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## Create your own teaching resources
Teacherbot 25 Aug, 13:39:32
Here is an example of a problem that involves the segment addition postulate, midpoint, and knowing the total length:
Problem: In a line segment AB, point M is the midpoint. The length of segment AB is 12 units. Point N is located on segment AB such that AN is 3 units longer than MN. Find the lengths of AN and NB.
Solution: Let’s assume that the length of MN is x units. Since M is the midpoint, the length of AM is also x units.
According to the segment addition postulate, the total length of AB is equal to the sum of the lengths of AN and NB. Therefore, we have:
AN + NB = AB
Since AB is given as 12 units, we can substitute it into the equation:
AN + NB = 12
We also know that AN is 3 units longer than MN. Since MN is x units, AN can be expressed as (x + 3) units.
Now, we can rewrite the equation using the given information:
(x + 3) + NB = 12
To find the value of NB, we need to isolate it on one side of the equation. Subtracting (x + 3) from both sides, we get:
NB = 12 - (x + 3) NB = 12 - x - 3 NB = 9 - x
So, the length of NB is 9 - x units.
Now, we can substitute the expression for NB into the equation AN + NB = 12:
AN + (9 - x) = 12
To find the value of AN, we need to isolate it on one side of the equation. Subtracting (9 - x) from both sides, we get:
AN = 12 - (9 - x) AN = 12 - 9 + x AN = 3 + x
So, the length of AN is 3 + x units.
Therefore, the lengths of AN and NB are 3 + x units and 9 - x units, respectively. |
# Tabular Integration (The tabular method) Examples |Where to use tabular integration by parts method|
Home > Math Article > Tabular Integration Method
Page Contents
## What is Tabular Integration method and when to use?
Tabular integration is a short method for integration to solve the integral problem quickly instead of using the lengthy and tedious process of integration by parts traditional method.
The advantage of the tabular integration method is that it can save huge time in solving the problem and gives the solution fairly accurate than the integration by parts method. This tabular method is fairly quick to memorize and easy to use.
This method is also called the DI method of Integration by parts.
## When can I use the Tabular integration method?
The tabular integration method can be applied to any function which is the product of two expressions, where one of the expressions can be differentiated until it gets zero, and another expression can be integrated simultaneously multiple times.
Suppose a function, f(x) = m(x).n(x), from given two expression, one of the expression, let’s take it m(x) should be differentiate multiple times until it reach to zero, and the another expression n(x) should be integrated simultaneously multiple times.
Note: Tabular method can also be used where neither of the expression differentiation goes to zero. Example – (∫e2x.sin3x dx) or (∫sin3x.cos4x dx), etc.
Take a look at where we can apply the tabular integration method.
• When Integrand is the product of Polynomial times and something that can be repeatedly integrated. (x10.cos x dx)
• Integrand multiple of power function and an exponential function. (∫x².ex dx)
• Integrand multiple of an exponential and trigonometry function. (∫ex.sin x dx)
• Integrand multiple of power function and a trigonometry function. (∫x³.tan x dx)
## How to use or apply the Tabular integration method and its formulas?
Let the understand by taking the example of the integral function
$\int _ { } ^ { } t ^ { 2 } \sin t dt$
Step 1 = First identify the polynomial function or the function which is differential. So, t² is a polynomial function and denote it as F(x)
F(x) = t²
Step 2 = Now take another product left in function which is integrable and denotes it as F(y)
F(y) = sint
Step 3 = Create a table and differentiate F(x) until it reaches 0 and simultaneously integrates F(y).
Step 4 = Now change the sign of every second term of the F(x) function to negative Or Add alternating “+” and “-” signs, starting with “+” on the derivatives side of the table.
### Tabular Integration Method
Now we have, F(x) = t2 and F(y) = sint
F(x) Differentiate F(y) Integration t² sin t 2t -cos t 2 -sin t 0 cos t
Now we will add alternating “+” and “-” signs, starting with “+” on the derivatives side of the table.
F(x) Differentiate F(y) Integration + t² sin t – 2t -cos t + 2 -sin t 0 cos t
Now Most important Steps in the tabular integration method to find the integral of a given function
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
So, $\int _ { } ^ { } t ^ { 2 } \sin t dt$ = t²(-cost) + (-2t)(-sint) + 2cost + C
= -t²cost + 2tsint + 2cost + C Answer
## Tabular Integration Method for solving Definite Integral with limits problems
The definite integral is those integrals that have upper and lower limits or bounds or boundaries. In short, definite integral has start and end values.
Now, How to solve definite integral problems by the Tabular Integration rather than the traditional method and save our time.
### Example 1:
Step 1 = First identify the polynomial function or the function which is differential. So, t³ is a polynomial function and denote it as F(x)
F(x) = t³
Step 2 = Now take another product left in function which is integrable and denotes it as F(y)
F(y) = sin t
Step 3 = Create a table and differentiate F(x) until you reach 0 and simultaneously integrate F(y).
Step 4 = Now change the sign of every second term of the F(x) function to negative Or add alternating “+” and “-” signs, starting with “+”:
Now we have, F(x) = t3 and F(y) = sint
F(x) Derivative Function F(y) Integration Function (+) t³ sin t (-) 3t² -cos t (+) 6t -sin t (-) 6 cos t 0 sin t
• Multiply F(x) with the first integration of F(y)
• Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int _ { 0 } ^ { \pi } t ^ { 3 } \sin t dt$ = t³( -cost ) $\Big| _ 0 ^ \pi$ – 3t² ( -sint ) $\Big| _ 0 ^ \pi$ + 6tcost $\Big| _ 0 ^ \pi$ – 6sint $\Big| _ 0 ^ \pi$
= -t³cost $\Big| _ 0 ^ \pi$ + 3t²sint $\Big| _ 0 ^ \pi$ + 6tcost $\Big| _ 0 ^ \pi$ – 6(sint)$\Big| _ 0 ^ \pi$
= [-π³cosπ – 0] + [3π².0 – 0] + [6πcosπ – o] + [-6(0 – 0]
= -π³(-1) + 6π(-1) + 0
So, Definite integral of the function $\int _ { 0 } ^ { \pi } t ^ { 3 } \sin t dt$ by tabular integration by parts method is π³ – 6π
### Example 2:
Solution: Using same procedure, Make F(x) = t and F(y) = Cos2t
Now construct the table and take appropriate action to solve these definite integral problems by the tabular method
F(x) Derivative Function F(y) Integration Function (+) t Cos2t (-) 1 (1/2)Sin2t (+) 0 -(1/4)Cos2t
Follow these steps for further solving definite integral by the tabular method
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int _ { 0 } ^ { \frac { \pi } { 4 } } t \cos2t dt$ = t/2Sin2t $\Big| _ 0 ^ \frac { \pi } { 4 }$ + 1/4Cos2t $\Big| _ 0 ^ \frac { \pi } { 4 }$
= $[ \frac { \pi } { 8 } \sin \frac { \pi } { 2 } – 0 ]$ + $\frac { 1 } { 4 } [ \cos \frac { \pi } { 2 } – \cos0 ]$
= $\frac { \pi } { 8 } + \frac { 1 } { 4 } ( 0 – 1 )$
= $\frac { \pi – 2 } { 8 }$ Answer
## Tabular Integration Examples with their Solutions
Let’s take a few more examples to understand the tabular integration method completely.
### Example 1:
Solving problems based on exponential and trigonometric functions using Tabular Integration.
Solution: Now you are thinking that there is no function in the above example that comes up 0 after derivating it endlessly. Then how can we use the tabular integration method here?
Don’t worry just follow the good approach.
Assume, F(x) = sin3t and F(y) = e2t
Construct tabular method table for integrating this problem
F(x) Derivative Function F(y) Integration Function Sin(3t) (+) e2t 3cos(3t) (-) 1/2(e2t) -9sin(3t) (+) 1/4(e2t)
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y).
3. Multiply the second derivative of F(x) with the second integration of F(y).
So, $\int _ { } ^ { } e ^ { 2t } \sin3t dt$ = $\frac { 1 } { 2 }$ e2t sin3t – $\frac { 3 } { 4 }$ e2t cos3t – $\frac { 9 } { 4 }$ $\int _ { } ^ { } e ^ { 2t } \sin3t dt$
$\frac { 13 } { 4 }$ $\int _ { } ^ { } e ^ { 2t } \sin3t dt$ = $\frac { 1 } { 2 }$ e2t sin3t – $\frac { 3 } { 4 }$ e2t cos3t + C
$\int _ { } ^ { } e ^ { 2t } \sin3t dt$ = $\frac { 1 } { 13 }$ e2t [ 2sin3t – 3cos3t ] + C Answer
### Example 2:
Solution: let’s give a shot at this integral problem by using the tabular integration method.
Look at the problem and choose a function that will give 0 value after derivating multiple times.
F(x) = t3 – 4t2 + 5t + 32 and F(y) = sint
Construct the table to solve this problem by Tabular integration method
F(x) Derivative function F(y) Integration function (+) t3 -4t2 + 5t + 32 sint (-) 3t2 – 8t + 5 -cost (+) 6t – 8 -sint (-) 6 cost 0 sint
Use same steps
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
So, we have $\int _ { } ^ { } (t ^ { 3 } – 4t ^ { 2 } + 5t + 32)(sint) dt$ = -(t3 – 4t2 + 5t + 32)(cost) + (3t2 – 8t + 5)(sint) + (6t – 8)(cost) – 6sint + C
Simplify form = (3t2 8t 1)sin(t) + (t3 + 4t2 + t40)cos(t) + C Answer
### Example 3:
Solving problems based on power and exponential function using Tabular integration.
Solution: F(x) = t and F(y) = e-t
Construct the table to solve this integral problem by tabular integration method
F(x) Derivative Function F(y) Integration Function (+) t5 e-t (-) 5t4 -e-t (+) 20t³ e-t (-) 60t² -e-t (+) 120t e-t (-) 120 -e-t (+) 0 e-t
Use same steps
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int _ { } ^ { } t ^ { 5 } e ^ { – t } dt$ = t5(-e-t) – 5t4(e-t) + 20t3(-e-t) – 60t2(e-t) + 120t(-e-t) – 120(e-t) + C
$\int _ { } ^ { } t ^ { 5 } e ^ { – t } dt$ = -t5(e-t) – 5t4(e-t) – 20t3(e-t) – 60t2(e-t) – 120t(e-t) – 120(e-t) + C
= (t5 + 5t4 + 20t3 + 60t2 + 120t + 120)et + C Answer
### Example 4:
Solving problem based on two trigonometric function using Tabular integration method
Solution: We know, in the above problem, both function sin3t and cos4t will not get zero regardless of how much time you can differentiate them.
So, take, any of the functions as F(x) and another one as F(y) and apply the Tabular integration method.
So, I assume, F(x) = Sin(3t) and F(y) = Cos(4t) [∴You can choose whatever you want]
Now construct the table for solving this by tabular method
F(x) Derivative Function F(y) Integration Function (+) Sin(3t) Cos(4t) (-) 3Cos(3t) $\frac { 1 } { 4 }$ Sin(4t) (+) -9Sin(3t) $– \frac { 1 } { 16 }$Cos(4t)
Use these steps solve this integral problem
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y).
3. Multiply the second derivative of F(x) with the second integration of F(y).
$\int _ { } ^ { } \sin3t \cos4t dt$ = $\frac { 1 } { 4 }$ Sin(3t)Sin(4t) + $\frac { 3 } { 16 }$ Cos(3t)Cos(4t) + $\frac { 9 } { 16 }$ $\int _ { } ^ { } \sin3t \cos4t dt$
$\frac { 7 } { 16 } \int _ { } ^ { } \sin3t \cos4t dt$ = $\frac { 1 } { 4 }$ Sin(3t)Sin(4t) + $\frac { 3 } { 16 }$ Cos(3t)Cos(4t) + C
$\int _ { } ^ { } \sin3t \cos4tdt$ = $\frac { 16 } { 7 }$ [ $\frac { 1 } { 4 }$ Sin(3t)Sin(4t) + $\frac { 3 } { 16 }$ Cos(3t)Cos(4t) ] + C
= [ $\frac { 4 } { 7 }$ Sin(3t)Sin(4t) + $\frac { 3 } { 7 }$ Cos(3t)Cos(4t) ] + C Answer
### Example 5:
Based on when you have only one function in the problem
Solution: We can solve this problem by assuming as other function 1 and then apply tabular integration method.
So, let F(x) = 80x³ and F(y) = 1
Construct the table to solve this problem by tabular integration method
F(x) Derivative Function F(y) Integration Function (+) 80x³ 1 (-) 240x² x (+) 480x x²/2 (-) 480 x³/6 (+) 0 x4/24
Use these steps to further solve this integral problem.
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
So, $\int _ { } ^ { } 80x ^ { 3 } dx$ = $80x ^ { 4 } – 240 \frac { x ^ { 4 } } { 2 } + 480 \frac { x ^ { 4 } } { 6 } – 480 \frac { x ^ { 4 } } { 24 }$
= $80x ^ { 4 } – 120x ^ { 4 } + 80x ^ { 4 } – 20x ^ { 4 }$
= $20x ^ { 4 } + C$ Answer
### Example 6:
Solution: Let F(x) = x³ and F(y) = e-x
Construct the table to solve this problem by tabular integration method
F(x) Derivative Function F(y) Integration Function (+) x³ e-x (-) 3x² – e-x (+) 6x e-x (-) 6 – e-x (+) 0 e-x
Use these steps to further solve this integral problem.
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int _ { } ^ { } x ^ { 3 } e ^ { – x } dt$ = +x3(-e-x) – 3x2(e-x) + 6x(-e-x) – 6(e-x) + C
= – x3.e-x – 3x2.e-x – 6x.e-x – 6.e-x + C Answer
### Example 7:
Solution: Let F(x) = x and F(y) = (x – 2)3/2
Construct the table to solve this problem by tabular integration method
F(x) Derivative Function F(y) Integration Function (+) x2 (x – 2)3/2 (-) 2x 2/5(x – 2)5/2 (+) 2 4/35(x – 2)7/2 (-) 0 8/315(x – 2)9/2
Use these steps to further solve this integral problem.
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int_{}^{} x^{2}(x – 2)^{\frac{3}{2}}dx$ = $\frac{2}{5}x^{2}(x-2)^{\frac{5}{2}} – \frac{8}{35}x(x-2)^{\frac{7}{2}}+ \frac{16}{315}(x-2)^{\frac{9}{2}} + C$ Answer
### Example 8:
Solution: Let F(x) = 6x and F(y) = cos(2x)
Construct the table to solve this problem by tabular integration
F(x) Derivative Function F(y) Integration Function (+) 6x3 cos(2x) (-) 18x2 sin(2x)/2 (+) 36x -cos(2x)/4 (-) 36 -sin(2x)/8 0 cos(2x)/16
Use these steps to further solve this integral problem.
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int_{}^{} 6x^{3}.cos(2x)dx$ = 3x3sin2x + (9/2)x2cos2x – (9/2)xsin2x – (9/4)cos2x + C Answer
### Example 9:
Solution: Let F(x) = 4x and F(y) = e-2x
Construct the table to solve this problem by tabular integration
F(x) Derivative Function F(y) Integration Function (+) 4x2 e-2x (-) 8x -e-2x/2 (+) 8 e-2x/4 (-) 0 -e-2x/8
Use these steps to further solve this integral problem.
1. Multiply F(x) with the first integration of F(y)
2. Multiply the first derivative of F(x) with the second integration of F(y)………and so on.
$\int_{}^{} 4x^{2}.e^{-2x}dx$ = -2x2e-2x – 2xe-2x – e-2x + C Answer
## FAQ
### What is the Tabular Integration?
The tabular Integration is also called the DI method to solve integration problems quickly by forming three columns, the first one for “Alternative sign”, the second column for “Derivative function” and the third column for “Integration function”.
### Integration by parts vs Tabular integration Method?
Integration by parts is the traditional method that is used to finds the integral of a product of functions and Tabular integration is a short technique to solve integral problems quickly by letting one of the functions can be differentiated multiple times and the other function can be integrated multiple times with ease.
### When do you can use the tabular integration?
Tabular integration can be used where one function is differentiated until it gets zero, and another function can be integrated simultaneously multiple times. Examples – ∫x6.sinx dx, ∫x3.e4x dx, etc.
Tabular integration can also be used where none of the functions gets zero when differentiated multiple times. Examples – ∫cosx.sinx dx, ∫ex.sin5x dx, etc.
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# Whole Number Exponents
## Distinguish bases and powers.
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Whole Number Exponents
### Let's Think About It
Credit: Seabamirum
Source: https://www.flickr.com/photos/seabamirum/2850890725
License: CC BY-NC 3.0
Horatio is a carpenter building a habitat for tigers at his local zoo. He has been told that the tiger’s enclosure needs to be at least 1720 cubic feet. Horatio has only ever built two animal habitats before, and he was hoping to use one of those as a model. One habitat was 93\begin{align*}9^3\end{align*} cubic feet, and the other was 123\begin{align*}12^3\end{align*} cubic feet. Which of Horatio’s previous habitat designs meets the requirements for the tiger habitat?
In this concept, you will learn how powers, bases, and exponents work together.
### Guidance
An exponent is a little number that shows how many times to multiply a certain number by itself.
Let's look at an example: 35\begin{align*}3^5\end{align*}
The large number, 4, is called the base
The small number, 2, is called the exponent. This number tells you how many times to multiply the base by itself.
When a base has an exponent of 2, such as 42\begin{align*}{4}^{2}\end{align*}, you read it as "four squared".
When a base has an exponent of 3, such as 63\begin{align*}{6}^{3}\end{align*}, you read it as "six cubed".
For all other exponents, you read the exponent as a power. For example,
35\begin{align*}3^5\end{align*} is read as "three to the fifth power".
27\begin{align*}2^7\end{align*} is read as "two to the seventh power".
59\begin{align*}5^9\end{align*} is read as "five to the ninth power".
An exponent tells you how many times the base should be multiplied by itself. If you look at a number with an exponent written out the long way, you can see why exponents are useful.
7×7×7=73
In this case, there are three factors of 7. Remember that factors are numbers or variables multiplied by each other.
5×5×5×5×5×5×5×5×5×5=510
In this case, there are ten factors of 5.
Exponents are a multiplication shortcut a lot like the way that multiplication is an addition shortcut.
### Guided Practice
Write out the factors of 35\begin{align*}3^5\end{align*}, then evaluate the product.
35=3×3×3×3×3
Once you have written out all the factors, simply multiply from left to right.
3×3=9×3=27×3=81×3=243
The answer is 243.
### Examples
#### Example 1
Write 63\begin{align*}6^3\end{align*} out in words.
First, identify the base number, which is the big number that will be multiplied by itself. In this case, the base is 6.
Next, consider the power, which is the small number that describes how many times the number will be multiplied. In this case, the power is 3, so the base will be multiplied by itself 3 times.
The solution is "6 times 6 times 6."
#### Example 2
Write out the factors of 45\begin{align*}4^5\end{align*}
First, remember that factors are numbers that are multiplied by each other to equal a certain value.
Next, identify the base number, which is the big number that will be multiplied by itself. In this case, the base is 4.
Then consider the power, which is the small number that describes how many times the number will be multiplied. In this case the power is 5, so the base will be multiplied by itself 5 times.
The solution is 45=4×4×4×4×4\begin{align*}4^5 = 4 \times 4 \times 4 \times 4 \times 4\end{align*}.
#### Example 3
Which is the base number: 910\begin{align*}9^{10}\end{align*}?
First, remember that the base is the big number that is multiplied by itself.
Then, remember that the power is the little number that describes how many times to multiply the base by itself.
The answer is that the base in the expression 910\begin{align*}9^{10}\end{align*} is 9 and the power is 10.
Credit: Cliff1066
Source: https://www.flickr.com/photos/nostri-imago/2854166745
License: CC BY-NC 3.0
Remember Horatio and his tiger housing project?
Horatio needs to figure out which dimensions are closest to 1720 cubic feet: 93 cubic feet or 123 cubic feet.
First, write out the factors for each expression.
93=9×9×9
123=12×12×12
Next, multiply.
9×912×12=81×9=729=144×12=1728
The answer is 123.
Horatio can use his old 123 cubic feet habitat design, which even gives the tigers a little extra room.
### Explore More
Write each power out in words.
1. 32\begin{align*}3^2\end{align*}
2. 55\begin{align*}5^5\end{align*}
3. 63\begin{align*}6^3\end{align*}
4. 26\begin{align*}2^6\end{align*}
5. 34\begin{align*}3^4\end{align*}
6. 74\begin{align*}7^4\end{align*}
7. 52\begin{align*}5^2\end{align*}
8. 24\begin{align*}2^4\end{align*}
9. 33\begin{align*}3^3\end{align*}
10. 93\begin{align*}9^3\end{align*}
Evaluate each expression (they are the same expressions as in problems 1 - 10).
11. 32\begin{align*}3^2\end{align*}
12. 55\begin{align*}5^5\end{align*}
13. 63\begin{align*}6^3\end{align*}
14. 26\begin{align*}2^6\end{align*}
15. 34\begin{align*}3^4\end{align*}
16. 74\begin{align*}7^4\end{align*}
17. 52\begin{align*}5^2\end{align*}
18. 24\begin{align*}2^4\end{align*}
19. 33\begin{align*}3^3\end{align*}
20. 93\begin{align*}9^3\end{align*}
### Vocabulary Language: English
Base
Base
When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Cubed
Cubed
The cube of a number is the number multiplied by itself three times. For example, "two-cubed" = $2^3 = 2 \times 2 \times 2 = 8$.
Power
Power
The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power".
Squared
Squared
Squared is the word used to refer to the exponent 2. For example, $5^2$ could be read as "5 squared". When a number is squared, the number is multiplied by itself.
Whole Numbers
Whole Numbers
The whole numbers are all positive counting numbers and zero. The whole numbers are 0, 1, 2, 3, ...
1. [1]^ Credit: Seabamirum; Source: https://www.flickr.com/photos/seabamirum/2850890725; License: CC BY-NC 3.0
2. [2]^ Credit: Cliff1066; Source: https://www.flickr.com/photos/nostri-imago/2854166745; License: CC BY-NC 3.0
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Home >> Median >>
## Define Median - find median from given data
Before you understand What is Median ?
You must know:
• What is Ascending Order ?
• What is Descending Order ?
Definition
In a given set of observation, after arranging in ascending or descending order, the middle observation which divides the set of observation in two equal parts is referred to as Median.
In order words, Median refers to the value which lies in the middle of the set of data, which is arranged in ascending or descending order.
** Note While finding Median only those cases are considered, where the value number of observation in a given set of data is ODD.
## Steps to find Median in a given set of observations or data:
• Step 1 Arrange the given set of observations or data in ascending or descending order.
• Step 2 Find the middle value, which divides the given data in two equal parts.
• Step 3 The resultant value is the Median of the data.
For further understanding lets try following examples:
Example 1 Following data represents weight of 11 students in a class. Find Median.
35, 29, 37, 38, 41, 36, 42, 47, 39, 40, 48
• Step 1 Arrange the given set of observations or data in ascending order and we get;
29, 35, 36, 37, 38, 39, 40, 41, 42, 47, 48
• Step 2 Find the middle value, which divides the given data in two equal parts.
29, 35, 36, 37, 38, 39, 40, 41, 42, 47, 48
You can see value 39 is the middle of set of data and divides the given data in a two equal groups comprising 5 values each.
• Step 3 Hence, 39 is the Median of the data.
Example 2 Following data represents score of 15 students in a science. Find Median.
75, 80, 91, 76, 82, 93, 78, 83, 94, 84, 96, 79, 87, 89, 90
• Step 1 Arrange the given set of observations or data in descending order and we get;
96, 94, 93, 91, 89, 90, 87, 84, 83, 82, 80, 79, 78, 76, 75
• Step 2 Find the middle value, which divides the given data in two equal parts.
96, 94, 93, 91, 89, 90, 87, 84, 83, 82, 80, 79, 78, 76, 75
You can see value 84 is the middle of set of data and divides the given data in a two equal groups comprising 7 values each.
• Step 3 Hence , 84 is the Median of the data. |
## RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1
Other Exercises
Question 1.
Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.
Solution:
Length of cuboid (l) = 80 cm
Height (h) = 20 cm
(i) ∴ Lateral surface area = 2h(l + b)
= 2 x 20(80 + 40) cm²
= 40 x 120 = 4800 cm²
(ii) Total surface area = 2(lb + bh + hl)
= 2(80 x 40 + 40 x 20 + 20 x 80) cm²
= 2(3200 + 800 + 1600) cm²
= 5600 x 2 = 11200 cm²
Question 2.
Find the lateral surface area and total surface area of a cube of edge 10 cm.
Solution:
Edge of cube (a) = 10 cm
(i) ∴ Lateral surface area = 4a²
= 4 x (10)² = 4 x 100 cm²= 400 cm²
(ii) Total surface area = 6a² = 6 x(10)² cm²
= 6 x 100 = 600 cm²
Question 3.
Find the ratio of the total surface area and lateral surface area of a cube.
Solution:
Let a be the edge of the cube, then Total surface area = 6a2²
and lateral surface area = 4a²
Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2
Question 4.
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require? [NCERT]
Solution:
Length of box (l) = 80 cm
and height (h) = 20 cm
∴ Total surface area = 2(lb + bh + hl)
= 2[80 x 40 + 40 x 20 + 20 x 80] cm²
= 2[3200 + 800 + 1600] cm² = 2 x 5600 = 11200 cm²
Size of paper sheet = 40 cm
∴ Area of one sheet = (40 cm)² = 1600 cm²
∴ No. of sheets required for the box = 11200 = 1600 = 7 sheets
Question 5.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m².
Solution:
Length of a room (l) = 5m
and height (h) = 3 m
∴ Area of 4 walls = 2(l + b) x h
= 2(5 + 4) x 3 = 6 x 9 = 54 m²
and area of ceiling = l x b = 5 x 4 = 20 m²
∴ Total area = 54 + 20 = 74 m2
Rate of white washing = 7.50 per m²
∴ Total cost = ₹74 x 7.50 = ₹555
Question 6.
Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let each side of a cube = a cm
Then surface area = 6a² cm²
and surface area of 3 such cubes = 3 x 6a² = 18a² cm²
By placing three cubes side by side we get a cuboid whose,
Length (l) = a x 3 = 3a
Height (h) = a
∴ Total surface area = 2(lb + bh + hf)
= 2[3a x a+a x a+a x 3a] cm²
= 2[3a² + a² + 3a²] = 14 a²
∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9
Question 7.
A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.
Solution:
Side of cube = 4 cm
But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64
Now surface area of one cube = 6 x (1)²
= 6 x 1=6 cm²
and surface area of 64 cubes = 6 x 64 cm² = 384 cm²
Question 8.
The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.
Solution:
Let h be the height of the room
Length (l) = 18 m
and width (b) = 12 m
Now surface area of floor and roof = 2 x lb = 2 x 18 x 12 m²
= 432 m²
and surface area of 4-walls = 2h (l + b)
= 2h(18 + 12) = 2 x 30h m² = 60h m²
∵ The surface are of 4-walls and area of floor and roof are equal
∴ 60h = 432
⇒ h = $$\frac { 432 }{ 60 }$$ = $$\frac { 72 }{ 10 }$$ m
∴ Height = 7.2m
Question 9.
Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if the cost of tiles is ₹360 per dozen. [NCERT]
Solution:
Edge of cubical tank = 1.5 m
∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m²
Area of floor = (1.5)² = 2.25 m²
∴ Total surface area = 9 + 2.25 = 11.25 m²
Edge of square tile = 25 m = 0.25 m²
∴ Area of 1 tile = (0.25)2 = .0625 m²
Question 10.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let edge of a cube = a
Total surface area = 6a2
By increasing edge at 50%,
Question 11.
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of ₹5 per metre sheet, sheet being 2 m wide.
Solution:
Length of iron tank (l) = 12 m
Depth (h) = 4 cm
Question 12.
Ravish wanted to make a temporary shelter for his car by making a box-like structure with tarpaulin that covers all the four sides and the top of tire car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make (he shelter of height 2.5 m with base dimensions 4 m x 3 m? [NCERT]
Solution:
Length of base (l) = 4m
Height (h) = 2.5 m
Question 13.
An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of ₹50 per sq. metre.
Solution:
Length of open wood box (L) = 1.48 m = 148 cm
Breadth (B) = 1.16 m = 116 cm
and height (H) = 8.3 dm = 83 cm
Thickness of wood = 3 cm
Question 14.
The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at ₹3.50 per square metre.
Solution:
Length of room (l) = 12.5 m
and height (h) = 7 m
∴ Total area of walls = 2h(l + b)
= 2 x 7[12.5 + 9] = 14 x 21.5 m² = 301 m²
Area of 2 doors of 2.5 m x 1.2 m
= 2 x 2.5 x 1.2 m² = 6 m²
and area of 4 window of 1.5 m x 1 m
= 4 x 1.5 x 1 = 6 m²
∴ Remaining area of walls = 301 – (6 + 6)
= 301 – 12 = 289 m²
Rate of painting the walls = ₹3.50 per m²
∴ Total cost = 289 x 3.50 = ₹1011.50
Question 15.
The paint in a certain container is sufficient to paint on area equal to 9.375 m2. How many bricks of dimension 22.5 cm x 10 cm x 7.5 cm can be painted out of this container? [NCERT]
Solution:
Area of place for painting = 9.375 m²
Dimension of one brick = 22.5 cm x 10 cm x 7.5 cm
∴ Surface area of one bricks = 2 (lb + bh + hl)
= 2[22.5 x 10 + 10 x 7.5 + 7.5 x 22.5] cm2
= 2[225 + 75 + 168.75]
= 2 x 468.75 cm² = 937.5 cm²
Question 16.
The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of ₹8 and ₹9.50 per m2 is ₹1248. Find the dimensions of the box.
Solution:
Ratio in the dimensions of a cuboidal box = 2 : 3 : 4
Let length (l) = 4x
and height (h) = 2x
∴ Total surface area = 2 [lb + bh + hl]
Question 17.
The cost of preparing the walls of a room 12 m long at the rate of ₹1.35 per square metre is ₹340.20 and the cost of matting the floor at 85 paise per square metre is ₹91.80. Find the height of the room.
Solution:
Cost of preparing walls of a room = ₹340.20
Question 18.
The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 metres. The cost of decorating its walls (including doors and windows) at ₹6.60 per square metre is ₹5082. Find the length and breadth of the room
Solution:
Ratio in length and breadth = 4:3
and height (h) = 5.5 m
Cost of decorating the walls of a room including doors and windows = ₹5082
Rate = ₹6.60 per m²
Question 19.
A wooden bookshelf has external dimensions as follows: Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf. [NCERT]
Solution:
Length (l) = 85 cm
and height (h) = 110 cm
Thickness of plank = 5 cm
Surface area to be polished
= [(100 x 85) + 2 (110 x 25) + 2 (85 x 25) + 2 (110 x 5) + 4 (75 x 5)]
= (9350 + 5500 + 4250 + 1100 + 1500) cm² = 21700 cm²
Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS
Other Exercises
Question 1.
In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
Solution:
Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle
∴ ∠AOB = 2∠APB = 2 x 70° = 140°
∠AOB + ∠ACB = 180° (Sum of the angles)
⇒ 140° +∠ACB = 180°
⇒ ∠ACB = 180° – 140° = 40°
∴ ∠ACB = 40°
Question 2.
In the figure, two congruent circles with centre O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.
Solution:
Two congruent circles with centres O and O’ intersect at A and B
∠AO’B = 50°
∵ OA = OB = O’A = 04B (Radii of the congruent circles)
Question 3.
In the figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = IT, AC and BD intersect at P. Then, find ∠DPC.
Solution:
∵ ABCD is a cyclic quadrilateral,
∴ ∠BAD + ∠BCD = 180°
⇒ 75° + ∠BCD – 180°
⇒ ∠BCD = 180°-75°= 105° and ∠ADC + ∠ABC = 180°
⇒ 77° + ∠ABC = 180°
⇒ ∠ABC = 180°-77°= 103°
∴ ∠DBC = ∠ABC – ∠ABD = 103° – 58° = 45°
∵ Arc AD subtends ∠ABD and ∠ACD in the same segment of the circle 3
∴ ∠ABD = ∠ACD = 58°
∴ ∠ACB = ∠BCD – ∠ACD = 105° – 58° = 47°
Now in ∆PBC,
Ext. ∠DPC = ∠PBC + ∠PCB
=∠DBC + ∠ACB = 45° + 47° = 92°
Hence ∠DPC = 92°
Question 4.
In the figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.
Solution:
In the figure, ∠AOB = 80°, ∠ABC = 30°
∵ Arc AB subtends ∠AOB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = $$\frac { 1 }{ 2 }$$∠AOB = $$\frac { 1 }{ 2 }$$ x 80° = 40°
In ∆OAB, OA = OB
∴ ∠OAB = ∠OBA
But ∠OAB + ∠OBA + ∠AOB = 180°
∴ ∠OAB + ∠OBA + 80° = 180°
⇒ ∠OAB + ∠OAB = 180° – 80° = 100°
∴ 2∠OAB = 100°
⇒ ∠OAB = $$\frac { { 100 }^{ \circ } }{ 2 }$$ = 50°
Similarly, in ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180°
∠BAC + 40° + 30° = 180°
⇒ ∠BAC = 180°-30°-40°
= 180°-70°= 110°
∴ ∠CAO = ∠BAC – ∠OAB
= 110°-50° = 60°
Question 5.
In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
Solution:
In the figure, ABCD is a parallelogram and
CDE is a straight line
∵ ABCD is a ||gm
∴ ∠A = ∠C
and ∠C = ∠ADE (Corresponding angles)
Similarly, ∠ABE = ∠BED (Alternate angles)
∵ arc BD subtends ∠BAD at the centre and
∠BED at the remaining part of the circle
Question 6.
In the figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
Solution:
In the figure, AB is the diameter of the circle such that ∠A = 35° and ∠Q = 25°, join OP.
Arc PB subtends ∠POB at the centre and
∠PAB at the remaining part of the circle
∴ ∠POB = 2∠PAB = 2 x 35° = 70°
Now in ∆OP,
OP = OB radii of the circle
∴ ∠OPB = ∠OBP = 70° (∵ ∠OPB + ∠OBP = 140°)
Now ∠APB = 90° (Angle in a semicircle)
∴ ∠BPQ = 90°
and in ∆PQB,
Ext. ∠PBR = ∠BPQ + ∠PQB
= 90° + 25°= 115°
∴ ∠PBR = 115°
Question 7.
In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
Solution:
In the figure, P and Q are the centres of two circles which intersect each other at C and B
ACD is a straight line ∠APB = 150°
Arc AB subtends ∠APB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = $$\frac { 1 }{ 2 }$$ ∠APB = $$\frac { 1 }{ 2 }$$ x 150° = 75°
But ∠ACB + ∠BCD = 180° (Linear pair)
⇒ 75° + ∠BCD = 180°
∠BCD = 180°-75°= 105°
Now arc BD subtends reflex ∠BQD at the centre and ∠BCD at the remaining part of the circle
Reflex ∠BQD = 2∠BCD = 2 x 105° = 210°
But ∠BQD + reflex ∠BQD = 360°
∴ ∠BQD+ 210° = 360°
∴ ∠BQD = 360° – 210° = 150°
Question 8.
In the figure, if O is circumcentre of ∆ABC then find the value of ∠OBC + ∠BAC.
Solution:
In the figure, join OC
∵ O is the circumcentre of ∆ABC
∴ OA = OB = OC
∵ ∠CAO = 60° (Proved)
∴ ∆OAC is an equilateral triangle
∴ ∠AOC = 60°
Now, ∠BOC = ∠BOA + ∠AOC
= 80° + 60° = 140°
and in ∆OBC, OB = OC
∠OCB = ∠OBC
But ∠OCB + ∠OBC = 180° – ∠BOC
= 180°- 140° = 40°
⇒ ∠OBC + ∠OBC = 40°
∴ ∠OBC = $$\frac { { 40 }^{ \circ } }{ 2 }$$ = 20°
∠BAC = OAB + ∠OAC = 50° + 60° = 110°
∴ ∠OBC + ∠BAC = 20° + 110° = 130°
Question 9.
In the AOC is a diameter of the circle and arc AXB = 1/2 arc BYC. Find ∠BOC.
Solution:
In the figure, AOC is diameter arc AxB = $$\frac { 1 }{ 2 }$$ arc BYC 1
∠AOB = $$\frac { 1 }{ 2 }$$ ∠BOC
⇒ ∠BOC = 2∠AOB
But ∠AOB + ∠BOC = 180°
⇒ ∠AOB + 2∠AOB = 180°
⇒ 3 ∠AOB = 180°
∴ ∠AOB = $$\frac { { 180 }^{ \circ } }{ 3 }$$ = 60°
∴ ∠BOC = 2 x 60° = 120°
Question 10.
In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
Solution:
In the figure, ABCD is a cyclic quadrilateral
CD is produced to E such that ∠ADE = 95°
O is the centre of the circle
⇒ ∠ADC + 95° = 180°
⇒ ∠ADC = 180°-95° = 85°
Now arc ABC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle
∵ ∠AOC = 2∠ADC = 2 x 85° = 170°
Now in ∆OAC,
∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ ∠OAC = ∠OCA (∵ OA = OC radii of circle)
∴ ∠OAC + ∠OAC + 170° = 180°
2∠OAC = 180°- 170°= 10°
∴ ∠OAC = $$\frac { { 10 }^{ \circ } }{ 2 }$$ = 5°
Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS
Other Exercises
Mark the correct alternative in each of the following:
Question 1.
The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is
(a) 225 cm²
(b) 225$$\sqrt { 3 }$$ cm²
(c) 225$$\sqrt { 2 }$$ cm²
(d) 240 cm²
Solution:
Sides of triangle and 16 cm, 30 cm, 34 cm
Question 2.
The base of an isosceles right triangle is 30 cm. Its area is
(a) 225 cm²
(b) 225$$\sqrt { 3 }$$ cm²
(c) 225$$\sqrt { 2 }$$ cm²
(d) 450 cm²
Solution:
Base of isosceles triangle ∆ABC = 30cm
Let each of equal sides = x
Then AB = AC = x
Now in right ∆ABC,
Question 3.
The sides of a triangle are 7cm, 9cm and 14cm. Its area is
(a) 12$$\sqrt { 5 }$$ cm²
(b) 12$$\sqrt { 3 }$$ cm²
(c) 24 $$\sqrt { 5 }$$ cm²
(d) 63 cm²
Solution:
Question 4.
The sides of a triangular field are 325 m, 300 m and 125 m. Its area is
(a) 18750 m²
(b) 37500 m²
(c) 97500 m²
(d) 48750 m²
Solution:
Sides of a triangular field are 325m, 300m, 125m
Question 5.
The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is
(a) 20 cm
(b) 30 cm
(c) 40 cm
(d) 50 cm
Solution:
The sides of a triangle are 50 cm, 78 cm, 112cm
Question 6.
The sides of a triangle are 11m, 60m and 61m. Altitude to the smallest side is
(a) 11m
(b) 66 m
(c) 50 m
(d) 60 m
Solution:
Sides of a triangle are 11m, 60m and 61m
Question 7.
The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is
Solution:
Sides of a triangle are 11 cm, 15 cm, 16 cm
Question 8.
The base and hypotenuse of a right triangle are respectively 5cm and 13cm long. Its area is
(a) 25 cm²
(b) 28 cm²
(c) 30 cm²
(d) 40 cm²
Solution:
In a right triangle base = 5 cm
base hypotenuse = 13 cm
Question 9.
The length of each side of an equilateral triangle of area 4 $$\sqrt { 3 }$$ cm², is
Solution:
Area of an equilateral triangle = 4$$\sqrt { 3 }$$ cm²
Let each side be = a
Question 10.
If the area of an isosceles right triangle is 8cm, what is the perimeter of the triangle.
(a) 8 + $$\sqrt { 2 }$$ cm²
(b) 8 + 4$$\sqrt { 2 }$$ cm²
(c) 4 + 8$$\sqrt { 2 }$$ cm²
(b) 12$$\sqrt { 2 }$$ cm²
Solution:
Let base = x
ABC an isosceles right triangle, which has 2 sides same
⇒ Height = x
Question 11.
The length of the sides of ∆ABC are consecutive integers. If ∆ABC has the same perimeter as an equilateral triangle with a side of length 9cm, what is the length of the shortest side of ∆ABC?
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
Side of an equilateral triangle = 9 cm
Its perimeter = 3 x 9 = 27 cm
Now perimeter of ∆ABC = 27 cm
and let its sides be x, x + 1, x +2
Question 12.
In the figure, the ratio of AD to DC is 3 to 2. If the area of ∆ABC is 40cm2, what is the area of ∆BDC?
(a) 16 cm²
(b) 24 cm²
(c) 30 cm²
(d) 36 cm²
Solution:
Ratio in AD : DC = 3:2
and area ∆ABC = 40 cm²
Question 13.
If the length of a median of an equilateral triangle is x cm, then its area is
Solution:
∵ The median of an equilateral triangle is the perpendicular to the base also,
∴ Let side of the triangle = a
Question 14.
If every side of a triangle is doubled, then increase in the area of the triangle is
(a) 100$$\sqrt { 2 }$$ %
(b) 200%
(c) 300%
(d) 400%
Solution:
Let the sides of the original triangle be a, b, c
Question 15.
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 1272 cm, then area of the triangle is
Solution:
A square and an equilateral triangle have equal perimeter
Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5
Other Exercises
Question 1.
In the figure, ∆ABC is an equilateral triangle. Find m ∠BEC.
Solution:
∵ ∆ABC is an equilateral triangle
∴ A = 60°
∵ ABEC is a cyclic quadrilateral
∴ ∠A + ∠E = 180° (Sum of opposite angles)
⇒ 60° + ∠E = 180°
⇒ ∠E = 180° – 60° = 120°
∴ m ∠BEC = 120°
Question 2.
In the figure, ∆PQR is an isosceles triangle with PQ = PR and m ∠PQR = 35°. Find m ∠QSR and m ∠QTR.
Solution:
In the figure, ∆PQR is an isosceles PQ = PR
∠PQR = 35°
∴ ∠PRQ = 35°
But ∠PQR + ∠PRQ + ∠QPR = 180° (Sum of angles of a triangle)
⇒ 35° + 35° + ∠QPR = 180°
⇒ 70° + ∠QPR = 180°
∴ ∠QPR = 180° – 70° = 110°
∵ ∠QSR = ∠QPR (Angle in the same segment of circles)
∴ ∠QSR = 110°
But PQTR is a cyclic quadrilateral
∴ ∠QTR + ∠QPR = 180°
⇒ ∠QTR + 110° = 180°
⇒ ∠QTR = 180° -110° = 70°
Hence ∠QTR = 70°
Question 3.
In the figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.
Solution:
In the figure, O is the centre of the circle ∠BOD =160°
∵ Arc BAD subtends ∠BOD is the angle at the centre and ∠BCD is on the other part of the circle
∴ ∠BCD = $$\frac { 1 }{ 2 }$$ ∠BOD
⇒ x = $$\frac { 1 }{ 2 }$$ x 160° = 80°
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180°
⇒ y + x = 180°
⇒ y + 80° = 180°
⇒ y =180°- 80° = 100°
∴ x = 80°, y = 100°
Question 4.
In the figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
Solution:
In a circle, ABCD is a cyclic quadrilateral ∠BCD = 100° and ∠ABD = 70°
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180° (Sum of opposite angles)
⇒ ∠A + 100°= 180°
∠A = 180°- 100° = 80°
Now in ∆ABD,
∠A + ∠ABD + ∠ADB = 180°
⇒ 80° + 70° + ∠ADB = 180°
∴ ∠ADB = 180°- 150° = 30°
Question 5.
If ABCD is a cyclic quadrilateral in which AD || BC. Prove that ∠B = ∠C.
Solution:
Given : ABCD is a cyclic quadrilateral in which AD || BC
To prove : ∠B = ∠C
Proof : ∵ AD || BC
∴ ∠A + ∠B = 180°
(Sum of cointerior angles)
But ∠A + ∠C = 180°
(Opposite angles of the cyclic quadrilateral)
∴ ∠A + ∠B = ∠A + ∠C
⇒ ∠B = ∠C
Hence ∠B = ∠C
Question 6.
In the figure, O is the centre of the circle. Find ∠CBD.
Solution:
Arc AC subtends ∠AOC at the centre and ∠APC at the remaining part of the circle
∴ ∠APC = $$\frac { 1 }{ 2 }$$ ∠AOC
= $$\frac { 1 }{ 2 }$$ x 100° = 50°
∴ ∠APC + ∠ABC = 180°
⇒ 50° + ∠ABC = 180° ⇒ ∠ABC =180°- 50°
∴ ∠ABC =130°
But ∠ABC + ∠CBD = 180° (Linear pair)
⇒ 130° + ∠CBD = 180°
⇒ ∠CBD = 180°- 130° = 50°
∴ ∠CBD = 50°
Question 7.
In the figure, AB and CD are diameiers of a circle with centre O. If ∠OBD = 50°, find ∠AOC.
Solution:
Two diameters AB and CD intersect each other at O. AC, CB and BD are joined
∠DBA = 50°
∠DBA and ∠DCA are in the same segment
∴ ∠DBA = ∠DCA = 50°
In ∆OAC, OA = OC (Radii of the circle)
∴ ∠OAC = ∠OCA = ∠DCA = 50°
and ∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ 50° + 50° + ∠AOC = 180°
⇒ 100° + ∠AOC = 180°
⇒ ∠AOC = 180° – 100° = 80°
Hence ∠AOC = 80°
Question 8.
On a semi circle with AB as diameter, a point C is taken so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).
Solution:
A semicircle with AB as diameter
∠ CAB = 30°
∠ACB = 90° (Angle in a semi circle)
But ∠CAB + ∠ACB + ∠ABC = 180°
⇒ 30° + 90° + ∠ABC – 180°
⇒ 120° + ∠ABC = 180°
∴ ∠ABC = 180°- 120° = 60°
Hence m ∠ACB = 90°
and m ∠ABC = 60°
Question 9.
In a cyclic quadrilateral ABCD, if AB || CD and ∠B = 70°, find the remaining angles.
Solution:
In a cyclic quadrilateral ABCD, AB || CD and ∠B = 70°
∵ ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180°
⇒ 70° + ∠D = 180°
⇒ ∠D = 180°-70° = 110°
∵ AB || CD
∴ ∠A + ∠D = 180° (Sum of cointerior angles)
∠A+ 110°= 180°
⇒ ∠A= 180°- 110° = 70°
Similarly, ∠B + ∠C = 180°
⇒ 70° + ∠C- 180° ‘
⇒ ∠C = 180°-70°= 110°
∴ ∠A = 70°, ∠C = 110°, ∠D = 110°
Question 10.
In a cyclic quadrilateral ABCD, if m ∠A = 3(m ∠C). Find m ∠A.
Solution:
In cyclic quadrilateral ABCD, m ∠A = 3(m ∠C)
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180°
⇒ 3 ∠C + ∠C = 180° ⇒ 4∠C = 180°
⇒ ∠C = $$\frac { { 180 }^{ \circ } }{ 4 }$$ = 45°
∴ ∠A = 3 x 45°= 135°
Hence m ∠A =135°
Question 11.
In the figure, O is the centre of the circle and ∠DAB = 50°. Calculate the values of x and y.
Solution:
In the figure, O is the centre of the circle ∠DAB = 50°
∵ ABCD is a cyclic quadrilateral
∴ ∠A + ∠C = 180°
⇒ 50° + y = 180°
⇒ y = 180° – 50° = 130°
In ∆OAB, OA = OB (Radii of the circle)
∴ ∠A = ∠OBA = 50°
∴ Ext. ∠DOB = ∠A + ∠OBA
x = 50° + 50° = 100°
∴ x= 100°, y= 130°
Question 12.
In the figure, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.
Solution:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)
60° + ∠ABC + 20° = 180°
∠ABC + 80° = 180°
∴ ∠ABC = 180° -80°= 100°
∵ ABCD is a cyclic quadrilateral,
∴ ∠ABC + ∠ADC = 180°
∴ ∠ADC = 180°- 100° = 80°
Question 13.
In the figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
Solution:
In a circle, ∆ABC is an equilateral triangle
∴ ∠A = 60°
∵ ∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 60°
∵ BECD is a cyclic quadrilateral
∴ ∠BDC + ∠BEC = 180°
⇒ 60° + ∠BEC = 180°
⇒ ∠BEC = 180°-60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120°
Question 14.
In the figure, O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.
Solution:
∠AEC and ∠ADC are in the same segment
∴ ∠AEC = ∠ADC = 30°
∴ z = 30°
∴ ∠B + ∠D = 180°
⇒ x + z = 180°
⇒ x + 30° = 180°
⇒ x = 180° – 30° = 150°
Arc AC subtends ∠AOB at the centre and ∠ADC at the remaining part of the circle
∴ ∠AOC = 2∠D = 2 x 30° = 60°
∴ y = 60°
Hence x = 150°, y – 60° and z = 30°
Question 15.
In the figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
Solution:
In the figure, two circles intersect each other at C and D
∠BAD = 78°, ∠DCF = x, ∠DEF = y
∴ Ext. ∠DCF = its interior opposite ∠BAD
⇒ x = 78°
∠DCF + ∠DEF = 180°
⇒ 78° + y = 180°
⇒ y = 180° – 78°
y = 102°
Hence x = 78°, and y- 102°
Question 16.
In a cyclic quadrilateral ABCD, if ∠A – ∠C = 60°, prove that the smaller of two is 60°.
Solution:
∠A – ∠C = 60°
But ∠A + ∠C = 180° (Sum of opposite angles)
Adding, 2∠A = 240° ⇒ ∠A = $$\frac { { 62 }^{ \circ } }{ 2 }$$ = 120° and subtracting
2∠C = 120° ⇒ ∠C = $$\frac { { 120 }^{ \circ } }{ 2 }$$ = 60°
∴ Smaller angle of the two is 60°.
Question 17.
In the figure, ABCD is a cyclic quadrilateral. Find the value of x.
Solution:
∠CDE + ∠CDA = 180° (Linear pair)
⇒ 80° + ∠CDA = 180°
⇒ ∠CDA = 180° – 80° = 100°
Ext. ∠ABF = Its interior opposite angle ∠CDA = 100°
∴ x = 100°
Question 18.
ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC =110° and ∠B AC = 50°. Find ∠DAC.
(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
(iii) ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
Solution:
(i) In the figure,
∠BAC = 50°
∵ ∠B + ∠D = 180° (Sum of opposite angles)
⇒ ∠B + 110° = 180°
∴ ∠B = 180°- 110° = 70°
Now in ∆ABC,
∠CAB + ∠ABC + ∠BCA = 180° (Sum of angles of a triangle)
⇒ 50° + 70° + ∠BCA = 180°
⇒ 120° + ∠BCA = 180°
⇒ ∠BCA = 180° – 120° = 60°
But ∠DAC = ∠BCA (Alternate angles)
∴ ∠DAC = 60°
Diagonals AC and BD are joined ∠DBC = 80°, ∠BAC = 40°
Arc DC subtends ∠DBC and ∠DAC in the same segment
∴ ∠DBC = ∠DAC = 80°
∴ ∠DAB = ∠DAC + ∠CAB = 80° + 40° = 120°
But ∠DAC + ∠BCD = 180° (Sum of opposite angles of a cyclic quad.)
⇒ 120° +∠BCD = 180°
⇒ ∠BCD = 180°- 120° = 60°
(iii) In the figure, ABCD is a cyclic quadrilateral BD is joined
∠BCD = 100°
and ∠ABD = 70°
∠A + ∠C = 180° (Sum of opposite angles of cyclic quad.)
∠A+ 100°= 180°
⇒ ∠A= 180°- 100°
∴ ∠A = 80°
Now in ∆ABD,
∠A + ∠ABD + ∠ADB = 180° (Sum of angles of a triangle)
⇒ 80° + 70° + ∠ADB = 180°
⇒ ∠ADB = 180°- 150° = 30°
Question 19.
Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.
Solution:
Given : ABCD is a rhombus. Four circles are drawn on the sides AB, BC, CD and DA respectively
To prove : The circles pass through the point of intersection of the diagonals of the rhombus ABCD
Proof: ABCD is a rhombus whose diagonals AC and BD intersect each other at O
∵ The diagonals of a rhombus bisect each other at right angles
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Now when ∠AOB = 90°
and a circle described on AB as diameter will pass through O
Similarly, the circles on BC, CD and DA as diameter, will also pass through O
Question 20.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that is diagonals are equal.
Solution:
Given : In cyclic quadrilateral ABCD, AB = CD
AC and BD are the diagonals
To prove : AC = BC
Proof: ∵ AB = CD
∴ arc AB = arc CD
Adding arc BC to both sides, then arc AB + arc BC = arc BC + arc CD
⇒ arc AC = arc BD
∴ AC = BD
Hence diagonal of the cyclic quadrilateral are equal.
Question 21.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
Solution:
Given : In ∆ABC, circles are drawn on sides AB and AC
To prove : Circles drawn on AB and AC intersect at D which lies on BC, the third side
Construction : Draw AD ⊥ BC
So, the circles drawn on sides AB and AC as diameter will pass through D
Hence circles drawn on two sides of a triangle pass through D, which lies on the third side.
Question 22.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Solution:
In the figure, ABCD is a trapezium in which AD || BC and ∠B = 70°
∴ ∠A + ∠B = 180° (Sum of cointerior angles)
⇒ ∠A + 70° = 180°
⇒ ∠A= 180°- 70° = 110°
∴ ∠A = 110°
But ∠A + ∠C = 180° and ∠B + ∠D = 180° (Sum of opposite angles of a cyclic quadrilateral)
∴ 110° + ∠C = 180°
⇒ ∠C = 180°- 110° = 70°
and 70° + ∠D = 180°
⇒ ∠D = 180° – 70° = 110°
∴ ∠A = 110°, ∠C = 70° and ∠D = 110°
Question 23.
In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
Solution:
In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are drawn ∠DBC = 55° and ∠BAC = 45°
∵ ∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 45°
Now in ABCD,
∠DBC + ∠BDC + ∠BCD = 180° (Sum of angles of a triangle)
⇒ 55° + 45° + ∠BCD = 180°
⇒ 100° + ∠BCD = 180°
⇒ ∠BCD = 180° – 100° = 80°
Hence ∠BCD = 80°
Question 24.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Solution:
Given : ABCD is a cyclic quadrilateral
To prove : The perpendicular bisectors of the sides are concurrent
Proof : ∵ Each side of the cyclic quadrilateral is a chord of the circle and perpendicular of a chord passes through the centre of the circle
Hence the perpendicular bisectors of each side will pass through the centre O
Hence the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent
Question 25.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Solution:
Given : ABCD is a cyclic rectangle and diagonals AC and BD intersect each other at O
To prove : O is the point of intersection is the centre of the circle.
Proof : Let O be the centre of the circle- circumscribing the rectangle ABCD
Since each angle of a rectangle is a right angle and AC is the chord of the circle
∴ AC will be the diameter of the circle Similarly, we can prove that diagonal BD is also the diameter of the circle
∴ The diameters of the circle pass through the centre
Hence the point of intersection of the diagonals of the rectangle is the centre of the circle.
Question 26.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:
(ii) EB = EC.
Solution:
Given : ABCD is a cyclic quadrilateral in which sides BA and CD are produced to meet at E and EA = ED
To prove :
(ii) EB = EC
Proof: ∵ EA = ED
∠EAD = ∠EDA (Angles opposite to equal sides)
Similarly Ext. ∠EDA = ∠B
∴ ∠B = ∠C
Now in ∆EBC,
∵ ∠B = ∠C
∴ EC = EB (Sides opposite to equal sides)
But these are corresponding angles
Question 27.
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Solution:
Given : A segment ACB shorter than a semicircle and an angle ∠ACB inscribed in it
To prove : ∠ACB < 90°
Construction : Join OA and OB
Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle ∴ ∠ACB = $$\frac { 1 }{ 2 }$$ ∠AOB But ∠AOB > 180° (Reflex angle)
∴ ∠ACB > $$\frac { 1 }{ 2 }$$ x [80°
⇒ ∠ACB > 90°
Question 28.
Prove that the angle in a segment greater than a semi-circle is less than a right angle
Solution:
Given : A segment ACB, greater than a semicircle with centre O and ∠ACB is described in it
To prove : ∠ACB < 90°
Construction : Join OA and OB
Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠ACB =$$\frac { 1 }{ 2 }$$ ∠AOB
But ∠AOB < 180° (A straight angle) 1
∴ ∠ACB < $$\frac { 1 }{ 2 }$$ x 180°
⇒ ∠ACB <90°
Hence ∠ACB < 90°
Question 29.
Prove that the line segment joining the mid-point of the hypotenuse of a rijght triangle to its opposite vertex is half of the hypotenuse.
Solution:
Given : In a right angled ∆ABC
∠B = 90°, D is the mid point of hypotenuse AC. DB is joined.
To prove : BD = $$\frac { 1 }{ 2 }$$ AC
Construction : Draw a circle with centre D and AC as diameter
Proof: ∵ ∠ABC = 90°
∴ The circle drawn on AC as diameter will pass through B
∴ BD is the radius of the circle
But AC is the diameter of the circle and D is mid point of AC
∴ AD = DC = BD
∴ BD= $$\frac { 1 }{ 2 }$$ AC
Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 are helpful to complete your math homework.
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## RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
Other Exercises
Question 1.
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
In ∆ABC,
Base BC = 5cm
Question 2.
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
Sides of triangle are 3 cm, 4cm and 5cm
Question 3.
Find the area of an isosceles triangle having the base x cm and one side y cm.
Solution:
In isosceles ∆ABC,
AB = AC = y cm
BC = x cm
Question 4.
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Each side of equilateral triangle (a) = 4cm
Question 5.
Find the area of an equilateral triangle having each side x cm.
Solution:
Question 6.
The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Solution:
Perimeter of the field = 144 m
Ratio in the sides = 3:4:5
Sum of ratios = 3 + 4 + 5 = 12
Question 7.
Find the area of an equilateral triangle having altitude h cm.
Solution:
Altitude of an equilateral triangle = h
Let side of equilateral triangle = x
Question 8.
Let ∆ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
Solution:
Let a, b, c be the sides of the original triangle
Hence area of new triangle = 4 x area of original triangle.
Question 9.
If each side of a triangle is doubled, then find percentage increase in its area.
Solution:
Sides of original triangle be a, b, c
Question 10.
If each side of an equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
Solution:
Let the sides of the original triangle be a, b, c and area ∆, then
Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS are helpful to complete your math homework.
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## RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
Other Exercises
Question 1.
Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm (NCERT)
Solution:
In the quadrilateral, AC is the diagonal which divides the figure into two triangles
Now in ∆ABC, AB = 3 cm, BC = 4cm, AC = 5cm
Question 2.
The sides of a quadrangular field taken in order are 26 m, 27 m, 7 m and 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.
Solution:
In quad. ABCD, AB = 26 m, BC = 27 m CD = 7m, DA = 24 m, ∠CDA = 90°
Join AC,
Question 3.
The sides of a quadrilateral taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.
Solution:
AB = 5m, BC = 12 m, CD = 14m,
DA = 15 m and ∠ABC = 90°
Join AC,
Now in right ∆ABC,
AC² = AB² + BC² = (5)² + (12)²
= 25 + 144 = 169 = (13)²
∴ AC = 13 m
Now area of right ∆ABC
Question 4.
A park, in shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy? (NCERT)
Solution:
AB = 9m, BC = 12m, CD = 5m and
DA = 8m, ∠C = 90°
Join BD,
Now in right ∆BCD,
BD² = BC²+ CD² = (12)² + (5)²
= 144 + 25 = 169 = (13)²
∴ BD = 13m
Question 5.
Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is 24m.
Solution:
Perimeter of rhombus ABCD = 80 m
Question 6.
A rhombus sheet whose perimeter = 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of ₹5 per m². Find the cost of painting.
Solution:
Perimeter of the rhombus shaped sheet = 32 m
∴ Length of each side = $$\frac { 32 }{ 4 }$$ = 8m
and length of one diagonal AC = 10 m
In ∆ABC, sides are 8m, 8m, 10m
Question 7.
