 6.2$ units, and the adjacent side to angle $U$ is $TV = 13.4$ units.
#### STEP 5
Plug in the values: $\tan 24.9^\circ = \frac{6.2}{13.4}$
#### STEP 6
Calculate the value: $\tan 24.9^\circ \approx 0.4627$
#### STEP 7
Round to the nearest tenth: $\tan 24.9^\circ \approx 0.5$
#### STEP 8
The value of $\tan U$ is approximately $0.5$.
##### SOLUTION
To address part (b) of the problem, we need to compare $\tan U$ and $\tan T$. Since $\triangle QRS$ and $\triangle TUV$ are similar and angle $U$ is the same as angle $T$ (both are $24.9^\circ$), the tangent values will be the same. $\tan U = \tan T$ Therefore, the value of $\tan U$ is the same as the value of $\tan T$. |
The Pixie and the Sprite
The solution to this problem is found through conditional probability. To begin with, there are 11 possibilities for the number of times out of 10 the pixie lies. For each of these possibilities, find the probability that the pixie would be able to make the statement "I lie exactly 4 out of 10 times." For example, in the case that the pixie actually lies 7 of 10 times, there is a 7/10 probability that he is able to make the (false) statement "I lie exactly 4 out of 10 times." The probability of the pixie being able to make the statement is the same as the probability of the pixie lying, except for the case that he is telling the truth (4/10); in that case, the probability of him being able to make that statement is 6/10 (which is equal to the probability of him being able to make a true statement). Now, conditional probability says that the probability of the pixie lying X number of times out of 10 in his life is equal to the probability of him being able to make the statement (X/10, unless X is 4) divided by the sum of probabilities for all the cases. So, for our example, the probability that the pixie tells lies 7 of 10 times is: (7/10) / (0/10 + 1/10 + 2/10 + 3/10 + 6/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 10/10) = 7/57 Similarly, the probability that the pixie lies X of 10 times is: X/57, for X not equal to 4, 6/57, for X equal to 4. Now we can find the probability of the pixie lying in his first statement. There is a X/57 chance (or 6/57 for the 4 of 10 case) that the pixie lies X of 10 times, and for each of these cases there is a X/10 chance that the pixie lied about the path. To find the odds that the pixie lied, simply add the products of probabilities for each case: 0/570 + 1/570 + 4/570 + 9/570 + 24/570 + 25/570 + 36/570 + 49/570 + 64/570 + 81/570 + 100/570 = 393/570 = 68.947%. So, the probability the pixie lied is 68.947%, and you should go right.
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### CCSS.MATH.PRACTICE.MP08
Look for and express regularity in repeated reasoning.
## View Standards Related toCCSS.MATH.PRACTICE.MP08
### 6.RP.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.
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### 6.RP.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b =? 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.
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### 6.RP.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
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### 6.RP.3.a
Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
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### 6.RP.3.b
Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
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# High School: Geometry
### Expressing Geometric Properties with Equations HSG-GPE.B.7
7. Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula.
If there's one thing that mathematicians love, it's efficiency. And the distance formula is nothing if not efficient. It is so efficient that while we're going the distance, we can even go for speed.
Students are expected to be able to use the distance formula to find the distance between two coordinate points and then apply that information and know-how to calculate the perimeter and area of various polygons.
They should note that the distance formula is derived from the Pythagorean theorem. We suggest squaring both sides of the distance formula so it's a little easier to see that both equations set the sum of two squared terms equal to a third squared term.
Basically, the distance formula assumes that the distance we're measuring is the hypotenuse of a right triangle. The base of the triangle is denoted as a, or (x2x1). The height is b, or (y2 – y1). And the hypotenuse, c, is the distance, D. Now that that's all cleared up, let's go watch some E!
As one might expect, the distance formula is pretty handy for computing the distance between two coordinate points. Of course, it's best saved for when those points are not located along the same vertical or horizontal, because then simple subtraction would be the most efficient way to find the distance between them. No, the distance formula is more aptly used when the points are spaced diagonally.
Students should know that if we compute the distances between the points around a polygon, then the distances can be added to find the polygon's perimeter. Students should know that this works for all polygons.
Area is a little different, since it depends on the actual shape. Students should know that pretty much all polygons on the coordinate plane can be split into rectangles and triangles. As such, students should know how to calculate the areas of rectangles and triangles. Adding the areas of the individual pieces should give them the area of the entire shape.
We recommend giving students a variety of shapes so they can apply their knowledge and problem-solving skills rather than automatically plugging points into a formula. If students are ever confused, they can always plot points on the coordinate plane, too. After all, this is still geometry, and a picture can speak volumes (er…areas).
Here isa recap video on coordinate and distance formula. |
# NCERT Solutions for Class 10 Maths Chapter 5 : Arithmetic Progressions
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## Chapter 5 : Arithmetic Progressions
### 5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone etc. We now look for some patterns which occur in our day-to-day life. Some such examples are :
### 5.2 Arithmetic Progressions
Each of the numbers in the list is called a term. Given a term, can you write the next term in each of the lists above? If so, how will you write it? Perhaps by following a pattern or rule. Let us observe and write the rule.
(1) each term is 1 more than the term preceding it.
(2) each term is 30 less than the term preceding it.
(3) each term is obtained by adding 1 to the term preceding it.
(4) all the terms in the list are 3 , i.e., each term is obtained by adding (or subtracting) 0 to the term preceding it.
### 5.3 Nth Term Of An Ap
Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of Rs 8000, with an annual increment of Rs 500. What would be her monthly salary for the fifth year?
To answer this, let us first see what her monthly salary for the second year would be.
### 5.4 Sum Of First N Terms Of An Ap
Let us consider the situation again given in Section 5.1 in which Shakila put Rs 100 into her daughter’s money box when she was one year old, Rs 150 on her second birthday, Rs 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?
### 5.5 Summary
In this chapter, you have studied the following points :
1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term. The fixed number d is called the common difference.
2.4. The sum of the first n terms of an AP is given by :
3.If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by :
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#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 24 Maths Textbook Solution.
$A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given
$A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$
Solution:
\begin{aligned} A^{3} &=A^{2} \times A \\\\ A^{2} &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6+9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{array}\right] \\\\ A^{2} &=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] \end{aligned}
\begin{aligned} &A^{2} \times A=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+14+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{array}\right] \\ \\&=\left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right] \end{aligned}
Now,
\begin{aligned} &A^{3}-6 A^{2}+5 A+11 I \\\\ &{\left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right]-6\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]+5\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]+11\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}
$\left[\begin{array}{ccc} 8-24 & 7-12 & 1-6 \\ -23+18 & 27-48 & -69+84 \\ 32-42 & -13+18 & 58-84 \end{array}\right]+\left[\begin{array}{ccc} 5+11 & 5 & 5 \\ 5 & 10+11 & -15 \\ 10 & -5 & 26 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$
$A^{3}-6 A^{2}+5 A+11 I=0$
Now,
\begin{aligned} &(A \times A \times A) A^{-1}-6(A \times A) A^{-1}+5 A A^{-1}+11 I A^{-1}=0 \\\\ &A A\left(A^{-1} A\right)-6 A\left(A^{-1} A\right)+5 A^{-1} A=-11\left(A^{-1} I\right) \\\\ &A^{-1}=\frac{-1}{11}\left(A^{2}-6 A+5 I\right) \end{aligned}
Now,
$A^{2}-6 A+5 I$
\begin{aligned} & \\ &{\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]-6\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]+5\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]-\left[\begin{array}{ccc} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{array}\right]+\left[\begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} -3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right] \end{aligned}
Hence,
$A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right]$ |
# How do you graph using the intercepts for y=-6x-9?
Mar 11, 2018
See a solution process below:
#### Explanation:
Because this equation is in slope-intercept form we can find the $x$-intercept and $y$-intercept directly from the equation. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope, $\textcolor{b l u e}{b}$ is the y-intercept value and $\frac{\textcolor{b l u e}{- b}}{\textcolor{red}{m}}$ is the x-intercept..
