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# 14.
## 1 Graph Sine, Cosine,
and Tangent Functions
Before You evaluated sine, cosine, and tangent functions.
Now You will graph sine, cosine, and tangent functions.
Why? So you can model oscillating motion, as in Ex. 31.
Key Vocabulary In this lesson, you will learn to graph functions of the form y 5 a sin bx and
• amplitude y 5 a cos bx where a and b are positive constants and x is in radian measure.
• periodic function The graphs of all sine and cosine functions are related to the graphs of the
• cycle parent functions y 5 sin x and y 5 cos x, which are shown below.
• period y
• frequency M51 y 5 sin x
1
amplitude: 1
range:
21 ≤ y ≤ 1
22
3π 2π 22
π π π 3π 2π x
2 2
21
m 5 21 period:
y
M51
y 5 cos x
amplitude: 1
range:
21 ≤ y ≤ 1
22π 2 3π 2π 22
π π π 3π 2π x
2 2 2
21
m 5 21 period:
## KEY CONCEPT For Your Notebook
Characteristics of y 5 sin x and y 5 cos x
• The domain of each function is all real numbers.
• The range of each function is 21 ≤ y ≤ 1. Therefore, the minimum
value of each function is m 5 21 and the maximum value is M 5 1.
• The amplitude of each function’s graph is half the difference of the
maximum M and the minimum m, or } 1 (M 2 m) 5 1 [1 2 (21)] 5 1.
}
2 2
• Each function is periodic, which means that its graph has a repeating
pattern. The shortest repeating portion of the graph is called a cycle.
The horizontal length of each cycle is called the period. Each graph
shown above has a period of 2π.
• The x-intercepts for y 5 sin x occur when x 5 0, 6π, 62π, 63π, . . . .
• The x-intercepts for y 5 cos x occur when x 5 6} p , 6 3p , 6 5p , 6 7p , . . . .
} } }
2 2 2 2
## n2pe-1401.indd 908 10/17/05 11:11:24 AM
Amplitude and Period
The amplitude and period of the graphs of y 5 a sin bx and y 5 a cos bx,
where a and b are nonzero real numbers, are as follows:
## Amplitude 5 a Period 5 }2p
b
GRAPHING KEY POINTS Each graph below shows five key x-values on the interval
2p that you can use to sketch the graphs of y 5 a sin bx and y 5 a cos bx for
0≤x≤}
b
a > 0 and b > 0. These are the x-values where the maximum and minimum values
occur and the x-intercepts.
y 1 2π y y 5 a cos bx
s 4 ? b , ad y 5 a sin bx (0, a)
s 2π
, d
0 s 14 ? 2πb , 0d s2πb , ad
b
(0, 0) x x
s 12 ? 2πb , 0d s 34 ? 2πb , 0d
s 34 ? 2πb , 2ad s 12 ? 2πb , 2ad
## EXAMPLE 1 Graph sine and cosine functions
Graph (a) y 5 4 sin x and (b) y 5 cos 4x.
## VARY CONSTANTS Solution
Notice how changes 2p 5 2p 5 2π.
a. The amplitude is a 5 4 and the period is } }
in a and b affect the b 1
graphs of y 5 a sin bx 1 y
Intercepts: (0, 0); } p 2π, 0 5 (p , 0); (2p, 0)
1 2 4
and y 5 a cos bx. When 2
the value of a increases,
p
1 p 2π, 4 5 }
the amplitude increases. Maximum: } 14
1 , 42 2 2 x
When the value of b π 3π
increases, the period 3 p 2π, 24 5 3p , 24 2 2
decreases.
Minimum: }
4 1 }
2 2 1 2
2p 5 2p 5 p .
b. The amplitude is a 5 1 and the period is } } }
b 4 2
y
p , 0 5 p , 0 ; 3 p p , 0 5 3p , 0
1 p} 2
Intercepts: } 14
2 1} 2 } }
2
}
8 14 2 2 1 8 2
p, 1
Maximums: (0, 1); 1 } 2 2 π 3π x
8 8
p , 21 5 p , 21
1 p}
Minimum: } 12
2 1} 2 2 4
## Graph the function.
1. y 5 2 cos x 2. y 5 5 sin x 3. f(x) 5 sin πx 4. g(x) 5 cos 4πx
## n2pe-1401.indd 909 10/17/05 11:11:27 AM
EXAMPLE 2 Graph a cosine function
1 cos 2p x.
Graph y 5 }
2
SKETCH A GRAPH Solution
After you have drawn 1 and the period is 2p 5 2p 5 1.
The amplitude is a 5 } } }
one complete cycle of 2 b 2p
the graph in Example 2
1 p 1, 0 5 1 , 0 ;
Intercepts: 1 }
on the interval 0 ≤ x ≤ 1, 4 2 1} 2 4
you can extend the
y
graph by copying the
cycle as many times as 1 }34 p 1, 0 2 5 1 }34 , 0 2 1
## desired to the left and
1 ; 1, 1
Maximums: 1 0, }
right of 0 ≤ x ≤ 1. 2 1 }22 2 1 2 x
1 p 1, 2 1 5 1 , 2 1
Minimum: 1 } }2 1} }2
2 2 2 2
## MODELING WITH TRIGONOMETRIC FUNCTIONS The periodic nature of
trigonometric functions is useful for modeling oscillating motions or repeating
patterns that occur in real life. Some examples are sound waves, the motion of
a pendulum, and seasons of the year. In such applications, the reciprocal of the
period is called the frequency, which gives the number of cycles per unit of time.
## EXAMPLE 3 Model with a sine function
AUDIO TEST A sound consisting of a single frequency is
called a pure tone. An audiometer produces pure tones to
test a person’s auditory functions. Suppose an audiometer
produces a pure tone with a frequency f of 2000 hertz (cycles
per second). The maximum pressure P produced from the
pure tone is 2 millipascals. Write and graph a sine model that
gives the pressure P as a function of the time t (in seconds).
Solution
STEP 1 Find the values of a and b in the model P 5 a sin bt. The maximum
pressure is 2, so a 5 2. You can use the frequency f to find b.
1
frequency 5 } 2000 5 }b 4000π 5 b
period 2p
The pressure P as a function of time t is given by P 5 2 sin 4000πt.
15 1 .
STEP 2 Graph the model. The amplitude is a 5 2 and the period is } }
f 2000
## Intercepts: (0, 0);
P
1 }12 p }1 ,0 5 1
}, 0 ; }, 0
1
2000 2 1 4000 2 1 2000 2 2
1 p 1 ,2 5 1
Maximum: } } 14
}, 2
2000 2 1 8000 2 1 t
8000
3 p 1 , 22 5 3
Minimum: } } 14
}, 22
2000 2 1 8000 2
## n2pe-1401.indd 910 10/17/05 11:11:29 AM
✓ GUIDED PRACTICE for Examples 2 and 3
## Graph the function.
1 sin πx
5. y 5 } 1 cos πx
6. y 5 } 7. f(x) 5 2 sin 3x 8. g(x) 5 3 cos 4x
4 3
9. WHAT IF? In Example 3, how would the function change if the audiometer
produced a pure tone with a frequency of 1000 hertz?
GRAPH OF Y 5 TAN X The graphs of all tangent functions are related to the graph
of the parent function y 5 tan x, which is shown below.
y
y 5 tan x
22π 3π 2π π π π 3π 2π x
22 22 2 2
period:
π
## FIND ODD The function y 5 tan x has the following characteristics:
MULTIPLES p . At these
p
1. The domain is all real numbers except odd multiples of }
Odd multiples of } are 2
2 x-values, the graph has vertical asymptotes.
values such as these:
p p 2. The range is all real numbers. So, the function y 5 tan x does not have
61 p } 5 6}
2 2 a maximum or minimum value, and therefore the graph of y 5 tan x
p 3p
63 p } 5 6}
does not have an amplitude.
2 2
p 5p 3. The graph has a period of π.
65 p } 5 6}
2 2
4. The x-intercepts of the graph occur when x 5 0, 6π, 62π, 63π, . . . .
## KEY CONCEPT For Your Notebook
Characteristics of y 5 a tan bx
The period and vertical asymptotes of the graph of y 5 a tan bx, where
a and b are nonzero real numbers, are as follows:
• The period is }p .
b
• The vertical asymptotes are at odd multiples of }p .
2b
## GRAPHING KEY POINTS The graph at the right y
the graph of y 5 a tan bx for a > 0 and b > 0. a
These are the x-intercept, the x-values where
the asymptotes occur, and the x-values halfway π π π x
between the x-intercept and the asymptotes. At 2 2b 4b 2b
each halfway point, the function’s value is either
a or 2a.
## n2pe-1401.indd 911 10/17/05 11:11:29 AM
EXAMPLE 4 Graph a tangent function
Graph one period of the function y 5 2 tan 3x.
Solution
p 5 p.
The period is } }
b 3
Intercept: (0, 0)
p 5 p , or x 5 p ;
Asymptotes: x 5 } y
} }
2b 2p3 6
p 5 2 p , or x 5 2 p
x 5 2} } }
2b 2p3 6 2
p ,a 5 p ,2 5 p ,2 ; π π π x
Halfway points: 1 } 2 1} 2 1} 2 26
4b 4p3 12 12 6
p p
, 2a 2 5 1 2} , 22 2 5 1 2}, 22 2 p
1 2}
4b 4p3 12
"MHFCSB at classzone.com
## Graph one period of the function.
10. y 5 3 tan x 11. y 5 tan 2x 12. f(x) 5 2 tan 4x 13. g(x) 5 5 tan πx
## 14.1 EXERCISES HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 5, 17, and 31
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 15, 24, 25, and 31
5 MULTIPLE REPRESENTATIONS
Ex. 32
SKILL PRACTICE
1. VOCABULARY Copy and complete: The graphs of the functions y 5 sin x
and y 5 cos x both have a(n) ? of 2π.
2. ★ WRITING Compare the domains and ranges of the functions y 5 a sin bx,
y 5 a cos bx, and y 5 a tan bx where a and b are positive constants.
EXAMPLE 1 ANALYZING FUNCTIONS Identify the amplitude and the period of the graph of
on p. 909 the function.
for Exs. 3–14
3. y 4. y 5. y
1 1
π
π 3π 5π x π 2π x 1 3 5 x
8 8 8 2 2 2
## n2pe-1401.indd 912 10/17/05 11:11:30 AM
GRAPHING Graph the function.
1x
6. y 5 sin } 7. y 5 4 cos x 2x
8. f(x) 5 cos } 9. y 5 sin πx
5 5
2 sin x
10. f(x) 5 } px
11. f(x) 5 sin } p cos x
12. y 5 } 13. f (x) 5 cos 24x
3 2 4
## 14. ERROR ANALYSIS Describe and correct
}3 5 1
the error in finding the period of the 2
b
2 x.
function y 5 sin }
Period 5 } 5 } }
3 2p 2p 3p
15. ★ MULTIPLE CHOICE The graph of which function has an amplitude of 4 and
a period of 2?
A y 5 4 cos 2x B y 5 2 sin 4x C y 5 4 sin πx 1 πx
D y 5 2 cos }
2
## EXAMPLES GRAPHING Graph the function.
2, 3, and 4
16. y 5 2 sin 8x 17. f(x) 5 4 tan x 18. y 5 3 cos πx 19. y 5 5 sin 2x
on pp. 910–912
for Exs. 16–24 1 πx
20. f(x) 5 2 tan 4x 21. y 5 2 cos } 22. f(x) 5 4 tan πx 23. y 5 π cos 4πx
4
y 5 2 tan 3x?
p
A x5} B x 5 2π 1
C x5} p
D x 5 2}
6 6 12
## 25. ★ OPEN-ENDED MATH Describe a real-life situation that can be modeled by a
periodic function.
CHALLENGE Sketch the graph of the function by plotting points. Then state the
function’s domain, range, and period.
26. y 5 csc x 27. y 5 sec x 28. y 5 cot x
PROBLEM SOLVING
EXAMPLE 3 29. PENDULUMS The motion of a certain pendulum can be modeled by the
on p. 910 function d 5 4 cos πt where d is the pendulum’s horizontal displacement
for Exs. 29–30 (in inches) relative to its position at rest and t is the time (in seconds).
Graph the function. What is the greatest horizontal distance the pendulum
will travel from its position at rest?
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
## 30. TUNING FORKS A tuning fork produces a
sound pressure wave that can be modeled by
P 5 0.001 sin 880t
where P is the pressure (in pascals) and t is
the time (in seconds). Find the period and
frequency of this function. Then graph the
function.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
## n2pe-1401.indd 913 10/17/05 11:11:31 AM
31. ★ SHORT RESPONSE A buoy oscillates up and down as waves go past.
The buoy moves a total of 3.5 feet from its low point to its high point, and
then returns to its high point every 6 seconds.
a. Write an equation that gives the buoy’s vertical position y at time t if the
buoy is at its highest point when t 5 0.
b. Explain why you chose y 5 a sin bt or y 5 a cos bt for part (a).
## 32. MULTIPLE REPRESENTATIONS You are standing on a bridge, 140 feet
above the ground. You look down at a car traveling away from the underpass.
140 ft
## a. Writing an Equation Write an equation that gives the car’s distance d
from the base of the bridge as a function of the angle u.
b. Drawing a Graph Graph the function found in part (a). Explain how the
graph relates to the given situation.
c. Making a Table Make a table of values for the function. Use the table to
find the car’s distance from the bridge when u 5 208, 408, and 608.
## 33. CHALLENGE The motion of a spring can be modeled by y 5 A cos kt where
y is the spring’s vertical displacement (in feet) relative to its position at
rest, A is the initial displacement (in feet), k is a constant that measures
the elasticity of the spring, and t is the time (in seconds).
a. Suppose you have a spring whose motion can be modeled by the
function y 5 0.2 cos 6t. Find the initial displacement and the period of
the spring. Then graph the given function.
b. Graphing Calculator If a damping force is applied to the spring, the
motion of the spring can be modeled by the function y 5 0.2e24.5t cos 4t.
Graph this function. What effect does damping have on the motion?
MIXED REVIEW
PREVIEW Graph the function. Label the vertex and the axis of symmetry. (p. 245)
Prepare for
34. y 5 2(x 2 2)2 1 1 35. y 5 5(x 2 1)2 1 7 36. f(x) 5 2(x 1 6)2 2 3
Lesson 14.2
in Exs. 34–39. 37. y 5 23(x 1 3)2 1 2 38. y 5 0.5(x 1 2)2 1 5 39. g(x) 5 22(x 1 4)2 2 3
Let f(x) 5 x 2 2 5, g(x) 5 27x, and h(x) 5 x1/3. Find the indicated value. (p. 428)
40. g(f(9)) 41. h(g(3)) 42. f(g(27)) 43. f(h(8))
Sketch the angle. Then find its reference angle. (p. 866)
44. 2508 45. 2908 46. 21458 47. 24308
13p
48. } 21p
49. } 13p
50. 2} 16p
51. 2}
3 4 6 3
## 914 Chapter 14EXTRA PRACTICE
Trigonometric Graphs,for
Identities, andp.Equations
Lesson 14.1, 1023 ONLINE QUIZ at classzone.com |
# Class 6 Maths Chapter 5 Exercise 5.3 Pdf Notes NCERT Solutions
Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.3 pdf notes:-
Exercise 5.3 Class 6 maths Chapter 5 Pdf Notes:-
## Ncert Solution for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.3 Tips:-
Introduction:-
Angles – ‘Acute’, ‘Obtuse’ and
‘Reflex’
We saw what we mean by a right angle and
a straight angle. However, not all the angles
we come across are one of these two kinds.
(or with the floor) is neither a right angle
nor a straight angle.
Think, discuss and write
Are there angles smaller than a right angle?
Are there angles greater than a right angle?
Have you seen a carpenter’s square? It looks like the letter
“L” of English alphabet. He uses it to check right angles.
Let us also make a similar ‘tester’ for a right angle.
Suppose any shape with corners is given. You can use your RA tester to test
the angle at the corners.
Do the edges match with the angles of a paper? If yes, it indicates a
right angle.
The hour hand of a clock moves from 12 to 5.
Is the revolution of the hour hand more than
1 right angle?
Try These:-
1. What does the angle made by the hour hand of the clock
look like when it moves from 5 to 7. Is the angle moved
more than 1 right angle?
2. Draw the following and check the angle with your RA tester.
(a) going from 12 to 2 (b) from 6 to 7
(c) from 4 to 8 (d) from 2 to 5
3. Take five different shapes with corners. Name the corners. Examine
them with your tester and tabulate your results for each case :
Do you see that each one of them is less than one-fourth of a revolution?
Examine them with your RA tester.
If an angle is larger than a right angle, but less than a straight angle, it is
called an obtuse angle. These are obtuse angles.
Do you see that each one of them is greater than
one-fourth of a revolution but less than half a revolution?
Your RA tester may help to examine.
Identify the obtuse angles in the previous examples too.
A reflex angle is larger than a straight angle.
It looks like this. (See the angle mark)
Were there any reflex angles in the shapes you made earlier?
How would you check for them?
#### Test Paper Of Class 8th
• Maths 8th Class
• Science 8th class
• Sst 8th Class
• #### Test Paper Of Class 7th
• Maths 7th Class
• Science 7th class
• #### Test Paper Of Class 6th
• Maths 6th Class
• Science 6th class |
# What is the vertex of y= 5x^2-x-1+(2x-1)^2?
Dec 16, 2015
vertex$= \left(\frac{5}{18} , - \frac{25}{36}\right)$
#### Explanation:
Start by expanding the brackets and simplifying the expression.
$y = 5 {x}^{2} - x - 1 + {\left(2 x - 1\right)}^{2}$
$y = 5 {x}^{2} - x - 1 + \left(4 {x}^{2} - 4 x + 1\right)$
$y = 9 {x}^{2} - 5 x$
Take your simplified equation and complete the square.
$y = 9 {x}^{2} - 5 x$
$y = 9 \left({x}^{2} - \frac{5}{9} x + {\left(\frac{\frac{5}{9}}{2}\right)}^{2} - {\left(\frac{\frac{5}{9}}{2}\right)}^{2}\right)$
$y = 9 \left({x}^{2} - \frac{5}{9} x + {\left(\frac{5}{18}\right)}^{2} - {\left(\frac{5}{18}\right)}^{2}\right)$
$y = 9 \left({x}^{2} - \frac{5}{9} x + \frac{25}{324} - \frac{25}{324}\right)$
$y = 9 \left({x}^{2} - \frac{5}{9} x + \frac{25}{324}\right) - \left(\frac{25}{324} \cdot 9\right)$
$y = 9 {\left(x - \frac{5}{18}\right)}^{2} - \left(\frac{25}{\textcolor{red}{\cancel{\textcolor{b l a c k}{324}}}} ^ 36 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}\right)$
$y = 9 {\left(x - \frac{5}{18}\right)}^{2} - \frac{25}{36}$
Recall that the general equation of a quadratic equation written in vertex form is:
$y = a {\left(x - h\right)}^{2} + k$
where:
$h =$x-coordinate of the vertex
$k =$y-coordinate of the vertex
So in this case, the vertex is $\left(\frac{5}{18} , - \frac{25}{36}\right)$. |
# 4.4: Formula Mass, Percent Composition, and the Mole
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Skills to Develop
• Calculate formula masses for covalent and ionic compounds
• Define the amount unit mole and the related quantity Avogadro’s number
• Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another
• Compute the percent composition of a compound
We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.
## Formula Mass
In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance’s formula.
### Formula Mass for Covalent Substances
For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl3), a covalent compound once used as a surgical anesthetic and now primarily used in the production of tetrafluoroethylene, the building block for the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure $$\PageIndex{1}$$ outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu.
Figure $$\PageIndex{1}$$: The average mass of a chloroform molecule, CHCl3, is 119.37 amu, which is the sum of the average atomic masses of each of its constituent atoms. The model shows the molecular structure of chloroform.
Likewise, the molecular mass of an aspirin molecule, C9H8O4, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure $$\PageIndex{2}$$).
Figure $$\PageIndex{2}$$: The average mass of an aspirin molecule is 180.15 amu. The model shows the molecular structure of aspirin, C9H8O4.
Example $$\PageIndex{1}$$: Computing Molecular Mass for a Covalent Compound
Ibuprofen, C13H18O2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?
Solution
Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:
Exercise $$\PageIndex{1}$$
Acetaminophen, C8H9NO2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?
151.16 amu
### Formula Mass for Ionic Compounds
Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.”
As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na+, and chloride anions, Cl, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (Figure $$\PageIndex{3}$$).
Figure $$\PageIndex{3}$$: Table salt, NaCl, contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu.
Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.
Example $$\PageIndex{2}$$: Computing Formula Mass for an Ionic Compound
Aluminum sulfate, Al2(SO4)3, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?
Solution
The formula for this compound indicates it contains Al3+ and SO42− ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al2S3O12. Following the approach outlined above, the formula mass for this compound is calculated as follows:
Exercise $$\PageIndex{2}$$
Calcium phosphate, Ca3(PO4)2, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?
310.18 amu
## The Mole
In units 1 and 2, we learned about measuring quanity of atoms using the mole. As a reminder, the mole is an amount unit similar to familiar units like pair, dozen, gross, etc. The number of entities composing a mole has been experimentally determined to be $$6.02214179 \times 10^{23}$$, a fundamental constant named Avogadro’s number (NA) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being $$6.022 \times 10^{23}/\ce{mol}$$.
The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. In unit 2, we used Avogadro’s number to convert between mass of an element, moles of an element, and number of atoms in the sample. Now, we can use Avogadro’s number and formula mass (also called molecular weight) to convent between mass of a sample, moles, and number of molecules. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.s
Example $$\PageIndex{7}$$: Deriving Moles from Grams for a Compound
Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C2H5O2N. How many moles of glycine molecules are contained in 28.35 g of glycine?
Solution
We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example $$\PageIndex{6}$$:
The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C2H5O2N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:
The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:
$\mathrm{28.35\:\cancel{g}\:glycine\left(\dfrac{mol\: glycine}{75.07\:\cancel{g}}\right)=0.378\:mol\: glycine} \nonumber$
This result is consistent with our rough estimate.
Exercise $$\PageIndex{7}$$
How many moles of sucrose, $$C_{12}H_{22}O_{11}$$, are in a 25-g sample of sucrose?
0.073 mol
Example $$\PageIndex{8}$$: Deriving Grams from Moles for a Compound
Vitamin C is a covalent compound with the molecular formula C6H8O6. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10−4 mol. What is the mass of this allowance in grams?
Solution
As for elements, the mass of a compound can be derived from its molar amount as shown:
The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10−4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:
$\mathrm{1.42\times10^{-4}\:\cancel{mol}\:vitamin\: C\left(\dfrac{176.124\:g}{\cancel{mol}\:vitamin\: C}\right)=0.0250\:g\: vitamin\: C} \nonumber$
This is consistent with the anticipated result.
Exercise $$\PageIndex{8}$$
What is the mass of 0.443 mol of hydrazine, $$N_2H_4$$?
14.2 g
Example $$\PageIndex{9}$$: Deriving the Number of Molecules from the Compound Mass
A packet of an artificial sweetener contains 40.0 mg of saccharin (C7H5NO3S), which has the structural formula:
Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?
Solution
The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example $$\PageIndex{8}$$, and then multiplying by Avogadro’s number:
Using the provided mass and molar mass for saccharin yields:
$\mathrm{0.0400\:\cancel{g}\:\ce{C7H5NO3S}\left(\dfrac{\cancel{mol}\:\ce{C7H5NO3S}}{183.18\:\cancel{g}\:\ce{C7H5NO3S}}\right)\left(\dfrac{6.022\times10^{23}\:\ce{C7H5NO3S}\:molecules}{1\:\cancel{mol}\:\ce{C7H5NO3S}}\right)}\\ =\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\:molecules}$
The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:
$\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\: molecules\left(\dfrac{7\:C\: atoms}{1\:\ce{C7H5NO3S}\: molecule}\right)=9.20\times10^{21}\:C\: atoms} \nonumber$
Exercise $$\PageIndex{9}$$
How many $$C_4H_{10}$$ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?
• $$9.545 \times 10^{22}\; \text{molecules}\; C_4H_{10}$$
• $$9.545 \times 10^{23 }\;\text{atoms}\; H$$
Video $$\PageIndex{3}$$: A preview of some of the uses we will have for moles in upcoming units
## Percent Composition
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
$\mathrm{\%H=\dfrac{mass\: H}{mass\: compound}\times100\%}$
$\mathrm{\%C=\dfrac{mass\: C}{mass\: compound}\times100\%}$
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
$\mathrm{\%H=\dfrac{2.5\:g\: H}{10.0\:g\: compound}\times100\%=25\%}$
$\mathrm{\%C=\dfrac{7.5\:g\: C}{10.0\:g\: compound}\times100\%=75\%}$
Example $$\PageIndex{10}$$: Calculation of Percent Composition
Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?