Find the area of a quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. (Take $$\sqrt { 3 }$$ = 1.73 )
Solution:
BCD is an equilateral triangle with side 26cm
In right ∆ABD,
(26)² = AB² + (24)²
⇒ 676 = AB² + 576
AB² = 676 – 576 = 100 = (10)²
∴ AB = 10cm
Now area of right ∆ABD,
Question 8.
Find the area of a quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.
Solution:
AB = 42 cm, BC = 21 cm, CD = 29cm DA = 34 cm, BD = 20 cm
Question 9.
The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.
Solution:
In ||gm ABCD,
AB = 34cm, BC = 20 cm
and AC = 42 cm
∵ The diagonal of a parallelogram divides into two triangles equal in area,
Now area of ∆ABC,
Question 10.
Find the area of the blades of the magnetic compass shown in figure. (Take $$\sqrt { 11 }$$ = 3.32).
Solution:
ABCD is a rhombus with each side 5cm and one diagonal 1cm
Diagonal BD divides into two equal triangles Now area of ∆ABD,
Question 11.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.
Solution:
Area of a triangle with same base and area of a 11gm with equal sides of triangle are 13, 14, 15 cm
Question 12.
Two parallel sides of a trapezium are 60cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.
Solution:
In trapezium ABCD, AB || DC
AB = 77cm, BC = 26 cm, CD 60cm DA = 25 cm
Through, C, draw CE || DA meeting AB at E
∴ AE = CD = 60 cm and EB = 77 – 60 = 17 cm,
CE = DA = 25 cm
Now area of ∆BCE, with sides 17 cm, 26 cm, 25 cm
Question 13.
Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9cm, CD = 12cm, ∠ACB = 90° and AC = 15cm.
Solution:
In right ΔABC, ∠ACB = 90°
AB² = AC² + BC²
(17)² = (15)²+ BC² = 289 = 225 + BC²
Question 14.
A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each types of paper needed to make the hand fan.
Solution:
In the figure, a hand fan has 5 isosceles and triangle. With sides 25 cm, 25 cm and 14 cm each.
Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 are helpful to complete your math homework.
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## RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
Other Exercises
Question 1.
Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.
Solution:
Sides of triangle are 120 cm, 150 cm, 200 cm
Question 2.
Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.
Solution:
Sides of a triangle are 9 cpi, 12 cm, 15 cm
Question 3.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Perimeter of a triangle = 42 cm
Two sides are 18 cm and 10 cm
Third side = 42 – (18 + 10)
= 42 – 28 = 14 cm
Question 4.
In a ∆ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ∆ABC and hence its altitude on AC.
Solution:
Sides of triangle ABC are AB = 15 cm, BC = 13 cm, AC = 14 cm
Question 5.
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. [NCERT]
Solution:
Perimeter of a triangle = 540 m
Ratio in sides = 25 : 17 : 12
Sum of ratios = 25 + 17 + 12 = 54
Question 6.
The perimeter of a triangle is 300 m. If its sides are in the ratio 3:5:7. Find the area of the triangle. [NCERT]
Solution:
Perimeter of a triangle = 300 m
Ratio in the sides = 3 : 5 : 7
∴ Sum of ratios = 3 + 5 + 7= 15
Question 7.
The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.
Solution:
Perimeter of a triangular field = 240 dm
Two sides are 78 dm and 50 dm
∴ Third side = 240 – (78 + 50)
= 240 – 128 = 112 dm
Question 8.
A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.
Solution:
Sides of a triangle are 35 cm, 54 cm, 61 cm
Question 9.
The lengths of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.
Solution:
Ratio in the sides of a triangle = 3:4:5
Question 10.
The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.
Solution:
Perimeter of an isosceles triangle = 42 cm
Base = $$\frac { 3 }{ 2 }$$ of its one of equal sides
Let each equal side = x, then 3
Base = $$\frac { 3 }{ 2 }$$ x
Question 11.
Find the area of the shaded region in figure.
Solution:
In ∆ABC, AC = 52 cm, BC = 48 cm
and in right ∆ADC, ∠D = 90°
AD = 12 cm, BD = 16 cm
∴ AB²=AD² + BD² (Pythagoras Theorem)
(12)² + (16)² = 144 + 256 = 400 = (20)²
∴ AB = 20 cm
Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4
Other Exercises
Question 1.
In the figure, O is the centre of the circle. If ∠APB = 50°, find ∠AOB and ∠OAB.
Solution:
Arc AB, subtends ∠AOB at the centre and ∠APB at the remaining part of the circle
∴∠AOB = 2∠APB = 2 x 50° = 100°
Join AB
∆AOB is an isosceles triangle in which
OA = OB
∴ ∠OAB = ∠OBA But ∠AOB = 100°
∴∠OAB + ∠OBA = 180° – 100° = 80°
⇒ 2∠OAB = 80°
80°
∴∠OAB = $$\frac { { 80 }^{ \circ } }{ 2 }$$ = 40°
Question 2.
In the figure, O is the centre of the circle. Find ∠BAC.
Solution:
In the circle with centre O
∠AOB = 80° and ∠AOC =110°
∴ ∠BOC = ∠AOB + ∠AOC
= 80°+ 110°= 190°
∴ Reflex ∠BOC = 360° – 190° = 170°
Now arc BEC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠BOC = 2∠BAC
⇒ 170° = 2∠BAC
⇒ ∠BAC = $$\frac { { 170 }^{ \circ } }{ 2 }$$ = 85°
∴ ∠BAC = 85°
Question 3.
If O is the centre of the circle, find the value of x in each of the following figures:
Solution:
(i) A circle with centre O
∠AOC = 135°
But ∠AOC + ∠COB = 180° (Linear pair)
⇒ 135° + ∠COB = 180°
⇒ ∠COB = 180°- 135° = 45°
Now arc BC subtends ∠BOC at the centre and ∠BPC at the remaining part of the circle
∴ ∠BOC = 2∠BPC
⇒ ∠BPC = $$\frac { 1 }{ 2 }$$∠BOC = $$\frac { 1 }{ 2 }$$ x 45° = $$\frac { { 45 }^{ \circ } }{ 2 }$$
∴ ∠BPC = 22 $$\frac { 1 }{ 2 }$$° or x = 22 $$\frac { 1 }{ 2 }$$°
(ii) ∵ CD and AB are the diameters of the circle with centre O
∠ABC = 40°
But in ∆OBC,
OB = OC (Radii of the circle)
∠OCB = ∠OBC – 40°
Now in ABCD,
∠ODB + ∠OCB + ∠CBD = 180° (Angles of a triangle)
⇒ x + 40° + 90° = 180°
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°
∴ x = 50°
(iii) In circle with centre O,
∠AOC = 120°, AB is produced to D
∵ ∠AOC = 120°
and ∠AOC + convex ∠AOC = 360°
⇒ 120° + convex ∠AOC = 360°
∴ Convex ∠AOC = 360° – 120° = 240°
∴ Arc APC Subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle
∴ ∠ABC = $$\frac { 1 }{ 2 }$$∠AOC = $$\frac { 1 }{ 2 }$$x 240° = 120°
But ∠ABC + ∠CBD = 180° (Linear pair)
⇒ 120° + x = 180°
⇒ x = 180° – 120° = 60°
∴ x = 60°
(iv) A circle with centre O and ∠CBD = 65°
But ∠ABC + ∠CBD = 180° (Linear pair)
⇒ ∠ABC + 65° = 180°
⇒ ∠ABC = 180°-65°= 115°
Now arc AEC subtends ∠x at the centre and ∠ABC at the remaining part of the circle
∴ ∠AOC = 2∠ABC
⇒ x = 2 x 115° = 230°
∴ x = 230°
(v) In circle with centre O
AB is chord of the circle, ∠OAB = 35°
In ∆OAB,
OA = OB (Radii of the circle)
∠OBA = ∠OAB = 35°
But in ∆OAB,
∠OAB + ∠OBA + ∠AOB = 180° (Angles of a triangle)
⇒ 35° + 35° + ∠AOB = 180°
⇒ 70° + ∠AOB = 180°
⇒ ∠AOB = 180°-70°= 110°
∴ Convex ∠AOB = 360° -110° = 250°
But arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∴∠ACB = $$\frac { 1 }{ 2 }$$∠AOB
⇒ x = $$\frac { 1 }{ 2 }$$ x 250° = 125°
∴ x= 125°
(vi) In the circle with centre O,
BOC is its diameter, ∠AOB = 60°
Arc AB subtends ∠AOB at the centre of the circle and ∠ACB at the remaining part of the circle
∴ ∠ACB = $$\frac { 1 }{ 2 }$$ ∠AOB
= $$\frac { 1 }{ 2 }$$ x 60° = 30°
But in ∆OAC,
OC = OA (Radii of the circle)
∴ ∠OAC = ∠OCA = ∠ACB
⇒ x = 30°
(vii) In the circle, ∠BAC and ∠BDC are in the same segment
∴ ∠BDC = ∠BAC = 50°
Now in ABCD,
∠DBC + ∠BCD + ∠BDC = 180° (Angles of a triangle)
⇒ 70° + x + 50° = 180°
⇒ x + 120° = 180° ⇒ x = 180° – 120° = 60°
∴ x = 60°
(viii) In circle with centre O,
∠OBD = 40°
AB and CD are diameters of the circle
∠DBA and ∠ACD are in the same segment
∴ ∠ACD = ∠DBA = 40°
In AOAC, OA = OC (Radii of the circle)
∴ ∠OAC = ∠OCA = 40°
and ∠OAC + ∠OCA + ∠AOC = 180° (Angles in a triangle)
⇒ 40° + 40° + x = 180°
⇒ x + 80° = 180° ⇒ x = 180° – 80° = 100°
∴ x = 100°
(ix) In the circle, ABCD is a cyclic quadrilateral ∠ADB = 32°, ∠DAC = 28° and ∠ABD = 50°
∠ABD and ∠ACD are in the same segment of a circle
∴ ∠ABD = ∠ACD ⇒ ∠ACD = 50°
⇒ ∠ACB = 32°
Now, ∠DCB = ∠ACD + ∠ACB
= 50° + 32° = 82°
∴ x = 82°
(x) In a circle,
∠BAC = 35°, ∠CBD = 65°
∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 35°
In ∆BCD,
∠BDC + ∠BCD + ∠CBD = 180° (Angles in a triangle)
⇒ 35° + x + 65° = 180°
⇒ x + 100° = 180°
⇒ x = 180° – 100° = 80°
∴ x = 80°
(xi) In the circle,
∠ABD and ∠ACD are in the same segment of a circle
∴ ∠ABD = ∠ACD = 40°
Now in ∆CPD,
∠CPD + ∠PCD + ∠PDC = 180° (Angles of a triangle)
110° + 40° + x = 180°
⇒ x + 150° = 180°
∴ x= 180°- 150° = 30°
(xii) In the circle, two diameters AC and BD intersect each other at O
∠BAC = 50°
In ∆OAB,
OA = OB (Radii of the circle)
∴ ∠OBA = ∠OAB = 52°
⇒ ∠ABD = 52°
But ∠ABD and ∠ACD are in the same segment of the circle
∴ ∠ABD = ∠ACD ⇒ 52° = x
∴ x = 52°
Question 4.
O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.
Solution:
Given : O is the circumcentre of ∆ABC.
OD ⊥ BC
OB is joined
To prove : ∠BOD = ∠A
Construction : Join OC.
Proof : Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle
∴ ∠BOC = 2∠A …(i)
In right ∆OBD and ∆OCD Side OD = OD (Common)
Hyp. OB = OC (Radii of the circle)
∴ ∆OBD ≅ ∆OCD (RHS criterion)
∴ ∠BOD = ∠COD = $$\frac { 1 }{ 2 }$$ ∠BOC
⇒ ∠BOC = 2∠BOD …(ii)
From (i) and (ii)
2∠BOD = 2∠A
∴∠BOD = ∠A
Question 5.
In the figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = BC.
Solution:
Given : In the figure, a circle with centre O OB is the bisector of ∠ABC
To prove : AB = BC
Construction : Draw OL ⊥ AB and OM ⊥ BC
Proof: In ∆OLB and ∆OMB,
∠1 = ∠2 (Given)
∠L = ∠M (Each = 90°)
OB = OB (Common)
∴ ∆OLB ≅ ∆OMB (AAS criterion)
∴ OL = OM (c.p.c.t.)
But these are distance from the centre and chords equidistant from the centre are equal
∴ Chord BA = BC
Hence AB = BC
Question 6.
In the figure, O and O’ are centres of two circles intersecting at B and C. ACD is a straight line, find x.
Solution:
In the figure, two circles with centres O and O’ intersect each other at B and C.
ACD is a line, ∠AOB = 130°
Arc AB subtends ∠AOB at the centre O and ∠ACB at the remaining part of the circle.
∴ ∠ACB =$$\frac { 1 }{ 2 }$$∠AOB
= $$\frac { 1 }{ 2 }$$ x 130° = 65°
But ∠ACB + ∠BCD = 180° (Linear pair)
⇒ 65° + ∠BCD = 180°
⇒ ∠BCD = 180°-65°= 115°
Now, arc BD subtends reflex ∠BO’D at the centre and ∠BCD at the remaining part of the circle
∴ ∠BO’D = 2∠BCD = 2 x 115° = 230°
But ∠BO’D + reflex ∠BO’D = 360° (Angles at a point)
⇒ x + 230° = 360°
⇒ x = 360° -230°= 130°
Hence x = 130°
Question 7.
In the figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
Solution:
Arc AB subtend ∠ACB and ∠ADB in the same segment of a circle
∴ ∠ACB = ∠ADB = 40°
In ∆PDB,
∠DPB + ∠PBD + ∠ADB = 180° (Sum of angles of a triangle)
⇒ 120° + ∠PBD + 40° = 180°
⇒ 160° + ∠PBD = 180°
⇒ ∠PBD = 180° – 160° = 20°
⇒ ∠CBD = 20°
Question 8.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
A circle with centre O, a chord AB = radius of the circle C and D are points on the minor and major arcs of the circle
∴ ∠ACB and ∠ADB are formed Now in ∆AOB,
OA = OB = AB (∵ AB = radii of the circle)
∴ ∆AOB is an equilateral triangle,
∴ ∠AOB = 60°
Now arc AB subtends ∠AOB at the centre and ∠ADB at the remainder part of the circle.
∴ ∠ADB = $$\frac { 1 }{ 2 }$$ ∠AOB = $$\frac { 1 }{ 2 }$$x 60° = 30°
Now ACBD is a cyclic quadrilateral,
∴ ∠ADB + ∠ACB = 180° (Sum of opposite angles of cyclic quad.)
⇒ 30° + ∠ACB = 180°
⇒ ∠ACB = 180° – 30° = 150°
∴ ∠ACB = 150°
Hence angles are 150° and 30°
Question 9.
In the figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.
Solution:
In circle with centre O and ∠AOC = 150°
But ∠AOC + reflex ∠AOC = 360°
∴ 150° + reflex ∠AOC = 360°
⇒ Reflex ∠AOC = 360° – 150° = 210°
Now arc AEC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.
Reflex ∠AOC = 2∠ABC
⇒ 210° = 2∠ABC
∴ ∠ABC = $$\frac { { 210 }^{ \circ } }{ 2 }$$ = 105°
Question 10.
In the figure, O is the centre of the circle, prove that ∠x = ∠y + ∠z.
Solution:
Given : In circle, O is centre
To prove : ∠x = ∠y + ∠z
Proof : ∵ ∠3 and ∠4 are in the same segment of the circle
∴ ∠3 = ∠4 …(i)
∵ Arc AB subtends ∠AOB at the centre and ∠3 at the remaining part of the circle
∴ ∠x = 2∠3 = ∠3 + ∠3 = ∠3 + ∠4 (∵ ∠3 = ∠4) …(ii)
In ∆ACE,
Ext. ∠y = ∠3 + ∠1
(Ext. is equal to sum of its interior opposite angles)
⇒ ∠3 – ∠y – ∠1 …(ii)
From (i) and (ii),
∠x = ∠y – ∠1 + ∠4 …(iii)
Ext. ∠4 = ∠1 + ∠z …(iv)
From (iii) and (iv)
∠x = ∠y-∠l + (∠1 + ∠z)
= ∠y – ∠1 + ∠1 + ∠z = ∠y + ∠z
Hence ∠x = ∠y + ∠z
Question 11.
In the figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.
Solution:
In the figure, O is the centre of the circle,
PQ is the diameter and ∠ROS = 40°
Now we have to find ∠RTS
Arc RS subtends ∠ROS at the centre and ∠RQS at the remaining part of the circle
∴ ∠RQS = $$\frac { 1 }{ 2 }$$ ∠ROS
= $$\frac { 1 }{ 2 }$$ x 40° = 20°
∵ ∠PRQ = 90° (Angle in a semi circle)
∴ ∠QRT = 180° – 90° = 90° (∵ PRT is a straight line)
Now in ∆RQT,
∠RQT + ∠QRT + ∠RTQ = 180° (Angles of a triangle)
⇒ 20° + 90° + ∠RTQ = 180°
⇒ ∠RTQ = 180° – 20° – 90° = 70° or ∠RTS = 70°
Hence ∠RTS = 70°
Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3
Other Exercises
Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
Hope given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2
Other Exercises
Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = $$\frac { 1 }{ 2 }$$ x 108°
∴ We shall bisect it.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.
Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =$$\frac { 1 }{ 2 }$$ x 90° = 45°.
Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ $$\frac { 1 }{ 2 }$$ ∠DCA + $$\frac { 1 }{ 2 }$$ ∠DCB = 180° x $$\frac { 1 }{ 2 }$$ = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x $$\frac { 1 }{ 2 }$$ = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + $$\frac { 1 }{ 2 }$$ x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 $$\frac { 1 }{ 2 }$$°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
(vi) 22 $$\frac { 1 }{ 2 }$$°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 $$\frac { 1 }{ 2 }$$°
Hope given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3
Other Exercises
Question 1.
Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha. [NCERT]
Solution:
∵ Distance between Isha and Ishita and Ishita and Nisha is same
∴ RS = SM = 24 m
∴They are equidistant from the centre
In right ∆ORL,
OL² = OR² – RL²
Hence distance between Ishita and Nisha = 38.4 m
Question 2.
A circular park of radius 40 m is situated in a colony. Three beys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find thelength of the string of each phone. [NCERT]
Solution:
Radius of circular park = 40 m
Ankur, Amit and Anand are sitting at equal distance to each other By joining them, an equilateral triangle ABC is formed produce BO to L which is perpendicular bisector of AC
∴ BL = 40 + 20 = 60 m (∵ O is centroid of ∆ABC also)
Let a be the side of ∆ABC
Hence the distance between each other = 40$$\sqrt { 3 }$$ m
Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
# Radius of a Circle Calculator
## Calculate the radius of a circle with this free calculator
• Created by John Harris
• Reviewed by Ramesh Agarwal
The radius of the circle is:
## What is the radius of a circle?
The circle's radius can be described as the distance from the center of the circle to any point on the circumference of that circle (half a circle's width). It is an important property of a circle and its value can be found with the help of a simple formula.
## How to find the radius of a circle?
In the most cases, we should be able to find the radius from a circle fairly easily. It is the distance from the center of the circle to a point anywhere on the outer perimeter of the circle (also referred to as the semi perimeter of circle).
When we cannot find the radius, we can normally find the diameter instead. The diameter is the longest distance that passes through the circle, from one side to the other.
The radius of a circle calculation requires us to first know the surface area or volume of the circle. Once we have the area, we can then use that within out equation. We must divide the area by Pi (3.14), then get the square root of the result.
## Why do we calculate radius?
The main reasons why we calculate the radius is to use the value in other equations. Using the radius we can perform calculations to measure the circumference or the area of the circle. Knowing the radius also enables us to measure the diameter of the circle, giving a clearer picture of the dimensions of the circle.
## What is the formula for radius of a circle?
The formula for the radius of a circle is:
`Radius = (Area / 3.14)2`
The formula for radius can be carried out by taking the area of the circle and dividing it by pi (3.14) and then getting the square root.
You can use the radius of a circle when solving an equation where the radius is needed, such as finding the area of the circle, finding the circumference of the circle, or finding the average radius of a group of circles.
Obtaining the radius value is possible when you only have limited information about the measurements of the circle. If you know the surface area or volume of the circle, the value can be used to help determine the radius.
## FAQS
The radius of a circle is the distance from the center of the circle to any point on its circumference. It is denoted by the letter 'r'.
You can find the radius of a circle if you know the diameter by dividing the diameter by 2. The formula is r = d/2, where r is the radius and d is the diameter.
The circumference of a circle (C) is related to its radius (r) by the formula C = 2πr, where π is the mathematical constant pi (approximately 3.14159).
You can calculate the area of a circle if you know the radius using the formula A = πr², where A is the area and r is the radius.
No, the radius of a circle is a positive distance. It cannot be negative or zero.
The radius of a circle is usually measured in units such as centimeters (cm), meters (m), inches (in), or any other unit of length. |
Question
# If 12 distinct balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is (a) $\dfrac{55}{3}{{\left( \dfrac{2}{3} \right)}^{11}}$ (b) $55{{\left( \dfrac{2}{3} \right)}^{11}}$(c) $22{{\left( \dfrac{1}{3} \right)}^{11}}$(d) $228{{\left( \dfrac{2}{3} \right)}^{11}}$
Hint: To solve this question, we will first find the number of ways in which the given 12 balls can go in any one of the three boxes. This will be our sample set. Then we will find the number of ways in which we can choose 3 balls out of 12. We will also find the number of ways in which we can choose one of the three boxes. Then the rest of the remaining balls can go in any of the two remaining boxes. We will know the number of ways in which the remaining balls can be placed. The product of the number of ways to choose 3 balls out of 12 and number of ways of choosing one box out of 3 and number of ways remaining balls can be placed will be equal to the number of ways exactly 3 balls will be placed in one of the boxes. Thus, we can find the probability as the ratio of number of ways to place exactly 3 balls in one of the boxes and number of ways placing 12 balls in 3 boxes randomly.
Let S be the event of placing 12 distinct balls in 3 boxes randomly.
Each ball can be placed in 3 ways and there are 12 such balls.
Thus, number of ways 12 balls can be placed in 3 boxes is n(S) = ${{3}^{12}}......\left( 1 \right)$
Now, let A be the event of placing 3 balls in exactly one of the boxes.
To find the number of elements in the event set A, first we have to choose any 3 balls from the available 12 balls.
This can be done in $^{12}{{C}_{3}}$ = 220 ways.
Then, we have to choose one of the available 3 boxes. This is done in $^{3}{{C}_{1}}=3$ ways.
The remaining 9 balls can go into 1 of the 2 boxes. Hence the number of ways 9 balls can be placed in 2 boxes is given as ${{2}^{9}}$.
Therefore, number of elements in event set A is n(A) = ${{2}^{9}}\times 3\times 220......\left( 2 \right)$
Thus, the probability of event A happening, that one of the boxes has exactly 3 boxes is given as $\Rightarrow P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
We shall find the values of n(A) and n(S) from (1) and (2).
\begin{align} & \Rightarrow P\left( A \right)=\dfrac{{{2}^{9}}\times 3\times 220}{{{3}^{12}}} \\ & \Rightarrow P\left( A \right)=\dfrac{{{2}^{9}}\times 4\times 55}{{{3}^{11}}} \\ & \Rightarrow P\left( A \right)=55\left( \dfrac{{{2}^{11}}}{{{3}^{11}}} \right) \\ \end{align}
So, the correct answer is “Option b”.
Note: If we have to choose r items from n available items, then it can be in $^{n}{{C}_{r}}$ ways, where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. |
# Lesson 12
Ordenemos números
## Warm-up: Conversación numérica: Restemos decenas (10 minutes)
### Narrative
The purpose of this Number Talk is to elicit strategies and understandings students have for mentally subtracting a multiple of 10 from a number. Building on their understanding of place value, students subtract tens from tens. These understandings help students develop fluency and will be helpful in later lessons when students will need to be able to subtract using strategies based on place value.
### Launch
• Display one expression.
• “Hagan una señal cuando tengan una respuesta y puedan explicar cómo la obtuvieron” // “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Encuentra mentalmente el valor de cada expresión.
• $$80 - 50$$
• $$87 - 50$$
• $$76 - 40$$
• $$66 - 30$$
### Activity Synthesis
• “¿Cómo les puede ayudar la tercera expresión a encontrar el valor de la última expresión?” // “How could the third expression help you find the value of the last expression?” (76 is 10 more than 66 and 30 is 10 less than 40, so the difference between the two numbers is the same.)
## Activity 1: ¿Quién está en desorden? (15 minutes)
### Narrative
The purpose of this activity is for students to analyze a mistake in ordering numbers (MP3). When placing numbers in order from least to greatest, students can compare using their understanding of place value. However, they see that, unlike comparing just two numbers, when comparing sets of numbers, there is more to keep track of. Students learn that a number line provides a linear representation to help organize numbers in sequence and visualize the relative distance between numbers.
MLR8 Discussion Supports. Display sentence frames to support partner discussion: “Estoy de acuerdo con _____ porque . . .” // “I agree with _____ because . . . “ and “Estoy en desacuerdo con ______ porque . . .” // “I disagree with _____ because . . . .”
• Groups of 2
### Activity
• “Kiran y Andre ordenaron algunos números de menor a mayor” // “Kiran and Andre put some numbers in order from least to greatest.”
• “Andre estaba en desacuerdo con Kiran, así que usó una recta numérica para justificar su respuesta. ¿Con quién están de acuerdo?” // “Andre disagreed with Kiran, so he used a number line to justify his answer. Whom do you agree with?”
• 3 minutes: independent work time
• “Discutan con un compañero. Usen todo lo que saben sobre el valor posicional o la recta numérica para justificar su razonamiento” // “Discuss with a partner using what you know about place value or the number line to justify your reasoning.”
• 5 minutes: partner work time
• Monitor for students who:
• use precise place value language to describe the correct placement of 269 and 272 in the list
• use the number line to explain that a list of numbers from least to greatest should match the placement of the numbers on the number line from left to right
### Student Facing
Kiran y Andre ordenaron de menor a mayor una lista de números.
Kiran
207, 217, 272, 269, 290
Andre
207, 217, 269, 272, 290
Andre estaba en desacuerdo con Kiran, así que usó una recta numérica para justificar su respuesta.
¿Con quién estás de acuerdo? ¿Por qué?
Prepárate para explicar cómo pensaste. Usa lo que sabes sobre el valor posicional o la recta numérica para justificar tu razonamiento.
### Activity Synthesis
• Invite previously selected students to share using precise place value language.
• Invite a student to explain using the number line.
• “Cuando ordenamos números, podemos usar lo que sabemos sobre el valor posicional. También podemos pensar sobre la secuencia de conteo y usar una recta numérica para ayudarnos a ver los números en orden” // “When we order numbers, we can use what we know about place value. We can also think about the counting sequence and use a number line to help us see the numbers in order.”
## Activity 2: Ordenemos números (20 minutes)
### Narrative
The purpose of this activity is for students to order numbers. Students estimate the location and label numbers on a number line, and then write them in order from least to greatest or greatest to least. For the third set of numbers, students may order the numbers using any method that makes sense to them. Students reflect on how the number line can help us organize numbers (MP5). Throughout the activity, monitor for the way students explain their reasoning based on place value and the relative position of numbers on the number line.
Representation: Access for Perception. Use index cards, clothespins, and string to demonstrate the number line. Give students the numbers on the index cards and have them physically act it out by finding their place on the number line.
Supports accessibility for: Memory, Organization, Conceptual Processing
• Groups of 2
### Activity
• “Ahora van a tener la oportunidad de ordenar números” // “Now you will have a chance to order numbers.”
• “Algunas veces los van a ordenar de menor a mayor y algunas veces lo harán de mayor a menor” // “Sometimes you will put them in order from least to greatest, and sometimes it will be from greatest to least.”
• 10 minutes: independent work time
• “Comparen con un compañero. Expliquen cómo pensaron” // “Compare with a partner. Explain your thinking.”
• 5 minutes: partner discussion
• Monitor for students who order the last set of numbers by:
• explaining their reasoning using precise place value language
• placing each number on the number line before ordering the numbers
### Student Facing
1. Estima la ubicación de 839, 765, 788, 815 y 719 en la recta numérica. Marca cada número con un punto y escribe debajo el número que representa.
Ordena los números de menor a mayor.
_______, _______, _______, _______, _______
2. Estima la ubicación de 199, 245, 173, 218 y 137 en la recta numérica. Marca cada número con un punto y escribe debajo el número que representa.
Ordena los números de mayor a menor.
_______, _______, _______, _______, _______
3. Ordena los números de menor a mayor.
545, 454, 405, 504, and 445
_______, _______, _______, _______, _______
Explica o muestra cómo pensaste. Si te ayuda, usa la recta numérica.
4. ¿Te ayudó más ordenar los números primero o ponerlos en la recta numérica primero? Explica.
### Student Response
If students locate the numbers on the number line, but reverse the order given in the problem, consider asking:
• “¿Cómo decidiste el orden de tus números?” // “How did you decide the order of your numbers?”
• “¿Tu lista muestra los números en orden de menor a mayor o de mayor a menor?” // “Does your list show the numbers in order from least to greatest or greatest to least?”
### Activity Synthesis
• Invite previously identified students to share their reasoning.
• “¿En qué se parecen los números que marcaron en la recta numérica a la lista de números que escribieron? ¿En qué son diferentes?” // “How are the numbers you labeled on the number line the same as the list of numbers you wrote? How are they different?” (For least to greatest, its the same. The difference is the number line shows the distance between each number, but the list of numbers shows them right next to each other.)
## Lesson Synthesis
### Lesson Synthesis
“En esta unidad, hemos usado diferentes representaciones para ayudarnos a pensar sobre números grandes. Piensen en todo el trabajo que hicimos con números hasta 1,000” // “During this unit we have used different representations to help us think about large numbers. Think about all the work we did with numbers up to 1,000.”
“¿Cuáles representaciones les ayudan a entender los números grandes y a compararlos entre ellos? ¿Usar un diagrama en base diez, examinar los dígitos o usar una recta numérica?” // “Which representations help you make sense of large numbers and compare them to one another? Using a base-ten diagram, looking at the digits, or using a number line?”
Share and record responses.
“Observé que algunos estudiantes prefieren una representación para el valor posicional y una diferente para comparar u ordenar números. Es bueno saber qué le funciona mejor a cada uno y cuándo usarlo” // “I noticed some students prefer one representation for place value, but a different one when comparing or ordering numbers. It is good to know what works best for you and when to use it.”
## Student Section Summary
### Student Facing
En esta sección aprendimos cómo comparar números de tres dígitos. Usamos rectas numéricas, el valor de los dígitos en numerales en base diez y diagramas en base diez como ayuda para comparar y explicar cómo pensamos.
Los diagramas nos ayudan a comparar números porque podemos ver y comparar centenas con centenas, decenas con decenas y unidades con unidades. Aprendimos que eso también se puede hacer con los dígitos.
La recta numérica muestra los números en orden, entonces podemos ver cuál número es el más grande basándonos en su ubicación.
También escribimos expresiones usando los símbolos $$>$$, $$<$$ y $$=$$.
$$432 > 424$$
$$424 < 432$$
432 es mayor que 424
424 es menor que 432 |
5.6 More Decimal and Fraction Operations
Unit Goals
• Students solve multi-step problems involving measurement conversions, line plots, and fraction operations, including addition and subtraction of fractions with unlike denominators. They also explain patterns when multiplying and dividing by powers of 10 and interpret multiplication as scaling by comparing products with factors.
Section A Goals
• Explain patterns when multiplying and dividing by powers of 10.
• Solve multi-step problems involving measurement conversions.
Section B Goals
• Add and subtract fractions with unlike denominators.
• Create line plots to display fractional measurement data, and use the information to solve problems.
• Solve problems involving addition and subtraction of fractions.
Section C Goals
• Interpret multiplication as scaling (resizing).
• Make generalizations about multiplying a whole number by a fraction greater than, less than and equal to 1.
Problem 1
Pre-unit
Practicing Standards: 4.NF.A.1
Find the number that makes each equation true. Explain or show your reasoning.
1. $$\frac{\boxed{\phantom{\frac{0}{000}}}}{12} = \frac{2}{3}$$
2. $$\frac{5}{6} = \frac{\boxed{\phantom{\frac{0}{000}}}}{12}$$
Problem 2
Pre-unit
Practicing Standards: 4.MD.A.1
1. The road around a lake is 15 kilometers long. How many meters is that?
2. The length of an alligator is 4 meters. How many centimeters is that?
Problem 3
Pre-unit
Practicing Standards: 4.NBT.A.1
The value of the 6 in 618,204 has how many times the value as the 6 in 563? Explain or show your reasoning.
Problem 4
Pre-unit
Practicing Standards: 4.NF.B.3.c
Find the value of each sum.
1. $$\frac{3}{8} + \frac{9}{8}$$
2. $$3\frac{1}{5} + \frac{3}{5}$$
3. $$2\frac{4}{10} + 1\frac{7}{10}$$
Problem 5
Pre-unit
Practicing Standards: 4.OA.A.1
Lin spent 5 minutes reading the story. Noah spent 3 times as long as Lin. How long did Noah spend reading the story? Explain or show your reasoning.
Problem 6
Pre-unit
Practicing Standards: 5.NF.B.4
1. Write a multiplication expression for the shaded area and find the value of the expression. Explain or show your reasoning.
2. Find the value of $$\frac{5}{7} \times \frac{10}{3}$$.
Problem 7
Pre-unit
Practicing Standards: 4.MD.B.4
The line plot shows the lengths of some toothpicks in inches.
1. How many measurements are there?
2. What is the difference between the longest and shortest toothpicks?
Problem 8
1. Write a multiplication equation relating the values of 0.5 and 0.05.
2. Write a division equation relating the values of 0.5 and 0.05.
Problem 9
Write each number using exponential notation.
1. $$10 \times 10 \times 10$$
2. $$100 \times 100$$
3. 100,000
4. 1,000,000,000
Problem 10
1. How many centimeters are in each measurement?
0.12 m
3.5 m
19 m
2. How many millimeters are in each measurement?
3 m
37 m
1,915 m
3. How does a whole number of meters change when it is converted to millimeters?
Problem 11
1. How many meters are in each measurement?
16 millimeters
1,375 millimeters
57 millimeters
2. How does a whole number of millimeters change when you express the measurement in meters?
Problem 12
A track is 366 meters around. An athlete runs 15 laps. How many kilometers is that? Explain or show your reasoning.
Problem 13
Clare drinks 8 glasses of water each day. There are 235 milliliters in each glass. How many liters of water does Clare drink each day? Explain or show your reasoning.
Problem 14
A track is 400 yards around. How many full laps does Tyler need to run if he wants to run at least 2 miles?
Problem 15
Exploration
1. Write each of these numbers using exponential notation.
1. 1,000,000,000 (the approximate population of Africa in 2009)
2. 100,000,000,000 (estimated number of stars in Milky Way)
3. 1,000,000,000,000 (amount of dollars added to United States debt each year recently)
4. 100,000,000,000,000 (denomination of a bill in Zimbabwe)
5. 1,000,000,000,000,000,000,000 (estimated number of stars in universe)
6. 10,000,000,000,000,000,000 (estimated number of grains of sand on the earth)
7. 100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000 (estimated number of atoms in the universe!)
2. How is exponential notation helpful for writing these numbers?
Problem 16
Exploration
You have a piggy bank with 1 kg of coins inside. Which coins would you like to be in the piggy bank? Explain or show your reasoning.
coin approximate weight (grams)
penny 2.5
nickel 5
dime 2.3
quarter 5.7
Problem 1
1. Find the value of each sum. Explain or show your reasoning.
1. $$\frac{5}{6} +\frac{2}{6}$$
2. $$\frac{5}{6} + \frac{2}{3}$$
2. How were the calculations the same? How were they different?
Problem 2
1. Explain why the expressions $$\frac{2}{3} - \frac{7}{12}$$ and $$\frac{8}{12} - \frac{7}{12}$$ are equivalent.
2. How is the expression $$\frac{8}{12} - \frac{7}{12}$$ helpful to find the value of $$\frac{2}{3} - \frac{7}{12}$$?
Problem 3
Find the value of each expression. Explain or show your reasoning.
1. $$\frac{1}{4} + \frac{1}{5}$$
2. $$\frac{10}{9} - \frac{3}{4}$$
Problem 4
1. Find the value of $$2\frac{3}{4} - \frac{1}{3}$$. Explain or show your reasoning.
2. Find the value of $$3\frac{2}{7} - \frac{4}{5}$$. Explain or show your reasoning.
Problem 5
Jada picked $$4 \frac{2}{3}$$ cups of blackberries. Andre picked $$3\frac{5}{8}$$ cups of blackberries.
1. How many cups of blackberries did Jada and Andre pick together? Explain or show your reasoning.
2. How many more cups of blackberries did Jada pick than Andre? Explain or show your reasoning.
Problem 6
Find the value of each expression. Explain or show your reasoning.
1. $$\frac{7}{8} + \frac{4}{13}$$
2. $$\frac{7}{8} - \frac{3}{20}$$
Problem 7
Here are the lengths of some pieces of ribbon measured in inches:
• $$3\frac{1}{4}$$
• $$4\frac{1}{8}$$
• $$3\frac{6}{8}$$
• $$3\frac{1}{8}$$
• $$2\frac{5}{8}$$
• $$3\frac{2}{4}$$
• $$3\frac{1}{4}$$
• $$3\frac{7}{8}$$
• $$4\frac{1}{8}$$
• $$3\frac{1}{2}$$
• $$2\frac{7}{8}$$
• $$4\frac{1}{8}$$
• $$3\frac{3}{4}$$
• $$3\frac{2}{8}$$
1. Complete the line plot with the ribbon lengths.
2. What is the sum of the lengths of the ribbons that measure more than 4 inches? Explain or show your reasoning.
Problem 8
Han is making a line plot of the seedlings his class grew. This is what he has done so far.
Use this information to complete the line plot. Explain or show your reasoning.
• There are 15 seedlings altogether.
• The tallest seedling is $$2\frac{1}{8}$$ taller than the shortest seedling.
• There are 3 seedlings of the shortest height.
Problem 9
Exploration
1. Put the numbers 2, 3, 4, and 5 in the four boxes so that the expression is as close to 1 as possible. $$\frac{\boxed{\phantom{\frac{0}{000}}}}{\boxed{\phantom{\frac{0}{000}}}} + \frac{\boxed{\phantom{\frac{0}{000}}}}{\boxed{\phantom{\frac{0}{000}}}}$$
2. Put the numbers 2, 3, 4, and 5 in the four boxes so that the expression is as close to 1 as possible. $$\frac{\boxed{\phantom{\frac{0}{000}}}}{\boxed{\phantom{\frac{0}{000}}}} - \frac{\boxed{\phantom{\frac{0}{000}}}}{\boxed{\phantom{\frac{0}{000}}}}$$
Problem 10
Exploration
Make a line plot of seedling heights so that each of these statements is true.
• There are 12 measurements.
• The largest measurement is $$2 \frac{3}{8}$$ inches more than the smallest measurement.
• The sum of the measurements is $$18\frac{3}{8}$$ inches.
Explain how you made the line plot.
Problem 1
1. Andre ran $$\frac{4}{5}$$ of a 7 mile trail. Did Andre run more or less than 7 miles? Explain or show your reasoning.
2. Clare ran $$\frac{\boxed{\phantom{\frac{0}{000}}}}{\Large{10}}$$ of a 7 mile trail. She ran more than 7 miles. Choose a number that could go in the box. Explain or show your reasoning.
Problem 2
The point J on the number line shows how many miles Jada ran. Label the points on the number line to show how far each of these students ran.
1. Clare ran $$\frac{8}{5}$$ as far as Jada.
2. Tyler ran $$\frac{4}{3}$$ as far as Jada.
3. Lin ran $$\frac{1}{2}$$ as far as Jada.
Problem 3
The point A is labeled on the number line.
Label each of these points on the number line.
• $$\frac{2}{5} \times \text{A}$$
• $$\frac{13}{10} \times \text{A}$$
• $$\frac{7}{4} \times \text{A}$$
Problem 4
Use the equation $$\frac{5}{7} = \left(1 - \frac{2}{7}\right)$$ to explain why $$\frac{5}{7} \times \frac{11}{3} < \frac{11}{3}$$.
Problem 5
Explain why multiplying a fraction by a number less than 1 makes the fraction smaller.
Problem 6
Exploration
A point P is labeled on the number line.
1. P is $$\frac{3}{4}$$ of a number A. Plot A on the number line. Explain or show your reasoning.
2. P is $$\frac{5}{9}$$ of a number B. Plot B on the number line. Explain or show your reasoning.
Problem 7
Exploration
1. About $$10^6$$ people live in Michigan. About $$10^4$$ of the people in Michigan live in Flint.
1. How many times as many people live in Michigan as in Flint?
2. How many times as many people live in Flint as in Michigan?
2. There are about $$10^{11}$$ stars in the Milky Way. There are about $$10^{21}$$ stars in the universe.
1. How many times as many stars are there in the universe than in the Milky Way?
2. How many times as many stars are there in the Milky Way than in the universe? |
# Unit 2 Reasoning with Equations and Inequalities.
## Presentation on theme: "Unit 2 Reasoning with Equations and Inequalities."— Presentation transcript:
Unit 2 Reasoning with Equations and Inequalities
Properties of Equality Properties are rules that allow you to balance, manipulate, and solve equations
Addition Property of Equality Adding the same number to both sides of an equation does not change the equality of the equation. If a = b, then a + c = b + c. Ex: If x=y, so x+2=y+2
Subtraction Property of Equality Subtracting the same number to both sides of an equation does not change the equality of the equation. If a = b, then a – c = b – c. Ex: x = y, so x – 4 = y – 4
Multiplication Property of Equality Multiplying both sides of the equation by the same number, other than 0, does not change the equality of the equation. If a = b, then ac = bc. Ex: x = y, so 3x = 3y
Division Property of Equality Dividing both sides of the equation by the same number, other than 0, does not change the equality of the equation. If a = b, then a/c = b/c. Ex: x = y, so x/7 = y/7
Reflexive Property of Equality A number is equal to itself. (Think mirror) a = a Ex: 4 = 4
Symmetric Property of Equality If numbers are equal, they will still be equal if the order is changed. If a = b, then b = a. Ex: x = 4, then 4 = x
Transitive Property of Equality If numbers are equal to the same number, then they are equal to each other. If a = b and b = c, then a = c. Ex: If x = 8 and y = 8, then x = y
Substitution Property of Equality If numbers are equal, then substituting one in for another does not change the equality of the equation. If a = b, then b may be substituted for a in any expression containing a. Ex: x = 5, then y = x + 6 is the same as y = 5 + 6.
Commutative Property Changing the order of addition or multiplication does not matter. “Commutative” comes from “commute” or “move around”, so the Commutative Property is the one that refers to moving stuff around.
Commutative Property of Addition a + b = b + a Ex: 1 + 9 = 9 + 1
Commutative Property of Multiplication a ∙ b = b ∙ a Ex: 8 ∙ 6 = 6 ∙ 8
Associative Property The change in grouping of three or more terms/factors does not change their sum or product. “Associative” comes from “associate” or “group”, so the Associative Property is the one that refers to grouping.
Associative Property of Addition a + (b + c) = (a + b) + c Ex: 1 + (7 + 9) = (1 + 7) + 9
Associative Property of Multiplication a ∙ (b ∙ c) = (a ∙ b) ∙ c Ex: 8 ∙ (3 ∙ 6) = (8 ∙ 3) ∙ 6
Distributive Property The product of a number and a sum is equal to the sum of the individual products of terms.
Distributive Property a ∙ (b + c) = a ∙ b + a ∙ c Ex: 5 ∙ (x + 6) = 5 ∙ x + 5 ∙ 6
Additive Identity Property The sum of any number and zero is always the original number. Adding nothing does not change the original number. a + 0 = a Ex: 4 + 0 = 4
Multiplicative Identity Property The product of any number and one is always the original number. Multiplying by one does not change the original number. a ∙ 1 = a Ex: 2 ∙ 1 = 2
Additive Inverse Property The sum of a number and its inverse (or opposite) is equal to zero. a + (-a) = 0 Ex: 2 + (-2) = 0
Multiplicative Inverse Property The product of any number and its reciprocal is equal to Ex:
Multiplicative Property of Zero The product of any number and zero is always zero. a ∙ 0 = 0 Ex: 298 ∙ 0 = 0
Exponential Property of Equality Ex:
Properties of Equality EquationSteps Original Equation Associative Property of Addition (1) Addition Property of Equality (2) Division Property of Equality
Properties of Equality EquationSteps Original Equation (3) Subtraction Property of Equality (4) Division Property of Equality
Properties of Equality EquationSteps Original Equation (5) Distributive Property Combine Like Terms (6) Addition Property of Equality (7) Division Property of Equality (8) Symmetric Property of Equality
Properties of Equality EquationSteps Original Equation Subtraction Property of Equality Division Property of Equality
Properties of Equality EquationSteps Original Equation Subtraction Property of Equality Addition Property of Equality
Properties of Equality EquationSteps Original Equation Combine Like Terms Addition Property of Equality Subtraction Property of Equality Division Property of Equality |
# Mean Value Theorem
The mean value theorem states that if f is a continuous function on the closed interval [a,b] and differentiable on the open interval, then there exists a number c in (a,b) such that the tangent at c is parallel to the secant line through the endpoints, that is, $f’(c) = \frac{f(b)-f(a)}{b-a}$. Figure 1.e.
Geometrically, it states that the slope of the secant line is the same as the slope of the tangent line at x = c. Physically, there is a point at which the average velocity is equal to instantaneous velocity.
Be cautious. If f is not differentiable, even at a single point, the result may not hold. For example, the function f(x) = |x| - 1 is continuous over [-1, 1], f(-1) = f(1) = 0, but f’(c) ≠ 0 ∀c ∈ (-1, 1)
Proof.
We will use Rolle’s Theorem. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that f’(c) = 0.
Let $g(x) = f(x) - \frac{f(b)-f(a)}{b-a}x$
$g(a) = f(a) - \frac{f(b)-f(a)}{b-a}a = f(a)(1+\frac{a}{b-a})- a\frac{f(b)}{b-a} = f(a)\frac{b}{b-a}- f(b)\frac{a}{b-a}$
$g(b) = f(b) - \frac{f(b)-f(a)}{b-a}b = f(b)(1-\frac{b}{b-a})+ b\frac{f(a)}{b-a} = f(b)\frac{-a}{b-a}+ f(a)\frac{b}{b-a} = g(a)$
Therefore, ∃c ε (a, b): g’(c) = 0
$g’(c) = 0 = f’(c) - \frac{f(b)-f(a)}{b-a}$
# Increasing and decreasing functions.
Let’s suppose f is continuous on the interval [a, b], and diferenciable on (a, b).
1. If f’(x) > 0 ∀x ∈ (a, b), then f is increasing on (a, b).
2. If f’(x) < 0 ∀x ∈ (a, b), then f is decreasing on (a, b).
3. If f’(x) = 0 ∀x ∈ (a, b), then f is constant on (a, b).
Proof: ∀ x1, x2 ∈ (a, b), x1 < x2, then $x_2-x_1>0$ and the Mean Value Theorem guarantees that we can find a c ∈ (x1, x2): $\frac{f(x_2)-f(x_1)}{x_2-x_1}=f’(c)$ ⇒ $f(x_2)=f(x_1)+f’(c)(x_2-x_1)$
1. If f’(x) > 0 ∀x ∈ (a, b) ⇒ f’(c) > 0 ⇒ f(x2) > f(x1) ⇒ f is increasing.
2. If f’(x) < 0 ∀x ∈ (a, b) ⇒ f’(c) < 0 ⇒ f(x2) < f(x1) ⇒ f is decreasing.
3. If f’(x) = 0 ∀x ∈ (a, b) ⇒ f’(c) = 0 ⇒ f(x2) = f(x1) ⇒ f is constant.
Exercise 1: ex > x + 1 ∀x >0. Let be f(x) = ex -x -1. f(0) = 0, f’(x)>0 ⇒ f is increasing ∀x >0 ⇒ f(x) > 0 ∀x >0 ⇒ ex > x + 1 ∀x >0
Exercise 2: ex > x + 1 + $\frac{x^{2}}{2},$∀x >0. Let be g(x) = ex -x -1 -$\frac{x^{2}}{2}$, g(0) = 0, g’(x) = ex -1 -x > 0 ∀x >0 (Exercise 1) ⇒ g is increasing ⇒ g(x) > 0 ∀x >0 ⇒ ex > x + 1 + $\frac{x^{2}}{2},$∀x >0
Futhermore, ex > x + 1 + $\frac{x^{2}}{2} + \frac{x^{3}}{3.2} + …$, but in the infinite they are both equal.
# Fermat’s Theorem
Fermat’s Theorem. If a function ƒ(x) is defined on the interval (a, b), it has a local extremum (relative extrema) on the interval at x = c, and ƒ′(c) exists, then x = c is a critical point of f(x) and f'(c) = 0 (Figure 1.a.)
Proof:
Let’s assume that f(x) has a relative maximum to do the proof. The proof for a relative minimum is pretty identical.
$f’(c)=\lim_{h \to 0}=\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}≤0$ because we are dividing f(c+h)-f(c)≤0 and h>0 (Figure 1.b.). Observe that f(c) ≥ f(c+h) for all h that are sufficiently close to zero (c is a relative maximum). Therefore, f’(c) ≤ 0.
$f’(c)=\lim_{h \to 0}=\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}≥0$ because we are dividing f(c+h)-f(c) ≤ 0 and h<0. Therefore, f’(c) ≥ 0. We have already shown that f’(c) ≤ 0 and f’(c) ≥ 0, so f’(c) = 0.
# Rolle’s Theorem
Rolle’s Theorem. It states that any real-valued differentiable function that has equal values at two distinct points must have at least one stationary point somewhere between them, that is, a point where the first derivative is zero. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that f'(c) = 0 (Figure 1.c, 1.d).
Proof. There are two cases.
• f(x) = k, ∀x in [a, b] where k is a constant. ⇒ f’(x) = 0 ∀x in [a, b], and we can take c to be any number in the interval.
• Since f is continuous on [a, b] by the Extreme Value Theorem we know that the function must have a maximum M and a minimum on the interval m.
M ≠ m. Otherwise, if M = m, then f is constant ∀x in [a, b].
Assume M ≠ m. If f(a)=M, then f(b) = f(a) = M ⇒ m have to occur in the open interval (a, b). Therefore, there is at least one maximum or minimum in the open interval (a, b).
Without lose of generosity, let’s assume that f has a maximum in the open interval (a, b), that is, f has a maximum at some c in (a, b) ⇒ [by Fermat’s Theorem and f is differentiable on the open interval (a, b), so f’(c) exists] f’(c) = 0.
# Extreme Value Theorem.
The maxima and minima of a function, known collectively as extrema are the largest and smallest value of the function, either within a given range (the local or relative extrema), or on the entire domain.
Definitions. f(x) has an absolute or global maximum at x = c if f(x)≤f(x) ∀x ∈ D, that is, for every x in the domain we are working on. f(x) has an absolute or global minimum at x = c if f(x)≥f(x) ∀x ∈ D.
f(x) has a relative global maximum at x = c if f(x)≤f(x) ∀x in the neighborhood of c, that is, for every x in (b-ε, b+ε). f(x) has a relative minimum at x = c if f(x)≥f(x) ∀x in the neighborhood of c.
Extreme Value Theorem. If f is a continuous function on an interval [a,b], then f has both a maximum and minimum values on [a,b]. It states that if a function is continuous on a closed interval, then the function must have a maximum and a minimum on the interval (Figure 1.f.)
Definition. Let S be a non-empty set of real numbers. A real number M is called an upper bound or majorant for S if M ≥ s ∀s ∈ S. A real number M is the least upper bound or supremum for S if M is both an upper bound for S and also M ≤ y for every upper bound y of S.
Lemma. Let S be a non-empty set of real numbers with a supremum M. Then, there is a sequence {sn} that converges to the supremum, that is, {sn}→M.
For every N consider M -1N. It is pretty obvious that M -1N < M. M is a supremum, so by definition there is some (aka some element in S better than M -1N) Snε S such that M -1N < sn ≤ M. By the Squeeze theorem, M -1N→M, M→M, then sn →M.
# Boundedness theorem
Definition. A function is bounded if for all x in the domain of f there is a real M such that |f(x)| ≤ M.
Boundedness theorem. If f(x) is continuous on [a,b] then it is bounded on [a,b].
Be careful, f(x) = 1x is not bounded on (0, 1) -Figure 1.g.-
Proof. Suppose the function is not bounded on [a,b] ⇒ ∃xn ∈ [a,b] such that |f(xn)| > n.
Because [a,b] is obviously bounded, the Bolzano–Weierstrass theorem implies there exists a convergent subsequence {xnk} of {xnk} with {xnk} → x0. As [a,b] is closed, it contains x0.
Since f is continuous and {xnk} → x0, {|f(xnk)|} → |f(x0)| and |f(x0)| is a real number.
However, |f(xn)| > n ⇒ [and every subsequence do so, too] |f(xnk)| > nk, which implies {f(xnk)} diverges. That’s a contradiction.
# The least-upper-bound property
The least-upper-bound property states thatany non-empty set of real numbers that has an upper bound must have a least upper bound or supreme in real numbers.
Note: It is false for ℚ, S = {x∈ℚ: x2 < 2} in $(-\sqrt{2}, \sqrt{2})$. This is true because of the completeness of the real numbers. It implies that there are no gaps or missing points" in the real number line.
# Bolzano–Weierstrass theorem
Bolzano–Weierstrass theorem states that each infinite bounded sequence in ℝn has a convergent subsequence.
Proof (Extreme Value Theorem).
Let’s show that the function has a maximum. To prove that the function has a minimum, use -f.
By the boundedness theorem, there is a C > 0, |f(x)| ≤ C ∀x ∈ [a,b]
Let S = {f(x) : x ∈ [a,b]}. Since f is bounded, so S is bounded. Hence (by the least-upper-bound property) S has a least upper bound M.
By our previous lemma, there is a sequence {yn} in S such that {yn} → M
{yn} = {f(xn)} → M, xn ∈ [a,b]
Xn is obviously bounded (between a and b), therefore by the Bolzano–Weierstrass theorem, there is a convergent subsequence, {xnk} → x0 ∈ [a,b].
Let’s prove f(x0)=M.
Since {xnk} → x0 and f is continuous ⇒ {f(xnk)} → f(x0). We already know that {yn} = {f(xn)} → M, so any subsequence {xnk} must also convert to M, {f(xnk)} → M, and therefore f(x0)=M.
f(x0) = M and M is the least upper bound or supreme of S = {f(x) : x ∈ [a,b]}, f(x0) = M ≥ f(x) ∀x ∈ [a,b], so M is the maximum of f in [a,b].
# Differential
The term differential is used in calculus to refer to an infinitesimal, that is, infinitely small, change in some varying quantity. It is possible to relate to infinite small changes of various variables to each other using derivatives. If y is a function of x, then the differential dy of y is related to dx by dy = f’(x)dx or equivalently $\frac{dy}{dx} = f’(x)$.
There is an interpretation of derivate as a ratio of infinitesimals. Basically, dx and dy replace $\Delta x$ and $\Delta y$.
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Algebra Word Problems: How to Tackle Them with Ease
Algebra word problems can be a real headache for students. These types of problems require not only a strong understanding of algebraic concepts but also the ability to interpret and solve complex word problems. In this article, we'll take a look at some tips and tricks to help you solve algebra word problems with ease.
First and foremost, it's important to understand the key concepts involved in solving algebra word problems. These concepts include variables, equations, and inequalities. Variables are symbols used to represent unknown quantities, while equations and inequalities are mathematical statements that describe the relationship between these unknown quantities.
Now, let's dive into some specific strategies for solving algebra word problems. One helpful approach is to break the problem down into smaller parts. Read the problem carefully and identify the key pieces of information, such as the values of the variables and any given equations or inequalities. Then, begin to organize this information into a system of equations or inequalities that can be solved.
Another useful strategy is to use LSI keywords to help you identify the underlying meaning of the problem. LSI keywords are related phrases or words that help search engines understand the context and meaning of a piece of content. In the case of algebra word problems, LSI keywords might include "rate," "distance," or "time."
For example, consider the following algebra word problem:
"John drove 60 miles in 2 hours. What was his average speed?"
Identify that this problem involves the concepts of "distance" and "time." We can then use these keywords to create an equation that relates these two values: distance = rate x time. In this case, we know that the distance is 60 miles and the time is 2 hours. We can then solve for the rate, which is the average speed: rate = distance / time = 60 / 2 = 30 miles per hour.
Underlying meaning of the problem, we can more easily create a system of equations or inequalities that can be solved. This can help to simplify the problem and make it easier to understand and solve.
In conclusion, algebra word problems can be challenging, but with the right strategies and a solid understanding of key concepts, they can be tackled with ease. By breaking the problem down into smaller parts, the underlying meaning, and creating a system of equations or inequalities, you can solve even the most complex algebra word problems. Good luck! |
# 2/3 Cup Plus 2/3 Cup Equals How Much?
The answer is still 4 / 3 which, when simplified, is 1 whole and 1 / 3. To work with adding fractions you have to remember that you can leave the denominator as the same number as long as they are the same and simply just add the numerators (top number).
So in this case, if you add the second 2 to the first 2 and you will get 4. You write this answer as a fraction and so since you leave the bottom number the same you are left with 4 / 3; if you divide 4 by 3 you will get 1.33 which is essentially 1 and 1 third.
Taking this specific example out of the equation, I will explain the general steps involved in adding fractions:
• First you need to understand the difference between the numerator and the denominator. The numerator is the top number of the fraction, and the denominator is the bottom number of the fraction
• Next you will need to add numerators and write the answer in line with them
• Then write out the denominator under the answer of the two numerators to make a new fraction
Here is a simple example of the above which should help you to understand the formula:
We will use the following sum: ½ + ½. You will probably already know the answer to this which will make it much easier to see how the equation works.
So first you add the numerators so this is:
1 + 1 = 2
Since the bottom two numbers are the same, you can leave this as it is and just put it below your numerator answer. To simplify this answer, you can divide the numbers you are left with so in this case 2 divided by 2, this equals 1 which is your answer (2 / 2 = 1).
thanked the writer. |
86 terms
# Math Accuplacer Practice
Math Accuplacer
###### PLAY
divisible by 3
if the sum of the digits is divisible by 3
Ex) 456 ---> 4+5+6=15, 15 is divisible by 3
divisible by 4
if the number named by the last 2 digits is divisible by 4
Ex) 3936 is divisible by 4 because 36 is divisible by 4
divisible by 6
if rules for 2 and 3 check out ---- 1) the number is even & 2) the sum of the digits is divisible by 3
divisible by 8
if the number named by the last three digits is divisible by 8
Ex) 13802120 ---> 120/8 = 15
divisible by 9
if the sum of the digits is divisible by 9
Ex) 3483 ---> 3+4+8+3=18, 18 is divisible by 9
divisible by 11
if the difference between the sum of the even-place digits and the sum of the odd-place digits is a multiple of 11
Ex) 928,193,926 ---> odd sum (9+8+9+9+6=41) even sum (2+1+3+2=8) ---> 41-8=33, 33 is divisible by 11
multiplying decimals
count the number of decimal places and add them
Ex) 45.67 x .987 (5 places) = 45.07629 (5 places)
dividing decimals
move the decimal point in the divisor and the dividend the same number of places to get whole numbers
Ex) 23.1 / .004 ----> 23.1 x 1000 / .004 x 1000 ----> 23100/4 ---> 5775
1/2
.5 OR 50%
1/4
.25 OR 25%
3/4
.75 OR 75%
1/5
.2 OR 20%
2/5
.4 OR 40%
3/5
.6 OR 60%
4/5
.8 OR 80%
1/6
.16 OR 16 2/3%
1/3
.33 OR 33 1/3%
2/3
.6 OR 66 2/3%
5/6
.83 OR 83 1/3%
1/8
.125 OR 12 1/2%
3/8
.375 OR 37 1/2%
5/8
.625 OR 62 1/2%
7/8
.875 OR 87 1/2%
1/16
.0625 OR 6 1/4%
3/16
.1875 OR 18 3/4%
5/16
.3125 OR 31 1/4%
7/16
.4375 OR 43 3/4%
9/16
.5625 OR 56 1/4%
11/16
.6875 OR 68 3/4%
13/16
.8125 OR 81 1/4%
15/16
93 3/4%
(x²)(x³)
= x²⁺³
(x³) / (x²)
= x³⁻² OR 1 / x²⁻³
(x²)³
= x²⋅³
(xy)³
= (x³)(y³)
(x/y)³
= (x³) / (y³)
x^⁻³
= 1 / x³
1 / x^-m
= x^m
x⁰
= 1 (x ≠ 0)
x^(p/r)
= x^(power/root)
4
8
2⁴
16
2⁵
32
2⁶
64
2⁷
128
2⁸
256
2⁹
512
2¹⁰
1024
9
27
3⁴
81
3⁵
243
3⁶
729
16
4⁴
256
25
125
36
216
49
64
81
10²
100
11²
121
12²
144
13²
169
14²
196
15²
225
16²
256
17²
289
18²
324
19²
361
20²
400
21²
441
22²
484
23²
529
24²
576
25²
625
26²
676
27²
729
28²
784
29²
841
30²
900
31²
961
32²
1024 |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 12/3 * 21/3 = 35/9 = 3 8/9 ≅ 3.8888889
Spelled result in words is thirty-five ninths (or three and eight ninths).