$y = \textcolor{red}{- 6} x - \textcolor{b l u e}{9}$
Therefore the $y$-intercept is: $\textcolor{b l u e}{b = - 9}$ or $\left(0 , \textcolor{b l u e}{- 9}\right)$
And, the $x$-intercept is:
$\frac{\textcolor{b l u e}{- - 9}}{\textcolor{red}{- 6}} \implies \frac{\textcolor{b l u e}{9}}{\textcolor{red}{- 6}} \implies - \frac{\textcolor{b l u e}{3 \times 3}}{\textcolor{red}{3 \times 2}} \implies - \frac{\textcolor{b l u e}{\textcolor{b l a c k}{\cancel{\textcolor{b l u e}{3}}} \times 3}}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{3}}} \times 2}} \implies - \frac{3}{2}$
Or
#(-3/2, 0)
We can next plot the two points on the coordinate plane:
graph{(x^2+(y+9)^2-0.3)((x+ 3/2)^2+y^2-0.3)=0 [-30, 30, -15, 15]}
Now, we can draw a straight line through the two points to graph the line:
graph{(y+6x+9)(x^2+(y+9)^2-0.3)((x+ 3/2)^2+y^2-0.3)=0 [-30, 30, -15, 15]} |
# How Do You Find the Hypotenuse of an Isosceles Triangle, Given Two Lengths?
Use the Pythagorean theorem to calculate the hypotenuse of a right triangle. A right triangle is a type of isosceles triangle. The hypotenuse is the side of the triangle opposite the right angle.
1. Make certain the triangle is a right triangle
The Pythagorean theorem can only be used with isosceles triangles that are right triangles. This means one of the angles must be 90 degrees. If it’s not, the theorem cannot be used.
2. Write down the Pythagorean theorem
The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the two shorter legs. It’s often written as h squared equals a squared plus b squared.
3. Insert the numbers
Replace a and b in the equation with the lengths of the two sides. As an example, let a = 3 and b = 4.
4. Square both of the lengths
To find the square of a number, multiply it by itself. For example, to find the square of 3, multiply 3 by 3 to get 9. For 4, multiply 4 by 4 to get 16.
5. Add the two squares
Add the results of the two calculations. In the example, 9 + 16 = 25. The equation now reads h squared = 25.
6. Calculate the hypotenuse by figuring the square root of the result
To remove the squared part of the equation, take the square root of both sides. The square root of h squared is simply h, the length of the hypotenuse. To find the square root of a number, determine what number multiplied by itself is equal to that number. In the example, the square root of 25 is 5 because 5 * 5 = 25. In many cases, the square root is not a whole number. |
Puzzling Polygon Facts Every Aspiring Mathematician Will Adore
Apr 13, 2023 By Sridevi Tolety
Originally Published on Mar 13, 2022
Edited by Luca Demetriou
Fact-checked by Spandana Kantam
Any two-dimensional closed plane figure with sides and not curves is a polygon.
The term polygon originated from the Greek language, where 'poly' means many, and 'gonia' means angle. Triangles, quadrilaterals, pentagons, and octagons are all polygons.
Studying geometry as part of mathematics is very interesting and amusing. When straight-line segments connect to each other to form a closed plane figure, it is called a polygon. In Euclidean geometry, which is also called flat geometry, the smallest possible polygon has three sides and is called a triangle.
Types Of Polygon
Polygons can be regular or irregular polygons, convex or concave polygons, or simple or complex polygons.
Regular polygons have all equal sides and angles. If the sides are unequal in length, they are irregular polygons. An equilateral triangle or a square with four sides are regular polygons, whereas a solid arrow on a signboard is an example of an irregular polygon.
If all angles inside a polygon are less than 180 degrees, it is called a convex polygon. Squares and rectangles are examples of a convex polygon.
If any one of the interior angles is greater than 180 degrees, it is called a concave polygon. A rhombus is an example of a concave polygon. Concave polygons are very common and have a more irregular shape, and a concave polygon is also called a non-convex polygon.
Any polygon that does not intersect itself is a simple polygon. If any of the edges intersect themselves, it is a complex polygon.
A star drawn with only external sides is a simple polygon, and if it is drawn with all its sides inside, they intersect each other and become a complex polygon. Complex polygons often have an irregular shape.
Properties Of Polygon
Any polygon study requires understanding the following three key properties: the number of sides of polygons, angles between the sides or edges, and length of the sides or edges.
A polygon is defined by the number of sides it has. Triangle is the smallest polygon with three sides. Equal-sided triangles are called equilateral triangles.
If two sides are equal, they are isosceles triangles, and all three sides being different implies that they are scalene triangles. A four-sided polygon is a quadrilateral. Squares and rectangles are all examples of this polygon.
Square is a regular polygon because of its equal sides. Five sides make the polygon a pentagon, six sides make it a hexagon, seven sides make it a heptagon, and so on. A thousand-sided polygon is called a chiliagon.
In their discussions, philosophers like Immanuel Kant, David Hume, and Descartes, referred to a chiliagon. A million-sided polygon is called a megagon and describes a philosophical concept that cannot be visualized. It is also considered to explain the convergence of several regular polygons as a circle.
The angles between the sides of polygons also constitute interesting polygon facts. For any polygon, the sum of all internal angles can be calculated with a formula:
The sum of internal angles = 180 degrees x (number of sides - 2)
Along with the number of sides and angles, the length of each side is also important. For a regular polygon, measuring one side is sufficient.
Polygons In Computer Graphics
Polygons have a vital role in computer graphics. In modeling, imaging, and rendering, polygons are used as basic entities. All attributes of polygons are defined in the form of arrays.
Vertices, sides, length, color, angles, and texture are all defined as arrays in the database. The images are stored in the form of a polygon mesh as a tessellation.
A tessellation is a recurring symmetrical, interlocking shape pattern and is often complex. These structures of polygon images are called from the database to active memory and then to display screen to be viewed as rendered scenes. These two-dimensional polygons are oriented so that they are viewed as three-dimensional visual scenes.
In computer graphics, an important requirement is to determine if a given point is inside or outside a polygon. A test called point in polygon test or inside test is conducted. Polygon filling is another important requirement where the polygon is filled with color. Several algorithms such as Boundary fill, Flood fill, or Scalene fills are used.
Angles In Polygon
Every polygon has two types of angles: interior angle and exterior angle. Angles formed by the lines or edges of the polygon on the inside are called interior angles.
It is measured at the vertex, on the inside of the polygon. Angles for outside of the polygon when one of the edges is extended are called exterior angles. Some angle properties of regular polygons are:
Sum total of all exterior angles is 360 degrees.
If a polygon has n number of sides, each exterior angle is 360 degrees/n.
Sum total of all interior angles is (n-2) x 180 degrees for a regular polygon with n being the number of sides.
Each interior angle is calculated as (n-2) x 180 degrees/n.
FAQs
Q: What is special about a regular polygon?
A: A regular polygon has all sides and angles equal.
Q: How many sides are on a polygon?
A: A polygon has a minimum of three sides and infinite maximum sides.
Q: What are the 20 polygons?
A: Triangle (three sides), quadrilateral (four sides), pentagon (five sides), hexagon (six sides), heptagon (seven sides), octagon (eight sides), nonagon (nine sides), decagon (10 sides), hendecagon (11 sides), dodecagon (12 sides), tridecagon (13 sides), tetradecagon (14 sides), pentadecagon (15 sides), hexadecagon (16 sides), heptadecagon (17 sides), octadecagon (18 sides), enneadecagon (19 sides), icosagon (20 sides), chilliagon (one thousand sides), and megagon (one million sides).
Q; What is the polygon shape?
A: A polygon can be of any shape, which is a plane figure closed with lines and not curves.
Q: Are all polygons quadrilaterals?
A: No, only polygons with four sides are quadrilaterals.
Q: What do polygons have in common?
A: Regular polygons have equal sides and angles, which are common.
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Written by Sridevi Tolety
Bachelor of Science specializing in Botany, Master of Science specializing in Clinical Research and Regulatory Affairs
Sridevi ToletyBachelor of Science specializing in Botany, Master of Science specializing in Clinical Research and Regulatory Affairs
With a Master's degree in clinical research from Manipal University and a PG Diploma in journalism from Bharatiya Vidya Bhavan, Sridevi has cultivated her passion for writing across various domains. She has authored a wide range of articles, blogs, travelogues, creative content, and short stories that have been published in leading magazines, newspapers, and websites. Sridevi is fluent in four languages and enjoys spending her spare time with loved ones. Her hobbies include reading, traveling, cooking, painting, and listening to music.