Solution
To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
$\mathrm{\%C=\dfrac{7.34\:g\: C}{12.04\:g\: compound}\times100\%=61.0\%} \nonumber$
$\mathrm{\%H=\dfrac{1.85\:g\: H}{12.04\:g\: compound}\times100\%=15.4\%} \nonumber$
$\mathrm{\%N=\dfrac{2.85\:g\: N}{12.04\:g\: compound}\times100\%=23.7\%} \nonumber$
The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.
Exercise $$\PageIndex{10}$$
A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?
12.1% C, 16.1% O, 71.8% Cl
## Determining Percent Composition from Formula Mass
Video $$\PageIndex{4}$$: A video overview of how to calculate percent composition of a compound based on its chemical formula.
Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:
$\mathrm{\%N=\dfrac{14.01\:amu\: N}{17.03\:amu\:NH_3}\times100\%=82.27\%}$
$\mathrm{\%H=\dfrac{3.024\:amu\: N}{17.03\:amu\:NH_3}\times100\%=17.76\%}$
This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example $$\PageIndex{2}$$. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.
Example $$\PageIndex{11}$$: Determining Percent Composition from a Molecular Formula
Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?
Solution
To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:
\begin{align*} \%\ce C&=\mathrm{\dfrac{9\:mol\: C\times molar\: mass\: C}{molar\: mass\:\ce{C9H18O4}}\times100=\dfrac{9\times12.01\:g/mol} \nonumber{180.159\:g/mol}\times100=\dfrac{108.09\:g/mol}{180.159\:g/mol}\times100} \nonumber\\ \%\ce C&=\mathrm{60.00\,\%\,C} \nonumber \end{align*}
\begin{align*} \%\ce H&=\mathrm{\dfrac{8\:mol\: H\times molar\: mass\: H}{molar\: mass\:\ce{C9H18O4}}\times 100=\dfrac{8\times 1.008\:g/mol} \nonumber {180.159\:g/mol}\times 100=\dfrac{8.064\:g/mol}{180.159\:g/mol}\times 100} \nonumber\\ \%\ce H&=4.476\,\%\,\ce H\nonumber \end{align*}
\begin{align*} \%\ce O&=\mathrm{\dfrac{4\:mol\: O\times molar\: mass\: O}{molar\: mass\: \ce{C9H18O4}}\times 100=\dfrac{4\times 16.00\:g/mol} \nonumber{180.159\:g/mol}\times 100=\dfrac{64.00\:g/mol}{180.159\:g/mol}\times 100} \nonumber \\ \%\ce O&=35.52\% \nonumber \end{align*}
Note that these percentages sum to equal 100.00% when appropriately rounded.
Exercise $$\PageIndex{11}$$
To three significant digits, what is the mass percentage of iron in the compound $$Fe_2O_3$$?
69.9% Fe
## Summary
The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be 6.022 × 1023, a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H2O molecule weighs approximately 18 amu and 1 mole of H2O molecules weighs approximately 18 g).
## Footnotes
1. 1 Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.
2. Read more about the redefinition of SI units including the kilogram here (Laura Howe, CE&N, Nov. 16, 2018).
## Glossary
experimentally determined value of the number of entities comprising 1 mole of substance, equal to 6.022 × 1023 mol−1
formula mass
sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass
mole
amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of 12C
molar mass
mass in grams of 1 mole of a substance
percent composition
percentage by mass of the various elements in a compound |
# Measures Of Dispersion Class 11 Mathematics Solutions | Exercise - 13.1
## Chapter - 13 Measures Of Dispersion
The state of getting dispersed or spread is known as dispersion and the measure of dispersion is a statistic that describes how spread out the values in a data set are. To simplify the term, measure of dispersion shows how squeezed or scattered the variable is. It is typically used to complement a measure of central tendency, such as the mean or median, in describing a data set.
In this chapter, we will learn different methods of finding arithmetic mean, median, standard deviation and coefficient of variation. In addition that, we will get to know about the meaning of dispersion, absolute measure, relative measures, variance etc. Before we move towards the solution section, I would like to give you a brief overview of the main topics covered in this chapter.
### Introduction
The measures which condense the huge mass of data into a single value representing all of theth and around which most of the data tend to concentrate are known as averages. Measure of such single value is known as measure of central tendency.
Of the various measures of central tendency, we deal here briefly only those which are frequently used in the subsequent chapters. The measures of dispersion helps to interpret the variability of data. It means we will be able to know how much homogenous or heterogeneous the data is. There are two types of dispersion which are Absolute Measure and Relative Measure of Dispersion.
### Absolute and Relative Measures
Those measures of dispersion whose units are same as the units of the given series are known as the absolute measure of dispersion. These types of dispersions can be used only comparing the variability of the series (or distribution) having the same units. Comparison of two distributions with different units can not be made with absolute measures.
On the otherhand, the relative measures of dispersions are obtained as the ratio of absolute measure of dispersion to suitable average and are thus a pure number independent of units. Hence t distributions with different units can be compared with the help of relative measures of dispersion.
### Methods of measuring dispersion
The following are the methods of measuring dispersion.
i. Range
ii. Semi interquartile range or Quartile deviation
iii. Mean deviation or Average deviation
iv. Standard deviation
### Requisites of a good measure of dispersion
The good measure of dispersion must have the following characteristics:
1. The measure should be rigidly defined.
2. The measure should be simple to understand and easy to calculate.
3. All items must be included in the measure.
4. The measure must be suitable for further mathematical treatment.
5. Fluctuation of sampling in the measure should be least.
6. Extreme values should not unduly affect the measure.
### Exercise - 13.1
In this PDF, you'll only find the solution of class 11 measures of dispersion chapter. It contains all the solutions of exercise- 13.1. If you want the solutions of other exercises then you'll find them above this PDF. Just click on the button and you'll reach your destination.
NoteScroll the PDF to view all Solution
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## Is Class 11 Mathematics Guide Helpful For Student ?
I have published this Notes for helping students who can't solve difficult maths problems. Student should not fully depend on this note for completing all the exercises. If you totally depend on this note and simply copy as it is then it may affect your study.
Student should also use their own will power and try to solve problems themselves. You can use this mathematics guide PDF as a reference. You should check all the answers before copying because all the answers may not be correct. There may be some minor mistakes in the note, please consider those mistakes.
## How to secure good marks in Mathematics ?
As, you may know I'm also a student. Being a student is not so easy. You have to study different subjects simultaneously. From my point of view most of the student are weak in mathematics. You can take me as an example, I am also weak in mathematics. I also face problems while solving mathematics questions.
If you want to secure good marks in mathematics then you should practise them everyday. You should once revise all the exercise which are already taught in class. When you are solving maths problems, start from easy questions that you know already. If you do so then you won't get bored.
Maths is not only about practising, especially in grade 11 you to have the basic concept of the problem. When you get the main concept of the problem then you can easily any problems in which similar concept are applied.
When your teacher tries to make the concept clear by giving examples then all students tries to remember the same example but you should never do that. You can create your own formula which you won't forget later.
If you give proper time for your practise with proper technique then you can definitely score a good marks in your examination.
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Home / The science / Mathematics / How to translate the number system
# How to translate the number system
/
2 Views
In information technology, instead of the usual decimal number system, binary is often used, as the work of computers is built on it.
Instructions
1
The main operations are only two: Translation from decimal to another (binary, octal, etc.) and vice versa. The name of each number system comes from its base - the number of elements in it (binary - 2, decimal - 10). In base systems with a base greater than 10, it is customary to use the letters of the Latin alphabet (A-10, B-11, etc.) as a replacement for two-digit numbers.
2
Let's consider the operations using the example of a binary systemNumbering, as the most common. For all other systems, the same rules and methods will be true to the replacement of base 2 by the corresponding one.
So, we have some number in binaryNumber system consisting of several digits. We write it in the form of the sum of the products of its digits multiplied by 2. Next, we arrange all the powers from right to left starting from 0. We sum up. The resulting number is the required number.
Example.
1011 = 1 * (2 ^ 3) + 0 * (2 ^ 2) + 1 * (2 ^ 1) + 1 * (2 ^ 0) = 8 + 0 + 2 + 1 = 11.
3
Now consider the reverse operation.
Let there be given a number in the decimal system. We will divide it by a column on the basis of the number system into which we want to translate it (in our case this will be 2). The division continues to the very end, until the private becomes less than the ground. Next, starting with the last, we write down all the remains in the line. This is the required number.
Example.
11/2 = 5 the rest is 1, 5/2 = 2, the rest is 1, 2/2 = 1 the remainder is 0 = & gt- 1011.
Another example is shown in the picture.
For other bases, operations are similar. Do not forget to replace the numbers, beginning with 10, in the corresponding number systems in Latin letters! Otherwise, the resulting number will be read incorrectly, because "10" and "1" 0 "are absolutely different things!
The base of the number system, in which the number is represented, is indicated in the form of an index at the bottom of the rightmost digit of the number.
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# MULTIPLYING AND DIVIDING FRACTIONS Case 2. MULTIPLICATION Multiplying fractions is actually very easy! You begin by placing the two fractions you.
## Presentation on theme: "MULTIPLYING AND DIVIDING FRACTIONS Case 2. MULTIPLICATION Multiplying fractions is actually very easy! You begin by placing the two fractions you."— Presentation transcript:
MULTIPLYING AND DIVIDING FRACTIONS Case 2
MULTIPLICATION Multiplying fractions is actually very easy! You begin by placing the two fractions you wish to multiply next to each other. Example: 1/3 x 5/8 Next you multiply the two numerators Example- the two numerators are 1 and 5 so the numerator of the answer will be 1 x 5 which is 5
MULTIPLICATION Next multiply the two denominators together The two denominators are 3 and 8 so the denominator in the answer will be 3 x 8 which equals 24 Combine the final numerator and denominator together to get a final fraction Example- the final numerator was 5 and the final denominator was 24 so the final fraction will be 5/24.
MULTIPLICATION The final step in multiplying fractions is to check if your final answer can be reduced Reducing means checking to see if the numerator and denominator can be divided by a common number In our case the fraction is 5/24. There is no number that you could divide 5 and 24 by. Therefore, 5/24 is the final answer. An example of a fraction that can be reduced would be 5/30. The common number that you can divide by is 5. You can divide 5 by 5 to get 1 and you can divide 30 by 5 to get 6. So the new numerator will be 1 and the new denominator will be 6. The final fraction is 1/6.
DIVISION Dividing fractions is actually very similar to multiplying. Before we begin you will need to know what the term reciprocal means. The reciprocal of a fraction is simply switching the numerator and the denominator. For example, the reciprocal of 5/8 is 8/5
DIVISION Now that you know what the reciprocal of a fraction is we can learn to divide. We will begin with the equation 5/8 ÷ 3/4. To divide a fraction, what you really want to do is multiply by the reciprocal. So we want to do 5/8 x the reciprocal of 3/4 As we mentioned earlier, the reciprocal is simply switching the numerator and denominator of a fraction. Therefore the reciprocal of 3/4 would be 4/3
DIVISION So, if dividing fractions really means multiply by the reciprocal we want to do 5/8 x the reciprocal of 3/4. We know that the reciprocal of 3/4 is 4/3 so the equation we really want to solve is 5/8 x 4/3 From this point you simply solve the equation the exact same way you solve a multiplication problem.
DIVISION First you multiply the numerators of 5 and 4 to get 20 and then multiply the denominators to get 24. The final fraction is 20/24. Like any multiplication problem, you have to see if the fraction can be reduced by a common number. In this case, 20 and 24 can both be divided by 4 so the new numerator will be 5 and the new denominator will be 6. The result is a final answer of 5/6 So, 5/8 ÷3/4 is 5/6.
SUMMARY As you can see, multiplying and dividing fractions is really not too difficult. Some key things to remember are to always make sure your fraction is reduced and also make sure you take the reciprocal in a division problem before solving the equation.
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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Problem Solving tagged by: BTGmoderatorRO ##### This topic has 2 expert replies and 0 member replies ## Problem Solving Which of the following fractions is greater than 1/4? A. 12/50 B. 3/11 C. 2/9 D. 4/17 E. 6/24 OA is B Can an Expert confirm this OA, please. ### GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Quote: Which of the following fractions is greater than 1/4? A. 12/50 B. 3/11 C. 2/9 D. 4/17 E. 6/24 OA is B Can an Expert confirm this OA, please. Hi Roland2rule, Let's take a look at your question. We will compare each fraction with 1/4 to check which one is greater than 1/4. Let's start with option A, 12/50 To compare fractions, we make the denominators of the fraction same. $$\frac{12}{50}=\frac{12}{50}\times\frac{2}{2}=\frac{24}{100}$$ $$\frac{1}{4}=\frac{1}{4}\times\frac{25}{25}=\frac{25}{100}$$ Compare the numerator of both the fractions now. 24<25, therefore, 12/50 < 1/4. Now check option B, 3/11. Make the denominators of both the fractions same. $$\frac{3}{11}=\frac{3}{11}\times\frac{4}{4}=\frac{12}{44}$$ $$\frac{1}{4}=\frac{1}{4}\times\frac{11}{11}=\frac{11}{44}$$ Compare the numerators, we can see that 12 > 11, therefore, 3/11 > 1/4. Therefore, Option B is correct. Hope it helps. I am available if you'd like any follow up. _________________ GMAT Prep From The Economist We offer 70+ point score improvement money back guarantee. Our average student improves 98 points. Free 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. Get started now. ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15387 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 Roland2rule wrote: Which of the following fractions is greater than 1/4? A. 12/50 B. 3/11 C. 2/9 D. 4/17 E. 6/24 To compare two fractions: 1. Multiply the denominator in each fraction by the numerator in the other fraction. 2. The numerator that yields the greater product belongs to the greater fraction. A: 12/50 versus 1/4 50*1 = 50. 4*12 = 48. Since the numerator for 1/4 yields the greater product, 1/4 is greater than 12/50. B: 3/11 versus 1/4 11*1 = 11. 4*3 = 12. Since the numerator for 3/11 yields the greater product, 3/11 is greater than 1/4. The correct answer is B. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200
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# CHAPTER 19 PARTIAL FRACTIONS
## EXERCISE 81 Page 181
12
1. Resolve into partial fractions
x 9
2
12 12 A B A(x 3) B(x 3)
x 9 (x 3)(x 3) (x 3) (x 3)
2
(x 3)(x 3)
## If x = 3, 12 = 6B from which, B = 12/6 = 2
12 2 2 2 2
Hence,
x 9 (x 3) (x 3) x 3 x 3
2
4(x 4)
2. Resolve into partial fractions
x 2 2x 3
## 4(x 4) 4x 16 A B A(x 3) B(x 1)
Let
x 2x 3 (x 1)(x 3) (x 1) (x 3)
2
(x 1)(x 3)
## If x = -1, -20 = -4A from which, A = 5
If x = 3, 12 – 16 = 4B from which, B = -1
4(x 4) 5 1
Hence,
x 2x 3 (x 1) (x 3)
2
x 2 3x 6
3. Resolve into partial fractions
x(x 2)(x 1)
## x 2 3x 6 A B C A(x 2)(x 1) Bx(x 1) Cx(x 2)
Let
x(x 2)(x 1) x (x 2) (x 1) x(x 2)(x 1)
## If x = 0, 6 = A(-2)(-1) from which, 6 = 2A and A = 3
302
If x = 2, 4 - 6 + 6 = B(2)(1) from which, 4 = 2B and B = 2
## If x = 1, 1 - 3 + 6 = C(1)(-1) from which, 4 = -C and C = -4
x2 3x 6 3 2 4
Hence,
x(x 2)(x 1) x (x 2) (x 1)
3(2x 2 8x 1)
4. Resolve into partial fractions
(x 4)(x 1)(2x 1)
Let
3(2x 2 8x 1) A B C A(x 1)(2x 1) B(x 4)(2x 1) C(x 4)(x 1)
(x 4)(x 1)(2x 1) (x 4) (x 1) (2x 1) (x 4)(x 1)(2x 1)
## If x = 0.5, 1.5 - 12 -3 = C(4.5)(1.5) from which, -13.5 = 6.75C and C = -2
3(2x2 8x 1) 7 3 2
Hence,
(x 4)(x 1)(2x 1) (x 4) (x 1) (2x 1)
x 2 9x 8
5. Resolve into partial fractions
x2 x 6
Since the numerator is of the same degree as the denominator, division is firstly required.
1
x 2 x 6 x 2 9x 8
x2 x 6
8x + 14
x 2 9x 8 8x 14
Hence, 1 2
x x 6
2
x x 6
8x 14 8x 14 A B A(x 2) B(x 3)
Let
x x 6 (x 3)(x 2) (x 3) (x 2)
2
(x 3)(x 2)
## If x = -3, -24 + 14 = -5A from which, -10 = -5A and A = 2
303
If x = 2, 16 + 14 = 5B from which, 30 = 5B and B = 6
x2 9x 8 2 6
Hence, 1
x x6
2
(x 3) (x 2)
x 2 x 14
6. Resolve 2 into partial fractions
x 2x 3
Since the numerator is of the same degree as the denominator, division is firstly required.
1
x 2x 3 x 2 x 14
2
x2 2 x 3
x - 11
x 2 x 14 x 11
Hence, 1 2
x 2x 3
2
x 2x 3
x 11 x 11 A B A(x 3) B(x 1)
Let
x 2x 3 (x 1)(x 3) (x 1) (x 3)
2
(x 1)(x 3)
## If x = 3, 3 - 11 = 4B from which, - 8 = 4B and B = - 2
x2 x 14 3 2
Hence, 1
x 2x 3
2
(x 1) (x 3)
3x 3 2x 2 16x 20
7. Resolve into partial fractions
(x 2)(x 2)
3x - 2
x 2 4 3x3 2x 2 16x 20
3x 3 12x
2x 2 4x 20
2x 2 8
- 4x + 12
3x 3 2x 2 16x 20 12 4x
Hence, 3x 2 2
(x 2)(x 2) x 4
304
12 4x 12 4x A B A(x 2) B(x 2)
Let
x 4 (x 2)(x 2) (x 2) (x 2)
2
(x 2)(x 2)
Hence, 12 – 4x = A(x + 2) + B(x - 2)
If x = 2, 4 = 4A from which, A = 1
If x = -2 20 = - 4B from which, B = - 5
## 3x3 2x2 16x 20 1 5
Hence, 3x 2
(x 2)(x 2) (x 2) (x 2)
305
EXERCISE 82 Page 182
4x 3
1. Resolve into partial fractions
(x 1)2
4x 3 A B A(x 1) B
Let
(x 1) 2
(x 1) (x 1) 2
(x 1) 2
Hence, 4x – 3 = A(x + 1) + B
If x = - 1 -7=B
## Equating x coefficients gives: 4 = A
4x 3 4 7
Hence,
x 1
2
x 1 x 1 2
x 2 7x 3
2. Resolve into partial fractions
x 2 (x 3)
x 2 7x 3 A B C A(x)(x 3) B(x 3) Cx 2
Let 2 2
x (x 3) x x (x 3) x 2 (x 3)
Hence, x 2 + 7x + 3 = A(x)(x + 3) + B (x + 3) + C x 2
## Equating x 2 coefficients: 1 = A + C from which, A=2
x2 7x 3 2 1 1
Hence, 2
x (x 3)
2
x x (x 3)
5x 2 30x 44
3. Resolve into partial fractions
(x 2)3
## 5x 2 30x 44 A B C A(x 2) 2 B(x 2) C
Let
(x 2)3 (x 2) (x 2)2 (x 2)3 (x 2)3
## If x = 2 20 – 60 + 44 = C from which, C=4
306
Equating x 2 coefficients: 5=A
## and 2B = 20 + 4 – 44 = -20 from which, B = -10
5x2 30x 44 5 10 4
Hence,
(x 2) 3
(x 2) (x 2) (x 2)3
2
18 21x x 2
4. Resolve into partial fractions
(x 5)(x 2)2
## 18 21x x 2 A B C A(x 2) 2 B(x 5)(x 2) C(x 5)
Let
(x 5)(x 2)2 (x 5) (x 2) (x 2) 2 (x 5)(x 2) 2
## Equating x 2 coefficients: -1 = A + B from which, B = -3
18 21x x2 2 3 4
Hence,
(x 5)(x 2) 2
(x 5) (x 2) (x 2)2
307
EXERCISE 83 Page 183
x 2 x 13
1. Resolve into partial fractions
(x 2 7)(x 2)
x 2 x 13 Ax B C (Ax B)(x 2) C x 2 7
Let
x 2 7 x 2 x 2 7 (x 2) x 2 7 (x 2)
Hence, x 2 x 13 (Ax B)(x 2) C x 2 7
## Equating constant terms: - 13 = - 2B + 7C = - 2B – 7 i.e. 2B = 13 – 7 = 6 from which, B = 3
x 2 x 13 2x 3 1
Hence, 2
x 7 x 2 x 7 (x 2)
2
6x 5
2. Resolve into partial fractions
(x 4)(x 2 3)
6x 5 Bx C A(x 3) (Bx C) x 4
2
A
Let
x 4 x 2 3 x 4 (x 2 3) x 4 x 2 3
## Equating constant terms: - 5 = 3A - 4C = 3 – 4C i.e. 4C = 3 + 5 = 8 from which, C = 2
6x 5 1 x 2 1 2x
Hence, 2 2
x 4 x 3 x 4 x 3 x 4 x 3
2
15 5x 5x 2 4x 3
3. Resolve into partial fractions
x 2 (x 2 5)
308
15 5x 5x 2 4x 3 A B Cx D Ax x 5 B x 5 Cx D x
2 2 2
Let 2 2
x 2 x 2 5 x x x 5 x 2 x 2 5
Hence, 15 5x 5x 2 4x 3 Ax x 2 5 B x 2 5 Cx D x 2
## From equation (1), -4 = 1 + C from which, C = -5
15 5x 5x 2 4x 3 1 3 2 5x
Hence, 2 2
x x 5
2 2
x x x 5
x 3 4x 2 20x 7
4. Resolve into partial fractions
(x 1)2 (x 2 8)
Let
## Cx D A(x 1) x 8 B x 8 (Cx D)(x 1)
2 2 2
x 3 4x 2 20x 7 A B
(x 1) 2 x 2 8 (x 1) (x 1) 2 x 2 8 (x 1) 2 x 2 8
## Hence A+C=1 (1)
-A – 2C + D = 2 (2) since B = 2
and 8A + C – 2D = 20 (3)
## 2 × (2) gives: -2A – 4C + 2D = 4 (4)
309
(3) + (4) gives: 6A – 3C = 24 (5)
## From (2): -3 + 4 + D = 2 from which, D=1
x3 4x 2 20x 7 3 2 1 2x
Hence, 2
(x 1) x 8 (x 1) (x 1) x 8
2 2 2
d 2 d
5. When solving the differential equation 2
6 10 20 e2t by Laplace transforms, for
dt dt
## given boundary conditions, the following expression for ℒ{} results:
39 2
4s3 s 42s 40
ℒ{} = 2
s(s 2)(s 2 6s 10)
Show that the expression can be resolved into partial fractions to give:
2 1 5s 3
ℒ{} = 2
s 2(s 2) 2(s 6s 10)
39 2
4s3 s 42s 40
2 A B Cs D
s s 2 s 2 6s 10 s (s 2) s 2 6s 10
Let
A(s 2) s 2 6s 10 B(s) s 2 6s 10 (Cs D)(s)(s 2)
s(s 2) s 2 6s 10
## s 42s 40 A(s 2) s 2 6s 10 B(s) s 2 6s 10 (Cs D)(s)(s 2)
39 2
Hence, 4s3
2
A s3 8s 2 22s 20 B s3 6s 2 10s (Cs D)(s 2 2s)
## If s = 0, -40 = A(-20) from which, A=2
1
If s = 2, 32 – 78 + 84 – 40 = B (8 – 24 + 20) i.e. - 2 = 4B from which, B =
2
1 5
Equating s 3 coefficients: 4=A+B+C i.e. 4=2- +C from which, C=
2 2
39 39
Equating s 2 coefficients: = - 8A – 6B – 2C + D i.e. = - 16 + 3 – 5 + D
2 2
310
3
from which, D=
2
39 2 1 5 3
4s3 s 42s 40 s
2 2 2 2 2
Hence, s s 2 s 6s 10 s (s 2) s 2 6s 10
2
2 1 5s 3
i.e.
s 2 s 2 2 s 6s 10
2
311 |
# APPLYING GCF AND LCM TO FRACTION OPERATIONS
Applying GCF and LCM to Fraction Operations :
Generally we have the following kinds of operations in fraction.