### How do you solve fractions step by step?
1. Conversion a mixed number 1 2/3 to a improper fraction: 1 2/3 = 1 2/3 = 1 · 3 + 2/3 = 3 + 2/3 = 5/3
To find new numerator:
a) Multiply the whole number 1 by the denominator 3. Whole number 1 equally 1 * 3/3 = 3/3
b) Add the answer from previous step 3 to the numerator 2. New numerator is 3 + 2 = 5
c) Write a previous answer (new numerator 5) over the denominator 3.
One and two thirds is five thirds
2. Conversion a mixed number 2 1/3 to a improper fraction: 2 1/3 = 2 1/3 = 2 · 3 + 1/3 = 6 + 1/3 = 7/3
To find new numerator:
a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/3 = 6/3
b) Add the answer from previous step 6 to the numerator 1. New numerator is 6 + 1 = 7
c) Write a previous answer (new numerator 7) over the denominator 3.
Two and one third is seven thirds
3. Multiple: 5/3 * 7/3 = 5 · 7/3 · 3 = 35/9
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(35, 9) = 1. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - five thirds multiplied by seven thirds = thirty-five ninths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Dividends
The three friends divided the win by the invested money. Karlos got three-eighths, John 320 permille, and the rest got Martin. Who got the most and which the least?
• Buing
Brother got to buy 240 CZK and could buy for 1/8 what he wanted. Could he pay the rest of the purchase for 200 CZK?
• Fractions
Sort fractions z1 = (6)/(11); z2 = (10)/(21); z3 = (19)/(22) by its size. Result write as three serial numbers 1,2,3.
• A laundry
Mr. Green washed 1/4 of his laundry. His son washed 3/7 of it. Who washed most of the laundry? How much of the laundry still needs to be washed?
• Roma ate
Roma ate 2/5 of a cake while Somya ate 3/7 of the same cake. Who ate more and by how much?
• Turtles 2
A box turtle hibernates in the sand at 11 5/8. A spotted turtle hibernates at 11 16/25 feet. Which turtle is deeper? Write answer as number 1 or 2.
• Math test
Brayden was solving some math problems for the math team. He answered 2 math problems. Matthew answered 3, John answered 1 reasoning. Matthew 1/2 times as many. Brayden said that 2/6. Is he correct? Why or why not? Be sure to explain your answer.
• Arrange
Arrange the following in descending order: 0.32, 2on5, 27%, 1 on 3
• Simplest form of a fraction
Which one of the following fraction after reducing in simplest form is not equal to 3/2? a) 15/20 b) 12/8 c) 27/18 d) 6/4
• Leo hiked
Leo hiked 6/7 of a kilometer. Jericho hiked 2/3 kilometer. Who covered a longer distance? How much longer?
• Sandy
Sandy, John and Marg baked pies for the Bake Sale. Sandy cut his pies into 6ths, John but his into 8ths and Marg cut hers into quarters. Sandy sold 11/6, John sold 1 3/8 pies and Marg sold 9/4 pies. Who sold the most pies? Who sold the fewest?
• Colored blocks
Tucker and his classmates placed colored blocks on a scale during a science lab. The brown block weighed 8.94 pounds, and the red block weighed 1.87 pounds. How much more did the brown block weigh than the red block?
• Torque
Torque and Mari each multiplied 1/8 inch times 5/8 inch. Tartaric 5/8 squares point inches. And Marie got 5/64 squared thought inches tall. Which student found a corrupt area? |
# What is the kinetic energy of an object with a mass of 6.25xx10^4 "g" having a speed of "26.3 m/s"?
Jul 2, 2015
The person's kinetic energy is equal to 21.6 kJ.
#### Explanation:
Kinetic energy is simply a measure of how much work needs to be done in order to accelerate an object from complete rest to a given velocity.
Mathematically, kinetic energy is defined as
${K}_{E} = \frac{1}{2} \cdot m \cdot {v}^{2}$, where
$m$ - the mass of the object - expressed in kilograms;
$v$ - the velocity of the object - expressed in meters per second.
So, convert the mass of the person from grams to kilograms
6.25 * 10^4cancel("g") * "1 kg"/(10^3cancel("g")) = "62.5 kg"
Now plug your values into the equation and solve for ${K}_{E}$
${K}_{E} = \frac{1}{2} \cdot {\text{62.5 kg" * 26.2""^2 "m"^2/"s}}^{2}$
K_E = 1/2 * 62.5 * 26.2""^2 * underbrace(("kg" * "m"^2)/"s"^2)_(color(blue)("=Joule"))
${K}_{E} = \text{21615. 3 J}$
Rounded to three sig figs and expressed in kilojoules, the answer will be
${K}_{E} = 21615.3 \cancel{\text{J") * "1 kJ"/(1000cancel("J")) = color(green)("21.6 kJ}}$
Jul 3, 2015
$\text{21600 J}$ (Rounded to three significant figures.)
#### Explanation:
$\text{KE} = \frac{1}{2} \cdot m \cdot {v}^{2}$, where KE is kinetic energy in Joules, $m$ is mass in kilograms, and ${v}^{2}$ is speed in m/s squared.
Known
mass = (6.25xx10^4cancel"g")xx(1"kg")/(1000cancel"g")="62.5 kg"
speed = $\text{26.3 m/s}$
Unknown
Kinetic energy
Solution
$\text{KE} = \frac{1}{2} \cdot m \cdot {v}^{2}$
"KE"=1/2*62.5"kg"*(26.3"m/s")^2
$\text{KE}$$= 21600 \text{kg·m"^2"/s"^2}$=$21600 \text{J}$
(Rounded to three significant figures.) |
Systems of Non-Linear Equations:
Solving Simple Systems
(page 3 of 6)
To find the exact solution to a system of equations, you must use algebra. Let's look at that first system again:
• Solve the following system algebraically:
• y = x2
y
= 8 – x2
Since I am looking for the intersection points, I am therefore looking for the points where the equations overlap, where they share the same values. That is, I am trying to find any spots where y = x2 equals y = 8 – x2:
y = x2 = y = 8 – x2
The algebra comes in when I manipulate useful bits of this last equation. I can pick out whichever parts I like. (They're all equal, after all -- at least at the intersection points, but the intersection points are the only points that I care about anyway!) So I can pick out any of the following:
y = x2
y = 8 – x2
y = y
x
2 = 8 – x2
Each of these sub-equations is true, but only the last one is usefully new and different:
x2 = 8 – x2
I can solve this for the x-values that make the equation true:
x2 = 8 – x2
2x2 = 8
x2 = 4
x = –2, +2
Then the solutions to the original system will occur when x = –2 and when x = +2.
What are the corresponding y-values? To find them, I plug the x-values back in to either of the two original equations. (It doesn't matter which one I pick because I only care about the points where the equations spit out the same values. So I can pick whichever equation I like better.) I'll plug the x-values into the first equation, because it's the simpler of the two:
x = –2:
y = x2
y = (–2)2 = 4
x = +2:
y = x2
y = (+2)2 = 4
Then the solutions (as we already knew) are (x, y) = (–2, 4) and (2, 4).
In this case, the solutions were "neat" values; no fractions or decimals. But solutions will not always be neat, so, while the pictures can be very useful for giving you a "feel" for what is going on, graphing is not as accurate as doing the algebra. Warning: Students are often taught nowadays to "round" absolutely everything, and are thus implicitly taught that all answers will be "neat" answers. But this is wrong; don't fall for it. For instance:
• Solve the following system:
• y = x2 + 3x + 2
y = 2x + 3
I can solve this in the same manner as we did on the previous problem. The "solution" to the system will be any point(s) that the lines share; that is, any point(s) where the x-value and corresponding y-value for y = x2 + 3x + 2 is the same as the x-value and corresponding y-value for y = 2x + 3; that is, where the lines overlap or intersect; that is, where y = x2 + 3x + 2 equals y = 2x + 3. Copyright © 2002-2011 Elizabeth Stapel All Rights Reserved
Looking at the graph of the system: ...I can see that there appear to be solutions at around (x, y) = (–1.5, –0.25) and (x, y) = (0.5, 4.25). But I cannot assume that this is the answer! The picture can give me a good idea, but only the algebra can give me the actual answer. I'll set the equations equal, and solve: x2 + 3x + 2 = 2x + 3x2 + x – 1 = 0 Using the Quadratic Formula gives me: Then I have one solution: ...which has a corresponding y-value of: The other solution (from the "±" in front of the square root) is: ....which gives me a y-value of: So the solutions are:
For purposes of graphing, the approximate solutions are:
(x, y) = (–1.62, –0.24) and (0.62, 4.24).
In other words, while our guess from the picture was close, it was not entirely correct. (However, if the algebra had given me answers that are far afield of these picture-based guesses, I would have been able to safely assume that I had messed up the math somewhere. In this way, the graph can be helpful for checking your work.)
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Cite this article as: Stapel, Elizabeth. "Systems of Non-Linear Equations: Solving Simple Systems." Purplemath. Available from http://www.purplemath.com/modules/syseqgen3.htm. Accessed [Date] [Month] 2016
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Rational Numbers in Terminating and Non-Terminating Decimals
Integers are positive and negative whole numbers including zero, such as {-3, -2, -1, 0, 1, 2, 3}.
When these whole numbers are written in the form of ratio of whole numbers it is known as rational numbers. So, rational numbers can be positive, negative or zero. So, a rational number can be expressed in the form of p/q where ‘p’ and ‘q’ are integers and ‘q’ is not equal to zero.
Rational Numbers in Decimal Fractions:
Rational numbers can be expressed in the form of decimal fractions. These rational numbers when converted into decimal fractions can be both terminating and non-terminating decimals.
Terminating decimals: Terminating decimals are those numbers which come to an end after few repetitions after decimal point.
Example: 0.5, 2.456, 123.456, etc. are all examples of terminating decimals.
Non terminating decimals: Non terminating decimals are those which keep on continuing after decimal point (i.e. they go on forever). They don’t come to end or if they do it is after a long interval.
For example:
π = (3.141592653589793238462643383279502884197169399375105820974.....) is an example of non terminating decimal as it keeps on continuing after decimal point.
If a rational number (≠ integer) can be expressed in the form $$\frac{p}{2^{n} × 5^{m}}$$, where p ∈ Z, n ∈ W and m ∈ W, the rational number will be a terminating decimal. Otherwise, the rational number will be a nonterminating, recurring decimal.
For example:
(i) $$\frac{5}{8}$$ = $$\frac{5}{2^{3} × 5^{0}}$$. So, $$\frac{5}{8}$$ is a terminating decimal.
(ii) $$\frac{9}{1280}$$ = $$\frac{9}{2^{8} × 5^{1}}$$. So, $$\frac{9}{1280}$$ is a terminating decimal.
(iii) $$\frac{4}{45}$$ = $$\frac{4}{3^{2} × 5^{1}}$$. Since it is not in the form $$\frac{p}{2^{n} × 5^{m}}$$, So, $$\frac{4}{45}$$ is a non-terminating, recurring decimal.
For example let us take the cases of conversion of rational numbers to terminating decimal fractions:
(i) $$\frac{1}{2}$$ is a rational fraction of form $$\frac{p}{q}$$. When this rational fraction is converted to decimal it becomes 0.5, which is a terminating decimal fraction.
(ii) $$\frac{1}{25}$$ is a rational fraction of form $$\frac{p}{q}$$. When this rational fraction is converted to decimal fraction it becomes 0.04, which is also an example of terminating decimal fraction.
(iii) $$\frac{2}{125}$$ is a rational fraction form $$\frac{p}{q}$$. When this rational fraction is converted to decimal fraction it becomes 0.016, which is an example of terminating decimal fraction.
Now let us have a look at conversion of rational numbers to non terminating decimals:
(i) $$\frac{1}{3}$$ is a rational fraction of form $$\frac{p}{q}$$. When we convert this rational fraction into decimal, it becomes 0.333333… which is a non terminating decimal.
(ii) $$\frac{1}{7}$$ is a rational fraction of form $$\frac{p}{q}$$. When we convert this rational fraction into decimal, it becomes 0.1428571428571… which is a non terminating decimal.
(iii) $$\frac{5}{6}$$ is a rational fraction of form $$\frac{p}{q}$$. When this is converted to decimal number it becomes 0.8333333… which is a non terminating decimal fraction.
Irrational Numbers:
We have different types of numbers in our number system such as whole numbers, real numbers, rational numbers, etc. Apart from these number systems we have Irrational Numbers. Irrational numbers are those which do not terminate and have no repeating pattern. Mr. Pythagoras was the first person to prove a number as irrational number. We know that all square roots of integers that don’t come out evenly are irrational. Another best example of an irrational number is ‘pi’ (ratio of circle’s circumference to its diameter).
π = (3.141592653589793238462643383279502884197169399375105820974.....)
First three hundred digits of ‘pi’ are non-repeating and non-terminating. So, we can say that ‘pi’ is an irrational number.
`
Rational Numbers
Rational Numbers
Decimal Representation of Rational Numbers
Rational Numbers in Terminating and Non-Terminating Decimals
Recurring Decimals as Rational Numbers
Laws of Algebra for Rational Numbers
Comparison between Two Rational Numbers
Rational Numbers Between Two Unequal Rational Numbers
Representation of Rational Numbers on Number Line
Problems on Rational numbers as Decimal Numbers
Problems Based On Recurring Decimals as Rational Numbers
Problems on Comparison Between Rational Numbers
Problems on Representation of Rational Numbers on Number Line
Worksheet on Comparison between Rational Numbers
Worksheet on Representation of Rational Numbers on the Number Line |
What is a Geometric Sequence?
••• DragonImages/iStock/GettyImages
Print
In a geometric sequence, each term is equal to the previous term times a constant, non-zero multiplier called the common factor. Geometric sequences can have a fixed number of terms, or they can be infinite. In either case, the terms of a geometric sequence can rapidly become very large, very negative or very close to zero. Compared to arithmetic sequences, the terms change much more quickly, but while infinite arithmetic sequences increase or decrease steadily, geometric sequences can approach zero, depending on the common factor.
A geometric sequence is an ordered list of numbers in which each term is the product of the previous term and a fixed, non-zero multiplier called the common factor. Each term of a geometric sequence is the geometric mean of the terms preceding and following it. Infinite geometric sequences with a common factor between +1 and −1 approach the limit of zero as terms are added while sequences with a common factor larger than +1 or smaller than −1 go to plus or minus infinity.
How Geometric Sequences Work
A geometric sequence is defined by its starting number a, the common factor r and the number of terms S. The corresponding general form of a geometric sequence is:
a, ar, ar^2, ar^3, ... , ar^{S-1}
The general formula for term n of a geometric sequence (i.e., any term within that sequence) is:
a_n = ar^{n-1}
The recursive formula, which defines a term with respect to the previous term, is:
a_n = ra_{n-1}
An example of a geometric sequence with starting number 3, common factor 2 and eight terms is 3, 6, 12, 24, 48, 96, 192, 384. Calculating the last term using the general form listed above, the term is:
a_8 = 3 × 2^{8-1} = 3 × 2^7 = 3 × 128 = 384
Using the general formula for term 4:
a_4 = 3 × 2^{4-1} = 3 × 2^3 = 3 × 8 = 24
If you want to use the recursive formula for term 5, then term 4 = 24, and a5 equals:
a_5= 2 × 24 = 48
Geometric Sequence Properties
Geometric sequences have special properties as far as the geometric mean is concerned. The geometric mean of two numbers is the square root of their product. For example, the geometric mean of 5 and 20 is 10 because the product 5 × 20 = 100 and the square root of 100 is 10.
In geometric sequences, each term is the geometric mean of the term before it and the term after it. For example, in the sequence 3, 6, 12 ... above, 6 is the geometric mean of 3 and 12, 12 is the geometric mean of 6 and 24, and 24 is the geometric mean of 12 and 48.
Other properties of geometric sequences depend on the common factor. If the common factor r is greater than 1, infinite geometric sequences will approach positive infinity. If r is between 0 and 1, the sequences will approach zero. If r is between zero and −1, the sequences will approach zero, but the terms will alternate between positive and negative values. If r is less than −1, the terms will trend toward both positive and negative infinity as they alternate between positive and negative values.
Geometric sequences and their properties are especially useful in scientific and mathematical models of real world processes. The use of specific sequences can help with the study of populations that grow at a fixed rate over given periods of time or investments that earn interest. The general and recursive formulas make it possible to predict accurate values in the future based on the starting point and the common factor. |
# Simplify the fraction (1/(3+x)-1/3)/x?
Posted on
We have to simplify (1/(3+x)-1/3)/x
(1/(3+x)-1/3)/x
=> [3 - (3+x)]/3(3+x)x
=> [3 - 3 - x]/3(3+x)x
=> [- x]/3(3+x)x
=> -1/3*(3 + x)
The simplified expression is -1/3*(3 + x)
Posted on
First, we'll perform the subtraction from numerator. To subtract the given fractions they must have the same denominator.
1/(3+x) - 1/3 = [3 - (3+x)]/3(3+x)
We'll remove the brackets:
1/(3+x) - 1/3 = (3-3-x)/3(3+x)
1/(3+x) - 1/3 = -x/3(3+x)
Now, we'll re-write the fraction:
[1/(3+x) - 1/3]/x = -x/3x(3+x)
We'll simplify and we'll get:
[1/(3+x) - 1/3]/x = -1/(9 + 3x)
The simplified fraction is -1/(9 + 3x). |
# Series: Success at the Core: Relevant and Real Content
Math.8.G.B.7
Common core State Standards
• Math: Math
• G: Geometry
• B: Understand and apply the Pythagorean Theorem
• 7:
Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
Challenging Students to Discover Pythagoras
Lesson Objective: Work on a challenging problem to discover the Pythagorean Theorem
Grade 8 / Math / Pythagoras
Math.8.G.B.7
#### Thought starters
1. What aspects of Mr. Ivy's practice most challenge your thinking about math instruction?
2. What strategies does the teacher use to prepare students for the challenge he poses?
3. How does the teacher guide students to discover (vs. memorize) the Pythagorean Theorem?
I loved this video because it touches on many essential elements of a successful lesson, especially introductory. I also liked the activity shared by Mr. Jim Paige.
Recommended (0)
What about coming from the direction of area as an exploration? Students start with a square of known area, say 10 cm X 10 cm. Then they create 2 areas from the square such that the sum of these areas equals the sum of the original square, say 80 square cm and 20 square cm. For each of these areas, they create squares of equivalent area. For 80 square cm, the side length would be 8.94 cm, and for 20 square cm, the side length is 4.47 cm. Students draw and cut out all three squares and use the sides of each square to make a triangle. The purpose would be to physically and mathematically establish the area of the squares relationship for a right triangle. Many different right triangles would be created because students will end up with different sized smaller squares than their classmates. The activity might prevent students from a simple (and incorrect) relationship between side lengths for right triangles.
Recommended (0)
And from across the proverbial pond, I like this way too https://www.teachingchannel.org/videos/teaching-pythagorean-theorem
Recommended (1)
#### School Details
Sylvester Middle School
16222 Sylvester Rd Sw
Burien WA 98166
Population: 658
Data Provided By:
#### Teachers
Breck Ivy
Teaching Practice
### Three Ways to Encourage Student Collaboration
All Grades / All Subjects / Collaboration
Teaching Practice
### Five Ways to Start Your Lessons
All Grades / All Subjects / Planning
Teaching Practice
### Three Ways to Practice Goal Setting with Your Students
All Grades / All Subjects / Engagement
Lesson Idea
### New Teacher Coaching Cycle: Looking Closely at Text
Grades 9-12 / ELA / Tch DIY
TCHERS' VOICE
Differentiation
TCHERS' VOICE
Class Culture
TCHERS' VOICE
Lesson Planning
TCHERS' VOICE
### Move Over Debate, It’s Time to Deliberate
Educating for Democracy |
# How do you write 9x=2y+13 in slope intercept form?
Oct 1, 2016
Slope intercept form is $y = m x + b$.
$9 x - 13 = 2 y$
$y = \frac{9}{2} x - \frac{13}{2}$
Hopefully this helps!
Oct 1, 2016
$y = \frac{9}{2} x - \frac{13}{2}$
#### Explanation:
The slope-intercept form for a linear equation is $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept.
In order to write the equation $9 x = 2 y + 13$ in slope-intercept form, solve for $y$.
$9 x = 2 y + 13$
Subtract $2 y$ from both sides.
$- 2 y + 9 x = 13$
Subtract $9 x$ from both sides.
$- 2 y = - 9 x + 13$
Divide both sides by $- 2$.
$y = \frac{- 9 x}{- 2} + \frac{13}{- 2}$
Simplify.
$y = \frac{9}{2} x - \frac{13}{2}$
$m = \frac{9}{2}$
$b = - \frac{13}{2}$
graph{y=9/2x-13/2 [-10.85, 9.15, -8.19, 1.81]} |
# If there is a box which has 7 red balls
• campa
In summary, the probability of picking three red balls from a box containing 7 red balls, 4 white balls, and 9 black balls is 7/228 or approximately 0.0307. The probability of picking 2 white and 1 red ball is 7/190 or about 0.0368. This can be achieved by picking the balls in the order of white, white, red or white, red, white, or red, white, white. The probability of picking a specific order of these balls is the same and can be calculated by multiplying the probabilities of each individual pick.
campa
If there is a box which has 7 red balls, 4 white balls and 9 black ball and you take 3 balls out what is the probability of the three balls being red and the probability of the balls being 2 white and 1 red
What is your sample space? What is the probability of picking one red ball? If you don't return that red ball to the box, what is your new sample space? What is the probability of picking another red ball? ...
I assume you are talking about "sampling without replacement"- that is you do NOT put a ball back into the box after noting its color.
1. In order to get "all three ball being red", the first ball must be red. There are a total of 7+ 4+ 9= 20 ball, 7 of them red so the probability of that is 7/20.
Once that has happened, there are 19 balls left, 6 of them red. The probability that the second ball you pick will be red is 6/19. The probability that the first two balls are both red is the product: (7/20)(6/19).
If you have done that, then there are 18 balls left, 5 of them red: the probability that the next ball is also red is 5/18. The probability of picking 3 red balls is
(7/20)(6/19)(5/18) which can be reduced (since 5/20= 1/4 and 6/18= 1/3) to
"2 white and one red" is just a little harder. One way for that to happen is to get "white, white, red" in that order. Since there are 4 white balls, the probability that the first ball will be white is 4/20= 1/5. If the first ball is white, then there are 19 balls left and 3 of them are white. The probability that the second ball is also white is 3/19. Assuming that the first two balls were white, there are now 18 balls left but 7 of them are still red. The probability that the third ball will be red is 7/18. The probability of gettin "white,white,red", in that order, is (1/5)(3/19)(7/18).
But drawing "white, red, white" would also give "2 white and 1 red". As before the probability that the first ball is white is 4/20= 1/5. Assuming the first ball is white then on the second draw there are 19 balls left and 7 of them are red. The probability that the second ball is red is 7/19. Finally, if the draws have happened that way there are 18 balls left, of which 3 are still white. The probability that the third ball is white is 3/18= 1/6. The probability of getting "white, red, white" in that order is (1/5)(7/19)(3/18). Notice that the denominators are exactly the same as in the previous case! Of course: that was always the total number of balls and it didn't matter what color was taken out. Even more, the numerators are the same- just their order is changed. That's because we are taking out the same balls- just changing the order. But multiplication is "commutative"- order doesn't matter. The point is that the probability of getting these balls in any specific order is the same: (1/5)(3/19)(7/18). To find the probability of getting "white, white, red" is any order, we just have to recognize that there are 3 "orders": (white, white, red), (white, red, white), and (red, white, white) so the probability of getting 2 white and 1 red ball, in any order is 3(1/5)(3/19)(7/18)= (1/5)(1/19)(7/2)= 7/190 or about 0.0368.
thanx that really helped
## 1. How many balls are in the box?
There are 7 balls in the box.
## 2. What color are the balls in the box?
The balls in the box are all red.
## 3. Is there any other color of balls in the box?
No, all the balls in the box are red.
## 4. What is the probability of picking a red ball from the box?
The probability of picking a red ball from the box is 100%, since all the balls in the box are red.
## 5. Can I add more balls to the box?
It depends on the size and capacity of the box. If there is enough space, you can add more balls to the box.
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# Recursion A recursive function is a function that calls itself either directly or indirectly through another function. The problems that can be solved.
## Presentation on theme: "Recursion A recursive function is a function that calls itself either directly or indirectly through another function. The problems that can be solved."— Presentation transcript:
Recursion A recursive function is a function that calls itself either directly or indirectly through another function. The problems that can be solved by recursion must have the following characteristics: - One or more simple cases of the problem have a straightforward, nonrecursive solution. - The other cases can be redefined in terms of problems that are closer to the simple cases. - By applying this redefinition process every time the recursive function is called, eventually the problem is reduced entirely to simple cases, which are relatively easy to solve.
Recursion Recursive algorithm: if this is a simple case solve it else redefine the problem using recursion Problem: Multiply 6 by 3. The problem can be split into two problems: - 1. Multiply 6 by 2. - 2. Add 6 to the result of problem 1. Now we can solve problem 2 because we know our addition tables. But we cannot solve problem 1. Problem 1 is closer to the simple case than the original problem was. The recursive approach splits one size-n problem into n size-1 problems.
Recursion Problem: Multiply 6 by 3. The problem can be split into two problems: - 1. Multiply 6 by 2. - 2. Add 6 to the result of problem 1. We can split problem 1 into two problems: - 1.1 Multiply 6 by 1 - 1.2 Add 6 to the result of problem 1.1 Problem 1.1 is the simple case. We can solve it because we know that any number multiplied by 1 gives us the original number. Here, we assumed that we know our addition tables. We can solve problem 1.2. So, problem1 is solved. We have already solved Problem 2. Therefore, we have solved the problem.
Recursive function multiply // Performs integer multiplication using + operator. // Pre: m and n are defined and n > 0 int multiply ( int m, int n ) { int ans; if (n == 1) ans = m; // simple case else ans = m + multiply ( m, n -1 ); // recursive step return (ans); }
Count the # of vowels in a string #include int count_vowel (char *input); int main() { char line[] = “Hello”; int num, count; printf(“num of vowels are %d\n”, count_vowels(line); return 0; }
Count the # of vowels in a string int count_vowels (char *string_ptr) { if (*str_ptr == ‘\0’) return 0; // simple case else { switch (*string_ptr) { // redefine problem using recursion case ‘a’ : case ‘e’ : case ‘i’ : case ‘o’ : case ‘u’: return (1 + count_vowels ( string_ptr + 1)); break; default : return count_vowels (string_ptr + 1); break; }
Tracing a Recursive Function Multiply(6, 3) m = 6, n = 1 1 == 1 is true ans is 6 return (ans) m = 6, n = 2 2 == 1 is false ans is 6 + multiply(6, 2) return (ans) Three activation frames corresponding to each call of the function are generated to solve the problem. m = 6, n = 3 3 == 1 is false ans is 6 + multiply(6, 2) return (ans) 6 12 18
Function reverse_input_words void reverse_input_words(int n) { char word[WORDSIZ]; if (n <= 1) // simple case { scanf(“%s”, word); printf(“%s\n”, word); } else // get and print the rest of the words { // in reverse order and print it. scanf(“%s”, word); reverse_input_words(n - 1); printf(“%s\n”, word); }
Tracing a Void Recursive Function Reverse_input_words (3) n = 1 ** 1 <= 1 is true scan “you” into word display “you” return n = 2 ** 2 <= 1 is false scan “are” into word reverse_input_words(1) display “are” return Three activation frames corresponding to each call of the function are generated to solve the problem. n = 3 ** 3 <= 1 is false scan “how” into word reverse_input_words(2) display “how” return A void function’s return occurs when the closing brace of the function body is encountered. ** word is undefined
Sequence of Events for the Function Call reverse_input_words with n = 3. Scan the first word (“How”) into word. Call reverse_input_words with n = 2. Scan the second word (“are”) into word. Call reverse_input_words with n = 1. Scan the third word (“you”) into word. Display the third word (”you”). Return from third call. Display the second word (“are”). Return from second call. Display the first word (“How”). Return from the original call.
Parameter and Local Variable Stack C uses the stack data structure to keep track of the values of n and word at any given point. In this structure, we add data items and remove them from the same end of the list. The last item stored is the first processed. When executes a call to the function, the system pushes the parameter value on the top of the parameter stack. Pushes a new undefined cell on top of the stack maintained for the local variable word. A return from the function pops each stack, removing the top value.
Example After first call to reverse_input_words The word “How” is stored in word just before the second call After the second call. The word “are” is scanned and stored in word just before the third call After the third call 3 ? 3How 2323 ? How 2323 are How 123123 you are How 123123 ? are How
Example During the execution of the function, the word “you” is scanned and stored in word, and “you”is printed immediately because n =1 The function return pops both stacks. After the first return: After the second return The third and last return exits the original function call. So there is no longer memory allocated for n and word. 123123 you are How 2323 Are How 3
System Stack We have used separate stacks for each parameter in the example. The compiler actually maintains a single system stack Whenever a call to a function occurs, all its parameters and local variables are pushed onto the stack along with the memory address of the calling statement. This address gives the computer the return point after execution of the function. Multiple copies of a function’s parameters may be saved on the stack, only one copy of the function body is in memory.
Recursive factorial Function // Compute n! using recursion // Pre: n >= 0 int factorial (int n) { int ans; if (n == 0) // simple case ans = 1; else // redefine problem using recursion ans = n * factorial (n - 1); return (ans); } !1 = 1 !4 = 4 * !3 = 4 * 3 * !2 = 4 * 3 * 2 * !1 = 4 * 3 * 2 * 1 = 24
Recursive fibonacci Function // Compute the nth Fibonacci number using recursion // Pre: n >= 0 int fibonacci (int n) { int ans; if (n == 1 || n == 2) // simple case ans = 1; else // redefine problem using recursion ans = fibonacci(n - 2) + fibonacci (n - 1); return (ans); } Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, ….
Towers of Hanoi Towers of Hanoi problem involves moving a specified number of disks that are all different sizes from one tower (or peg) to another. Problem: Move n disks from peg A to peg C using peg B as an intermediate peg. Conditions: 1. Only one disk at a time may be moved, and this disk must be the top disk on a peg. 2. A larger disk cannot be placed on top of a smaller disk.
Tower of Hanoi (A) (B)(C) (A)(B)(C) 1 2 3 Start Goal
Tower of Hanoi (A) (B)(C) 1 Simple case: single disk. Just move the disk 1 from peg A to peg C
Tower of Hanoi (A) (B)(C) (A)(B) (C) 1 2 3 Move the top (n -1) disks from peg A to peg B. Top (n - 1) disks
Tower of Hanoi (A) (B) (C) (A)(B) (C) Move the bottom disk (1) from peg A to peg C. 1
Tower of Hanoi (A) (B) (C) (A)(B) (C) Move the top (n - 1) disks from peg B to peg C. Top (n - 1) disks
Data Requirements Problem Inputs: int n char from_peg char to_peg char aux_peg Problem Outputs: A list of instructions for moving n disks from from_peg to to_peg using aux_peg as an intermediate peg.
Algorithm 1. if n is 1 then 2. Move disk 1 from the from_peg to the to_peg. else 3. Move n - 1 disks from the from_peg to the auxiliary peg using the to_peg. 4. Move disk n from the from_peg to the to_peg. 5. Move n - 1 disks from the auxiliary peg to the to_peg using the from_peg.
Recursive Function Tower // Displays instruction for moving n disks from from_peg to to_peg // using aux_peg as an intermediate peg. void tower ( char from_peg, char to_peg, char aux_peg, int n) { if (n == 1) // simple case printf(“Move disk 1 from peg %c to peg %c\n”, from_peg, to_peg); else // redefine the problem using recursion { tower(from_peg, aux_peg, to_peg, n -1); printf(“Move disk %d from peg %c to peg %c\n”, n, from_peg, to_peg); tower(aux_peg, to_peg, from_peg, n - 1); }
Output of tower Function Output generated by tower (‘A’, ‘C’, ‘B’, 3): Move disk 1 from A to C Move disk 2 from A to B Move disk 1 from C to B Move disk 3 from A to C Move disk 1 from B to A Move disk 2 from B to C Move disk 1 from A to C n - disk problem requires 2 n - 1 moves.
Recursion Vs. Iteration Both iteration and recursion are based on a control structure: iteration uses a repetition structure and recursion uses a selection structure. Both iteration and recursion involve repetition. Iteration explicitly uses a repetition structure and recursion achieves repetition through repeated function calls. Iteration and recursion each involve a termination test. Iteration terminates when the loop-continuation condition fails. Recursion terminates when the base case is reached. Both iteration and recursion can occur infinitely.An infinite loop can occurs with iteration if the loop-continuation test never becomes false. Infinite recursion occurs if the recursive step does not reduce the problem each time in a manner that converges on the base case.
Recursion Vs. Iteration Recursion has many negatives. It repeatedly invokes the mechanism, and consequently the overhead, of function calls.This can be expensive in both processor time and memory space.. Each recursive call causes another copy of the function (actually only the function’s variables) to be created. This can consume considerable memory. Iteration normally occurs within a function so the overhead of repeated function calls and extra memory assignment is omitted. So why choose recursion? In many instances, the use of recursion enables us to specify a very natural, simple solution to a problem that would otherwise be very difficult to solve (e.g. Tower of Hanoi). For this reason, recursion is an important and powerful tool in problem solving and programming.
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Angles In a Alternate SegmentThe angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment.
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Example: In right ABC, a circle with side AB as diameter is drawn to intersect the hypotenuse AC at P. Prove that the tangent to the circle at P bisects the side BC.
Solution: Let O be the centre of the given circle.
Suppose the tangent at P meets BC at Q. Join BP. You have to prove that BQ = QC. Since BPQ and BAP = BAC are angles in the alternate segments of chord BP. BPQ = BAC ...(i) Since AB is a diameter of the circle and angle in a semi-circle is right angle. APB = 90 BPC = 90 ...[APB and BPC are linear pairs] ...(ii) Since A ABC is a right triangle, right angled at B. BAC + BCA = 90 ...(iii) From equations (ii) and (iii), you get BAC + BCA = BPC BAC + BCA = BPQ + QPC ...[BPC = BPQ + QPC] BAC + BCA = BAC + QPC ...[BPQ = BAC] BCA = QPC QCA = QPC ...[BCA = QCA] Thus, in A PQC, you have QCA = QPC PQ = QC [ Sides opposite to equal angles are equal] ...(iv) Since tangents from an exterior point to a circle are equal in length. QP = QB i.e. PQ = QB ...(v) From equations (iv) and (v), you get BQ = QC. Thus, Q is a point on BC such that BQ = QC. Hence, PQ bisects BC.
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# Mastering Long Division
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## Danielle Biddle
on 11 August 2013
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#### Transcript of Mastering Long Division
Step 1:
Look at the question. Can the divisor go into the first number of the dividend?
Step 2:
We have figured out that 4 goes into 24 six times.
Step 3:
What are we trying to figure out in the problem below?
That's right! The question is - how many times can 4 go into 24?
This is a question you should all be able to answer without using long division, but it is a good start to help us understand and start thinking about the process o f long division .
Now the question is, how does long division work?
If the answer is yes, write how many times it can go in above the first number of the dividend.
If the answer is no, you need to look at the next number and see if the divisor can go into the double digit number.
Let's try step 1
Can 4 go into 2?
No. What do we do next?
If you said to look at the next number in the dividend, you are correct!
Can 4 go into 24?
How many times?
Well done! 4 goes into 24 six times.
Copy this problem into your book. Place the 6 where you think it should go, then show the person sitting next to you. Do you both have the 6 in the same place? Tell your partner what you think you do next.
If you placed the 6 above the 4, you are correct!
Now for the next step of long division!
We need to multiply 6 and 4. Then we write the product under the 24 from the dividend. This helps us see if there is anything left over or if the divisor went in evenly.
The product of 6x4 is 24. To find out if we have anything left over we subtract the product from the numbers of the dividend we were looking at. (24 -24)
In this case, when we multiply 6 and 4 we get the same number as the two numbers of the dividend (24), so we have nothing left over.
What do we do next? Tell your partner what you think the next step is.
Mastering Long Division
Look at the next number in the dividen. What is it?
If you said 6, you are correct! Now we need to drag the 6 down to make sure we are considering any leftover numbers from the last step.
6
24
-
_____
0
6
Were there any leftover numbers in the last step?
If you said no, you are correct!
After dragging the 6 down, we can see that there was nothing left over in the last step so we just have 6. This means we have to look at how many times 4 goes into 6.
6
24
-
_____
0
6
1
4 can go into 6 one time. What do we do next?
6
24
-
_____
0
Repeat:
6
24
-
_____
0
6
1
Keep in mind that the question is asking you how many times 4 goes into 24,652. With such a big number, long division helps us figure out the answer by doing a few smaller steps. If we look at how many times 4 goes into the individual numbers (or pairs of numbers), it makes the problem much more manageable.
Continue using the same method of multiplying 4 by the number of times it goes into the dividend in each new step. Write that number under the last step, and then subtract to find the difference.
If 4 goes into 6 one time we need to multiply 4x1 and write the product under the 06 from the last step. Then we subtract 4 from 6 and are left with 2. Can 4 go into 2? What do we do next?
4
_____
-
2
Finish the problem and compare with your partner.
What happens if the divisor doesn't go into the dividend evenly?
If you were thinking about remainders, you are spot on!
Remainders can be shown in a variety of ways. Tell your partner one way to show a remainder. Can you think of any others?
Remainders can be shown by using an 'r'. This tells us what is left over because there wasn't enough to make another whole and fit into the dividend evenly. Using the 'r' to express what is left is the simplest way to show that the divisor didn't go into the dividend evenly.
A more complex way to show a remainder is by using decimals. This tells us exactly how many times the divisor goes into the dividend.
Another complex way to express a remainder is by using fractions. Did you get all three ways?
If a divisor doesn't go into the dividend evenly, I would like you to be able to express the remainder by using decimals.
It is a little tricky at first, but once you get it you'll be set for life!
Let's give it a go!
We'll start with something not too tricky and work our way up to more complex long division problems.
Now that you are experts with long division the first bit shouldn't be too tricky.
We know that 4 can't go into 2, so we have to see how many times it can go into 26.
From the earlier problem we know that 4 can go into 24 six times evenly.
6
24
-
2
___
____
4 goes into 26 six times with a remainder of 2. In the earlier problem we learned that the next step is to bring down the next number of the dividend and put it next to the 2. However, there aren't any more numbers to bring down.
What do we do now? Talk to your partner and think about what we need to do next.
6
24
-
2
___
____
Since there aren't any more numbers for us to bring down, we have to make some.
How could we add numbers without changing the value of the number we have?
By making the number a decimal!
Every whole number could be written with a decimal behind it and as many zeroes as you would like to write out.
For example: 2 is the same as 2.000000
Let's turn 26 into a decimal and try to solve the problem as normal. Don't forget that when add a decimal to a number you also need to bring the decimal point up to the line where the quotient is, or you won't get the correct answer.
6
24
-
2
___
____
.
.
Don't forget this decimal point!
000
0
Now we can find out how many times 4 goes into 20.
Finish solving this problem with your partner. Check with another pair to see if you got the correct answer.
You are on your way to being a long division expert! |
## Precalculus (6th Edition) Blitzer
Let ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{1}}$ is true, and 2. Let us assume that the ${{S}_{k}}$ be true for some integer $k$. Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $n$. Firstly, show that ${{S}_{1}}$ is true. For ${{S}_{1}}$ one has And write ${{S}_{1}}$ by taking the its first term on left and substituting n with $1$ on the right. \begin{align} & {{S}_{1}}: \\ & {{\left( \frac{a}{b} \right)}^{1}}=\frac{{{a}^{1}}}{{{b}^{1}}} \\ & =\frac{a}{b} \end{align} Therefore, the statement ${{S}_{1}}$ is true. Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction. ${{S}_{k}}:{{\left( \frac{a}{b} \right)}^{k}}=\frac{{{a}^{k}}}{{{b}^{k}}}$ Now, one has to prove ${{S}_{k+1}}$ is true ${{S}_{k+1}}:{{\left( \frac{a}{b} \right)}^{k+1}}=\frac{{{a}^{k+1}}}{{{b}^{k+1}}}$ Since, one assume that ${{S}_{k}}$ is true, ${{\left( \frac{a}{b} \right)}^{k}}=\frac{{{a}^{k}}}{{{b}^{k}}}$ And multiply $\left( \frac{a}{b} \right)$ on the both sides, \begin{align} & {{\left( \frac{a}{b} \right)}^{k}}\left( \frac{a}{b} \right)=\frac{{{a}^{k}}}{{{b}^{k}}}\left( \frac{a}{b} \right) \\ & {{\left( \frac{a}{b} \right)}^{k+1}}=\frac{{{a}^{k+1}}}{{{b}^{k+1}}} \\ \end{align} So, the final statement is ${{S}_{k+1}}$. Thus, by the method of principal of mathematical induction, the statement ${{\left( \frac{a}{b} \right)}^{n}}=\frac{{{a}^{n}}}{{{b}^{n}}}$ is true for every positive integer $n$. |
# Texas Go Math Grade 5 Lesson 6.5 Answer Key Use Multiplication
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 6.5 Answer Key Use Multiplication.
## Texas Go Math Grade 5 Lesson 6.5 Answer Key Use Multiplication
Unlock the Problem
Erica makes 6 submarine sandwiches and cuts each sandwich into thirds. How many $$\frac{1}{3}$$-size sandwich pieces does she have?
What do I need to find?
I need to find __________________
What information am I given?
I need to use the size of each __________ of sandwich and the number of ___________ she cuts.
Plan
What s my plan or strategy?
I can _______________ to organize the information from the problem.
Then I can use the organized information to find ___________________
Solve
Since Erica cuts 6 submarine sandwiches, my diagram needs to show 6 rectangles to represent the sandwiches. I can divide each of the 6 rectangles into thirds.
To find the total number of thirds in the 6 rectangles, I can multiply the number of thirds in each rectangle by the number of rectangles.
6 ÷ $$\frac{1}{3}$$ = 6 × _________ = ___________
So, Erica has __________ one-third-size sandwich pieces.
I need to find the number of 1/3 size sandwich pieces does Eric has after she cuts 6 sandwiches into thirds.
I need to use the size of each piece of sandwich and the number of sandwiches she cuts.
Plan
I can draw a diagram to organize the information from the problem.
Then I can use the organized information to find the number of 1/3 size sandwich pieces does Eric has after she cuts 6 sandwiches into thirds.
Solve
Since Erica cuts 6 submarine sandwiches, my diagram needs to show 6 rectangles to represent the sandwiches. I can divide each of the 6 rectangles into thirds.
To find the total number of thirds in the 6 rectangles, I can multiply the number of thirds in each rectangle by the number of rectangles.
6 ÷ 1/3 = 6 × 3 = 18
So, Erica has 18 one-third-size sandwich pieces.
Math Talk
Mathematical Processes
I can multiply the quotient and the divisor to see if the product is equal to the dividend.
Try Another Problem
Roberto is cutting 3 blueberry pies into halves to give to his neighbors. How many neighbors will get a $$\frac{1}{2}$$-size pie piece?
What do I need to find?
What information am I given?
Plan
What is my plan or strategy?
Solve
So, _________ neighbors will get a $$\frac{1}{2}$$-size pie piece.
I need to find how many neighbors will get 1/2 of a pie.
The information given is that Roberto is cutting 3 blueberry pies into halves to give to his neighbors. Some neighbors will get a 1/2 size pie piece.
Plan
I can draw a diagram to organize the information from the problem. Then i can use the diagram to find the number of neighbors that will get 1/2 of a pie.
Solve
So, 6 neighbors will get a 1/2 -size pie piece.
Explain how the diagram you drew for the division problem helps you write a multiplication sentence.
Since Roberto is cutting 3 pies, my diagram needs to show 3 circles to represent the pies. I can divide each of the circles into halves.
To find the total number of halves in the 3 circles, I can multiply the number of halves in each circle by the number of circles.
Share and Show
Question 1.
A chef has 5 blocks of butter. Each block weighs 1 pound. She cuts each block into fourths. How many $$\frac{1}{4}$$pound pieces of butter does the chef have?
First, draw rectangles to represent the blocks of butter.
Then, divide each rectangle into fourths.
Finally, multiply the number of fourths in each block by the number of blocks.
So, the chef has _________ one-fourth-pound pieces of butter.
So, the chef has 20 one-fourth-pound pieces of butter.
Explanation:
A chef has 5 blocks of butter. Each block weighs 1 pound. She cuts each block into fourths. First we have to draw rectangles to represent the blocks of butter as we can observe in the above image. Then, we have to divide each rectangle into fourths. Multiply the number of fourths in each block by the number of blocks. Divide 5 by 1/4 the result is 20. So, the chef has 20 one-fourth-pound pieces of butter.
Question 2.
What if the chef had 3 blocks of butter and cut the blocks into thirds? How many $$\frac{1}{3}$$-pound pieces of butter would the chef have?
So, the chef has 9 one-third pound pieces of butter.
Explanation:
A chef has 3 blocks of butter. Each block weighs 1 pound. She cuts each block into thirds. First we have to draw rectangles to represent the blocks of butter as we can observe in the above image. Then, we have to divide each rectangle into thirds. Multiply the number of thirds in each block by the number of blocks. Divide 3 by 1/3 the result is 9. So, the chef has 9 one-third-pound pieces of butter.
Problem Solving
Question 3.
H.O.T. Multi-Step Julie makes a drawing that is $$\frac{1}{4}$$ the size of the original drawing. Sahil makes a drawing that is $$\frac{1}{3}$$ the size of the original. A tree in the original drawing is 12 inches tall. What will be the difference between the height of the tree in Julie and Sahil’s drawings?
1/4 x 12 = 3 inches
The height of the tree in Julie’s drawing is 3 inches.
1/3 x 12 = 4 inches
The height of the tree in Sahil’s drawing is 4 inches.
4 inches – 3 inches = 1 inch
The difference between the height of the tree in Julie and Sahil’s drawings is 1 inch.
Explanation:
A tree in the original drawing is 12 inches tall. Julie makes a drawing that is 1/4 the size of the original drawing. Multiply 1/4 with 12 the product is 3. The height of the tree in Julie’s drawing is 3 inches. Sahil makes a drawing that is1/3 the size of the original. Multiply 1/3 with 12 the product is 4. The height of the tree in Sahil’s drawing is 4 inches. Subtract 3 inches from 4 inches the difference is 1 inch. The difference between the height of the tree in Julie and Sahil’s drawings is 1 inch.
Question 4.
H.O.T. Use Tools Brianna has a sheet of paper that is 6 feet long. She cuts the length of paper into sixths and then cuts the length of each of these $$\frac{1}{6}$$-pieces into thirds. How many pieces does she have? How many inches long is each piece?
Question 5.
A structure is made out of foam cubes that are each 1/4 foot tall. The height of the structure is 8 feet. How many cubes were used to build the structure?
(A) 2 cubes
(B) 16 cubes
(C) 8 cubes
(D) 32 cubes
8 ÷ 1/4 = 32
To build the structure 32 cubes are used.
So, option D is correct.
Explanation:
A structure is made out of foam cubes that are each 1/4 foot tall. The height of the structure is 8 feet. Divide 8 by 1/4 the product is 32. So, 32 cubes are used to build the structure. Draw a circle for option D.
Question 6.
Use Diagrams Terrance needs to divide 9 by $$\frac{1}{2}$$. How can he find the quotient?
(A) Draw 9 rectangles and divide each in halves. Count the halves.
(B) Draw 9 rectangles and shade $$\frac{1}{2}$$ of each. Count the shaded parts.
(C) Shade a rectangle to show $$\frac{1}{2}$$, and then divide the shaded part into 9 parts. Find the amount shaded.
(D) Draw a rectangle with a length of 9 and a width of $$\frac{1}{2}$$, Find the area.
Option A is correct.
Explanation:
Terrance needs to divide 9 by 1/2. First draw 9 rectangles and divide each rectangle into two halves. Then count the halves. There are 18 halves. So, draw a circle to option A.
Question 7.
Multi-Step Dorothy has a ribbon that is 3 feet long, and a ribbon that is 8 feet long. She needs to cut the ribbon into pieces that are $$\frac{1}{4}$$ foot long. How many pieces will she have in all?
(A) 12 pieces
(B) 32 pieces
(C) 11 pieces
(D) 44 pieces
3 ÷ 1/4 = 3 x 4 = 12
She have 12 pieces.
8 ÷ 1/4 = 8 x 4 = 32
12 + 32 = 44 pieces.
So, option D is correct.
Explanation:
Dorothy has a ribbon that is 3 feet long, and a ribbon that is 8 feet long. She needs to cut the ribbon into pieces that are 1/4 foot long. First divide 3 by 1/4 the result is 12 pieces. Next divide 8 by 1/4 the result is 32 pieces. Add 12 pieces with 32 pieces the sum is 44 pieces. She have 44 pieces in all. So, draw a circle for option D.
Texas Test Prep
Question 8.
Adrian made 3 carrot cakes. He cut each cake into fourths. How many $$\frac{1}{4}$$-size cake pieces does he have?
(A) 16
(B) 12
(C) 1
(D) 1$$\frac{1}{3}$$
3 ÷ 1/4 = 3 x 4 = 12
He have 12 one-fourth size cake pieces.
So, option B is correct.
Explanation:
Adrian made 3 carrot cakes. He cut each cake into fourths. Divide 3 by 1/4 the result is 12. He have 12 one-fourth size cake pieces. So, draw a circle for option B.
### Texas Go Math Grade 5 Lesson 6.5 Homework and Practice Answer Key
Question 1.
Julian wants to put a border around two sides of his rectangular shaped garden. He has a 12-foot piece of lumber that will be cut into $$\frac{1}{3}$$-foot pieces. How many $$\frac{1}{3}$$-foot pieces can Julian cut from the lumber?
12 ÷ 1/3 = 12 x 3 = 36
Julian can cut 36 foot pieces from the lumber.
Explanation:
Julian wants to put a border around two sides of his rectangular shaped garden. He has a 12-foot piece of lumber that will be cut into 1/3-foot pieces. Divide 12 by 1/3 the result is 36. So, Julian can cut 36 foot pieces from the lumber.
Question 2.
The camp counselors bought a 6-pound bag of raisins for an afternoon snack for the campers. If the counselors package the raisins into $$\frac{1}{8}$$-pound individual servings, how many individual servings can they make?
6 ÷ 1/8 = 6 x 8 = 48
They can make 48 individual servings.
Explanation:
The camp counselors bought a 6-pound bag of raisins for an afternoon snack for the campers. The counselors package the raisins into 1/8-pound individual servings. Divide 6 by 1/8 the result is 48. So, they can make 48 individual servings.
Question 3.
Kim has 5 yards of denim. She cuts each yard of denim into thirds. How many $$\frac{1}{3}$$-yard pieces of denim does she have?
5 ÷ 1/3 = 5 x 3 = 15
She have 15 pieces with 1/3-yard pieces of denim.
Explanation:
Kim has 5 yards of denim. She cuts each yard of denim into thirds. Divide 5 by 1/3 the result is 15. She have 15 pieces with 1/3-yard pieces of denim.
Question 4.
If it takes Fran $$\frac{1}{5}$$ of an hour to paint one section of the fence, how many sections can she paint in 2 hours?
1/5 x 2 = 2/5
She can paint 2/5 sections in 2 hours.
Explanation:
Fran takes 1/5 of an hour to paint one section of the fence. Multiply 1/5 with 2 the product is 2/5. So, she can paint 2/5 sections in 2 hours.
Problem Solving
Question 5.
You plan to sell pumpkin pie slices at the fall festival. If you cut 4 pies into $$\frac{1}{8}$$-size pieces, how many slices will you have to sell? Describe your strategy for solving the problem.
4 ÷ 1/8 = 4 x 8 = 32
I have to sell 32 slices.
Explanation:
I plan to sell pumpkin pie slices at the fall festival. If I cut 4 pies into 1/8-size pieces. Divide 4 by 1/8 the result is 32. So, I have to sell 32 slices.
Question 6.
Marcel has 6 strips of paper. Each strip is 1 foot long. He folds each strip into $$\frac{1}{4}$$-foot sections. Describe how you can draw a diagram to find the number of sections Marcel made when he folded the paper strips.
Lesson Check
Question 7.
The students in Mrs. Lester’s fifth-grade class are planting seeds in individual pots. Mrs. Lester supplied 8 pounds of potting soil. If each pot holds $$\frac{1}{4}$$ pound of soil, how many pots can the students fill?
(A) 12
(B) 32
(C) 16
(D) 36
8 ÷ 1/4 = 8 x 4 = 32
The student can fill 32 pots.
So, option B is correct.
Explanation:
The students in Mrs. Lester’s fifth-grade class are planting seeds in individual pots. Mrs. Lester supplied 8 pounds of potting soil. Each pot holds 1/4 pound of soil. Divide 8 by 1/4 the result is 32. The student can fill 32 pots. So draw a circle for option B.
Question 8.
Jake’s Cafe receives a 10-pound delivery of ground beef each day. How many $$\frac{1}{4}$$-pound hamburgers can Jake make each day?
(A) 40
(B) 14
(C) 44
(D) 20
10 ÷ 1/4 = 10 x 4 = 40
Jake makes 40 hamburgers each day.
So, option A is correct.
Explanation:
Jake’s Cafe receives a 10-pound delivery of ground beef each day. Divide 10 by 1/4 the result is 40. Jake makes 40 hamburgers each day. So draw a circle for option A.
Question 9.
Mrs. Miller is cutting 2 pizzas into eighths for students. How many students will get a $$\frac{1}{8}$$-size slice of pizza?
(A) 7
(B) 8
(C) 16
(D) 6
2 ÷ 1/8 = 2 x 8 = 16
16 students got 1/8 size slice of pizza.
So, option C is correct.
Explanation:
Mrs. Miller is cutting 2 pizzas into eighths for students. Divide 2 by 1/8 the result is 16. So, 16 students got 1/8 size slice of pizza. Draw a circle for option C.
Question 10.
A home building company purchases 24 acres of land and divides it into $$\frac{1}{2}$$-acre plots for sale. How many $$\frac{1}{2}$$-acre plots will be for sale?
(A) 6
(B) 24
(C) 12
(D) 48
24 ÷ 1/2 = 24 x 2 = 48
48 plots with 1/2 acre are sale.
So, option D is correct.
Explanation:
A home building company purchases 24 acres of land and divides it into 1/2-acre plots for sale. Divide 24 by 1/2 the result is 48. So, 48 plots with 1/2 acre are sale. Draw a circle for option D.
Question 11.
Multi-Step Cleo bought 10 feet of ribbon for costumes. He used 2 feet of ribbon last week. He cut the remaining ribbon into $$\frac{1}{6}$$-foot pieces. How many pieces of ribbon does Cleo have?
(A) 48
(B) 60
(C) 72
(D) 12
10 – 2 = 8
8 ÷ 1/6 = 8 x 6 = 48
Cleo have 48 pieces of ribbon.
So, option A is correct.
Explanation:
Cleo bought 10 feet of ribbon for costumes. He used 2 feet of ribbon last week. Subtract 2 feet from 10 feet the difference is 8 feet. He cut the remaining 8 feet ribbon into 1/6-foot pieces. Divide 8 by 1/6 the result is 48. So draw a circle for option A.
Question 12.
Multi-Step Reese brought 2 sandwiches for her school lunch. Each sandwich was cut into fourths. After she ate lunch, Reese has 3 pieces of sandwich left. How many pieces did Reese eat?
(A) 5
(B) 3
(C) 2
(D) 1 |
# 7Bayes’ Theorem With LEGO
The chapter starts with a small change in Bayes’ theorem:
Theorem 7.1 (Bayes’ Theorem (Variation 2)) $P(A \mid B) = \frac{P(B \mid A) \times P(A)}{P(B)} \tag{7.1}$
In contrast with Equation 6.3 the order of the terms of the multiplication in the nominator is changed. The meaning of the formula is identical, but sometimes changing the terms around can help clarify different approaches to problems. (By the way: This is the form of the formula that is most used in the literature.)
There is no new statistical content taught in this chapter The author tries to solidify the mathematics behinds the intuition by using LEGO bricks to visualize Bayes’ theorem.
## 7.1 Exercises
Try answering the following questions to see if you have a solid understanding of how we can use Bayes’ Theorem to reason about conditional probabilities. The solutions can be found at https://nostarch.com/learnbayes/.
### 7.1.1 Exercise 7-1
Kansas City, despite its name, sits on the border of two US states: Missouri and Kansas. The Kansas City metropolitan area consists of 15 counties, 9 in Missouri and 6 in Kansas. The entire state of Kansas has 105 counties and Missouri has 114. Use Bayes’ theorem to calculate the probability that a relative who just moved to a county in the Kansas City metropolitan area also lives in a county in Kansas. Make sure to show $$P(Kansas)$$ (assuming your relative either lives in Kansas or Missouri), $$P(\text{Kansas City metropolitan area})$$, and $$P(\text{Kansas City metropolitan area}| Kansas)$$.
US states: 219 counties = 105 Kansas, 114 Missouri Kansas City: 15 counties = 6 Kansas, 9 Missouri
Intuitively: $$\frac{6}{15} = \frac{2}{5}$$
\begin{align*} P(\text{County in Kansas} | \text{Kansas City}) = \frac{P(\text{Kansas City} | \text{County in Kansas}) \times P(\text{County in Kansas})}{P(\text{Kansas City})} \\ = \frac{\frac{6}{105} \times \frac{105}{219}}{\frac{15}{219}} = \frac{\frac{6}{219}}{\frac{15}{219}} = \frac{6}{15} = \frac{2}{5} \end{align*}
### 7.1.2 Exercise 7-2
A deck of cards has 52 cards with suits that are either red or black. There are four aces in a deck of cards: two red and two black. You remove a red ace from the deck and shuffle the cards. Your friend pulls a black card. What is the probability that it is an ace?
Cards: 51 cards = 26 black, 25 red Aces: 3 aces = 2 black, 1 red $$P(ace | \text{black card})$$?
\begin{align*} P(ace | \text{black card}) = \frac{P(\text{black card} | ace) \times P(ace))}{P(\text{black card})} \\ = \frac{\frac{2}{3} \times \frac{3}{51}}{\frac{26}{51}} = \frac{\frac{6}{153}} {\frac{26}{51}} = \frac{6 \times 51} {26 \times 153} = \frac{6}{26} \times \frac{1}{3} = \frac{2}{26} = \frac{1}{13} \end{align*} Or more easily without Bayes’ rule: We have 26 black cards with 2 aces = $$P(\frac{2}{26}) = \frac{1}{13}$$ |
Commutative and Associative Properties
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# Commutative and Associative Properties - PowerPoint PPT Presentation
Commutative and Associative Properties. Commutative and Associative Properties. Commutative Property – changing the order in which you add or multiply numbers does not change the sum or product.
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Commutative and Associative Properties
• Commutative Property – changing the order in which you add or multiply numbers does not change the sum or product.