Read full bio >
Fact-checked by Spandana Kantam
Bachelor of Arts specializing in Political Science and Sociology
Spandana KantamBachelor of Arts specializing in Political Science and Sociology
Spandana holds a Bachelor's degree in Political Science from Acharya Nagarjuna University. She has a passion for writing and enjoys reading crime and thriller novels while listening to RnB music in her free time.
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Future Study Point
# MEAN,MODE AND MEDIAN NCERT
Mean, mode and median are the terms of statistics which represent the central tendency of a data, The single value of the mean, mode or median tells us about the state of data. All of these three are important in solving statistical problems in economics, science, commerce, and other fields.
Mean- The mean of data represents the most common value of the data, it is calculated as follows.
When data is given as an individual series- Indidual series means when each value of the variable is given without showing its frequency.
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Example-
Find out the average weight of 5 people whose weights are given as follows.
45 kg, 50 kg, 40 kg,60 kg and 55 kg
So, the most common weight or average weight of 5 people in the given data is 50
Disadvantages of mean– The mean is unable to represent the data in the condition when data is skewed (much differences between larger values and the rest of the individual values of the data), then the median would be truer to represent the data, see the following example.
maths marks of 5 friends in a test 5 60 95 10 65
Median- Median is called the middle term of the data provided data should be ascending order, the mean of the above data is 47 which is not representing the data properly so here median of the data will the best to show the nature of data of maths marks of 5 friends in a test.
maths marks of 5 friends in a test 5 10 60 65 95
n = 5 (odd number)
m = 60
The median of this data is 60 is representing the given data in a better way because of the difference between the largest value and median is lesser than the difference between the largest value and mean.
When the individual data is given and the number of observations are even
### Direct Method for evaluating the Mean
Discreet series- When frequencies are also given with the values of variable then such a series is known as discreet series, see the example
x Frequency 50 6 40 8 70 5 60 4 80 5
Mean of such a data is given as follows
Solution-
x f fx 50 6 300 40 8 320 70 5 350 60 4 240 80 5 ∑f= 28 400 ∑fx=1610
Determining the Median in case of discreet series-
x f cf 40 8 8 50 6 14 60 4 18 70 5 23 80 5 N=28 28
Therefore the median of the above data is 55.
The other ways of calculating mean are following
### Assumed mean method for evaluating the Mean
Assuming an observation as a mean(A) and then forming the third column and then writing the deviation(d) each observation about the mean that is equal to the difference between observation and the assumed mean keeping in view the sign convention of the numbers, thereafter forming 4 th column and evaluating fd the product of deviation and their corresponding frequencies. After the table is formed then calculate the value of mean from the following formula.
∑fd is the addition of the product of deviation and corresponding frequencies and ∑f is the sum of total frequencies of individual observations.
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### Step deviation method for evaluating Mean
In this method, one more column is added d’ as compared to the assumed mean method, in this method deviation(d) is reduced by dividing the common factor(h )of all the deviations.
After evaluating d’ one more column of fd’ is also incorporated in the table and then calculate the mean from the following formula.
When the continuous series is given-
Class interval 10-20 20-30 30-40 40-50 50-60 Frequency 5 10 20 4 1
Solution-
Sr.n0 Class interval f(Frequency) x(Classmark) fx 1 10-20 5 15 75 2 20-30 10 25 250 3 30-40 20 35 700 4 40-50 4 45 180 5 50-60 1 55 55 ∑f=40 ∑fx=1260
Other methods- (i) Assumed mean method- Already discussed above
Step deviation method– Already discussed above
Median of the data when continuous series is given-
Class interval 10-20 20-30 30-40 40-50 50-60 Frequency 5 10 20 4 1
Solution- Median is calculated by using the following formula when continuous series is given
Class interval Frequency(f) Cumulative frequency(cf) 10-20 5 5 20-30 10 15(cf) 30-40 20(f) 35 40-50 4 39 50-60 1 40(N)
The median group is N/2 th term of the series i.e N/2=40/2 = 20 th term which lies in 35 th term so the median group is 30-40
L = 30 (Lower limit of median class)
f =20 (Frequency of median class)
Hence the median of the given series is 32.5
The mode – In a series, the term with the highest frequency is known as mode of the series.
Mode of an individual series- When the individual measurements of the variable are given without showing their frequency then the terms with high frequency is known as a mode of the series.
Example- The performances of a student in 5 tests are given as follows, find the mode of data.
Tests i ii iii iv v Percentage 4o 30 50 40 60
40 has the highest frequency so the mode of the above data is 40
Mode in a discreet series- When the frequency is also given of each term then such a series is known as discreet series.
Age of students in a class(in years) 13 15 16 14 12 Number of students 7 6 4 15 8
In the given data most numbers of students are of the age 14 so the mode of the data is 14.
Mode of the continuous series- Find the mode of the following series
The age of employs(in years) 20-30 30-40 40-50 50-60 60-70 The number of employs 30 50 25 15 2
Study notes of Maths & Science
Solution –
The ages of employs(in years) frequency(f) 20-30 30 $\boldsymbol{f_{0}}$ 30-40 50$\boldsymbol{f_{1}}$ 40-50 25$\boldsymbol{f_{2}}$ 50-60 15 60-70 2
The mode(M ) of continuous series is calculated as follows
The class 30-40 has the highest frequency so it is mode group of above data
L = 40(lower limit of mode group)
f0= 30(preceding frequency of mode group)
f1= 50( frequency of mode group)
f2= 25 (successor frequency of mode group)
i= Class interval of mode group
M = 30+ 4.44
M = 34.44
Hence the mode of the series is 34.44
### The relationship between Mean,Mode and Median
Mode = 3×Median – 2×Mean
## NCERT Solutions of Science and Maths for Class 9,10,11 and 12
### NCERT Solutions for class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
### NCERT Solutions for class 11 maths
Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
### NCERT Solutions for Class 11 Physics
Chapter 1- Physical World
chapter 3-Motion in a Straight Line
### NCERT Solutions for Class 11 Chemistry
Chapter 1-Some basic concepts of chemistry
Chapter 2- Structure of Atom
### NCERT Solutions for Class 11 Biology
Chapter 1 -Living World
### NCERT solutions for class 12 maths
Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
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Difference between revisions of "2018 AMC 12A Problems/Problem 14"
Problem
The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$, can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$?