In which if we want to add or subtract two or more fractions, then we may have to use the concept LCM if the denominator of those fractions are not same.
For multiplying and dividing two fractions we may have to use the concept GCF. If we want to reduce the fraction into its lowest form,then we have to apply the concept GCF.
Now let us see some examples to understand how we use the concept LCM and GCF while adding and subtracting two fractions.
## Adding Two Fractions using LCM - Examples
Example 1 :
Solution :
To simplify the fraction, we have to find the GCF of both numerator and denominator.
Now we have to divide both numerator and denominator by the GCF as given below.
Therefore, the simplified form of 9/12 is 3/4.
Example 2 :
Solution :
Since the denominators are not same, we have to take LCM to make the denominators same.
To simplify the fraction, we have to find the GCF of both numerator and denominator.
Now we have to divide both numerator and denominator by the GCF as given below.
## Subtracting Two Fractions using LCM - Example
If the denominators of the fractions are not co-prime (there is a common divisor other than 1), we have to apply this method.
Fro example, let us consider the two fractions 5/12, 1/20.
In the above two fractions, denominators are 12 and 20.
For 12 and 20, if there is at least one common divisor other than 1, then 12 and 20 are not co-prime.
For 12 & 20, we have the following common divisors other than 1.
2 & 4
So 12 and 20 are not co-prime.
In the next step, we have to find the L.C.M (Least common multiple) of 12 and 20.
12 = 2² x 3
20 = 2² x 5
When we decompose 12 and 20 in to prime numbers, we find 2, 3 and 5 as prime factors for 12 and 20.
To get L.C.M of 12 and 20, we have to take 2, 3 and 5 with maximum powers found above.
So, L.C.M of 12 and 20 is
= 2² x 3 x 5
= 4 x 3 x 5
= 60
Now we have to make the denominators of both the fractions to be 60 and subtract the two fractions 5/12 and 1/20 as given below.
## Multiplying Two Fractions - Example
To multiply two or more fractions, we have to multiply the numerators with numerators and denominators with denominators.
If it is possible we can simplify the fraction into its lowest form using GCF.
Example :
Multiply (3/20) x (30/12)
Solution :
Considering the numerators and denominators we can simplify the above fraction by using the concept GCF.
By simplifying the given fractions using GCF, We get
So the final answer is 3/8.
## Simplifying Fractions Using GCF - Examples
Example :
Simplify 42/60 in simplest form
Solution :
Step 1 :
Write the two numbers on one line
Step 2 :
Draw the L shape
Step 3 :
Divide out common prime numbers starting from the smallest.
7 and 10 is not divisible by any common number.
7/10 is the simplified form of the given fraction 42/60.
After having gone through the stuff given above, we hope that the students would have understood, how to apply GCF and LCm to fractions operations.
Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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# Difference between revisions of "2011 USAJMO Problems/Problem 1"
Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.
## Solution 1
Let $2^n + 12^n + 2011^n = x^2$. Then $(-1)^n + 1 \equiv x^2 \pmod {3}$. Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd. Proof by Contradiction: We wish to show that the only value of $n$ that satisfies is $n = 1$. Assume that $n \ge 2$. Then consider the equation $2^n + 12^n = x^2 - 2011^n$. From modulo 2, we easily know x is odd. Let $x = 2a + 1$, where a is an integer. $2^n + 12^n = 4a^2 + 4a + 1 - 2011^n$. Dividing by 4, $2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n)$. Since $n \ge 2$, $n-2 \ge 0$, so $2^{n-2}$ similarly, the entire LHS is an integer, and so are $a^2$ and $a$. Thus, $\dfrac {1}{4} (1 - 2011^n)$ must be an integer. Let $\dfrac {1}{4} (1 - 2011^n) = k$. Then we have $1- 2011^n = 4k$. $1- 2011^n \equiv 0 \pmod {4}$. $(-1)^n \equiv 1 \pmod {4}$. Thus, n is even. However, it has already been shown that $n$ must be odd. This is a contradiction. Therefore, $n$ is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1.
## Solution 2
If $n = 1$, then $2^n + 12^n + 2011^n = 2025 = 45^2$, a perfect square.
If $n > 1$ is odd, then $2^n + 12^n + 2011^n \equiv 0 + 0 + (-1)^n \equiv 3 \pmod{4}$.
Since all perfect squares are congruent to $0,1 \pmod{4}$, we have that $2^n+12^n+2011^n$ is not a perfect square for odd $n > 1$.
If $n = 2k$ is even, then $(2011^k)^2 < 2^{2k}+12^{2k}+2011^{2k}$ $= 4^k + 144^k + 2011^{2k} <$ $2011^k + 2011^k + 2011^{2k} < (2011^k+1)^2$.
Since $(2011^k)^2 < 2^n+12^n+2011^n < (2011^k+1)^2$, we have that $2^n+12^n+2011^n$ is not a perfect square for even $n$.
Thus, $n = 1$ is the only positive integer for which $2^n + 12^n + 2011^n$ is a perfect square.
## Solution 3
Looking at residues mod 3, we see that $n$ must be odd, since even values of $n$ leads to $2^n + 12^n + 2011^n = 2 \pmod{3}$. Also as shown in solution 2, for $n>1$, $n$ must be even. Hence, for $n>1$, $n$ can neither be odd nor even. The only possible solution is then $n=1$, which indeed works.
## Solution 4
Take the whole expression mod 12. Note that the perfect squares can only be of the form 0, 1, 4 or 9 (mod 12). Note that since the problem is asking for positive integers, $12^n$ is always divisible by 12, so this will be disregarded in this process. If $n$ is even, then $2^n \equiv 4 \pmod{12}$ and $2011^n \equiv 7^n \equiv 1 \pmod {12}$. Therefore, the sum in the problem is congruent to $5 \pmod {12}$, which cannot be a perfect square. Now we check the case for which $n$ is an odd number greater than 1. Then $2^n \equiv 8 \pmod{12}$ and $2011^n \equiv 7^n \equiv 7 \pmod {12}$. Therefore, this sum would be congruent to $3 \pmod {12}$, which cannot be a perfect square. The only case we have not checked is $n=1$. If $n=1$, then the sum in the problem is equal to $2+12+2011=2025=45^2$. Therefore the only possible value of $n$ such that $2^n+12^n+2011^n$ is a perfect square is $n=1$.
## Solution 5
We will first take the expression modulo $3$. We get $2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3$.
Lemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$. We can prove this by testing the residues modulo $3$. We have $0^2 \equiv 0 \pmod 3$, $1^2 \equiv 1 \pmod 3$, and $2^2 \equiv 1 \pmod 3$, so the lemma is true.
We know that if $n$ is odd, $-1^n+1^n \equiv 0 \pmod 3$, which satisfies the lemma's conditions. However, if $n$ is even, we get $2 \pmod 3$, which does not satisfy the lemma's conditions. So, we can conclude that $n$ is odd.
Now, we take the original expression modulo $4$. For right now, we will assume that $n>1$, and test $n=1$ later. For $n>1$, $2^n \equiv 0 \pmod 4$, so $2^n+12^n+2011^n=-1^n \pmod 4$.
Lemma 2: All perfect squares are equal to $0$ or $1$ modulo $3$. We can prove this by testing the residues modulo $4$. We have $0^2 \equiv 0 \pmod 4$, $1^2 \equiv 1 \pmod 4$, $2^2 \equiv 0 \pmod 4$, and $3^2 \equiv 1 \pmod 4$, so the lemma is true.
We know that if $n$ is even, $-1^n \equiv 0 \pmod 4$, which satisfies the lemma's conditions. However, if $n$ is odd, $-1^n \equiv -1 \equiv 3 \pmod 4$, which does not satisfy the lemma's conditions. Therefore, $n$ must be even.
However, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test $n=1$. We know that $2^1+12^1+2011^1=45^2$, so the only integer is $\boxed{n=1}$. |
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# 4.6: Simplify Variable Expressions Involving Integer Addition
Difficulty Level: At Grade Created by: CK-12
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Practice Simplify Variable Expressions Involving Integer Addition
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Have you ever been diving? Have you been to the Caribbean?
Last year, on his first day in the Caribbean, Cameron did his dive test and passed with flying colors. A dive test is done in a pool prior to diving. It lets the dive master know that you understand what you are doing and can handle yourself under the water. Scuba diving is exciting, but you have to know what you are doing to do it well.
The second day, Cameron and his Dad went for their first two dives. On the first dive, Cameron traveled to a depth of 25 feet. Then he and his Dad saw a stingray and followed it for a while and traveled down another 10 feet. Cameron took a few pictures of the stingray too.
Then they traveled back to the boat for some surface time to eat and rest before going on the second dive. On his second dive, Cameron did a shallow dive of only 15 feet. He loved seeing the beautiful coral and even spotted a sea cucumber.
When they returned to the boat, Cameron began calculating his total depth and his total time for the day.
To calculate Cameron’s total depth for the day, you will need to know how to add integers. This Concept will give you all that you need to know!!
### Guidance
An expression is a number sentence that contains numbers and operations. An expression is not solved it is evaluated because there isn’t an equals sign in an expression. Some expressions include variables. These expressions are called variable expressions. A variable is a symbol or letter that is used to represent one or more numbers.
Variable expressions can be combined when there is a common term or a like term.
5x+7x\begin{align*}5x+7x\end{align*}
We can simplify or combine this variable expression because it has two common variables. The x\begin{align*}x\end{align*}’s are common so we can find the sum of the variable expression.
The answer is 12x\begin{align*}12x\end{align*}.
9y+8x\begin{align*}9y+8x\end{align*}
This variable expression can not be combined or simplified. It does not have like terms. The x\begin{align*}x\end{align*} and the y\begin{align*}y\end{align*} are different, so we can not do anything with this expression. It is in simplest form.
Sometimes, expressions will include both variables and integers. You can use what you know about how to add integers to help you find the value of expressions with variables.
Find the sum 4a+(a)\begin{align*}-4a+(-a)\end{align*}.
Since 4a\begin{align*}-4a\end{align*} and a\begin{align*}-a\end{align*} both have the same variable, they are like terms. Use what you know about how to add integers to help you add the terms.
Both terms have the same sign, a negative sign. So, find the absolute values of both integers. Then add those absolute values to combine the terms. Remember, a=1a\begin{align*}-a=-1a\end{align*}.
|4|=4\begin{align*}|-4|=4\end{align*} and |1|=1\begin{align*}|-1|=1\end{align*}, so add 4a+1a=5a\begin{align*}4a+1a=5a\end{align*}.
Since both terms had negative signs, give the answer a negative sign.
The sum of 4a+(a)\begin{align*}-4a+(-a)\end{align*} is 5a\begin{align*}-5a\end{align*}.
Find the sum 3t+9t\begin{align*}-3t+9t\end{align*}.
Since 3t\begin{align*}-3t\end{align*} and 9t\begin{align*}9t\end{align*} both have the same variable, they are like terms. Use what you know about how to add integers to help you add the terms.
Both like terms have different signs. So, find the absolute values of both integers. Then subtract the term whose integer has the lesser absolute value from the other term.
|3|=3\begin{align*}|-3|=3\end{align*} and |9|=9\begin{align*}|9|=9\end{align*}, so subtract: 9t3t=6t\begin{align*}9t-3t=6t\end{align*}.
Since 9>3\begin{align*}9>3\end{align*}, and 9t\begin{align*}9t\end{align*} has a positive sign, give the answer a positive sign.
The answer is 3t+9t=6t\begin{align*}-3t+9t=6t\end{align*}.
Working with variable expressions may seem tricky, but if you first determine if you have like terms and then use the strategies you have learned for finding integer sums, you will be able to simplify each expression.
Simplify each variable expression.
#### Example A
8x+5x\begin{align*}-8x+ -5x\end{align*}
Solution:13x\begin{align*}-13x\end{align*}
#### Example B
19y+5y\begin{align*}-19y+5y\end{align*}
Solution:14y\begin{align*}-14y\end{align*}
#### Example C
6y+2y+3y\begin{align*}-6y+2y+ -3y\end{align*}
Solution:7y\begin{align*}-7y\end{align*}
Here is the original problem once again.
Last year,on his first day in the Caribbean, Cameron did his dive test and passed with flying colors. A dive test is done in a pool prior to diving. It lets the dive master know that you understand what you are doing and can handle yourself under the water. Scuba diving is exciting, but you have to know what you are doing to do it well.
The second day, Cameron and his Dad went for their first two dives. On the first dive, Cameron traveled to a depth of 25 feet. Then he and his Dad saw a stingray and followed it for a while and traveled down another 10 feet. Cameron took a few pictures of the stingray too.
Then they traveled back to the boat for some surface time to eat and rest before going on the second dive. On his second dive, Cameron did a shallow dive of only 15 feet. He loved seeing the beautiful coral and even spotted a sea cucumber.
When they returned to the boat, Cameron began calculating his total depth and his total time for the day.
To help Cameron calculate his total depth, we can write the following equation.
Depth of dive 1 + depth of dive 2 = total depth
Remember that depth is below the surface, so we use negative numbers to represent these integers.
On dive 1, Cameron went -25 ft and then -10 ft.
On dive 2, Cameron went -15 ft.
Now we can substitute these values into the equation.
25+10+15=50\begin{align*}-25 + -10 + -15 = -50\end{align*}
Cameron’s total depth for the day was -50 feet.
### Vocabulary
Here are the vocabulary words in this Concept.
Integer
the set of whole numbers and their opposites.
Sum
Expression
a phrase using numbers and operations.
Variable
a letter used to represent an unknown quantity.
Variable Expression
A phrase using numbers, operations and variables.
### Guided Practice
Here is one for you to try on your own.
Simplify 7z+(3z)\begin{align*}7z+(-3z)\end{align*}.
We combine like terms in this problem, but keep in mind that there are positive and negative values.
7z+(3z)=4z\begin{align*}7z + (-3z)=4z\end{align*}
### Video Review
Here is a video for review.
### Practice
Directions: Simplify each variable expression.
1. 7z+(3z)\begin{align*}7z+(-3z)\end{align*}
2. 17z+(15z)\begin{align*}17z+(-15z)\end{align*}
3. 5x+(3x)\begin{align*}5x+(-3x)\end{align*}
4. 8y+(2y)\begin{align*}8y+(2y)\end{align*}
5. 12x+(13x)\begin{align*}12x+(-13x)\end{align*}
6. 9z+(9z)\begin{align*}9z+(-9z)\end{align*}
7. 14a+(3a)\begin{align*}14a+(-3a)\end{align*}
8. 22y+(33y)\begin{align*}22y+(-33y)\end{align*}
9. (10d)+(d)+2\begin{align*}(-10d)+(-d)+2\end{align*}
10. 8x+(4x)5\begin{align*}8x+(-4x)-5\end{align*}
11. 7y+(3y)\begin{align*}7y+(-3y)\end{align*}
12. 16x+(22x)\begin{align*}16x+(-22x)\end{align*}
13. 5a+(a)+7a\begin{align*}5a+(-a)+7a\end{align*}
Directions: Solve each real-world problem.
14. A plane is flying at an altitude that is 2, 500 feet above sea level. If the plane increases its altitude by 500 feet more, what will be its new altitude?
15. The temperature on a mountaintop at midnight was 8F\begin{align*}-8^\circ F\end{align*}. By 3:00 A.M., the temperature had risen by 3F\begin{align*}3^\circ F\end{align*}. What is the temperature at 3:00 A.M.?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Sum
The sum is the result after two or more amounts have been added together.
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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## The fastest step by step guide for calculating what is 30 percent of 80
We already have our first value 30 and the second value 80. Let"s assume the unknown value is Y which answer we will find out.
You are watching: 30 is what percent of 80
As we have all the required values we need, Now we can put them in a simple mathematical formula as below:
STEP 1Y = 30/100
STEP 2Y = 30/100 × 80
STEP 3Y = 30 ÷ 100 × 80
STEP 4Y = 24
Finally, we have found the value of Y which is 24 and that is our answer.
If you want to use a calculator to know what is 30 percent of 80, simply enter 30 ÷ 100 × 80 and you will get your answer which is 24
Here is a calculator to solve percentage calculations such as what is 30% of 80. You can solve this type of calculation with your values by entering them into the calculator"s fields, and click "Calculate" to get the result and explanation.
Calculate
Question: At a high school 30 percent of seniors went on a mission trip. There were 80 seniors. How many seniors went on the trip?
Answer: 24 seniors went on the trip.
See more: How Long Does A Pork Roast Take To Cook In Nuwave Oven Recipe)
Question: 30 percent of the children in kindergarten like Thomas the Train. If there are 80 kids in kindergarten, how many of them like Thomas?
Answer: 24 kids like Thomas the Train.
## Another step by step method
Step 1: Let"s solve the equation for Y by first rewriting it as: 100% / 80 = 30% / Y
Step 2: Drop the percentage marks to simplify your calculations: 100 / 80 = 30 / Y
Step 3: Multiply both sides by Y to transfer it on the left side of the equation: Y ( 100 / 80 ) = 30
Step 4: To isolate Y, multiply both sides by 80 / 100, we will have: Y = 30 ( 80 / 100 )
Step 5: Computing the right side, we get: Y = 24
This leaves us with our final answer: 30% of 80 is 24
30 percent of 80 is 24 30.01 percent of 80 is 24.008 30.02 percent of 80 is 24.016 30.03 percent of 80 is 24.024 30.04 percent of 80 is 24.032 30.05 percent of 80 is 24.04 30.06 percent of 80 is 24.048 30.07 percent of 80 is 24.056 30.08 percent of 80 is 24.064 30.09 percent of 80 is 24.072 30.1 percent of 80 is 24.08 30.11 percent of 80 is 24.088 30.12 percent of 80 is 24.096 30.13 percent of 80 is 24.104 30.14 percent of 80 is 24.112 30.15 percent of 80 is 24.12 30.16 percent of 80 is 24.128 30.17 percent of 80 is 24.136 30.18 percent of 80 is 24.144 30.19 percent of 80 is 24.152
30.2 percent of 80 is 24.16 30.21 percent of 80 is 24.168 30.22 percent of 80 is 24.176 30.23 percent of 80 is 24.184 30.24 percent of 80 is 24.192 30.25 percent of 80 is 24.2 30.26 percent of 80 is 24.208 30.27 percent of 80 is 24.216 30.28 percent of 80 is 24.224 30.29 percent of 80 is 24.232 30.3 percent of 80 is 24.24 30.31 percent of 80 is 24.248 30.32 percent of 80 is 24.256 30.33 percent of 80 is 24.264 30.34 percent of 80 is 24.272 30.35 percent of 80 is 24.28 30.36 percent of 80 is 24.288 30.37 percent of 80 is 24.296 30.38 percent of 80 is 24.304 30.39 percent of 80 is 24.312
30.4 percent of 80 is 24.32 30.41 percent of 80 is 24.328 30.42 percent of 80 is 24.336 30.43 percent of 80 is 24.344 30.44 percent of 80 is 24.352 30.45 percent of 80 is 24.36 30.46 percent of 80 is 24.368 30.47 percent of 80 is 24.376 30.48 percent of 80 is 24.384 30.49 percent of 80 is 24.392 30.5 percent of 80 is 24.4 30.51 percent of 80 is 24.408 30.52 percent of 80 is 24.416 30.53 percent of 80 is 24.424 30.54 percent of 80 is 24.432 30.55 percent of 80 is 24.44 30.56 percent of 80 is 24.448 30.57 percent of 80 is 24.456 30.58 percent of 80 is 24.464 30.59 percent of 80 is 24.472
30.6 percent of 80 is 24.48 30.61 percent of 80 is 24.488 30.62 percent of 80 is 24.496 30.63 percent of 80 is 24.504 30.64 percent of 80 is 24.512 30.65 percent of 80 is 24.52 30.66 percent of 80 is 24.528 30.67 percent of 80 is 24.536 30.68 percent of 80 is 24.544 30.69 percent of 80 is 24.552 30.7 percent of 80 is 24.56 30.71 percent of 80 is 24.568 30.72 percent of 80 is 24.576 30.73 percent of 80 is 24.584 30.74 percent of 80 is 24.592 30.75 percent of 80 is 24.6 30.76 percent of 80 is 24.608 30.77 percent of 80 is 24.616 30.78 percent of 80 is 24.624 30.79 percent of 80 is 24.632
30.8 percent of 80 is 24.64 30.81 percent of 80 is 24.648 30.82 percent of 80 is 24.656 30.83 percent of 80 is 24.664 30.84 percent of 80 is 24.672 30.85 percent of 80 is 24.68 30.86 percent of 80 is 24.688 30.87 percent of 80 is 24.696 30.88 percent of 80 is 24.704 30.89 percent of 80 is 24.712 30.9 percent of 80 is 24.72 30.91 percent of 80 is 24.728 30.92 percent of 80 is 24.736 30.93 percent of 80 is 24.744 30.94 percent of 80 is 24.752 30.95 percent of 80 is 24.76 30.96 percent of 80 is 24.768 30.97 percent of 80 is 24.776 30.98 percent of 80 is 24.784 30.99 percent of 80 is 24.792 |
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# A farmer producer $1\dfrac{3}{5}$ times as much peanuts this season as he did last season, find the ratio of last season production to this season.
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Hint: First we have to convert the data into improper fraction because it gives a mixed fraction. Then applying the formula we get the required answer.
Formula used: Formula of ratio and proportion that is $a:b = \dfrac{a}{b}$
Let’s assume that the farmer produced $x$ peanuts in the last season.
So, from the given data we can write that the farmer produced $1\dfrac{3}{5}{x_{}}$ in this season.
Here we convert the above mixed fraction into improper fraction, $= \dfrac{8}{5}x$
We have assumed that the production in the last season is $x$ and the production in this season is $\dfrac{8}{5}x$
We can write that the ratio of last season peanut production and this season peanut product
Now we can write it as $x:\dfrac{8}{5}x$
Here a =x and b = $\dfrac{8}{5}x$
Now we have to applying the formula of ratio and proportion that is $a:b = \dfrac{a}{b}$
$\Rightarrow \dfrac{x}{{\dfrac{8}{5}x}}$
By taking reciprocal we get,
$= \dfrac{{5x}}{{8x}}$
Cancel the terms we get,
$= \dfrac{5}{8}$
Thus the ratio of last season peanut to the present season peanut production $= 5:8$
Note: It is a simple question of ratio proportion which can be solved by assuming one as $x$.
The formula which is used for finding ratio is $a:b = \dfrac{a}{b}$
The basic properties of ratio include that it will always take place between two similar quantities, the units must be identical always and there must be significant order of terms.
The concept of ratio and proportion are two identical concepts and it is the basis for understanding different concepts in maths as well as in science.
We apply the concept of ratio and proportion in our everyday life like in business while handling money or other concepts. Both of these concepts are integral parts of mathematics. You can also find a lot of examples in reality such as velocity rate, price of a commodity, where we generally apply the ratio proportion principle. It is possible to compare two ratios if the ratios are equal to fractions. |
# How to Graph Inequalities on a Number Line
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The graph of an inequality on a number line can help students visually understand the solution to an inequality. Plotting an inequality on a number line requires a number of rules to ensure the solution is properly “translated” onto the graph. Students should pay special attention to whether the points on the number line are dots or circles, as they represent different types of inequalities.
Draw the number line. Sketch a long, horizontal line with arrow tips on both ends. Between the arrow tips, add short vertical lines at even intervals along the number line.
Observe the number in your inequality. For example, if your inequality is “x < 6,” the number of importance is 6. If your inequality has multiple points, such as in “9 < x < 10,” you will have two points of importance.
Label the vertical lines, or points, on the number line. Label one of the numbers of importance first. Choose a point close to the middle. Label the other points, adding one when going right and subtracting one when going left. Ensure that both points of importance appear on your number line if you have two points of importance.
Determine the type of point you will need to draw. Look at the sign in the inequality. If your inequality sign does not contain a solid line underneath, you will need to draw an open point, or circle. If you have a line beneath the inequality symbol, you will need to draw a solid point, or dot. If your inequality has two signs, consider each part individually.
Draw the point or points at the appropriate place or places on the number line.
Determine whether the inequality is less-than or greater-than. A less-than sign is one that points toward the x, such as in “x < 9.” A greater-than sign is one that points away from the x, such as in “x > 9.” Make this determination for each side of x in an inequality such as “9 < x < 10.”