• Associative Property – changing the grouping of numbers when adding or multiplying does not change their sum or product.
• Grouping symbols are typically parentheses (),but can include brackets [] or Braces {}.
Commutative and Associative Properties
Commutative Properties
Commutative Property of addition - (Order)
For any numbers a and b , a + b = b + a
45 + 5 = 5 + 45
50 = 50
Commutative Property of multiplication - (order)
For any numbers a and b , a b = b a
6 8 = 8 6
48 = 48
Commutative and Associative Properties
Associative Properties
Associative Property of addition - (grouping symbols)
For any numbers a, b, and c,
(a + b) + c = a + (b + c)
(2 + 4) + 5 = 2 + (4 + 5)
(6) + 5 = 2 + (9)
11 = 11
Associative Property of multiplication - (grouping symbols)
For any numbers a, b, and c,
(ab)c = a (bc)
(2 3) 5 = 2 (3 5)
(6) 5 = 2 (15)
30 = 30
Commutative and Associative Properties
Commutative and associative properties are very helpful to solve problems using mental math strategies
Rewrite the problem by grouping numbers that can be formed easily. (Associative property)
This process may change the order in which the original problem was introduced. (Commutative property)
Solve: 18 + 13 + 16 + 27 + 22 + 24
(18 + 22) + (16 + 24) + (13 + 27)
(40) + (40) + (40) = 120
Commutative and Associative Properties
Commutative and associative properties are very helpful to solve problems using mental math strategies
Solve: 4 7 25
Rewrite the problem by changing the order in which the original problem was introduced. (Commutative property)
4 25 7
Group numbers that can be formed easily. (Associative property)
(4 25) 7
(100) 7 = 700 |
Related Articles
# Lines and Angles
Lines and Angles are the basic terms used in the Geometry. They provide a base for understanding all the concepts of geometry. We define a line as a 1-D figure which can be extended to infinity in opposite directions, whereas an angle is defined as the opening created by joining two or more lines. An angle is measured in degrees or in radians depending on the concept of the problem.
All the geometrical figures have lines and angles and having an understanding of them helps us to better understand the world of Geometry. In this article, we will learn about lines, angles, their types, properties, and others in detail.,
## Definition of Lines and Angles
We already know that lines and angles are the base shape of geometry and the knowledge of these helps us to better understand the concept of geometry. The basic definition of lines and angles is that the lines are 1-D figures that can be extended infinitely in the opposite direction, whereas an Angle is defined as the degree of mouth open when two lines intersect. If the wide space between the two intersecting lines is more then the angle between them is more.
These concepts are highly used to define various terms and are very helpful for students to study Geometry. Now let’s learn about them in detail.
## What are Lines?
Line is defined as a one-dimensional figure that can be extended infinitely. It can extend in both directions and the length of a line is infinite. We can also define a line as the collection of infinitely many points that join together to form a continuous figure.
A line does not have any starting point or an endpoint. If a line has a starting point and a end point then it is called a line segment, whereas if a line only has a starting point but no end point then it is called a ray.
## Types of Lines and Angles
In this topic, we will learn all the different types of lines and angles classified in Geometry.
### Types of Lines
We can classify the lines on the basis of their endpoint and the starting point as,
• Line Segment
• Ray
Lines can also be categorized as,
• Parallel Lines
• Perpendicular Lines
• Transversal
Now let’s learn about them in detail.
### Line Segment
Line segment is a part of a line that has two endpoints. It is the shortest distance between two points and has a fixed length and can’t be extended further. A line segment AB is shown in the image added below:
### Ray
Ray is a line that has a starting point or end point and moves to infinity in one direction. A ray OA is shown in the image added below. Here O is starting point and is moving towards A.
### Perpendicular Lines
When two lines form a right angle to each other and meet at a single point then they are called Perpendicular lines. Two perpendicular lines AB and CD are shown in the image added below:
### Parallel Lines
Parallel lines are those lines that do not meet each other on a plane at any point and do not intersect with each other. The distance between any two points of the parallel lines is fixed. Two parallel lines l and m are shown in the image added below:
### Transversal Lines
When two given lines intersect each other at a distinct point, they will be called transversal lines. Line n is transversal to line l and line m as shown in the image added below:
### Properties of Lines
Various properties of the lines are,
• If three or more than three points lie in the same line then they are called the collinear points.
• Two lines are called parallel lines if the distance between them is always constant.
• If the line intersects at the right angles then they are called the perpendicular lines.
## What are Angles?
When the end points of two rays meet at a common point the figure so formed is called the angle. An angle is measured either in degrees or in radians and we can easily convert a degree to radian. We use ∠it to represent an angle.
There are various types of angles depending upon their measure. They are discussed below,
### Types of Angles
There are various types of lines and angles in geometry based on the measurements and different scenarios. Let us learn here all those lines and angles along with their definitions.
• Acute Angle
• Obtuse Angle
• Right Angle
• Straight Angle
• Reflex Angle
• Complete Angle
### Acute Angle
When the angle is less than a right angle, then it is called an acute angle. It measures between 0 degrees to 90 degrees. The image added below shows an acute angle:
### Obtuse Angle
When the measure of the angle is more than a right angle, then it is called an obtuse angle. It measures greater than 90 degrees. The image added below shows an obtuse angle:
### Right Angle
When the angle measures exactly 90 degrees, then it is called a right angle. The image added below shows a right angle:
### Straight Angle
If the measure of an angle is 180 degrees then the angle so formed is called the straight angle. The image added below shows a straight angle.
### Reflex Angle
When the measurement of the angle is greater than 180° and less than 360° then it is called Reflex Angle. The image added below shows Reflex Angle
### Complete Angle
When the measurement of the angle is 360° then it is called Complete Angle. The image added below shows the Complete angle.
We can also categorize angles as,
• Supplementary angles
• Complementary angles
• Vertically opposite angles
### Complementary Angles
When two angles sum to 90°, they are called complementary angles. Two complementary angles AOB and BOC are shown in the image below.
### Supplementary Angles
When two angles sum up to 180° they will be called Supplementary angles. Two complementary angles PMN and QMN are shown in the image below.
When two angles have a common side and a common vertex and the remaining two sides lie on alternate sides of the common arm then they are called Adjacent Angles. Two adjacent angles A and B are shown in the image below.
### Vertically Opposite Angles
When two angles that are opposite two each other and two lines intersect each other at a common point are called vertically opposite angles. Two angles AOB and COD are shown in the image added below.
∠AOD= ∠BOC
∠AOB= ∠DOC
### Vertical Angles are Equal
The vertical opposite angles are always equal to each other. The image added below shows the pair of equal vertically opposite angles
• ∠MON = ∠POQ
• ∠MOP = ∠NOQ
We can prove this as,
To Prove:
• ∠MON = ∠POQ
• ∠MOP =∠NOQ
Proof:
∠MOP+ ∠MON = 180°
∠MOP+ ∠POQ = 180°
Therefore:
∠MOP+ ∠MON = ∠MOP+ ∠POQ
Now subtracting ∠MOP at both sides
∠MOP + ∠MON – ∠MOP = ∠MOP + ∠POQ – ∠MOP
∠MON = ∠POQ
Similarly,
∠MON + ∠NOQ = 180°
∠POQ + ∠NOQ = 180°
Therefore,
∠MOP+ ∠NOQ = ∠POQ+ ∠NOQ
Subtracting ∠NOQ at both sides
∠MOP + ∠NOQ – ∠NOQ = ∠POQ + ∠NOQ – ∠NOQ
∠MOP = ∠POQ
Thus,
• ∠MON = ∠POQ
• ∠MOP =∠NOQ
## Angles in a Triangle Sum Up to 180°
Sum of all the angles in any triangle is 180°. This is proved below, Suppose we have a triangle ABC as shown in the image below:
To Prove: ∠A + ∠B + ∠C = 180°
Proof:
Draw a line parallel to BC and pass through vertices A of the triangle.
Now as both lines PQ and BC are parallel, AB and AC are transversal.
So:
• ∠PAB = ∠ABC (Alternate Interior Angles)…(i)
• ∠QAC = ∠ACB (Alternate Interior Angles)…(ii)
Now, ∠PAB + ∠BAC + ∠QAC = 180° (Linear Pair)…(iii)
From eq (i), (ii) and (iii)
∠ABC + ∠BAC + ∠ACB = 180°
∠A + ∠B + ∠C = 180°
Proved.
## Properties of Lines and Angles
In this section, we will learn about some general properties of lines and angles
### Properties of Lines
There are the following properties of line
• Line has only one dimension i.e. length. It does not have breadth and height.
• A line has infinite points on it.
• Three points lying on a line are called collinear points
### Properties of Angles
There are the following properties of angles
• Angles tell about how much a person has rotated from his position.
• Angles are formed when two lines meet and they are called arms of the angle.
## Examples on Lines and Angles
Example 1: Find the reflex angle of ∠x, if the value of ∠x is 75 degrees.
Solution:
Let the reflex angle of ∠x be ∠y.
Now, according to the properties of lines and angles, the sum of an angle and its reflex angle is 360°.
Thus,
∠x + ∠y = 360°
75° + ∠y = 360°
∠y = 360° − 75°
∠y = 285°
Thus, the reflex angle of 75° is 285°.
Example 2: Find the complementary angle of ∠x, if the value of ∠x is 75 degrees.
Solution:
Let the complementary angle of ∠x be ∠y.
Now, according to the properties of lines and angles, the sum of an angle and its complementary angle is 90°.
Thus,
∠x + ∠y = 90°
75° + ∠y = 90°
∠y = 90° − 75°
∠y = 15°
Thus, the complementary angle of 75° is 15°.
Example 3: Find the supplementary angle of ∠x, if the value of ∠x is 75 degrees.
Solution:
Let the supplementary angle of ∠x be ∠y.
Now, according to the properties of lines and angles, the sum of an angle and its supplementary angle is 180°.
Thus,
∠x + ∠y = 180°
75° + ∠y = 180°
∠y = 180° − 75°
∠y = 105°
Thus, the supplementary angle of 75° is 105°.
Example 4: Find the value of ∠A and ∠B if ∠A = 4x and ∠B = 6x are adjacent angles and they form a straight line.
Solution:
According to the properties of lines and angles, the sum of the adjacent linear angles formed by a line is 180°.
Thus,
∠A + ∠B = 180°
4x + 6x = 180°
10x = 180°
x = 180°/10 = 18°
Thus,
• ∠A = 4x = 4×18 = 72°
• ∠B = 6x = 6×18 = 108°
## FAQs on Lines and Angles
### Q1: What are Lines in Geometry?
We define a line in geometry as a one-dimensional figure that extends infinitely in opposite directions. A line can be vertical or horizontal with respect to the reference line or plane. As the line can extend infinitely the length of the line is infinity.
### Q2: What are Angles in Geometry?
An angle in geometry is defined as the shape formed when two lines meet one another at some point. We measure in degrees and radians. The symbol ∠is used to represent the radian.
### Q3: What are the types of Angles?
There are five types of angles that are,
• Acute Angle
• Right Angle
• Obtuse Angle
• Straight Angle
• Reflex Angle
### Q4: What are the types of Lines?
The different types of lines are:
• Horizontal lines
• Vertical lines
• Parallel lines
• Perpendicular lines |
# Probability Questions for 6th Grade Math Students
Probability is the study of how likely something is to happen. Students study probability in 6th grade and continue learning about probability in later grades. Standardized math tests can include probability questions, so extra practice can be beneficial for your child. Use the sample questions below as a guide to create your own practice.
## How to Create 6th Grade Probability Questions
Probability and statistics are often studied together. Statistics analyzes how frequently something happened in the past, while probability predicts the likelihood of future events. In 6th grade, students study how data is collected for statistical problems. They also learn how to graph the data collected. Based on the statistical data, 6th graders begin to make probability predictions.
Because of the nature of probability, many questions involving probability are word problems. If your child is having difficulty finding the answers to these questions, remind her to pick out the important information from the question and ignore the unimportant information. For instance, the first sentence in many word problems is not pertinent to the problem itself. Instead, she should pay attention to the numbers and how they relate to one another. Using this information, your child can then create a proportion to find the probability.
## Probability Problems by Subject
### Preference
In your class, 30 students were asked if they prefer action movies or drama movies. There were 21 students who preferred action, while the remaining 9 students liked dramas. There are 650 students in the whole school. How many would you expect to prefer action movies?
To solve this problem, your child should make a proportion. In the initial survey, 21 out of 30 students preferred action. So the proportional equation should look like this: 21/30 = x/650. After solving this equation, your child should find that 455 students are expected to prefer action movies.
### Luck
Relating the concept of probability to luck can help your child become interested in the subject matter. Try making it a game by using a deck of cards or flipping a coin. Practice doesn't always have to involve a worksheet. Ask your child, 'If you were to draw one card at random, what is the likelihood that you will draw a red card?'
To solve this problem, first determine how many red cards are in a deck. Half of the cards are red, so 26 of the 52-card deck are red. As a result, the probability is 26/52, which is reduced to 1/2. There is a 50% chance that a red card will be drawn.
### Real World
Increase your student's interest in probability by applying it to the real world. For ten days, watch the weather forecast with your child and keep track of its accuracy. To calculate the probability, divide the number of correct weather reports by ten. For instance, if the forecasts were accurate 4 out of the 10 days, then there is a 40% probability that the weather reports will be accurate on any given day.
Did you find this useful? If so, please let others know!
## Other Articles You May Be Interested In
• Elementary Math Help: 5th and 6th Grade
Mathematics classes get increasingly difficult towards the end of elementary school. Learn how to help your fifth and sixth grade students improve their math skills.
• MIND Games Lead to Math Gains
Imagine a math teaching tool so effective that it need only be employed twice per week for less than an hour to result in huge proficiency gains. Impossible, you say? Not so...and MIND Research Institute has the virtual penguin to prove it.
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"New" Ways of Working with Triangles
Hans-Georg Weigand and James P. Dildine
The following links allow an exploration into new ways of working with triangles that can be afforded to us through technology.
Working with flexible triangles: We want you to explore the locus of special points of a triangle. We solve problems in five steps Circumcenter Center of gravity Orthocenter
Go ahead with the following problems. Try to solve them first by your own. Afterward you will see the solution.
First: Given is the triangle ABC shown below (The point C should be on the parallel!). Point C moves along a line that is parallel to the line segment AB and the line the segment is contained in.
Problem 1: First step: You see the CIRCUMCENTER of a triangle ABC. If you move point C along the line parallel to base AC... What locus of points is created if C moves along the line parallel to the base. Do a hand draft of this locus.
Second step: PICTURE were you can do a hand draft.
Third step: Next PICTURE (without perpendicular bisectors): Now you can move the point C. Also animate .....
Forth step: Why do you think that the formation created is as such? Write it down: TEXTBOX.
Fifth step: If you don't know it, you may also look at the next PICTURE (with perpendicular bisectors): Do you have an idea now, why. ...
????There is a possibility to go on e. g.
Problem 2: Now point C moves along a circle (or on another curve). ......
.........
Problem 3: Now we consider a quadrilateral (4-sided polygon) ABCD. Construct the perpendicular bisectors of the quadrilateral ABCD
This is a completely different problem....But this can lead to the classification of quadrilaterals.
-----------------------------------
The CIRCUMCENTER of a triangle results at the intersection of the perpendicualr bisectors of he sides of triangle ABC. You can construct the cirumcenter .....Back
----------------------------------------
Problem II: You see CENTER OF GRAVITY of a triangle ABC. If you move point C along the line parallel to base AC... What locus of points is created if C moves along the line parallel to the base. Do a hand draft of this locus.
PICTURE were you can do a hand draft.
Next PICTURE (without perpendicular bisectors): Now you can move the point C. Also animate .....
Why do you think that the formation created is as such? Write it down: TEXTBOX.
If you don't know it: Do you know any property concerning the bisectors and the center of gravity.
Do you know what a dialation is? Now we do a dialation......
Problem 2: Now we consider a quadrilateral (4-sided polygon) ABCD. Construct the perpendicular bisectors of the quadrilateral ABCD
This is a completely different problem....But this can lead to the classification of quadrilaterals.
-----------------------------------
The five steps:
First step: The problem
Second step: Enactive level: Do it by hand
Third step: Do a computer animation
Forth step: Explain the computer picture
Fifth step: The computer as a help for explaining the problem.
Back
----------------------------------
The CIRCUMCENTER of a triangle results at the intersection of the perpendicualr bisectors of the sides of triangle ABC.
You can construct the cirumcenter .....
---------------------------------------------
The CENTER OF GRAVITY OF A TRIANGLE is the intersection of the lines that make up the bisectors of sides triangle ABC.
If you move point B along the line parallel to the base AC then...What locus of points does this intersection point create? Draw a
picture on your own first and then use the interactive simulator here.
Back
Circumcenter Exploration
Center of Gravity (Median Center)
Orthocenter Exploration
Back |
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What is the smallest number that, when divided by $35,56$ and $91$ leaves remainder of $7$ in each case?
Last updated date: 19th Sep 2024
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Hint: To find the smallest number which when divided by $35,56,91$ leaves remainder $7$ in each case, calculate the LCM of $35,56,91$ and add $7$ to the LCM of $35,56,91$.
We have to find the smallest number which when divided by $35,56,91$ leaves remainder $7$ in each case. To do so, we will firstly evaluate the smallest number which is a multiple of each of $35,56,91$.
Thus, we will evaluate the LCM of $35,56,91$.
To find the LCM of three numbers, we will use the prime factorization method. We will write the prime factorization of each of the numbers in exponential form. Then, we will align the common prime factor base whenever possible. For the numbers with a common prime factor base, select the prime number that has the highest power. The prime factor with highest power implies that it occurs the most in the list. If a distinct prime factor has no matching prime factor base in the list, include this factor with its exponent in the collection of numbers. Multiply all the numbers which were collected earlier to get the LCM of three numbers.
Factorizing each of the numbers, we have $35=5\times 7,56={{2}^{3}}\times 7,91=7\times 13$ .
Thus, the LCM of $35,56,91$ is $7\times 5\times {{2}^{3}}\times 13=3640$.
The LCM of $35,56,91$ is $3640$.
We have to find the smallest number which leaves a remainder $7$ when divided by $35,56,91$, add $7$ to the LCM of numbers $35,56,91$.
Thus, we have $3640+7=3647$.
Hence, the smallest number which leaves a remainder $7$ when divided by $35,56,91$ is $3647$.
Note: Least Common Multiple (LCM) of any two numbers is the smallest positive number that is divisible by both the numbers. We can also find LCM of two numbers using list method. We should carefully calculate the LCM of three numbers; otherwise we will get an incorrect answer. |
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The Ultimate College Algebra Guide
We love this basic college algebra guide simply because the answers aren’t always obvious in math. So here’s a go-to library of resources for you to get bigger results and better grades.
```How to Factor Polynomials
How to Turn a Decimal Into A Fraction
How to Solve a Linear Equation
How to Solve a Linear Function
Quadratic Equation: What All Those Letters Mean```
College algebra can really help you live life with more success. Not only does it challenge you to expand your mind, but college algebra introduces life skills as well. For example, it can give you a guide to measure long it will take to repay a college loan.
But if you just want to pass the class right now, we’ve compiled a step-by-step college algebra guide to making your algebra homework as easy as possible.
## How to Factor Polynomials
If there is a constant in college algebra, it is the letter X. In algebra, X is the source of many a mental math meltdown. But with this college algebra guide, it doesn’t have to be that way! Here are the steps to factoring polynomials:
1. Put x factors in descending order of x based on negative and positive numbers as well as the factors. Negative and big factors come first.
2. Simplify them by adding like-factors together: all of the x2, then all the x, then the numbers without x.
3. Look for common factors in all groups. For example, if these groups all have even numbered coefficients know they are divisible by four.
4. Use FOIL to break down the polynomial even further. This stands for “first, outside, inside, last”.
## How to Turn a Decimal Into A Fraction
This process is pretty straightforward and we compiled the steps for figuring it out in college algebra and beyond:
1. Count the number of digits in the decimal.
2. Move the decimal point to the right until it becomes a whole number. This is the numerator.
3. Add one “0” for ever digit you counted in step 1 and make it the denominator after 1. Ex: 2 decimal digits means 100.
4. Reduce the fraction.
Check out this guide to see turning a decimal into a fraction with real examples.
## How to Solve a Linear Equation
1. Isolate the variable you’re instructed to solve for.
2. Everything you do to one side must be done to the other.
If you’re new at this, the following article explains the three steps to solve a linear equation with more detail.
## How to Solve a Linear Function
Basically, isolate your variable, do everything you did to one side of the equation to the other, and simplify your terms to their smallest values.
1. Make sure you’re actually looking at a linear function. It always starts with “f(x)=“.
2. Substitute variables for values.
## Quadratic Equation: What All Those Letters Mean
Even mathematicians admit that the quadratic formula can look messy and overwhelming. Here’s how we break it down.
1. When you see c, you are looking at a quadratic.
2. The “x” corresponds with a specific line on a graph, usually the horizontal line.
3. The vertical line of the graph, or “y” is understood as the “0”.
4. The “a”, “b”, and “c” elements are simply place marks for numbers. |
Trying to read a string of binary 1's and 0's can seem a daunting task. However, with a bit of logic we can figure out what they mean. Humans have adapted to use a base ten number system simply because we have ten fingers. Computers, on the other hand (no pun intended), have only two "fingers"--on and off or one and zero. Therefore, the base two number system has been created.[1]
Method 1
Method 1 of 2:
### With Exponents
1. 1
Find a binary number you want to convert. We'll use this as an example: 101010.
2. 2
Multiply each binary digit by two to the power of its place number. Remember, binary is read from right to left.[2] The rightmost place number being zero.
3. 3
Add all the results together. Let's go from right to left.[3]
• 0 × 20 = 0
• 1 × 21 = 2
• 0 × 22 = 0
• 1 × 23 = 8
• 0 × 24 = 0
• 1 × 25 = 32
• Total = 42
4. 4
Try another example. Let's use 101. Here is the same method as above, but in a slightly different format. You may find this format easier to understand.[4]
• 101= (1X2) power of 2 + (0X2) power of 1 + (1X2) power 0
• 101= (2X2) + (0X0) + (1)
• 101= 4 + 0 + 1
• 101= 5
• The 'zero' is not a number, but its place value must be noted.
Method 2
Method 2 of 2:
### Slot Value
1. 1
Find your number. The example we'll use is 00101010.
2. 2
Read from right to left. With each slot, the values are doubled. The first digit from the right has a value of 1, the second is a 2, then a 4, and so on.[5]
3. 3
Add the values of the ones. The zeros are assigned their correlating number, but those numbers are not added.
• So, in this example, add 2, 8, and 32. The result is 42.
• There is a 'no' on 1, a 'yes' on 2, a 'no' on 4, a 'yes' on 8, a 'no' on 16, a 'yes' on 32, a 'no' on 64 and a 'no' on 128. "Yes" means to add, "no" is to skip. You can stop at the last one-digit.
4. 4
Translate the value into letters or punctuation marks.[6] In addition, you can convert numbers from binary to decimal or convert from decimal to binary.
• In punctuation marks, the 42 equals an asterisk (*). Click here for a chart.
## Community Q&A
Search
• Question
How can I turn binary into a sentence?
You translate a string of binary code to decimals and then translate the decimals into html code. For example, 111011= 123, 123= { . Write multiple strings of binary code and you'll get a sentence.
• Question
Can I learn to read binary if I am frightened of math?
You will need some basic understanding of math (addition and multiplication) or a calculator.
• Question
How do I add two lines of binary numbers together?
It really depends on the method you're using. If you're looking for a sentence conversion, then the computer automatically reads in sets of eight e.g 00111111 (?), 00111101 (=), 01000001 (A) etc. However, the process the computer goes through is much more complicated than that. Essentially, every five digits is equal to either a symbol, number, or letter.
200 characters left
## Tips
• The numbers we deal with today have a place value. Assuming we are working with whole numbers, the right-most digit is the one's place, the next right-most digit is the ten's place, then hundred's, and so on. The place value for binary numbers go from one's, two's, four's, eight's, and so on.[7]
⧼thumbs_response⧽
• Binary counts just like normal numbers. The rightmost digit increments by one until it cannot increase any more (in this case from 0 to 1) and then increments the next digit to the left by one and starts again at zero.
⧼thumbs_response⧽
Submit a Tip
All tip submissions are carefully reviewed before being published
Thanks for submitting a tip for review!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 52 people, some anonymous, worked to edit and improve it over time. This article has been viewed 1,008,436 times.
Co-authors: 52
Updated: November 18, 2022
Views: 1,008,436
Article SummaryX
To read binary, find a number that you want to read, and remember to count the places from right to left. Then, multiply each digit by 2 to the power of its place number. For example, if the 3rd place from the right is a 1, you would multiply 1 by 2 to the power of 3 to get 8. Once you have an answer for each place, add the numbers together from right to left. For example, 101 would translate to the number 9. For tips on using other techniques, like exponents or slot value, read on! |
# Calculus 1 : How to find solutions to differential equations
## Example Questions
1 2 7 8 9 10 11 12 13 15 Next →
### Example Question #395 : Equations
Find the derivative of the function .
Explanation:
To find the derivative of the function, you must use the power rule
.
Using this rule, the correct answer
is obtained.
### Example Question #1441 : Functions
Find the derivative of the function .
Explanation:
To take the derivative of the first two terms, you use the power rule
.
To take the derivative of sin(x), there is the identity that,
.
Using all of the rules and applying,
.
### Example Question #397 : Equations
Determine the solution to the following: with .
The equation is not separable.
Explanation:
Solving a differential equation tends to be quite simple if it is separable. A separable equation is one that can be rewritten in the following form
.
From here we can manipulate the equation
and then you can integrate both sides.
We can rewrite the equation as
and we therefore see that it is separable.
Next we manipulate it to get the the y terms on the left hand side and the x terms on the right hand side like the following
.
The next step is to integrate both sides
.
This gives us
,
since the c's are just arbitrary constants we can combine them to get
.
Since we want to write this as an equation in terms of y we will apply e to both sides.
Therefore we get
.
Since the problem gives us an initial condition we solve for c.
Therefore
so c=4.
Thus the solution is
.
### Example Question #398 : Equations
Solve the following differentiable equation:
with a condition of .
Explanation:
When given a first order differential equation to solve, one of the first things to check is that it is homogenous.Homogenous equations are ones in which, given
given any t.
If an equation is homogenous you can do a substitution of and this new equation will be separable.
The first step is to check if the following equation is homogenous,
.
Next we show that the equation is homogenous
.
So we now replace y with xz. This gives us the following new equation:
.
This is a separable equation which we manipulate and integrate both sides
.
Thus we get the equation
,
then plugging y back in we get
.
So our family of equations is
yet plugging in the given condition gives us that c=0.
Thus the final solution
.
1 2 7 8 9 10 11 12 13 15 Next → |
# RD Sharma Solutions for Class 6 Maths Chapter 19: Geometrical Constructions Exercise 19.6
In this exercise, students learn how to construct angles with the help of the ruler and compasses. The solutions for the exercise-wise problems are designed by faculty at BYJU’S based on the latest CBSE guidelines. The students can use the solutions in order to understand other methods which can be used to solve the complex problems in a shorter duration. To improve time management which is an important aspect from the exam point of view, students can download RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions Exercise 19.6 PDF which are available here.
## RD Sharma Solutions for Class 6 Maths Chapter 19: Geometrical Constructions Exercise 19.6 Download PDF
### Exercise 19.6 page: 19.15
1. Construct an angle of 60o with the help of compasses and bisect it by paper folding.
Solution:
Construct a ray OA
Taking O as centre and convenient radius, construct an arc which cuts the ray OA at the point P.
Taking P as centre and same radius, construct another arc which cuts the previous arc at the point Q.
Construct OQ and extend it to the point B.
Here, ∠AOB is the required angle of 60o
Now cut the part of paper as sector OPQ
Fold the part of paper such that the line segments OP and OQ coincides.
The angle which is made at point O is the required angle which is half of ∠AOB.
2. Construct the following angles with the help of ruler and compasses only:
(i) 30o
(ii) 90o
(iii) 45o
(iv) 135o
(v) 150o
(vi) 105o
Solution:
(i) 30o
Construct a ray OA.
Taking O as centre and with convenient radius, construct an arc which cuts OA at the point P.
Taking P as centre and same radius, construct an arc which cuts the previous arc at the point Q.
Considering P and Q as centres and radius which is more than half of PQ construct two arcs which cuts each other and name it as point R.
Construct OR and extend it to the point B.
Here, ∠AOB is the required angle of 30o
(ii) 90o
Construct a ray OA.
Taking O as centre and with convenient radius, construct an arc which cuts OA at the point P.
Taking P as centre and same radius, construct an arc which cuts the previous arc at the point Q.
Taking Q as centre and same radius, construct an arc which cuts the previous arc at the point R.
Considering Q and R as centres and radius which is more than half of QR construct two arcs which cuts each other and name it as point S.
Construct OS and extend it to the point B from the ray OB.
Here, ∠AOB is the required angle of 90o.
(iii) 45o
In order to draw an angle of 45o, construct an angle of 90o and bisect it.
Construct ∠AOB = 90o where OA and OB are the rays which intersect the arc at the points P and T
Taking P and T as centres and radius which is more than half of PT, construct two arcs which cuts each other and name it as point X
Construct OX and extend it to the point C to form the ray OC
Here, ∠AOC is the required angle of 45o.
(iv) 135o
Construct a line AB and mark a point O in the middle of AB.
Taking O as centre and convenient radius, construct an arc which cuts the line AB at the points P and Q.
Construct an angle of 90o on the ray OB as ∠BOC = 90o where OC cuts the arc at the point R
Taking Q and R as centres and radius which is more than half of QR, construct two arcs which cuts each other and name it as point S.
Construct OS and extend it to the point D to form the ray OD.
Here, ∠BOD is the required angle of 135o.
(v) 150o
Construct a line AB and mark a point O in the middle of AB.
Taking O as centre and convenient radius construct an arc which cuts the line AB at the points P and Q.
Taking Q as centre and same radius construct an arc which cuts the previous arc and name it as point R.
Taking R as centre and same radius construct an arc which cuts the previous arc and name it as point S.
Taking P and S as centres and radius which is more than half of PS, construct two arcs which cuts each other and name it as point T.
Construct OT and extend it to the point C to form the ray OC.
Here, ∠BOC is the required angle of 150o.
(vi) 105o
Construct a ray OA and make ∠AOB = 90o and ∠AOC = 120o
Bisect ∠BOC and get the ray OD.
Here, ∠AOD is the required angle of 105o.
3. Construct a rectangle whose adjacent sides are 8 cm and 3 cm.
Solution:
Construct a line segment AB of length 8 cm.
Draw ∠BAX = 90o at A and ∠ABY = 90o at B
With the help of compass and ruler, mark a point D on the ray AX where AD = 3 cm
In the same way mark the point C on the ray BY where BC = 3 cm
Construct the line segment CD
Hence, ABCD is the required rectangle. |
IPAT geometry
Very elementary geometry is tested in the IBM IPAT. Most of it just involves the algebraic manipulations of basic geometry formulas. So let’s dig right in with a few examples but before that here is a refresher of the formulas you will encounter in the IPAT.
Square Perimeter = 4 x a Area = a2 where a is the side length Rectangle Perimeter = 2(l+b) Area = l x b where l is the length and b is the width of the rectanle Circle Perimeter = 2 x pi x r Area = pi x r2 where r is radius of circle Triangle Perimeter = Sum of all sides Area = 1/2 x height x base Cylinder Curved area = 2 x pi x radius x height Total area = top circle area + bottom circle area + curved area Volume = pi x radius2 x height Cuboid Surface area = 2 x ( length x base + base x height + height x length) Volume = length x base x height Cone Curved area = pi x radius x slant height slant height2 = radius2+height2 Volume = 1/3 x pi x radius2 x height Sphere Surface area = 4 x pi x radius2 Volume = 4/3 x pi x radius3 Pythagoras Theorem In a right triangle, Longest side2 = base2 + height2
Let’s try out two examples which have come in the IPAT.
Q) If the length of a rectangle is increased by 7 cm and its width increased by 5 cm, the resulting rectangle is 100 cm2 greater in area than before. If the perimeter of the rectangle is 20 cm, what is the length of the original rectangle?
A] 4.4 cm
B] 3.6 cm
C] 3.2 cm
D] 2.5 cm
E] 2.6 cm
Solution
Solution – D] 2.5 cm
Let l be length of the rectangle and its width be w. Converting the information given to us into linear equations:
New rectangle area – old rectangle area = 100 cm, therefore (l+7)(w+5)-(l)(w) = 100 cm2
7w+5l = 65
Perimeter of original rectangle = 20 cm
2(l+w) = 20 therefore l+w = 10
Hence we are left with, l+w = 10 and 5l+7w = 65
Solving the two equations we have length of original rectangle = 2.5 cm
Q) If the dimensions of a rectangle are in the ration 7:5 and the total perimeter reduction achieved by a shrinking length and width in their original ratio has been 30 cm. What is the mathematical difference between the old width and new width?
A] 4.02 cm
B] 6.25 cm
C] 7.50 cm
D] 8.80 cm
E] 9.40 cm
Solution
Solution – B] 6.25 cm
Change in perimeter = 30 cm
Change in[ 2(l+b) ] = 30 cm
2 x Change in (l+b) = 30 cm
Change in (l+b) = 15 cm
Since the reduction in dimensions follow the original ratio, 7: 5, change in width = 15 cm x 5/(5+7) = 15 x 5/12 = 6.25 cm which is our answer.
We wanted to give you a quick taste of the type of problems to expect in the IPAT geometry section. As always, for more practice do check out our comprehensive IPAT word problem guide here which supports a full length practice format to train yourself before the test. |
Get Started by Finding a Local Center
# December 2015 / January 2016: Tips & Techniques
Dec 1, 2015
Conversion and Units of Measure: Yards, Meters, Miles, and Furlongs!
3,000 yards. 2,800 meters. 1.5 miles. 9 furlongs.
On first glance, one may take the easy way out and assume that they’re all equal in length. Guess what… they’re not! We’ll walk you through a couple of word problems that work with conversion and these units of measure and show you some strategies you can apply to similar types of problems at school!
Upper Elementary:
Arrange 3,000 yards, 2,800 meters, 1.5 miles, and 9 furlongs in order from least to greatest. (Hint: Convert into feet!)
At Mathnasium, we have what we refer to as the Law of SAMEness. The Law of SAMEness states that we can only compare things that have the same name. So in order to compare 3,000 yards and 2,800 meters, for instance, we have to give both amounts the same name, which is why we express both in terms of feet.
But why feet? Simple. Yards, meters, miles, and furlongs convert more or less seamlessly into feet.
Here’s a conversion chart to get you started:
1 yard = 3 feet
1 meter = 3.28 feet
1 mile = 5280 feet
1 furlong = 660 feet
So…
3,000 yards = 3,000 x 3 = 9,000 feet
2,800 meters = 2,800 x 3.28 = 9,184 feet
1.5 miles = 1.5 x 5,280 = 7,920 feet
9 furlongs = 9 x 660 = 5,940 feet
And in order from least to greatest: 5,940 feet, 7,920 feet, 9,000 feet, 9,184 feet ... which is: 9 furlongs, 1.5 miles, 3,000 yards, 2800 meters.
Middle School:
After arranging 3,000 yards, 2,800 meters, 1.5 miles, and 9 furlongs in order from least to greatest, tell how many times bigger each number is than its predecessor.
When we want to know “how many times bigger” a number is compared to a smaller number, we want to find out how many of the smaller number fits in the bigger number.
So, we divide.
7,920 ÷ 5,940 = 1.333… times bigger (1.333… “5,940s” fit in 7,920.)
9,000 ÷ 7,920 = 1.13636… times bigger (1.13636… “7,920s” fit in 9,000.)
9,184 ÷ 9,000 = 1.02 times bigger (1.02 “9,000s” fit into 9,184.)
## OUR METHOD WORKS
Mathnasium meets your child where they are and helps them with the customized program they need, for any level of mathematics. |
2009Homework#1solution-1
# 2009Homework#1solution-1 - Name NetID HADM 2222 Fall 2009...
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Name: ___________________________________ NetID: _______________________ Prof. Q. Ma, HADM 2222 Fall 2009 1/2 HADM 2222 Fall 2009, Prof. Q. Ma Homework assignment #1 Solutions 1. Interest First City Bank pays 6% simple interest on its savings account balances, whereas Second City Bank pays 6% interest compounded annually. If you made a \$5,000 deposit in each bank, how much more money would you earn from your Second City Bank account at the end of 10 years? Solution: The simple interest per year is: \$5,000 × .06 = \$300 So after 10 years you will have: \$300 × 10 = \$3,000 in interest. The total balance will be \$5,000 + 3,000 = \$8,000 With compound interest we use the future value formula: FV = PV(1 + r ) t ; FV = \$5,000(1.06) 10 = \$8,954.24 The difference is: \$8,954.24 – 8,000 = \$954.24 2. Assume the total cost of a college education will be \$280,000 when your child enters college in 18 years. You presently have \$50,000 to invest. What annual rate of interest must you earn on your investment to cover the cost of your child’s college education? Solution: FV = PV(1 + r ) t Solving for r , we get: r = (FV / PV) 1 / t – 1 r = (\$280,000 / \$50,000) 1/18 – 1 = 10.04% 3. You’re trying to save to buy a new \$170,000 Ferrari. You have \$40,000 today that can be invested at
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# 1.1 Real numbers: algebra essentials (Page 8/35)
Page 8 / 35
List the constants and variables for each algebraic expression.
1. $2\pi r\left(r+h\right)$
2. 2( L + W )
3. $4{y}^{3}+y$
Constants Variables
a. $2\pi r\left(r+h\right)$ $2,\pi$ $r,h$
b. 2(L + W) 2 L, W
c. $\text{\hspace{0.17em}}4{y}^{3}+y$ 4 $y$
## Evaluating an algebraic expression at different values
Evaluate the expression $\text{\hspace{0.17em}}2x-7\text{\hspace{0.17em}}$ for each value for x.
1. $\text{\hspace{0.17em}}x=0$
2. $\text{\hspace{0.17em}}x=1$
3. $\text{\hspace{0.17em}}x=\frac{1}{2}$
4. $\text{\hspace{0.17em}}x=-4$
1. Substitute 0 for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}\hfill 2x-7& =& 2\left(0\right)-7\\ & =& 0-7\hfill \\ & =& -7\hfill \end{array}$
2. Substitute 1 for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}2x-7& =& 2\left(1\right)-7\hfill \\ & =& 2-7\hfill \\ & =& -5\hfill \end{array}$
3. Substitute $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}\hfill 2x-7& =& 2\left(\frac{1}{2}\right)-7\hfill \\ & =& 1-7\hfill \\ & =& -6\hfill \end{array}$
4. Substitute $\text{\hspace{0.17em}}-4\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}\hfill 2x-7& =& 2\left(-4\right)-7\\ & =& -8-7\hfill \\ & =& -15\hfill \end{array}$
Evaluate the expression $\text{\hspace{0.17em}}11-3y\text{\hspace{0.17em}}$ for each value for y.
1. $\text{\hspace{0.17em}}y=2$
2. $\text{\hspace{0.17em}}y=0$
3. $\text{\hspace{0.17em}}y=\frac{2}{3}$
4. $\text{\hspace{0.17em}}y=-5$
1. 5;
2. 11;
3. 9;
4. 26
## Evaluating algebraic expressions
Evaluate each expression for the given values.
1. $\text{\hspace{0.17em}}x+5\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x=-5$
2. $\text{\hspace{0.17em}}\frac{t}{2t-1}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}t=10$
3. $\text{\hspace{0.17em}}\frac{4}{3}\pi {r}^{3}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}r=5$
4. $\text{\hspace{0.17em}}a+ab+b\text{\hspace{0.17em}}$ for $a=11,b=-8$
5. $\text{\hspace{0.17em}}\sqrt{2{m}^{3}{n}^{2}}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}m=2,n=3$
1. Substitute $\text{\hspace{0.17em}}-5\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}\hfill x+5& =& \left(-5\right)+5\hfill \\ & =& 0\hfill \end{array}$
2. Substitute 10 for $\text{\hspace{0.17em}}t.$
$\begin{array}{ccc}\hfill \frac{t}{2t-1}& =& \frac{\left(10\right)}{2\left(10\right)-1}\hfill \\ & =& \frac{10}{20-1}\hfill \\ & =& \frac{10}{19}\hfill \end{array}$
3. Substitute 5 for $\text{\hspace{0.17em}}r.$
$\begin{array}{ccc}\hfill \frac{4}{3}\pi {r}^{3}& =& \frac{4}{3}\pi {\left(5\right)}^{3}\\ & =& \frac{4}{3}\pi \left(125\right)\hfill \\ & =& \frac{500}{3}\pi \hfill \end{array}$
4. Substitute 11 for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and –8 for $\text{\hspace{0.17em}}b.$
$\begin{array}{ccc}\hfill a+ab+b& =& \left(11\right)+\left(11\right)\left(-8\right)+\left(-8\right)\\ & =& 11-88-8\hfill \\ & =& -85\hfill \end{array}$
5. Substitute 2 for $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ and 3 for $\text{\hspace{0.17em}}n.$
$\begin{array}{ccc}\hfill \sqrt{2{m}^{3}{n}^{2}}& =& \sqrt{2{\left(2\right)}^{3}{\left(3\right)}^{2}}\hfill \\ & =& \sqrt{2\left(8\right)\left(9\right)}\hfill \\ & =& \sqrt{144}\hfill \\ & =& 12\hfill \end{array}$
Evaluate each expression for the given values.
1. $\text{\hspace{0.17em}}\frac{y+3}{y-3}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}y=5$
2. $\text{\hspace{0.17em}}7-2t\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}t=-2$
3. $\text{\hspace{0.17em}}\frac{1}{3}\pi {r}^{2}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}r=11$
4. $\text{\hspace{0.17em}}{\left({p}^{2}q\right)}^{3}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}p=-2,q=3$
5. $\text{\hspace{0.17em}}4\left(m-n\right)-5\left(n-m\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}m=\frac{2}{3},n=\frac{1}{3}$
1. 4;
2. 11;
3. $\text{\hspace{0.17em}}\frac{121}{3}\pi$ ;
4. 1728;
5. 3
## Formulas
An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation $\text{\hspace{0.17em}}2x+1=7\text{\hspace{0.17em}}$ has the unique solution of 3 because when we substitute 3 for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the equation, we obtain the true statement $\text{\hspace{0.17em}}2\left(3\right)+1=7.$
A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ of a circle in terms of the radius $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ of the circle: $\text{\hspace{0.17em}}A=\pi {r}^{2}.\text{\hspace{0.17em}}$ For any value of $\text{\hspace{0.17em}}r,$ the area $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ can be found by evaluating the expression $\text{\hspace{0.17em}}\pi {r}^{2}.$
## Using a formula
A right circular cylinder with radius $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ and height $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ has the surface area $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ (in square units) given by the formula $\text{\hspace{0.17em}}S=2\pi r\left(r+h\right).\text{\hspace{0.17em}}$ See [link] . Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of $\text{\hspace{0.17em}}\pi .$
Evaluate the expression $\text{\hspace{0.17em}}2\pi r\left(r+h\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}r=6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h=9.$
$\begin{array}{ccc}\hfill S& =& 2\pi r\left(r+h\right)\hfill \\ & =& 2\pi \left(6\right)\left[\left(6\right)+\left(9\right)\right]\hfill \\ & =& 2\pi \left(6\right)\left(15\right)\hfill \\ & =& 180\pi \hfill \end{array}$
The surface area is $\text{\hspace{0.17em}}180\pi \text{\hspace{0.17em}}$ square inches.
A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm 2 ) is found to be $\text{\hspace{0.17em}}A=\left(L+16\right)\left(W+16\right)-L\cdot W.\text{\hspace{0.17em}}$ See [link] . Find the area of a matte for a photograph with length 32 cm and width 24 cm.
1,152 cm 2
## Simplifying algebraic expressions
Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions.
## Simplifying algebraic expressions
Simplify each algebraic expression.
1. $3x-2y+x-3y-7$
2. $2r-5\left(3-r\right)+4$
3. $\left(4t-\frac{5}{4}s\right)-\left(\frac{2}{3}t+2s\right)$
4. $2mn-5m+3mn+n$
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3 |
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Found in: Page 7
### Pre-algebra
Book edition Common Core Edition
Author(s) Ron Larson, Laurie Boswell, Timothy D. Kanold, Lee Stiff
Pages 183 pages
ISBN 9780547587776
# Evaluate the experssion when $a=2.5, b=15,\text{and}c=3.5$localid="1648024266631" $bc$
The value of the expression is $52.5$.
See the step by step solution
## Step-1 – Apply the concept of product
When two or more numbers are multiplied together, it gives a number. The numbers that are being multiplied are called the factors. And the result of the multiplication of numbers is called the product of those numbers.
## Step-2 – Example for the product
In the product $2·3·4=24$, the numbers 2, 3, and 4 are being multiplied to give the product 24. So, the numbers 2, 3, and 4 are the factors in this equation and 24 is the product of the numbers.
## Step-3 – Evaluate the value of expression
To evaluate the expression $bc$, substitute the values of b as 15 and value of c as 3.5 and multiply the numbers.
To find the product of the decimal numbers 15 and 3.5, first multiply the numbers normally without decimal point. Observe that the first number has no decimal while the second number has one digit after decimal, so place the decimal point in the final product at the left of last digit from right end.
The product of the numbers can be computed as shown below,
$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}15\\ \underset{¯}{×\text{\hspace{0.17em}}3.5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}75\\ \underset{¯}{\text{\hspace{0.17em}}45×\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\\ 52.5\end{array}$
Thus, the value of the expression is $52.5$. |
# Printable Multiplication Charts Up To 20
Learning multiplication right after counting, addition, and subtraction is good. Children discover arithmetic through a all-natural progression. This progress of understanding arithmetic is often the subsequent: counting, addition, subtraction, multiplication, and finally section. This declaration brings about the question why understand arithmetic within this series? Furthermore, why learn multiplication following counting, addition, and subtraction before division?
## The next details respond to these questions:
1. Children discover counting initially by associating visual objects because of their hands. A real instance: How many apples exist in the basket? A lot more abstract case in point is just how old are you currently?
2. From counting phone numbers, another rational step is addition accompanied by subtraction. Addition and subtraction tables can be quite useful training aids for kids since they are visual instruments making the changeover from counting much easier.
3. Which ought to be discovered up coming, multiplication or section? Multiplication is shorthand for addition. At this point, young children have got a organization understanding of addition. As a result, multiplication may be the following plausible kind of arithmetic to find out.
## Assess basic principles of multiplication. Also, look at the essentials utilizing a multiplication table.
Allow us to review a multiplication example. Utilizing a Multiplication Table, flourish 4 times 3 and acquire a response 12: 4 by 3 = 12. The intersection of row 3 and column several of your Multiplication Table is 12; 12 may be the response. For kids beginning to understand multiplication, this is certainly simple. They may use addition to eliminate the problem therefore affirming that multiplication is shorthand for addition. Instance: 4 by 3 = 4 4 4 = 12. It is an excellent overview of the Multiplication Table. An added gain, the Multiplication Table is aesthetic and mirrors back to understanding addition.
## Where can we start learning multiplication while using Multiplication Table?
1. First, get informed about the table.
2. Get started with multiplying by a single. Start off at row number 1. Go on to column primary. The intersection of row one and line the initial one is the perfect solution: one.
3. Perform repeatedly these methods for multiplying by 1. Multiply row one by columns one particular by way of a dozen. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Recurring these steps for multiplying by two. Grow row two by columns one particular via five. The replies are 2, 4, 6, 8, and 10 correspondingly.
5. We will leap ahead. Repeat these techniques for multiplying by 5 various. Flourish row several by columns 1 through a dozen. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now let us boost the level of problems. Repeat these methods for multiplying by about three. Grow row a few by columns one particular by means of a dozen. The answers are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are comfortable with multiplication so far, consider using a check. Remedy these multiplication problems in your thoughts after which assess your responses on the Multiplication Table: increase six and two, multiply 9 and a few, flourish one and eleven, increase four and several, and multiply 7 and two. The trouble replies are 12, 27, 11, 16, and 14 correspondingly.
If you got 4 out from 5 difficulties proper, design your personal multiplication checks. Compute the responses in your thoughts, and appearance them making use of the Multiplication Table. |
# Question Video: Finding the Volume of a Cone given Its Height and Base Radius Mathematics • 7th Grade
Determine the volume of the given solid to the nearest hundredth.
02:31
### Video Transcript
Determine the volume of the given solid to the nearest hundredth.
Here, we can see the solid is a combination of two cones stuck together. We have this blue cone and we also have this pink cone. So in order to find the volume of this solid, we need to add together the volume of the blue cone and the volume of the pink cone. And since they are both cones, we need the volume of a cone formula. And it is one-third times 𝜋 times the radius squared times the height of the cone.
This means for each cone, we need to find the radius and the height. A radius will go from the center of the circle to a point on the circle. And here they share the same radius and it is 21 millimetres. Another last measurement that we need to find is the height of each cone. And these ones are actually different. The height of the blue cone can be found on the left. It is perpendicular to the circular base and it is 23 millimetres. And now the height of the pink cone is 26 millimetres.
And now, we can begin multiplying. However, we need to round to the nearest hundredth. So we want to be as exact as possible. So notice that we have 𝜋 twice in our equation. So instead of multiplying by 𝜋 and having to round two times, that’s gonna cause a bigger rounding error. So what we will do is leave 𝜋 until the very last step and then multiply by 𝜋 at the very end. So we’re only rounding once.
So we will multiply these numbers, but leave 𝜋 out — just write it next to the number — and the same here. So we have 3381𝜋 plus 3822𝜋, which is equal to 7203𝜋. And now, we multiply by 𝜋, which is approximately 22628.8918.
However, we need to round to the nearest hundredth. So we either need to keep the nine a nine or round up. So we look at the one and one is less than five. So we will keep the nine a nine, making our volume 22628.89 millimetres cubed. |
GCSE Maths Quiz
# Straight Line (F)
In the Linear Equations quizzes you practised solving equations using algebra. This GCSE Maths quiz should help you understand the link between the algebra and the geometry of the straight line.
When a straight line is plotted on an x-y grid there are two key features that we can obtain, which will allow us to write down the equation of the straight line. This equation is the rule that connects the x- and the y-values.
The first feature is the gradient, which is a measure of how steep the line is. Recall that the gradient is ‘change in height over change in distance’. To calculate the gradient, plot two points on the line that pass directly through the crosshairs of the grid.
Form a right-angled triangle, then count how many units up or down – this is change in height – and how many units left or right, which is change in distance. Then you need to decide if it is positive (goes up from left to right) or negative (goes down from left to right). Gradient is denoted using the letter m.
The other feature is the y-axis intercept. This is the number on the y-axis (vertical) that the line passes through, and is represented using the letter c.
You obtain the equation of the line by substituting the values for m and c into the general equation of the straight line, y = mx + c.
The equation of horizontal and vertical lines often catches people out. On a horizontal line, the y-value of any point is the same. This means the equation is of the form y = …. The same logic applies to a vertical line – the x-value of any point on the line will be the same, so its equation is x = ….
Did you know...
You can play all the teacher-written quizzes on our site for just £9.95 per month. Click the button to sign up or read more.
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# Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(b)
Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(b) will help students to clear their doubts quickly.
## Intermediate 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(b)
I.
Question 1.
Find the angle between the lines represented by 2x² + xy – 6y² + 7y – 2 = 0.
Solution:
Comparing with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
a = 2, 2g = 0, g = 0
b = – 6, 2f = 7, f = 7/2
c = – 2, 2h = 1, h = 1/2
Angle between the lines is given by
Question 2.
Prove that the equation 2x² + 3xy – 2y² + 3x + y + 1 = 0 represents a pair of perpendicular lines.
Solution:
Given a = 2, b = -2
a + b = 0 ⇒ cos α = 0 ⇒ α = π/2
∴ The given lines are perpendicular.
II.
Question 1.
Prove that the equation 3x² + 7xy + 2y² + 5x + 5y + 2 = 0 represents a pair of straight lines and find the co-ordinates of the’point of intersection.
Solution:
The given equation is
3x² + 7xy + 2y² + 5x + 5y + 2 = 0
Comparing a = 3 2f = 5 ⇒ f = $$\frac{5}{2}$$
b = 2 2g = 5 ⇒ g = $$\frac{5}{2}$$
c = 2 2h = 7 ⇒ h = $$\frac{7}{2}$$
∆ = abc + 2fgh – af² – bg² – ch²
= 3(2)(2) + 2.$$\frac{5}{2}$$.$$\frac{5}{2}$$.$$\frac{7}{2}$$ – 3.$$\frac{25}{4}$$ – 2.$$\frac{25}{4}$$ – 2.$$\frac{49}{4}$$ = 0
= $$\frac{1}{4}$$(48 + 175 – 75 – 50 – 98)
= $$\frac{1}{2}$$(223 – 223) = 0
∴ The given equation represents a pair of lines point of intersection is
Question 2.
Find the value of k, if the equation 2x² + kxy – 6y² + 3x + y + l = 0 represents a pair of straight lines. Find the point of intersection of the lines and the angle between the straight lines for this value of k.
Solution:
The given equation is
2x² + kxy – by² + 3x + y+ 1 = 0
a = 2 2f = 1 ⇒ f = $$\frac{1}{2}$$
b = -6 2g = 3 ⇒ g = $$\frac{3}{2}$$
c = 1 2h = k ⇒ h = $$\frac{k}{2}$$
The given equation represents a pair of straight lines abc + 2fgh – af² – bg² – ch² = 0
-12 + 2.$$\frac{1}{2}$$.$$\frac{3}{2}$$(+$$\frac{k}{2}$$ -2.$$\frac{1}{4}$$ + 6.$$\frac{9}{4}$$ – $$\frac{k^{2}}{4}$$ = 0
– 48 + 3k – 2 + 54 – k² = 0
-k² + 3k + 4 = 0 ⇒ k² – 3k – 4 = 0
(k – 4) (k + 1) = 0
k = 4 or – 1.
Case (i): k = – 1
Point of’intersection is $$\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)$$
Point of intersection is ($$\frac{-5}{7}$$, $$\frac{1}{7}$$)
Angle between the lines
Case (ii): k = 4
Point of intersection is P(-$$\frac{5}{8}$$, –$$\frac{1}{8}$$)
Question 3.
Show that the equation x² – y² – x + 3y – 2 = 0 represents a pair of perpendicular lines and find their equations.
Solution:
Comparing a= 1, f = $$\frac{3}{2}$$
b = -1, g = –$$\frac{1}{2}$$
c = -2, h = 0
abc + 2fgh – af² – bg² – ch²
= 1 (-1) (-2) + 0- 1.$$\frac{9}{4}$$ + 1.$$\frac{1}{4}$$ + 0
= + 2 – $$\frac{9}{4}$$ + $$\frac{1}{4}$$ = 0
h² – ab = 0 – 1(-1) = 1 > 0,
f² – be = $$\frac{9}{4}$$ – 2 = $$\frac{1}{4}$$ = 1 > 0
g² – ac = $$\frac{1}{4}$$ + 2 = $$\frac{9}{4}$$ > 0
a + b = 1 – 1 = 0
The given equation represent a pair of per-pendicular lines
Let x² – y² – x + 3y – 2
= (x + y + c1) (x – y + c2)
Equating the co-efficients of x
⇒ c1 + c2 = – 1
Equating the co-efficients of y
⇒ – c1 + c2 = 3
Adding 2c2 = 2 ⇒ c2 = 1
c1 + c2 = – 1 ⇒ c1 + 1 = – 1
c1 = – 2
Equations of the lines are x + y – 2 = 0 and x – y + 1 = 0
Question 4.
Show that the lines x² + 2xy – 35y² – 4x + 44y – 12 = 0 are 5x + 2y – 8 = 0 are concurrent.
Solution:
Equations of the given lines are
x² + 2xy- 35y² -4x + 44y- 12 = 0
a = 1, f = 22
b = – 35, g = – 2
c = – 12, h = 1
Point or intersection is $$\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)$$
Point of intersection of the given lines is P($$\frac{4}{3}$$, $$\frac{2}{3}$$)
5x + 2y – 8 = 5.$$\frac{4}{3}$$ + 2.$$\frac{2}{3}$$ – 8
= $$\frac{20+4-24}{3}$$ = 0
P lies on the third line 5x + 2y – 8 = 0
∴ The given lines are concurrent.
Question 5.
Find the distances between the following pairs of parallel straight lines :
i) 9x² – 6xy + y² + 18x – 6y + 8 = 0
Solution:
Distance between parallel lines = $$2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}$$
ii) x² + 2√3xy + 3y² – 3x – 3√3y – 4 = 0
Solution:
Distance between parallel lines = $$2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}$$
Question 5.
Show that the two pairs of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4y – 1 = 0 form a square.
Solution:
Combined equation of CA and CB is
3x² + 8xy – 3y²
(x + 3y) (3x – y) = 0
3x – y = 0, x + 3y = 0
Equation of OA is 3x – y = 0 ………. (1)
Equation of OB is x + 3y = 0 ……….(2)
Combined equation of CA and CB is
3x² + 8xy – 3y² + 2x – 4y + 1 =0
Let 3x² + 8xy – 3y² + 2x – 4y + 1 = (3x – y + c1) (x + 3y + c2)
Equating the co-efficients of
x, we get c1 + 3c2 = 2
y, we have 3c1 + c2 = -4
Equation of BC is 3x – y – 1 = 0 ………. (3)
Equation of AC is x + 3y + 1 = 0 ………. (4)
OA and BC differ by a constant ⇒ OA parallel to BC
OB and CA differ by a constant ⇒ OB parallel to AC
From combined equation of OA and OB
OACB is a rectangle a + b = 3 – 3 = 0
OA = Length of the -L lar from O to AC = $$\frac{|0+0+1|}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}$$
OB = Length of the perpendicular from O to BC = $$\frac{|0+0-1|}{\sqrt{9+1}}=\frac{1}{\sqrt{10}}$$
OA = OB and OACB is a rectangle
OACB is a square.
III.
Question 1.
Find the product of the lengths of the perpendiculars drawn from (2, 1) upon the lines
12x² + 25xy + 12y² + 10x + 11y + 2 = 0
Solution:
Combined equation of AB, AC is
12x² + 25xy+ 12y² + 10x + 11y + 2 = 0
12x² + 25xy +12y²
= 12x² + 16xy + 9xy + 12y² = 0
= 4x (3x + 4y) + 3y (3x + 4y)
= (3x + 4y) (4x + 3y)
Let 12x² + 25xy + 12y² + 10x + 11y + 2
= (3x + 4y + c1) (4x + 3y + c2)
Equating the co-efficients of x, we get
4c1 + 3c2 = 10 ………. (1)
Equating the co-efficients of y, we get
3c1 + 4c2 = 11 ………… (2)
i.e., 4c1 + 3c2 – 10 = 0
3c1 + 4c2 – 11 = 0
Equation of AB is 3x + 4y + 1 = 0
Equation of AC is 4x + 3y + 2 = 0
PQ = Length of the perpendicular from P(2, 1) on
AB = $$\frac{6+4+1}{\sqrt{9+16}}=\frac{11}{5}$$
PQ = Length of the perpendiculars from P(2, 1) on
AC = $$\frac{|8+3+2|}{\sqrt{16+9}}=\frac{13}{5}$$
Product of the length of the perpendiculars
= PQ × PR= $$\frac{11}{5}$$ × $$\frac{13}{5}$$ = $$\frac{143}{25}$$
Question 2.
Show that the straight lines y² – 4y + 3 = 0 and
x² + 4xy + 4y² + 5x + 10y + 4 = 0 form a parallelogram and find the lengths of its sides.
Solution:
Equation of the first pair of lines is
y² – 4y + 3 = 0
(y – 1) (y – 3) = 0
y – 1 = 0 or y – 3 = 0
Equation of AB is y – 1 = 0 ……….. (1)
Equation of CD is y – 3 = 0 ………. (2)
Equations of AB and CD differ by a constant.
∴ AB and CD are parallel.