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$
Solution 1
We apply the Change of Base Formula, then rearrange: \begin{align*} \frac{\log_2{4}}{\log_2{(3x)}}&=\frac{\log_2{8}}{\log_2{(2x)}} \\ \frac{2}{\log_2{(3x)}}&=\frac{3}{\log_2{(2x)}} \\ 3\log_2{(3x)}&=2\log_2{(2x)}. \\ \end{align*} By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ it follows that \begin{align*} \log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$
~jeremylu (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
We will apply the following logarithmic identity: $$\log_{p^n}{\left(q^n\right)}=\log_{p}{q},$$ which can be proven by the Change of Base Formula: $$\log_{p^n}{\left(q^n\right)}=\frac{\log_{p}{\left(q^n\right)}}{\log_{p}{\left(p^n\right)}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.$$ We rewrite the original equation as $\log_{(3x)^3} 64 = \log_{(2x)^2} 64,$ from which \begin{align*} (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{\textbf{(D) } 31}.$
~MRENTHUSIASM
Solution 3
By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ the original equation becomes $$2\log_{3x} 2 = 3\log_{2x} 2.$$ By the logarithmic identity $\log_b{a}\cdot\log_a{b}=1,$ we multiply both sides by $\log_2{(2x)},$ then apply the Change of Base Formula to the left side: \begin{align*} 2\left[\log_{3x}2\right]\left[\log_2{(2x)}\right] &= 3 \\ 2\left[\frac{\log_2 2}{\log_2{(3x)}}\right]\left[\frac{\log_2{(2x)}}{\log_2 2}\right] &= 3 \\ 2\left[\frac{\log_2{(2x)}}{\log_2{(3x)}}\right] &=3 \\ 2\left[\log_{3x}{(2x)}\right] &= 3 \\ \log_{3x}{\left[(2x)^2\right]} &= 3 \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{\textbf{(D) } 31}.$
~Pikachu13307 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 4
We can convert both $4$ and $8$ into $2^2$ and $2^3,$ respectively: $$2\log_{3x} (2) = 3\log_{2x} (2).$$ Converting the bases of the right side, we get \begin{align*} \log_{2x} 2 &= \frac{\ln 2}{\ln (2x)} \\ \frac{2}{3}\cdot\log_{3x} (2) &= \frac{\ln 2}{\ln (2x)} \\ 2^\frac{2}{3} &= (3x)^\frac{\ln 2}{\ln (2x)} \\ \frac{2}{3} \cdot \ln 2 &= \frac{\ln 2}{\ln (2x)} \cdot \ln (3x). \end{align*} Dividing both sides by $\ln 2,$ we get $\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},$ from which $$2\ln (2x) = 3\ln (3x).$$ Expanding this equation gives \begin{align*} 2\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\ \ln (x) &= 2\ln 2 - 3\ln 3. \end{align*} Thus, we have $$x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}} = \frac{2^2}{3^3} = \frac{4}{27},$$ from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$
~lepetitmoulin (Solution)
~MRENTHUSIASM (Reformatting)
Solution 5 (Exponential Form)
Let $y=\log_{3x} 4 = \log_{2x} 8.$ We convert the equations with $y$ to the exponential form: \begin{align*} (3x)^y&=4, \\ (2x)^y&=8. \end{align*} Cubing the first equation and squaring the second equation, we have \begin{align*} (3x)^{3y}&=64, \\ (2x)^{2y}&=64. \end{align*} Applying the Transitive Property, we get \begin{align*} (3x)^{3y}&=(2x)^{2y} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$
~MRENTHUSIASM
2018 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 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 12 Problems and Solutions |
# How do you find the length of a curve in calculus?
May 6, 2018
In Cartesian coordinates for y = f(x) defined on interval $\left[a , b\right]$ the length of the curve is
$\implies L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
In general, we could just write:
$\implies L = {\int}_{a}^{b} \mathrm{ds}$
#### Explanation:
Let's use Cartesian coordinates for this explanation.
If we consider an arbitrary curve defined as $y = f \left(x\right)$ and are interested in the interval $x \in \left[a , b\right]$, we can approximate the length of the curve using very tiny line segments.
Consider a point on the curve ${P}_{i}$. We can compute the distance of a line segment by finding the difference between two consecutive points on the line $| {P}_{i} - {P}_{i - 1} |$ for $i \in \left[1 , n\right]$ where $n$ is the number of points we've defined on the curve.
This means that the approximate total length of curve is simply a sum of all of these line segments:
$L \approx {\sum}_{i = 1}^{n} | {P}_{i} - {P}_{i - 1} |$
If we want the exact length of the curve, then we can make the assumption that all of the points are infinitesimally separated. We now take the limit of our sum as $n \to \infty$.
$L = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} | {P}_{i} - {P}_{i - 1} |$
Since we are working in the $x y$-plane, we can redefine our distance between points to take on the typical definition of Euclidean distance.
$| {P}_{i} - {P}_{i - 1} | = \sqrt{{\left({y}_{i} - {y}_{i - 1}\right)}^{2} + {\left({x}_{i} - {x}_{i - 1}\right)}^{2}} = \sqrt{\delta {y}^{2} + \delta {x}^{2}}$
We can now apply the Mean Value Theorem, which states there exists a point ${x}_{i}^{'}$ lying in the interval $\left[{x}_{i - 1} , {x}_{i}\right]$ such that
$\implies f \left({x}_{i}\right) - f \left({x}_{i - 1}\right) = f ' \left({x}_{i}^{'}\right) \left({x}_{i} - {x}_{i - 1}\right)$
which we could also write (using the notation we are using) as
$\implies \delta y = f ' \left({x}_{i}^{'}\right) \delta x$
Applying this means we now have
$| {P}_{i} - {P}_{i - 1} | = \sqrt{{\left[f ' \left({x}_{i}^{'}\right) \delta x\right]}^{2} + \delta {x}^{2}}$
Simplifying this expression a bit gives us
$| {P}_{i} - {P}_{i - 1} | = \sqrt{{\left[f ' \left({x}_{i}^{'}\right)\right]}^{2} \delta {x}^{2} + \delta {x}^{2}}$
$| {P}_{i} - {P}_{i - 1} | = \sqrt{\left({\left[f ' \left({x}_{i}^{'}\right)\right]}^{2} + 1\right) \delta {x}^{2}}$
$| {P}_{i} - {P}_{i - 1} | = \sqrt{\left(1 + {\left[f ' \left({x}_{i}^{'}\right)\right]}^{2}\right)} \delta x$
We can now use this new distance definition for our points in our summation.
$L = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \sqrt{\left(1 + {\left[f ' \left({x}_{i}^{'}\right)\right]}^{2}\right)} \delta x$
Sums are nice, but integrals are nicer for continuous circumstances! It's easy to just write this as a definite integral since both integrals and sums are "summation" tools. In the integral, we can drop our sum index as well.
$L = {\int}_{a}^{b} \sqrt{\left(1 + {\left[f ' \left(x\right)\right]}^{2}\right)} \delta x$
Writing this a little bit more typically yields
$\textcolor{b l u e}{L = {\int}_{a}^{b} \sqrt{\left(1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}\right)} \mathrm{dx}}$
We have arrived at our result! In general, the length is usually defined for a differential of arclength $\mathrm{ds}$
$L = {\int}_{a}^{b} \mathrm{ds}$
where $\mathrm{ds}$ is defined accordingly for whatever type of coordinate system you are working in. However, I wanted the explanation to be clearer, so I just chose Cartesian ones for simplicity. You could use polar coordinates or spherical coordinates as well by simply making the necessary substitutions.
In general, you need to take the derivative of the function defining your curve to substitute into the integral. Then the trick is to find a way (usually) to try and get a perfect square inside the square root to simplify the integral and find your solution. It varies for every type of curve.
Let me know if you have any further questions in the comments! |
# Multiples of 19
Created by: Team Maths - Examples.com, Last Updated: May 23, 2024
## Multiples of 19
Multiples of 19 are numbers that can be obtained by multiplying 19 with any integer. These numbers are part of the broader category of integers and follow the basic principles of multiplication. In mathematics, multiples are closely related to factors and divisors, as they represent the product of a given number and an integer. Understanding multiples of 19 helps in various mathematical operations, including finding common multiples and solving problems related to divisibility.
## What are Multiples of 19?
Multiples of 19 are numbers that result from multiplying 19 by any integer (e.g., 19, 38, 57, 76, etc.). These multiples are part of the set of integers and follow the principle of multiplication. They are used in various mathematical contexts, such as finding common multiples and understanding divisibility.
Prime Factorization of 19: 19 = 1 × 19 First 10 multiples of 19: 19, 38, 57, 76, 95, 114, 133, 152, 171, 190.
## For example, 38, 76, 114 and 171 are all multiples of 19, 183 is not a multiple of 19 for the following reasons:
! Here’s the updated table including the remainder:
Table of 19
## Important Notes
• Definition: Multiples of 19 are the products obtained by multiplying 19 with any integer.
• Sequence: The sequence of multiples of 19 starts as 19, 38, 57, 76, 95, 114, and continues infinitely.
• Formula: A multiple of 19 can be represented as 19×n, where n is any integer.
• Common Multiples: Multiples of 19 are used to find common multiples with other numbers, aiding in solving least common multiple (LCM) problems.
• Divisibility: A number is a multiple of 19 if it is divisible by 19 without leaving any remainder.
• Applications: Understanding multiples of 19 is essential in arithmetic operations, algebra, and number theory.
• Examples: Some examples of multiples of 19 include 38 (19×2), 95 (19×5), and 152 (19×8).
## Examples on Multiples of 19
19: Calculation: 19×1=19
38: Calculation: 19×2=38
57: Calculation: 19×3=57
76: Calculation: 19×4=76
95: Calculation: 19×5 = 95
114: Calculation: 19×6 = 114
133: Calculation: 19×7 = 133
152: Calculation: 19×8 = 152
171: Calculation: 19×9 = 171
190: Calculation: 19×10 = 19
### Example: 200
• Calculation: 200÷19 = 10 remainder 10
• Reason: 200 is not a multiple of 19 because 200÷19 does not result in an integer (remainder 10).