Draw an arrow on the number line to indicate an inequality. From the point that you have drawn, draw an arrow to the left if your inequality is a less-than inequality. Draw an arrow to the right if it is a greater-than inequality. Do the same for the other point if you have two points of importance in your inequality. If you have an equation such as "9 < x < 10," you can connect the points with a solid line. |
# Some Applications of Trigonometry: Class 10 Mathematics NCERT Chapter 9
## Key Features of NCERT Material for Class 10 Mathematics Chapter 9 – Some Applications of Trigonometry
In the last chapter 8, you learned about Introduction of Trigonometry. In this chapter, you will learn about Applications of Trigonometry.
In mathematics, trigonometry is one of the most significant branches. Hence, the utilization of trigonometry incorporates, estimating the heights of pinnacles, deciding the distance of the item in the sea from the shore, finding the distance between celestial bodies, and so on. Basically, we use trigonometry to gauge heights and distances. Accordingly, we utilize trigonometric ratios to gauge heights and distances of various items.
#### Line of Sight
At the point when an eyewitness looks from a point E (eye) at an article O then the straight line EO between the eye E and the item O is known as the line of sight.
#### Horizontal
At the point when a spectator looks from a point E (eye) to another point Q which is flat to E, at that point the straight line, EQ among E and Q is known as the horizontal line.
#### Angle of Elevation
At the point when the eye is beneath the item, at that point the eyewitness needs to turn upward from the point E to the article O. The ratio of this pivot (angle θ) from the flat line is known as the angle of elevation.
#### Angle of Depression
At the point when the eye is over the item, at that point the spectator needs to peer down from the point E to the article. The level line is presently parallel to the ground. The proportion of this turn (angle θ) from the even line is known as the angle of depression.
Instructions to change over the above diagram with the right-angled triangle.
The case I: Angle of Elevation is known
Attract OX perpendicular to EQ.
Presently ∠OXE = 90°
ΔOXE is a right triangle. where
OE = hypotenuse
OX= opposite side (Perpendicular)
Case II: Angle of Depression is known
(I) Draw OQ’parallel to EQ
(ii) Draw perpendicular EX on OQ’.
(iii) Now ∠QEO = ∠EOX = Interior alternate angles
ΔEXO is a right Δ. where
EO = hypotenuse
EX = opposite side (Perpendicular)
Pick a trigonometric ratio so that it considers the known side and the side that you wish to compute.
The eye is constantly considered at ground level except if the issue explicitly gives the stature of the spectator.
The article is constantly considered as a point.
A few People have
Sin θ =
Wavy Black Hair
Cos θ =
Turning Permanent Black.
Tan θ =
#### Points to remember :
• The angle of elevation is numerically equivalent to the angle of depression. They are consistently acute angles.
• Both the angles are estimated with the even level.
• To decide the not known side of the right triangle where one side is known and the acute angle is known. So we consolidate both the sides with trigonometric ratios of the triangle. For example
#### Figuring Heights and Distances
To, figure heights and distances, we can utilize trigonometric ratios.
Stage 1: Draw a line graph comparing to the issue.
Stage 2: Mark every known tallness, distances and angles to find lengths by factors.
Stage 3: Use the estimations of different trigonometric ratios of the angles to acquire the obscure lengths from the known lengths.
Estimating the distances of Celestial bodies with the assistance of trigonometry
Enormous distances can be estimated by the parallax technique. The parallax angle is a large portion of the angle between two lines of sights when an item is seen from two distinct positions. Realizing the parallax angle and the distance between the two positions, huge distances can be estimated. |
# AP Statistics Curriculum 2007 Normal Std
## General Advance-Placement (AP) Statistics Curriculum - Standard Normal Variables and Experiments
### Standard Normal Distribution
The Standard Normal Distribution is a continuous distribution with the following density:
• Standard Normal density function $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}.$
• Standard Normal cumulative distribution function $\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2 \over 2} \over \sqrt{2 \pi}} dx}.$
Note that the following exact areas are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin:
• The area: -1 < x < 1 = 0.8413 - 0.1587 = 0.6826
• The area: -2.0 < x < 2.0 = 0.9772 - 0.0228 = 0.9544
• The area: -3.0 < x < 3.0 = 0.9987 - 0.0013 = 0.9974
• Note that the inflection points (f''(x) = 0)of the Standard Normal density function are $\pm$ 1.
• The Standard Normal distribution is also a special case of the more general normal distribution where the mean is set to zero and a variance to one. The Standard Normal distribution is often called the bell curve because the graph of its probability density resembles a bell.
### Experiments
Suppose we decide to test the state of 100 used batteries. To do that, we connect each battery to a volt-meter by randomly attaching the positive (+) and negative (-) battery terminals to the corresponding volt-meter's connections. Electrical current always flows from + to -, i.e., the current goes in the direction of the voltage drop. Depending upon which way the battery is connected to the volt-meter we can observe positive or negative voltage recordings (voltage is just a difference, which forces current to flow from higher to the lower voltage.) Denote Xi={measured voltage for battery i} - this is random variable with mean of 0 and unitary variance. Assume the distribution of all Xi is Standard Normal, $X_i \sim N(0,1)$. Use the Normal Distribution (with mean=0 and variance=1) in the SOCR Distribution applet to address the following questions. This Distributions help-page may be useful in understanding SOCR Distribution Applet. How many batteries, from the sample of 100, can we expect to have?
• Absolute Voltage > 1? P(X>1) = 0.1586, thus we expect 15-16 batteries to have voltage exceeding 1.
• |Absolute Voltage| > 1? P(|X|>1) = 1- 0.682689=0.3173, thus we expect 31-32 batteries to have absolute voltage exceeding 1.
• Voltage < -2? P(X<-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than -2.
• Voltage <= -2? P(X<=-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than or equal to -2.
• -1.7537 < Voltage < 0.8465? P(-1.7537 < X < 0.8465) = 0.761622, thus we expect 76 batteries to have voltage in this range. |
calculus
# Calculus: Methods for Solving Limits with Explanations, Practice Questions, and Answers [AP Calculus, Calculus 101, Math]
In this article, we will talk about the methods for actually solving or evaluating limits. There are practice questions included, labeled PRACTICE, and they are there for you to test your understanding of the different methods. The answers are at the very bottom. Enjoy!
### Method 1: Direct Substitution
The first method you should try when given a limit is direct substitution. This is by far the easiest of the methods for analyzing limits, and can save a lot of time. All you need to do is simply plug the desired x-value into the expression.
In other words, you can use this property:
For instance, say you wanted to evaluate the limit as x approaches 2 of 2x + 3. This could be written as
To evaluate this, you could simply plug 2 into the expression 2x + 3, and you would get 7 as your answer.
Here’s one that would require a little more thought.
Evaluate the limit as x approaches 0 of 1/x2.
When we try plugging in 0 for x, we get 1/0, which is clearly undefined. Uh oh, we’re stuck. Well, not really – we can look back to our study of infinite limits to help us. If we think about taking smaller and smaller values of x, the denominator, x2, gets smaller and smaller, getting super close to 0. Therefore, our expression 1/ x2 takes on larger and larger positive values without bound, generating a vertical asymptote. Noting that the left and right limits are the same, we can say:
Important!
As always, the left- and right- sided limits must be equal for the double-sided limit to exist. Don’t forget this as you explore more complicated methods for evaluating limits!
Clearly direct substitution is quick and easy, so you may wonder why we cannot always use it. Take a look at the following problem:
Trying our substitution tactic again, we find that the limit evaluates to 0/0. So, we conclude that, just like in the problem above, the limit evaluates to positive or negative infinity, right? WRONG!
When we were looking at 1/x2 as x approached 0, direct substitution told us that our limit evaluated to 1/0, and from this we were able to conclude that there was an asymptote. We were able to do this, because the numerator was a nonzero constant value, namely 1, and thus even as our denominator kept getting smaller and smaller, our numerator did not change.
However, in this example, our limit evaluated to 0/0, and thus we don’t actually know whether our function is going to blow up to a very large or small number. We also cannot conclude that the limit is equal to 1, because 0/0 is undefined. For all we know, our numerator and denominator could have approached 0 at the same or different rates, and we must use more advanced methods to evaluate the limit.
PRACTICE
### Method 2: Factoring
Factoring can be very useful when evaluating funky limits like the one we saw above, where we got 0/0 as our answer. In fact, if you ever get 0/0 as an answer and you are taking the limit of a rational function, there is a good chance you will need some kind of factoring to help you out. Recall that a rational function is a ratio of 2 polynomials (the one in the denominator should be nonzero).
Let’s take another look at the problem from earlier.
We know that simple substitution results in 0/0, so this time, we factor the numerator in hopes of finding some terms that cancel. Fortunately, we are not disappointed.
We can now cancel the (x-2) term that appears in the numerator and denominator, which gets rid of our division-by-zero we had earlier. Now we can finish with substitution.
Now you try.
PRACTICE
### Method 3: Conjugates
Sometimes, substitution and factoring do not work, and you will need to find an alternative solution. If you see a bunch of radical symbols in the expression you need to take the limit of, you may want to consider using conjugates. By multiplying the numerator and denominator of an expression by the conjugate of a radical expression, you can create factors that cancel and a nicer looking expression. Remember that since you are multiplying the numerator and denominator by the same thing, you are really just multiplying the entire expression by 1.
For instance, say you wanted to evaluate the following:
Simple substitution yields 0/0 which we know we don’t want, and it’s unclear how we would factor this expression (although there is a way to factor the numerator, and I would encourage you to try it). Seeing a radical in the denominator, we multiply by the conjugate of our denominator. We then use difference of squares to clean up our denominator and cancel the common factor of (x-1).
Notice that we just used direct substitution at the end to finish it up.
Important!
This example demonstrates the general method for evaluating limits of expressions with radicals in the numerator or denominator.
1. Multiply the numerator AND denominator by the conjugate of the numerator OR denominator (depending on which one has the radical expression)
2. Simplify using difference of squares
3. Cancel common factors
4. Use substitution to evaluate
PRACTICE
P1. Does not exist (DNE); the left and right sided limits don’t match up
P2. Substitute 1 into the expression to get 4
P3. Factor the numerator, cancel the (x-4) term, and use substitution to get 4
P4. Multiply the numerator and denominator by the conjugate of the expression in the numerator, factor the denominator using difference of squares, cancel the (x- 25) term, and substitute x = 25 into the expression to get 1/500
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Categories: calculus, education, math, stem
### 3 replies »
1. Anonymous says:
Amazing article on calculus and limits!
Liked by 1 person
2. Daniel says:
Excellent calculus lesson! Haven’t touched my maths in ages. I’m certain this will be quite useful for the young one, like my grandson. Cheers.
Liked by 2 people |
# Lesson 15
Multiplying Rational Numbers
Let’s get more practice multiplying signed numbers.
### 15.1: Which One Doesn’t Belong: Expressions
Which expression doesn’t belong?
$$7.9x$$
$$7.9\boldcdot (\text- 10)$$
$$7.9 + x$$
$$\text-79$$
### 15.2: Rational Numbers Multiplication Grid
Look at the patterns along the rows and columns and continue those patterns to complete the table. When you have filled in all the boxes you can see, click on the "More Boxes" button.
What does this tell you about multiplication by a negative?
### 15.3: Card Sort: Matching Expressions
Your teacher will give you cards with multiplication expressions on them. Match the expressions that are equal to each other. There will be 3 cards in each group.
### 15.4: Row Game: Multiplying Rational Numbers
Evaluate the expressions in one of the columns. Your partner will work on the other column. Check in with your partner after you finish each row. Your answers in each row should be the same. If your answers aren’t the same, work together to find the error.
column A column B
$$790\div 10$$ $$(7.9)\boldcdot 10$$
$$\text- \frac67 \boldcdot 7$$ $$(0.1) \boldcdot \text- 60$$
$$(2.1) \boldcdot \text- 2$$ $$(\text-8.4) \boldcdot\frac12$$
$$(2.5) \boldcdot (\text-3.25)$$ $$\text{-} \frac52 \boldcdot \frac{13}{4}$$
$$\text-10 \boldcdot (3.2) \boldcdot (\text-7.3)$$ $$5\boldcdot (\text-1.6) \boldcdot (\text-29.2)$$
A sequence of rational numbers is made by starting with 1, and from then on, each term is one more than the reciprocal of the previous term. Evaluate the first few expressions in the sequence. Can you find any patterns? Find the 10th term in this sequence.
$$\displaystyle 1\qquad\quad 1+\frac{1}{1}\qquad\quad 1+\frac{1}{1+1}\qquad\quad 1+\frac{1}{1+\frac{1}{1+1}} \qquad \quad 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+1}}}\qquad\quad \dots$$
### Summary
We can think of $$3\boldcdot 5$$ as $$5 + 5 + 5$$, which has a value of 15.
We can think of $$3\boldcdot (\text-5)$$ as $$\text-5 + \text-5 + \text-5$$, which has a value of -15.
We know we can multiply positive numbers in any order: $$3\boldcdot 5=5\boldcdot 3$$
If we can multiply signed numbers in any order, then $$(\text-5)\boldcdot 3$$ would also equal -15.
Now let’s think about multiplying two negatives.
We can find $$\text-5\boldcdot (3+\text-3)$$ in two ways:
• Applying the distributive property:
$$\text-5\boldcdot 3 + \text-5\boldcdot (\text-3)$$
• Adding the numbers in parentheses:
$$\text-5\boldcdot (0) = 0$$
This means that these expressions must be equal.
$$\text-5\boldcdot 3 + \text-5\boldcdot (\text-3) = 0$$
Multiplying the first two numbers gives
$$\text-15 + \text-5\boldcdot (\text-3) = 0$$
Which means that
$$\text-5\boldcdot (\text-3) = 15$$
There was nothing special about these particular numbers. This always works!
• A positive times a positive is always positive.
For example, $$\frac35 \boldcdot \frac78 = \frac{21}{40}$$.
• A negative times a negative is also positive.
For example, $$\text-\frac35 \boldcdot \text-\frac78 = \frac{21}{40}$$.
• A negative times a positive or a positive times a negative is always negative.
For example, $$\frac35 \boldcdot \text-\frac78 = \text-\frac35 \boldcdot \frac78 = \text-\frac{21}{40}$$.
• A negative times a negative times a negative is also negative.
For example, $$\text-3 \boldcdot \text-4 \boldcdot \text-5 = \text-60$$. |
# Ex.5.3 Q4 Arithmetic Progressions Solution - NCERT Maths Class 10
Go back to 'Ex.5.3'
## Question
How many terms of the AP. $$9, 17, 25 \dots$$ must be taken to give a sum of $$636$$?
Video Solution
Arithmetic Progressions
Ex 5.3 | Question 4
## Text Solution
What is Known?
The AP and sum.
What is Unknown?
Number of terms.
Reasoning:
Sum of the first $$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.
Steps:
Given,
• First term, $$a = 9$$
• Common difference, $$d = 17 - 9 = 8$$
• Sum up to nth terms, $${S_n} = 636$$
We know that sum of $$n$$ terms of AP
\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\636 &= \frac{n}{2}\left[ {2 \!\times \!9 \!+\! \left( {n - 1} \right)8} \right]\\636 &= \frac{n}{2}\left[ {18 + 8n - 8} \right]\\636& = \frac{n}{2}\left[ {10 + 8n} \right]\\636& = n\left[ {5 + 4n} \right]\\636& = 5n + 4{n^2}\\4{n^2} + 5n - 636 &= 0\\\begin{bmatrix} 4{n^2} + 53n - \\48n - 636\end{bmatrix} &= 0\\\begin{bmatrix}n\left( {4n + 53} \right) -\\ 12\left( {4n + 53} \right)\end{bmatrix} &= 0\4n + 53)(n - 12) &= 0\end{align} Either \(4n + 53 = 0 or $$n - 12 = 0$$
$$n = - \frac{{53}}{4}$$ or $$n = 12$$
$$n$$ cannot be $$\frac{{ - 53}}{4}$$. As the number of terms can neither be negative nor fractional, therefore, $$n = 12$$
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# What is the side splitter formula?
The side splitter formula is a formula used to solve various mathematical equations involving an equation that can be expressed in the form ab=c, where a and b are integers, and c is an integer or a rational number.
To solve this equation, the side splitter formula takes into account the cases when one side of the equation is greater than the other, and divides both sides of the equation by the lesser number. The result of this division is then expressed as a/b, where a is now the smaller number divided by the larger number, and b is the larger number divided by itself.
This division allows us to simplify complex equations and arrive at the final solution quickly. The side splitter formula is most often used to find the greatest common factor (GCF), lowest common multiple (LCM), and other factors involved in solving equations.
Additionally, the side splitter formula can also be used to find the square root of a number, determine the value of fractions, and solve equations that involve factors such as exponents.
## What is the formula for similar triangles?
The formula for similar triangles states that the corresponding sides of two triangles are proportional. This means that if two triangles have corresponding sides in the same proportion (i. e. all three sides of one triangle are twice as long as the corresponding sides of another triangle), then the two triangles are similar.
The formula for similar triangles may also be stated as follows: If two triangles have corresponding angles equal, then the triangles are similar. This means that if all three angles of one triangle are equal to the corresponding angles of another triangle, then the two triangles are similar.
Additionally, the ratio of the lengths of two corresponding sides of two similar triangles is always equal and is called the “scale factor”.
## What is the SAS similarity theorem?
The SAS (Side-Angle-Side) similarity theorem states that when two triangles have three pairwise congruent sides, they must also be similar triangles, meaning that their respective interior angles must also be congruent.
The converse of this theorem is also seen to be true, meaning that if two triangles have three pairwise congruent angles, then the two triangles must also be similar.
The SAS similarity theorem is an important theorem in geometry, as it is used to prove that two triangles are similar, provided that certain criteria is met. The theorem states that if two triangles have three pairwise congruent sides, then the two triangles must be similar, or have the same angles.
This is important, as it means that if the angles of one triangle are known, then so too are the angles of the other triangle. It is also useful in helping to calculate the lengths of sides of a triangle, once the angles and lengths of other sides are known.
The SAS similarity theorem also has several important corollaries, including that if two pairs of corresponding angles in two triangles are congruent, then the two triangles must be similar. Similarly, if one pair of corresponding angles in two triangles are congruent, and the sides opposite to these two angles are proportional, then the two triangles must also be similar.
Overall, the SAS similarity theorem is an important theorem in geometry, as it allows us to compare two triangles, and draw conclusions about their fundamental properties. It is essential when dealing with problems that involve triangles, as it helps us to better understand the properties of these shapes.
## How do you solve proportionality theorem?
Proportionality theorem states that if two similar shapes are equiangular, then their corresponding sides are proportional. To solve this theorem, the following steps should be taken:
1. Identify the two similar shapes and make sure they are equiangular (i.e. they have the same number of interior angles).
2. Calculate the ratios of the corresponding sides.
3. Set up a ratio table in which the corresponding sides of the two shapes are written side by side.
4. Check to make sure the ratios are equal.
5. If the ratios are not equal, then the two shapes are not similar, and the theorem does not apply.
6. If the ratios are equal, then the theorem holds true and the corresponding sides are proportional.
## How do you prove the side splitter theorem?
The Side Splitter Theorem states that if a triangle has an altitude drawn to its longest side, the two parts of the side will provide lengths which sum to the length of the third side of the triangle.
To prove this theorem, we can use basic Euclidean geometry.
To begin, let us assume we have a triangle ABC with an altitude h drawn to its longest side AC. We can then divide the side AC into two parts, AB and BC. Now, let us examine each side length of the triangle ABC.
Using basic geometry, we can see that angle A is a right angle, and thus must equal 90°. This implies that triangle ABC is a right triangle, with the hypotenuse being AC. We can also see from the geometry that angle B and angle C are complementary angles, which means that they equal 180°.
Using the Pythagorean theorem, we can then calculate the lengths of all three sides of the triangle ABC. Using the Pythagorean theorem, we can calculate that the length of AC is equal to the square root of the sum of the squares of the lengths of AB and BC.
We can then substitute in the values of AB and BC into the equation and solve for the length of AC. We then have the proof of the side splitter theorem – the lengths of the two parts of the triangle ABC’s longest side add up to the length of the third side.
## How to find the missing side of a triangle using proportions?
Finding the missing side of a triangle using proportions requires setting up a proportion. To do this, start by finding the known sides of your triangle and labeling them. For example, let’s say that we have a triangle in which side “a” has a length of 10 units, side “b” has a length of 8 units, and we want to determine the length of side “c”.
We can then set up our proportion like this:
a:b = 10:8 = x:c
Where “x” is the unknown side length of the triangle. To solve for “x”, first use inverse operations to solve for “c”:
a:b = x:c
10:8 = x:c
x = 10*c/8
Now simply plug in the known values for “a” and “b” to solve for “x”, the unknown side:
x = 10*c/8
x = 10*8/8
x = 10
Therefore, the missing side length of the triangle is 10 units.
## What is a 30 60 90 triangle?
A 30 60 90 triangle is a special right triangle with angles measuring 30°, 60°, and 90° and sides in the ratio of 1:√3:2. This means that the shortest side, the one opposite to 30°, will be half the length of the longest side, the one opposite to 90°.
The side opposite to 60° will be √3/2 times the length of the longest side. These relationships between the lengths of the sides of the triangle allow for a variety of calculations that can be made with a 30 60 90 triangle.
For example, given enough information about the lengths of two sides, the length of the remaining side can be calculated. Furthermore, the measure of each interior angle can also be determined from the corresponding sides.
For example, from knowing the lengths of the two shortest sides, the measure of the 60° angle can be calculated.
## What is SAS SSS ASA AAS?
SAS, SSS, ASA, and AAS are all forms of triangle congruence postulates or theorems. SAS stands for Side-Angle-Side, which states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
SSS stands for Side-Side-Side, which states that if all three sides of one triangle are congruent to the three sides of another triangle, then the two triangles are congruent. ASA stands for Angle-Side-Angle, which states that if two angles and one side in one triangle are congruent to two angles and one side in another triangle, then the two triangles are congruent.
Lastly, AAS stands for Angle-Angle-Side, which states that if two angles and the non-included side in one triangle are congruent to two angles and the non-included side of another triangle, then the two triangles are congruent.
All four of these postulates are important principles in geometry, as they allow us to determine whether two triangles are congruent or not.
## How do you solve a split triangle?
Solving a split triangle is a fascinating topic and there are several ways to approach it.
First, you need to determine if there are any right triangles or other special right triangle patterns present in the split triangle. For example, if one of the sides of the split triangle is a right angle, then it can easily be solved with the Pythagorean theorem (A2 + B2 = C2).
Next, you need to determine the angle measures of the split triangle. This can be done by examining the length of the sides and then solving for the angles using the Law of Cosines (c2 = a2 + b2 – 2ab × cos(C)).
Then you can use the Law of Sines (a/sin A =b/sin B = c/sin C) to find out the angles of the split triangle.
Finally, you can use these angle measures and the length of the sides to solve for the area of the split triangle. For example, if you know the length of all three sides, you can use Heron’s formula to calculate the area (area = sqrt[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2).
By following these steps, you will be able to solve the split triangle with confidence.
## How do you solve parallel triangles with similar lines?
If two triangles have three pairs of parallel lines in them, then they are considered to be parallel triangles with similar lines. To solve this type of problem, the first step is to identify the angles and the sides of both triangles, as this will help you determine if the triangles really do have similar lines.
Once you have identified the angles and sides of each triangle, you can then use the Side-Side-Side congruence rule to determine if the triangles have similar lines. This rule states that if all three sides of one triangle are equal in length to all three sides of another triangle, then the two triangles are considered to be similar.
If the triangles are similar, then you can use the Angles-Angles-Angles congruence rule to find the missing parts of each triangle. This rule states that if two triangles have the same corresponding angles, then the triangles are similar.
Using this rule, you can find the missing lengths and angles of both triangles.
Once you have identified all of the angles and sides of both triangles, you can then use the Proportion Rule to solve for the measurement of any unknown sides or angles. The Proportion Rule states that if two sides of similar triangles are proportional, then the angles opposite those sides are equal.
This means that if you know the measure of one side of a triangle, you can use the proportion to solve for the unknown side of the other triangle.
By using the Side-Side-Side, Angles-Angles-Angles and Proportion rules, you can easily solve parallel triangles with similar lines.
## What is splitter function?
The splitter function is a mathematical operation that divides a numerical value or a sequence of values into multiple parts. It is used in a variety of contexts, from programming to statistics. In programming, a splitter function might be used to break up a longer sequence of numbers into shorter segments.
This can be useful for analyzing data or for producing graphical representations. In statistics, a given numerical value can be split into various subgroups to facilitate comparison and analysis. Splitter functions can also be used to break up a larger numerical value into several smaller ones.
For example, if a given number is 1000, it can be split into two components: 500 and 500. This allows for more accurate analysis of any given number. Ultimately, the usage of a splitter function can provide more information on a given value or sequence of values, allowing for more insight into any given data set.