Equation of the second pair of lines is
x² + 4xy + 4y² + 5x + 10y + 4 = 0
(x + 2y)² + 5(x + 2y) + 4 = 0
(x + 2y)² + 4 (x + 2y) + (x + 2y) + 4 = 0
(x + 2y)(x + 2y + 4) + 1(x + 2y + 4) = 0
(x + 2y + 1) (x + 2y + 4) = 0
x + 2y + 1 = 0, x + 2y + 4 = 0
Equation of AD is x + 2y + 1 = 0 ……… (3)
Equation of BC is x + 2y + 4 = 0 ……… (4)
Solving (1), (3) x + 2 + 1 = 0
x = – 3
Co-ordinates of A are (-3, 1)
Solving (2), (3) x + 6 + 1 = 0
x = -7
Co-ordinates of DC are (-7, 3)
Solving (1), (4) x + 2 + 4 = 0
x = – 6
Co-ordinates of B are (-6, 1)
Lengths of the sides of the parallelogram are 3, 2√5.
Question 3.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 is $$\frac{|c|}{\sqrt{(a-b)^{2}+4 h^{2}}}$$
Solution:
Let ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents the lines
l1x + m1y + n1 = 0 ……….. (1)
l2x + m2y + n2 = 0 ………. (2)
⇒ ax² + 2hxy + by² + 2gx + 2fy + c ≡ (l1x + m1y + n1) (l2x + m2y + n2)
l1l2 = a, m1m2 = b, l1m2 + l2m1 = 2h,
l1n2 + l2n1 = 2g, m1n2 + m2n1 = 2f, n1n2 = c
Ir distance from origin to (1) is = $$\frac{\left|n_{1}\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}$$
Irdistance from origin to (2) is = $$\frac{\left|n_{2}\right|}{\sqrt{I_{2}^{2}+m_{2}^{2}}}$$
Product of perpendiculars
Question 4.
If the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of intersecting lines, then show that the square of the distance of their point of intersection from the origin is $$\frac{c(a+b)-f^{2}-g^{2}}{a b-h^{2}}$$. Also slow that the square of this distance is $$\frac{f^{2}+g^{2}}{h^{2}+b^{2}}$$ if the given lines are perpendicular.
Solution:
Let the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0
represent the lines
l1x + m1y + n1 = 0 …….. (1)
l2x + m2y + n2 = 0 …….. (2)
(l1x + m1y + n1)(l2x + m2y + n2) = ax² + 2hxy + by² + 2gx + 2fy + c
l1l2 = a, m1 m2 = b, n1n2 = c
l1m2 + l2m1 = 2h, l1n2 + l12n1 = 2g, m1n2 + m2n1 = 2f
Solving (1) and (2)
$$\frac{x}{m_{1} n_{2}-m_{2} n_{1}}=\frac{y}{I_{2} n_{1}-l_{1} n_{2}}=\frac{1}{l_{1} m_{2}-I_{2} m_{1}}$$
The point of intersection, P
If the given pair of lines are perpendicular, then a + b = 0
∴ a = -b
$$\mathrm{OP}^{2}=\frac{0-r^{2}-g^{2}}{(-b) b-h^{2}}=\frac{r^{2}+g^{2}}{h^{2}+b^{2}}$$ |
# Lesson video
In progress...
Hello! My name is Ms. Parnham and in this lesson we're going to learn how to subtract mixed numbers.
Subtracting mixed numbers follows the same rules as subtracting fractions.
Let's take a look at an example.
The first step is to rewrite these mixed numbers as improper fractions.
So 15/4 subtracts 7/6.
Now we need to look at the denominators and determine the lowest common multiple.
This is 12.
So we rewrite these fractions as 12ths.
45/12, subtract 14/12, is 31/12.
But this is an improper fraction, so we rewrite this as a mixed number of 2 and 7/12.
In this next example, we start by converting the mixed numbers to improper fractions and then we find the lowest common multiple of 15 and 6, which is 30.
So these can be rewritten as equivalent fractions of 172/30, subtract 115/30.
This gives us an answer of 57/30.
But this is an improper fraction, so we rewrite this as a mixed number of 1 and 27/30.
But that's not quite the final answer, because 27/30 can be simplified further.
So the final answer is 1 and 9/10.
Here are some questions for you to try.
Pause the video to complete the task and restart the video when you're finished.
Did you notice that the last two parts of this question could be simplified? So in part E, if you have 1 and 3/12 and for part F, if you have 3 and 6/15, then you're mathematically correct, but you need to cancel common factors in the fraction parts of these numbers in order to get your answer as simple as possible.
Here is another question for you to try.
Pause the video to complete the task, and restart the video when you're finished.
Notice that Jack made a little mistake.
He either assumed that subtraction is commutative, or he missed the sign of the fraction and it meant that his answer is incorrect.
Now let's look at an example involving negative mixed numbers.
Just like before, we need to convert these mixed numbers into improper fractions.
So we can rewrite these as as 9/4 subtract -7/6.
The lowest common multiple of 4 and 6 is 12, so we rewrite this as 27/12 subtract -14/12.
This gives us an answer of 41/12.
This is an improper fraction, so we convert it to a mixed number, 3 and 5/12.
In the next example, just like before, we convert these mixed numbers to improper fractions.
So we have -22/15 subtract 29/6.
The lowest common multiple of 15 and 6 is 30.
So rewriting these as equivalent fractions of 30 gives us -44/30 subtract 145/30.
This is -189/30.
This simplifies to -6 and 9/30, but this can be simplified further because 9/30 is equivalent to 3/10, so this is -6 and 3/10.
Here are some questions for you to try.
Pause the video to complete the task and restart the video when you're finished. |
# 정은's question at Yahoo! Answers regarding depth of water in trough when half full
#### MarkFL
Staff member
Here is the question:
Integration using trapezium rule?
The diagram shows the cross section of a water-trough. One side is vertical and the other side slopes. Calculate the depth of water in the trough when it is exactly half-full. (Hint: this means the areas of two trapezia must be equal.)
Above link is shows the diagram of this question.
The answer to this question= 27.19cm
I have posted a link there to this topic so the OP can see my work.
#### MarkFL
Staff member
Hello 정은,
This problem does not require integration, or the trapezium rule to approximate a definite integral. While it could be used, it is simpler to just use the given hint.
To compute the areas of the trapezia in the cross-section of the trough above and below the water line, we need to know the width $w$ of the trough at the water line as a function of the depth of the water, which we are calling $d$. All linear measures are in cm.
We know this width increases linearly, and we know two points of the form $(d,w)$:
$(0,70)$ and $(50,100)$
and so the slope of the linear function is:
$$\displaystyle m=\frac{\Delta w}{\Delta h}=\frac{100-70}{50-0}=\frac{3}{5}$$
Thus, using the slope, and the first point in the point-slope formula, we get:
$$\displaystyle w-70=\frac{3}{5}(d-0)$$
$$\displaystyle w=\frac{3}{5}d+70$$
Now, using the formula for the area of a trapezium:
$$\displaystyle A=\frac{h}{2}(B+b)$$
We find the area $A_1$ of the trapezium below the water line is:
$$\displaystyle A_1=\frac{d}{2}\left(\frac{3}{5}d+70+70 \right)=\frac{d}{2}\left(\frac{3}{5}d+140 \right)$$
We find the area $A_2$ of the trapezium above the water line is:
$$\displaystyle A_2=\frac{50-d}{2}\left(100+\frac{3}{5}d+70 \right)=\frac{50-d}{2}\left(\frac{3}{5}d+170 \right)$$
Equating the two areas, we have:
$$\displaystyle \frac{d}{2}\left(\frac{3}{5}d+140 \right)=\frac{50-d}{2}\left(\frac{3}{5}d+170 \right)$$
Multiplying through by $10$ we get:
$$\displaystyle d(3d+700)=(50-d)(3d+850)$$
Distributing and arranging in standard quadratic form, we obtain:
$$\displaystyle 3d^2+700d-21250=0$$
Application of the quadratic formula, and discarding the negative root, there results:
$$\displaystyle d=\frac{25}{3}\left(\sqrt{298}-14 \right)\approx27.1889708469339$$
#### MarkFL
Staff member
Suppose we wish to generalize a bit, and let the width of the trapezoidal cross-section of a trough at the base and top be $w_1$ and $w_2$ respectively, where $w_1<w_2$. The depth of the trough we can call $h$, and we will, as before, let $d$ be the depth when the trough is $$\displaystyle k$$ full, where $0\le k\le1$.
To find the width $w$ of the trough at $d$, we note we have the points:
$$\displaystyle \left(0,w_1 \right),\,\left(h,w_2 \right)$$
and so the slope of the linear width function is:
$$\displaystyle m=\frac{w_2-w_1}{h}$$
and the point-slope formula gives us:
$$\displaystyle w=\frac{w_2-w_1}{h}d+w_1$$
Now, in order for the trough to be $$\displaystyle k$$ full, we require:
$$\displaystyle \frac{d}{2}\left(\frac{w_2-w_1}{h}d+2w_1 \right)=\frac{kh}{2}\left(w_1+w_2 \right)$$
Multiplying through by $2h$, and arranging in standard quadratic form in $d$, we obtain:
$$\displaystyle \left(w_2-w_1 \right)d^2+2hw_1d-kh^2\left(w_1+w_2 \right)=0$$
$$\displaystyle d=\frac{h\left(\sqrt{w_1^2+k\left(w_2^2-w_1^2 \right)}-w_1 \right)}{w_2-w_1}$$ |
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Math Central Quandaries & Queries
My question: Why is the value of a trigonometric function, the same, for an angle over 90 degrees and its reference angle? How are the angle and its reference related? Do they both form a triangle that has equal sides?
Hi Tim,
The value of a trigonometric function of an angle over 90 degrees (or less than zero degrees) is not always the same as the value of its reference angle, but they can be easily evaluated. Since any trigonometric function can be written in terms of the sine and cosine functions I am going to deal wit these functions only.
Suppose you want the sine and cosine of an angle t which is between 0 and 90 degrees. Draw a circle with centre at the origin and radius 1. Draw a line segment starting at the origin and at an angle of t degrees, measured counterclockwise from the x-axis. Let this line segment intersect the circle at the point B as in the diagram. Give B the coordinates (x, y).
From the diagram
cos(t) = |CA|/|BC| = |CA| = x and
sin(t) = |AB|/|BC| = |AB| = y.
This illustrates the way that the sine and cosine function are defined for all angles t.
For any number t draw a lie segment starting at the origin and at an angle of t degrees, measured counterclockwise from the x-axis. Let B with coordinates (x, y) be the point where this line segment intersects the circle with radius one. Define
cos(t) = x and
sin(t) = y.
(If t is negative then measure t degrees clockwise from the x-axis.)
Here is an example where t is between 90 and 180 degrees. Again draw the line segment CB at an angle of t degrees measured counterclockwise for the x-axis and give B the coordinates (x, y). Then cos(t) = x and sin(t) = y.
Draw in the reference angle r and the point D where the line segment at angle r degrees counterclockwise from the x-axis intersects the circle of radius 1. From the diagram, since B has coordinates (x, y), D has coordinates (-x,y). Hence
cos(t) = x = - cos(r) and
sin(t) = y = sin(r).
I hope this helps,
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |
Class 8 Maths MCQ – Applications of Compound Interest Formula – 2
This set of Class 8 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Compound Interest Formula – 2”.
1. A company increased the production of cars from 15625 in 2002 to 27000 in 2005. Find the annual rate of growth of production of the cars.
a) 15%
b) 2%
c) 10%
d) 20%
Explanation: Let the annual rate of growth be R% per annum. Then,
⇒ 27000 = 15625 (1 + $$\frac{R}{100})^3$$
⇒ $$\frac{27000}{15625}$$ = (1 + $$\frac{R}{100})^3$$
⇒ ($$\frac{30}{25})^3$$ = (1 + $$\frac{R}{100})^3$$
⇒ $$\frac{30}{25}$$ = 1 + $$\frac{R}{100}$$
⇒ $$\frac{1}{5}$$ = $$\frac{R}{100}$$
⇒ R = 20%.
2. Fe-57 decays at the constant rate in such a way that it reduces to 50% in 5568 years. Find the age of an old rod which the iron (Fe) is only 12.5% of the original.
a) 16700
b) 16705
c) 16704
d) 16000
Explanation: Let the rate of decay be R% per annum and the age of rod be n years. Let the original amount of iron in the rod be P. Then in 5568 years the amount left is $$\frac{P}{2}$$
⇒ $$\frac{P}{2}$$ = P (1 – $$\frac{R}{100})^{5568}$$
⇒ $$\frac{1}{2}$$ = (1 – $$\frac{R}{100})^{5568}$$
After n years, the iron left in the rod is 12.5% of P
⇒ $$\frac{P}{8}$$ = P (1 – $$\frac{R}{100})^n$$
⇒ $$\frac{1}{8}$$ = (1 – $$\frac{R}{100})^n$$
⇒ ($$\frac{1}{2})^3$$ = (1 – $$\frac{R}{100})^n$$
⇒ (1 – $$\frac{R}{100})^{16704}$$ = (1 – $$\frac{R}{100})^n$$
⇒ n = 5568 × 3 = 16704 years.
Hence the age of the rod is 16704 years.
3. The present population of a village is 10000. If it increases at the rate of 2% per annum, find the population after 2 years.
a) 10404
b) 10004
c) 10402
d) 10400
Explanation: Initial population = 10000
⇒ Let after 2 years be P.
⇒ P = 10000 (1 + $$\frac{2}{100})^2$$ = 10000 ($$\frac{51}{50})^2$$ = 10404.
4. The initial population of a country is 1000000. If the birth rate and the death rate is 10% and 4% respectively, then find the population of the country after 2 years.
a) 1126000
b) 1023600
c) 1123000
d) 1123600
Explanation: Initial population = 1000000
Increment in population = 10%
Decrement in population = 4%
Net growth in population = 10-4 = 6%
Population after 2 years = 1000000 (1 + $$\frac{6}{100})^2$$ = 1000000 ($$\frac{53}{50})^2$$ = 1123600.
5. Two years ago the population of a village was 4000. If the annual increase during the two successive years be at the rate of 2% & 4%, find the present population.
a) 4230
b) 4320
c) 4240
d) 4420
Explanation: Present population = 4000 (1 + $$\frac{2}{100}$$)(1 + $$\frac{4}{100}$$) = 4000 ($$\frac{51}{50}$$)($$\frac{26}{25}$$) = 4243.20 ≈ 4240.
6. The annual rate of growth in population of a certain city is 10%. If its present population is 532400, find the population 3 years ago.
a) 404000
b) 300000
c) 400400
d) 400000
Explanation: Population 3 years ago = P (1 + $$\frac{R}{100})^3$$
⇒ 532400 = P (1 + $$\frac{10}{100})^3$$.
⇒ P = $$\frac{532400\times10\times10\times10}{11\times11\times11}$$
⇒ P = 400000.
7. Vidhya started a business with initial investment of 500000. She incurred a loss of 2% in the first year and in the second year she gained the profit of 10%. Calculate her net profit earned.
a) 39000
b) 38000
c) 41000
d) 40000
Explanation: Initial investment = 500000
Profit = 500000 (1 – $$\frac{2}{100}$$)(1 + $$\frac{10}{100}$$) = 500000 ($$\frac{49}{50}$$)($$\frac{11}{10}$$) = 539000
⇒ Net profit earned = 500000 – 539000 = 39000.
8. An apartment of two floors is constructed at the cost of Rs. 5000000. It is depreciating at the rate of 20% per annum. Find its value 3 years after construction.
a) 2400000
b) 2460000
c) 2500000
d) 2560000
Explanation: V0 = Rs. 5000000
Value after 3 years = V0 (1 – $$\frac{R}{100})^3$$ = 5000000 (1 – $$\frac{20}{100})^3$$ = 5000000 ($$\frac{4}{5})^3$$ = Rs. 2560000.
9. A new two-wheeler costs Rs. 90000. Its price depreciates at the rate of 10% a year during first two years and the rate of 25% a year thereafter. Find the cost of the two-wheeler after 3 years.
a) 55000
b) 54675
c) 50460
d) 54000
Explanation: Cost of the vehicle = Rs 90000
Rate of depreciation in the first two years = 10%
Rate of depreciation in the third year = 25%
⇒ Price of the vehicle after 3 years = 90000 (1 – $$\frac{10}{100})^2$$ (1 – $$\frac{25}{100}$$) = 90000 ($$\frac{9}{10})^2$$($$\frac{3}{4}$$) = Rs. 54675.
10. The present price of a cycle is Rs 8000. If its value is decreasing every year by 10% then find the price before 3 years.
a) 10977
b) 10976
c) 10975
d) 10974
Explanation: Let the price of the cycle be Rs. P before 3 years. Then its present value is Rs P (1 – $$\frac{10}{100})^3$$
But the present price is Rs. 8000
⇒ 8000 = P (1 – $$\frac{10}{100})^3$$
⇒ 8000 = P ($$\frac{9}{10})^3$$
⇒ P = 10974.
11. The value of a gadget worth Rs.10000 is depreciating at the rate of 10% per annum. In how many years will its value be reduced to Rs 6561?
a) 5
b) 3
c) 4
d) 2
Explanation: Present value = Rs. 10000
Depreciated value = Rs. 6561
Rate of depreciation = 10%
⇒ 6561 = 10000 (1 – $$\frac{10}{100})^n$$
⇒ $$\frac{6561}{10000} = (\frac{9}{10})^n$$
⇒ ($$\frac{9}{10})^4$$ = ($$\frac{9}{10})^n$$
⇒ n = 4.
12. The value of a land increases every year at the rate of 8%. If its value at the end of 3 years be Rs 6000000, find its original value at the beginning of these years.
a) 4762994
b) 3762994
c) 3000000
d) 4000000
Explanation: R = 8% and n = 3
⇒ 6000000 = P (1 + $$\frac{8}{100})^3$$
⇒ 6000000 = P ($$\frac{27}{25})^3$$
⇒ P = 4762994.
13. Max bought a gadget for Rs. 24000. If the cost of the gadget after 2 years depreciates to Rs. 20757.60, find the rate of depreciation.
a) 20%
b) 5%
c) 15%
d) 7%
Explanation: Let R be the rate of depreciation
⇒ 20757.60 = 24000 (1 – $$\frac{R}{100})^2$$
⇒ $$\frac{20757.60}{24000}$$ = (1 – $$\frac{R}{100})^2$$
⇒ $$\frac{93^2}{100^2}$$ = (1 – $$\frac{R}{100})^2$$
⇒ $$\frac{93}{100}$$ = 1 – $$\frac{R}{100}$$
⇒ R = 7%.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] |
# Problem of the Week Problem B and Solution Sven’s Gym-Cans
## Problem
Recently, Sven decided to start exercising at home instead of going to the gym. He does not have the nice weights that they have at the gym, but he does have different containers of food that he can use as weights. Sven has containers of soup, beans, breadcrumbs, jam, and peanuts. He also has one can whose label fell off that he calls the Mystery Can.
Sven used his balance scale and came up with the following discoveries:
• Two cans of soup have the same mass as one can of beans.
• Five containers of breadcrumbs have the same mass as one jar of jam.
• One can of beans and two containers of bread crumbs together have the same mass as the container of peanuts.
• The Mystery Can and one can of soup together have the same mass as the container of peanuts.
Sven’s friend, Rob, who has perfect estimation skills, determined that the Mystery Can has a mass of $$580$$ grams and the can of beans has a mass of $$640$$ grams. Assuming that Rob is correct, determine the mass of each container, in grams. To do so, you may find it helpful to set up algebraic equations to show the relationships among the cans’ masses.
## Solution
First we will write algebraic equations to show the relationships given by Sven’s four discoveries.
To avoid a lot of writing, we will use the following variables: $$S$$ represents the mass, in grams, of one can of soup, $$P$$ represents the mass, in grams, of one container of peanuts, $$B$$ represents the mass, in grams, of one can of beans, $$C$$ represents the mass, in grams, of one container of breadcrumbs, $$J$$ represents the mass, in grams, of one jar of jam, and $$M$$ represents the mass, in grams, of the Mystery Can.
Now we can write an equation for each of Sven’s four discoveries.
• $$2\times S = B$$
• $$5\times C = J$$
• $$B+2\times C = P$$
• $$M+S = P$$
We are given a mass of $$640$$ grams for one can of beans. Thus the first equation tells us that $$2\times S = 640$$. Since $$2\times 320=640$$, we can conclude that $$S = 320$$.
We are also given that the mass of the Mystery Can is $$580$$ grams. Using the fourth equation we get $$M+S=580+320=900$$. Therefore $$P=900$$.
Using the third equation we get $$640+2\times C = 900$$. Subtracting $$640$$ from both sides of this equation gives us $$2\times C = 260$$. Since $$2\times 130=260$$, it follows that $$C=130$$.
Finally, using the second equation we get $$5\times 130=J$$. Thus $$J=650$$.
Therefore, the mass of each container is as follows.
Container Mass (grams)
can of soup $$320$$
container of peanuts $$900$$
container of breadcrumbs $$130$$
jar of jam $$650$$ |
## Constrained Extrema
In practice, we wish to optimize a function considering some existing constraints.
• In economics and engineering, the constrain may be due to limited funds, materials, or energy.
• If we wish to find the distance from a point $$P=(x_0,y_0)$$ to a line $$ax+by=c$$, we should find the minimum value of $$d(x,y)=\sqrt{(x-x_0)^2+(y-y_0)^2}$$ while $$(x,y)$$ satisfies $$ax+by=c$$.
• Suppose $$T(x,y,z)$$ represents the temperature in space, and we want to find the maximum temperature on a surface given by $$g(x,y,z)=0$$.
When the constraint (also called the side condition) equation is given, we can solve it for one of the variables, say $$z=\phi(x,y)$$, and replace it in the function $$T$$. Then the problem would reduce to finding the extremum value of the function $$T(x,y,\phi(x,y))$$, which now depends only on two independent variables. We have already applied this method in this example, where we optimized the value of the function on the boundary of a region. To practice how to use this method, let’s consider the following example.
Example 1
We want to make a rectangular box without a top, and of given volume $$V$$. If the least amount of material is to be used, determine the design specifications.
Solution
Let
$$x=$$ length of the box
$$y=$$ width of the box
$$z=$$ height of the box
where $$x,y$$ and $$z$$ are in the interval $$(0,\infty)$$. The specified volume is $V=xyz$ The amount of material for constructing this box is proportional to its surface area $S=xy+2xz+2yz.$ So we want to minimize $$S(x,y,z)$$ subject to $$xyz=V$$.
From $$xyz=V$$ we obtain $$z=\frac{V}{xy}$$ and then we plug it into the formula of $$S$$: $S=xy+2\frac{V}{y}+2\frac{V}{x}.$ Now $$S$$ is expressed as a function of only two variables. To determine the relative extremum, we differentiate and set the partial derivatives equal to zero:
$\frac{\partial S}{\partial x}=y-\frac{2V}{x^2}=0, \quad \frac{\partial S}{\partial y}=x-\frac{2V}{y^2}=0$
or equivalently:
$\left\{\begin{array}{l} x^2y=2V\\ \\ xy^2=2V \end{array}\right.$ If we divide the first question by the second one, we obtain $$x/y=1$$. Therefore: $x^2y=x^2 x=x^3=2V \Rightarrow x=y=(2V)^{1/3}$ From these values of $$x$$ and $$y$$, we get $z=\frac{V}{xy}=\frac{(2V)^{1/3}}{2}$ Using the second derivative test, we can show that these values of $$x,y$$ and $$z$$ give a relative minimum of $$S$$. Because $$S\to \infty$$ as either $$x\to 0+$$, $$y\to 0+$$, $$x\to \infty$$, or $$y\to \infty$$, we can conclude that the relative minimum is also the absolute minimum.
## Lagrange Multipliers
When the constraint is given implicitly by $$g(x,y,z)=c$$, it is not always possible or easy to solve the constraint equation for one of the variables (express $$x, y$$ or $$z$$ as a function of the remaining variables). The problem can be more complicated when there is more than one constraint. In such cases, an alternative procedure is a method called Lagrange multipliers.
To explain this method, let’s start with an example. Suppose we want to find the shortest and longest distance between the point $$(1,-1)$$ and the curve $$C$$
$g(x,y)=\frac{(x-y-2)^2}{18}+\frac{(x+y)^2}{32}=1.$
The distance between a point $$(x,y)$$ and $$(1,-1)$$ is given by $f(x,y)=\sqrt{(x-1)^2+(y+1)^2}.$ So we want to maximize and minimize $$f(x,y)$$ subject to $$g(x,y)=1$$. First let’s sketch the curve $$C$$ and some level curves of $$f$$ (Fig. 1).
To extremize $$f(x,y)$$ subject to $$g(x,y)=1$$, we have to find the largest and smallest value of $$k$$ such that the level curve $$f(x,y)=k$$ intersects $$g(x,y)=1$$. Among the level curves that intersect $$g(x,y)=1$$, the minimum value of $$f(x,y)$$ occurs at the points $$P$$ and $$Q$$ where $$f(x,y)$$ has a value of 3 (see Fig. 1). At these points, the constraint curve $$g(x,y)=1$$ and the level curve $$f(x,y)=3$$ just touch each other; in other words, $$f=3$$ and $$g=1$$ have a common tangent line at $$P$$ and a common tangent at $$Q$$.
Note that $$g(x,y)=1$$ is the level curve of $$z=g(x,y)$$. Because at each point $$\overrightarrow{\nabla} f$$ is perpendicular to the level curves of $$f$$ and similarly$$\overrightarrow{\nabla} g$$ is perpendicular to the level curves of $$g$$, a common tangent at $$P$$ means that $$\overrightarrow{\nabla} f(P)$$ and $$\overrightarrow{\nabla} g(P)$$ are parallel. That is, there is a number $$\lambda_1$$ such that
$\overrightarrow{\nabla} f(P)=\lambda_1 \overrightarrow{\nabla} g(P).$
Similarly, there is a number $$\lambda_2$$ such that
$\overrightarrow{\nabla} f(Q)=\lambda_2\overrightarrow{\nabla} g(Q).$
We also observe that the maximum value of $$f(x,y)$$ subject to $$g(x,y)=1$$ occurs where the constraint curve and a level curve (here $$f=4$$) touch each other (In Fig. 1, they are denoted by$$A$$ and $$B$$). Thus $\overrightarrow{\nabla} f(A)=\mu_1\overrightarrow{\nabla} g(A),$ and $\overrightarrow{\nabla} f(B)=\mu_2 \overrightarrow{\nabla} g(B),$ for some $$\mu_1$$ and $$\mu_2$$. Therefore, to find the maximum or minimum of $$f(x,y)$$ subject to the constraint $$g(x,y)=1$$ , we look for a point $$\textbf{x}_0$$ such that
$\overrightarrow{\nabla} f(\textbf{x}_0)=\lambda \overrightarrow{\nabla} g(\textbf{x}_0)$
for some $$\lambda$$. This is the method of Lagrange multiplier. But why is this true?
Suppose the constraint curve $$C$$ is parameterized by some functions:1 $x=X(t),\quad y=Y(t)$ If, in the equation of $$f$$, $$x$$ and $$y$$ are replaced by $$X(t)$$ and $$Y(t)$$, then the distance between $$(1,-1)$$ and the points on $$C$$ becomes a function of $$t$$ $F(t)=f(x(t),y(t)).$ Therefore, the extreme values of the distance occur where $$F'(t)=0$$. From the chain rule we know
$F'(t)=\frac{dF}{dt}=\frac{\partial f}{\partial x}\frac{dX}{dt}+\frac{\partial f}{\partial y}\frac{dY}{dx}$
We can write the equation $$F'(t)=0$$ as
$\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\boldsymbol{\cdot}\left(\frac{dX}{dt},\frac{dY}{dt}\right)=0$
This means at the extreme points the gradient vector $$(f_x,f_y)$$ is perpendicular to $$(X'(t),Y'(t))$$. Recall that $$(X'(t),Y'(t))$$ is tangent to the curve $$C$$.
On the other hand $$C$$ is a level curve of $$g$$. Therefore, at each point $$\overrightarrow{\nabla} g$$ is perpendicular to $$C$$. Therefore at the extreme points, $$\overrightarrow{\nabla} g$$ and $$\overrightarrow{\nabla} f$$ are parallel. The simplest version of the method of Lagrange multipliers is as follows.
Theorem 1. Suppose $$U\subseteq \mathbb{R}^n$$ is an open set and $$f:U\to\mathbb{R}$$ and $$g:U\to\mathbb{R}$$ are two continuously differentiable functions2. Let $$S=\left\{\mathbf{x}\in U|\ g(\mathbf{x})=c\right\}$$ be the level set for $$g$$ with value $$c$$. If $$f |S$$ denoting “$$f$$ restricted to $$S$$,” has a relative extremum on $$S$$ at $$\mathbf{x}_0\in U$$, and $$\overrightarrow{\nabla} g(\mathbf{x}_0)\neq\mathbf{0}$$, then there exists a real number $$\lambda$$ such that
$\overrightarrow{\nabla} f(\mathbf{x}_0)=\lambda \overrightarrow{\nabla} g(\mathbf{x}_0).$
Let $$\mathbf{r}:I\subseteq\mathbb{R}\to\mathbb{R}^n$$ be a differentiable curve on the level set $$S$$ such that $$\mathbf{r}(t_0)=\mathbf{x}_0$$, $$\mathbf{r}'(t_0)\neq \mathbf{0}$$, and $$\mathbf{r}(t)\in S$$ for every $$t\in I$$. Then $$g(\mathbf{r}(t))=c$$, so the chain rule gives
$\overrightarrow{\nabla} g(\underbrace{\mathbf{r}(t_0)}_{=\mathbf{x}_0})\boldsymbol{\cdot}\mathbf{r}'(t_0)=0.$
Because $$f|S$$ attains a relative maximum or minimum at $$\mathbf{x}_0$$, the function $$h=f\circ\mathbf{r}:I\to\mathbb{R}$$ attains a relative maximum or minimum at $$t_0$$. Hence $$h'(t_0)=0$$ and according to the chain rule we have:
$h'(t_0)=\overrightarrow{\nabla} f(\mathbf{x}_0)\boldsymbol{\cdot}\mathbf{r}'(t_0)=0.$
Thus the vectors $$\overrightarrow{\nabla} f(\mathbf{x}_0)$$ and $$\overrightarrow{\nabla} g(\mathbf{x}_0)\neq \mathbf{0}$$ are both normal to the nonzero vector $$\mathbf{r}'(t_0)$$ and are therefore parallel; that is, $$\overrightarrow{\nabla} f(\mathbf{x}_0)=\lambda \overrightarrow{\nabla} g(\mathbf{x}_0)$$ for some $$\lambda$$. $$\blacksquare$$
• The number $$\lambda$$ in the above theorem is called a Lagrange multiplier.
• $$\lambda$$ might be zero.
Note that to find the extremum of $$f|S$$, we have $$n+1$$ unknowns ($$n$$ components of $$\mathbf{x}_0$$ and $$\lambda$$) and $$n+1$$ equations:
\label{Eq:LagrangeEQ} \left.\begin{align} \frac{\partial f}{\partial x_1}(\mathbf{x}_0)&=\lambda \frac{\partial g}{\partial x_1}(\mathbf{x}_0)\\ &\vdots\\ \frac{\partial f}{\partial x_n}(\mathbf{x}_0)&=\frac{\partial g}{\partial x_n}(\mathbf{x}_0)\\ g(\mathbf{x}_0)&=c \end{align}\right\} \quad (n+1) \text{ equations} \tag{i}
Example 2
Find the extrema of the function $$f(x,y)=xy$$ subject to the constraint $$\frac{1}{4}x^2+\frac{1}{9}y^2=1$$.
Solution
Let $$g(x,y)=\frac{1}{4}x^2+\frac{1}{9}y^2$$, so $$S$$ consists of all points $$(x,y)$$ such that $$g(x,y)=1$$. We have:
$\overrightarrow{\nabla} f(x,y)=(y,x),\quad \overrightarrow{\nabla} g(x,y)=\left(\frac{1}{2}x,\frac{2}{9}y\right)$
Note that $$\overrightarrow{\nabla} g=\mathbf{0}$$ if and only if $$(x,y)=(0,0)$$. Thus if $$f$$ subject to the constraint has an extremum at $$(x_0,y_0)$$, we must have $$\overrightarrow{\nabla} f(x_0,y_0)=\lambda \overrightarrow{\nabla} g(x_0,y_0)$$. Equations i take on the form:
$y_0=\frac{1}{2}\lambda x_0, \quad x_0=\frac{2}{9}\lambda y_0, \quad \frac{1}{4}x_0^2+\frac{1}{9}y_0^2=1.$
There are different ways to solve the above system of equations. One way is to replace $$y_0$$ from the first equation in the second equation
$x_0=\frac{2}{9}\lambda y_0=\frac{2\lambda}{9}(\frac{\lambda}{2} x_0)=\frac{1}{9} \lambda^2 x_0$
$x_0-x_0\frac{\lambda^2}{9}=x_0(1-\frac{\lambda^2}{9})=0\Rightarrow x=0 \ \text{or}\ \lambda=\pm 3.$
If $$x=0$$, then substituting in $$g(x_0,y_0)=1$$, we obtain $\frac{1}{4} 0^2+\frac{1}{9}y_0^2=1\Rightarrow y_0=\pm 3.$ If $$\lambda=\pm 3$$, it follows from $$y_0=\frac{1}{2}\lambda x_0$$ that $$y_0=\pm \frac{3}{2} x_0$$. Substituting into $$g(x_0,y_0)$$, we get:
$\frac{1}{4}x_0^2+\frac{1}{9}\frac{9}{4}x_0^2=1 \Rightarrow \frac{1}{\sqrt{2}}x_0^2=1\Rightarrow x_0=\pm \sqrt{2}.$
If $$x_0=\sqrt{2}$$, then $$y_0=\pm \frac{3}{2}\sqrt{2}$$, and if $$x_0=-\sqrt{2}$$, then $$y_0=\pm\frac{3}{2}\sqrt{2}$$.
Now we list the points we found and the values of $$f$$ at these points:
$\large&space;(x_0,y_0)$ $$(0,3)$$ $$(0,-3)$$ $$(\sqrt{2},\frac{3\sqrt{2}}{2})$$ $$(\sqrt{2},-\frac{3\sqrt{2}}{2})$$ $$(\sqrt{2},-\frac{3\sqrt{2}}{2})$$ $$(-\sqrt{2},-\frac{3\sqrt{2}}{2})$$
$$f(x_0,y_0)=x_0y_0$$ $$0$$ $$0$$ $$3$$ $$-3$$ $$-3$$ $$3$$
The function $$f$$ attains its maximum value 3 at $$(\sqrt{2},3\sqrt{2}/2)$$ and $$(-\sqrt{2},-3\sqrt{2}/2)$$, and its minimum value -3 at $$(-\sqrt{2},3\sqrt{2}/2)$$ and $$(-\sqrt{2},3\sqrt{2}/2)$$. Note that at these points the level curve $$g=1$$ is tangent to one of the level curves of $$f$$ (see Fig. 2).
(a) (b) Figure 2. (a) The graph of $$f(x,y)$$; the blue curve shows $$f$$ restricted to the ellipse $$g(x,y)=1$$. (b) The level curves of $$f$$ and $$g$$; the extrema of $$f$$ subject to the constraint $$g(x,y)=1$$ occur at points where the level curves of $$f$$ are tangent to the level curve $$g(x,y)=1$$.
Let’s go back to the example of making an open box and try to solve it using the method of Lagrange multipliers.
Example 3
We want to make a rectangular box without a top, and of given volume $$V$$. If the least amount of material is to be used, determine the design specifications using Lagrange multipliers.
Solution
Again Let $$x, y$$ and $$z$$ be the length, width, and height of the box, respectively. We want to minimize $S=xy+2xz+2yz$ subject to $$xyz=V$$, where $$V$$ is the given volume.
If we let $$g(x,y,z)=xyz$$, then
$\overrightarrow{\nabla} S=(y+2z, x+2z, 2x+2y),\quad \overrightarrow{\nabla} g=(yz, xz, xy)$
Note that $$\overrightarrow{\nabla} g=\mathbf{0}$$ if and only if $$(x,y,z)=(0,0,0)$$ which is not on the level surface $$g(x,y,z)=V$$. So if $$S$$ has an extremum at $$(x_0,y_0,z_0)$$, we have
$\overrightarrow{\nabla} S(x_0,y_0,z_0)=\lambda \overrightarrow{\nabla} g(x_0,y_0,z_0).$
Hence our equations become
\begin{align} \left\{\begin{array}{l c c} y+2z=\lambda yz & & \text{(i)}\\ \\ x+2z=\lambda xz & & \text{(ii)}\\ \\ 2x+2y=\lambda xy& & \text{(iii)}\\ \\ xyz=V& & \text{(iv)} \end{array}\right.\end{align}
Because $$x,y$$ and $$z$$ are nonzero, if we divide the first equation by $$yz$$, the second equation by $$xz$$ and the third one $$xy$$, we get:
\begin{align} \left\{\begin{array}{l c c} \dfrac{1}{z}+\dfrac{2}{y}=\lambda & & \text{(i^\prime)}\\ \\ \dfrac{1}{z}+\dfrac{2}{x}=\lambda & & \text{(ii^\prime)}\\ \\ \dfrac{2}{y}+\dfrac{2}{x}=\lambda& & \text{(iii^\prime)}\\ \\ xyz=V& & \text{(iv^\prime)} \end{array}\right.\end{align}
Then
$\text{Eq. (i^\prime)},\ \ \text{Eq. (ii^\prime)}\quad \Rightarrow \quad x=y$
$\text{Eq. (ii^\prime)},\ \ \text{Eq (iii^\prime)}, \ \text{and}\ x=y\quad \Rightarrow \quad z=\frac{1}{2} x$
$x=y=2z, \ \ \text{Eq (4)} \Rightarrow \frac{1}{2}x^3=V\quad \Rightarrow\quad x=y=2z=(2V)^{1/3}.$
This is the result we previously obtained in Example 1.
The following example shows an application of Lagrange multipliers in economics and when there are many variables.
Example 4
Assume there are $$n$$ commodities with prices per unit $$p_1,\cdots,p_n$$. Assume we have $$L$$ dollars to spend on these commodities. This means we have the constraint $$p_1 x_1+\cdots p_n x_n=L$$ where $$x_1,\cdots,x_n$$ are the amounts of the commodities. Assume the utility is given by the Cobb-Douglas function $$U=f(x_1,\cdots,x_n)=K x_1^{\alpha_1}\cdots x_n^{\alpha_n}$$ where $$K, \alpha_1,\cdots,\alpha_n$$ are positive constants. Find the maximum utility we can achieve.
Solution
Let $$g(x_1,\cdots,x_n)=p_1 x_1+\cdots+p_n x_n$$. So we seek for the maximum value of $$f(x_1,\cdots,x_n)$$ subject to $$g(x_1,\cdots,x_n)=L$$. Because $$x_i\geq 0$$ (for $$i=1,\cdots,n$$) the constraint is a part of the level set $$g(x_1,\cdots,x_n)=L$$. We note that when the amount of a commodity $$x_i$$ approaches zero, the utility approaches zero. Hence, the utility takes on its maximum value when $$x_i$$’s are positive. Because $$\overrightarrow{\nabla} g(x_1,\cdots,x_n)=(p_1,\cdots,p_n)\neq \mathbf{0}$$, if $$f$$ attains its maximum at a point, we have $$\overrightarrow{\nabla} f(x_1,\cdots,x_n)=\lambda \overrightarrow{\nabla} g(x_1,\cdots,x_n)$$; that is,
$\left\{\begin{array}{l} \alpha_1 K x_1^{\alpha_1-1}x_2^{\alpha_2}\cdots,x_n^{\alpha_n}=\lambda p_1\\ \hspace{1cm}\vdots\\ \alpha_n K x_1^{\alpha_1}x_2^{\alpha_2}\cdots,x_n^{\alpha_n-1}=\lambda p_n \end{array}\right.$
If we multiply both sides of the first equation by $$x_1$$, the second equation by $$x_2$$, $$\cdots$$, and the $$n$$-th equation by $$x_n$$, we get:
$\left\{\begin{array}{l} \alpha_1 f(x_1,\cdots,x_n)=\lambda p_1 x_1\\ \hspace{1cm}\vdots\\ \alpha_n f(x_1,\cdots,x_n)=\lambda p_n x_n \end{array}\right.$
Therefore: $\frac{p_1}{\alpha_1}x_1=\cdots=\frac{p_n}{\alpha_n} x_n$ This result makes sense because it says the amount of commodity $$x_i$$ is proportional to its power and inversely proportional to its price, $$p_i$$.
Let $$\frac{\alpha_i}{p_i} x_i=C$$, then
$p_1 x_1+\cdots+p_n x_n=p_1\frac{\alpha_1}{p_1} C+\cdots+p_n\frac{\alpha_1}{p_n} C=L$
or
$C(\alpha_1+\cdots+\alpha_n)=C\sum_{i=1}^n \alpha_i=L\Rightarrow C=\frac{L}{\sum_{i=1}^n \alpha_i}.$
Consequently,
$x_i=\frac{\alpha_i}{p_i}\frac{L}{\sum_{i=1}^n \alpha_n}.$ Finally we conclude the maximum value of $$f$$ is:
$K\left(\frac{\alpha_i L}{p_i \sum_{i=1}^n \alpha_i}\right)^{\alpha_1} \cdots \left(\frac{\alpha_n L}{p_n \sum_{i=1}^n \alpha_i}\right)^{\alpha_n}=K\left(\frac{L}{\sum_{i=1}^n \alpha_i}\right)^{\sum_{i=1}^n\alpha_i} \left(\frac{\alpha_1}{p_1}\right)^{\alpha_1} \cdots \left(\frac{\alpha_n}{p_n}\right)^{\alpha_n}$
The following example shows when maximizing or minimizing a function subject to a constraint, we should also investigate the points where the gradient of the constraint is zero $$\overrightarrow{\nabla} g(\mathbf{x})=\mathbf{0}$$.
Example 5
Find the nearest point on the curve $$y^3-x^2=0$$ to the point $$(0,-1)$$.
Solution
If we plot the level curve
$$g(x,y)=y^3-x^2=0$$ (Fig. 3), it is clear that $$(0,0)$$ is the nearest point on this curve to the point $$(0,-1)$$, and the distance is 1. If we want to use the method of Lagrange multipliers, we need to minimize the square of distance $$f(x,y)=(x-0)^2+(y-(-1))^2$$ subject to the constraint $$g(x,y)=y^3-x^2=0$$.
Here
$\overrightarrow{\nabla} f(x,y)=(2x,2y+2),\quad \overrightarrow{\nabla} g(x,y)=(-2x,3y^2).$
Equations i take on the form
$\left\{\begin{array}{l} 2x_0=-2\lambda x_0\\ 2y_0+2=3\lambda y_0^2\\ y_0^3-x_0^2=0 \end{array}\right.$
However the solution, $$(x_0,y_0)=(0,0)$$, does not satisfy the second equation. The reason is attributed to the fact that at this point the gradient of $$g$$ is $$\mathbf{0}$$, $$\overrightarrow{\nabla} g(0,0)=(0,0)$$, so we cannot use Theorem 1. In addition, the level curve is not smooth at $$(0,0)$$.
If a function $$f$$ subject to a constraint $$g(\mathbf{x})=c$$ attains its extreme value at a point $$\mathbf{x}_0$$, then $$\mathbf{x}_0$$ is one of the four types of the point:
1. $$\mathbf{x}_0$$ is a point where $$\overrightarrow{\nabla} f(\mathbf{x}_0)=\lambda \overrightarrow{\nabla} g(\mathbf{x}_0)$$,
2. $$\mathbf{x}_0$$ is a point where $$\overrightarrow{\nabla} g(\mathbf{x}_0)=\mathbf{0}$$,
3. $$\mathbf{x}_0$$ is a rough point of $$f$$ or $$g$$ where $$\overrightarrow{\nabla} f(\mathbf{x}_0)$$ or $$\overrightarrow{\nabla} g(\mathbf{x}_0)$$ does not exist, or
4. $$\mathbf{x}_0$$ is on the boundary of the level set $$g(\mathbf{x})=c$$.
## Multiple Constraints
The method of Lagrange multipliers extends to the case of multiple constraints: say to $$f(x,y,z)$$ subject to two constraints
$\label{Eq:Lagrange_2Constraint} g_1(x,y,z)=c_1,\quad \text{and} \quad g_2(x,y,z)=c_2.\tag{ii}$
Geometrically this means we seek the extreme values of $$f(x,y,z)$$ on a curve $$C$$ formed by the intersection of two level surfaces $$g_1(x,y,z)=c_1$$ and $$g_2(x,y,z)=c_2$$ (Fig. 4).
Suppose $$f$$ subject to these constraints, $$f|C$$, has an extremum at $$P=(x_0,y_0,z_0)$$, and the functions $$f, g_1$$, and $$g_2$$ have continuous first partial derivatives near $$P$$. Analogous to the case of one constraint, we can argue that $$\overrightarrow{\nabla} f$$ is perpendicular to $$C$$ at $$P$$. Additionally because the gradient is perpendicular to the level surface, both $$\overrightarrow{\nabla} g_1$$ and $$\overrightarrow{\nabla} g_2$$ are perpendicular to $$C$$. If $$\overrightarrow{\nabla} g_1(x_0,y_0,z_0)$$ and $$\overrightarrow{\nabla} g_2(x_0,y_0,z_0)$$ are neither zero vectors nor collinear vectors, then $$\overrightarrow{\nabla} f(x_0,y_0,z_0)$$ lie in a plane spanned by the two vectors $$\overrightarrow{\nabla} g_1(x_0,y_0,z_0)$$ and $$\overrightarrow{\nabla} g_2(x_0,y_0,z_0)$$ and hence can be expressed as a linear combination of these two vectors, say:
\begin{align} \label{Eq:Lagrange_2Multipliers} \bbox[#F2F2F2,5px,border:2px solid black]{\overrightarrow{\nabla} f(x_0,y_0,z_0)=\lambda_1 \overrightarrow{\nabla} g_1(x_0,y_0,z_0)+\lambda_2 \overrightarrow{\nabla} g_2(x_0,y_0,z_0)}\tag{iii}\end{align}
In this method, Equations ii and iii must be solved simultaneously for five unknowns $$x_0,y_0,z_0,\lambda_1$$ and $$\lambda_2$$:
$\left\{\begin{array}{l} \dfrac{\partial f}{\partial x}(x_0,y_0,z_0)=\lambda_1 \dfrac{\partial g_1}{\partial x}(x_0,y_0,z_0)+\lambda_2 \dfrac{\partial g_2}{\partial x}(x_0,y_0,z_0)\\ \\ \dfrac{\partial f}{\partial y}(x_0,y_0,z_0)=\lambda_1 \dfrac{\partial g_1}{\partial y}(x_0,y_0,z_0)+\lambda_2 \dfrac{\partial g_2}{\partial y}(x_0,y_0,z_0)\\ \\ \dfrac{\partial f}{\partial z}(x_0,y_0,z_0)=\lambda_1 \dfrac{\partial g_1}{\partial z}(x_0,y_0,z_0)+\lambda_2 \dfrac{\partial g_2}{\partial z}(x_0,y_0,z_0)\\ \\ g_1(x,y,z)=c_1\\ \\ g_2(x,y,z)=c_2 \end{array}\right.$
Example 6
Find the extreme points of the function $$f(x,y,z)=x+4y-2z$$ on the curve of the intersection of the cylinder $$x^2+y^2=4$$ and the plane $$2y-z=3$$ (Fig. 5).
Solution
Here we are given two constraints:
$g_1(x,y,z)=x^2+y^2=4,\quad g_2(x,y,z)=2y-z=3.$ Note that
$\overrightarrow{\nabla} f=(1,4,-2),\quad \overrightarrow{\nabla} g_1=(2x,2y,0),\quad\text{and}\quad \overrightarrow{\nabla} g_2=(0,2,-1).$
The vector $$\overrightarrow{\nabla} g_1=(0,0,0)$$ only when $$x=0$$ and $$y=0$$, which clearly does not satisfy the constraint $$g_1=4$$. Thus, the two vectors $$\overrightarrow{\nabla} g_1$$ and $$\overrightarrow{\nabla} g_2$$ are clearly not parallel. Therefore, any constrained critical point $$(x_0,y_0,z_0)$$ must satisfy
$\overrightarrow{\nabla} f(x_0,y_0,z_0)=\lambda_1 \overrightarrow{\nabla} g_1(x_0,y_0,z_0)+\lambda_2 \overrightarrow{\nabla} g_2(x_0,y_0,z_0).$
It follows that we must solve the following system of equations for five unknowns
$$x,y,z,\lambda_1$$ and $$\lambda_2$$:
$\left\{\begin{array}{l} 1=2x\lambda_1 +0\\ 4=2y\lambda_1+2\lambda_2\\ -2=0-\lambda_2\\ x^2+y^2=4\\ 2y-z=3 \end{array}\right.$
From the third equation, we know $$\lambda_2=2$$. So the second equation becomes $$2y\lambda_1=0$$. Because from the first equation, $$\lambda_1\neq 0$$, it follows from $$2y\lambda_1=0$$ that $$y=0$$. If we substitute $$y=0$$ in the fourth and fifth equations, we get $$x=\pm 2$$ and $$z=1$$. Therefore, $$f$$ may have extrema at $$(\pm 2,0,1)$$.
The condition $$x^2+y^2=4$$ implies $$-2\leq x,y\leq 2$$. Then it follows from $$2y-z=3$$ that $$-7\leq z\leq 1$$. This means the constraint set, $$S$$, is bounded. Because the constraint set $$S$$ is closed and bounded, and $$f$$ is a continuous function, $$f|S$$ assumes its maximum and minimum ( Theorem 1 in the previous section). Here we have only two potentials, therefore one of them $$(2,0,1)$$ is the maximum point and the other one $$(-2,0,1)$$ is the minimum point.
By similar reasoning, we obtain equations for minimizing or maximizing
$$f(x_1,\cdots,x_n)$$ subject to several constraints
\begin{align} \label{Eq:m-constraints} \left\{\begin{array}{l} g_1(x_1,\cdots,x_n)=c_1\\ g_2(x_1,\cdots,x_n)=c_2\\ \vdots\\ g_m(x_1,\cdots,x_n)=c_m \end{array}\right.\tag{iv}\end{align}
where $$m<n$$. Assume $$f(x_1,\cdots,x_n)$$ has a relative extremum at $$\mathbf{x}_0$$ when the variables are restricted by the constraint equations iv. If $$f, g_1,\cdots,$$ and $$g_m$$ have continuous first partial derivatives at all points near $$\mathbf{x}_0$$ and if each $$\overrightarrow{\nabla} g_i(\mathbf{x}_0)$$ is not a linear combination of the other $$\overrightarrow{\nabla} g_j(\mathbf{x}_0)$$ ($$j\neq i$$), then there exists $$m$$ real numbers $$\lambda_1,\cdots, \lambda_m$$ such that
$\bbox[#F2F2F2,5px,border:2px solid black]{\overrightarrow{\nabla} f(\mathbf{x}_0)=\lambda_1\overrightarrow{\nabla} g_1(\mathbf{x}_0)+\cdots+\lambda_m \overrightarrow{\nabla} g_m(\mathbf{x}_0).}$
1 For instance, in this specific example,
$\begin{cases} x=X(t)=\dfrac{3}{\sqrt{2}}\cos t+2\sqrt{2} \sin t\\ \\ y=Y(t)=-\dfrac{3}{\sqrt{2}}\cos t+2\sqrt{2}\sin t \end{cases}\qquad (0\leq t\leq2\pi)$
2 In other words, $$f$$ and $$g$$ have continuous partial derivatives. |
## Online Encylopedia and Dictionary Research Site
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# Law of cosines
In trigonometry, the law of cosines is a statement about arbitrary triangles which generalizes the Pythagorean theorem by correcting it with a term proportional to the cosine of the opposing angle. Let a, b, and c be the sides of the triangle and A, B, and C the angles opposite those sides. Then,
$c^2 = a^2 + b^2 - 2ab \cos C . \;$
This formula is useful for computing the third side of a triangle when two sides' and their enclosed angle's values are known, and in computing the angles of a triangle if all three sides' values are known.
The law of cosines also shows that
$c^2 = a^2 + b^2 \;$ iff $\cos C = 0 . \;$
The statement cos C = 0 implies that C is a right angle, since a and b are positive. In other words, this is the Pythagorean theorem and its converse. Although the law of cosines is a broader statement of the Pythagorean theorem, it isn't a proof of the Pythagorean theorem, because the law of cosines derivation given below depends on the Pythagorean theorem.
Contents
## Derivation (for acute angles)
Triangle
Let a, b, and c be the sides of the triangle and A, B, and C the angles opposite those sides. Draw a line from angle B that makes a right angle with the opposite side, b. If the length of that line is x, then $\sin C = \frac{x}{a} , \;$ which implies $x=a \sin C . \;$
That is, the length of this line is $a \sin C. \;$ Similarly, the length of the part of b that connects the foot point of the new line and angle C is $a \cos C. \;$ The remaining length of b is $b - a \cos C. \;$ This makes two right triangles, one with legs $a \sin C , \;$ $b - a \cos C , \;$ and hypotenuse c. Therefore, according to the Pythagorean theorem:
$c^2 = (a \sin C)^2 + (b - a \cos C)^2 \;$
$= a^2 \sin^2 C + b^2 - 2 ab \cos C + a^2 \cos^2 C \;$
$= a^2 (\sin^2 C + \cos^2 C) + b^2 - 2ab \cos C \;$
$=a^2+b^2-2ab\cos C\;$
because
$\sin^2 C + \cos^2 C=1. \;$
## Law of cosines using vectors
Using vectors and vector dot products, we can easily prove the law of cosines. If we have a triangle with vertices A, B, and C whose sides are the vectors a, b, and c, we know that:
$\mathbf{a = b - c} \;$
since
$\mathbf{(b - c)\cdot (b - c) = b\cdot b - 2 b\cdot c + c\cdot c}. \;$
Using the dot product, we simplify this into
$\mathbf{|a|^2 = |b|^2 + |c|^2 - 2 |b||c|}\cos \theta. \;$ |
### Geoboards
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
### Polydron
This activity investigates how you might make squares and pentominoes from Polydron.
If you had 36 cubes, what different cuboids could you make?
# A Square Deal
##### Age 7 to 11 Challenge Level:
Thank you to everyone who sent in correct answers to our magic square problem. As usual, here we will give some examples of those of you who gave us great explanations as to how you approached it.
Gitty and Gabriel from Bournemouth Jewish Day School wrote an extremely clear account:
We wrote down the square numbers, prime numbers, triangle numbers, cube numbers, powers of two, palindromic numbers, factors of $100$, upside down numbers and the median. All our answers had to be $25$ or less.
We drew a $5$ x $5$ grid and put in the median as $13$ in c3.
Next we found a5 because it is palindromic, prime and upside down, $11$.
Then we found d1 because it is palindromic, $22$.
After that we found a1 because it is the only prime and triangular number, $3$.
Next was d3 because it is square, triangular, cube, upside down and a factor of $100$, $1$.
Now b2 must be $8$ because it is the only cube left. It is also upside down and a power of two.
Then e5 as it is only said to be prime, $23$.
As there is only one space left in the diagonal a1 to e5, we have $47$ so d4 must be $65 - 47 = 18$. $18$ is not found in any group.
e4 came next as 6 - triangular.
Then we found e2 because it is prime and a factor of $100$ i.e. $2$.
a2 is a factor of $100$ that is not in any other group, $20$.
a3 is a prime number and nothing else, $7$.
c4 is prime and a factor of $100$, $5$.
We can now find column a because only a4 is missing. $65 - 41 = 24$.
Now we can find row 4, $65 - 53 = 12$ (b4).
We then spotted a pattern on the diagonal a5 to e1. We discovered that it was $11, 12, 13$ so we tried $14$ and $15$ and we checked and found that e1 has to be triangular and so can be $15$ and then d2 can be $14$. Also $11 + 12 + 13 + 14 + 15 = 65$.
We can now find c2 to complete row 2, $65 - 44 = 21$ and it is also triangular.
d5 is $10$ because $65 - 55 = 10$ and $10$ is also triangular.
Next e3 is $19$ because $65 - 46 = 19$ and e3 is also odd and prime.
Next b3 is $25$ because $65 - 40 = 25$ and $25$ is square and a factor of $100$.
Only b1, c1, b5 and c5 are left. c5 is not square so it must be $17$ as it is prime.
This means that c1 must be $65 - 56 = 9$ and $9$ is odd and square .
b5 is square and a factor of $100, 4$.
So b1 is $16$, square and a power of $2$.
We checked by adding all the rows and columns and they made $65$ each.
This is what the square looks like:
Bronya from Tattingstone School used a very systematic approach too, although slightly different from Gitty and Gabriel's:
I know straight away that c3 = $13$ because $13$ is the median of all the numbers.
I also know that d3 had to be $1$ as it had to be odd, it was a factor of $100$ and it looked the same upside-down. The numbers that looked the same upside-down were $1, 8 and 11$. I know it couldn't be $8$ as $8$ is even. It couldn't be $11$ either as $11$ isn't a factor of $100$.
I also worked out that a5 = $11$ because it was a prime number, it was a palindromic number and it was a number that looked the same upside-down.
This told me that d3 = $8$ as it was the only number left that looked the same upside-down.
Also it meant that d1 = $22$ because $22$ is the only other palindromic number left.
I worked out that b3 = $25$ because b3 was odd, it was a perfect square and it was a factor of $100$. The perfect squares left were $4, 9, 16 and 25$ but only $25$ was a factor of $100$.
I know that c4 = 5 because it was a factor of $100$ and it was a prime number which left me with $2$ or $5$ as I had already used $25$. I also know that it had to be odd.
That meant that e2 = $2$ because like c4, the number was a prime and a factor of $100$ but as a definite answer e2 was also a power of $2$.
I carried on like this until I had filled in the whole square.
I listed the numbers that fitted with the fact:
Perfect Squares: $1, 4, 9, 16, 25$
Prime Numbers: $2, 3, 5, 7, 11, 13, 17, 19, 23$
Triangular Numbers: $1, 3, 6, 10, 15, 21$
Perfect Cubes: $1, 8$
Powers of $2$: $2, 4, 8, 16$
Palindromic Numbers: $11, 22$
Factors of $100$: $1, 2, 4, 5, 10, 20, 25$
Median: $13$
Numbers that are the same upside-down: $1, 8, 11$
This is a very good strategy, writing down a list like this. Gemma from Hillside Community Primary School also did this in order to help her get the solution.
Katelyn from Kilvington Girls Grammar in Australia wrote to us with some handy hints for how to start:
Rule up a graph $25$ x $10$
In the $10$ vertical lines write an abbreviation for the attributes required.
In the $25$ horizontal lines write the numbers $1 - 25$.
Tick what attributes each number has.
Rule up an identical $5$ x $5$ grid, shown in the problem.
From the clues given, write what attributes each square requires.
Transfer the numbers with all of the required attributes into the appropriate square. This will reduce the number of possibilities per square.
Fill in the squares with only one possible answer. Then eliminate these numbers from other squares.
When you have at least $3$ numbers in a line, add these numbers together and subtract the total from $65$.
Now select the appropriate numbers from the selection in each square tocomplete the row.
Continue crossing out the numbers as you select them. This will reduce thepossible selection in other squares.
Well done also to Claire from St Ives, Haslemere, Duncan from Balgowan Primary School, Samuel from Bispham Drive Junior School, Stefan from Sacred Heart Roman Catholic Primary School and Suzy who was working from home, who all got the correct answer. |
Lesson Video: Speed | Nagwa Lesson Video: Speed | Nagwa
# Lesson Video: Speed Science • Third Year of Preparatory School
## Join Nagwa Classes
In this video, we will learn how to determine the speed of an object that moves a distance in a time.
11:49
### Video Transcript
In this video, we will learn how to determine the speed of an object that moves a distance in a time.
To get started, say that we’re standing at a particular spot and decide to move to another spot over here. And also we decide to time ourselves as we move. So we take a step, then another, and another, step, step, step, step, step, step, step, step, step until finally we arrive. We traveled a distance of 15 steps, and we did it in a time of 15 seconds. We can say that we moved at a certain speed. That speed equals the distance we moved divided by the time taken to move that distance. In general, a speed is always a distance divided by a time. Let’s look at a quick example.