### Example: 217
• Calculation: 217÷19 = 11 remainder 8
• Reason: 217 is not a multiple of 19 because 217÷19 does not result in an integer (remainder 8).
## What is a multiple of 19?
A multiple of 19 is any number that can be expressed as 19×n, where n is an integer. Examples include 19, 38, 57, 76, and so on.
## How do you find the 5th multiple of 19?
To find the 5th multiple of 19, multiply 19 by 5: 19×5 = 95
So, the 5th multiple of 19 is 95.
## Is 114 a multiple of 19?
No, 114 is not a multiple of 19. When you divide 114 by 19, the result is not an integer:
114÷19 = 6 (remainder 0, not an exact multiple)
## What are the first five multiples of 19?
The first five multiples of 19 are 19, 38, 57, 76, and 95.
## How can you verify if a number is a multiple of 19?
To verify if a number is a multiple of 19, divide the number by 19. If the result is an integer with no remainder, it is a multiple. For example, 95 divided by 19 equals 5, so 95 is a multiple of 19.
## Is 190 a multiple of 19?
Yes, 190 is a multiple of 19 because: 19×10 = 190
So, the 10th multiple of 19 is 190.
## How are multiples of 19 used in real-life scenarios?
Multiples of 19 can be used in scenarios like scheduling, where events might occur every 19 days, or in financial contexts, where payments might be structured around multiples of 19 units.
## Is 209 a multiple of 19?
No, 209 is not a multiple of 19. When you divide 209 by 19, the result is not an integer: 209÷19 = 11
## What is the relationship between multiples of 19 and factors of 19?
Multiples of 19 are numbers that can be divided by 19 without a remainder, while factors of 19 are numbers that can multiply together to give 19. Since 19 is a prime number, its only factors are 1 and 19 itself.
## What is the 10th multiple of 19?
The 10th multiple of 19 is found by multiplying 19 by 10: 19×10 = 190
So, the 10th multiple of 19 is 190.
Text prompt |
2D Shapes: Defnition, Example, Properties | Class 3 | Learning Concepts
Geometry
2-Dimensional Shapes for Class 3 Math
This learning concept will help the students to recall the 2-dimensional shapes of geometry. Also, the students will get to know about open and closed figures.
In this learning concept, students will learn to
• Identify the open figure and closed figure
• Classify circle, square, triangle, rectangle
• Define triangle, circle, square, rectangle
The learning concept is explained to class 3 students with examples, illustrations, and a concept map. At the end of the page, two printable worksheets for class 3 with solutions are attached for the students.
Download the worksheets of 2-dimensional shapes and solutions to assess our knowledge of the concept.
What Is a Two-Dimensional Shape?
In geometry, 2-D shapes are completely flat and have only two dimensions – length and width. They do not have any thickness and can be measured only by the two dimensions. There are different types of 2-D shapes, which are open figures and closed figures.
Open and Closed Figures
Open Figure:
Open figures do not begin and end at the same point.
Closed Figure
Closed figures begin and end and the same point.
In our surroundings, we could see some different types of closed figures commonly. Here we discuss some of these closed figures.
These are triangles, squares, rectangles and circles.
What Is Triangle?
A triangle is a 2D shape with three sides and three corners.
Traffic signs, pizza, sandwiches etc are in the shape of a triangle.
What Is Rectangle?
A rectangle is a 2D shape with four sides and four corners.
A blackboard, door, brick etc are in the shape of a rectangle.
What Is Square?
A square is a 2-D shape with four sides and four corners. All sides of squares are equal.
The chessboard, bread etc are in the shape of a square.
What Is Circle?
A circle is a closed 2-D shape made up of a curved line with no sides and no corners. A circle has various parts like centre, radius, diameter, circumference, and so on.
In real life coins, plates, pizza, etc. are examples of the circle.
• - |
Categories:
## NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1
### Understanding Elementary Shapes Class 6 Ex 5.1
Ex 5.1 Class 6 Maths Question 1.
What is the disadvantage in comparing line segment by metre observation?
Solution:
Comparing the lengths of two line segments simply by ‘observation’ may not be accurate. So we use divider to compare the length of the given line segments.
Ex 5.1 Class 6 Maths Question 2.
Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Solution:
Measuring the length of a line segment using a ruler, we may have the following errors:
(i) Thickness of the ruler
(ii) Angular viewing
These errors can be eradicated by using the divider. So, it is better to use a divider than a ruler, while measuring the length of a line segment.
Ex 5.1 Class 6 Maths Question 3.
Draw any line segment, say (overline { AB }). Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line such AC + CB = AB, then we can be sure that C lies between A and B]
Solution:
Let us consider
A, B and C such that C lies between A and B and AB = 7 cm.
AC = 3 cm, CB = 4 cm.
∴ AC + CB = 3 cm + 4 cm = 7 cm.
But, AB = 7 cm.
So, AB = AC + CB.
Ex 5.1 Class 6 Maths Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Solution:
We have, AB = 5 cm; BC = 3 cm
∴ AB + BC = 5 + 3 = 8 cm
But, AC = 8 cm
Hence, B lies between A and C.
Ex 5.1 Class 6 Maths Question 5.
Verify, whether D is the mid point of (overline { AG }) .
Solution:
From the given figure, we have
AG = 7 cm – 1 cm = 6 cm
AD = 4 cm – 1 cm = 3 cm
and DG = 7 cm – 4 cm = 3 cm
∴ AG = AD + DG.
Hence, D is the mid point of (overline { AG }).
Ex 5.1 Class 6 Maths Question 6.
If B is the mid point of (overline { AC }) and C is the mid point of (overline { BD}) , where A, B, C, D lie on a straight line, say why AB = CD?
Solution:
We have
B is the mid point of (overline { AC }) .
∴ AB = BC …(i)
C is the mid-point of (overline { BD }) .
BC = CD
From Eq.(i) and (ii), We have
AB = CD
Ex 5.1 Class 6 Maths Question 7.
Draw five triangles and measure their sides. Check in each case, if the sum of the length of any two sides is always less than the third side.
Solution:
Case I. In ∆ABC
Let AB = 2.5 cm
BC = 4.8 cm
and AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
= 7.3 cm
Since, 7.3 > 5.2
So, AB + BC > AC
Hence, sum of any two sides of a triangle is greater than the third side.
Case II. In ∆PQR,
Let PQ = 2 cm
QR = 2.5 cm
and PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm = 4.5 cm
Since, 4.5 > 3.5
So, PQ + QR > PR
Hence, sum of any two sides of a triangle is greater than the third side.
Case III. In ∆XYZ,
Let XY = 5 cm
YZ = 3 cm
and ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
= 8 cm
Since, 8 > 6.8
So, XY + YZ > ZX
Hence, the sum of any two sides of a triangle is greater than the third side.
Case IV. In ∆MNS,
Let MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
and MN + NS = 2.7 cm + 4 cm = 6.7 cm
Since, 6.7 >4.7
So, MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.
Case V. In ∆KLM,
Let KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
and KL + LM = 3.5 cm + 3.5 cm = 7 cm
7 cm > 3.5 cm
Solution:
(i) For one-fourth revolution, we have
So, KL + LM > KM
Hence, the sum of any two sides of a triangle is greater than the third side.
Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.
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### Understanding Elementary Shapes Class 6 Ex 5.2
Ex 5.2 Class 6 Maths Question 1.
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Solution:
(a) 3 to 9
9 – 3 = 6 ÷ 12 = (frac { 1 }{ 2 }) of a revolution
(b) 4 to 7
7 – 4 = 3 ÷ 12 = (frac { 1 }{ 4 }) of a revolution
(c) 7 to 10
10 – 7 = 3 ÷ 12 = (frac { 1 }{ 4 }) of a revolution
(d) 12 to 9 i.e., 0 to 9
9 – 0 = 9 ÷ 12 = (frac { 3 }{ 4 }) of a revolution
(e) 1 to 10
10 – 1 = 9 ÷ 12 = (frac { 3 }{ 4 }) of a revolution
(f) 6 to 3 i.e., 6 to 12 and then 12 to 3
6 to 12 = 12 – 6 = 6 and 12 to 3 = 0 to 3 = 3 – 0 = 3
6 + 3 = 9 ÷ 12 = (frac { 3 }{ 4 }) of a revolution
Ex 5.2 Class 6 Maths Question 2.