## Does the AAA theorem work?
Yes, the AAA theorem (or Aronszajn-Donoghue-McCarthy Theorem) works. This theorem is a fundamental result of mathematical logic that states that any countable, recursive, α-recursive ordinal does not contain a countably infinite descending chain of ordinals.
This theorem was proved by mathematicians Abraham Aronszajn, William Donoghue, and Stephen C. McCarthy, who also wrote the paper that introduced the theorem.
The theorem states that if there is a countable recursive and α-recursive ordinal, then there are no infinite descending chains of ordinals. This implies that any recursive and α-recursive ordinal is well-ordered, meaning that the set of all its elements can be placed into a sequence in which each element is less than all its predecessors.
Moreover, the ordinal is total, meaning that there is no element which is greater than all its predecessors. This result has several intriguing implications, and it has been used to prove other theorems in mathematical logic.
In conclusion, the AAA theorem works, and it has important implications in mathematical logic.
## How do you find the perimeter of a 45 45 90 triangle?
To find the perimeter of a 45-45-90 triangle, you will need to use the Pythagorean Theorem to calculate the length of the hypotenuse, then add this to the lengths of the two legs of the triangle.
To use the Pythagorean Theorem, we use a2 + b2 = c2 where a and b are the lengths of the two legs of the triangle and c is the length of the hypotenuse. Since a 45-45-90 triangle has two legs of the same length and an angle of 45 degrees, we can take the length of one leg (x) and square it.
So, if x is the length of one leg of the triangle, then both legs will be x and the hypotenuse (c) will be x√2. Thus, using the Pythagorean Theorem we can calculate the hypotenuse: x2 + x2 = (x√2)2, and so x2 + x2 = 2×2 and x = √2x.
To calculate the perimeter, we now need to add the lengths of the legs and the hypotenuse. The two legs have a length of x and the hypotenuse has a length of x√2, therefore the perimeter is (2x + x√2).
Therefore, to find the perimeter of a 45-45-90 triangle, use (2x + x√2), where x is the length of one of the legs of the triangle.
## What are the 3 ways to prove two triangles are similar?
The three ways to prove two triangles are similar are by using
1) the Angle-Angle (AA) postulate,
2) the Side-Angle-Side (SAS) theorem, and
3) the Side-Side-Side (SSS) theorem.
The Angle-Angle (AA) postulate states that if two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.
The Side-Angle-Side (SAS) theorem states that if two sides of one triangle are proportional to two sides of another triangle and the included angle between those sides is the same, then the two triangles are similar.
The Side-Side-Side (SSS) theorem states that if the three sides of one triangle are proportional to the three sides of another triangle, then the two triangles are similar.
## What do parallel lines mean on a triangle?
Parallel lines on a triangle refer to two or more lines that are always the same distance apart and never intersect. These lines help to define the shape and size of the triangle. They also help to give it structure and symmetry which is important if you want to use the triangle for math or geometric problems.
The various types of triangles are classified by their parallel lines and angles. Examples include isosceles triangles which have two lines parallel to each other and two equal angles, equilateral triangles which have all three lines parallel to each other and three equal angles, and scalene triangles which have no parallel lines or equal angles.
Categories FAQ |
# Science:Math Exam Resources/Courses/MATH200/December 2013/Question 06/Solution 1
Like the hint suggests, we need to change the order of integration. First, we must begin by setting up the new bounds. We can begin by drawing the domain of integration.
${\displaystyle -1\leq x\leq 0}$
${\displaystyle -2\leq y\leq 2x}$
In order to change the order of integration, we must find the bounds of ${\displaystyle x}$ in terms of ${\displaystyle y}$ and the bounds of ${\displaystyle y}$ in terms of numbers.
We can observe from the picture of the domain of integration that the bounds are equivalent to
${\displaystyle {\frac {y}{2}}\leq x\leq 0}$
${\displaystyle -2\leq y\leq 0}$
Using this, we can change the order of integration.
${\displaystyle \int _{-1}^{0}\int _{-2}^{2x}e^{y^{2}}dydx=\int _{-2}^{0}\int _{\frac {y}{2}}^{0}e^{y^{2}}dxdy=\int _{-2}^{0}(xe^{y^{2}}|_{x={\frac {y}{2}}}^{x=0})dy=\int _{-2}^{0}-{\frac {y}{2}}e^{y^{2}}dy\color {red}{=}}$
Let ${\displaystyle u=y^{2}}$ and ${\displaystyle du=2ydy}$
${\displaystyle \color {red}{=}}$ ${\displaystyle -{\frac {1}{4}}\int _{4}^{0}e^{u}du={\frac {1}{4}}\int _{0}^{4}e^{u}du={\frac {1}{4}}(e^{4}-e^{0})=\color {blue}{\frac {1}{4}}(e^{4}-1)}$ |
# Assume that adults with smartphones are randomly selected in meetings and classes. Find the probability of them using smartphones in classes or meetings.
This question aims to find the probability of adults using smartphones in meetings or classes when phone users are randomly selected.
One of the largest smartphone manufacturers LG surveyed smartphone usage among adults in the social environment like meetings and classes and it was found that 54% of the adults use smartphones in meetings and classes.
Assuming a certain number of smartphone users are selected randomly, we can find the probability of these users using smartphones. If we select 8 adult smartphone users randomly in meetings or classes, we can easily find the probability of 6 smartphone users.
Probability is defined as the number of chances in which an event can occur randomly. It gives the possible outcomes of the occurrence of an event.
There is various kind of probabilities. Some of them are theoretical probability, experimental probability, and axiomatic probability.
The given data is as follows:
$p = 54 %$
$p = \frac { 54 } { 100 } = 0 . 54$
$n = 8$
Where p is the percentage of smartphone users and n is the total number of randomly selected users.
Binomial probability is the type of probability that takes two outcomes of an event. One of the two outcomes is success which is more likely expected while the other outcome is a failure.
The formula of binomial probability is:
$P ( X = x ) = \frac { n ! } { x ! ( n – x ) ! } . p ^ x. ( 1 – p ) ^ { n – x }$
By putting values in the formula:
$P ( X = 6 ) = \frac { 8 ! } { 6 ! ( 8 – 6 ) ! } . 0 . 54 ^ 6 . ( 1 – 0 . 54 ) ^ { 8 – 6 }$
$P ( X = 6 ) = \frac { 8 ! } { 6 ! ( 2 ) ! } . 0 . 54 ^ 6 . ( 1 – 0 . 54 ) ^ { 2 }$
$P ( X = 6 ) = 28 . 0 . 54 ^ 6 . 0 . 46 ^ 2$
$P ( X = 6 ) \approx 0 . 1469$
## Numerical Solution
The probability of adults using smartphones in meetings or classes is approximately $0.1469 %$.
## Example
Samsung surveyed the users of smartphones and found that 44% of adults use smartphones in social gatherings. Find the probability of 6 adult users out of 8 randomly selected users.
$P ( X = 6 ) = \frac { 8 ! } { 6 ! ( 8 – 6 ) ! } . 0 . 44 ^ 6 . ( 1 – 0 . 44 ) ^ { 8 – 6 }$
$P ( X = 6 ) = 28 . 0 . 44 ^ 6 . 0 . 56 ^ 2$
$P ( X = 6 ) \approx 0 . 0637$
The probability of Samsung users out of 8 users is $0. 637 %$
Image/Mathematical drawings are created in Geogebra. |
Solving Systems of Equations using Substitution
# Solving Systems of Equations using Substitution
## Solving Systems of Equations using Substitution
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Solving Systems of Equations using Substitution HCPS Henrico County Public Schools Designed by Vicki Hiner- Godwin High School
2. Lesson Objective: • Solve systems of equations by substitution method.
3. Assignment: • pp. 450-451 #5-30 Skip 9 • Fractions on some of these, so don’t freak out.
4. Solving Systems of Equations using Substitution Steps: 1. Solve One equation for One variable( y= ; x= ; a=) 2. Substitute equation from step one into other equation (get an equation with only one variable) 3. Solve for the first variable. 4. Go back and use the found variable in step 3 to find second variable. 5. Check the solution in both equations of the system. WHAT DO WE DO???? CLICK TO SEE AN EXAMPLE
5. GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP1: y=4x (Already solved for y) STEP 2: Substitute into second equation: 3x + y = -21 becomes:
6. GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP1: y=4x (Already solved for y) STEP 2: Substitute into second equation: 3x + y = -21 becomes: 3x +4x =-21
7. GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP 3: Solve for the variable 3x + 4x=-21 7x=-21 x=-3
8. GIVEN EXAMPLE: y= 4x 3x+y=-21 STEP 4: Solve for the other variable use x=-3 and y=4x y=4x and x = -3 therefore: y=4(-3) or y = -12 Solution to the system is (-3,-12)
9. GIVEN EXAMPLE: y= 4x 3x+y=-21 Check solution ( -3,-12) y=4x -12=4(-3) -12=-12 3x+y=-21 3(-3)+(-12)=-21 -9+(-12)=-21 -21=-21
10. Solving Systems of Equations using Substitution Review Steps --Questions? Steps: 1. Solve One equation for One variable( y= ; x= ; a=) 2. Substitute equation from step one into other equation (get an equation with only one variable) 3. Solve for the first variable. 4. Go back and use the found variable in step 3 to find second variable. 5. Check the solution in both equations of the system.
11. GIVEN EXAMPLE: x + y=10 5x - y=2 STEP1: Solve for y x + y = 10 y = -x +10 STEP 2: Substitute into second equation: 5x - y = 2 becomes:
12. GIVEN EXAMPLE: x + y=10 5x - y=2 STEP1: Solve for Y x + y = 10 y = -x +10 STEP 2: Substitute into second equation: 5x - y = 2 becomes: 5x - (-x+10) = 2
13. GIVEN EXAMPLE: x + y=10 5x - y=2 STEP 3: Solve for the variable 5x-(-x+10)=2 5x+x-10=2 6x-10=2 6x=12 x=2
14. GIVEN EXAMPLE: x + y=10 5x - y=2 STEP 4: Solve for the other variable use x=2 and x+y=10 x=2 and x+y = 10 therefore: 2+y=10 and y = 8 Solution to the system is (2,8)
15. GIVEN EXAMPLE: x + y=10 5x - y=2 Check solution (2,8) 5x-y=2 5(2)-(8)=2 10-8=2 2=2 x + y=10 2+8=10 10=10 |
# How do you write the point slope form of the equation given (0,2) and (2,1)?
Aug 10, 2017
The equation is $y - 2 = - \frac{1}{2} \left(x - 0\right)$
#### Explanation:
Point slope form of equation is $y - {y}_{1} = m \left(x - {x}_{1}\right)$, which represents equation of a line passing through point $\left({x}_{1} , {y}_{1}\right)$ and having a slope $m$.
As slope of line joining $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$, slope of line joining $\left(0 , 2\right)$ and $\left(2 , 1\right)$ is $\frac{1 - 2}{2 - 0} = - \frac{1}{2}$.
Hence, slope of desired line is $- \frac{1}{2}$ and as it passed through $\left(0 , 2\right)$, point slope form of its equation is
$y - 2 = - \frac{1}{2} \left(x - 0\right)$
or $2 y - 4 = - x$ or $x + 2 y - 4 = 0$ |
# ISEE Middle Level Math : Data Analysis
## Example Questions
### Example Question #61 : Data Analysis And Probability
What is the range of the set below?
Explanation:
The range of a set is the difference between the largest number and the smallest number in the set. Given that 16 is the largest number and 3 is the smallest, the range is equal to 16 minus 3. The result is 13, which is the correct answer.
### Example Question #61 : Data Analysis
Find the range of the following set of numbers:
Explanation:
To solve, simply find the difference between the smallest and the largest number.
In this particular case the largest number is 15 and the smallest number is 2.
Thus,
### Example Question #61 : Data Analysis
If the maximum value in a set is and the range of the set is , then what would the minimum value be?
Explanation:
The range of a set is the maximum minus the minimum. We know the maximum is and the range is so you can subtract the range from the maximum value to find you minimum value. The minimum would then be .
### Example Question #61 : Range
How would the range of a set change if ALL the values are decreased by ?
Decreases
Cannot be determined
Increases
Stays the same
Stays the same
Explanation:
The range of a set is the difference between the largest value and the smallest value. If both values are changed by the same amount, then the difference between will not change and thus the range is the same.
### Example Question #63 : Data Analysis
Find the range of the following data set:
Explanation:
Find the mode of the following data set:
First, let's rearrange our numbers in increasing order:
Next, identify the range by subtracting the first number from the last number:
So our range is 185
### Example Question #64 : Data Analysis
Use the following set to answer the question.
Find the range.
Explanation:
To find the range of a data set, you find the difference of the smallest and largest number within the set. So, given the set
the smallest number is 1 and the largest number is 12. The difference of those two numbers is 11.
Therefore, the range of the data set is .
### Example Question #62 : Data Analysis And Probability
Find the range of the following data set:
Explanation:
Find the range of the following data set:
To find the range, simply find the difference between the largest and smallest terms in our series:
So our range is 111
### Example Question #65 : Data Analysis
Use the following data set to answer the question:
Find the range.
Explanation:
To find the range of a data set, we will find the difference of the smallest and largest number within the set. So, given the set
we can see the smallest number is and the largest number is . To find the difference of these two numbers, we will subtract them. So,
Therefore, the range of the data set is .
### Example Question #67 : Range
Find the range in this set of numbers:
Explanation:
To solve, first order the numbers from least to greatest:
Then, subtract the smallest number from the largest:
### Example Question #62 : Data Analysis And Probability
Find the range in this set of numbers: |
# Average Help : Videos | Worksheets | Word Problems
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### Probability and Statistics
in Probability and Statistics.
# Average
An average is a single value that is meant to typify a list of values. If all the numbers in the list are the same, then this number should be used. If the numbers are not all the same, an easy way to get a representative value from a list is to randomly pick any number from the list. However, the word 'average' is usually reserved for more sophisticated methods that are generally found to be more useful.
The most common type of average is the arithmetic mean, often simply called the mean. The arithmetic mean of two numbers, such as 2 and 8, is obtained by finding a value A such that 2 + 8 = A + A. It is then simple to find that A = (2 + 8)/2 = 5. Switching the order of 2 and 8 to read 8 and 2 does not change the resulting value obtained for A. The mean 5 is not less than the minimum 2 nor greater than the maximum 8. If we increase the number of terms in the list for which we want an average, we get, for example, that the arithmetic mean of 2, 8, and 11 is found by solving for the value of A in the equation 2 + 8 + 11 = A + A + A. It is simple to find that A = (2 + 8 + 11)/3 = 7.
Again, changing the order of the three members of the list does not change the result: A = (8 + 11 + 2)/3 = 7, and that 7 is between 2 and 11. This summation method is easily generalized for lists with any number of elements. However, the mean of a list of integers is not necessarily an integer. "The average family has 1.7 children" is a jarring way of making a statement that is more appropriately expressed by "the average number of children in the collection of families examined is 1.7".
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# Finding The Average Speed Video Clip
Finding The Average Speed
Video Clip Length: 1 minute 3 seconds
Video Clip Views: 19199
average, distance, speed, statistics, time
# Finding The Average Of Decimal Values Containing Hundredths And Thousandths Places Video Clip
Finding The Average Of Decimal Values Containing Hundredths And Thousandths Places
Video Clip Length: 5 minutes 9 seconds
Video Clip Views: 16983
average, decimals, number sense, numbers, statistics
# Finding The Average Of Decimal Values Containing Hundredths Places Video Clip
Finding The Average Of Decimal Values Containing Hundredths Places
Video Clip Length: 4 minutes 20 seconds
Video Clip Views: 9780
average, decimals, number sense, numbers, statistics
# Finding The Average Of Decimal Values Containing Hundredths And Thousandths Places 2 Video Clip
Finding The Average Of Decimal Values Containing Hundredths And Thousandths Places 2
Video Clip Length: 4 minutes 56 seconds
Video Clip Views: 9489
average, decimals, number sense, numbers, statistics
# How Others Use Our Site
Remediation for my chemistry students who are, on the average, 10th graders.
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# A line segment has endpoints at (3 ,7 ) and (4 ,5). If the line segment is rotated about the origin by (pi )/2 , translated vertically by -1 , and reflected about the y-axis, what will the line segment's new endpoints be?
Jul 14, 2017
$\left(7 , 2\right) \text{ and } \left(5 , 3\right)$
#### Explanation:
$\text{since there are 3 transformations to be performed}$
$\text{label the endpoints " A(3,7)" and } B \left(4 , 5\right)$
$\textcolor{b l u e}{\text{First transformation}}$
$\text{under a rotation about the origin of } \frac{\pi}{2}$
• " a point "(x,y)to(-y,x)
$\Rightarrow A \left(3 , 7\right) \to A ' \left(- 7 , 3\right) , B \left(4 , 5\right) \to B ' \left(- 5 , 4\right)$
$\textcolor{b l u e}{\text{Second transformation}}$
$\text{under a translation of } \left(\begin{matrix}0 \\ - 1\end{matrix}\right)$
• " a point "(x,y)to(x,y-1)
$\Rightarrow A ' \left(- 7 , 3\right) \to A ' ' \left(- 7 , 2\right)$
$\Rightarrow B ' \left(- 5 , 4\right) \to B ' ' \left(- 5 , 3\right)$
$\textcolor{b l u e}{\text{Third transformation}}$
$\text{under a reflection in the y-axis}$
• " a point "(x,y)to(-x,y)
$\Rightarrow A ' ' \left(- 7 , 2\right) \to A ' ' ' \left(7 , 2\right)$
$\Rightarrow B ' ' \left(- 5 , 3\right) \to B ' ' ' \left(5 , 3\right)$
$\text{after all 3 transformations}$
$\left(3 , 7\right) \to \left(7 , 2\right) \text{ and } \left(4 , 5\right) \to \left(5 , 3\right)$ |
# Grade 12 How to Find Zeros of Polynomials Questions with Detailed Solutions
How to find the zeros of polynomials using factoring, division of polynomials and the rational zeros theorem. Grade 12 math questions are presented along with detailed solutions and graphical interpretations.
Polynomial p is defined by $p(x) = x^3+5x^2-2x-24$ has a zero at x = 2. Factor p completely and find its zeros. Solution p(x) has a zero at x = 2 and therefore x - 2 is a factor of p(x). Divide p(x) by x - 2 p(x) / (x - 2) = (x3 + 5 x2 - 2 x - 24) / (x - 2) = x2 + 7 x + 12 Using the division above, p(x) may now be written in factored form as follows: p(x) = (x - 2)(x2 + 7 x + 12) Factor the quadratic expression x2 + 7 x + 12. p(x) = (x - 2)(x + 3)(x + 4) The zeros are found by solving the equation. p(x) = (x - 2)(x + 3)(x + 4) = 0 For p(x) to be equal to zero, we need to have x - 2 = 0 , or x + 3 = 0 , or x + 4 = 0 Solve each of the above equations to obtain the zeros of p(x). x = 2 , x = - 3 and x = - 4 The polynomial $p(x)=3x^4+5x^3-17x^2-25x+10$ has irrational zeros at x = ~+mn~ √5. Find the other zeros. Solution Zeros at x = ~+mn~ √5, correspond to the factors. (x - √5) and (x + √5) Hence polynomial p(x) may be written as p(x) = (x - √5)(x + √5) Q(x) = (x2 - 5)Q(x) Find Q(x) using long division of polynomials Q(x) = p(x) / (x2 - 5) = (3 x4 + 5 x3 - 17 x2 - 25 x + 10) / (x2 - 5) = 3 x2 + 5 x - 2 Factor Q(x) = 3 x2 + 5 x - 2 Q(x) = 3 x2 + 5 x - 2 = (3x - 1)(x + 2) Factor p(x) completely p(x) = (x - √5)(x + √5)(3x - 1)(x + 2) Set each of the factors of p(x) to zero to find the zeros. x = ~+mn~√ 5 , x = 1 / 3 , x = - 2 Polynomial p is given by $p(x) = x^4 - 2x^3 - 2x^2 + 6x - 3$ a) Show that x = 1 is a zero of multiplicity 2. b) Find all zeros of p. c) Sketch a possible graph for p. Solution a) If x = 1 is a zero of multiplicity 2, then (x - 1)2 is a factor of p(x) and a division of p(x) by (x - 1)2 must give a remainder equal to 0. A long division gives p(x) / (x - 1)2 = (x4 - 2x3 - 2x2 + 6x - 3) / (x - 1)2 = x2 - 3 The remainder in the division of p(x) by (x - 1)2 is equal to 0 and therefore x = 1 is a zero of multiplicity 2. b) Using the division above, p(x) may now be written in factored form as follows p(x) = (x - 1)2(x2 - 3) Factor the quadratic expression x2 - 3. p(x) = (x - 1)2 (x - √3) (x + √3) The zeros are found by solving the equation. p(x) = (x - 1)2 (x - √3) (x + √3) = 0 For p(x) to be equal to zero, we need to have (x - 1)2 = 0 , or (x - √3) = 0 , or (x + √3) = 0 Solve each of the above equations to obtain the zeros of p(x). x = 1 (multiplicity 2) , x = √3 and x = - √3 c) With the help of the factored form of p(x) and its zeros found above, we now make a table of signs. . We use the zeros of p(x) which graphically are shown as x intercepts, the table of signs and the y intercept (0 , -3) to complete the graph as shown below. . Use the Rational Zeros Theorem to determine all rational zeros of the polynomial $p(x) = 6x^3-13x^2+x+2$. Solution Rational Zeros Theorem: If p(x) is a polynomial with integer coefficients and if m / n (in lower terms) is a zero of p(x), then m is a factor of the constant term 2 of p(x) and n is a factor of the leading 6 coefficient of p(x). Find factors of 2 and 6. factors of 2: ~+mn~ 1 , ~+mn~ 2 factors of 6: ~+mn~ 1 , ~+mn~ 2 , ~+mn~ 3 , ~+mn~ 6 possible zeros: divide factors of 2 by factors of 6: ~+mn~ 1 , ~+mn~ 1 / 2 , ~+mn~ 1 / 3 , ~+mn~ 1 / 6 , ~+mn~ 2 , ~+mn~ 2 / 3 Because of the large list of possible zeros, we graph the polynomial and guess the zeros from the location of the x intercepts. Below is the graph of the the given polynomial p(x) and we can easily see that the zeros are close to -1/3, 1/2 and 2. . We now calculate p(-1/3), p(1/2) and p(2) to finally check if these are the exact zeros of p(x). p(-1/3) = 6(-1/3)^3 - 13(-1/3)^2 + (-1/3) + 2 = 0 p(1/2) = 6(1/2)^3 - 13(1/2)^2 + (1/2) + 2 = 0 p(2) = 6(2)^3 - 13(2)^2 + (2) + 2 = 0 We have used the rational zeros theorem and the graph of the given polynomial to determine the 3 zeros of the given polynomial which are -1/3, 1/2 and 2.
More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
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# How do you find the domain and range of g(x) = (2x – 3)/( 6x - 12)?
##### 1 Answer
Aug 11, 2017
$x \in \mathbb{R} , x \ne 2$
$y \in \mathbb{R} , y \ne \frac{1}{3}$
#### Explanation:
The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.
$\text{solve "6x-12=0rArrx=2larrcolor(red)" excluded value}$
To find any excluded values in the range rearrange g(x) making x the subject.
$g \left(x\right) = y = \frac{2 x - 3}{6 x - 12}$
$\Rightarrow y \left(6 x - 12\right) = 2 x - 3 \leftarrow \textcolor{b l u e}{\text{ cross-multiplying}}$
$\Rightarrow 6 x y - 12 y = 2 x - 3$
$\Rightarrow 6 x y - 2 x = 12 y - 3$
$\Rightarrow x \left(6 y - 2\right) = 12 y - 3$
$\Rightarrow x = \frac{12 y - 3}{6 y - 2}$
$\text{equate denominator to zero for excluded value}$
$6 y - 2 = 0 \Rightarrow y = \frac{1}{3} \leftarrow \textcolor{red}{\text{ excluded value}}$
$\text{domain is } x \in \mathbb{R} , x \ne 2$
$\text{range is } y \in \mathbb{R} , y \ne \frac{1}{3}$ |
# Les maths
Unit 8: Multiplication and Division of Decimals
Unit 8: Multiplication and Division of Decimals I Can Statements
I can use front-end estimation to place the decimal when I need to find the product of a decimal number and a whole number.
I can use front-end estimation to place the decimal when I need to find the quotient of a decimal number and a whole number.
I can use different estimation strategies (front-end estimation, compatible numbers, rounding and compensation) to predict products and quotients of decimal numbers.
I can solve problems using multiplication and division of decimal numbers.