Which of the following is the correct formula for the speed of an object? (A) Speed equals distance moved multiplied by time moved for. (B) Speed equals distance moved divided by time moved for. (C) Speed equals time moved for divided by distance moved.
To see which formula is correct, let’s think about an object in motion. The faster this object moves, the greater its speed will be. A higher speed means the object moves a greater distance in the same amount of time. Or the object could move the same distance but take less time. We can say the object’s speed increases in two different cases: first, when it moves a greater distance in the same amount of time and second when it moves the same distance in less time. Looking at our answer choices, the only one that is true for both of these conditions is option (B) speed equals distance moved divided by time moved for.
Let’s now return to our situation of walking from one point to another. Remember that we moved from the start to the end in 15 steps, and it took us 15 seconds. Let’s say that each one of those steps covered one meter of distance. This means that after our first step, we had traveled one meter. Then after the second step, we had traveled a total distance of two meters, then three meters after the third step, and so on, with the person taking 15 steps. So the total distance moved is 15 meters. But then, remember that at each one of these steps, one more second of time had passed. When we had traveled one meter, one second had elapsed, then two seconds at two meters, three seconds at three meters, and so on. For every second of time passed, we moved the same distance, one meter.
Whenever motion happens at equal distance intervals over equal times, that motion happens at what we call uniform speed. This means the speed is always the same. That is, the object moving always travels equal intervals of distance in equal times.
Let’s now consider another example.
Fill in the blank: If an object moves with a uniform speed, then it moves blank distances in blank times. (A) Unequal, equal; (B) unequal, unequal; (C) equal, equal; (D) equal, unequal.
We’re talking here about an object that moves at uniform speed. Say that this is our object, and then it’s moving to the right. Since the object moves with uniform speed, that means every time it moves a certain distance to the right, that moving always takes the same amount of time. In other words, moving equal distances requires equal amounts of time. Our completed sentence reads as follows: If an object moves with a uniform speed, then it moves equal distances in equal times.
So far, we’ve seen that this sentence is true. An object’s speed equals the distance it moves divided by the time taken. Another way to say this is using an equation. The equation has to do with speed, distance, and time. This sentence tells us that speed is equal to distance divided by time. This equation says the same thing that the sentence says, and we can even write it in a shorter way. We can represent the quantity speed by the letter 𝑣. Distance we’ll represent with a 𝑑 and time with a 𝑡. Then, we can rewrite this equation. Using symbols, we can say that speed 𝑣 equals distance 𝑑 divided by time 𝑡. This equation in symbols says the same thing as the equation in words. And that says the same thing as this sentence.
Let’s now get some practice using this equation for speed.
A bike moves uniformly. The position of the bike at two different times is shown. What is the speed of the bike?
In this diagram, we see the bicycle at zero seconds and two seconds. Over this time, it’s traveled a distance of 10 meters. We want to solve for the speed of the bike. We’ll call the speed 𝑣. In general, speed is given by this equation. The speed of an object 𝑣 is equal to distance it travels, 𝑑, divided by the time taken to travel that distance 𝑡. With our bicycle, we know the distance it travels. That’s 10 meters, so 𝑑 is 10 meters. Then, what about the time 𝑡? We see the bike’s position at zero seconds and two seconds. That means the total time is two seconds.
Now that we know 𝑑 and 𝑡, we can use our equation to solve for speed. 𝑣 is equal to 10 meters divided by two seconds. In other words, it’s 10 divided by two meters per second. 10 divided by two is five. So the speed of the bike is five meters per second. The magnitude of this speed is five, and the units are meters per second.
Let’s now look at another example.
An aircraft moves at a speed of 630 kilometers per hour for eight hours. How far does the aircraft move?
Let’s say that this is our aircraft. We know how fast it’s traveling, 630 kilometers per hour. And we know for how long it travels, eight hours. We want to know how far it moves, its total distance. We can think of it like this. If the aircraft moves this far in one hour, then in eight hours, it will travel this distance. We want to know this distance that we’ll call 𝑑. It’s related to the aircraft’s speed and the time passed. In general, the speed, 𝑣, of an object equals the distance it travels, 𝑑, divided by the time, 𝑡. In our case, the speed 𝑣 is 630 kilometers per hour. The time 𝑡 is eight hours.
But look at this. This equation lets us solve for speed. However, it’s distance that we want to know. What we’ll do is rearrange this equation. We multiply both sides by the time 𝑡. Since we’re multiplying each side by the same value, we’re not changing the equation. By doing this though, the 𝑡 in the numerator on the right and in the denominator cancel out. We end up with this equation. Distance 𝑑 equals time 𝑡 multiplied by speed 𝑣. And we know 𝑡 and 𝑣, so we can substitute those values in. The distance this aircraft moves is eight hours times 630 kilometers per hour.
And notice something here with the units. In our time of eight hours, we have a unit of hours in the numerator. And this time multiplies a speed with a unit of hours in the denominator. That means when we calculate 𝑑, these units of hours will cancel out. Our final answer will have units of kilometers. When we multiply eight by 630, the result is 5040. And as we said, this has units of kilometers. This distance is how far the aircraft moves.
Let’s look at one more example.
A train travels 160 meters at a uniform speed of eight meters per second. The position of the train at two different times is shown. How much time does the train take to travel the distance between the positions?
In this diagram, we see the train moving a distance of 160 meters while traveling at a uniform speed of eight meters per second. We want to know how long it takes for the train to travel this distance. In general, the speed of an object, 𝑣, is equal to the distance it moves, 𝑑, divided by the time taken to travel that distance, 𝑡.
In our example, we know the train’s speed and distance traveled. Its speed is eight meters per second, and the distance is 160 meters. What we want to solve for is time 𝑡. To begin doing that, let’s rearrange this equation. First, we’ll multiply both sides by the time 𝑡. On the right side of the equation, 𝑡 now appears in the numerator and the denominator. That means it cancels out. And that gives us this equation. And our next step will be to divide both sides by 𝑣. We do this because we want the time 𝑡 all by itself on one side. Notice what happens on the left. The 𝑣 in the numerator and the denominator cancel one another out.
Now, we have an equation where 𝑡 is the subject. It’s equal to distance divided by speed. We know the distance and speed for our train, so we can plug those values in. The time taken for the train equals 160 meters divided by eight meters per second. Eight divided into 160 equals 20. And in the units of meters divided by meters per second, the units of meters will cancel out. We’re left only with units of seconds. The time the train takes to travel 160 meters at a speed of eight meters per second is 20 seconds.
Let’s finish up this lesson by reminding ourselves of a few key points. In this lesson, we saw that speed equals the distance an object travels divided by the time taken to travel that distance. Written as an equation, 𝑣 equals 𝑑 divided by 𝑡, where 𝑣 is speed, 𝑑 is distance, and 𝑡 is time. An object moving with uniform speed travels equal distances in equal times. Lastly, we saw that the equation 𝑣 equals 𝑑 divided by 𝑡 means that 𝑑 equals 𝑣 times 𝑡 and 𝑡 equals 𝑑 divided by 𝑣.
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### Section 5-4 : Cross Product
In this final section of this chapter we will look at the cross product of two vectors. We should note that the cross product requires both of the vectors to be three dimensional vectors.
Also, before getting into how to compute these we should point out a major difference between dot products and cross products. The result of a dot product is a number and the result of a cross product is a vector! Be careful not to confuse the two.
So, let’s start with the two vectors $$\vec a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle$$ and $$\vec b = \left\langle {{b_1},{b_2},{b_3}} \right\rangle$$ then the cross product is given by the formula,
$\vec a \times \vec b = \left\langle {{a_2}{b_3} - {a_3}{b_2},{a_3}{b_1} - {a_1}{b_3},{a_1}{b_2} - {a_2}{b_1}} \right\rangle$
This is not an easy formula to remember. There are two ways to derive this formula. Both of them use the fact that the cross product is really the determinant of a 3x3 matrix. If you don’t know what this is that is don’t worry about it. You don’t need to know anything about matrices or determinants to use either of the methods. The notation for the determinant is as follows,
$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|$
The first row is the standard basis vectors and must appear in the order given here. The second row is the components of $$\vec a$$ and the third row is the components of $$\vec b$$. Now, let’s take a look at the different methods for getting the formula.
The first method uses the Method of Cofactors. If you don’t know the method of cofactors that is fine, the result is all that we need. Here is the formula.
$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}}{{a_2}}&{{a_3}}\\{{b_2}}&{{b_3}}\end{array}} \right|\vec i - \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_3}}\\{{b_1}}&{{b_3}}\end{array}} \right|\vec j + \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}\\{{b_1}}&{{b_2}}\end{array}} \right|\vec k$
where,
$\left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right| = ad - bc$
This formula is not as difficult to remember as it might at first appear to be. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors.
The second method is slightly easier; however, many textbooks don’t cover this method as it will only work on 3x3 determinants. This method says to take the determinant as listed above and then copy the first two columns onto the end as shown below.
$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\{{a_1}}&{{a_2}}\\{{b_1}}&{{b_2}}\end{array}$
We now have three diagonals that move from left to right and three diagonals that move from right to left. We multiply along each diagonal and add those that move from left to right and subtract those that move from right to left.
This is best seen in an example. We’ll also use this example to illustrate a fact about cross products.
Example 1 If $$\vec a = \left\langle {2,1, - 1} \right\rangle$$ and $$\vec b = \left\langle { - 3,4,1} \right\rangle$$ compute each of the following.
1. $$\vec a \times \vec b$$
2. $$\vec b \times \vec a$$
Show All Solutions Hide All Solutions
a $$\vec a \times \vec b$$ Show Solution
Here is the computation for this one.
\begin{align*}\vec a \times \vec b & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\2&1&{ - 1}\\{ - 3}&4&1\end{array}} \right|\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\2&1\\{ - 3}&4\end{array}\\ & = \vec i\left( 1 \right)\left( 1 \right) + \vec j\left( { - 1} \right)\left( { - 3} \right) + \vec k\left( 2 \right)\left( 4 \right) - \vec j\left( 2 \right)\left( 1 \right) - \vec i\left( { - 1} \right)\left( 4 \right) - \vec k\left( 1 \right)\left( { - 3} \right)\\ & = 5\vec i + \vec j + 11\vec k\end{align*}
b $$\vec b \times \vec a$$ Show Solution
And here is the computation for this one.
\begin{align*}\vec b \times \vec a & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 3}&4&1\\2&1&{ - 1}\end{array}} \right|\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\{ - 3}&4\\2&1\end{array}\\ & = \vec i\left( 4 \right)\left( { - 1} \right) + \vec j\left( 1 \right)\left( 2 \right) + \vec k\left( { - 3} \right)\left( 1 \right) - \vec j\left( { - 3} \right)\left( { - 1} \right) - \vec i\left( 1 \right)\left( 1 \right) - \vec k\left( 4 \right)\left( 2 \right)\\ & = - 5\vec i - \vec j - 11\vec k\end{align*}
Notice that switching the order of the vectors in the cross product simply changed all the signs in the result. Note as well that this means that the two cross products will point in exactly opposite directions since they only differ by a sign. We’ll formalize up this fact shortly when we list several facts.
There is also a geometric interpretation of the cross product. First we will let $$\theta$$ be the angle between the two vectors $$\vec a$$ and $$\vec b$$and assume that $$0 \le \theta \le \pi$$, then we have the following fact,
$$$\left\| {\vec a \times \vec b} \right\| = \left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|\,\sin \theta \label{eq:eq1}$$$
and the following figure.
There should be a natural question at this point. How did we know that the cross product pointed in the direction that we’ve given it here?
First, as this figure implies, the cross product is orthogonal to both of the original vectors. This will always be the case with one exception that we’ll get to in a second.
Second, we knew that it pointed in the upward direction (in this case) by the “right hand rule”. This says that if we take our right hand, start at $$\vec a$$ and rotate our fingers towards $$\vec b$$our thumb will point in the direction of the cross product. Therefore, if we’d sketched in $$\vec b \times \vec a$$ above we would have gotten a vector in the downward direction.
Example 2 A plane is defined by any three points that are in the plane. If a plane contains the points $$P = \left( {1,0,0} \right)$$, $$Q = \left( {1,1,1} \right)$$ and $$R = \left( {2, - 1,3} \right)$$ find a vector that is orthogonal to the plane.
Show Solution
The one way that we know to get an orthogonal vector is to take a cross product. So, if we could find two vectors that we knew were in the plane and took the cross product of these two vectors we know that the cross product would be orthogonal to both the vectors. However, since both the vectors are in the plane the cross product would then also be orthogonal to the plane.
So, we need two vectors that are in the plane. This is where the points come into the problem. Since all three points lie in the plane any vector between them must also be in the plane. There are many ways to get two vectors between these points. We will use the following two,
\begin{align*}\overrightarrow {PQ} & = \left\langle {1 - 1,1 - 0,1 - 0} \right\rangle = \left\langle {0,1,1} \right\rangle \\ \overrightarrow {PR} & = \left\langle {2 - 1, - 1 - 0,3 - 0} \right\rangle = \left\langle {1, - 1,3} \right\rangle \end{align*}
The cross product of these two vectors will be orthogonal to the plane. So, let’s find the cross product.
\begin{align*}\overrightarrow {PQ} \times \overrightarrow {PR} & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\0&1&1\\1&{ - 1}&3\end{array}} \right|\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\0&1\\1&{ - 1}\end{array}\\ & = 4\vec i + \vec j - \vec k\end{align*}
So, the vector $$4\vec i + \vec j - \vec k$$ will be orthogonal to the plane containing the three points.
Now, let’s address the one time where the cross product will not be orthogonal to the original vectors. If the two vectors, $$\vec a$$ and $$\vec b$$, are parallel then the angle between them is either 0 or 180 degrees. From $$\eqref{eq:eq1}$$ this implies that,
$\left\| {\vec a \times \vec b} \right\| = 0$
From a fact about the magnitude we saw in the first section we know that this implies
$\vec a \times \vec b = \vec 0$
In other words, it won’t be orthogonal to the original vectors since we have the zero vector. This does give us another test for parallel vectors however.
#### Fact
If $$\vec a \times \vec b = \vec 0$$ then $$\vec a$$ and $$\vec b$$ will be parallel vectors.
Let’s also formalize up the fact about the cross product being orthogonal to the original vectors.
#### Fact
Provided $$\vec a \times \vec b \ne \vec 0$$ then $$\vec a \times \vec b$$ is orthogonal to both $$\vec a$$ and $$\vec b$$.
Here are some nice properties about the cross product.
#### Properties
If $$\vec u$$, $$\vec v$$ and $$\vec w$$ are vectors and $$c$$ is a number then,
\begin{align*} & \vec u \times \vec v = - \vec v \times \vec u & \hspace{0.75in} & \left( {c\vec u} \right) \times \vec v = \vec u \times \left( {c\vec v} \right) = c\left( {\vec u \times \vec v} \right)\\ &\vec u \times \left( {\vec v + \vec w} \right) = \vec u \times \vec v + \vec u \times \vec w & \hspace{0.75in} & \vec u\centerdot \left( {\vec v \times \vec w} \right) = \left( {\vec u \times \vec v} \right)\centerdot \vec w\\ & \vec u\centerdot \left( {\vec v \times \vec w} \right) = \left| {\begin{array}{*{20}{c}}{{u_1}}&{{u_2}}&{{u_3}}\\{{v_1}}&{{v_2}}&{{v_3}}\\{{w_1}}&{{w_2}}&{{w_3}}\end{array}} \right| & & \end{align*}
The determinant in the last fact is computed in the same way that the cross product is computed. We will see an example of this computation shortly.
There are a couple of geometric applications to the cross product as well. Suppose we have three vectors $$\vec a$$, $$\vec b$$ and $$\vec c$$ and we form the three dimensional figure shown below.
The area of the parallelogram (two dimensional front of this object) is given by,
${\rm{Area}} = \left\| {\vec a \times \vec b} \right\|$
and the volume of the parallelepiped (the whole three dimensional object) is given by,
${\rm{Volume}} = \left| {\vec a\centerdot \left( {\vec b \times \vec c} \right)} \right|$
Note that the absolute value bars are required since the quantity could be negative and volume isn’t negative.
We can use this volume fact to determine if three vectors lie in the same plane or not. If three vectors lie in the same plane then the volume of the parallelepiped will be zero.
Example 3 Determine if the three vectors $$\vec a = \left\langle {1,4, - 7} \right\rangle$$, $$\vec b = \left\langle {2, - 1,4} \right\rangle$$ and $$\vec c = \left\langle {0, - 9,18} \right\rangle$$ lie in the same plane or not.
Show Solution
So, as we noted prior to this example all we need to do is compute the volume of the parallelepiped formed by these three vectors. If the volume is zero they lie in the same plane and if the volume isn’t zero they don’t lie in the same plane.
\begin{align*}\vec a\centerdot \left( {\vec b \times \vec c} \right) & = \left| {\begin{array}{*{20}{c}}1&4&{ - 7}\\2&{ - 1}&4\\0&{ - 9}&{18}\end{array}} \right|\,\,\,\,\begin{array}{*{20}{c}}1&4\\2&{ - 1}\\0&{ - 9}\end{array}\\ & = \left( 1 \right)\left( { - 1} \right)\left( {18} \right) + \left( 4 \right)\left( 4 \right)\left( 0 \right) + \left( { - 7} \right)\left( 2 \right)\left( { - 9} \right) - \\ & \hspace{0.75in}\left( 4 \right)\left( 2 \right)\left( {18} \right) - \left( 1 \right)\left( 4 \right)\left( { - 9} \right) - \left( { - 7} \right)\left( { - 1} \right)\left( 0 \right)\\ & = - 18 + 126 - 144 + 36\\ & = 0\end{align*}
So, the volume is zero and so they lie in the same plane. |
# What The Numbers Say About Gabourey Sidibe (03/27/2020)
How will Gabourey Sidibe get by on 03/27/2020 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is for entertainment purposes only – don’t get too worked up about the result. I will first find the destiny number for Gabourey Sidibe, and then something similar to the life path number, which we will calculate for today (03/27/2020). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology experts.
PATH NUMBER FOR 03/27/2020: We will take the month (03), the day (27) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 27 we do 2 + 7 = 9. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 9 + 4 = 16. This still isn’t a single-digit number, so we will add its digits together again: 1 + 6 = 7. Now we have a single-digit number: 7 is the path number for 03/27/2020.
DESTINY NUMBER FOR Gabourey Sidibe: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Gabourey Sidibe we have the letters G (7), a (1), b (2), o (6), u (3), r (9), e (5), y (7), S (1), i (9), d (4), i (9), b (2) and e (5). Adding all of that up (yes, this can get tiring) gives 70. This still isn’t a single-digit number, so we will add its digits together again: 7 + 0 = 7. Now we have a single-digit number: 7 is the destiny number for Gabourey Sidibe.
CONCLUSION: The difference between the path number for today (7) and destiny number for Gabourey Sidibe (7) is 0. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is of questionable accuracy. If you want a reading that people really swear by, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
# GHSGT Number/Comp and Algebra Review - PowerPoint PPT Presentation
GHSGT Number/Comp and Algebra Review
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GHSGT Number/Comp and Algebra Review
## GHSGT Number/Comp and Algebra Review
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1. GHSGT Number/Comp and Algebra Review How to use this review: Play the PPT from the beginning and try each question. Press ‘Enter’ to determine the correct answer. Look for boxes that highlight the main idea of the concept. Return to the SHS GHSGT Prep page to try more review questions. Pass the FIRST time!
2. Number and Computation17% of test
3. Percents, Fractions, Decimals Percent to decimal – decimal point two places to left 50.0 % becomes 0.50 Decimal to percent – decimal point two places to right 1.37 becomes 137. % Fraction to decimal – divide numerator by denominator becomes
4. Represent 45% as a fraction. • A)1/5 • B)2/10 • C)9/20 • D)45/50
5. Express 0.456 as a percent. • A).00456% • B)4.56% • C)45.6% • D)456%
6. Using Percents Interest = principal × rate × time Mr. Sam borrowed \$1000 for 6 months at an annual interest rate of 27 percent. How much interest did he pay?
7. Exponents Multiply terms with exponents by adding exponents Divide terms with exponents by subtracting exponents
8. Exponents • Simplify: p-2p5 • A)p-10 • B)p-3 • C)p3 • D)p7
9. Scientific Notation Huge Numbers Tiny Numbers
10. Estimates • You have a calculator – use it ! • Main idea • Round the numbers in the problem so you can do the math in your head…(or on your fingers) • 12,357 becomes 12,000 • 3.12 becomes 3
11. Properties of Numbers • Remember the Names • Commutative “commute”3 + 5 = 5 + 3 • Associative “associate at a party in different groups” 4(2x + 3) = (2x + 3)4 • Distributive 3(5 + 1) = 3 × 5 + 3 × 1 • Identity “mirror #” (Addition) -3 + 0 = -3 • Identity “mirror #” (Multiplication) 5 × 1 = 5
12. Rules for Integers Use the calculator!!! • Adding • 2 positives = positive 2 negatives = negative • Subtracting • Subtraction is just adding the opposite sign • 3 – -5 = 3 + -(-5) = 3 + 5 = 8 • MultiplyingandDividing 2 positives or 2 negative = positive 1 positive and 1 negative = negative
13. Order of Operations P Parenthesis E Exponents M, D Multiply & divide A, S Add & subtract
14. Simplify: 2 3 + 4 - 62 • A) -26 • B) -22 • C) -2 • D) 10
15. Simplify Algebraic Expressions Combine like terms
16. Algebra Review29% of test
17. Simplify • Add: (3t - 3) + (6t2 + 4t + 13) + (-8t2 + 11t - 1)? • A)-2t2 +18t + 9 • B) -2t4 + 182 + 9 • C)2t4 -18t – 9 • D) 25t7
18. Evaluate Algebraic Expressions Evaluate an algebraic expression means to substitute a number for each variable and then carry out the indicated operations – remember to follow the order of operations.
19. If r = -6, s = -2, t = -3 v = 10, and w = -1/2 then wv|-t - s| = • A) -25 • B) -5 • C) 5 • D) 25
20. Write Algebraic Expressions Translate word for word into algebraic symbols The sum of a and b multiplied by two-thirds A number divided by two is equal to two-thirds of another number
21. Which represents the sentence twenty less than triple a number is thirty? • 20 – 3x = 30 • 20 – 3x > 30 • 3x - 20 < 30 • 3x – 20 = 30
22. Solve Equations Keep it balancedPerform the opposite operations
23. Solve the equation for x. • -16 • 16 • -32/3 • 32/3
24. Solve for x • Solve the quadratic equation: 4x2 - 9 = 0 • A) 3/2 • B) -3/2 • C) 9/4 and -9/4 • D) 3/2and -3/2 If you’re clueless, just use your calculator to plug in all answers for x! Remember x2 equations have TWO solutions!
25. Linear Equations- intercept method y intercept (x = 0) x intercept (y = 0)
26. Linear Equations – Slope positive undefined negative zero
27. -3 0 Solve and Graph Inequalities Same as equations EXCEPT when multiply or divide by a a negative number flip the inequality
28. 6x - 2(x + 1) > 0 • Solve the inequality. • A) x > 2 • B) x < 2 • C) x < 1/2 • D) x > 1/2
29. Solve. • Solve the inequality.-2x + 5 ≥ 3x + 20 • A) x ≤ 3 • B) x≥ 3 • C) x ≤ -3 • D) x≥ -3
30. Systems of Linear Equations Solution Main Idea The solution to a system of equations is the point where the 2 lines intersect When you substitute the solution, it must make BOTH equations true !!
31. Identify the solution to the system of equations graphed. • (0, 1) • (1 -1) • (0, -3) • (-1, 1)
32. Systems of Equations – Substitution substitute
33. x = y - 3x + 3y = 13 • What is the solution to the system of equations? A) (1, 4) • B) (4, 1) • C) (2.5, 5.5) • D) (5.5, 2.5)
34. Systems of Equations – Elimination eliminate y add both equations
35. 2x - 3y = 16 5x - 3y = 13 What is the solution to the system of equations? A) (-1, -6) B) (-6, -1) C) (5, -2) D) (5, 4)
36. A BCD
37. Note: This PPT was trimmed from an original document on the web by Hillgrove HS in Cobb County, GA. I have made particular additions and subtractions due to our school’s needs. The original home of the PPT is: http://www.cobbk12.org/schools/Hillgrove/GHSGT%20REVEIWS/Math%20GHSGT.htm Mr. White 3/2009 |
# Lesson 8
Multiplying Expressions
These materials, when encountered before Algebra 1, Unit 7, Lesson 8 support success in that lesson.
## 8.1: Math Talk: Combining the Similar Numbers (5 minutes)
### Warm-up
The purpose of this Math Talk is to elicit strategies and understandings students have for combining the same number in a variety of ways. These understandings help students develop fluency and will be helpful later in the supported Algebra 1 lesson when students will need to be able to factor quadratic expressions.
### Launch
Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole class discussion.
### Student Facing
Evaluate mentally.
$$100 \boldcdot 100$$
$$\text{-}3 \boldcdot 3$$
$$\text{-}300 + 300$$
$$1,\!279 + \text{-}1,\!279$$
### Activity Synthesis
Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:
• “Who can restate $$\underline{\hspace{.5in}}$$’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to $$\underline{\hspace{.5in}}$$’s strategy?”
• “Do you agree or disagree? Why?”
## 8.2: A Method for Multiplying (15 minutes)
### Activity
In this activity, students learn and practice a method for multiplying two values of the form $$(10^n - a)(10^n+a)$$. Through the discussion, students should begin to recognize that two of the terms from the expanded form are opposites. In the associated algebra lesson, students examine quadratic expressions that can be factored into the form $$(x-a)(x+a)$$.
### Student Facing
Here is a method for multiplying 97 and 103:
97 is $$100 - 3$$
103 is $$100 + 3$$
So $$97 \boldcdot 103 = (100-3)(100+3)$$
100 -3 100 10,000 -300 3 300 -9
1. Explain how this diagram is used to compute $$97 \boldcdot 103 = 9,\!991$$.
2. Draw a similar diagram that helps you mentally compute $$(30+1)(30-1)$$. What is the result? What multiplication problem did you just solve?
3. Use this method to compute:
1. $$7 \boldcdot 13$$
2. $$102 \boldcdot 98$$
3. $$995 \boldcdot 1,\!005$$
4. Create a challenge problem for your partner, that could use this method. Create a diagram that shows the answer before giving the problem to your partner.
### Activity Synthesis
The purpose of the discussion is to notice the pattern that two of the cells in the table are opposites when the multiplication is done for numbers in this form. Select students to share their solutions and challenge problems. Ensure that students are working with numbers in the form $$(10^n - a)(10^n+a)$$.
• “What do you notice about the values inside the table after you multiply the appropriate values?” (The number on the bottom left and top right are opposites.)
• “Although this method can be used to multiply any numbers, why is this setup easier to work with than something like $$97 \boldcdot 105$$?” (With other values, there will not be 2 parts of the table that are opposites, so the entire table would need to be filled in and summed. When the values are like the ones in the activity, only the top left and bottom right of the table need to be combined.)
## 8.3: Find the Missing Pieces (20 minutes)
### Activity
In this activity, students find missing parts of a multiplication diagram. This strategy will be useful in the associated Algebra 1 lesson when they factor polynomials. Students look for and make use of structure (MP7) when they use tables to understand the relationships between 2 numbers.
### Student Facing
Complete each diagram. Write some equivalent expressions based on the diagram.
1. 10 5 10 100 45
2. 7 10 -7 -70
3. $$x$$ 8 $$x$$ -8
4. $$a$$ -9 $$\text{-}9a$$ 9
5. $$b$$ $$\frac12$$ $$b$$ $$b^2$$ $$\text{-}\frac{1}{4}$$
6. 7 $$c$$ $$\text{-}c^2$$ 7 49 |
How do you simplify (4y-9)-(3y+10)?
Apr 16, 2016
$y - 19$
Explanation:
A place where students commonly make mistakes with is the terms after the subtraction sign, in this case $3 y + 10$.
Pretend that there is a negative 1 in front of $\left(3 y + 10\right)$. So you would have: $- 1 \left(3 y + 10\right)$
Distribute that negative one to the terms in the parenthesis, to get $- 1 \left(3 y\right) + - 1 \left(10\right) = - 3 y - 10$
Combine that with the remaining expression and then simplify.
$4 y - 9 - 3 y - 10$
Combine like terms:
$\left(4 y - 3 y\right) = y$
$\left(- 9 - 10\right) = - 19$
Put those together to get the answer of:
$y - 19$ |
# Warm Up 09.28.11 Week 7 1) What is the postulate? A B C D m∠ ADB + m ∠ BDC = m ∠ ADC 2) If ∠ 4 and ∠ 5 are a linear pair and ∠ 4 = 79⁰. What is m ∠ 5?
## Presentation on theme: "Warm Up 09.28.11 Week 7 1) What is the postulate? A B C D m∠ ADB + m ∠ BDC = m ∠ ADC 2) If ∠ 4 and ∠ 5 are a linear pair and ∠ 4 = 79⁰. What is m ∠ 5?"— Presentation transcript:
Warm Up 09.28.11 Week 7 1) What is the postulate? A B C D m∠ ADB + m ∠ BDC = m ∠ ADC 2) If ∠ 4 and ∠ 5 are a linear pair and ∠ 4 = 79⁰. What is m ∠ 5?
Geometry 2.4 Day 1 I will use the properties of algebra to write reasons for steps in solving equations. Properties of Equality Addition: If a = b then a + c = b + c Subtraction: If a = b then a - c = b - c
Multiplication: Division: If a = b then ac = bc If a = b then cc = a b Properties of Equality
Substitution: Distributive: If a = b then b can replace a. If a( b + c ) = ab + ac Properties of Equality
Ex 1 55z – 3( 9z + 12 ) = -64 Solve and give reasons: StepReason 55z – 3( 9z + 12 ) = -64 Given 55z - 27z - 36 = -64Distributive property 28z - 36 = -64Simplify 28z = -28 Addition property of equality z = -1 Division property of equality
Properties of equality for Segment length Reflexive: Symmetric: AB = AB Transitive: If AB = CD, then CD = AB If AB = CD, and CD = EF, then AB = EF. Segment Length
Properties of equality for angle measure Reflexive: Symmetric: Transitive: m∠ A = m ∠ A If m∠ A = m ∠ B, then m ∠ B = m∠ A If m∠ A = m ∠ B, and m ∠ B = m∠ C then m∠ A = m∠ C Angle Measure
Ex 2 m ∠ 1 + m ∠ 2 = 132⁰ m ∠ 2 = 105⁰ 2 1 StepReason m ∠ 1 + m ∠ 2 = 132⁰ Given m ∠ 2 = 105⁰ Given show that m ∠ 1 = 27⁰
Ex 1 m ∠ 1 + m ∠ 2 = 132⁰ m ∠ 2 = 105⁰ 2 1 StepReason m ∠ 1 + m ∠ 2 = 132⁰ Given m ∠ 2 = 105⁰ Given show that m ∠ 1 = 27⁰
Ex 3 m ∠ 1 + m ∠ 2 = 132⁰ m ∠ 2 = 105⁰ 2 1 StepReason m ∠ 1 + m ∠ 2 = 132⁰ Given m ∠ 2 = 105⁰ Given m ∠ 1 + 105⁰ = 132⁰ Substitution property of Eq. m ∠ 1 = 27⁰ Subtraction property of Eq. show that m ∠ 1 = 27⁰
Do: 1 5x + 7y = 58 Assignment: Textbook Page 99, 10 - 27 All What is the property when 5x + 7y = 58 5( 27 ) + 7y = 58 ? given and x = 27
Download ppt "Warm Up 09.28.11 Week 7 1) What is the postulate? A B C D m∠ ADB + m ∠ BDC = m ∠ ADC 2) If ∠ 4 and ∠ 5 are a linear pair and ∠ 4 = 79⁰. What is m ∠ 5?"
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# Chapter 7: Algebraic Expressions Exercise – 7.2
### Question: 1
(i) 3x and 7x
(ii) -5xy and 9xy
### Solution:
We have
(i) 3x + 7x = (3 + 7) x = 10x
(ii) -5xy + 9xy = (-5 + 9)xy = 4xy
### Question: 2
Simplify each of the following:
(i) 7x3y +9yx3
(ii) 12a2b + 3ba2
### Solution:
Simplifying the given expressions, we have
(i) 7x3y + 9yx3 = (7 + 9)x3y = 16x3y
(ii) 12a2b + 3ba2 = (12 + 3)a2b =15a2b
### Question: 3
(i) 7abc, -5abc, 9abc, -8abc
(ii) 2x2y, - 4x2y, 6x2y, -5x2y
### Solution:
Adding the given terms, we have
(i) 7abc + (-5abc) + (9abc) + (-8abc)
= 7abc – 5abc + 9abc – 8abc
= (7 – 5 + 9 – 8)abc
= (16 – 13)abc
= 3abc
(ii) 2x2y +(-4x2y) + (6x2y) + (-5x2y)
= 2x2y - 4x2y + 6x2y - 5x2y
= (2- 4 + 6 - 5) x 2y
= (8 - 9) x 2y
= -x2y
### Question: 4
(i) x3 -2x2y + 3xy2- y3, 2x3- 5xy2 + 3x2y - 4y3
(ii) a4 - 2a3b + 3ab3 + 4a2b2 + 3b4, - 2a4 - 5ab3 + 7a3b - 6a2b2 + b4
### Solution:
Adding the given expressions, we have
(i) x3 -2x2y + 3xy2- y3, 2x3- 5xy2 + 3x2y - 4y3
Collecting positive and negative like terms together, we get
x3 +2x3 - 2x2y + 3x2y + 3xy2 - 5xy2 - y3- 4y3
= 3x3 + x2y - 2xy2 - 5y3
(ii) a4 - 2a3b + 3ab3 + 4a2b2 + 3b4, - 2a4 - 5ab3 + 7a3b - 6a2b2 + b4
a4 - 2a3b + 3ab3 + 4a2b2 + 3b4 - 2a4 - 5ab3 + 7a3b - 6a2b2 + b4
Collecting positive and negative like terms together, we get
a4 - 2a4- 2a3b + 7a3b + 3ab3 - 5ab3 + 4a2b2 - 6a2b2 + 3b4 + b4
= - a4 + 5a3b - 2ab3 - 2a2b2 + 4b4
### Question: 5
(i) 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b
(ii) 5x3 + 7 + 6x - 5x2, 2x2 – 8 - 9x, 4x - 2x2 + 3 x 3, 3 x 3 - 9x - x2 and x - x2 - x3 - 4
### Solution:
(i) Required expression = (8a – 6ab + 5b) + (–6a – ab – 8b) + (–4a + 2ab + 3b)
Collecting positive and negative like terms together, we get
8a – 6a – 4a – 6ab – ab + 2ab + 5b – 8b + 3b
= 8a – 10a – 7ab + 2ab + 8b – 8b
= –2a – 5ab
(ii) Required expression = (5 x 3 + 7+ 6x - 5x2) + (2 x 2 – 8 - 9x) + (4x - 2x2 + 3 x 3) + (3 x 3 - 9x-x2) + (x - x2 - x3 - 4)
Collecting positive and negative like terms together, we get
5x3 + 3x3 + 3x3 - x3 - 5x2 + 2x2 - 2x2- x2 - x2 + 6x - 9x + 4x - 9x + x + 7 – 8 - 4
= 10x3 - 7x2 - 7x - 5
### Question: 6
(i) x – 3y – 2z
5x + 7y – 8z
3x – 2y + 5z
(ii) 4ab – 5bc + 7ca
–3ab + 2bc – 3ca
5ab – 3bc + 4ca
### Solution:
(i) Required expression = (x – 3y – 2z) + (5x + 7y – 8z) + (3x – 2y + 5z)
Collecting positive and negative like terms together, we get
x + 5x + 3x – 3y + 7y – 2y – 2z – 8z + 5z
= 9x – 5y + 7y – 10z + 5z
= 9x + 2y – 5z
(ii) Required expression = (4ab – 5bc + 7ca) + (–3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)
Collecting positive and negative like terms together, we get
4ab – 3ab + 5ab – 5bc + 2bc – 3bc + 7ca – 3ca + 4ca
= 9ab – 3ab – 8bc + 2bc + 11ca – 3ca
= 6ab – 6bc + 8ca
### Question: 7
Add 2x2 - 3x + 1 to the sum of 3x2 - 2x and 3x + 7.
### Solution:
Sum of 3x2 - 2x and 3x + 7
= (3x2 - 2x) + (3x +7)
=3x2 - 2x + 3x + 7
= (3x2 + x + 7)
Now, required expression = 2x2 - 3x + 1+ (3x2 + x + 7)
= 2x2 + 3x2 - 3x + x + 1 + 7
= 5x2 - 2x + 8
### Question: 8
Add x2 + 2xy + y2 to the sum of x2 - 3y2and 2x2 - y2 + 9.
### Question: 9
Add a3+ b3 - 3 to the sum of 2a3 - 3b3 - 3ab + 7 and -a3 + b3 + 3ab - 9.
### Question: 10
Subtract:
(i) 7a2b from 3a2b
(ii) 4xy from -3xy
### Solution:
(i) Required expression = 3a2b -7a2b
= (3 -7)a2b
= - 4a2b
(ii) Required expression = –3xy – 4xy
= –7xy
### Question: 11
Subtract:
(i) - 4x from 3y
(ii) - 2x from – 5y
### Solution:
(i) Required expression = (3y) – (–4x)
= 3y + 4x
(ii) Required expression = (-5y) – (–2x)
= –5y + 2x
### Question: 12
Subtract:
(i) 6x3 −7x2 + 5x − 3 from 4 − 5x + 6x2 − 8x3
(ii) − x2 −3z from 5x2 – y + z + 7
(iii) x3 + 2x2y + 6xy2 − y3 from y3−3xy2−4x2y
### Question: 13
From
(i) p3 – 4 + 3p2, take away 5p2 − 3p3 + p − 6
(ii) 7 + x − x2, take away 9 + x + 3x2 + 7x3
(iii) 1− 5y2, take away y3 + 7y2 + y + 1
(iv) x3 − 5x2 + 3x + 1, take away 6x2 − 4x3 + 5 + 3x
### Question: 14
From the sum of 3x2 − 5x + 2 and − 5x2 − 8x + 9 subtract 4x2 − 7x + 9.
### Question: 15
Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7.
### Solution:
Sum of (13x – 4y + 7z) and (–6z + 6x + 3y)
= (13x – 4y + 7z) + (–6z + 6x + 3y)
= (13x – 4y + 7z – 6z + 6x + 3y)
= (13x + 6x – 4y + 3y + 7z – 6z)
= (19x – y + z)
Sum of (6x – 4y – 4z) and (2x + 4y – 7)
= (6x – 4y – 4z) + (2x + 4y – 7)
= (6x – 4y – 4z + 2x + 4y – 7)
= (6x + 2x – 4z – 7)
= (8x – 4z – 7)
Now, required expression = (8x – 4z – 7) – (19x – y + z)
= 8x – 4z – 7 – 19x + y – z
= 8x – 19x + y – 4z – z – 7
= –11x + y – 5z – 7
### Question: 16
From the sum of x2 + 3y2 − 6xy, 2x2 − y2 + 8xy, y2 + 8 and x2 − 3xy subtract −3x2 + 4y2 – xy + x – y + 3.
### Question: 17
What should be added to xy – 3yz + 4zx to get 4xy – 3zx + 4yz + 7?
### Solution:
The required expression can be got by subtracting xy – 3yz + 4zx from 4xy – 3zx + 4yz + 7.
Therefore, required expression = (4xy – 3zx + 4yz + 7) – (xy – 3yz + 4zx)
= 4xy – 3zx + 4yz + 7 – xy + 3yz – 4zx
= 4xy – xy – 3zx – 4zx + 4yz + 3yz + 7
= 3xy – 7zx + 7yz + 7
### Question: 18
What should be subtracted from x2 – xy + y2 – x + y + 3 to obtain −x2 + 3y2 − 4xy + 1?
### Question: 19
How much is x – 2y + 3z greater than 3x + 5y – 7?
### Solution:
Required expression = (x – 2y + 3z) – (3x + 5y – 7)
= x – 2y + 3z – 3x – 5y + 7
Collecting positive and negative like terms together, we get
x – 3x – 2y + 5y + 3z + 7
= –2x – 7y + 3z + 7
### Question: 20
How much is x2 − 2xy + 3y2 less than 2x2 − 3y2 + xy?
### Question: 21
How much does a2 − 3ab + 2b2 exceed 2a2 − 7ab + 9b2?
### Question: 22
What must be added to 12x3 − 4x2 + 3x − 7 to make the sum x3 + 2x2 − 3x + 2?
### Question: 23
If P = 7x2 + 5xy − 9y2, Q = 4y2 − 3x2 − 6xy and R = −4x2 + xy + 5y2, show that P + Q + R = 0.
### Question: 24
If P = a2 − b2 + 2ab, Q = a2 + 4b2 − 6ab, R = b2 + b, S = a2 − 4ab and T = −2a2 + b2 – ab + a. Find P + Q + R + S – T.
### Solution:
You Might Like to Refer: |
# Math Insight
### Limits at infinity
Next, let's consider $$\lim_{x\rightarrow \infty} {2x+3\over 5-x}$$ The hazard here is that $\infty$ is not a number that we can do arithmetic with in the normal way. Don't even try it. So we can't really just ‘plug in’ $\infty$ to the expression to see what we get.
On the other hand, what we really mean anyway is not that $x$ ‘becomes infinite’ in some mystical sense, but rather that it just ‘gets larger and larger’. In this context, the crucial observation is that, as $x$ gets larger and larger, $1/x$ gets smaller and smaller (going to $0$). Thus, just based on what we want this all to mean, $$\lim_{x\rightarrow \infty} {1\over x}=0$$ $$\lim_{x\rightarrow \infty} {1\over x^2}=0$$ $$\lim_{x\rightarrow \infty} {1\over x^3}=0$$ and so on.
This is the essential idea for evaluating simple kinds of limits as $x\rightarrow\infty$: rearrange the whole thing so that everything is expressed in terms of $1/x$ instead of $x$, and then realize that $$\lim_{x\rightarrow \infty}\;\;\;\;\hbox{ is the same as }\;\;\;\;\lim_{{1\over x}\rightarrow 0}$$
So, in the example above, divide numerator and denominator both by the largest power of $x$ appearing anywhere: $$\lim_{x\rightarrow \infty} {2x+3\over 5-x}= \lim_{x\rightarrow \infty} {2+{3\over x} \over {5\over x}-1}= \lim_{y\rightarrow 0} {2+3y\over 5y-1}={2+3\cdot 0\over 5\cdot 0-1}=-2$$
The point is that we called $1/x$ by a new name, ‘$y$’, and rewrote the original limit as $x\rightarrow \infty$ as a limit as $y\rightarrow 0$. Since $0$ is a genuine number that we can do arithmetic with, this brought us back to ordinary everyday arithmetic. Of course, it was necessary to rewrite the thing we were taking the limit of in terms of $1/x$ (renamed ‘$y$’).
Notice that this is an example of a situation where we used the letter ‘$y$’ for something other than the name or value of the vertical coordinate.
#### Exercises
1. Find $\lim _{x\rightarrow\infty} { x+1 \over x^2+3 }$.
2. Find $\lim _{x\rightarrow\infty} { x^2+3 \over x+1 }$.
3. Find $\lim _{x\rightarrow\infty} { x^2+3 \over 3x^2+x+1 }$.
4. Find $\lim _{x\rightarrow\infty} { 1-x^2 \over 5x^2+x+1 }$. |
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# If one root of the polynomial $k{{x}^{2}}-15x+18=0$ is 2, then find the value of k?
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381.6k+ views
Hint:
We are given that x = 2 is the root of the given equation $k{{x}^{2}}-15x+18=0$. So, x = 2 will satisfy this equation. Substitute x = 2 in the equation and find the value of k.
Here, we are given that one of the roots of the polynomial $k{{x}^{2}}-15x+18=0$ is 2 then we have to find the value of k.
First of all, we must know that roots or zeros of the given equation satisfy it or in other words, we can say that by substituting the value of roots in place of a variable in any equation, the equation becomes zero. Here variable is x, in terms of which that equation of polynomial is written.
Let us consider the polynomial given in this question.
$k{{x}^{2}}-15x+18=0....\left( i \right)$
We are given that 2 is the root of this polynomial. This means that x = 2 will satisfy this polynomial.
By substituting x = 2 in equation (i), we get,
$k{{\left( 2 \right)}^{2}}-15\left( 2 \right)+18=0$
By simplifying the above equation, we get,
\begin{align} & 4k-30+18=0 \\ & \Rightarrow 4k-12=0 \\ \end{align}
By adding 12 on both sides of the above equation, we get,
$4k-12+12=12$
Or, $4k=12$
By dividing 4 on both sides of the equation, we get,
$\dfrac{4k}{4}=\dfrac{12k}{4}$
Or $k=\dfrac{12}{4}$
By simplifying the RHS of the above equation, we get,
$\Rightarrow k=3$
So, we get the value of k equal to 3.
Note:
In this question, we can cross-check our answer as follows:
Let us consider the polynomial $k{{x}^{2}}-15x+18=0$
By substituting k = 3 and x = 2 in the above polynomial, we get,
$\left( 3 \right){{\left( 2 \right)}^{2}}-15\left( 2 \right)+18=0$
By simplifying the above equation, we get,
$3\left( 4 \right)-30+18=0$
$\Rightarrow 12-30+18=0$
$\Rightarrow 0=0$
LHS = RHS
Since, LHS = RHS, therefore our answer is correct. |
# How to do this math problem step by step
This can be a great way to check your work or to see How to do this math problem step by step. Keep reading to learn more!
## How can we do this math problem step by step
Read on for some helpful advice on How to do this math problem step by step easily and effectively. The binomial solver can be used to solve linear equations, quadratic equations, and polynomial equations. The binomial solver is a versatile tool that can be used to solve many different types of equations. The binomial solver is a useful tool for solving equations that contain two variables.
For example, if you have the equation 2^x=8, you can take the logarithm of both sides to get: log(2^x)=log(8). This can be rewritten as: x*log(2)=log(8). Now all you need to do is solve for x, and you're done! With a little practice, solving for exponents will become second nature.
How to solve for roots: There are several ways to solve for roots, or zeros, of a polynomial function. The most common method is factoring. To factor a polynomial, one expands it into the product of two linear factors. This can be done by grouping terms, by difference of squares, or by completing the square. If the polynomial cannot be factored, then one may use synthetic division to divide it by a linear term. Another method that may be used is graphing. Graphing can show where the function intersects the x-axis, known as the zeros of the function. Graphing can also give an approximate zero if graphed on a graphing calculator or computer software with accuracy parameters. Finally, numerical methods may be used to find precise zeros of a polynomial function. These include Newton's Method, the Bisection Method, and secant lines. Knowing how to solve for roots is important in solving many real-world problems.
It is usually written with an equals sign (=) like this: 4 + 5 = 9. This equation says that the answer to 4 + 5 (9) is equal to 9. So, an equation is like a puzzle, and solving it means finding the value of the missing piece. In the above example, the missing piece is the number 4 (because 4 + 5 = 9). To solve an equation, you need to figure out what goes in the blank space. In other words, you need to find the value of the variable. In algebra, variables are often represented by letters like x or y. So, an equation like 2x + 3 = 7 can be read as "two times x plus three equals seven." To solve this equation, you would need to figure out what number multiplied by 2 and added to 3 would give you 7. In this case, it would be x = 2 because 2 * 2 + 3 = 7. Of course, there are many different types of equations, and some can be quite challenging to solve. But with a little practice, you'll be solving equations like a pro in no time!
The base is typically 10, but it can also be other values, such as 2 or e. Once the base is determined, one can use algebra to solve for the unknown variable. For example, if the equation is log_10(x)=2, then one can solve for x by raising 10 to the 2nd power, which gives a value of 100. Logarithmic functions are powerful tools that can be used to solve a variety of problems. With a little practice, anyone can learn how to solve them.
## Math solver you can trust
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# Formulas For Arrangements
## Arrangements Formulas :
In arrangements questions, we have to arrange a group of persons fulfilling certain conditions. This is also written as sitting arrangement or sitting arrangement reasoning at some places.
PrepInsta will provide you Formulas to solve Arrangements Question easily. Go through the page in detailed to know more about Arrangement Formulas.
Here we can classify these problems into 4 types:
• Type 1 Linear Arrangement
• Type 2 Double Row Arrangement
• Type 3 Circular Arrangement
• Type 4 Rectangular Arrangement
## Formulas For Arrangements
#### Type 1. Linear Arrangement:
Here the arrangement of the persons is linear i.e. you have to arrange them in a line. Here generally a single row of arrangement is formed.
#### Type 2. Double Row Arrangement:
In these questions, there will be two groups of persons. You have to arrange one group in one row and the other group in other row. The persons in these rows normally face each other.
#### Type 3. Circular Arrangement:
In the circular seating arrangement questions, you have to arrange the persons around a circular table etc. fulfilling certain conditions.
#### Type 4. Rectangular Arrangement:
These arrangements are almost similar to the circular arrangements; the only difference is that the persons are sitting around a rectangular table.
Formulas for Arrangements
Eleven friends M, N, O, P, Q, R, S, T, U, V and W are sitting in the first row of the stadium watching a cricket match.
T is to the immediate left of P and third to the right of U.
V is the immediate neighbour of M and N and third to the left of S.
M is the second to the right of Q, who is at one of the ends.
R is sitting next to the right of P and P is second to the right of O.
Question 1:
Who is sitting in the center of the row?
(A) N
(B) O
(C) S
(D) U
Solution: Option D
The arrangement of the persons is
Q W M V N U S O T P R
U is sitting in the center of the row.
Question 2:
Which of the following people are sitting to the right of S?
(A) OTPQ
(B) OTPR
(C) UNVM
(D) UOTPR
Solution: Option B
The arrangement of the persons is
Q W M V N U S O T P R
OTPR are sitting to the right of S.
Question 3:
Which of the following statements is true with respect to the above arrangement?
1. There are three persons sitting between P and S
2. W is between M and V.
3. N is sitting between V and U
4. S and O are neighbours sitting to the immediate right of T
(A) 1
(B) 2
(C) 3
(D) 4
Solution: Option C
The arrangement of the persons is
Q W M V N U S O T P R
Hence N is sitting between V and U.
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Here is the most efficient way to answer GMAT Quant questions that involve calculations based on the median of a set of values.
Consider the following question from a GMAT Tutor lesson:
If in a certain sequence of consecutive multiples of 50, the median is 625, and the greatest term is 950, how many terms that are smaller than 625 are there in the sequence?
A) 6
B) 7
C) 8
D) 12
E) 13
## How to use the median to solve this question
Remember that by definition, the median is equidistant from the first and last values in a sequence.
• Calculate the difference between the last value and median
• 950 – 625 = 325
• Subtract 325 from the median to get the first value in the sequence
• 625 – 325 = 300
The multiples of 50 that are smaller than 625 are therefore 300, 350, 400, 450, 500, 550 and 600 – a total of 7. Hence, B is the answer.
## How to spot the trap in these questions
If you were not paying attention to detail, you may have thought that 625 was a value in the sequence. You could quickly have calculated that there are 13 multiples of 50 between 300 and 950 (a difference of 650). You may have then thought that there would be six multiples of 50 below the median and 6 above the median. In that case you would have answered A.
Note that while 625 is the median of the sequence, it is not actually a value in the sequence. It is the average of the 6th and 7th values in the sequence – the average of 600 and 650. You would recognize this quickly because 625 is not a multiple of 50.
Be particularly careful on test day if you have a tendency to quickly land on answer choices like A. The test makers are often trying to get you to answer questions too quickly. When they are doing that, they will often put the incorrect answer choice right in front of your eyes – Answer A.
This was a sample of the in-depth instruction that GMAT Tutor offers about solving questions in the GMAT Quant section. For complete and interactive lessons and online tutor support, subscribe to one of GMAT Tutor's top-rated GMAT prep plans. Commitment-free trials are available for seven days. |
1.2 Physical quantities and units (Page 6/18)
Page 6 / 18
(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows:
which are obviously not the desired units of km/h.
(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units.
(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect.
(4) Next, check whether the answer is reasonable. Let us consider some information from the problem—if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.
Solution for (b)
There are several ways to convert the average speed into meters per second.
(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed—one to convert hours to seconds, and another to convert kilometers to meters.
(2) Multiplying by these yields
$\text{Average speed}=\text{30}\text{.}0\frac{\text{km}}{\text{h}}×\frac{1\phantom{\rule{0.25em}{0ex}}\text{h}}{\text{3,600 s}}×\frac{1,\text{000}\phantom{\rule{0.25em}{0ex}}\text{m}}{\text{1 km}}\text{,}$
$\text{Average speed}=8\text{.}\text{33}\frac{\text{m}}{\text{s}}\text{.}$
Discussion for (b)
If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s.
You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces? The module Accuracy, Precision, and Significant Figures will help you answer these questions.
Nonstandard units
While there are numerous types of units that we are all familiar with, there are others that are much more obscure. For example, a firkin is a unit of volume that was once used to measure beer. One firkin equals about 34 liters. To learn more about nonstandard units, use a dictionary or encyclopedia to research different “weights and measures.” Take note of any unusual units, such as a barleycorn, that are not listed in the text. Think about how the unit is defined and state its relationship to SI units.
Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a hummingbird to beat its wings once. Which fundamental unit should the scientist use to describe the measurement? Which factor of 10 is the scientist likely to use to describe the motion precisely? Identify the metric prefix that corresponds to this factor of 10.
The scientist will measure the time between each movement using the fundamental unit of seconds. Because the wings beat so fast, the scientist will probably need to measure in milliseconds, or ${\text{10}}^{-3}$ seconds. (50 beats per second corresponds to 20 milliseconds per beat.)
One cubic centimeter is equal to one milliliter. What does this tell you about the different units in the SI metric system?
The fundamental unit of length (meter) is probably used to create the derived unit of volume (liter). The measure of a milliliter is dependent on the measure of a centimeter.
Summary
• Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements.
• Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four fundamental units.
• The four fundamental units we will use in this text are the meter (for length), the kilogram (for mass), the second (for time), and the ampere (for electric current). These units are part of the metric system, which uses powers of 10 to relate quantities over the vast ranges encountered in nature.
• The four fundamental units are abbreviated as follows: meter, m; kilogram, kg; second, s; and ampere, A. The metric system also uses a standard set of prefixes to denote each order of magnitude greater than or lesser than the fundamental unit itself.
• Unit conversions involve changing a value expressed in one type of unit to another type of unit. This is done by using conversion factors, which are ratios relating equal quantities of different units.
Conceptual questions
Identify some advantages of metric units.
Problems&Exercises
The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?
1. $\text{27}\text{.}\text{8 m/s}$
2. $\text{62}\text{.}\text{1 mph}$
A car is traveling at a speed of $\text{33 m/s}$ . (a) What is its speed in kilometers per hour? (b) Is it exceeding the $\text{90 km/h}$ speed limit?
Show that $1\text{.}\text{0 m/s}=3\text{.}\text{6 km/h}$ . Hint: Show the explicit steps involved in converting $1\text{.}\text{0 m/s}=3\text{.}\text{6 km/h.}$
$\frac{\text{1.0 m}}{s}=\frac{1\text{.}\text{0 m}}{s}×\frac{\text{3600 s}}{\text{1 hr}}×\frac{1 km}{\text{1000 m}}$
$=3\text{.}\text{6 km/h}$ .
American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)
length: $\text{377 ft}$ ; width: ; .
What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)
Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)
$8\text{.}\text{847 km}$
The speed of sound is measured to be $\text{342 m/s}$ on a certain day. What is this in km/h?
Tectonic plates are large segments of the Earth's crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1 s at this speed? (b) What is its speed in kilometers per million years?
(a)
(b) $\text{40 km/My}$
(a) Refer to [link] to determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilometers per second. (b) What is this in meters per second?
Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
for the answer to complete, the units need specified why
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
Igor
what is angular velocity
angular velocity can be defined as the rate of change in radian over seconds.
Fidelis
Why does earth exert only a tiny downward pull?
hello
Islam
Why is light bright?
an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
thanks so much. i undersooth well
what is physics
is the study of matter in relation to energy
Kintu
physics can be defined as the natural science that deals with the study of motion through space,time along with its related concepts which are energy and force
Fidelis
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
what is power?
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
did you solve?
Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
what do we call velocity
Kings
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
hi
Godfred
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag |
Quicker Maths
## Trick to Find Square of Numbers from 51 to 59
Posted on August 11, 2011
I’ll share with you one simple method of finding the square of numbers between 50 and 60. Like many other Vedic Mathematics methods, in this method also, we will get the answer in two parts. Since the numbers are in 50s and square of 50 is 2500, we will just use 25 in our calculations, ignoring the zeros.
1. To get the first part of the answer, add the digit at the units place to 25 and write the sum
2. To get the second part, calculate the square of units place digit and write it
It’ll be easier to understand this with an example.
To find the square of 57 –
First part: 25 + digit at units unit in 57 = 25 + 7 = 32
Second part: square of 7 = 49
Combining both the parts – 3249 is the answer.
Another way to look at it can explain you the logic behind this technique – 57 can be written as
(50+7)^2
= 50^2 +2*50*7 + 7^2
= 2500 + 100*7 + 7^2
=100*(25+7) + 7^2
You can replace 7 by any other number in unit’s place and get answer for it.
You may also like:
#### Posted by Vineet Patawari
1. to find square of any no btn 100-150
eg (112)2
1. double unit and tens digit 124
2. multiply by 100 124*100=12400
4. that’s it
2. to find square of any no btn 50-86
eg (62)2
1.62-50=12
2. 25+12=37
3. 37*100=3700
4. 3700 + (12)2=3700+144=3844
5. square of 62=3844
3. to find square of any no btn 36-50
eg (39)2
1. 50-39=11
2. 25-11=14
3. 14*100=1400
4. 1400 + (11)2=1400+121=1521
5. square of 39=1521
4. Square of any two digit number:
(ab)=(a^2+carry)(2ab+carry)(b^2)
Note: Always start from right end.
Example:
63^2=(6^2)(2*6*3)(3^2)
=(36+3)(6)(9)
=3969.
This way u calculate square of any 2 digit number.
5. how to calculate the square of nos from 91-99??
6. how to calculate the square of nos from 91-99?
• Why 91-99 any number, very easy, read my proverb in Hindi dt 24th of FEb 2012:
Rule: Add and substract equal number to make a number in tens. Multiply one with another and add square of the number added/substracted
eg 91^2 = ( 91-1)x (91+1)= (90×92) +1×1 = 8281
Try with any number. Read Rule again in Hindi:
????? ????? ???? ????,
?? ?????? ?? ???? ?????
?? ???? ?? ???? ?? ???? ???,
??? ???? ?? ???? ???? Rule from ” Forget Forgetting” Book
• Sir,
Can u pls tell in english, for those who could’t able to read/understand hindi..
-Skr
7. ???? ?? ??? ?? ????: ???? ???? ?? ????? ????? ???? ????,
?? ?????? ?? ???? ?????
?? ???? ?? ???? ?? ???? ???,
??? ???? ?? ???? ????
??????: 77 ?? ???? ????? ????
77-7=70 ? 77+7=84
77×84=5880 ? ????? 7 ?? ???? ???? 49 ???? ?? ????? 5929 ? ??????? (????? ?????)
???: 7 ?? ?????? ? ?????, ???? ?? ???? ????? ?? ??? ??: ” ?????? ??? ?????” ?????? ?? ???? ???? ?? ??????
8. Cud u plz…tel me thd trick to find the square of the number from 61 to 69 ,from 71 to 79 ,from 81 to 89 nd from 91 to 99
9. Thanx
10. Thanx 4 research
11. For 41 to 49 we shd remember “15″ & “50″.
To find the square of 42 –
First part: 15 + digit at units unit in 42 = 15 + 2 = 17
Second part: square of (50-42) i.e., suare of 8 =64
Combining both the parts – 1764 is the answer
12. thats cool bhai…
13. same you have to do in 41 to 49 number’s squares
• For squares of 41 to 49, the same logic will be used with a small modification….
Let us find the square of 46
Step 1:
First place square + unit place
(4×4) + 6 = 16 + 6 = 22
Step 2:
Unit place square
6×6 = 36
Hence 2236.