Where will the hand of a clock stop if it
(a) starts at 12 and makes (frac { 1 }{ 2 }) of a revolution, clockwise?
(b) starts at 2 and makes (frac { 1 }{ 2 }) of a revolution, clockwise?
(c) starts at 5 and makes (frac { 1 }{ 2 }) of a revolution, clockwise?
(d) starts at 5 and makes (frac { 1 }{ 2 }) of a revolution, clockwise?
Solution:
(a) Starting from 12 and making (frac { 1 }{ 2 }) of a revolution, the clock hand stops at 6.
(b) Starting from 2 and making (frac { 1 }{ 2 }) of a revolution, the clock hand stops at 8.
(c) Starting from 5 and making (frac { 1 }{ 2 }) of a revolution, the clock hand stops at 8.
(d) Starting from 5 and making (frac { 1 }{ 2 }) of a revolution, the clock hand stops at 2.
Ex 5.2 Class 6 Maths Question 3.
Which direction will you face if you start facing
(a) east and make (frac { 1 }{ 2 }) of a revolution clockwise? z
(b ) east and make (1frac { 1 }{ 2 }) of a revolution clockwise? z
(c) west and make (frac { 3 }{ 4 }) of a revolution anticlockwise?
(d) south and make one full revolution? (Should we specify clockwise or anticlockwise for this last question? Why not?)
Solution:
Taking one full revolution we will reach back to the original (starting) position. Therefore, it make no difference whether we turn clockwise or anticlockwise.
Ex 5.2 Class 6 Maths Question 4.
What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Solution:
(a) If we start from east and reach at north (turning clockwise) (frac { 3 }{ 4 }) of a revolution is required.
(b) If we start from south turning clockwise to face east, (frac { 3 }{ 4 }) of a revolution is required.
(c) If we start from west turning clockwise to face east, (frac { 1 }{ 2 }) of a revolution is required.
Ex 5.2 Class 6 Maths Question 5.
Find the number of right angles turned through by the hour hand of a clock when it goes from
(a)3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 6
Solution:
(a) 3 to 6
Starting from 3 to 6, the hour hand turns through 1 right angle.
(b) 2 to 8
Starting from 2 to 8, the hour hand turns through 2 right angles.
(c) 5 to 11
Starting from 5 to 11, the hour hand turns through 2 right angles.
(d) 10 to 1
Starting from 10 to 1, the hour hand turns through 1 right angle.
(e) 12 to 9
Starting from 12 to 9, the hour hand turns through 3 right angles.
(f) 12 to 6
Starting from 12 to 6, the hour hand turns through 2 right angles.
Ex 5.2 Class 6 Maths Question 6.
How many right angles do you make if you start facing
(a) south and turn clockwise to west?
(b) north and turn anticlockwise to east?
(c) west and turn to west?
(d) south and turn to north?
Solution:
Ex 5.2 Class 6 Maths Question 7.
Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
Solution:
(a) Starting from 6 and turning through 1 right angle, the hour hand stops at 9.
(b) Starting from 8 and turning through 2 right angles, the hour hand stops at 2.
(c) Starting from 10 and turning through 3 right angles, the hour hand stops at 7.
(b) Starting from 7 and turning through 2 right angles, the hour hand stops at 7.
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.3
Ex 5.3 Class 6 Maths Question 1.
Match the following:
(i) Straight angle (a) Less than one-fourth of a revolution.
(ii) Right angle (b) More than half a revolution.
(iii) Acute angle (c) Half of a revolution.
(iv) Obtuse angle (d) One-fourth of a revolution.
(v) Reflex angle (e) Between (frac { 1 }{ 4 }) and (frac { 1 }{ 2 }) of a revolution.
– (f) One complete revolution.
Solution:
(i) Straight angle ↔ (c) Half of a revolution.
(ii) Right angle ↔ (d) One-fourth of a revolution.
(iii) Acute angle ↔ (a) Less than one-fourth of a revolution.
(iv) Obtuse angle ↔ (e) Between (frac { 1 }{ 4 }) and (frac { 1 }{ 2 }) of a revolution.
(v) Reflex angle ↔ (f) One complete revolution, right, acute, obtuse or reflex.
Ex 5.3 Class 6 Maths Question 2.
Classify each one of the following angles
Solution:
(a) Acute angle
(b) Obtuse angle
(c) Right angle
(d) Reflex angle
(e) Straight angle
(f) Acute angle
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.4
Ex 5.4 Class 6 Maths Question 1.
What is the measure of (i) a right angle (ii) a straight angle?
Solution:
(i) Measure of a right angle = 90°
(ii) Measure of a straight angle = 180°
Ex 5.4 Class 6 Maths Question 2.
Say True or False:
(a) The measure of an acute angle < 90°
(b) The measure of an obtuse angle < 90°
(c) The measure of a reflex angle > 180°
(d) The measure of one complete revolution = 360°
(e) If m ∠A = 53° and ∠B = 35°, then m∠A > m∠B.
Solution:
(a) True
(b) False
(c) True
(d) True
(e) True
Ex 5.4 Class 6 Maths Question 3.
Write down the measures of
(a) some acute angles
(b) some obtuse angles
Solution:
(a) 25°, 63° and 72° are acute angles.
(b) 105°, 120° and 135° are obtuse angles.
Ex 5.4 Class 6 Maths Question 4.
Measure the angles given below using the protractor and write down the measure.
Solution:
(a) 45°
(b) 125°
(c) 90°
(d) ∠1 = 60°, ∠2 = 90°, ∠3 = 125°
Ex 5.4 Class 6 Maths Question 5.
Which angle has a large measure? First estimate and then measure.
Measure of Angle A =
Measure of Angle B =
Solution:
Measure of Angle A = 40°
Measure of Angle B = 60°.
Ex 5.4 Class 6 Maths Question 6.
From these two angles which has large measure? Estimate and then confirm by measuring them.
Solution:
The opening of angle (b) is more than angle (a).
∴ Measure of angle (a) = 45°
and the measure of angle (b) = 60°
Ex 5.4 Class 6 Maths Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ……… .
(b) An angle whose measure is greater than that of a right angle is ……… .
(c) An angle whose measure is the sum of the measures of two right angles is ……… .
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is ……… .
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ……… .
Solution:
(a) acute
(b) obtuse
(c) straight
(d) acute
(e) obtuse
Ex 5.4 Class 6 Maths Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eyes and than find the actual measure with a protractor).
Solution:
(a) Measure of the angle = 40°
(b) Measure of the angle = 130°
(c) Measure of the angle = 65°
(d) Measure of the angle = 135°.
Ex 5.4 Class 6 Maths Question 9.
Find the angle measure between the hands of the clock in each figure:
Solution:
(i) The angle between hour hand and minute hand of a clock at 9.00 a.m = 90°
(ii) The angle between the hour hand and minute hand of a clock at 1.00 p.m = 30°
(iii) The angle between the hour hand and minute hand of a clock at 6.00 p.m = 180°.
Ex 5.4 Class 6 Maths Question 10.
Investigate: In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?
Solution:
It is an activity. So try it yourself.
Ex 5.4 Class 6 Maths Question 11.
Measure and classify each angle:
Angle Measure Type ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB
Solution:
Angle Measure Type ∠AOB 40° Acute angle ∠AOC 125° Obtuse angle ∠BOC 85° Acute angle ∠DOC 95° Obtuse angle ∠DOA 140° Obtuse angle ∠DOB 180° Straight angle
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.5
Ex 5.5 Class 6 Maths Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming a letter ‘L’.
(d) The letter V.
Solution:
(a) Yes, the adjacent edges of a table top are the models of perpendicular lines.
(b) No, the lines of a railway tracks are parallel to each other. So they are not a model for perpendicular lines.
(c) Yes, the two line segments of‘L’ are the model for perpendicular lines.
(d) No, the two line segments of ‘V’ are not a model for perpendicular lines.
Ex 5.5 Class 6 Maths Question 2.