STUDENTS NEED TO PRACTICE THEIR MULTIPLICATION AND DIVISION FACTS! IF THEY DO NOT KNOW THEIR FACTS THEY WILL STRUGGLE. FACTS CANNOT BE TAUGHT, THEY ARE TO BE MEMORIZED.
https://nl.mathgames.com/skill/6.31-multiply-decimals-with-numbers-up-to-1000
https://nl.mathgames.com/skill/6.33-divide-decimals-by-whole-numbers
https://nl.mathgames.com/skill/6.34-divide-decimals-by-whole-numbers
or any multiplication or division game.
Unit 7: Fractions I Can Statements (Chapter 7 in textbook)
I can use models to show that an improper fraction is greater than 1.
I can show improper fractions as mixed numbers.
I can show mixed numbers as improper fractions.
I can place fractions, mixed numbers and improper fractions on a number line and explain why I placed them where I did.
I can show an improper fraction using pictures, numbers and models (pattern blocks, snap cubes, fraction pizza, etc.)
I can show a mixed number using pictures, numbers and models (pattern blocks, snap cubes, fraction pizza, etc.)
Unit 6 - Ratio and Percents (Chapter 6)
I Can Statements:
6N5.1 I can make or draw a ratio.
6N5.2 I can write a ratio from a model or picture.
6N5.3 I can write a ratio in different ways.
6N5.4 I can find examples of ratios from everyday life and write them in ratio form.
6N5.5 I can explain the part/whole and part/part ratios of a set.
6N5.6 I can show that I understand equivalent ratios.
6N5.7 I can solve a ratio problem.
Nelson - Try it out! (extra practice)
Ratio Game
Equivalent Ratios
Spy Guys - Ratios
Ration Proportion and Percent Games
The following link has a number of percent games at varying levels of difficulty which will help you prepare for the test: http://www.maths-games.org/percentage-games.html
The game Percent Shopping is particularly challenging, however, it demonstrates how we use percent in everyday life.
Enjoy!
Unit 5 - Motion Geometry
I Can Statements...
I can show that a shape is congruent to its image.
I can transform a shape repeatedly.
I can rotate a shape repeatedly.
I can reflect a shape repeatedly.
I can perform two or more transformations of a figure.
I can draw and describe the shape of an image after several transformations..
I can describe how a shape is transformed to create its image.
I can perform one or more transformations of a figure and record the coordinates.
I can look at a design that has been transformed and tell what its original shape was and what transformations were required to make the design.
I can create a design using one or more figures and describe what transformations were used.
I can name the coordinates of the vertices of the given shape.
I can transform a shape and name the coordinates of the vertices of its image.
I can describe how a shape was transformed.
http://www.learnalberta.ca/content/me5l/html/Math5.html?launch=true
http://www.learnalberta.ca/content/mesg/html/math6web/index.html?page=lessons&lesson=m6lessonshell17.swf
http://www.harcourtschool.com/activity/icy_slides_flips_turns/
http://www.sheppardsoftware.com/mathgames/geometry/shapeshoot/TranslateShapesShoot.htm
Patterns: "I can" statements
6PR1. Demonstrate an understanding of the relationships within tables of values to solve problems.
• I can identify and use a table of values that shows a pattern
• I can complete a table of values by knowing the patterns in the table
• I can tell how the numbers in one column of a table of values relate to the numbers in the other column
• I can represent the relationship in a table of values using pictures, words, or manipulatives
• I can predict and check the value of an unknown term by using a table of values
• I can create a rule to describe the relationship between two columns of numbers in a table of values.
• I can identify missing numbers or rules in a table of values.
• I can identify errors in a given table of values.
• I can describe the pattern in each column or row of a table of values.
• I can create a table of values to solve problem and describe the pattern
6PR3. Represent generalizations arising from number relationships, using equations with letter variables. [C, CN, PS, R, V]
• I can write a mathematical expression to describe the relationship in a table
• I can represent a pattern rule using a simple mathematical expression (4d or 2n + 1)
6PR4 Demonstrate and explain the meaning of preservation of equality, concretely and pictorially.[C, CN, PS, R, V]
• I can model equations on a pan balance
• I can write examples of equal equations (3b = 12 is the same as 3b+5 =17 )
Number Patterns:
http://www.scweb4free.com/patt2.html
Describing Relationships in Tables:
Function Machine (input and output)
Find the Function
What's the Rule?
Function Machine Two
Chairs Around The Table
Writing Expressions:
Writing Variable Expressions to Represent Word Problems
Expressions with One Variable
Expressions Millionaire
Equal Equations:
http://nlvm.usu.edu/en/nav/frames_asid_201_g_4_t_2.html?open=instructions
UNIT 2
Dear Parents/Guardians,
The unit we are currently studying in Math is Unit 2: Number Relationships. In this unit students will find the factors and multiples of various numbers. They will also be learning about prime and composite numbers. One way you can help your child succeed in this unit is by practicing multiplication and division facts with them. Students should be able to quickly name the multiples of the numbers from 0 to 12. Flash cards or games, such as those found at www.mathplayground.com, can make this process more enjoyable. We are also exploring negative integers and the order of operations. If your child requires any additional practice they may use the questions from Chapter 3 of the textbook (please note that the unit number and the chapter in the textbook are not the same). Thank you so much for supporting your child’s learning and development.
http://www.learnalberta.ca/content/mf5ed/html/math5.html?golesson=2
http://slideplayer.fr/slide/3683774/
Sincerely,
Mme Morgan
UNIT 1
Good afternoon Parents/Guardians,
I wanted to give you a better look at what students need to know for this unit. As well, if students need any extra practice they may use any questions from Chapter 2 of our textbook. *** We will only briefly look at decimal numbers in this unit as we will revisit them in another unit later on.
Unit 1
Students should:
• Demonstrate an understanding of place value, including numbers that are greater than one million and less than one thousandth.
• Solve problems involving large whole numbers and decimal numbers.
Students will be able to:
• explain how the pattern of the place value system, i.e., the repetition of ones, tens and hundreds within each period, makes it possible to read and write numerals for numbers of any magnitude.
• provide examples of where large and small numbers are used; e.g., media, science, medicine,
• identify which operation is necessary to solve a given problem, and solve it.
• estimate the solution to, and solve, a given problem.
• determine the reasonableness of an answer.
In order for students to stay on top of their studies, I would ask that students please review their Math for a few minutes each night at home. I also recommend searching online for fun games and helpful websites as well as use YouTube for videos.
Thank you again for your continued support!
Sincerely,
Mme Morgan
Unit 9 - Les mesures
I can...
- identify, classify, measure, and draw five different types of angles
- identify the sum of the interior angles of a triangle and a polygon
- calculate and create a formula for the perimeter, area, or volume of a figure with some dimensions given
http://www.learnalberta.ca/content/mesg/html/math6web/index.html?page=lessons&lesson=m6lessonshell16.swf
http://www.logicieleducatif.fr/math/geometrie/geometrie.php
http://www.jeuxmaths.fr/exercices-de-maths-angles.html
http://www.learnalberta.ca/content/mesg/html/math6web/index.html?page=lessons&lesson=m6lessonshell16.swf
### Unit 8 - Multiplication and Division of Decimals
I Can Statements
I can predict products and quotients of decimals, using estimation strategies.
I can solve problems that include the multiplication and division of decimals
I can place a decimal point in a product using front-end estimation
I can correct errors of decimals point placement just be looking at it.
I can place a decimal point in a quotient using front-end estimation.
Approximation
Rounding decimals
Multiplying decimals by whole numbers
Useful Videos
Multi-digit multiplication
Spyguys - decimals
Spyguys - solving problems with decimals
Games
Unité 7 - Les fractions
6N4.1 I can show that an improper fraction represents a number greater than 1.
6N4.2 I can represent an improper fraction with manipulatives, pictures, words and with numbers.
6N4.3 I can show improper fractions as mixed numbers.
6N4.4 I can represent mixed number with manipulatives, with pictures, words and with numbers.
6N4.5 I can show mixed numbers as improper fractions.
6N4. 6 I can place fractions, mixed numbers and fractions on a number line and explain why I placed each one where I did.
Fractotron
Fraction Switch
Equivalent Fraction Match
Fact Monster: Improper and Mixed Number Pizzas
Unit 6 - Ratio and Percents (Chapter 6 in the text)
I Can Statements:
6N5.1 I can make or draw a ratio.
6N5.2 I can write a ratio from a model or picture.
6N5.3 I can write a ratio in different ways.
6N5.4 I can find examples of ratios from everyday life and write them in ratio form.
6N5.5 I can explain the part/whole and part/part ratios of a set.
6N5.6 I can show that I understand equivalent ratios.
6N5.7 I can solve a ratio problem.
Nelson - Try it out! (extra practice)
Ratio Game
Equivalent Ratios
Spy Guys - Ratios
Ration Proportion and Percent Games
http://www.maths-games.org/percentage-games.html (The game Percent Shopping is particularly challenging, however, it demonstrates how we use percent in everyday life.)
https://www.mathplayground.com/ASB_RatioBlaster.html
Enjoy!
Unit 5 - Transformations
I Can Statements…
I can show that a shape is congruent to its image.
I can transform a shape repeatedly.
I can rotate a shape repeatedly.
I can reflect a shape repeatedly.
I can perform two or more transformations of a figure.
I can draw and describe the shape of an image after several transformations..
I can describe how a shape is transformed to create its image.
I can perform one or more transformations of a figure and record the coordinates.
I can look at a design that has been transformed and tell what its original shape was and what transformations were required to make the design.
I can create a design using one or more figures and describe what transformations were used.
I can name the coordinates of the vertices of the given shape.
I can transform a shape and name the coordinates of the vertices of its image.
I can describe how a shape was transformed.
http://www.learnalberta.ca/content/me5l/html/Math5.html?launch=true
http://www.learnalberta.ca/content/mesg/html/math6web/index.html?page=lessons&lesson=m6lessonshell17.swf
http://www.harcourtschool.com/activity/icy_slides_flips_turns/
http://www.sheppardsoftware.com/mathgames/geometry/shapeshoot/TranslateShapesShoot.htm
Interesting videos
Unit 3 - Patterns
Patterns: I can statements
(Chapter 1 in textbook)
6PR1. Demonstrate an understanding of the relationships within tables of values to solve problems.
• I can identify and use a table of values that shows a pattern
• I can complete a table of values by knowing the patterns in the table
• I can tell how the numbers in one column of a table of values relate to the numbers in the other column
• I can represent the relationship in a table of values using pictures, words, or manipulatives
• I can predict and check the value of an unknown term by using a table of values
• I can create a rule to describe the relationship between two columns of numbers in a table of values.
• I can identify missing numbers or rules in a table of values.
• I can identify errors in a given table of values.
• I can describe the pattern in each column or row of a table of values.
• I can create a table of values to solve problem and describe the pattern
6PR3. Represent generalizations arising from number relationships, using equations with letter variables. [C, CN, PS, R, V]
• I can write a mathematical expression to describe the relationship in a table
• I can represent a pattern rule using a simple mathematical expression (4d or 2n + 1)
6PR4 Demonstrate and explain the meaning of preservation of equality, concretely and pictorially.[C, CN, PS, R, V]
• I can model equations on a pan balance
• I can write examples of equal equations (3b = 12 is the same as 3b+5 =17 )
Number Patterns:
http://www.scweb4free.com/patt2.html
Describing Relationships in Tables:
Function Machine (input and output)
Find the Function
What's the Rule?
Function Machine Two
Chairs Around The Table
Writing Expressions:
Writing Variable Expressions to Represent Word Problems
Expressions with One Variable
Expressions Millionaire
Equal Equations:
http://nlvm.usu.edu/en/nav/frames_asid_201_g_4_t_2.html?open=instructions
Dear Parents/Guardians,
The unit we are currently studying in Math is Unit 2: Number Relationships. In this unit students will find the factors and multiples of various numbers. They will also be learning about prime and composite numbers. One way you can help your child succeed in this unit is by practicing multiplication and division facts with them. Students should be able to quickly name the multiples of the numbers from 0 to 12. Flash cards or games, such as those found at www.mathplayground.com, can make this process more enjoyable. We are also exploring negative integers and the order of operations. If your child requires any additional practice they may use the questions from Chapter 3 of the textbook (please note that the unit number and the chapter in the textbook are not the same). Thank you so much for supporting your child’s learning and development.
http://www.learnalberta.ca/content/mf5ed/html/math5.html?golesson=2
http://slideplayer.fr/slide/3683774/
Sincerely,
Mme Morgan
Good afternoon Parents/Guardians,
I wanted to give you a better look at what students need to know for this unit. As well, if students need any extra practice they may use any questions from Chapter 2 of our textbook.
Unit 1
Students should:
• Demonstrate an understanding of place value, including numbers that are greater than one million and less than one thousandth.
• Solve problems involving large whole numbers and decimal numbers.
Students will be able to:
• explain how the pattern of the place value system, i.e., the repetition of ones, tens and hundreds within each period, makes it possible to read and write numerals for numbers of any magnitude.
• provide examples of where large and small numbers are used; e.g., media, science, medicine,
• identify which operation is necessary to solve a given problem, and solve it.
• estimate the solution to, and solve, a given problem.
• determine the reasonableness of an answer.
In order for students to stay on top of their studies, I would ask that students please review their Math for a few minutes each night at home. I also recommend searching online for fun games and helpful websites as well as use YouTube for videos.
Thank you again for your continued support!
Sincerely,
Mme Morgan
Les liens utiles pour le perimètre et l'aire
http://keepschool.com/fiches-de-cours/primaire/math/aires-volumes-perimetres.pdf
http://www.alloprof.qc.ca/BV/pages/m1199.aspx
Unit 9 (Ch 8 of textbook)
Quiz 1:
• Identifying angles (review of gr 4 and 5)
• Mesuring angles
• Drawing angles
• Finding the missing angle of a triangle
• Finding the missing angle of a quadrilateral
Please see games below post. A quick google search of these topics also yields more games and helpful sites!
This video is very helpful in seeing how we multiply decimal numbers using visuals.
This video from Learn Alberta provides in depth examples of multiplying and dividing decimals. We have viewed this a few times in class and it has been very helpful.
### Unit 8 - Multiplication and Division of Decimals
I Can Statements
I can predict products and quotients of decimals, using estimation strategies.
I can solve problems that include the multiplication and division of decimals
I can place a decimal point in a product using front-end estimation
I can correct errors of decimals point placement just be looking at it.
I can place a decimal point in a quotient using front-end estimation.
Approximation
Rounding decimals
Multiplying decimals by whole numbers
Useful Videos
Multi-digit multiplication
Spyguys - decimals
Spyguys - solving problems with decimals
Unité 7 - Les fractions: Test le vendredi 16 mars 2018
6N4.1 I can show that an improper fraction represents a number greater than 1.
Je peux démontrer qu’une fraction impropre représente un nombre supérieur à 1.
6N4.2 I can represent an improper fraction with manipulatives, pictures, words and with numbers.
Je peux représenter une fraction impropre de façon concrete, avec des images, mots, et nombres.
6N4.3 I can show improper fractions as mixed numbers.
Je peux exprimer des fractions impropres sous forme de nombres fractionnaires.
6N4.4 I can represent mixed number with manipulatives, with pictures, words and with numbers.
Je peux représenter un nombre fractionnaire de façon concrete, avec des images, mots et nombres.
6N4.5 I can show mixed numbers as improper fractions.
Je peux démontrer des nombres fractionnaires sous forme de fractions impropres.
6N4. 6 I can place fractions, mixed numbers and fractions on a number line and explain why I placed each one where I did.
Je peux placer les fractions,des nombres fractionnaires et des fractions impropres, sur une droite numérique et expliquer les stratégies utilisées pour en déterminer leur position.
Fractotron
Fraction Switch
Equivalent Fraction Match
Fact Monster: Improper and Mixed Number Pizzas
Notes from Friday, February 23rd, 2018 Remember: For review students can work on pages 203-204.
Pour trouver 10% d’un nombre, on divise le nombre par 10.
10% de 80 = 8
10% de 120 = 12
10% de 10 = 1
Pour trouver 25% d’un nombre, on divise le nombre par 4.
25% de 60 = 15
25% de 12 = 3
25% de 120 = 30
Pour trouver 50% d’un nombre, on divise le nombre par 2.
50% de 70 = 35
50% de 140 = 70
50% de 4 = 2
Pour trouver 5% d’un nombre, on divise 10% en 2.
5% de 20 = ?
10% de 20 = 2 alors 5% de 20 = 1
5% de 60 = ?
10% de 60 = 6 alors 5% de 60 = 3
Pour trouver les autres % on peut combiner les autre pourcentages que nous connaissons:
35% = 25% + 10%
60% = 50% + 10%
90% = 100% - 10%
40% = 50% - 10% OU 10% + 10% + 10% + 10%
75% = 25% + 25% + 25%
30% = 10% + 10% + 10%
Par exemple:
60% de 80 = ?
50% de 80 = 40
10% de 80 = 8
alors...60% de 80 = 48
30% de 40 = ?
10% de 40 = 4
10% de 40 = 4
10% de 40 = 4
alors...30% de 40 = 12
Notes from Tuesday, February 20th, 2018 (This will help with homework)
Unit 6 - Ratio and Percents (Chapter 6)
I Can Statements:
6N5.1 I can make or draw a ratio.
6N5.2 I can write a ratio from a model or picture.
6N5.3 I can write a ratio in different ways.
6N5.4 I can find examples of ratios from everyday life and write them in ratio form.
6N5.5 I can explain the part/whole and part/part ratios of a set.
6N5.6 I can show that I understand equivalent ratios.
6N5.7 I can solve a ratio problem.
6N6 I can show that I understand percent using pictures, numbers, and words.
6N6.6 I can show a percentage as a fraction and a decimal.
6N6.7 I can solve a problem involving percents.
Nelson - Try it out! (extra practice)
Ratio Game
Equivalent Ratios
Spy Guys - Ratios
Ration Proportion and Percent Games
The following link has a number of percent games at varying levels of difficulty which will help you prepare for the test: http://www.maths-games.org/percentage-games.html
The game Percent Shopping is particularly challenging, however, it demonstrates how we use percent in everyday life.
Enjoy!
Unit 5 - Motion Geometry (Chapter 5 in the textbook)
I Can Statements...
I can show that a shape is congruent to its image.
I can transform a shape repeatedly.
I can rotate a shape repeatedly.
I can reflect a shape repeatedly.
I can perform two or more transformations of a figure.
I can draw and describe the shape of an image after several transformations..
I can describe how a shape is transformed to create its image.
I can perform one or more transformations of a figure and record the coordinates.
I can look at a design that has been transformed and tell what its original shape was and what transformations were required to make the design.
I can create a design using one or more figures and describe what transformations were used.
I can name the coordinates of the vertices of the given shape.
I can transform a shape and name the coordinates of the vertices of its image.
I can describe how a shape was transformed.
http://www.learnalberta.ca/content/me5l/html/Math5.html?launch=true
http://www.learnalberta.ca/content/mesg/html/math6web/index.html?page=lessons&lesson=m6lessonshell17.swf
http://www.harcourtschool.com/activity/icy_slides_flips_turns/
http://www.sheppardsoftware.com/mathgames/geometry/shapeshoot/TranslateShapesShoot.htm |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.2: Inequalities Using Multiplication and Division
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
At the end of this lesson, students will be able to:
• Solve an inequality using multiplication.
• Solve an inequality using division.
• Multiply or divide an inequality by a negative number.
## Vocabulary
Terms introduced in this lesson:
inequality notation
set notation
interval notation
solution graph
open and closed brackets
including/not including
reversing the inequality
## Teaching Strategies and Tips
Use a simple example to present the four ways of expressing the solution set of an inequality:
Solve for x\begin{align*}x\end{align*}:
3x3x3x242438\begin{align*} 3x & \le -24\\ \frac{\cancel{3}x} {\cancel{3}} & \le \frac{-24} {3}\\ x & \le -8\end{align*}
Solution:
• Inequality notation: x8\begin{align*}x \le -8\end{align*}.
• Set notation: The answer is {x|x\begin{align*}\left \{x | x \right .\end{align*} a real number, x8}\begin{align*}\left . x \le -8 \right \}\end{align*}.
• Interval notation: (,8]\begin{align*}( - \infty, - 8 ]\end{align*}. A combination of parentheses and brackets are used. Because numbers less than 8\begin{align*}–8\end{align*} are solutions, negative infinity is used.
• Solution graph:
The answer is expressed as a closed circle (solid dot) at 8\begin{align*}-8\end{align*} and shaded to the left.
Tips on set notation:
• The vertical bar separates the variable from the condition that is used to describe the set. It is read as “such that.”
• {x|x\begin{align*}\left \{x | x \right .\end{align*} a real number, x8}\begin{align*}\left . x \le -8 \right \}\end{align*} is read as “the set of all values of x\begin{align*}x\end{align*}, such that x\begin{align*}x\end{align*} is a real number less than or equal to 2\begin{align*}2\end{align*}.”
• Show students that set notation takes the general form: {\begin{align*}\left \{ \right .\end{align*} “variable” such that “condition is true” }\begin{align*}\left . \right \}\end{align*}.
• Students may not see the value of using set notation for the set {x|x\begin{align*}\left \{x | x \right .\end{align*} a real number, x8}\begin{align*}\left . x \le -8 \right \}\end{align*} when inequality notation suffices. Suggest that although shading a number line or expressing an answer in interval notation may be easier in this example, set notation has advantages in other examples.
Discrete sets are easily described using set notation:
The number of pets belonging to students in class {0,1,2,3,4}\begin{align*}\left \{0, 1, 2, 3, 4 \right \}\end{align*}
The solutions to an equation {1,2}\begin{align*}\left \{–1, 2 \right \}\end{align*}.
The set of prime numbers {2,3,5,7,11,13,17,}\begin{align*}\left \{2, 3, 5, 7, 11, 13, 17, \ldots \right \}\end{align*}.
Tips on interval notation:
• This is the only form of solution that uses the infinity symbol.
• Point out that infinity is always paired with parentheses and never a bracket. The reason for this is because infinity is not a number and only a concept. (There is no end to the number line, so it cannot be included in the solution set.)
• The two extreme cases of interval notation are (a,a)\begin{align*}(a, a)\end{align*} which represents the single number a\begin{align*}a\end{align*}, and (,)\begin{align*}(-\infty, \infty)\end{align*} which represents all real numbers.
Use Example 2 to show that inequality signs change direction when multiplying or dividing by negative numbers.
Explain why the rule is necessary.
1. The number line is constructed so that the negative side is a reflection of the positive side. Multiplying or dividing a number by a negative is equivalent to reflecting it across the origin, as through a mirror.
2. For any two positive real numbers on the number line, one will be further to the right than the other. Multiplying or dividing them by a negative has the effect of reflecting them across the origin. The rightmost number on the positive side of the origin becomes the leftmost number on the negative side of the origin.
3. Therefore, if a\begin{align*}a\end{align*} is greater than b\begin{align*}b\end{align*}, then a\begin{align*}-a\end{align*} is less than b\begin{align*}-b\end{align*}. This means that the inequality sign changes direction when multiplying or dividing by a negative.
4. In the case that one number is negative and the other positive, a simpler argument holds.
## Error Troubleshooting
General Tip. Remind students to reverse the inequality sign when multiplying or dividing an inequality by a negative number.
Because the right side of the inequality in Example 2b has a fraction, suggest that students multiply both sides by 1/9\begin{align*}-1/9\end{align*} to avoid dividing the fraction.
6x(16)(6x)x<29<(16)(29)>127The direction of the inequality is changed.\begin{align*}-6x & < \frac{2} {9}\\ \left (-\frac{1} {6}\right ) (-6x) & < \left (- \frac{1} {6}\right ) \left (\frac{2} {9}\right ) && \text{The direction of the inequality is changed.}\\ x & > - \frac{1} {27}\end{align*}
In Examples 2b and 2d, remind students that the direction of the inequality does not change on account of the negative sign on the right side of the inequality. For example, when solving for x\begin{align*}x\end{align*} in 12x>30\begin{align*}12x > –30\end{align*}, do not change the direction of the inequality.
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### Course: Class 6 (Old)>Unit 10
Lesson 2: Proportion
# Worked example: Solving proportions
Learn the reasoning behind solving proportions. We'll put some algebra to work to get our answers, too. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• In the first example on how to find the proportion, Sal said to multiply 8x10/8 to get 10. He said he made the calculation on his head, but I wonder what steps should I follow to get that answer? The 10/8 looks obvious after he gave it away but if he hadn't I don't think I could have find it on my own. At least I don't know how to do it at the moment.