Step 3: (Modification)
Now to get the answer, multiply unit place with 20 and deduct that number from the number which we got in step 2
i.e. 2236-(6*20) = 2236-120 = 2116
Therefore 46×46 = 2116. |
# Angle of Banking Calculator
Created by Davide Borchia
Reviewed by Anna Szczepanek, PhD and Rijk de Wet
Last updated: Jun 05, 2023
A turning road is not flat, it's tilted: with our angle of banking calculator, you will discover why!
Some simple physical considerations and some math will be enough to understand something that seems trivial but has consequences for our safety when driving: tilted turns. With a world going faster and faster by the day, a slightly oblique road can make the difference between staying on the road and not. Here you will learn:
• How a vehicle behaves in a turn;
• How to solve the problem of turns at high speed;
• The mathematical derivation of the angle of banking equation;
• What is the angle of banking in aircraft;
• How to use our angle of banking calculator.
## Turning a vehicle
When turning a vehicle, like a car, we move it from a straight trajectory. Physics tells us that we need to apply a force in the desired direction: in the case of a turn, this is the centripetal force, which points to the turn's center.
🔎 Centripetal comes from the Latin words centrum (center) and petere (to look for).
Consider now a turn on a flat surface and the absence of forces besides gravity and friction. The only force that can counteract the centripetal one is friction. The frictional force $F_f$ is a function of the mass $m$ of the object and the friction coefficient $\mu$:
$F_\text{f}=\mu\cdot m\cdot g$
To keep the car on track, the force due to friction must be bigger (or equal, at least) to the centripetal force:
$F_\text{f} = \mu\cdot m\cdot g > \frac{m\cdot v^2}{r}$
Above, $v$ is the vehicle's speed and $r$ is the radius of the turn.
How can we facilitate things? Enter the tilted turn.
## Tilted turns, a.k.a. how not to lose control
Suppose you tilt the road at a certain angle $\theta$, the angle of banking. In that case, the forces acting on the car will break into components, and by carefully tuning the angle, engineers can design the safest road for a given speed.
We can consider two situations: with and without friction. In the first one, the only forces acting on the vehicle are gravity and centripetal force: there is no friction to help to steer.
🔎 Remember that centrifugal force doesn't exist: it is an apparent force that comes from inertia. The magnitude of this apparent force equals the one of the centripetal force, which, pointing inward, causes the turning.
We will study the tilted turn using the formalism of the inclined plane — the same we met at the inclined plane calculator — but from a different perspective. Consider the normal force $F_N$, perpendicular to the plane, and consider its vertical and horizontal components. What's the normal force? Glad you asked: our normal force calculator is here to help!
Notice that usually, we consider the normal force a component itself.
The two components (obtained with the sine and cosine, we met those at the trigonometric functions calculator) are:
• The vertical component, which is nothing but the weight of the vehicle;
• The horizontal component, which equals the centripetal force.
Let's consider the horizontal one first:
$F_{\text{N}}\sin{(\theta)}=\frac{m\cdot v^2}{r}$
And the vertical:
$F_{\text{N}}\cos{(\theta)} = m\cdot g$
Rearranging them to isolate $F_{N}$ results in the equality:
$\frac{m\cdot v^2}{r}\cdot\frac{1}{\sin{(\theta)}}=\frac{m\cdot g}{\cos{(\theta)}}$
Now we can find the velocity at which we can engage a turn with a specified range and angle of banking.
$v=\sqrt{r\cdot g\cdot \tan{(\theta)}}$
Notice how the mass cancels out — this often happens in physics. The daily experience suggests that the mass is important, but this is often the result of real-life, more complex phenomena. If you ignore outside forces and assume spherical cows, the mass disappears from the equation, like in a free fall.
🔎 All of the results on this page depend somehow on Newton's second law: find out more with our Newton's second law calculator!
## Tilted turns with a side of friction
Let's introduce friction in our tilted turn! We will write the formulas first and then take a closer look at them.
We can now see a friction force parallel to the turn. We can decompose that one too in horizontal and vertical components, which will eventually contribute to the total force acting on the vehicle. When friction appears, we can identify a range of speeds at which the vehicle can approach the turn. When driving at the maximum speed, the friction points inward: going above that speed would make the vehicle skid out of the turn. The minimal speed sees the vehicle "falling" toward the center of the turn: the friction points outward.
Let's analyze the maximum speed case first.
The horizontal component, the one contributing to the centripetal force, is:
$\footnotesize F_{\text{N}}\sin{(\theta)}+\mu\cdot F_{\text{N}}\cos{(\theta)}=\frac{m\cdot v^2}{r}$
The centripetal force now receives two contributions: as a result, the angle can be smaller while keeping the speed unvaried.
To isolate the speed, we need to consider the vertical component too: in this case, the friction force adds a downward contribution:
$\footnotesize F_{\text{N}}\cos{(\theta)} =\mu\cdot F_{\text{N}}\sin{(\theta)} + m\cdot g$
To calculate the minimum speed equations, simply invert the signs of the friction terms. Now rearrange the equations to isolate the speed as we've seen in the frictionless case:
$v=\sqrt{\frac{r\cdot g \cdot (\tan{(\theta)}\pm\mu)}{1\mp\mu\tan{(\theta)}}}$
Where choosing $+$ at the numerator gives the maximum speed, whereas $-$ returns the minimum one. Remember to choose the correct sign at the denominator, too!
🙋 Notice how by setting $\mu=0$, we can simplify the equation for $v$ to the frictionless case's one.
## In the sky: the angle of banking of an aircraft
What if it's not the road to tilt, but the vehicle? That's what happens with planes. The set of equations to calculate the angle of banking for an aircraft is slightly different.
First thing: we need to think in terms of lift, the vertical force that counteracts the weight of a plane. The lift point upwards and equals the aircraft's weight in level flight. In a turn, the lift changes direction, pointing towards the center of the turn, thus creating the component of centripetal force we need.
The sequence of actions needed to steer an aircraft is:
• Roll the aircraft of an angle $\theta$, lowering the wing in the direction of the center of the turn;
• Pull the yoke.
Congrats, you now know how to steer a plane!
Pulling on the cloche increases the lift, which now outmatches the weight. The excess horizontal component contributes to the centripetal force.
The two components of the lift are:
$\begin{split} L_{\text{v}}&=m\cdot g\cos{(\theta)}\\ L_{\text{h}}&=\frac{m\cdot v^2}{r} \sin{(\theta)} \end{split}$
Since an aircraft can't really get thrown off the road — there are no roads in the sky — the airspeed is usually specified in the calculation instead of the radius, leaving the latter as the unknown of the equation. The angle of banking equation is:
$r=\frac{v^2}{g\tan{(\theta)}}$
## How to find the angle of banking?
To find the angle of banking, you need to reverse the equations used to find either the speed or turning radius.
For a vehicle on a fixed-design road, the equation for the angle of banking is:
$\theta = \arctan{\left(\frac{v^2 \mp r \cdot g \cdot \mu}{r\cdot g \pm v^2\cdot \mu}\right)},$
where $\arctan$ is the inverse tangent function. For an aircraft, the angle of banking equation is:
$\theta = \arctan{\left(\frac{r\cdot g}{v^2}\right)}$
## Real-life examples
A usual value for the tilt of a curve on highways is 7% — if snow is unlikely. This angle corresponds to $\theta=4\degree$. The radius of the turn is $500\ \text{m}$. It didn't rain, so we get a nice friction coefficient $\mu = 0.7$.
Let's first assume a flat road: what is the maximum speed we can go before falling off the tracks? Set $\theta = 0\degree$.
$\begin{split} v&=\sqrt{500\cdot g \cdot 0.7}\\ &= 58.6\ \text{m}/\text{s} \\ &= 211\ \text{km}/\text{h} \end{split}$
Rather high, isn't it? Engineers are known to play safe!
What about the minimum speed? It is undefined since you can't fall from a flat road! Let's check what happens when we tilt the road.
$\begin{gather*} \small v=\sqrt{\frac{500\cdot g \cdot (\tan{(4\degree)}+0.7)}{1-0.7\tan{(4\degree)}}} =\\ \small = 63.0\ \frac{\text{m}}{\text{s}} = 226.8\ \frac{\text{km}}{\text{h}} \end{gather*}$
Well, that's a noticeable increase! Let's check the minimum speed now: again undefined. That's because of the relatively low bank angle. Take a look at the circuits of NASCAR races: the tilt is noticeable there, and the risk of sliding down is not negligible!
What about planes? Let's say you are flying in an SR71, at the of $3,951\ \text{km}/\text{s}$ — more than three times the speed of sound! If you want to turn remaining inside the territory of Ohio, you have to keep your turning radius at $150\ \text{km}$. Let's find the banking angle!
$\small \theta = \arctan{\left(\frac{150,\!000\cdot g}{1097.5^2}\right)} = 50.7\degree$
This is a lot! But we guess it's not that hard to go fast with those engines!
## How to use our angle of banking calculator
Everything you read in the article is in our angle of banking calculator!
First, choose if you are a car or an airplane. Erm... that's if you are even studying a problem involving a car or an airplane. This will give you the right set of equations to work on.
Next choice, the direction: choose if you want to find the speed (or radius) or the angle of banking.
Now you are all set! Insert the values you know in our calculator, and we will do the math for you. 😀
We hope our angle of banking calculator helped you, both in your physics problems and in satisfying your curiosity!
## FAQ
### What is the angle of banking?
The angle of banking is the angle at which a road is tilted to guarantee safety for cars driving on it at high speed. This intelligent solution uses simple physics to trade some of the friction force for an increased centripetal force, thus allowing for higher speeds.
### How to find the angle of banking?
To find the angle of banking, you have to know the speed and the radius of the turn, plus the friction coefficient. Considering the maximum allowed speed on a turn, that's how you find the angle of banking:
θ = arctan((v² - r × g × μ) / (r × g + v² × μ)),
where:
• v is the speed;
• r is the turning radius;
• g the gravitational parameter; and
• μ the friction coefficient.
### What is the advantage of the angle of banking?
Tilting a road at a certain angle allows increasing the maximum speed allowed on that particular turn. The increase in maximum speed may be negligible in normal conditions, but when the weather is adverse, a tilted road can counterbalance the decrease in traction due to the reduced friction between tires and road.
### Why do bikers lean during a turn?
Bikers and cyclists lean when making a turn — and they lean more as the turn gets "smaller" — to increase their centripetal force. It is a natural action that allows performing the turn more safely and at higher speeds!
Davide Borchia
Where are we turning?
On the ground
What to calculate?
Velocity
Maximum or minimum speed?
Maximum
ft
Banking angle (θ)
deg
Speed (v)
mph
Friction coefficient (μ)
Tire, dry asphalt
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# How do you solve 6h^2+17h+12=0?
May 20, 2015
$6 {h}^{2} + 17 h + 12 = 0$
We can Split the Middle Term of this equation to factorise it and then find the solution for the equation
In this technique, if we have to factorise an expression like $a {h}^{2} + b h + c$, we need to think of 2 numbers such that:
${N}_{1} \times {N}_{2} = a \times c = 6 \times 12 = 72$
and
${N}_{1} + {N}_{2} = b = 17$
After trying out a few numbers we get ${N}_{1} = 9$ and ${N}_{2} = 8$
$9 \times 8 = 72$, and $9 + 8 = 17$
$6 {h}^{2} + 17 h + 12 = 6 {h}^{2} + 9 h + 8 h + 12$
$= 3 h \left(2 h + 3\right) + 4 \left(2 h + 3\right)$
$= \left(3 h + 4\right) \left(2 h + 3\right)$
the solution for the equation is :
$\textcolor{b l u e}{h} = - \frac{4}{3}$
$\textcolor{b l u e}{h} = - \frac{3}{2}$
May 20, 2015
The answer is $h = - \frac{3}{2} , - \frac{4}{3}$
Problem: Solve $6 {h}^{2} + 17 h + 12 = 0$ .
Multiply $6$ times $12$ to get $72$.
Find two numbers that when multiplied equal $72$ and when added equal $17$.
Numbers $8$ and $9$ meet the criteria.
Write the equation substituting $8 h$ and $9 h$ for $17 h$.
$6 {h}^{2} + 8 h + 9 h + 12 = 0$
Separate the first two terms from the second two terms.
$\left(6 {h}^{2} + 8 h\right) + \left(9 h + 12\right)$
Factor out the GCF from both sets of terms.
$2 h \left(3 h + 4\right) + 3 \left(3 h + 4\right) = 0$
Factor out the GCF $\left(3 h + 4\right)$ .
$\left(3 h + 4\right) \left(2 h + 3\right) = 0$
Solve for $h$.
$3 h + 4 = 0$
$3 h = - 4$
$h = - \frac{4}{3}$
$2 h + 3 = 0$
$2 h = - 3$
$h = - \frac{3}{2}$
$h = - \frac{3}{2} , - \frac{4}{3}$ |
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### Course: Multivariable calculus>Unit 3
Lesson 4: Optimizing multivariable functions (articles)
# Examples: Second partial derivative test
Practice using the second partial derivative test
## Prepare for the slog
I have a challenge for you.
In this article, you can walk through two examples of finding maxima and minima in multivariable functions. In modern applications, most of the steps involved in solving these sorts of problems would be performed by a computer. However, the only way to test that you really understand how the second partial derivative test is used is to walk through it yourself, at least once.
After all, you may one day need to write the program to tell a computer how to do this, which requires somewhat of an intimate knowledge of all the steps involved. Moreover, it is a good way to become more fluent with partial derivatives.
So my challenge to you is this: try entering the answer to each step as you move through the article to test your own understanding.
## The statement of the second partial derivative test (for reference)
Start by finding a point $\left({x}_{0},{y}_{0}\right)$ where both partial derivatives of $f$ are $0$.
$\begin{array}{r}{f}_{x}\left({x}_{0},{y}_{0}\right)=0\\ \\ {f}_{y}\left({x}_{0},{y}_{0}\right)=0\end{array}$
The second partial derivative test tells us how to determine if $\left({x}_{0},{y}_{0}\right)$ is a local maximum, local minimum, or saddle point. Start by computing this term:
$H={f}_{xx}\left({x}_{0},{y}_{0}\right){f}_{yy}\left({x}_{0},{y}_{0}\right)-\left({f}_{xy}\left({x}_{0},{y}_{0}\right){\right)}^{2}$
where ${f}_{xx}$, ${f}_{yy}$ and ${f}_{xy}$ are the second partial derivatives of $f$.
If $H<0$, then $f$ has a neither minimum nor maximum at $\left({x}_{0},{y}_{0}\right)$, but instead has a saddle point.
If $H>0$, then $f$ definitely has either a maximum or minimum at $\left({x}_{0},{y}_{0}\right)$, and we must look at the sign of ${f}_{xx}\left({x}_{0},{y}_{0}\right)$ to figure out which one it is.
• If ${f}_{xx}\left({x}_{0},{y}_{0}\right)>0$, then $f$ has a local minimum.
• If ${f}_{xx}\left({x}_{0},{y}_{0}\right)<0$, then $f$ has a local maximum.
If $H=0$, the second derivatives alone cannot tell us whether $f$ has a local minimum or maximum.
## Example 1: All of the stable points!
Problem: Find all the stable points (also called critical points) of the function
$\begin{array}{r}{x}^{4}-4{x}^{2}+{y}^{2}\end{array}$
And determine whether each one gives a local maximum, local minimum, or a saddle point.
## Step 1: Find all stable points
The stable points are all the pairs $\left({x}_{0},{y}_{0}\right)$ where both partial derivatives equal $0$. First, compute each partial derivative
${f}_{x}\left(x,y\right)=$
${f}_{y}\left(x,y\right)=$
Next, find all the points $\left({x}_{0},{y}_{0}\right)$ where both partial derivatives are $0$, which is to say, solve the system of equations
$\begin{array}{rl}{f}_{x}\left({x}_{0},{y}_{0}\right)& =0\\ \\ {f}_{y}\left({x}_{0},{y}_{0}\right)& =0\end{array}$
Which of the following pairs satisfying the system of equations?
## Step 2: Apply second derivative test
To start, find all three second partial derivatives of $f\left(x,y\right)={x}^{4}-4{x}^{2}+{y}^{2}$
${f}_{xx}\left(x,y\right)=$
${f}_{yy}\left(x,y\right)=$
${f}_{xy}\left(x,y\right)=$
The expression we care about for the second partial derivative test is
${f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-\left({f}_{xy}\left(x,y\right){\right)}^{2}$
If we apply the second derivatives we just found, what does this expression become (as a function of $x$ and $y$)?
${f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-\left({f}_{xy}\left(x,y\right){\right)}^{2}=$
To apply the second derivative test, we plug in each of our stable points to this expression and see if it becomes positive or negative.
• Stable point 1:
At $\left(x,y\right)=\left(0,0\right)$, the expression evaluates as
$\begin{array}{r}\phantom{\rule{1em}{0ex}}24{x}^{2}-16=24\left(0{\right)}^{2}-16=-16\end{array}$
This is negative, so according to the second partial derivative test, the point $\left(0,0\right)$ is a
• Stable point 2: At $\left({x}_{0},{y}_{0}\right)=\left(\sqrt{2},0\right)$, the expression becomes
$\begin{array}{rl}24{x}^{2}-16& =24\left(\sqrt{2}{\right)}^{2}-16\\ \\ & =48-16\\ \\ & =32\end{array}$
This is positive. Also,
$\begin{array}{rl}{f}_{xx}\left(\sqrt{2},0\right)& =12\left(\sqrt{2}{\right)}^{2}-8\\ \\ & =24-8\\ \\ & =16\end{array}$
Therefore, the point $\left(\sqrt{2},0\right)$ must be a
• Stable point 3: We could plug in the point $\left(-\sqrt{2},0\right)$ just as we have with the other stable points, but we could also notice that the function $f\left(x,y\right)={x}^{4}-4{x}^{2}+{y}^{2}$ is symmetric, in the sense that replacing $x$ with $-x$ will yield the same expression:
$\left(-x{\right)}^{4}-4\left(-x{\right)}^{2}+{y}^{2}={x}^{4}-4{x}^{2}+{y}^{2}$
Therefore the point $\left(-\sqrt{2},0\right)$ will have precisely the same behavior as $\left(\sqrt{2},0\right)$
Here is a clip of the graph of $f\left(x,y\right)$ rotating, where the two local minima are clear, and we can see that the point at the origin is indeed a saddle point.
## Example 2: Getting more intricate
Let's not sugarcoat things; optimization problems can get long. Very long.
Problem: Find all the stable points (also called critical points) of the function.
$\begin{array}{r}f\left(x,y\right)={x}^{2}y-{y}^{2}x-{x}^{2}-{y}^{2}\end{array}$
And determine whether each one gives a local maximum, local minimum, or a saddle point.
## Step 1: Find stable points
We need to find where both partial derivatives are zero, so start by finding both partial derivatives of $f\left(x,y\right)={x}^{2}y-{y}^{2}x-{x}^{2}-{y}^{2}$
${f}_{x}\left(x,y\right)=$
${f}_{y}\left(x,y\right)=$
So we must solve the system of equations
$\begin{array}{rl}2xy-{y}^{2}-2x& =0\\ \\ {x}^{2}-2xy-2y& =0\end{array}$
In the real world, when you come across a system of equations, you should almost certainly use a computer to solve it. For the sake of practice, though, and to see that optimization problems are not always that simple, let's do something crazy and actually work it out ourselves.
• Solve one equation to get $y$ in terms of $x$.
• Plug that into the other expression to get an equation with only $x$.
• Solve for $x$.
• Plug each solution for $x$ into both equations and solve for $y$.
• Check which resulting $\left(x,y\right)$ pairs actually solve the expression.
This can be a real mess since you might use the quadratic formula to solve for $y$ treating $x$ as a constant, and plug that nasty expression in elsewhere. Otherwise, you might find yourself solving a degree $4$ equation, which aside from being a pain gives quite a few solutions to plug in.
In this particular system, the equations feel very symmetric, which is an indication that adding/subtracting them might make things simpler. Indeed, if we add them together, we get
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}2xy-{y}^{2}-2x& =0\\ \\ +\phantom{\rule{1em}{0ex}}{x}^{2}-2xy-2y& =0\\ \\ \\ \\ {x}^{2}-{y}^{2}-2\left(x+y\right)& =0\\ \\ \left(x+y\right)\left(x-y\right)-2\left(x+y\right)& =0\\ \\ \left(x+y\right)\left(x-y-2\right)& =0\end{array}$
What does this equation imply about the relationship between $x$ and $y$? (Express each answer as an equation involving the variables $x$ and $y$)
Either
or
Each of these possibilities lets us write $x$ in terms of $y$, which in turn lets us write one of our equations purely in terms of $y$.
For example, if you plug in the relation $x=-y$ to the first expression $2xy-{y}^{2}-2x$, you can get a quadratic expression purely in terms of $y$. What are the roots of this expression?
and
Since this arose from assuming $x=-y$, the corresponding $x$ values are $x=-0$ and $x=-\frac{2}{3}$ respectively. This gives us our first two solution pairs:
$\begin{array}{rl}& \left(x,y\right)=\overline{)\left(0,0\right)}\\ \\ & \left(x,y\right)=\overline{)\left(-\frac{2}{3},\frac{2}{3}\right)}\end{array}$
Alternatively, if we consider the case where $x=y+2$. Again, when we plug this relation into the expression $2xy-{y}^{2}-2x$, we have a quadratic expression purely in terms of $y$. What are the roots of this expression?
and
Because we found these under the assumption that $x=y+2$, the corresponding values of $x$ are
$\begin{array}{r}x=2-1+\sqrt{5}=1+\sqrt{5}\\ \\ x=2-1-\sqrt{5}=1-\sqrt{5}\end{array}$
This gives two more solution pairs:
$\begin{array}{r}\left(x,y\right)=\overline{)\left(1+\sqrt{5},-1+\sqrt{5}\right)}\\ \\ \left(x,y\right)=\overline{)\left(1-\sqrt{5},-1-\sqrt{5}\right)}\end{array}$
We've now exhausted all possibilities since we initially found that either $x=-y$ or $x=y+2$, and we completely solved the equations resulting from each assumption.
## Step 2: Apply second derivative test
Man, that was already a lot of work for one example, and we're not even halfway done! Now we have to apply the second derivative test to each one of these. First, find all of the second derivatives of our function
$\begin{array}{r}f\left(x,y\right)={x}^{2}y-{y}^{2}x-{x}^{2}-{y}^{2}\end{array}$
${f}_{xx}\left(x,y\right)=$
${f}_{yy}\left(x,y\right)=$
${f}_{xy}\left(x,y\right)=$
According to the second derivative test, to analyze whether each of our stable points is a local maximum or minimum, we plug them into the expression
$\begin{array}{r}{f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-\left({f}_{xy}\left(x,y\right){\right)}^{2}\end{array}$
What does this expression become when we apply the second derivatives you just found?
Since we only care about whether this expression is positive or negative, we can divide everything by $4$ to make things a bit simpler.
$\left(y-1\right)\left(-x-1\right)-\left(x-y{\right)}^{2}\phantom{\rule{1em}{0ex}}←\text{Key expression}$
Now we see what the sign of this expression is for each of our stable points.
• Stable point $\left(0,0\right)$:
At the point $\left(x,y\right)=\left(0,0\right)$, the key expression above evaluates to
.
From this we conclude that $\left(0,0\right)$ is
Now,
• Stable point $\left(-\frac{2}{3},\frac{2}{3}\right)$:
At the point $\left(x,y\right)=\left(-\frac{2}{3},\frac{2}{3}\right)$, the key expression above evaluates to
.
From this we conclude that $\left(-\frac{2}{3},\frac{2}{3}\right)$ is
Now,
• Stable point $\left(1+\sqrt{5},-1+\sqrt{5}\right)$:
At the point $\left(x,y\right)=\left(1+\sqrt{5},-1+\sqrt{5}\right)$, the key expression above evaluates to
.
From this we conclude that $\left(1+\sqrt{5},-1+\sqrt{5}\right)$ is
Now,
• Stable point $\left(1-\sqrt{5},-1-\sqrt{5}\right)$:
The arithmetic here is almost identical to the previous case.
Here is a short clip of the graph of $f\left(x,y\right)={x}^{2}y-{y}^{2}x-{x}^{2}-{y}^{2}$ rotating, where you can see the three saddle points and the one local maximum at the origin.
## Pat yourself on the back
These are long problems, so if you actually worked through them, give yourself some hearty congratulations!
## Want to join the conversation?
• In step 1 of example 2, the caption for the second input box should read f_y(x, y) rather than f_x(x,y)
• In step 2 of example 1, it asks for fxx, fyy, and then fxx again, but the third one should be fxy.
• In Step 2 of Example 2, third stable point (1+√5, -1+√5), the answers wrongly relate to the second stable point (-2/3, 2/3).
• if more than 2 variable wat do?
• How would one come up with the second equation? (or was it a random symmetry-inclined guess?)
• In problem 2, isn't (2,2) also a solution? If you just do a little rearranging you get:
2x(y - 1) - y^2 = 0 and x^2 - 2y(x-1) = 0
(2,2) works in both equations. Clearly, I've made some sorta logic error here, so where did I go wrong?
(1 vote)
• You just made a mistake in rearranging your second equation. It should be x^2-2y(x+1)=0. (2,2) is not a solution.
(1 vote)
• In the "Or we could get clever" hint, that symmetry does not hold everywhere. The graph clearly shows that this symmetry is only available when the x and y have the same sign (quadrants 1 and 3). So we cannot say that f(x,y) = f(-x,-y) without qualification. The symmetry only holds in one diagonal direction.
(1 vote) |
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In the quadratic equation form ax2+bx+c=0, x stands for the unknown value and a, b and c the numbers that are known such that a will never be equal to zero. If in case a is zero, then that equation is not a quadratic equation but a linear equation. a, b and c are the coefficients of the equation which can be distinguished by referring to them as the constant, the linear or the quadratic coefficient or the free term. A quadratic equation can be solved using methods like factoring by the quadratic formula, by graphing or by completing the square. To solve a quadratic equation in one variable can be done through the factoring technique together with the zero factor property as below:
• Begin by writing the quadratic equation in a standard form.
• Factor the quadratic polynomial to a product of linear factors.
• Use Zero Factor Property to make each factor equal to zero.
• Find the solution to each of the resulting linear equations. The resulting solution is the solution of the original quadratic equation.
The quadratic equation is also known as a univariate because it involves only one unknown. The quadratic equation only has powers of x which are non-negative integers and this makes it a polynomial equation and in particular, it is a second-degree polynomial equation because its greatest power is two.
The second method of solving a quadratic equation is by completing the square. The completing by square method is used to come up with a new formula for solving a quadratic equation. The formula is known as the quadratic formula. The quadratic formula’s mathematical proof is.
One property of this equation form is that it gives one valid root when a = 0, while the other root has division by zero, since when a = 0, the quadratic equation changes to a linear equation which has one root. By contrast, the formula contains division by zero in both cases. With the polynomial expansion, you can easily tell that this equation is equivalent to the quadratic equation.
The third method of solving a quadratic equation is by graphing. The graph of a quadratic equation is a parabola that opens up when the leading coefficient is positive and it opens down when the leading is a negative. The x intercepts are crucial and they should be gotten, plotted and labeled. In any function, the x intercepts are gotten by getting the real zeros of that function and the zeros of any given function can be gotten by solving that equation that results from f(x)=0. In a quadratic function, the function f of the equation that results from f(x) =0 can be solved by factoring along with the zero factor property or with the quadratic formula. The vertex of a parabola is also crucial and can be gotten, plotted and labeled. In the case of a quadratic function f, the equation resulting from f(x) = 0 is always solvable with the quadratic formula or by factoring in along with the Zero Factor Property.
The first step in graphing a quadratic equation is determining the form of the equation given. A quadratic equation can be expressed in three forms. These forms are the quadratic form, vertex form and standard form. Any of the forms can be used in graphing but the process of graphing each of them is a bit different. The standard form is the form where the equation is written as f(x) = ax2+bx+c whereby a, b and c are real numbers and a is not a zero. Examples of standard form equations are f(x) = x2 +2x+1 and f(x) =9 x2+10x-8. A vertex form is the form whereby the equation is expressed as f(x) =a(x-h) 2 +k whereby a, h and k are real numbers and a is not zero. This equation form is known as a vertex form because h and k will directly give you a central point of the parabola at point (h, k). Examples of vertex form equations are -3(x-5) 2+1 and 9(x-4) 2+18. To graph any of this quadratic equation forms you will first have to get the vertex of the parabola which is the middle point (h, k) at the tip of the curve. The vertex coordinates in the standard form are given by k=f(h) and h=-b/2a. In the vertex equation form, h and k are gotten directly from the equation.
The second step in graphing a quadratic equation is to define the variables. To solve a quadratic equation, the variables a, b and c or a, h and k must be defined. Common algebra problems will give you a quadratic equation with variables already filled in a vertex or standard form. An example of a standard equation form with variables is f(x)= 2x2+16x+39, so a is 2, b is 16 and c is 39. An example of a vertex equation form with variables is f(x)=4(x-5) 2+12, a is 4, b is 16 and k is 12. The third step is to find h. In the vertex form, h is already provided. In standard form h is calculated by h =-b/2a and so in f(x)= 2x2+16x+39 h will be -16/2(2). After solving you will get h=-4. In the vertex form f(x)=4(x-5) 2+12, h is equal to 5.
The third step is to find k. In the vertex form, k is already known just like h but for standard form k is equal to f(h). This means that you can get k in the standard form equation by replacing every x with the value of h. So k=2(-4) 2 +16(-4)+39, k=2(16)-64+39, k=32-64+39 =7. The value of k in the vertex quadratic equation form is 12.
The fifth step is to plot the vertex. The parabola vertex is at point (h, k). The h represents the x coordinate and the k represents the y coordinate. The vertex is the middle point of the parabola. In the standard equation form, the vertex will be at point (-4, 7). This point will be plotted on the graph and labeled. In the vertex equation form, the vertex is at (5, 12).
The sixth step is to draw the parabola axis. The axis of symmetry in a parabola is a line that runs through the middle and divides the parabola into half. Across the axis, the right side of a parabola will mirror the left side of the parabola. For the quadratic equation of the form ax2+bx+c or a(x-h) 2+k, the axis is the line which is vertical and passes through the vertex. In the standard form equation, the axis is the line that is parallel to the y axis and passes through the point (-4, 7). This line is not part of the parabola but it shows you how a parabola curves symmetrically.
The seventh step is to get the direction of opening. After knowing the axis and vertex of the parabola determine if the parabola opens downwards or upwards. If a is positive, the parabola opens upwards but if a is negative the parabola will open downwards. For the standard form equation f(x)= 2x2+16x+39, the parabola will open upwards since a=2(positive), and in the vertex equation form f(x)=4(x-5) 2+12, the parabola will also open upwards since a=4(positive).
The eight step if necessary, you can find and plot the x intercepts. The x intercepts are the two points where the parabola meets the x axis. Not all parabolas have x intercepts. If the parabola is a vertex that opens upwards and has the vertex above the x axis or if it opens downwards and has a vertex below the x axis, it will not have x intercepts. You can also find and plot the y intercept too. To find the y intercept, you will set x to zero and then solve the equation for y or f(x). This will give you the value of y when the parabola passes through the y axis. And unlike the x intercept, standard parabolas can have one y intercept and for the standard quadratic equation forms, the y intercept is at y=c.
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# 1999 JBMO Problems/Problem 4
## Problem 4
Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$).
## Solution
Its easy to see that $B'$, $C'$, $D$ are collinear (since $\angle B'DC = \angle C'DC$ = 90^\circ$). Applying the sine rule in triangle$ABC$, we get$\frac{\sin BAD }{ \sin CAD} = \frac{BD }{ DC}.$Since$BAB'D$and$CC'AD$are cyclic quadrilaterals,$\angle BAD \angle BB'D$and$\angle CAD = \angle CC'D.$So,$\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}$and$\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.$Thus,$BB' = CC'$(the circumcircles$\mathcal{C}_1,\mathcal{C}_2$are congruent). From right triangles$ (Error compiling LaTeX. ! Missing $inserted.)BB'A$and$CC'A$, we have AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2} So$AC' = AB'.$Since$M$is the midpoint of$B'C'$,$AM$is perpendicular to$B'C'$and hence$AM$is parallel to$BC$. So area of$[MBC] = [ABC]$and hence is independent of position of$D$on$BC$. |
# FIND CENTER AND RADIUS OF CIRCLE FROM STANDARD FORM
Equation of circle in standard form :
x2 + y2 + 2gx + 2fy + c = 0
Here (-g, -f) is radius. and radius = √g2 + f2 - c
Note :
Using completing the square method, we can convert the given equation from standard form to (x - h)2 + (y - k)2 = r2, we get center and radius.
Problem 1 :
x2 + 10x + y2 – 6y = -18
The graph of the equation shown above is a circle. What is the radius of the circle?
A) 3 B) 4 C) 5 D) 9
Solution :
x2 + 10x + y2 – 6y = -18
x2 + 2x⋅5 + y2 – 2y⋅3 = -18
To complete the formula
x2 + 2x⋅5 + 52 - 52 + y2 – 2y⋅3 + 32 - 32 = -18
(x + 5)2 + (y - 3)2 - 25 - 9 = -18
(x + 5)2 + (y - 3)2 = -18 + 25 + 9
(x + 5)2 + (y - 3)2 = 16
(x + 5)2 + (y - 3)2 = 42
(x - (-5))2 + (y - 3)2 = 42
Comparing with
(x−h)2 + (y−k)2 = r2
(h, k) is (-5, 3) and radius = 4
So, option B is correct.
Problem 2 :
x2 + 18x + y2 – 8y = -48
The graph of the equation shown above is a circle. What is the radius of the circle?
A) 4 B) 5 C) 6 D) 7
Solution :
x2 + 18x + y2 – 8y = -48
x2 + 2x⋅9 + y2 – 2y⋅4 = -48
To complete the formula
x2 + 2x⋅9 + 92 - 92+ y2 – 2y⋅4 + 4- 42 = -48
(x + 9)2 + (y - 4)2 - 81 - 16 = -48
(x + 9)2 + (y - 4)2 - 97 = -48
(x - (-9))2 + (y - 4)2 = -48 + 97
(x - (-9))2 + (y - 4)2 = 49
(x - (-9))2 + (y - 4)2 = 72
Problem 3 :
x2 - 4x + y2 + 6y = 113
The graph of the equation shown above is a circle. What is the coordinate point of the center of the circle?
A) (13, 10) B) (4, 13) C) (-4, 6) D) (2, -3)
Solution :
x2 - 4x + y2 + 6y = 113
x2 - 2x⋅2 + y2 + 2y⋅3 = 113
x2 - 2x⋅2 + 22 - 22 + y2 + 2y⋅3 + 32 - 32 = 113
(x - 2)2 - 4 + (y + 3)2 - 9 = 113
(x - 2)2 + (y + 3)2 - 13 = 113
(x - 2)2 + (y + 3)2 = 113 + 13
(x - 2)2 + (y + 3)2 = 126
h = 2 and k = -3
So, the center of the circle is (2, -3).
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Preschool Math: Dry Erase Dice Game
Recently I found this gem at the Dollar Store.
You write the numbers on this die with a dry erase marker. This is gold because it means you can change the numbers at any time. In elementary you could add the number zero or teen numbers. My little preschooler is focusing on the numbers one to five so I changed the six to a smiley face.
Now I just needed a game to play 🙂 Joseph loves games! At first I thought it was because he really liked winning (like his mom). But soon I saw he didn’t mind losing either. What he loves is the undivided attention. He is often careful to point out that his one year old sister “cannot play because she’s too little.”
In math we are working on learning the numbers one to five. But we are not focusing on counting. Instead, we are spending our time on the more important skill of understanding the amount each number represents. So far he can correctly show me groups of one, two, or three objects when asked. That is more important to me than if he can count to twenty. We’re going to keep practicing until he has a solid grasp of four and five too. This game was a great way to make that practice fun.
So without further ado…the amazing dry erase dice game!
You will need:
• 1 die
• A dry erase marker
• 2 plates or bowls
• 20 beads, beans, or other type of counter
1. Mark the die with the numbers 1 to 5. Practice saying the names of the numbers. Instead of a 6 put a smiley face. When a player rolls a smiley face they have to do something similar (we jumped up and stuck out our tongues).
2. Each player starts with 10 beads on their plate.
3. When a player rolls the die they take that many beads from the other player. They should carefully count them in the center before putting them on their plate. That way if they made a mistake you can catch it.
4. Play continues for 10 turn or until one player wins all the beads. Whatever you have the patience for 🙂
That’s it! Super simple but we spent so much time practice our numbers this morning because he was engaged in the game. The beads worked really well for us because he liked trying to steal certain colors from me. |
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# TNPSC – HCF & LCM
## TNPSC – HCF & LCM
### HCF & LCM:
#### Factors and Multiples:
If a number x divides another number y exactly (without leaving any remainder), then x is a factor of y and y is a multiple of x.
or
Factors Set of numbers which exactly divides the given number
Mulitples Set of numbers which are exactly divisible by the given number
For example, If the number is 8, then {8, 4, 2, 1} is the set of factors, while {8, 16, 24, 32, …..} is the set of multiples of 8.
#### Common Multiple:
A common multiple of two or more numbers is a number which is completely divisible (without leaving remainder) by each of them.
For example: We can obtain common multiples of 3, 5 and 10 as follows:
à Multiples of 3 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 ….}
à Multiples of 5 = {5, 10, 15, 20, 25, 30, 35, 40…}
à Multiples of 10 = {10, 20, 30, 40……}
Therefore, common multiples of 3, 5 and 10 = {30, 60, 90, 120 ….}
#### Least Common Multiple (LCM)
The LCM of two or more given numbers is the least number to be exactly divisible by each of them.
For example, We can obtain LCM of 4 and 12 as follows
Multiples of 4 = {4,8,12,16,20,24,32,36,….}
Multiples of 12 = {12,24,36,48,60,72,……}
Common multiples of 4 and 12 = 12, 24, 36, ……
LCM of 4 and 12 = 12
#### Methods to calculate LCM:
There are two methods to find the LCM of two or more numbers which are explained below.
1. Prime Factorisation Method
Following are the steps to obtain LCM through prime factorisation method.
Step 1: Resolve the given numbers into their prime factors.
Step 2: Find the product of all the prime factors (with highest powers) that occur in the given numbers.
Step 3: This product of all the prime factors (with highest powers) is the required LCM.
Example 1: Find the LCM of 8, 12 and 15.
Soln.
Factors of 8 = 2*2*2 = 23
Factors of 12 = 2*2*3 = 22*31
Factors of 15 = 3*5 = 31*51
Here, the prime factors that occur in the given numbers are 2, 3 and 5 and their highest powers are 23, 31 and 51.
Required LCM = 23*31*51 = 8*3*5 = 120
2. Division Method:
Following are the steps to obtain LCM through division method.
Step 1: Write down the given numbers in a line, separating them by commas.
Step 2: Divide them by a prime numbers which exactly divides atleast any two of the given numbers.
Step 3: Write down the quotients and the undivided numbers in a line below the 1st.
Step 4: Repeat the process until you get a line of numbers which are prime to one another.
Step 5: The product of all divisors and the numbers in the last line will be the required LCM.
Example 2: What will be the LCM of 15, 24, 32, 45?
Soln. LCM of 15, 24, 32 and 45 is calculated as
Required LCM = 2*2*2*3*5*4*3 = 1440
Note: Start division with the least prime number.
#### Common Factor:
A common factor of two or more numbers is that particular number which divides each of them exactlyh.
For example: We can obtain common factors of 12, 48, 54 and 63 as follows:
Factors of 12 = 12, 6, 4, 3, 2, 1
Factors of 48 = 48, 24, 16, 12, 8, 6, 4, 3, 2, 1
Factors of 54 = 54, 27, 18, 9, 6, 3, 2, 1
Factors of 63 = 63, 21, 9, 7, 3, 1
Common factors of 12, 48, 54 and 63 = 3
#### Highest Common Factor:
HCF of two or more numbers is the greatest number which divides each of them exactly. For example 6 is the HCF of 12 and 18 as there is no number greater than 6 that divides both 12 and 18. Similarly, 3 is the highest common factor of 6 and 9.
HCF is also known as ‘Highest common Divisor’ (HCD) and ‘Greatest Common Measure’ (GCM).
### Methods to Calculate HCF:
There are two methods to calculate the HCF of two or more numbers which are explained below.
1. Prime Factorisation Method:
Following are the steps for calculating HCF through prime factorisation method.
Step 1: Resolve the given numbers into their prime factors.
Step 2: Find the product of all the prime factors (with least power) common to all the numbers.
Step 3: the product of common prime factors (with the least powers) gives HCF.
Example 3: Find the HCF of 24, 30 and 42.
Soln. Resolving 24, 30 and 42 into their prime factors,
Factors of 24 = 2*2*2*3* = 23*31
Factors of 30 = 2*3*5 = 21*31*51
Factors of 42 = 2*3*7 = 21*31*71
The product of common prime factors with the least power = 21*31 = 6
So, HCF of 24, 30 and 42 = 6.
2. Division Method:
Following are the seeps to obtain HCF through division method.
Step 1: Divide the larger number by the division method.
Step 2: Divide the divisor by the remainder
Step 3: Repeat the step 2 till the remainder becomes zero. The last divisor is the required HCF.
Note: To calculate the HCF of more than two numbers, calculate the HCF of first two numbers then take the third number and HCF of first two numbers and calculate their HCF and so on. The resulting HCF will be the required HCF of numbers.
Example 4: Find the HCF of 26 and 455.
Soln.
26)455(17
26
——-
195
182
————-
13)26(2
26
——-
X
Required HCF = 13
Remember:
To find the HCF of given numbers you can divide the numbers by their lowest possible difference. If these numbers are divisible by this difference, then this difference itself is the HCF of the given numbers any other factor of this difference will be its HCF.
Example 5: Find the HCF of 30, 42 and 135
Soln. We can notice that the difference between 30 and 42 is less than difference between 135 and 42 or 30.
Difference between 30 and 42 is 12, but 12 does don’t divide 30, 42 and 135 completely.
Factors of 12 = 12, 6, 4, 3, 2, 1
Clearly, 3 is the highest factor, which divides all the three numbers completely.
Therefore, 3 is the HCF of 30, 42 and 135.
#### Method to calculate LCM and HCF fractions:
The LCM and HCF can be obtained by the following formulae.
=> LCM of fractions = LCM of numerators / HCF of denominators
=> HCF of fractions = HCF of numerators / LCM of denominators
Note:
=> All the fractions must be in their lowest terms. If they are not in their lowest terms, then conversion in the lowest form is required before finding the HCF or LCM.
=> The required HCF of two or more fractions is the highest fraction which exactly divides each of the fractions.
=> The required LCM of two or more fractions is the least fraction / integer which is exactly divisible by each of them.
=> The HCF of numbers of fractions is always a fraction but this is not true in case of LCM.
Example 6: Calculate the LCM of 72/250, 126/75 and 162/165.
Soln. Here, 72/250 = 36/125, 126/75 = 42/25 and 162/165 = 54/55.
According to the formula,
Required LCM = LCM of 36, 42 and 54 / HCF of 125, 25 and 55 = 756/5 = 151 1/5.
Example 7: Find the HCF of 36/51 and 3 9/17.
Soln. Here, 36/51 = 12/17 and 3 9/17 = 60/17
Now, we have to find the HCF of 12/17 and 60/17
According to the formula,
HCF of fractions = HCF of numerators / LCM of denominators
= HCF of 12 and 60 / LCM of 17 and 17 = 12/17
Note:
The LCM and HCF of decimals can also be obtained from the above formulae (by converting the decimal into fraction).
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ss_10_3
# ss_10_3 - Section 10.3 Matrices Matrices have many uses...
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Section 10.3 Matrices Matrices have many uses, most of which will be discussed in later sections. In this section, we will use matrices to help us solve systems of linear equations. A matrix is simply a rectangular array of numbers. Several notations are used for matrices. We will use only the following notation. Examples of matrices: 1 2 - 3 - 1 - 3 - 1 2 2 1 2 3 3 2 1 2 1 0 4 - 1 2 3 3 - 1 4 - 2 The following is common notation for a matrix with m rows and n columns. A matrix with m rows and n columns is said to be of order ”m by n”, often written ”m x n”: a 11 a 12 · · · a 1 j · · · a 1 n a 21 a 22 · · · a 2 j · · · a 2 n · · · a i 1 a i 2 · · · a ij · · · a in · · · a m 1 a m 2 · · · a mj · · · a mn a ij is the entry in the i-th row and j-th column of the matrix A Matrices associated with linear systems Two types of matrices are commonly used for solving linear systems. They are coefficient matrices and augmented matrices . These two types of matrices are shown in the following example. Example. The coefficient matrix, A , and the augmented matrix, B , for the following system are shown below. 2 x - y + z = 3 x - y = 2 y + 2 z = - 3 A = 2 - 1 1 1 - 1 0 0 1 2 Coefficient matrix B = 2 - 1 1 | 3 1 - 1 0 | 2 0 1 2 | - 3 Augmented matrix Example. Solve the following system using matrices. 2 x - y + z = 3 x - y = 2 y + 2 z = - 3 Solution. We begin with the augmented matrix for the system. Then we use elementary row operations to simplify the augmented matrix to one having zeros below the main diagonal of the matrix. The simplified matrix is the matrix of a system equivalent to the original system.
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# “What is the Measure of Circumscribed ∠x? Explained with Options!”
Geometry is an interesting and essential branch of mathematics that deals with the study of shapes, sizes, positions, and dimensions of objects in a space. One of the fundamental concepts in geometry is the measurement of angles. In this article, we will be discussing a particular type of angle known as the circumscribed angle, specifically the measurement of circumscribed ∠x.
What is a Circumscribed Angle?
A circumscribed angle refers to an angle formed by two tangents drawn to a circle from the same point on the circle’s circumference. The vertex of the angle lies outside the circle, and the sides of the angle intersect the circle at two points. In other words, it is an angle formed by two chords of a circle that intersect at a point on the circumference.
Circumscribed angles are an essential concept in geometry, and they have a lot of practical applications. They are used in various fields such as engineering, architecture, and physics, among others.
Now that we understand what a circumscribed angle is let’s take a look at how to measure circumscribed ∠x.
How to Measure Circumscribed ∠x?
To measure the circumscribed ∠x, we need to use a theorem called the “Inscribed Angle Theorem.” This theorem states that the measure of an inscribed angle is half the measure of the intercepted arc. That is,
∠x = 1/2(arc AB + arc CD)
Where AB and CD are the intercepted arcs in the diagram below:
[image]
Let’s take a closer look at each of the intercepted arcs.
Intercepted Arc AB:
In the diagram above, Arc AB is the arc intercepted by ∠AEB. To calculate the measure of arc AB, we need to know the measure of the central angle that subtends it. In other words, if we can find the measure of ∠AOB, we can calculate the measure of arc AB using the formula:
arc AB = (central angle/360) * circumference of the circle
We know that ∠AOB is a right angle, which means it measures 90 degrees. The circumference of the circle is given by the formula,
circumference = 2πr
Where r is the radius of the circle. In our case, the radius is 5cm, so the circumference is:
circumference = 2π(5) = 10π cm
Therefore, the measure of arc AB is:
arc AB = (90/360) * 10π = (1/4) * 10π = (5/2)π cm
Intercepted Arc CD:
Arc CD is the arc intercepted by ∠CED. To calculate the measure of arc CD, we follow the same process as above. The central angle ∠COD is a straight line, so it measures 180 degrees. The circumference of the circle is still 10π cm.
Therefore, the measure of arc CD is:
arc CD = (180/360) * 10π = (1/2) * 10π = 5π cm
Now that we have calculated the measures of arcs AB and CD, we can substitute them into the formula for the measure of ∠x:
∠x = 1/2(arc AB + arc CD)
= 1/2[(5/2)π + 5π]
= 1/2[(15/2)π]
= (15/4)π
Hence, the measure of circumscribed ∠x is (15/4)π radians or approximately 11.78 radians.
Options for Measuring Circumscribed ∠x
It’s worth noting that there are different methods of measuring circumscribed angles depending on the given information or situation. Let’s take a look at some of them:
1. Measuring Central Angles:
One way to measure circumscribed angles is by using central angles. If we know the measure of the central angle that intersects the circle at two points, we can easily calculate the measure of the inscribed angle using the Inscribed Angle Theorem.
2. Measuring Intercepted Arcs:
Another way to measure circumscribed angles is by measuring intercepted arcs. We can use the formula for the measure of an arc to calculate the intercepted arc’s length, and then use the Inscribed Angle Theorem to find the measure of the corresponding inscribed angle.
3. Measuring Tangents:
We can also measure circumscribed angles by measuring tangents. If we know the length of the tangent lines drawn from a point outside the circle to the circle, we can calculate the distance between the two points of intersection of the tangent lines with the circle. This distance corresponds to the length of the intercepted arc, which we can use to find the measure of the circumscribed angle.
Conclusion
In conclusion, the measurement of circumscribed ∠x is an important concept in geometry that has practical applications in different fields of study. To measure circumscribed angles, we need to use the Inscribed Angle Theorem, which states that the measure of an inscribed angle is half the measure of the intercepted arc. There are several options for measuring circumscribed angles, including measuring central angles, intercepted arcs, and tangents. Understanding the concept of circumscribed angles and their measurements is crucial in solving geometric problems and conducting accurate calculations in various applications. |
# 66 Times Table
Hey there, math enthusiasts! Get ready to embark on an exhilarating mathematical journey as we dive into the marvelous world of the 66 times table.
As your math teacher and partner in this adventure, I assure you that this multiplication voyage will be filled with excitement, discovery, and a whole lot of fun! So, buckle up and let's explore the captivating universe of the 66 times table!
## Setting the Stage
Before we jump right in, let's start by understanding the fundamentals. The 66 times table involves multiplying numbers by 66.
Now, you might be thinking, "Whoa, that sounds challenging!" But don't worry, we'll break it down step by step, and you'll see just how fascinating and approachable it can be.
## Let's Begin
To kickstart our exploration, let's take a closer look at the 66 times table using smaller numbers. By doing so, we'll build a solid foundation before tackling larger calculations. Let's dive in and examine the table up to 10:
• 1 x 66 = 66
• 2 x 66 = 132
• 3 x 66 = 198
• 4 x 66 = 264
• 5 x 66 = 330
• 6 x 66 = 396
• 7 x 66 = 462
• 8 x 66 = 528
• 9 x 66 = 594
• 10 x 66 = 660
## Spotting the Patterns
Patterns are like hidden treasures waiting to be discovered in the world of multiplication tables.
In the 66 times table, we can uncover some interesting patterns that will make our journey even more enjoyable.
### Pattern 1: Double Trouble!
As we observe the products in the 66 times table, we notice a doubling pattern. Each subsequent product is obtained by doubling the previous one.
For instance, 66, 132, 264, 528, and so on. This doubling pattern continues throughout the table, making it easier for us to find the products quickly.
### Pattern 2: The Zeros at the End
When we multiply any number ending in zero by 66, the resulting product will have the same digits in the units and tens places, followed by two zeros.
For example, 30 x 66 = 1980 (30 at the beginning, zero at the end). This pattern remains consistent across the entire table, simplifying calculations involving numbers that end in zero.
### Putting It All Together
Now that we've unraveled the patterns and structure of the 66 times table, it's time to put our knowledge into action.
By recognizing the doubling pattern and the rule for numbers ending in zero, you'll become more confident and proficient in multiplying numbers by 66.
## Sixty-six Multiplication Table
Read, Repeat and Learn Sixty-six times table and Check yourself by giving a test below
## Table of 66
66 Times table Test
## How much is 66 multiplied by other numbers?
@2024 PrintableMultiplicationTable.net |
# Find the remainder using remainder theorem find the remainder when $x^3-6x^2+2x-4$ is divided by $1-\frac{3x}{2}$.
Given: $x^3-6x^2+2x-4$ is divided by $1-\frac{3x}{2}$.
To do: To find the remainder using remainder theorem find the remainder when $x^3-6x^2+2x-4$ is divided by $1-\frac{3x}{2}$.
Solution:
Let $f( x)=x^3-6x^2+2x-4$ and $g( x)=1-\frac{3x}{2}$
Let $g( x)=1-\frac{3x}{2}=0$
$\Rightarrow \frac{3x}{2}=1$
$\Rightarrow 3x=2$
$\Rightarrow x=\frac{2}{3}$, on substituting this value in $f( x)$.
$f( \frac{2}{3})=( \frac{2}{3})^3-6( \frac{2}{3})^2+2( \frac{2}{3})-4$
$\Rightarrow f( \frac{2}{3})=( \frac{2\times2\times2}{3\times3\times3})-6( \frac{2\times2}{3\times3})+2( \frac{2}{3})-4$
$f( \frac{2}{3})=( \frac{8}{27})-6( \frac{4}{9})+2( \frac{2}{3})-4$
$f( \frac{2}{3})=( \frac{8}{27})-( \frac{24}{9})+( \frac{4}{3})-4$
$f( \frac{2}{3})=( \frac{8}{27})-( \frac{24\times3}{9\times3})+( \frac{4\times9}{3\times9})-4\times\frac{27}{27}$
$f( \frac{2}{3})=( \frac{8}{27})-( \frac{72}{27})+( \frac{36}{27})-\frac{108}{27}$
$f( \frac{2}{3})=( \frac{8-72+36-108}{27})$
$f( \frac{2}{3})=( \frac{44-180}{27})$
$f( \frac{2}{3})=( \frac{-136}{27})$
Thus, the remainder is $( \frac{-136}{27})$.
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Updated on: 10-Oct-2022
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# Video: AQA GCSE Mathematics Foundation Tier Pack 3 • Paper 3 • Question 4
Solve −4𝑡 = 20. Circle your answer. [A] 𝑡 = −1/5 [B] 𝑡 = 24 [C] 𝑡 =5 [D] 𝑡 = −5
02:45
### Video Transcript
Solve negative four 𝑡 equals 20. Circle your answer. Is it 𝑡 is equal to negative one-fifth, 𝑡 equals 24, 𝑡 equals five, or 𝑡 equals negative five?
We can solve this equation using one step. We do this by dividing both sides of the equation by negative four. This is because negative four 𝑡 means negative four multiplied by 𝑡. The opposite or inverse operation to multiplying by negative four is dividing by negative four. When we divide two positive numbers, we get a positive answer. Dividing a positive by a negative or a negative by a positive gives us a negative answer. And dividing a negative number by another negative number gives us a positive answer.
Using this bottom rule, the left-hand side of the equation becomes positive one 𝑡 as negative four 𝑡 divided by negative four is equal to positive one 𝑡. Dividing four by four gives us one. And the two negative signs become a positive. One 𝑡 can just be written as 𝑡. 20 divided by negative four is equal to negative five as 20 divided by four is five. And dividing a positive number by a negative number gives a negative answer. The solution to the equation negative four 𝑡 equals 20 is 𝑡 equals negative five. So we need to circle this answer.
We can check our answer by substituting this value of 𝑡 back into the original equation. Negative four multiplied by 𝑡 equals 20. Substituting in 𝑡 equals negative five gives us negative four multiplied by negative five. This is equal to 20 as multiplying two negative numbers gives us a positive answer. And four multiplied by five is 20.
Of the four options, 𝑡 equals negative one-fifth, 24, five, and negative five, the correct answer is negative five. Another way of thinking about this problem is to work out what we need to multiply negative four by to give us 20. The answer to this is negative five. |
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### Grade 8 - Mathematics7.23 Special Products: (a+b)2 = a2+2ab+b2
#### Formula: (a+b)2 = a2+2ab+b2
Example:
Expand (3x+4y)2
Solution:
(3x+4y)2 is in the form of (a+b)2, we have 3x for a and 4y for b
=(3x+4y)2 = (3x)2+2.3x.4y+(4y)2
= 9x2+24xy+16y2
Directions: Solve the following problems. Also write at least ten examples of your own.
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### Grade 8 - Mathematics7.23 Special Products: (a+b)2 = a2+2ab+b2
Q 1: Expand (5x+2y)25x2+20xy+4y225x2+20xy+4y225x2+10xy+4y225x2+20xy+2y2 Q 2: Expand (3x+2y)29x2+12xy+2y29x2+6xy+4y29x2+12xy+4y23x2+12xy+4y2 Q 3: Expand (2x+3y)24x2+6xy+9y22x2+12xy+9y24x2+12xy+3y24x2+12xy+9y2 Q 4: Expand (7x+2y)249x2+28xy+2y249x2+14xy+4y27x2+28xy+4y249x2+28xy+4y2 Q 5: Expand (2x+4y)22x2+16xy+16y24x2+8xy+16y24x2+16xy+16y24x2+16xy+4y2 Q 6: Expand (3x+5y)29x2+30xy+5y29x2+15xy+25y29x2+30xy+25y23x2+30xy+25y2 Q 7: Expand (8x+3y)28x2+48xy+9y264x2+48xy+3y264x2+48xy+9y264x2+24xy+9y2 Q 8: Expand (4x+7y)216x2+56xy+49y216x2+28xy+49y216x2+56xy+7y24x2+56xy+49y2 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
##### Basic Math & Pre-Algebra For Dummies
Every whole number greater than 1 has a prime factorization — that is, the list of prime numbers (including repeats) that equal that number when multiplied together. For example, here are the prime factorizations of 14, 20, and 300:
14 = 2 × 7
20 = 2 × 2 × 5
300 = 2 × 2 × 2 × 3 × 5
Factor trees are a useful tool for finding the prime factorization of a number. For example, to find the prime factorization 30, first find a factor pair — a pair of numbers that, when multiplied together, equals 30:
Note that 3 is circled because it's a prime number. But 10 is not prime, so you can continue the tree, finding a factor pair for 10:
This time, both 2 and 5 are circled because they're each prime numbers. When the bottom numbers on the factor tree are all prime, you have your answer: 30 = 2 × 3 × 5.
## Using prime factorization to find the GCF
You can use prime factorization to find the greatest common factor (GCF) of a set of numbers. This method often works better for large numbers, when generating lists of all factors can be time-consuming.
Here's how to find the GCF of a set of numbers, using prime factorization:
1. List the prime factors of each number.
2. Circle every common prime factor — that is, every prime factor that's a factor of every number in the set.
3. Multiply all the circled numbers.
The result is the GCF.
For example, suppose you want to find the GCF of 28, 42, and 70. Step 1 says to list the prime factors of each number. Step 2 says to circle every prime factor that's common to all three numbers:
As you can see, the numbers 2 and 7 are common factors of all three numbers, so multiply these two numbers as follows:
2 × 7 = 14
Thus, the GCF of 28, 42, and 70 is 14.
Knowing how to find the GCF of a set of numbers is important when you begin reducing fractions to lowest terms.
## Using prime factorization to find the LCM
One method for finding the least common multiple (LCM) of a set of numbers is to use the prime factorizations of those numbers. Here's how:
1. List the prime factors of each number.
Suppose you want to find the LCM of 18 and 24. List the prime factors of each number:
18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3
2. For each prime number listed, underline the most repeated occurrence of this number in any prime factorization.
The number 2 appears once in the prime factorization of 18 but three times in that of 24, so underline the three 2s:
18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3
Similarly, the number 3 appears twice in the prime factorization of 18 but only once in that of 24, so underline the two 3s:
18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3
3. Multiply all the underlined numbers.
Here's the product:
2 × 2 × 2 × 3 × 3 = 72
So the LCM of 18 and 24 is 72. This solution checks out because
18 × 4 = 72
24 × 3 = 72
Mark Zegarelli is a professional writer with degrees in both English and Math from Rutgers University. He has earned his living for many years writing vast quantities of logic puzzles, a hefty chunk of software documentation, and the occasional book or film review. Along the way, he’s also paid a few bills doing housecleaning, decorative painting, and (for ten hours) retail sales. He likes writing best, though. |
# 1969 AHSME Problems/Problem 28
## Problem
Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$, such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$. Then $n$ is:
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 4\quad \text{(E) } \infty$
## Solution
$[asy] draw(circle((0,0),50)); draw((-50,0)--(-10,20)--(50,0)--(-50,0)); draw((-10,20)--(30,40)--(50,0),dotted); dot((-50,0)); label("A",(-50,0),W); dot((50,0)); label("B",(50,0),E); dot((-10,20)); label("P",(-10,20),S); dot((30,40)); label("C",(30,40),NE); [/asy]$
Let $A$ and $B$ be points on diameter. Extend $AP$, and mark intersection with circle as point $C$.