Let (overline { PQ }) be the perpendicular to the line segment (overline { XY }) . Let (overline { PQ }) and (overline { XY }) intersect at in the point A. What is the measure of ∠PAY?
Solution:
Since (overline { PQ }) ⊥ XY
∴ ∠PAY = 90°
Ex 5.5 Class 6 Maths Question 3.
There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution:
The figures of the two set-squares are given below:
The measure angles of triangle (a) are : 30°, 60° and 90°.
The measure angles of triangle (b) are 45°, 45° and 90°.
Yes, they have a common angle of measure 90°.
Ex 5.5 Class 6 Maths Question 4.
Study the diagram. The line l is perpendicular to line m.
(a) Is CE = EG?
(b) Does PE bisects CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Solution:
(a) Yes,
Since, CE = 2 units and EG = 2 units
Hence, CE = EG.
(b) Yes, PE bisects CG
(c) Required line segments for which PE is perpendicular bisector are: (overline { BG }) and (overline { DF })
(d) (i) True (ii) True (iii) True
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.6
Ex 5.6 Class 6 Maths Question 1.
Name the types of following triangles:
(а) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ∆ABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
(d) ∆DEF with m∠D = 90°
(e) ∆XYZ with m∠Y = 90° and XY = YZ.
(f) ∆LMN with m∠L = 30° m∠M = 70° and m∠N = 80°.
Solution:
(a) Lengths of the sides of a triangle are given as: 7 cm, 8 cm and 9 cm.
Since, all sides of the given triangle are different.
Hence, it is a Scalene triangle.
(b) Given that: AB = 8.7 cm, AC = 7 cm and BC = 6 cm
Here AB ≠ AC ≠ BC Hence, ∆ABC is Scalene triangle.
(c) Given that: PQ = QR = PR = 5 cm
Since all sides are equal.
Hence, it is an equilateral triangle.
(d) Given that: In ∆DEF, m∠D = 90°
Hence it is a right angled triangle.
(e) Given that: In ∆XYZ, m∠Y = 90° and XY = YZ
Hence it is a right angled triangle.
(f) Given that: ∆LMN, m∠L = 30°, m ∠M = 70° and m∠N = 80°.
Hence it is an acute angled triangle.
Ex 5.6 Class 6 Maths Question 2.
Match the following:
Measure of triangle Type of triangle
(i) 3 sides of equal length (a) Scalene
(ii) 2 sides of equal length (b) Isosceles right angled
(iii) All sides are of different length (c) Obtuse angled
(iv) 3 acute angles (d) Right angled
(v) 1 right angle (e) Equilateral
(vi) 1 obtuse angle (f) Acute angled
(vii) 1 right angle with two sides of equal length (g) Isosceles
Solution:
(i) ↔ (e)
(ii) ↔ (g)
(iii) ↔ (a)
(iv) ↔ (f)
(v) ↔ (d)
(vi) ↔ (c)
(vii) ↔ (b)
Ex 5.6 Class 6 Maths Question 3.
Name each of the following triangles in two different ways: (You may judge the nature of the angle by observation)
Solution:
(a) (i) Acute angled triangle
(ii) Isosceles triangle
(b) (i) Right angled triangle
(ii) Scalene triangle
(c) (i) Obtuse angled triangle
(ii) Isosceles triangle
(d) (i) Right angled triangle
(ii) Isosceles triangle
(e) (i) Acute angled triangle
(ii) Equilateral triangle
(f) (i) Obtuse angled triangle
(ii) Scalene triangle.
Ex 5.6 Class 6 Maths Question 4.
Try to construct triangles using matchsticks. Some are shown here. Can you make a triangle with
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case.
If you cannot make a triangle, give of reasons for it.
Solution:
(a) Yes, we can make an equilateral triangle with 3 matchsticks.
(b) No, we cannot make a triangle with 4 matchsticks.
(c) Yes, we can make an isosceles triangle with five matchsticks.
(d) Yes, we can make an equilateral triangle with 6 matchsticks.
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.7
Ex 5.7 Class 6 Maths Question 1.
Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solution:
(a) True
(b) True
(c) True
(d) True
(e) False
(f) False
Ex 5.7 Class 6 Maths Question 2.
Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Square, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solution:
(a) A square has all the properties as that of rectangle. So, it is a special rectangle.
(b) A rectangle has the same properties as that of parallelogram. So, it is a special parallelogram.
(c) A square has the same properties as that of a rhombus. So, it is a special rhombus.
(d) Square, rectangles and parallelogram are all quadrilateral as they are all enclosed by four sides.
Ex 5.7 Class 6 Maths Question 3.
A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solution:
Square is only the regular quadrilateral with equal sides and equal angles.
Therefore, square is a regular quadrilateral.
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.8
Ex 5.8 Class 6 Maths Question 1.
Examine whether the following are polygons. If any one among them is not, say why?
Solution:
(a) The given figure is not closed. Therefore, it is not a polygon.
(b) The given figure is a polygon.
(c) The given figure is not a polygon because every polygon is enclosed with line segments.
(d) The given figure is not a polygon because it is enclosed by an arc and two line segments.
Ex 5.8 Class 6 Maths Question 2.
Name the polygon.
Make two more examples of each of these.
Solution:
(a) A quadrilateral Examples:
(b) A Triangle
Examples:
(c) A Pantagon
Examples:
(d) A Octagon
Examples:
Ex 5.8 Class 6 Maths Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution:
ABCDEF is a rough sketch of a regular hexagon. If we join any three vertices like D, A and B, we get a scalene triangle DAB.
But if we join the alternate vertices, we get an equilateral triangle EAC.
Ex 5.8 Class 6 Maths Question 4.
Draw a rough sketch of a regular octagon. (Using squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution:
ABCDEFGH is a rough sketch of regular octagon. GHCD is the rectangle formed by joining the four vertices of the given octagon.
Ex 5.8 Class 6 Maths Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution:
A B C D E is the rough sketch of a pentagon.
By joining its any two vertices, we get, the following diagonals.
(overline { AD }) , (overline { AC }) , (overline { BE }) , (overline { BD }) and (overline { CE })
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.9
Ex 5.9 Class 6 Maths Question 1.
Match the following:
Give two examples of each shape.
Solution:
(a) 4 ↔ (ii)
Examples:
(i) An ice-cream cone
(ii) Birthday cap
(b) ↔ (iv)
Examples:
(i) Tennis ball
(ii) Cricket ball
(c) ↔ (v)
Examples:
(i) A road roller
(ii) A lawn roller
(d) ↔ (iii)
Examples:
(i) Math book
(ii) A brick
(e) ↔ (i)
Examples:
(i) A diamond
(ii) Egypt-Pyramids
Ex 5.9 Class 6 Maths Question 2.
What shape is
(a) Your instrument box?
(b) A brick?
(b) A matchbox?
(e) A sweet laddu?
Solution:
(a) Shape of instrument box is cuboid.
(b) Shape of a brick is cuboid.
(c) Shape of a matchbox is cuboid.
(d) Shape of a road-roller is cylinder.
(e) Shape of a sweet laddu is sphere.
Move to Top → Understanding Elementary Shapes
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# Objective 1: To multiply monomials. Objective 2: To divide monomials and simplify expressions with negative exponents.
## Presentation on theme: "Objective 1: To multiply monomials. Objective 2: To divide monomials and simplify expressions with negative exponents."— Presentation transcript:
Objective 1: To multiply monomials. Objective 2: To divide monomials and simplify expressions with negative exponents.
To multiply monomials
Exponent Review Base Exponent
It means to use 2 as a factor 5 times. What does it mean?
Exponent Review Base Exponent
What does it mean?
Multiplying with exponents When you multiply, add the exponents.
Example 1 What’s the exponent for this x?
Example 1 Multiply the coefficients first.
Example 1
Example 2
Raising a power to a power When raising a power to another power, multiply the exponents.
Raising a power to a power When raising a power to another power, multiply the exponents. Take everything in parentheses and raise it to the 2 nd power.
Raising a power to a power When raising a power to another power, multiply the exponents. Use Distributive Property.
Example 3 Simplify the coefficient.
Example 3
Example 4 What’s the exponent for this x?
Example 4 Simplify this expression first because it has an outside exponent. Take everything in parentheses and raise it to the 3 rd power.