• Well, let's see...we're basically asking 8 times what = 10, right? So, in algebra terms, 8 x n = 10 or 8n = 10. Divide both sides by 8 to get the n by itself and you get n = 10/8. Does that make sense?
• Is there any easier way to do it? Like a way without using common core? Because this is way to confusing!
• In that first example that Sal gave you, try checking for fractions to simplify first. You can see that the first fraction `8/36` simplifies to `2/9`.
``2 10-- = --9 n``
Now, we need to ask ourselves this question: "2 times what equals 10?" And to answer that question, 2 times 5 equals 10. So, we now know to multiply 5 on the denominator of the first fraction to find n. 9 times 5 equals 45, so `n = 45`.
There is another way: it's to cross-multiply and then solve the equation. But you won't learn about solving equations until much later in Pre-Algebra: https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-intro-equations/v/variables-expressions-and-equations
Let's use Sal's example again:
``2 10-- = --9 n``
Try drawing an 'X' with your fingers on the proportion above. That's how we'll cross-multiply it. The equation will ultimately look like this: `2 ⋅ n = 10 ⋅ 9`. Simplify the equation and you get: `2n = 90`. To solve it, you just divide 2 on both sides:
``2n = 90÷2 = ÷2n = 45``
I hope this helped!
• Did anyone else hear breathing in between while he was talking? :(
• i did :( i was genuinely worried but then i realised its probably because he's the one narrating almost every khan academy video so i do think its a case of tired-of-talking
• i don't even know what he said during those whole 7 minutes
• same :/ normally all his videos help me but this one kinda was a lil you know confusing. to help tho i'd just use the community's help and look at the comments :)
• I was doing the "Solving Proportions" and the question was like "12/7 = k/8", and whatever I did I kept getting 13.7142857143, but it was wrong. I don't understand.
• Your answer is technically correct, but I think they want you to leave it as a fraction (96/7) or a mixed number (13 5/7)
• Is there anytime the answer is 0?
• maby if you had 0/0 = y/0
• The video is a bit confusing, and I'm struggling to transfer this to solving the questions for "Solving Proportions". For example in the question:
4/z = 12/5
I understand that you begin by multiplying by z.
z * 4/z = 12/5*z
--> 4 = 12/5*z
After this, the solution set asks you to multiply both sides by 5/12, the opposite fraction of the right side. Why is that?
And how does multiplying a fraction with its opposite give you one?
• This concept is based upon the Inverse Property of Multiplication that says:
Any number multiplied by its reciprocal = 1
For example: 12/5 * 5/12 = 60/60 = 1
If you find it easier, you can do cross multiplication. This is where you multiply along each diagonal of the proportion.
4/z = 12/5
12(z) = 4(5)
12z = 20
Then, divide by 12: z = 20/12 = 5/3
Hope this helps.
• i am pretty confused on how to solve problems like 7/3 = 4/t
• All you have to do is cross multiply.
7/3 * 4/t = 12/7t
12/7=1 1/3
• Err... Did Sal show 5 different ways of solving one problem?
• Yes he told many ways and some of them were confusing. The simplest way I think is Cross multiplication. Here's how to do it.
Lets take a number 3/12=4/X
1. First cross multiply the digits i.e, multiply 12 x 4 and 3 x X
2. 12 x 4 = 48 and 3 x X = 3X
3.Next, According to question 3/12=4/X so, 48 will be equal to 3X
4.48=3X
5. Now I need to find the value for X So in order to isolate X I need to divide 3X by 3 (In order to find the value for X)
6. 48= [4X/3] If I divide one side by 3 then I must also divide the other side by 3 (That is the rule of the Equation).
So, we get [48/3]=[3X/3]
7. Next lets solve it 48/3 is 16 and 3X/3 is X
8. Now we get 16=X
9. Finally, the X value is 16
10. So X=16
Hope this helps. Feel free to Comment me if you have any questions :) |
# Probability: Important Questions and Preparation Tips
Get the CBSE Class 10th Mathematics, Probability: Important Questions & Preparation Tips.
Created On: Feb 23, 2015 12:43 IST
Modified On: Jun 5, 2015 10:27 IST
Get the CBSE Class 10th Mathematics, Probability: Important Questions & Preparation Tips. This will help with a precise description of the whole chapter along with a gist of each topic covered in the chapter. The Questions mentioned are exclusively topic oriented and are famous for its repetition in many Board papers. Consider the below mentioned points and questions at the time of preparation.
Important Points:
In a statement when you find an element of uncertainty, it is measured by the means of Probability.
Experiment: An activity which ends in a well defined outcome.
Trial: An activity performed once, which results in one or several outcomes.
Random Experiment: An experiment in which all possible outcomes are known in advance, but the exactness of the result cannot be predicted.
Event: An outcome of a random experiment.
Sample Space: The set of all possible outcomes in an experiment.
Equally likely outcomes: The results of a random experiment are said to be equally likely if the different outcomes have the same chance of occurrence, that is, there is no reason to expect one outcome in preference to another.
Experimental Probability: They are based only on estimates which are acquired as a result of actual experiments and adequate recordings of the happening of the event. These estimates might change if the experiment is performed once again.
Theoretical Probability: It is what you expect to happen as a result of an experiment, but never actually happens. The theoretical probability of an event E, written as P (E), is defined as
where we assume that the outcomes of the experiment are equally likely.
The probability of a sure event or a certain event is 1 and that the probability of an impossible event is 0.
The probability of an event E is a number P(E) such that
0 ≤ P (E) ≤ 1
Elementary Event: An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.
Important Questions:
• Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail. (1 mark)
• Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match? (2 mark)
• It is known that a box of 800 electric bulbs contains 16 defective bulbs. One bulb is picked up at random from this box. What is the probability that it is a non-defective bulb. (3 mark)
• A bag contains five red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag. (3 mark)
• Suppose you drop a die at random on the rectangular region as shown in figure. What is the probability that it will land inside a circle of 70 cm diameter. (5 mark)
• Nidhi and nisha are two friends. What is the probability that both will have: (5 mark)
• Same birthday
• Different birthday (ignore the leap year)
You can also find:
Syllabus | NCERT Solution | Online Test | Practice Papers
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10
Q:
# Paint needs to be thinned to a ratio of 2 parts paint to 1.5 parts water. The painter has by mistake added water so that he has 6 litres of paint which is half water and half paint. What must he add to make the proportions of the mixture correct?
A) 1 litre paint B) 1 litre water C) ½ litre water and one litre paint D) ½ litre paint and one litre water
Explanation:
At the moment the paint is 3 liters of paint and 3 liters of water. We need to add to this to make the new ratio 2 liters paint to 1.5 liters water. As this is equivalent to 4 : 3 we can see that we have the right amount of water, and just need 1 liter of paint to make it correct.
Q:
One year ago the ratio between Maneela’s and Shanthi’s salary was 3 : 4. The ratios of their individual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Maneela, now is?
A) Rs. 1600 B) Rs. 1700 C) Rs. 1800 D) Rs. 1900
Explanation:
Let the salaries of Maneela and Shanthi one year before be M1, S1 & now be M2, S2 respectively.
Then, from the given data,
M1/S1 = 3/4 .....(1)
M1/M2 = 4/5 .....(2)
S1/S2 = 2/3 .....(3)
and M2 + S2 = 4160 .....(4)
Solving all these eqtns, we get M2 = Rs. 1600.
18 3425
Q:
The ratio of Pens and Pencils in a shop is 3 : 2 respectively. The average number of Pens and Pencils is 180. What is the number of Pencils in the shop?
A) 444 B) 344 C) 244 D) 144
Explanation:
Given ratio of pens and pencils = 3 :2
Number of Pens = 3x
Number of Pencils = 2x
Average number of pencils & Pens = 180
5x = 360
=> x = 72
Hence, the number of pencils = 2x = 72 x 2 = 144.
18 2965
Q:
The ratio of boys and girls in a school is 9:5.If the total number of students in the school is 1050.Then number of boys is
Let the ratio be 'R'
Total number of students = 1050
Then,
9R + 5R = 1050
14R = 1050
=> R = 75
Hence, the number of boys = 9R = 9 x 75 = 675
3728
Q:
3 : 12 :: 5 : ?
A) 17 B) 30 C) 26 D) 32
Explanation:
22 3051
Q:
The ratio of the incomes of Pavan and Amar is 4 : 3 and the ratio of their expenditures are 3:2. If each person saves Rs. 1889, then find the income of Pavan?
A) 6548 B) 5667 C) 7556 D) 8457
Explanation:
Let ratio of the incomes of Pavan and Amar be 4x and 3x
and Ratio of their expenditures be 3y and 2y
4x - 3y = 1889 ......... I
and
3x - 2y = 1889 ...........II
I and II
y = 1889
and x = 1889
Pavan's income = 7556
15 3415
Q:
Maneela lent Rs. 8000 partly at the rate of 5% and partly at the rate of 6% per annum simple interest. The total interest she get after 2 years is Rs. 820, then in which ratio will Rs. 8000 is to be divided?
A) 7:1 B) 13:5 C) 15:7 D) 2:7
Explanation:
Maneela lent Rs. 8000 in two parts,
15 2788
Q:
The ratio of male and female in a city is 7 : 8 respectively and percentage of children among male and female is 25 and 20 respectively. If number of adult females is 156800, what is the total population of the city?
A) 4,12,480 B) 3,67,500 C) 5,44,700 D) 2,98,948
Explanation:
Let the total population be 'p'
Given ratio of male and female in a city is 7 : 8
In that percentage of children among male and female is 25% and 20%
=> Adults male and female % = 75% & 80%
But given adult females is = 156800
=> 80%(8p/15) = 156800
=> 80 x 8p/15 x 100 = 156800
=> p = 156800 x 15 x 100/80 x 8
=> p = 367500
Therefore, the total population of the city = p = 367500
29 4759
Q:
Two numbers are in the ratio 3 : 7. If 6 be added to each of them, then they are in the ratio 5 : 9. Find the numbers ?
A) 11 & 17 B) 7 & 17 C) 9 & 21 D) 13 & 23
Explanation:
Let the two numbers be x and y
Given x : y = 3 : 7 .....(1)
Now, x+6 : y+6 = 5 : 9 .....(2)
From (1), x = 3y/7
From (2), 5y - 9x = 24
=> 5y - 9(3y/7) = 24
=> y = 21
=> From(1), x = 9
Hence, the two numbers be 9 and 21 |
875 views
### Expand and simplify (5x – 2y)(3x – 4y)
Expand and simplify (5x – 2y)(3x – 4y)
In this question, we are asked to 'expand' out the brackets. What this means is we need each term in one set of brackets (5x and -2y) to be multiplied by each term in the other set of brackets (3x and -4y).
So let's take this step-by-step.
We're going to use a simple technique which will help us in any future questions where we need to expand brackets. This is the FOIL technique:
F = first
O = outside
I = inside
L = last
1) We start from of FOIL by multiplying together both of our first terms.
These are 5x and 3x because if we look at each bracket separately, these are the two individual terms that come first:
5x*3x = 15x^2
(5 times 3 = 15 and x times x = x^2)
2) Next we use of FOIL by multiplying together the terms on the outside of the brackets.
These are 5x and -4y because when both brackets are side by side, we can see that these terms are on the outer part of the equation.
5x*-4y = -20xy
(5 times -4 = -20 and x times y = xy)
3) Then we use of FOIL by multiplying together the terms on the inside of the brackets.
These are -2y and 3x because when both brackets are side by side, we can see that these terms are on the inner part of the equation.
-2y*3x = -6xy
(-2 times 3 = -6 and x times y = xy)
4) Finally we use L of FOIL by multiplying together the last terms of the brackets.
These are -2y and -4y because if we look at each bracket separately, these are the two individual terms that come last:
-2y*-4y = 8y^2
(-2 times -4 = 8 and y times y = y^2)
5) Now let's look at what terms we're left with after each of those 4 steps using our FOIL technique:
15x^2 - 20xy - 6xy + 8y^2
Our last step is to 'simplify' our answer by collecting any like terms. These are terms which have the same algebraic ending when we ignore the number in front of them.
So in this case, we have xy terms which we can collect: -20xy-6xy = -26xy.
6) Our final answer will be 15x^2 - 26xy + 8y^2.
REMEMBER:
Do check your signs when you look over your answers and your working. It's easy to make silly mistakes when you multiply a negative term with another negative term, for example (the answer will be positive).
1 year ago
Answered by Jamie, a GCSE Maths tutor with MyTutor
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#### Other GCSE Maths questions
Factorise 4xy-6xz
Sarah’s collection contains dresses, skirts and blouses. If the ratio of dresses to skirts is 7 to 4 and the ratio of skirts to blouses is 7 to 2, what is the ratio of dresses to blouses?
How do I solve x-6=15?
Factorize x³-x
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# Congruent Circles Theorem
Two circles are congruent circles if and only if they have the same radius.
Based on the above diagram, the theorem can be written as follows.
### Proof
The biconditional statement will be proved separately.
## If Two Circles Are Congruent, Then They Have the Same Radius
Consider two congruent circles and and a point on each one.
Because the distance from the center to a point on the circle is the same for both circles. Therefore, which implies that both circles have the same radius. That is,
It has been proved that if two circles are congruent, then they have the same radius.
## If Two Circles Have the Same Radius, Then They Are Congruent
Consider now two circles with the same radius.
Next, can be translated so that point is mapped onto point The image of is which is a circle centered at Since the circles have the same radius, this translation maps onto
Because a rigid motion maps one circle onto the other, it is concluded that both circles are congruent. |
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# 7.5: Volume of Prisms
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Find the volume of a prism.
## Volume of a Right Rectangular Prism
Volume is a measure of how much space a 3-dimensional figure occupies.
This means that the volume tells you how much a 3-dimensional figure can hold.
What does volume represent?
Volume is the space inside a 3-dimensional solid.
One way to understand volume is to compare it to surface area.
We can use real-life examples of objects to compare volume to surface area:
A fish tank:
Surface area = the glass used to build the outside of the tank
Volume = the water inside the tank
A pillow:
Surface area = the fabric used to make the pillowcase
Volume = the feathers or stuffing inside the pillow
Can you think of other examples?
The basic unit of volume is the cubic unit — cubic centimeter, cubic inch, cubic meter, cubic foot, and so on. Each basic cubic unit has a measure of 1 for its length, width, and height:
______________________ is the measure of space inside a solid object.
The basic unit of volume is a ________________________ unit.
In calculating volume, it is important to know that if 2 polyhedrons (or solids) are congruent, then their volumes are congruent also.
A right rectangular prism is a prism with rectangular bases and the angle between each base and its rectangular lateral sides is also a right angle. You can recognize a right rectangular prism by its “box” shape, like in the diagram below.
The volume of a solid is the sum of the volumes of all of its non-overlapping parts. Using this, we can find the volume of a right rectangular prism by counting boxes.
The box below measures 2 units in height, 4 units in width, and 3 units in depth. Each layer has (24\begin{align*}2 \cdot 4\end{align*}) cubes or 8 cubes.
Together, you get 3 groups of (24\begin{align*}2 \cdot 4\end{align*}) so the total volume is:
V=243=24\begin{align*}V & = 2 \cdot 4 \cdot 3\\ &= 24\end{align*}
The volume is 24 cubic units.
This same pattern is true for any right rectangular prism.
Volume is given by the formula:
VolumeV=lengthwidthheight=lwh\begin{align*}\text{Volume}& = \text{length} \cdot \text{width} \cdot \text{height}\\ V & = l \cdot w \cdot h\end{align*}
You can calculate the volume of any right rectangular prism by multiplying the ________________________ of the solid, the ________________________ , and its _________________________ .
Example 1
Find the volume of this prism:
Use the formula for volume of a right rectangular prism:
VVV=lwh==560\begin{align*}V &= l \cdot w \cdot h\\ V &= \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ V &= 560\end{align*}
So the volume of this rectangular prism is 560 cubic units.
1. In your own words, what is volume?
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
2. True/False: An appropriate unit for the answer to a volume problem is cubic inches.
3. True/False: If 2 solids are congruent, then their volumes are the same.
4. True/False: Volume is calculated by taking the sum of the length, the width, and the height of a solid.
## Volume of a Right Prism
Looking at the volume of right prisms with the same height and different bases, you can see a pattern. The computed area of each base is given below. The height of all 3 solids is the same, 10.
Putting the data for each solid into a table, we get:
Solid Height Area of base Volume
Rectangle 10 300 3000
Trapezoid 10 140 1400
Triangle 10 170 1700
The relationship in each case is clear: when you multiply the height of the solid by the area of its base, you get the volume. This relationship can be proven to establish the following formula for any right prism.
The volume of a right prism is:
V=Bh\begin{align*}V = Bh\end{align*}
where B\begin{align*}B\end{align*} is the area of the base of the 3-dimensional figure and h\begin{align*}h\end{align*} is the prism’s height (also called altitude)
To find the volume of a right prism, you ______________________________ the area of its ________________________ by the __________________________ of the prism.
Example 2
Find the volume of the prism with a triangular equilateral base and the dimensions shown in centimeters.
To find the volume, first find the area of the base. In this diagram, the base is actually facing forwards instead of on the bottom. The base is an equilateral triangle as the directions say, so we use the area of a triangle formula:
A=12 bh\begin{align*}A = \frac{1}{2}\ bh\end{align*}
The height (or altitude) of the triangle is 10.38 cm. Each side of the triangle measures 12 cm. So the triangle has the following area:
A=12 bh=12(12)(10.38)=62.28 cm2\begin{align*}A &= \frac{1}{2}\ bh\\ &= \frac{1}{2} (12)(10.38)\\ &= 62.28 \ cm^2\end{align*}
Now use the formula for the volume of the prism, V=Bh\begin{align*}V = Bh\end{align*}, where B\begin{align*}B\end{align*} is the area of the base (the area of the triangle) and h\begin{align*}h\end{align*} is the height of the prism.
Remember that the "height" of the prism is the distance between the bases, so in this case the height of the prism is 15 cm. Imagine that the prism is lying on its side.
V=Bh=(62.28)(15)=934.2\begin{align*}V& = Bh\\ &= (62.28)(15)\\ &= 934.2\end{align*}
Thus, the volume of the prism is 934.2 cm3\begin{align*}934.2 \ cm^3\end{align*} (or cubic centimeters).
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Question
Find the equation of the given conic. horizantal ellipse with centre (5,1), major diameter 10, minor diameter 8
1. thienan
The equation of the given conic, horizontal ellipse is (x – 5)²/5² + (y – 1)²/4² = 1 or 16x² + 25y² – 160x – 50y + 325 = 0.
In mathematics, an ellipse is the location of points in a plane so that their separation from a fixed point has a set ratio of “e” to their separation from a fixed line (less than 1). The conic section, which is the intersection of a cone with a plane that does not intersect the base of the cone, includes the ellipse. The eccentricity of the ellipse is symbolized by the constant ratio “e,” the fixed point is known as the focus, and the fixed line is known as the directrix (d).
The ellipse formula equation aids in the algebraic representation of an ellipse. You may use the following formula to determine an ellipse’s equation:
The ellipse’s equation with a center at (0, 0) is as follows: x²/a² + y²/b² = 1.
An ellipse with a center at (h,k) has the following equation: (x – h)²/a² + (y – k)²/(b²) = 1, where a and b are the semi-major and semi-minor diameters.
In the question, we are given that the center (h,k) = (5,1), the major diameter = 10, or, a = 10/2 = 5, and the minor diameter = 8, or, b = 8/2 = 4.
Substituting in the equation, we get:
(x – 5)²/5² + (y – 1)²/4² = 1, in the general form, and,
16(x² + 25 – 10x) + 25(y² + 1 – 2y) = 100,
or, 16x² + 400 – 160x + 25y² + 25 – 50y = 100,
or, 16x² + 25y² – 160x – 50y + 325 = 0.
Thus, the equation of the given conic, horizontal ellipse is (x – 5)²/5² + (y – 1)²/4² = 1 or 16x² + 25y² – 160x – 50y + 325 = 0. |
# How do you graph 10x+6y<30 and 5x-10y>30?
Jul 11, 2015
Graph system of linear equations in 2 variables:
(1) 10 x + 6y - 30 < 0
(2) 5x - 10 y - 30 > 0
#### Explanation:
First graph Line (1) -> 10 x + 6y - 30 = 0 by its 2 intercepts.
Make x = 0 -> y = 5; Make y = 0, --> x = 3.
Find the solution set of (1). Use origin O as test point. Substitute x = 0 and y = 0 into (1), we get: -30 < 0. It is true, then the area containing O is the solution set. Color or shade it.
Next, graph Line (2) -> 5x - 10y - 30 = 0 by its 2 intercepts.
Use O as test point. We get: - 30 > 0. Not true. Then, the area, not containing O, is the solution set. Color or shade it.
The compound solution set is the commonly shared area.
graph{10x + 6y - 30 = 0 [-10, 10, -5, 5]}
graph{5x - 10y - 30 = 0 [-10, 10, -5, 5]} |
1. Home
3. Scale Factors and Similarity
Similar polygons
You’re given this project for art class. You’re asked to draw your most favorite building in town in a short bond paper. The teacher specified that you need to scale the drawing to the size of the paper. Ordinarily you would just draw what you see not really paying attention to the actual scale of the drawing and later on realize that you are able to draw the building in a much smaller scale. If you want to know how to scale correctly then you better look at this chapter thoroughly.
Scaling isn’t just in drawing, but also in making models for buildings, for machines and a lot of other things. The smaller versions are created to represent what the life-size version would look like. To give you more idea about scaling then you should check out Scale City, this video shows small figures in the real world.
Now, in this chapter, we would start with learning the enlargement and reduction of a particular subject using a scale factor. Earlier you were asked to scale your drawing to the size of the paper, which means the scale factor of your drawing should be based on the size of the paper. For this first part of this chapter, we will be using a grid paper to fully grasp what scaling really means. We will learn how to adjust sizes and also to solve for the scale factor.
In 4.2 we will then apply the scale factor in order to make a copy of a particular object that would be of proportion to the original. Knowing whether you are able to make a proportional copy would teach you that this characteristic of proportionality can be applied in mathematical problems especially those involving Geometry. We are able to check whether your drawings are good scaling of the original by looking at the proportion of the sizes.
Applying what we will learn from the first parts of the chapter, in 4.3 we would be learning how to determine whether a pair of triangles is proportional to each other by looking at corresponding sides and corresponding angles. Knowing the Pythagorean theorem and it’s application, plus looking at the measurements of the corresponding sides and angles given, we would be able to determine if they are proportional with each other.
After that, we would then branch out to looking at similar polygons. In 4.4 we would be learning all about similar polygons. These shapes have proportional sides and angles, much like how similar triangles are.
Similar polygons
In this lesson, we will learn how to determine whether a pair of polygons is similar to each other. Once we know that the polygons are similar, we can calculate unknowns such as, side lengths, scale factors, and surface areas. |
# Interactive Real Analysis - part of MathCS.org
Next | Previous | Glossary | Map | Discussion
## Ratio Test
Consider the series . Then
• if lim sup | a n+1 / a n | < 1 then the series converges absolutely.
• if there exists an N such that | a n+1 / a n | 1 for all n > N then the series diverges.
• if lim sup | a n+1 / a n | = 1, this test gives no information
Note that the second condition is true if lim | an+1 / an | exists and is strictly bigger than 1.
The ratio test is easier to use than the Root test . However, there are series for which the ratio tests gives no information, but the root test will. In that sense, the ratio test is weaker than the root test. In addition, the ratio test can be proved using the root test, but not visa versa.
Using the lim sup rather than the regular limit has the advantage that we don't have to worry about existence of the limit. However, if the regular limit exists, the lim sup yields the same number. Therefore, we loose nothing by looking at the limit superior.
Example 4.2.16: Does Euler's series converge ? Use the ratio test to test the series for convergence. Compare with the same example using the root test. Are the following statements equivalent ? If not, which statement is stronger ? There exists an N such that | a n+1 / a n | 1 for all n > N lim sup | a n+1 / a n | 1 Give an alternative proof of the ratio test using the root test (therefore showing that the root test is stronger than the ratio test). Use the ratio test and the divergence test to show that for any fixed number a an/n! = 0
### Proof:
The proof is very similar to the proof of the root test:
Assume that lim sup | a n+1 / a n | < 1: because of the properties of the lim sup, we know that there exists > 0 and N > 1 such that
| a n+1 / a n | < 1 -
for n > N. Multiplying both sides by a n we obtain
| a n+1 | < (1- ) | a n |
for n > N. Therefore, we also have
| a n+2 | < (1- ) | a n+1 | < (1 - ) 2 | a n |
for n > N. Repeating this procedure, we get, eventually, that
| a k | < (1 - ) k-N | a N |
for k > N. But the terms on the right hand side form a convergent geometric series, indexed using the variable k, where N is some constant integer. Hence, by the comparison test the series with terms on the left-hand side will converge absolutely.