Because $AB$ is a diameter, $\angle ACB = 90^\circ$. Also, by Exterior Angle Theorem, $\angle ACB + \angle CBP = \angle APB$, so $\angle APB > \angle ACB$, making $\angle APB$ an obtuse angle.
By the Law of Cosines, $AP^2 + BP^2 - 2 \cdot AP \cdot BP \cdot \cos{\angle APB} = 4$. Since $AP^2 + BP^2 = 3$, substitute and simplify to get $\cos{\angle APB} = -\frac{1}{2 \cdot AP \cdot BP}$. This equation has infinite solutions because for every $AP$ and $BP$, where $AP + BP \ge 2$ and $AP$ and $BP$ are both less than $2$, there can be an obtuse angle that satisfies the equation, so the answer is $\boxed{\textbf{(E)}}$. |
# Lesson 1
Story Problems and Expressions
## Warm-up: Notice and Wonder: A Library (10 minutes)
### Narrative
The purpose of this warm-up is to elicit the idea that math is found everywhere in our world. Students look for mathematical situations in a picture of a library, which will be helpful when students solve story problems about the library in later activities. While students may notice and wonder many things about this image, noticing numbers or quantities in the image are the important discussion points.
### Launch
• Groups of 2
• Display the image.
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
What do you notice?
What do you wonder?
### Activity Synthesis
• “Tell a math story based on the picture.” (There were two books on the shelf. Then the librarian put two more books on the shelf. How many books are on the shelf now? There are six pictures hanging from the ceiling. Three pictures fall down. How many pictures are still hanging from the ceiling?)
## Activity 1: The Library (20 minutes)
### Narrative
The purpose of this activity is for students to make sense of the structure of a Take From, Result Unknown story problem. Students have access to connecting cubes or two-color counters and 10-frames, which they may choose to use strategically to represent and solve the problem (MP5). Some students may apply the ideas from the previous unit where they related subtraction to counting back. As students share their methods, the teacher records their thinking. It is important for the teacher to focus attention on the expression during the activity synthesis. Although some students may attempt to write equations and do so accurately, this representation will be the focus in a later lesson.
Students keep their books closed for the launch of the activity, as the teacher displays and reads the problem.
MLR8 Discussion Supports. Synthesis: Display sentence frames to support whole-class discussion: “_____ and _____ are the same/alike because . . .” and “_____ and _____ are different because . . .” Encourage students to use these sentence frames to compare and contrast representations.
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• Give students access to 10-frames and connecting cubes or two-color counters.
• Display the image from the warm-up.
• “This is a picture of a library. Talk to your partner about what you know and what you wonder about libraries.”
• 3 minutes: partner discussion
• Share and record what students know and wonder about libraries.
• "We are going to solve a lot of story problems about libraries."
• Display and read the numberless story.
• 30 seconds: quiet think time
• 1 minute: partner discussion
• Share responses.
• If not already mentioned, ask, “Are there more or fewer kids at the library after some go home?”
### Activity
• Ask students to open their books.
• Read the problem with numbers.
• 2 minutes: independent work time
• “Share your thinking with your partner.”
• 2 minutes: partner discussion
• Monitor for students who solve or represent the problem in the following ways:
• objects
• drawings
• count back
• an expression ($$9 - 2$$)
### Student Facing
1. Some kids were at the library.
Then some of the kids went home.
What do you notice?
What do you wonder?
2. There were 9 kids at the library.
Then 2 of the kids went home.
How many kids are at the library now?
Show your thinking using drawings, numbers, or words.
### Advancing Student Thinking
If students use more or fewer than nine objects to represent the kids at the library, consider asking:
• “How did you choose how many connecting cubes to take out?”
• “There are nine kids at the library. Which connecting cubes show the nine kids? Which connecting cubes show the two kids that went home?”
### Activity Synthesis
• Invite previously identified students to share in the order listed above.
• “How are these representations the same? How are they different?”
• If needed, display $$9 - 2$$.
• “How does this expression match the story problem?”
## Activity 2: Story Problems about the Library (20 minutes)
### Narrative
The purpose of this activity is for students to solve Add To and Take From, Result Unknown problems in a way that makes sense to them (MP1). The problems use the same numbers in order to encourage students to think about the action in the problem and how it relates to operations. Students may represent the problems with objects or drawings, and count all, count on, or count back to solve the problems. Students write expressions and some may attempt to write equations.
During the lesson synthesis, students are re-introduced to equations. They notice how the answer to the question is represented in the equation. The teacher draws a box around the answer to signify that this part of the equation is the answer to the question.
Action and Expression: Internalize Executive Functions. Invite students to plan a method, including the tools they will use, for solving the story problems. If time allows, invite students to share their plan with a partner before they begin.
Supports accessibility for: Conceptual Processing, Organization
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• Give students access to 10-frames and connecting cubes or counters.
### Activity
• Read the story problems.
• 6 minutes: independent work time
• “Share your thinking with your partner. Be sure that you both agree on the answer.”
• 4 minutes: partner discussion
### Student Facing
1. 5 books were on a shelf.
Clare put 2 more books on the shelf.
How many books are on the shelf now?
Show your thinking using drawings, numbers, or words.
Expression: ________________________________
2. 6 books were stacked up on the table.
4 of the books fell on the floor.
How many books are still on the table?
Show your thinking using drawings, numbers, or words.
Expression: ________________________________
3. 6 kids were listening to a story.
4 more kids joined the group.
How many kids are listening to the story now?
Show your thinking using drawings, numbers, or words.
Expression: ________________________________
4. There were 5 computers turned on. The librarian turned 2 of the computers off.
How many computers are still on?
Show your thinking using drawings, numbers, or words.
Expression: ________________________________
### Activity Synthesis
• Review solutions and expressions for both story problems with the numbers 5 and 2.
• “How does the expression match the story?”
• If needed ask, "What does the 5 represent? What does the 2 represent?”
• “How are these problems the same? How are they different?” (They have the same numbers, but different answers. Problem 1 is addition, problem 4 is subtraction.)
## Lesson Synthesis
### Lesson Synthesis
Display the problem about books on a shelf.
“Today we wrote expressions to represent story problems. For this problem we wrote the expression 5 + 2. Then we solved the problem. We can write an equation that shows the answer. The equation “$$5 + 2 = 7$$” tells us that 5 plus 2 is the same amount as 7.”
“What does the 7 represent?” (the total number of books on the shelf, 5 books + 2 books)
“When we write an equation, we can draw a box around the number that shows the answer to the question. That means the equation would be written as $$5 + 2 = \boxed{7}$$.“ |
# Mean, Median and Mode
Mean, Median and Mode
by Jessica Wesaquate and Andrea Rogers
Strand:
Number
Three
Students will be able to view how a tipi raising is performed.
Students will be able to create estimations based on their observations
of the tipi raising video clips.
Materials:
tipi-raising videos, graph paper, pencils, math logs/duo tangs/journals
Video Clips:
As the teacher you may choose to use the video clips that demonstrate the tipi-raising done with Elder Glen Anaquod using a Saulteaux perspective. Or you may choose the video clips that demonstrate the tipi raising done with Tim Haywahe using a Nakota perspective. Depending on your area, it may be appropriate to choose one over the other. If time permits, showing them both tipi raisings is a good opportunity to compare and contrast different traditions and teachings.
Introduction:
As a class, you are going to show the students the tipi raising videos. Start with showing them the clip that demonstrates them measuring the first poles on the canvas. As they are viewing the videos, have students make estimations to how many people they think could fit comfortably in this tipi? They should record this in their math logs/duo tangs/journals.
Step Two:
Play the next clips. Pause the clip when the canvas is 1/2 to ¾ around the poles.
Allow students to re-think their estimation. Do they still want to go with their original estimation or make a new estimation? Have them record their estimation and state whether it is new or if it remained the same.
Step Three:
Watch the remainder of the videos and pause the last slide where the students can see the complete tipi. This is their last opportunity to either keep their estimation, or create a new one, and record.
Introduce or review what mean, median and mode are with your students. Choose five students’ numbers to work with and record these on the board (students should have recorded three numbers each). Have students work individually to determine what the mean, median and mode are of those numbers. Review as a class.
Mean:
Average
Median:
arrange the numbers from lowest to highest and the median is the middle number
Activity:
On construction paper, have students trace their hand (fingers together and thumb close in) and then cut it out. This will be used as a non-standard measuring tool for items around the classroom. If you have cultural items available to measure, great, but if not this can be used for basically anything. For example, if you have a rain stick in your classroom have students measure how many handprints long it is. Remember to share background on the meaning of the rain stick so they understand its significance. Other items you can measure are things like their desks, tables, drawers, windows, and etcetera.
You can set up stations for this activity so that students are not trying to measure the same thing all at the same time, also so that you can take anecdotal records on the way they measure items*, behaviors, other.
*Do they measure items with the hand vertically or horizontally, do they use the width of their hand cut-out or the length of it?
Mode:
the number that occurs most often
Optional Activities:
Have the students do some graphing with this information. Students can create a line or bar graph. Have them note where they see the most common guesses. Have the class make a consensus on how many people can fit into this tipi.
Find a place outdoors around the school where you can measure a diameter of anywhere for 10 to 25 feet using broken branches, or other objects Mother Nature provides us with (to represent the tipi) to explore their estimations. You can also use masking tape if needed. Have students sit inside the circle, how many fit comfortably?
Aboriginal Perspectives is supported by the University of Regina, the Imperial Oil Foundation, the Canadian Mathematical Society and the Pacific Institute for the Mathematical Sciences.
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Did you recognize that 53 is composed as 35 in that is hexadecimal representation, i m sorry is the number v the very same digits together 53 in reversed order? In this lesson, we will certainly calculate the determinants of 53, prime components of 53, and also factors that 53 in pairs together with solved instances for a much better understanding.
You are watching: What is the prime factorization of 53
Factors of 53: 1 and 53Prime administer of 53: 53 = 531
1 What space the factors of 53? 2 How to Calculate determinants of 53? 3 Factors of 53 by prime Factorization 4 Factors of 53 in Pairs 5 FAQs on factors of 53
## What space the factors of 53?
The factors of 53 room numbers (integers) that division 53 without any type of remainder. For example, 53 is a element of itself because 53 divides itself without any remainder.
Interestingly, 1, i m sorry is the quotient of the over division, is also a factor of 53. Examine if you get 0 together the remainder on separating 53 through 1 using long division.
Tips and also Tricks:
A factor of a number is always less than or equal to the number.1 and also the number chin are always the factors of a number.
## How to calculation the components of 53?
In the previous section, us learned that factors can be found using division. Since multiplication is the reverse process of division, us can uncover the factors using multiplication as well.To find the factors of 53, we compose 53 together a product of any kind of two numbers in every the feasible ways. The number that us multiply to gain 53 are dubbed the determinants of 53.
53 cannot be created as the product of any type of two numbers other than these 2 ways.
Hence, the factors that 53 are 1 and 53.
Explore components using illustrations and also interactive examples.
## Factors that 53 by prime Factorization
Prime administer of 53 is expressing it as the product of element numbers. Yet as we questioned in the ahead section, 53 has actually only two factors, 1 and also 53.Thus, 53 chin is a element number.
So the element factorization the 53 is 53 =531Also, we know that 1 is a factor of every number. Thus, factors the 53 room 1 and also 53.Now the we have done the element factorization of our number, we have the right to multiply them and also get the various other factors. Can you try and find out if every the factors are spanned or not? and as you might have already guessed it, because that prime numbers, there space no other factors. But how carry out we find the factors of a composite number using prime factorization? Let"s discover this v some examples.
See more: Sprint Htc One M7 Sim Unlock The Htc One M7, M8, And M9, How To Sim Unlock Htc One Sprint, All Carriers
Challenging Questions:
Liam desires to tree 24 rose plants, 30 lily plants, and 42 jasmine tree in his garden. If he wants to put the same number of plants in every row and also each row has actually only one kind of plant, what is the greatest variety of plants he can put in one row?
## Factors that 53 in Pairs
The pair components of 53 are derived by writing 53 as a product of two numbers in all possible ways. In each product, both multiplicands are the factors of 53. |
# How do you find the standard deviation formula?
## How do you find the standard deviation formula?
To calculate the standard deviation of those numbers:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result.
3. Then work out the mean of those squared differences.
4. Take the square root of that and we are done!
### How do we calculate probability?
Divide the number of events by the number of possible outcomes. This will give us the probability of a single event occurring. In the case of rolling a 3 on a die, the number of events is 1 (there’s only a single 3 on each die), and the number of outcomes is 6.
#### What is the formula for calculating probability?
Divide the number of events by the number of possible outcomes.
1. Determine a single event with a single outcome.
2. Identify the total number of outcomes that can occur.
3. Divide the number of events by the number of possible outcomes.
4. Determine each event you will calculate.
5. Calculate the probability of each event.
How do you find standard deviation with N and P?
The mean of the distribution (μx) is equal to n * P . The variance (σ2x) is n * P * ( 1 – P ). The standard deviation (σx) is sqrt[ n * P * ( 1 – P ) ].
How can you determine the standard deviation with probability?
Like data, probability distributions have standard deviations. To calculate the standard deviation ( σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root.
## What are the assumptions of standard deviation?
For the within-subject standard deviation, it is assumed that the size of the deviation is not related to the magnitude of the measurement . This can be assessed graphically, by plotting the individual subject’s standard deviations against their means.
### How do you calculate standard deviation of a variable?
There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together.
#### What are the steps of standard deviation?
The steps to calculating the standard deviation are: Calculate the mean of the data set (x-bar or 1. μ) Subtract the mean from each value in the data set2. Square the differences found in step 23. Add up the squared differences found in step 34.
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# Word Problems on Addition of Mixed Fractions | Adding Mixed Numbers Word Problems
Mixed Fractions are the ones having a Whole Number and a Fraction. Learn how to solve problems on adding mixed fractions by availing our quick resource on Word Problems on Addition of Mixed Fractions. Try to solve the Questions on Adding Mixed Numbers available on your own before you cross-check with the respective Solutions and explanations. For better understanding, we have provided the Examples on Mixed Numbers Addition step by step. Understand the Problem Solving Strategy Used and Solve the Related Problems on your own.
Also, Refer:
Example 1.
Ram walked 1$$\frac { 3 }{ 5 }$$Â km on Monday, 3 $$\frac { 1 }{ 6 }$$ km on Tuesday, and 2 $$\frac { 1 }{ 12 }$$ Â km on Wednesday. Find out the distance he walked in all?
Solution:
Ram walked on monday= 1 $$\frac { 3 }{ 5 }$$
Ram walked on tuesday=3 $$\frac { 1 }{ 6 }$$
Ram walked on wednesday=2 $$\frac { 1 }{ 12 }$$
The distance he walked in all days= 1 $$\frac { 3 }{ 5 }$$ + 3$$\frac { 1 }{ 6 }$$ + 2 $$\frac { 1 }{ 12 }$$
=$$\frac { 8 }{ 5 }$$ +$$\frac { 19 }{ 6 }$$ + $$\frac { 25 }{ 12 }$$
=$$\frac {8 × 12 }{ 5 ×12 }$$ +$$\frac {19 × 10 }{ 6 ×10 }$$ +$$\frac {25 ×5  }{ 12 ×5 }$$ ( LCM of 5,6,12 is 60)
= $$\frac { 96 }{ 60 }$$+$$\frac { 190 }{ 60 }$$1 +$$\frac { 125 }{ 60 }$$
=$$\frac { 411 }{ 60 }$$
=$$\frac { 137 }{ 20 }$$ =6 $$\frac { 17 }{ 20 }$$
Hence, Ram walked 6 $$\frac { 17 }{ 20 }$$ km.
Example 2.
Laxman bought 2 $$\frac { 1 }{ 2 }$$ kg wheat, 4$$\frac { 1 }{ 4 }$$Â of sugar, 10 $$\frac { 1 }{ 2 }$$ kg rice. How many kg of items did he buy?
Solution:
Laxman bought wheat= 2 $$\frac { 1 }{ 2 }$$ kg
Laxman bought sugar=4 $$\frac { 1 }{ 4 }$$ kg
Laxman bought rice=10 $$\frac { 1 }{ 2 }$$ kg
Total no. of kg of items Laxman bought = 2 $$\frac { 1 }{ 2 }$$ + 4 $$\frac { 1 }{ 4 }$$ + 10 $$\frac { 1 }{ 2 }$$
=$$\frac { 5 }{ 2 }$$ +$$\frac { 17 }{ 4 }$$ +$$\frac { 21 }{ 2 }$$
=$$\frac { 5 ×2}{ 2×2 }$$ + $$\frac { 17}{ 4 }$$  +$$\frac { 21 ×2 }{ 2×2 }$$
= $$\frac { 10 }{ 4 }$$ +$$\frac { 17 }{ 4 }$$ + $$\frac { 42 }{ 4 }$$
=Â $$\frac { 69 }{ 4 }$$ =17 $$\frac { 1 }{ 4 }$$ .
Therefore, Laxman bought 17 $$\frac { 1 }{ 4 }$$ kg of items.
Example 3.
Rajesh donated 19 $$\frac { 1 }{ 2 }$$ kg of rice, 25 $$\frac { 1 }{ 4 }$$ kg of wheat, and 11 $$\frac { 3 }{ 4 }$$ of oil. Find how many kgs of groceries he donated?
Solution:
Rajesh donated rice=19 $$\frac { 1 }{ 2 }$$
Rajesh donated wheat=25 $$\frac { 1 }{ 4 }$$
Rajesh donated oil=11$$\frac { 3}{ 4 }$$
Total no. of kg of groceries Rajesh donated =19 $$\frac { 1 }{ 2 }$$+ 25 $$\frac { 1 }{ 4 }$$ + 11 $$\frac { 3 }{ 4 }$$
=$$\frac { 39 }{ 2 }$$+$$\frac { 101 }{ 4 }$$+$$\frac { 47 }{ 4 }$$
= $$\frac { 39 × 2 }{ 2 × 2 }$$+$$\frac { 101 }{ 4 }$$+ $$\frac { 47 }{ 4 }$$
=$$\frac { 78 }{ 4 }$$+$$\frac { 101 }{ 4 }$$+ $$\frac { 47 }{ 4 }$$47/4
=$$\frac { 226 }{ 4 }$$=$$\frac { 113 }{ 2 }$$=56$$\frac {1 }{ 2 }$$
Hence, Rajesh donated 56 $$\frac {1 }{ 2 }$$ kgs of groceries.
Example 4.
Giri went on walking 3 $$\frac {1 }{ 2 }$$Â km, and then cycling by 4 $$\frac {1 }{ 4 }$$ km and then he took a lift and traveled 8 $$\frac {3 }{ 8 }$$ km. Find out how much distance he traveled?
Solution:
Giri went on walking =3 $$\frac {1 }{ 2 }$$Â km
Giri went on cycling=4 $$\frac {1 }{ 4 }$$ km
Giri took a lift and travelled=8 $$\frac {3}{ 8 }$$ km
The total distance he travelled= 3 $$\frac {1 }{ 2 }$$+4 $$\frac {1 }{ 4 }$$ + 8 $$\frac {3}{ 8 }$$
=$$\frac {7}{ 2 }$$+$$\frac {9 }{ 4 }$$+$$\frac {67 }{ 8 }$$
=$$\frac {7 × 4 }{ 2 ×4 }$$ + $$\frac {9 × 2 }{ 4 × 2}$$  + $$\frac {67}{ 8 }$$ (LCM of 2,4,8 is 8)
=$$\frac {28}{ 8 }$$+$$\frac {18}{ 8 }$$+ $$\frac {67}{ 8 }$$
=$$\frac {113}{ 8 }$$=14 $$\frac {1}{ 8 }$$
Example 5.
Sameera went to a market and bought mangoes 3 $$\frac {3}{ 4 }$$ and her sister bought 4 $$\frac {5}{ 8 }$$ mangoes. Find how many kgs of mangoes bought by Sameera and her sister?
Solution:
Sameera bought mangoes= 3 $$\frac {3}{ 4 }$$
Sameera sister bought mangoes=4 $$\frac {5}{ 8 }$$
Total no. of kg of mangoes bought by Sameera and her sister=3 $$\frac {3}{ 4 }$$Â + 4 $$\frac {5}{ 8 }$$
= $$\frac {15}{ 4 }$$ + $$\frac {29}{ 8 }$$
= $$\frac {15 × 2}{ 4× 2}$$+$$\frac {29}{ 8 }$$
=Â $$\frac {30}{ 8 }$$+$$\frac {29}{ 8 }$$=$$\frac {59}{ 8 }$$
=7$$\frac {3}{ 8 }$$
Therefore, Sameera and her sister bought 7$$\frac {3}{ 8 }$$Â kg of mangoes.
Example 6.
For a birthday party, Akhil distributed 14 $$\frac {1}{ 2 }$$ liters of Pepsi, 10 $$\frac {3}{ 4 }$$ liters of sprite, and 12 $$\frac {5}{ 8 }$$ liters of thum sup. Find how many liters of cool drinks he distributed at the birthday party?
Solution:
Akhil distributed pepsi= 14 $$\frac {1}{ 2 }$$
Akhil distributed sprite=10 $$\frac {3}{ 4 }$$
Akhil distributed thums up=12 $$\frac {5}{ 8 }$$
Total no. of liters of cool drinks Akhil distributed in the party=14 $$\frac {1}{ 2 }$$ +10 $$\frac {3}{ 4 }$$ +12 $$\frac {5}{ 8 }$$
=$$\frac {29}{ 2 }$$+$$\frac {43}{ 4 }$$+$$\frac {101}{ 8 }$$
=$$\frac {29 ×4}{ 2 ×4 }$$+$$\frac {43 ×2}{ 4 ×2 }$$+$$\frac {101}{ 8 }$$
=$$\frac {116}{ 8 }$$+ $$\frac {86}{ 8 }$$+$$\frac {101}{ 8 }$$
=$$\frac {303}{ 8 }$$=37 $$\frac {7}{ 8 }$$
Akhil distributed 37 $$\frac {7}{ 8 }$$ liters of cool drinks at the birthday party.
Example 7.
In a competition between two friends, Anjana prepared 20 $$\frac {1}{ 2 }$$ lit of orange juice and Sanjana prepared 22 $$\frac {5}{ 8 }$$ of orange juice. Find how many liters of juice are prepared by both of them?
Solution:
Anjana prepared orange juice= 20 $$\frac {1}{ 2 }$$
Sanjana prepared orange juice=22 $$\frac {5}{ 8}$$
No. of liters of juice prepared by both of them= 20 $$\frac {1}{ 2 }$$ + 22 $$\frac {5}{ 8 }$$
= $$\frac {41}{ 2 }$$+ $$\frac {181}{ 8 }$$
=$$\frac {41 ×4}{ 2 ×4 }$$ +$$\frac {181}{ 8 }$$
=$$\frac {164}{ 8 }$$+$$\frac {181}{ 8 }$$=$$\frac {345}{ 8 }$$
=43$$\frac {1}{ 8 }$$
Anjana and sanjana both prepared 43 $$\frac {1}{ 8 }$$ liters of juice.
Example 8.Â
Sriram has wires of length 6 $$\frac {3}{ 4 }$$Â m, 10 $$\frac {5}{ 8 }$$. Find the length of both the wires?
Solution:
First wire length= 6 $$\frac {3}{ 4 }$$
Second wire length=10 $$\frac {5}{ 8 }$$
Length of both the wires= 6 $$\frac {3}{ 4 }$$Â +10 $$\frac {5}{ 8 }$$
=$$\frac {27}{ 4 }$$Â +$$\frac {85}{ 8 }$$
= $$\frac {27 × 2}{ 4 × 2 }$$ +$$\frac {85}{ 8 }$$
=$$\frac {54}{ 8 }$$Â +$$\frac {85}{ 8 }$$ =$$\frac {139}{ 8 }$$
=17 $$\frac {3}{ 8 }$$
The length of both wires is 17 $$\frac {3}{ 8 }$$ .
Example 9.
Bhaskar spent 8 $$\frac {1}{ 2 }$$ hours on work, 2 $$\frac {1}{ 2}$$ hours on the playground. He spent 1 $$\frac {3}{ 4 }$$on walking. Find how much time he spent on the day?
Solution:
Bhaskar spent on work=8 $$\frac {1}{ 2 }$$
Bhaskar spent on playground=2 $$\frac {1}{ 2 }$$
Bhaskar spent on walking=1 $$\frac {3}{ 4}$$
Bhaskar spent on the day= 8 $$\frac {1}{ 2 }$$ + 2 $$\frac {1}{ 2 }$$ + 1$$\frac {3}{ 4 }$$
= $$\frac {17}{ 2 }$$ +$$\frac {5}{ 2 }$$ +$$\frac {7}{ 4 }$$
= $$\frac {17 × 2}{ 2 ×2 }$$ + $$\frac { 5× 2}{ 2 ×2 }$$  +$$\frac {7}{ 4 }$$
=$$\frac {34}{ 4 }$$ + $$\frac {10}{ 4 }$$ +$$\frac {7}{ 4 }$$
=$$\frac {51}{ 4 }$$ =12 $$\frac {3}{ 4 }$$
Therefore, Bhaskar spent 12 $$\frac {3}{ 4 }$$Â hours a day.
Example 10.
Srikrishna bought 5 $$\frac {1}{ 4 }$$ of vegetables, 20 $$\frac {3}{ 4 }$$Â kg of groceries, and 1 $$\frac {1}{ 2}$$Â kg of chicken. Find how many kgs he bought?
Solution:
Srikrishna bought vegetables= 5 $$\frac {1}{ 4 }$$ kg
Srikrishna bought groceries=20 $$\frac {3}{ 4 }$$Â Â kg
Srikrishna bought chicken= 1 $$\frac {1}{ 2 }$$Â kg
Srikrishna bought all=5 $$\frac {1}{ 4 }$$ +20 $$\frac {3}{ 4 }$$ +1 $$\frac {1}{ 2 }$$
= $$\frac {21}{ 4 }$$ +$$\frac {83}{ 4 }$$ +$$\frac {3}{ 2 }$$
=$$\frac {21}{ 4 }$$ +$$\frac {83}{ 4 }$$+ $$\frac {3 ×2}{ 2 × 2 }$$
=$$\frac {21}{ 4 }$$ +$$\frac {83}{ 4 }$$ + $$\frac {6}{ 4 }$$
=$$\frac {110}{ 4 }$$=$$\frac {55}{ 2 }$$=27 $$\frac {1}{ 2 }$$
Srikrishna bought 27 $$\frac {1}{ 2 }$$ kg.
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# PSAT Math : How to find rate
## Example Questions
← Previous 1 3
### Example Question #1 : How To Find Rate
Bob can build a house in 3 days. Gary can build a house in 5 days. How long does it take them to build a house together?
5/4 days
3/2 days
2 days
15/8 days
4 days
15/8 days
Explanation:
DO NOT pick 4 days, which would be the middle number between Bob and Gary's rates of 3 and 5 days respectively. The middle rate is the answer that students always want to pick, so the SAT will provide it as an answer to trick you!
Let's think about this intuitively before we actually solve it, so hopefully you won't be tempted to pick a trick answer ever again! Bob can build the house in 3 days if he works by himself, so with someone else helping him, it has to take less than 3 days to build the house! This will always be true. Never pick the middle rate on a combined rates problem like this!
Now let's look at the problem computationally. Bob can build a house in 3 days, so he builds 1/3 of a house in 1 day. Similarly, Gary can build a house in 5 days, so he builds 1/5 of a house in 1 day. Then together they build 1/3 + 1/5 = 5/15 + 3/15 = 8/15 of the house in 1 day.
Now, just as we did to see how much house Gary and Bob can build separately in one day, we can take the reciprocal of 8/15 to see how many days it takes them to build a house together. (When we took the reciprocal for Bob, 3 days/1 house = 1/3 house per day.) The reciprocal of 8/15 is 15/8, so they took 15/8 days to build the house together. 15/8 days is almost 2 days, which seems like a reasonable answer. Make sure your answer choices make sense when you are solving word problems!
### Example Question #505 : Arithmetic
A family is on a road trip from Cleveland to Virginia Beach, totaling 600 miles. If the first half of the trip is completed in 6.5 hours and the second half of the trip is completed in 5.5 hours, what is the average speed in miles per hour of the whole trip?
65 mph
50 mph
55 mph
60 mph
45 mph
50 mph
Explanation:
Take the total distance travelled (600 miles) and divide it by the total time travelled (6.5 hrs + 5.5 hrs = 12 hours) = 50 miles/hour
### Example Question #506 : Arithmetic
Two electric cars begin moving on circular tracks at exactly 1:00pm. If the first car takes 30 minutes to complete a loop and the second car takes 40 minutes, what is the next time they will both be at the starting point?
2:40 p.m.
3:30 p.m.
4:00 p.m.
3:00 p.m.
1:35 p.m.
3:00 p.m.
Explanation:
Call the cars “Car A” and “Car B”.
The least common multiple for the travel time of Car A and Car B is 120. We get the LCM by factoring. Car A’s travel time gives us 3 * 2 * 5; Car B’s time gives us 2 * 2 * 2 * 5. The smallest number that accommodates all factors of both travel times is 2 * 2 * 2 * 3 * 5, or 120. There are 60 minutes in an hour, so 120 minutes equals two hours. Two hours after 1:00pm is 3:00pm.
### Example Question #1 : How To Find Rate
If Jon is driving his car at ten feet per second, how many feet does he travel in 30 minutes?
18,000
600
1800
12,000
5,800
18,000
Explanation:
If Jon is driving at 10 feet per second he covers 10 * 60 feet in one minute (600 ft/min). In order to determine how far he travels in thirty minutes we must multiply 10 * 60 * 30 feet in 30 minutes.
### Example Question #508 : Arithmetic
An arrow is launched at 10 meters per second. If the arrow flies at a constant velocity for an hour, how far has the arrow gone?
36,000 meters
3600 meters
600 meters
100 meters
36,000 meters
Explanation:
There are 60 seconds in a minute and 60 minutes in an hour, therefore 3600 seconds in an hour. The arrow will travel 3600x10= 36,000 meters in an hour.
### Example Question #509 : Arithmetic
If Jack ran at an average rate of 7 miles per hour for a 21 mile course, and Sam ran half as fast for the same distance, how much longer did it take for Sam to run the course than Jack?
3 hours
2.5 hours
1 hour
2 hours
4 hours
3 hours
Explanation:
Using the rate formula: Distance = Rate x Time,
Since Jack’s speed was 7 mph, Jack completed the course in 3 hours
21 = 7 x t
t = 3
Sam’s speed was half of Jack’s speed: 7/2 = 3.5
21 = 3.5 x t
t = 6
Therefore it took Sam 3 hours longer to run the course.
### Example Question #21 : Proportion / Ratio / Rate
If a pail collects x ounces of dripping water every 15 minutes, how many ounces will it collect in h hours?
4xh
15xh
xh
15x/h
4x/h
4xh
Explanation:
Algebraic solution: First, convert minutes to hours.
60/15 = 4, so there are 4 15-minute increments in each hour. Therefore, 4x ounces of water are collected each hour. Multiply by h to get 4xh as the solution
Plug-in method: Just choose numbers.
x = 2
h = 3
If 2 ounces drip in 15 minutes, how many ounces will drip in one hour?
2/15 = x/60
15x = 120
x = 8
If 8 ounces drip in one hour, how many ounces will drip in 3 hours? (remember we chose that h = 3)
3 x 8 = 24
This is the answer we are looking for.
Plug x = 2, and h = 3 into each answer choice, to determine which will work. Remember you must plug into every answer choice in case more than one works. In that case, choose different values for x and h, and plug into only the choices that worked the first time.
### Example Question #1 : How To Find Rate
Mary can make 20 snowballs in an hour. Mark can make 15 snowballs in 30 minutes. If they work together, how long will it take them to make 150 snowballs?
2.5 hours
120 minutes
3 hours
185 minutes
3 hours
Explanation:
If Mark makes 15 snowballs in 30 minutes, he can make 30 snowballs in an hour. Working together they can 50 snowballs in one hour. 150 snowballs divided by the amount they can make in one hour (50) will give us the total time it will take them to make 150 snowballs. In this case, 3 hours.
### Example Question #2 : How To Find Rate
Car X used 4 gallons of gas in one week, and gets 10 miles to the gallon. If car Y went the same number of miles but only gets 8 miles to the gallon, how much gas did car Y use?
5 gallons
8 gallons
10 gallons
4 gallons
5 gallons
Explanation:
We first use the data for car X to conclude that car X went 40 miles (4gallons*10mi/gallon). We then use 40 miles for car Y, and divide 40 by 8, to give us 5 gallons of gas.
### Example Question #1 : How To Find Rate
Bob and Sally are doing chores. It takes them 10 hours to do one of their chores. Assuming everyone works at the same rate, how many of their friends would they need to get to help them to do their chores in 2 hours?
10
8
None of the above
5 |
Convert from decimal come fraction. Transform 2.3333 to Fraction. Decimal to fraction chart and also calculator. Writes any kind of decimal number as a fraction.
You are watching: What is 2.3333 as a fraction
## How to transform a Decimal come a portion - Steps
Step 1: create down the decimal together a fraction of one (decimal/1);Step 2: If the decimal is not a entirety number, multiply both top and also bottom by 10 until you get an interger in ~ the numerator.
Learn more reading the examples listed below or use our self-explaining calculator above
## Convert decimal 0.05 come a fraction
0.05 = 1/20 together a fraction
### Step by step Solution
To convert the decimal 0.05 come a fraction follow these steps:
Step 1: compose down the number together a portion of one:
0.05 = 0.05/1
Step 2: multiply both top and bottom by 10 because that every number after the decimal point:
As we have 2 number after the decimal point, us multiply both numerator and also denominator through 100. So,
0.05/1 = (0.05 × 100)/(1 × 100) = 5/100.
Step 3: simplify (or reduce) the fraction:
5/100 = 1/20 when reduced to the most basic form.
## What is 0.45 as a fraction?
0.45 = 9/20 together a fraction
### Step by action Solution
To convert the decimal 0.45 come a portion follow these steps:
Step 1: compose down the number as a fraction of one:
0.45 = 0.45/1
Step 2: multiply both top and bottom by 10 for every number after ~ the decimal point:
As we have 2 number after the decimal point, we multiply both numerator and also denominator by 100. So,
0.45/1 = (0.45 × 100)/(1 × 100) = 45/100.
Step 3: leveling (or reduce) the fraction:
45/100 = 9/20 when reduced to the easiest form.
## Equivalent portion for 1.3 percent
1.3 = 13/10 = 13/10 as a fraction
### Step by step Solution
To transform the decimal 1.3 come a fraction follow these steps:
Step 1: compose down the number as a fraction of one:
1.3 = 1.3/1
Step 2: main point both top and bottom by 10 because that every number ~ the decimal point:
As we have 1 number after the decimal point, we multiply both numerator and also denominator through 10. So,
1.3/1 = (1.3 × 10)/(1 × 10) = 13/10.
(This portion is alread reduced, we can"t minimize it any type of further).
As the numerator is better than the denominator, we have actually an not correct fraction, so we can additionally express it together a blended NUMBER, thus 13/10 is additionally equal to 1 3/10 once expressed together a blended number.
## Conversion table: fraction to decimal inches and millimeter equivalence
To convert fractions come decimals and millimeters and also vice-versa use this formula: 1 customs = 25.4 mm exactly, therefore ...To convert from customs to millimeter main point inch worth by 25.4.To convert from millimeter inch division millimeter worth by 25.4.
an easier means to carry out it is to use the table below. How?
### Example 1
Convert 1 1/32" come mm: uncover 1 1/32 and read to the ideal under mm column! friend will discover 26.1938.
### Example 2
Convert 0.875 decimal inches come inches (fraction form).Look down the decimal obelisk until you discover 0.875, then review to the left to discover 7/8 inchesor relocate to the right column to discover the mm value!
fractioninchesmm
1/640.01560.3969
1/320.03130.7938
3/640.04691.1906
1/160.06251.5875
5/640.07811.9844
3/320.09382.3813
7/640.10942.7781
1/80.12503.1750
9/640.14063.5719
5/320.15633.9688
11/640.17194.3656
3/160.18754.7625
13/640.20315.1594
7/320.21885.5563
15/640.23445.9531
1/40.25006.3500
17/640.26566.7469
9/320.28137.1438
19/640.29697.5406
5/160.31257.9375
21/640.32818.3344
11/320.34388.7313
23/640.35949.1281
3/80.37509.5250
25/640.39069.9219
13/320.406310.3188
27/640.421910.7156
7/160.437511.1125
29/640.453111.5094
15/320.468811.9063
31/640.484412.3031
1/20.500012.7000
33/640.515613.0969
17/320.531313.4938
35/640.546913.8906
9/160.562514.2875
37/640.578114.6844
19/320.593815.0813
39/640.609415.4781
5/80.625015.8750
41/640.640616.2719
21/320.656316.6688
43/640.671917.0656
11/160.687517.4625
45/640.703117.8594
23/320.718818.2563
47/640.734418.6531
3/40.750019.0500
49/640.765619.4469
25/320.781319.8438
51/640.796920.2406
13/160.812520.6375
53/640.828121.0344
27/320.843821.4313
55/640.859421.8281
7/80.875022.2250
57/640.890622.6219
29/320.906323.0188
59/640.921923.4156
15/160.937523.8125
61/640.953124.2094
31/320.968824.6063
63/640.984425.0031
11.000025.4000
fractioninchesmm
1 1/641.015625.7969
1 1/321.031326.1938
1 3/641.046926.5906
1 1/161.062526.9875
1 5/641.078127.3844
1 3/321.093827.7813
1 7/641.109428.1781
1 1/81.125028.5750
1 9/641.140628.9719
1 5/321.156329.3688
1 11/641.171929.7656
1 3/161.187530.1625
1 13/641.203130.5594
1 7/321.218830.9563
1 15/641.234431.3531
1 1/41.250031.7500
1 17/641.265632.1469
1 9/321.281332.5438
1 19/641.296932.9406
1 5/161.312533.3375
1 21/641.328133.7344
1 11/321.343834.1313
1 23/641.359434.5281
1 3/81.375034.9250
1 25/641.390635.3219
1 13/321.406335.7188
1 27/641.421936.1156
1 7/161.437536.5125
1 29/641.453136.9094
1 15/321.468837.3063
1 31/641.484437.7031
1 1/21.500038.1000
1 33/641.515638.4969
1 17/321.531338.8938
1 35/641.546939.2906
1 9/161.562539.6875
1 37/641.578140.0844
1 19/321.593840.4813
1 39/641.609440.8781
1 5/81.625041.2750
1 41/641.640641.6719
1 21/321.656342.0688
1 43/641.671942.4656
1 11/161.687542.8625
1 45/641.703143.2594
1 23/321.718843.6563
1 47/641.734444.0531
1 3/41.750044.4500
1 49/641.765644.8469
1 25/321.781345.2438
1 51/641.796945.6406
1 13/161.812546.0375
1 53/641.828146.4344
1 27/321.843846.8313
1 55/641.859447.2281
1 7/81.875047.6250
1 57/641.890648.0219
1 29/321.906348.4188
1 59/641.921948.8156
1 15/161.937549.2125
1 61/641.953149.6094
1 31/321.968850.0063
1 63/641.984450.4031
22.000050.8000
fractioninchesmm
2 1/642.015651.1969
2 1/322.031351.5938
2 3/642.046951.9906
2 1/162.062552.3875
2 5/642.078152.7844
2 3/322.093853.1813
2 7/642.109453.5781
2 1/82.125053.9750
2 9/642.140654.3719
2 5/322.156354.7688
2 11/642.171955.1656
2 3/162.187555.5625
2 13/642.203155.9594
2 7/322.218856.3563
2 15/642.234456.7531
2 1/42.250057.1500
2 17/642.265657.5469
2 9/322.281357.9438
2 19/642.296958.3406
2 5/162.312558.7375
2 21/642.328159.1344
2 11/322.343859.5313
2 23/642.359459.9281
2 3/82.375060.3250
2 25/642.390660.7219
2 13/322.406361.1188
2 27/642.421961.5156
2 7/162.437561.9125
2 29/642.453162.3094
2 15/322.468862.7063
2 31/642.484463.1031
2 1/22.500063.5000
2 33/642.515663.8969
2 17/322.531364.2938
2 35/642.546964.6906
2 9/162.562565.0875
2 37/642.578165.4844
2 19/322.593865.8813
2 39/642.609466.2781
2 5/82.625066.6750
2 41/642.640667.0719
2 21/322.656367.4688
2 43/642.671967.8656
2 11/162.687568.2625
2 45/642.703168.6594
2 23/322.718869.0563
2 47/642.734469.4531
2 3/42.750069.8500
2 49/642.765670.2469
2 25/322.781370.6438
2 51/642.796971.0406
2 13/162.812571.4375
2 53/642.828171.8344
2 27/322.843872.2313
2 55/642.859472.6281
2 7/82.875073.0250
2 57/642.890673.4219
2 29/322.906373.8188
2 59/642.921974.2156
2 15/162.937574.6125
2 61/642.953175.0094
2 31/322.968875.4063
2 63/642.984475.8031
33.000076.2000
### Decimal to fraction Calculator
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# HOW TO DETERMINE WHETHER THE RELATION IS A FUNCTION
## About "How to determine whether the relation is a function"
How to determine whether the relation is a function :
If a relation has to be a function, it has to satisfy the following conditions.
(i) Domain of f is A.
(ii) For each x ∈ A, there is only one y ∈ B such that
(x, y) f .
## How to check if the arrow diagram represent a function ?
Let us look into the following examples to understand the above concept.
Example 1 :
Does the following arrow diagrams represent a function? Explain.
Solution :
Every element in A has a unique image. Hence it is a function.
Example 2 :
Does the following arrow diagrams represent a function? Explain.
Solution :
C has two images namely 20 and 40. Hence, it is not a function.
Example 3 :
Let X = { 1, 2, 3, 4 }. Examine whether each of the relations given below is a function from X to X or not. Explain
f = { (2, 3), (1, 4), (2, 1), (3, 2), (4, 4) }
Solution :
To check whether the relation forms a function, it has to satisfy the above two conditions.
(i) Domain of f = {1, 2, 3, 4} = Set X
(ii) Every element in X must be associated with different elements. Here 2 is associated with two different elements 3 and 1.
Hence it is not a function.
Example 4 :
Which of the following relations are functions from A = { 1, 4, 9, 16 } to B = { –1, 2, –3, –4, 5, 6 }? In case of a function, write down its range.
f = { (1, –1), (4, 2), (9, –3), (16, –4) }
Solution :
To check whether the relation forms a function, it has to satisfy the above two conditions.
(i) Domain of f = {1, 4, 9, 16} = Set A
(ii) Each element in A is associated with elements in B.
Hence it is not a function.
Range = {-1, 2, -3, -4}
Example 5 :
If X = { 1, 2, 3, 4, 5 }, Y = { 1, 3, 5, 7, 9 } determine which of the following relations from X to Y are functions? Give reason for your answer. In case of a function, write down its domain, range and codomain.
R = { (1, 1), (2, 1), (3, 3), (4, 3), (5, 5) }
Solution :
To check whether the relation is a function, it has to satisfy the above two conditions.
(i) Domain of R = {1, 2, 3, 4, 5} = Set X
(ii) Every element in X has image in Y.
Hence it is not a function.
Domain = {1, 2, 3, 4, 5}
Range = {1, 3, 5}
Co domain = { 1, 3, 5, 7, 9 }
Example 6 :
Let A = { 10, 11, 12, 13, 14 }; B = { 0, 1, 2, 3, 5 } determine the following relations from X to Y are functions? Give reason for your answer.
f = { (10, 1), (11, 2), (12, 3), (13, 5), (14, 3) }
Solution :
To check whether the relation forms a function, it has to satisfy the above two conditions.
(i) Domain of f = {10, 11, 12, 13, 14} = Set A
(ii) Every element in A has image in B.
Hence it is not a function.
Let us see the next example on "How to determine whether the relation is a function".
Example 7 :
Does the following arrow diagrams represent a function? Explain.
Solution :
Every element in P has a unique image. Hence it is a function.
Example 8 :
Does the following arrow diagrams represent a function? Explain.
Solution :
Every element in P has a unique image. Hence it is a function.
After having gone through the stuff given above, we hope that the students would have understood "How to determine whether the relation is a function".
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# How would you find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2)?
Nov 28, 2016
$\left(\frac{1}{\sqrt{33}}\right) \left(\begin{matrix}- 5 \\ - 2 \\ - 2\end{matrix}\right)$
#### Explanation:
If point P is (2,4,4) and Q is (-3,2,2), the vector PQ would be (-5,-2,-2). To find the unit vector, divide vector PQ by its magnitude. $| | \vec{P Q} | |$ would be sqrt((-5)^2 +(-2)^2 + (-2)^2)) =sqrt33. Hence unit vector would be
$\left(\frac{1}{\sqrt{33}}\right) \left(\begin{matrix}- 5 \\ - 2 \\ - 2\end{matrix}\right)$
Nov 28, 2016
#### Explanation:
Given the points $\left({x}_{0} , {y}_{0} , {z}_{0}\right) \mathmr{and} \left({x}_{1} , {y}_{1} , {z}_{1}\right)$
You make a vector in the direction from one point to another by subtracting each starting coordinate from its respective ending coordinate:
$\overline{V} = \left({x}_{1} - {x}_{0}\right) \hat{i} + \left({y}_{1} - {y}_{0}\right) \hat{j} + \left({z}_{1} - {z}_{0}\right) \hat{k}$
To make it a unit vector, $\hat{V}$, you divide each component of the vector by the magnitude, $| \overline{V} |$:
$| \overline{V} | = \sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}$
Therefore the general form for a unit vector is:
$\hat{V} = \frac{{x}_{1} - {x}_{0}}{\sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}} \hat{i} + \frac{{y}_{1} - {y}_{0}}{\sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}} \hat{j} + \frac{{z}_{1} - {z}_{0}}{\sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}} \hat{k}$
For the given points $\left(2 , 4 , 4\right)$ and $\left(- 3 , 2 , 2\right)$
Compute the magnitude
$| \overline{V} | = \sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}} =$
$\sqrt{{\left(- 3 - 2\right)}^{2} + {\left(2 - 4\right)}^{2} + {\left(2 - 4\right)}^{2}} =$
$\sqrt{33}$
We need to divide which it the same as multiplying by the reciprocal, $\frac{\sqrt{33}}{33}$
Substitute into the unit vector general form:
$\hat{V} = \left(- 3 - 2\right) \frac{\sqrt{33}}{33} \hat{i} + \left(2 - 4\right) \frac{\sqrt{33}}{33} \hat{j} + \left(2 - 4\right) \frac{\sqrt{33}}{33} \hat{k}$
Simplify:
$\hat{V} = - \frac{5 \sqrt{33}}{33} \hat{i} - \frac{2 \sqrt{33}}{33} \hat{j} - \frac{2 \sqrt{33}}{33} \hat{k}$ |
# Multiplying Mixed Fractions
("Mixed Fractions" are also called "Mixed Numbers")
To multiply Mixed Fractions:
### Example: What is 138 × 3 ?
Think of Pizzas.
138 is 1 pizza and 3 eighths of another pizza.
First, convert the mixed fraction (138) to an improper fraction (118):
Cut the whole pizza into eighths and how many eighths do you have in total? 1 lot of 8, plus the 3 eighths = 8+3 = 11 eighths.
Now multiply that by 3:
138 × 3 = 118 × 31 = 338 You have 33 eighths.
And, lastly, convert to a mixed fraction (only because the original fraction was in that form):
33 eighths is 4 whole pizzas (4×8=32) and 1 eighth left over.
And this is what it looks like in one line:
138 × 3 = 118 × 31 = 338 = 418
### Another Example: What is 112 × 215 ?
Do the steps from above:
1. convert to Improper Fractions
2. Multiply the Fractions
3. convert the result back to Mixed Fractions
### Step, by step it is:
Convert Mixed to Improper Fractions:
112 = 22+12 = 32
215 = 105+15 = 115
Multiply the fractions (multiply the top numbers, multiply bottom numbers):
32 × 115 = 3 × 112 × 5 = 3310
Convert to a mixed number
3310 = 3310
If you are clever you can do it all in one line like this:
112 × 215 = 32 × 115 = 3310 = 3310
### One More Example: What is 314 × 313 ?
Convert Mixed to Improper Fractions:
314 = 134
313 = 103
Multiply
134 × 103 = 13012
Convert to a mixed number:
13012 = 101012
And simplify:
101012 = 1056
Here it is in one line:
314 × 313 = 134 × 103 = 13012 = 101012 = 1056
### This One Has Negatives: What is −159 × −217 ?
Convert Mixed to Improper Fractions:
159 = 99 + 59 = 149
217 = 147 + 17 = 157
Then multiply the Improper Fractions (note that negative times negative gives positive):
−149 × −157 = −14 × −159 × 7 = 21063
We can simplify now. Here we use two steps, first by 7 (21 and 63 are both multiples of 7), then again by 3. But it could be done in one step by dividing by 21:
21063 = 309 = 103
Finally convert to a Mixed Fraction (because that was the style of the question):
103 = (9+1)3 = 93 + 13 = 313
940, 941, 942, 1418, 943, 1419, 1420, 1421, 3573, 3574 |
# GSEB Solutions Class 8 Maths Chapter 12 Exponents and Powers Intext Questions
Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 12 Exponents and Powers Intext Questions and Answers.
## Gujarat Board Textbook Solutions Class 8 Maths Chapter 12 Exponents and Powers
Try These (Page 194)
Question 1.
Find the multiplicative inverse of the following?
1. 2-4
2. 10-5
3. 7-2
4. 5-3
5. 100-100
Solution:
1. The multiplicative inverse of 2-4 is 24 ponents.
2. The multiplicative inverse of 10-5 is 105
3. The multipicative inverse of 7-2 is 72
4. The multipicative inverse of 5-3 is 52
5. The multipicative inverse of 10-100 is 10100
Try These (Page 194)
Question 1.
Expand the following number using exponents?
(i) 1025.63
(ii) 1256.249
Solution:
Try These (Page 195)
Question 1.
Simpfy and write in exponential form?
1. (-2)-3 Ć (-2)-4
2. p3 Ć p-10
3. 32 Ć 3-5 Ć 36
Solution:
1. (-2)3 Ć (-2)-4 = (-2)(-3)+(-4) [āµam Ć an]
= (-2)-7 or $$\frac{1}{(-2)^{7}}$$
2. p3 Ć p-10 = (p)3+(-10) = (p)-7 or $$\frac{1}{(10)^{7}}$$
3. 32 Ć 3-5 Ć 36 = 32+(-5)+6 = 38-5 = 33
Law II: $$\frac{a^{m}}{a^{n}}$$ = am-n
Example: 5-1 + 5-2 = 5-1-(-2) = 5-1+2 = 51 or 5
Law III: (am)n = amn
Example: (9-1)-3 = 9(-1)Ć(-3) = 93
Law IV: am Ć bm = (ab)m
Example: 2-4 Ć 3-4 = (2 Ć 3)-4 = 6-4 or $$\frac{1}{6^{4}}$$
Law V: $$\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}$$
Example: $$\frac{3^{-5}}{7^{-5}}$$ = $$\frac{3}{7}$$-5 or $$\frac{7}{3}$$5
Law VI: a0 = 1
Example:
(i) (-38)0 = 1
(ii) (32456)0 = 1
Try These (Page 199)
Question 1.
Write the following numbers in standard form?
1. 0.000000564
2. 0.0000021
3. 15240000
Solution:
1. 0.000000564 = $$\frac{564}{1000000000}$$
= $$\frac{5.64}{10^{9}} \times 10^{2}$$
= $$\frac{5.64}{10^{7}}$$
= 5.64 Ć 10-7
2. 0.0000021 = $$\frac{21}{10000000}$$
= $$\frac{2.1 \times 10}{10000000}=\frac{2.1}{1000000}$$
= 2.1 Ć 10-6
ā“ 0.0000021 = 2.1 Ć 10-6
3. 15240000 = 1524 Ć 1000
= 1.524 Ć 1000 Ć 1000
= 1.524 Ć 103 Ć 104
= 1.524 Ć 107
ā“ 15240000 = 1.524 Ć 107
Question 2.
Write all the facts given in the standard form?
Solution:
A number is said to be in the standard form
when it is written as k Ć 10n,
where 1 ā¤ k < 10 and ‘n’ is an integer
A number expressed as the product of a number between 1 and 10 and an integral power of 10.
Example: Compare the size of a red blood cell which is 0.000007 m to that of a plant cell which is 0.0000129 m.
Solution:
Size of red blood cell = 0.000007 m
= $$\frac{7}{1000000}$$
= 7 Ć 10-6 m
Size of the plant cell = 0.0000129 m
Thus, the size of a red blood cell is half of the plant cell size. |
Where is the mean formula?
How to calculate the mean
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are.
Where is the mean math
A mean in math is the average of a data set, found by adding all numbers together and then dividing the sum of the numbers by the number of numbers. For example, with the data set: 8, 9, 5, 6, 7, the mean is 7, as 8 + 9 + 5 + 6 + 7 = 35, 35/5 = 7.
Where is the mean in a data set
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set. Created by Sal Khan.
Where is the mean median and mode
To find the mean, add up the values in the data set and then divide by the number of values that you added. To find the median, list the values of the data set in numerical order and identify which value appears in the middle of the list. To find the mode, identify which value in the data set occurs most often.
What is the mean mode formula
How to Calculate the Mean Using Mean Median Mode Formula If the set of 'n' number of observations is given then the mean can be easily calculated by using a general mean median mode formula that is, Mean = {Sum of Observations} ÷ {Total number of Observations}.
Why do we calculate the mean
The mean represents the average value in a dataset. The mean is important because it gives us an idea of where the center value is located in a dataset. The mean is also important because it carries a piece of information from every observation in a dataset.
Where is the mean in statistics
Then out divided by how many of them I have looks like I have a total of nothing. Now. You may have to use a calculator if you have lots of different data points but that's okay.
Is the mean the median
The mean is the number you get by dividing the sum of a set of values by the number of values in the set. In contrast, the median is the middle number in a set of values when those values are arranged from smallest to largest. The mode of a set of values is the most frequently repeated value in the set.
Why do we find the mean of data
The mean represents the average value in a dataset. The mean is important because it gives us an idea of where the center value is located in a dataset. The mean is also important because it carries a piece of information from every observation in a dataset.
How do you find the mode
How Do I Calculate the Mode Calculating the mode is fairly straightforward. Place all numbers in a given set in order; this can be from lowest to highest or highest to lowest, and then count how many times each number appears in the set. The one that appears the most is the mode.
Where are the mean and median located on a normal distribution
the center
mean and median are equal; both located at the center of the distribution. ≈68%approximately equals, 68, percent of the data falls within 1 standard deviation of the mean. ≈95%approximately equals, 95, percent of the data falls within 2 standard deviations of the mean.
What is the mean or median formula
The mean (informally, the “average“) is found by adding all of the numbers together and dividing by the number of items in the set: 10 + 10 + 20 + 40 + 70 / 5 = 30. The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.
What is the formula for mean example
To calculate, just add up all the given numbers then divide by how many numbers are given. Example: What is the mean of 3, 5, 9, 5, 7, 2 5.16 is the answer.
Why do we find mean and median
Mean, median, and mode are different measures of center in a numerical data set. They each try to summarize a dataset with a single number to represent a "typical" data point from the dataset. Mean: The "average" number; found by adding all data points and dividing by the number of data points.
What is the mean in statistics
The mean is the average or the most common value in a collection of numbers. In statistics, it is a measure of central tendency of a probability distribution along median and mode. It is also referred to as an expected value. It is a statistical concept that carries a major significance in finance.
Where can we locate the mean in the normal curve
We can find both the mean μ and standard deviation σ by eye on a normal curve. The mean μ is the center of symmetry for the curve. To find σ , start at the center of the curve and run a pencil outward.
Is the mean always the median
The mean is affected by outliers that do not influence the mean. Therefore, when the distribution of data is skewed to the left, the mean is often less than the median. When the distribution is skewed to the right, the mean is often greater than the median.
Is mean or median the middle
The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers. The "median" is the "middle" value in the list of numbers.
How can we find mean in statistics
How do I find the mean You can find the mean, or average, of a data set in two simple steps: Find the sum of the values by adding them all up. Divide the sum by the number of values in the data set.
How to find a median
Finding the median
To find the median: Arrange the data points from smallest to largest. If the number of data points is odd, the median is the middle data point in the list. If the number of data points is even, the median is the average of the two middle data points in the list.
Where is the mean located in a normal distribution
center
mean and median are equal; both located at the center of the distribution.
Where do you locate the mean in the normal curve
We can find both the mean μ and standard deviation σ by eye on a normal curve. The mean μ is the center of symmetry for the curve. To find σ , start at the center of the curve and run a pencil outward.
How to find the median
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
How do we find mean in statistics
Now. You may have to use a calculator if you have lots of different data points but that's okay. Well. I added everything up I got a total of 74 divided by 9. Now.
What is the basic formula of mean
Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers. Mean = (Sum of all the observations/Total number of observations) |
Home » Partial Fractions » Page 2
# Evaluate the integral of 1 / (x2 – 1)2
Compute the following integral.
First, we write
Then we use partial fractions, writing
This gives us the equation
Evaluating at and we obtain the values for and ,
Using these values of and and evaluating at and we have
Solving this system for and we get
Thus,
# Evaluate the integral of (x-3) / (x3 + 3x2 + 2x)
Compute the following integral.
The denominator of the integrand factors as
Therefore, we can use partial fractions as follows,
This gives us the equation
Evaluating at , , and we obtain
Therefore, we have
# Evaluate the integral of 1 / ((x2 – 4x + 4)(x2 – 4x + 5))
Compute the following integral.
In the denominator we have
Then we use partial fractions,
This gives us the equation
We evaluate at to obtain a value for ,
Then using this value of and evaluating at , and to obtain
Solving this system of equations we obtain
Therefore, we have
# Evaluate the integral of x2 / (x2 + x – 6)
Compute the following integral.
First, we have
Therefore,
We use partial fractions to evaluate the integral on the right. To that end, we write
This gives us the equation
Evaluating at and we then have
Therefore,
# Evaluate the integral of 1 / (x3 – x)
Compute the following integral.
Since factors as we have the following,
This gives us the equation
Substituting the values , and we obtain
Therefore, we have
# Evaluate the following integral 1 / ((x+1)(x+2)2(x+3)3)
Compute the following integral.
First, we use partial fraction decomposition. We write
This gives us the equation
First, we substitute the values , , and which gives us
Substituting these values of , and into our equation we have
Now, we substitute the values , and to obtain the equations
Solving this system we obtain the values
Therefore, we have the following,
# Evaluate the integral of 1 / (x (x2 + 1)2)
Compute the following integral.
To evaluate this we use partial fractions. First, we write
This gives us the equation
First, we evaluate at to find that . Using this value of we obtain the equation
Equating like powers of this gives us the system of equations:
We solve this system to obtain,
Therefore,
# Evaluate the following integral (x+2) / (x2 + x)
Compute the following integral.
We have
To evaluate the second integral on the right we use partial fractions. Write,
This gives us the equations
Letting and we have
Therefore, we have
# Evaluate the integral of x4 / (x4 +5x2 + 4)
Compute the following integral.
First, we simplify the integrand and then use partial fractions,
To evaluate the second integral we use partial fractions. We have
Therefore, we write
Then we have the equation
Equating like powers of we then have four equations and four unknowns:
Solving this system we find , and . Therefore, we have
# Evaluate the integral of (8x3 + 7) / ((x+1)(2x+1)3)
Compute the following integral.
Since the denominator is already factored into linear terms we can proceed directly with the partial fraction decomposition. We write,
This gives us the equation
First, we can find the values of and by evaluating at and , respectively. This gives us
Then using these values of and we evaluate at and (these are just convenient values, there isn’t anything special about them) to obtain the two equations
Solving these two equations we obtain and . Therefore, we have the following partial fraction decomposition:
We can now evaluate the integral, |
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