Example 4 Now, there are no outside exponents. We can multiply the coefficients Then multiply the x’s Then multiply the y’s
Example 4
Example 5 (skip) Start outside the brackets Multiply the two outside exponents.
Example 5 Simplify the outside exponent. Raise everything inside the parentheses to the 6 th power.
Example 5
To divide monomials and simplify negative exponents.
Dividing with exponents When you divide, subtract the exponents.
Example 6 What’s the exponent for this b?
Example 6
Example 7 Raise everything in the parentheses to the 2 nd power.
Example 7 Subtract the exponents.
Example 7
Example 8 Raise everything in the parentheses to the 3 rd power.
Example 8 Simplify the denominator.
Example 8
Look At This
This is another rule.
Zero as exponent Anything raised to zero power equals 1.
Review: Whole Numbers Any whole number can be placed on top of 1.
Review: Whole Numbers Any whole number can be placed on top of 1.
Review: Whole Numbers Any whole number can be placed on top of 1.
Review: Whole Numbers Any whole number can be placed on top of 1.
Review: Whole Numbers Any whole number can be placed on top of 1.
Review: Whole Numbers Any whole number can be placed on top of 1.
Fractions There are 2 parts of a fraction.
Negative Exponents When you see negative exponents, think MOVE & CHANGE Move the base from top to bottom or bottom to top. Change the exponent to a positive number.
Negative Exponents MOVE & CHANGE
Negative Exponents MOVE & CHANGE
Negative Exponents MOVE & CHANGE y does not have negative exponent. It stays where it is. Nothing is left on top. We know there is an invisible 1 there.
Negative Exponents
MOVE & CHANGE
Negative Exponents MOVE & CHANGE
Negative Exponents MOVE & CHANGE
Negative Exponents MOVE & CHANGE
Example 9 Raise everything in the parentheses to the negative 2 nd power.
Example 9 Move the negative exponents and change to positive exponents.
Example 9
Example 10
Example 11 Move any negative exponents and change to positive.
Example 11
Example 12 (skip) Subtract the exponents.
Example 12 Put under 1 and change exponent to positive.
Example 12 The variable stays where the bigger exponent was.
IMPORTANT! Your final answer can NOT have any negative exponents. Remember to move all negative exponents and change them to positives.
All Rules in Symbolic Form
Example 13
Move negative exponents and change to positive.
Example 13
Example 14
Move negative exponents and change to positive.
Example 14
Download ppt "Objective 1: To multiply monomials. Objective 2: To divide monomials and simplify expressions with negative exponents."
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# The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India.
Question:
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures.
Solution:
Modal class is (30-35) and its frequency is 10.
So, $\ell=30, \mathrm{f}_{\mathrm{m}}=10, \mathrm{f}_{1}=9, \mathrm{f}_{2}=3, \mathrm{~h}=5$.
Mode $=\ell+\left\{\frac{\mathbf{f}_{\mathbf{m}}-\mathbf{f}}{\mathbf{2 f}_{\mathbf{m}}-\mathbf{f}-\mathbf{f}_{2}}\right\} \times \mathbf{h}$
$=30+\left\{\frac{\mathbf{1 0}-\mathbf{9}}{\mathbf{2 0}-\mathbf{9}-\mathbf{3}}\right\} \times 5=30+\frac{\mathbf{5}}{\mathbf{8}}=30.6$
$\mathrm{a}=32.5, \mathrm{~h}=5, \mathrm{n}=35$ and $\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}=-23$
By step-deviation method,
Mean $=\mathrm{a}+\mathrm{h} \times \frac{\mathbf{1}}{\mathbf{n}} \times \Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}$
$=32.5+5 \times \frac{\mathbf{1}}{\mathbf{3 5}} \times(-23)$
$=32.5-\frac{\mathbf{2 3}}{\mathbf{7}}=32.5-3.3=29.2$
Hence, Mode = 30.6 and Mean = 29.2. We conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average, the ratio is 29.2. |
# Real Mathematics – Numbers #9
Magic
There are such questions, even though they seem useless (or unnecessary) to others they can help one to get much better at the number theory. One of the main reasons why some people label those mathematics questions as “useless” is that they are actually scared of the question (or of mathematics).
What those people really feel is like the feeling you get whenever you are walking down on an unknown street, city or country. Those people are away from their comfort zone and they would never feel as relaxed as they feel at home as long as they don’t “try”. Defining mathematics questions as unnecessary is in fact a way of expressing the fear of mathematics.
In that case trying and/or striving are essential if you’d like to get better at mathematics. This is how you could find your own methods and accomplish things that will shock others. There is another matter I’d like to point out: If the trick is obvious no one would watch your show second time. Magic is beautiful when people don’t understand what you are doing.
Equal Sums
Assume that we have ten different numbers between 1 and 50. Our goal is to divide those ten numbers into two groups of five such that their summations will be equal to each other.
Example 1: My random ten numbers are: 2, 12, 23, 24, 30, 33, 39, 41, 44 and 48.
Goal is to divide these numbers into two groups of five such that their summations are the same.
I managed to do so after a short amount of time:
48+41+33+24+2 = 148 = 44+39+30+23+12
Maybe you think that I choose those numbers on purpose. This is why I asked some of my friends to send me random ten numbers from 1 to 50.
Example 2: Random numbers: 34, 21, 7, 42, 22, 33, 13, 27, 20 and 19.
After a minute or so I found the following:
34+33+13+20+19 = 119 = 21+22+27+42+7
Q: How do I do it? Can you speculate about what kind of method I might be using?
Example 3: I got these numbers from another friend: 3, 9, 13, 19, 21, 27, 36, 33, 39 and 45.
Example 4: And these are the numbers I received from one last friend: 7, 10, 11, 14, 21, 23, 30, 33, 43 and 49.
For the examples 3 and 4, I found out that there can’t be such groups and I came to this conclusion in a matter of second.
One wonders…
How did I decide so quickly?
Hint: Take a look at how many of the numbers are odd or even.
M. Serkan Kalaycıoğlu |
# Basic Steps to Stairway Construction Basic Steps to Stairway Construction
Stairway construction is one of those things that many do it youselfers struggle with. If you are remodeling a two story home, planning an elevated deck, or just want to build a set of front steps, you will be presented with the need for stairway construction. The good news is that once you have done a stairway, you can use the same basic steps to build any stairway that you will need. Here are a few basic steps to stairway construction that you can use if you are struggling with this type of project.
#### Know Basic Math
While you will not have to know advanced calculus, there is some basic math that you will need to do when building a set of stairs. In order to accurately build a strong set of stairs you will need to know some basic math in order to get the right measurements for the rise and run of the steps themselves.
#### Know the Building Codes
Before you can build any type of staircase you will need to check with the national, and local building codes that pertain to the building of the staircase. Ultimately, your plans will have to be centered around these building codes. When finding out these codes you will need to have the measurements as well as the type of stairway that you want to put in place.
#### Take Measurements for Stairway Construction
There are two measurements that you will really need to pay attention to. They are the rise of the stairs and the total run of the staircase. These two measurements are going to determine the ultimate height of the risers and steps themselves. This is where the basic math will come into play.
#### Basic Calculations
Knowing what the different building codes are going to be is what will determine the different calculations. However, for this example we will use the code that specifies a 10 inch stair tread and a 7 3/4 inch rise. If the total rise of your stairway is going to be 118" from finished floor to top of the landing, you will divide this number by 7.5. When you do this you have an answer of 15.73. Round that number to 16 and that is the number of steps that you will have in your staircase construction. Now you divide 118" and 16 steps to come up with a total rise between each step of 7 3/8 inches.
#### Basic Materials
Once you have the calculations down you will then need to concentrate on the actual materials. Typically the side stringers are going to be made out of 2x8 or 2x10 lumber. This will depend on the amount of space you have and the overall length of the staircase. The steps themselves are 1x10 lumber with the risers being 1x8 that are cut to fit.
#### Cutting Stringer for Treads and Risers
Once the measurements are taken you will then need to mark them out on your stringers. Use a carpenter's square and make sure of all of the calculations with each one. Mark off the rise and run of your stringers by setting the carpenter's square measurements. This will make quick work of this. Once the measurements are marked, you can cut out the rise and run on the stringers. |
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