The proof for the second case if left as an exercise.
Next | Previous | Glossary | Map | Discussion |
$$\require{cancel}$$
# 2.6: Conic Sections
We have so far defined an ellipse, a parabola and a hyperbola without any reference to a cone. Many readers will know that a plane section of a cone is either an ellipse, a parabola or a hyperbola, depending on whether the angle that the plane makes with the base of the cone is less than, equal to or greater than the angle that the generator of the cone makes with its base. However, given the definitions of the ellipse, parabola and hyperbola that we have given, proof is required that they are in fact conic sections. It might also be mentioned at this point that a plane section of a circular cylinder is also an ellipse. Also, of course, if the plane is parallel with the base of the cone, or perpendicular to the axis of the cylinder, the ellipse reduces to a circle.
$$\text{FIGURE II.36}$$
A simple and remarkable proof can be given in the classical Euclidean "Given. Required. Construction. Proof. Q.E.D." style.
Proof
Given: A cone and a plane such that the angle that the plane makes with the base of the cone is less than the angle that the generator of the cone makes with its base, and the plane cuts the cone in a closed curve $$\text{K}$$. Figure $$\text{II.36}$$.
Required: To prove that $$\text{K}$$ is an ellipse.
Construction: Construct a sphere above the plane, which touches the cone internally in the circle $$\text{C}_1$$ and the plane at the point $$\text{F}_1$$ . Construct also a sphere below the plane, which touches the cone internally in the circle $$\text{C}_2$$ and the plane at the point $$\text{F}_2$$.
Join a point $$\text{P}$$ on the curve $$\text{K}$$ to $$\text{F}_1$$ and to $$\text{F}_2$$.
Draw the generator that passes through the point $$\text{P}$$ and which intersects $$\text{C}_1$$ at $$\text{Q}_1$$ and $$\text{C}_2$$ at $$\text{Q}_2$$.
Proof: $$\text{PF}_1 = \text{PQ}_1$$ (Tangents to a sphere from an external point.)
$$\text{PF}_2 = \text{PQ}_2$$ (Tangents to a sphere from an external point.)
$$\therefore \text{PF}_1 + \text{PF}_2 = \text{PQ}_1 + \text{PQ}_2 = \text{Q}_1 \text{Q}_2$$
and $$\text{Q}_1\text{Q}_2$$ is independent of the position of $$\text{P}$$, since it is the distance between the circles $$\text{C}_1$$ and $$\text{C}_2$$ measured along a generator.
$\therefore \ \text{K is an ellipse and } \text{F}_1 \text{ and } \text{F}_2 \text{ are its foci}. \tag{Q.E.D.}$
A similar argument will show that a plane section of a cylinder is also an ellipse.
The reader can also devise drawings that will show that a plane section of a cone parallel to a generator is a parabola, and that a plane steeper than a generator cuts the cone in a hyperbola. The drawings are easiest to do with paper, pencil, compass and ruler, and will require some ingenuity. While I have seen the above proof for an ellipse in several books, I have not seen the corresponding proofs for a parabola and a hyperbola, but they can indeed be done, and the reader should find it an interesting challenge. If the reader can use a computer to make the drawings and can do better than my poor effort with figure $$\text{II.36}$$, s/he is pretty good with a computer, which is a sign of a misspent youth. |
21
Q:
# The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they meet?
A) 10 a.m B) 10.30 a.m C) 11 a.m D) 11.30 a.m
Explanation:
Suppose they meet x hrs after 8 a.m
then,
[Distance moved by first in x hrs] + [Distance moved by second in (x-1) hrs] = 330.
Therefore, 60x + 75(x-1) = 330.
=> x=3.
So,they meet at (8+3) i.e, 11a.m.
Q:
Ashwin fires two bullets from the same place at an interval of 15 minutes but Rahul sitting in a bus approaching the place hears the second sound 14 minutes 30 seconds after the first. If sound travels at the speed of 330 meter per second, what is the approximate speed of bus?
A) 330/29 m/s B) 330 x 30 m/s C) 330/14 m/s D) 330/900 m/s
Explanation:
Second gun shot take 30 sec to reach rahul imples distance between two.
given speed of sound = 330 m/s
Now, distance = 330 m/s x 30 sec
Hence, speed of the bus = d/t = 330x30/(14x60 + 30) = 330/29 m/s.
0 11
Q:
Important Time and Distance formulas with examples.
1. How to find Speed(s) if distance(d) & time(t) is given:
Ex: Find speed if a person travels 4 kms in 2 hrs?
Speed = D/T = 4/2 = 2 kmph.
2. Similarly, we can find distance (d) if speed (s) & time (t) is given by
Distance (D) = Speed (S) x Time (T)
Ex : Find distance if a person with a speed of 2 kmph in 2 hrs?
Distance D = S X T = 2 x 2 = 4 kms.
3. Similarly, we can find time (t) if speed (s) & distance (d) is given by
Time (T) =
Ex : Find in what time a person travels 4 kms with a speed of 2 kmph?
Time T = D/S = 4/2 = 2 hrs.
4. How to convert km/hr into m/sec :
Ex : Convert 36 kmph into m/sec?
36 kmph = 36 x 5/18 = 10 m/sec
5. How to convert m/sec into km/hr :
.
Ex : Convert 10 m/sec into km/hr?
10 m/sec = 10 x 18/5 = 36 kmph.
6. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is b : a.
7. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km/hr . Then, the average speed during the whole journey is $\frac{\mathbf{2}\mathbf{xy}}{\mathbf{x}\mathbf{+}\mathbf{y}}$ km/hr.
83
Q:
Two trains are running with speeds 30 kmph and 58 kmph respectively in the same direction. A man in the slower train passes the faster train in 18 seconds. Find the length of the faster train?
A) 105 mts B) 115 mts C) 120 mts D) 140 mts
Explanation:
Speeds of two trains = 30 kmph and 58 kmph
=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s
Given a man takes time to cross length of faster train = 18 sec
Now, required Length of faster train = speed x time = 70/9 x 18 = 140 mts.
0 99
Q:
A train-A passes a stationary train B and a pole in 24 sec and 9 sec respectively. If the speed of train A is 48 kmph, what is the length of train B?
A) 200 mts B) 180 mts C) 160 mts D) 145 mts
Explanation:
Length of train A = 48 x 9 x 5/18 = 120 mts
Length of train B = 48 x 24 x 5/18 - 120
=> 320 - 120 = 200 mts.
1 248
Q:
Tilak rides on a cycle to a place at speed of 22 kmph and comes back at a speed of 20 kmph. If the time taken by him in the second case is 36 min. more than that of the first case, what is the total distance travelled by him (in km)?
A) 132 km B) 264 km C) 134 km D) 236 km
Explanation:
Let the distance travelled by Tilak in first case or second case = d kms
Now, from the given data,
d/20 = d/22 + 36 min
=> d/20 = d/22 + 3/5 hrs
=> d = 132 km.
Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.
0 184
Q:
A train covers 180 km distance in 5 hours. Another train covers the same distance in 1 hour less. What is the difference in the distances covered by these trains in one hour if they are moving in the same direction?
A) 15 kms B) 9 kms C) 6 kms D) 18 kms
Explanation:
The first train covers 180 kms in 5 hrs
=> Speed = 180/5 = 36 kmph
Now the second train covers the same distance in 1 hour less than the first train => 4 hrs
=> Speed of the second train = 180/4 = 45 kmph
Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.
4 362
Q:
Chennai express left Hyderabad for Chennai at 14 : 30 hours, travelling at a speed of 60 kmph and Charminar Express left Hyderabad for Chennai on the same day at 16 : 30 hours, travelling at a speed of 80 kmph. How far away from Hyderabad will the two trains meet?
A) 360 kms B) 480 kms C) 520 kms D) 240 kms
Explanation:
Now, the distance covered by Chennai express in 2 hrs = 60 x 2 = 120 kms
Let the Charminar Express takes 't' hrs to catch Chennai express
=> 80 x t = 60 x (2 + t)
=> 80 t = 120 + 60t
=> t = 6 hrs
Therefore, the distance away from Hyderabad the two trains meet = 80 x 6 = 480 kms.
6 299
Q:
A man covers half of his journey by train at 60 km/hr, half of the remaining by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during the entire journey?
A) 32 kmph B) 20 kmph C) 18 kmph D) 24 kmph |
# Dividing Polynomials
In this worksheet, students divide polynomial expressions by the given term.
Key stage: KS 4
Curriculum topic: Algebra
Curriculum subtopic: Simplify/Manipulate Algebraic Expressions/Fractions
Difficulty level:
### QUESTION 1 of 10
A polynomial expression is one that contains several algebraic terms, such as
3x4 - 5x3 + 2x2 + 3x - 2
It is common to write a polynomial expression with terms in decreasing order (x4, x3, x2 etc).
When we divide a polynomial expression by a given term, we divide each term in turn.
Example
Divide by 2x3:
8x14 - 6x8 + 2x4 - 4x3
We divide each term in turn, remembering to subtract indices and divide coefficients.
8x14 - 6x8 + 2x4 - 4x3 2x3 2x3 2x3 2x3
This simplifies to
4x11 - 3x5 + x - 2
Divide by x
3x4 - 4x3 + 2x2 - 5x
3x4 - 4x3 - 2x2 + x
3x3 - 4x2 + 2x - 5
3x4 - 4x3 - 2x1 + 5x
Divide by x
4x4 - 5x3 + 3x2 - 3x
4x4 - 3x3 - 2x2 + x
4x3 - 5x2 + 3x + 3
4x3 - 5x2 + 3x - 3
Divide by 2
4x4 - 6x3 + 10x2 - 2x
8x6 - 2x3 + 5x2 - x
2x4 - 3x3 + 5x2 - x
2x4 - 12x2 + 3x - 1
Divide by 3
24x6 - 6x3 + 15x2 - 3x
8x6 - 2x3 + 5x2 - x
8x5 - 2x3 + 5x2 - 1
6x6 - 2x2 + 3x - 1
Divide by 2x
4x4 - 6x3 + 10x2 - 2x
8x6 - 2x3 + 5x2 - x
2x4 - 3x3 + 5x2 - x
2x3 - 3x2 + 5x - 1
Divide by 5x
40x5 - 35x4 - 10x2 - 20x
8x4 - 7x3 - 2x - 4
8x5 - 7x3 + 5x2 - 4x
10x4 - 7x2 + 5x - 4x
Divide by x2
4x5 - 5x4 - 10x3 - 2x2
4x3 - 10x3 - 2x - 2x
4x3 - 5x2 - 10x - 2
3x3 - 5x2 - 2x - 2
Divide by 5x2
40x5 - 35x4 - 10x3 - 20x2
8x4 - 7x3 - 2x - 4
8x5 - 7x3 + 5x2 - 4x
8x3 - 7x2 - 2x - 4
Divide by 5x2
40x7 + 35x4 - 15x3 - 20x2
8x4 + 7x3 - 2x - 4
8x5 + 7x2 - 3x - 4
8x3 + 7x2 -3x - 4
Divide by 4x3
40x7 + 36x6 - 16x4 - 20x3
10x4 + 9x3 - 4x - 5
36x5 + 9x2 - 4x - 5x
10x4 + 9x2 -12x - 5
• Question 1
Divide by x
3x4 - 4x3 + 2x2 - 5x
3x3 - 4x2 + 2x - 5
EDDIE SAYS
3x4 - 4x3 + 2x2 - 5x x x x x
• Question 2
Divide by x
4x4 - 5x3 + 3x2 - 3x
4x3 - 5x2 + 3x - 3
EDDIE SAYS
4x4 - 5x3 + 3x2 - 3x x x x x
• Question 3
Divide by 2
4x4 - 6x3 + 10x2 - 2x
2x4 - 3x3 + 5x2 - x
EDDIE SAYS
4x4 - 6x3 + 10x2 - 2x 2 2 2 2
• Question 4
Divide by 3
24x6 - 6x3 + 15x2 - 3x
8x6 - 2x3 + 5x2 - x
EDDIE SAYS
24x6 - 6x3 + 15x2 - 3x 3 3 3 3
• Question 5
Divide by 2x
4x4 - 6x3 + 10x2 - 2x
2x3 - 3x2 + 5x - 1
EDDIE SAYS
4x4 - 6x3 + 10x2 - 2x 2x 2x 2x 2x
• Question 6
Divide by 5x
40x5 - 35x4 - 10x2 - 20x
8x4 - 7x3 - 2x - 4
EDDIE SAYS
40x5 - 35x4 - 10x2 - 20x 5x 5x 5x 5x
• Question 7
Divide by x2
4x5 - 5x4 - 10x3 - 2x2
4x3 - 5x2 - 10x - 2
EDDIE SAYS
4x5 - 5x4 - 10x3 - 2x2 x2 x2 x2 x2
• Question 8
Divide by 5x2
40x5 - 35x4 - 10x3 - 20x2
8x3 - 7x2 - 2x - 4
EDDIE SAYS
40x5 - 35x4 - 10x3 - 20x2 5x2 5x2 5x2 5x2
• Question 9
Divide by 5x2
40x7 + 35x4 - 15x3 - 20x2
8x5 + 7x2 - 3x - 4
EDDIE SAYS
40x7 + 35x4 - 15x3 - 20x2 5x2 5x2 5x2 5x2
• Question 10
Divide by 4x3
40x7 + 36x6 - 16x4 - 20x3
10x4 + 9x3 - 4x - 5
EDDIE SAYS
40x7 + 36x6 - 16x4 - 20x3 4x3 4x3 4x3 4x3
---- OR ----
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# Build a Fraction_建立一個分數
下載 嵌入 關閉 嵌入模擬教學執行複本 利用 HTML 來嵌入模擬教學執行複本。您可以在 HTML 中改變已嵌入模擬教學的寬度及高度。 嵌入啓動模擬教學之圖樣
原始模擬教學與翻譯
### 標題
• Fractions
• Equivalent Fractions
• Mixed Numbers
### 描述
Amber Chang (2012)
### 學習目標
• Build equivalent fractions using numbers and pictures
• Compare fractions using numbers and patterns
• Recognize equivalent simplified and unsimplified fractions
• Note: Build A Fraction expands on ideas from the Fraction Intro and Fraction Matcher sims or can be used as a stand-alone tool
### 標準對齊
#### 共用核心 - 數學
1.G.A.3
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
2.G.A.3
Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
3.G.A.2
Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into 4 parts with equal area, and describe the area of each part as 1/4 of the area of the shape.
3.NF.A.1
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
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# 2003 AIME I Problems/Problem 7
## Problem
Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s.$
## Solution 1 (Pythagorean Theorem)
$[asy] size(220); pointpen = black; pathpen = black + linewidth(0.7); pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle); D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7)); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2); [/asy]$
Denote the height of $\triangle ACD$ as $h$, $x = AD = CD$, and $y = BD$. Using the Pythagorean theorem, we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$. Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$. The LHS is difference of squares, so $(x + y)(x - y) = 189$. As both $x,\ y$ are integers, $x+y,\ x-y$ must be integral divisors of $189$.
The pairs of divisors of $189$ are $(1,189)\ (3,63)\ (7,27)\ (9,21)$. This yields the four potential sets for $(x,y)$ as $(95,94)\ (33,30)\ (17,10)\ (15,6)$. The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of $\triangle ACD$ is equal to $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}$.
## Solution 2 (Stewart's Theorem)
Let $AD=c$ and $BD=d$, then by Stewart's Theorem we have:
$30d^2+21*9*30=9c^2+21c^2=30c^2$. After simplifying:
$d^2-c^2=189$.
The solution follows as above.
## Solution 3 (Law of Cosines)
Drop an altitude from point $D$ to side $AC$. Let the intersection point be $E$. Since triangle $ADC$ is isosceles, AE is half of $AC$, or $15$. Then, label side AD as $x$. Since $AED$ is a right triangle, you can figure out $\cos A$ with adjacent divided by hypotenuse, which in this case is $AE$ divided by $x$, or $\frac{15}{x}$. Now we apply law of cosines. Label $BD$ as $y$. Applying law of cosines, $y^2 = x^2+9^2- 2 \cdot x \cdot 9 \cdot \cos A$. Since $\cos A$ is equal to $\frac{15}{x}$, $y^2 = x^2+9^2- 2 \cdot x \cdot 9 \cdot \frac{15}{x}$, which can be simplified to $x^2-y^2=189$. The solution proceeds as the first solution does.
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Prime Factorization Of 625
Prime Factorization Of 625. A composite number is an integer that can be divided by at least another natural number, besides itself and 1, without leaving a remainder (divided exactly). To get the prime factorisation of 625, we have to start with dividing it by primes 625 ÷ 125 = 5 125 ÷ 25 = 5 25 ÷ 5 = 5 5 ÷ 1 = 5 so here he prime factorisation of 625 = 5 x 5 x 5 x 5 = 5 4 we can. 25 ÷ 5 = 5. To get the prime factorisation of 625, we have to start with dividing it by primes.
Calculate how many prime numbers are smaller than the number of your choice as an example, if you want the prime factorization calculator to determine how many primes are before the. Factors of 625 are all the numbers that are multiplied to get 625 as the product. Now square root of 625:
To get the prime factorisation of 625, we have to start with dividing it by primes.
To find the gcf of numbers using factoring list out all the factors of each number. But still, 25 is a composite. Now square root of 625:
The Prime Factorization Of 625 = 5 4.
5 × 5 × 5 × 5. All the prime numbers that are used to divide in the prime factor tree are the prime factors of 625. After finding the smallest prime factor of the number 625, which. 5 | 625 5 | 125 5 | 25 5 | 5 | 1.
In Number Theory, The Prime Factors Of A Positive Integer Are The Prime Numbers That Divide That Integer Exactly.
The smallest prime number which can divide 625 without a remainder is 5.
Kesimpulan dari Prime Factorization Of 625.
The prime factorization of 625 is 5 x 5 x 5 x 5 or, in index form (in other words, using exponents),. So the first calculation step would look like: Prime factorization shown below home prime number calc prime factorization of 625 what is the prime factorization of 625 [solved] answer the prime factors of 625: |
# Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1
## Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1
Question 1.
If the ordered pairs (x2 – 3x,y2 + 4y) and (-2,5) are equal, then find x and y.
Solution:
(x2 – 3x, y2 + 4y) = (-2,5)
x2 – 3x = -2
x2 – 3x + 2 = 0
(x – 2)(x – 1) = 0
Question 2.
The Cartesian product A × A has 9 elements among which (-1,0) and (0,1) are found. Find the set A and the remaining elements of A×A.
Solution:
A = {-1,0, 1},B = {1,0,-1}
A × B = {(-1,1), (-1,0), (-1,-1), (0,1), (0, 0), (0,-1), (1,1), (1,0), (1,-1)}
Question 3.
Given that
(i) f(0)
(ii) f(3)
(iii) f(a + 1) in terms of a.(Given that a > 0)
Solution:
(i) f(0) = 4
(ii) f(3) = $$\sqrt { 3-1 }$$ = $$\sqrt { 2 }$$
(iii) f(a+ 1) = $$\sqrt { a+1-1 }$$ = $$\sqrt { a }$$
Question 4.
Let A = {9,1O,11,12,13,14,15,16,17} and let f: A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Solution:
A = {9,10,11, 12, 13, 14,15,16,17}
f: A → N
f(n) = the highest prime factor of n ∈ A
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13,13), (14, 7), (15, 5), (16,2), (17, 17)}
Range = {3,5, 11, 13,7,2, 17}
= {2,3,5,7,11,13,17}
Question 5.
Find the domain of the function
Solution:
Question 6.
If f(x)= x2, g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).
Solution:
f(x) = x2
g(x) = 3x
h(x) = x – 2
(fog)oh = x – 2
LHS = fo(goh)
fog = f(g(x)) = f(3x) = (3x)2 = 9x2
(fog)oh = (fog) h(x) = (fog) (x – 2)
= 9(x – 2)2 = 9(x2 – 4x + 4)
= 9x2 – 36x + 36 …..(1)
RHS = fo(goh)
(goh) = g(h(x)) = g(x- 2)
= 3(x – 2) = 3x – 6
fo(goh) = f(3x – 6) = (3x – 6)2
= 9x2 – 36x + 36 ……(2)
(1) = (2)
LHS = RHS
(fog)oh = fo(goh) is proved.
Question 7.
A = {1, 2} and B = {1, 2, 3, 4} , C = {5, 6} and D = {5, 6, 7, 8} . Verify whether A × C is a subset of B × D?
Solution:
A = {1,2),B = (1,2,3,4)
C = {5,6},D = {5,6,7,8)
(A × C) ⊂ (B × D) It is proved.
Question 8.
If f(x) =$$\frac { x-1 }{ x+1 }$$, x ≠ 1 show that f(f(x)) = – $$\frac { 1 }{ x }$$, Provided x ≠ 0.
Solution:
Hence it is proved.
Question 9.
The function/and g are defined by f(x) = 6x + 8; g(x) = $$\frac { x-2 }{ 3 }$$.
(i) Calculate the value of gg ($$\frac { 1 }{ 2 }$$)
(ii) Write an expression for g f(x) in its simplest form.
Solution:
Question 10.
Write the domain of the following real functions
Solution:
(i) f(x) = $$\frac { 2x+1 }{ x-9 }$$
The denominator should not be zero as the function is a real function.
∴ The domain = R – {9}
(ii) p(x) = $$\frac{-5}{4 x^{2}+1}$$
The domain is R.
(iii) g(x) = $$\sqrt { x-2 }$$
The domain = (2, ∝)
(iv) h(x) = x + 6
The domain is R. |
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# Discrete Random Variables
## A variable having a countable number of values
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Discrete Random Variables
What do you think the probability is that it will rain on Monday? You watched the weather forecast yesterday and the meteorologist said that the probability is 100% certain that it will rain on Monday and about 75% chance of rain on Tuesday. Could the probability of rain ever be described by 200%? What about 12%?
### Watch This
First watch this video to learn about discrete random variables.
Then watch this video to see some examples.
Watch this video for more help.
### Guidance
In previous Concepts, we looked at the mathematics involved in probability events. We looked at examples of event $A$ occurring if event $B$ had occurred (conditional events), of event $B$ being affected by the outcome of event $A$ (dependent events), and of event $A$ and event $B$ not being affected by each other (independent events). We also looked at examples where events cannot occur at the same time (mutually exclusive events), or when events were not mutually exclusive and there was some overlap, so that we had to account for the double counting (mutually inclusive events). If you recall, we used Venn Diagrams (below), tree diagrams, and even tables to help organize information in order to simplify the mathematics for the probability calculations.
Our examination of probability, however, began with a look at the English language. Although there are a number of differences in what terms mean in mathematics and English, there are a lot of similarities as well. We saw this with the terms independent and dependent. In this and the following Concepts, we are going to learn about variables. In particular, we are going to look at discrete random variables. When you see the sequence of words discrete random variables , it may, at first, send a shiver down your spine, but let’s look at the words individually and see if we can "simplify" the sequence!
The term discrete, in English, means to constitute a separate thing or to be related to unconnected parts. In mathematics, we use the term discrete when we are talking about pieces of data that are not connected. Random, in English, means to lack any plan or to be without any prearranged order. In mathematics, the definition is the same. Random events are fair, meaning that there is no way to tell what outcome will occur. In the English language, the term variable means to be likely to change or subject to variation. In mathematics, the term variable means to have no fixed quantitative value.
Now that we have seen the 3 terms separately, let’s combine them and see if we can come up with a definition of a discrete random variable. We can say that discrete variables have values that are unconnected to each other and have variations within the values. Think about the last time you went to the mall. Suppose you were walking through the parking lot and were recording how many cars were made by Ford. The variable is the number of Ford cars you see. Therefore, since each car is either a Ford or it is not, the variable is discrete. Also, random variables are simply quantities that take on different values depending on chance, or probability. Thus, if you randomly selected 20 cars from the parking lot and determined whether or not each was manufactured by Ford, you would then have a discrete random variable.
Now let’s define discrete random variables. Discrete random variables represent the number of distinct values that can be counted of an event. For example, when Robert was randomly chosen from all the students in his classroom and asked how many siblings there are in his family, he said that he has 6 sisters. Joanne picked a random bag of jelly beans at the store, and only 15 of 250 jelly beans were green. When randomly selecting from the most popular movies, Jillian found that Iron Man 2 grossed 3.5 million dollars in sales on its opening weekend. Jack, walking with his mom through the parking lot, randomly selected 10 cars on his way up to the mall entrance and found that only 2 were Ford vehicles.
The probability of a discrete random variable can range anywhere from 0 to 1. The less likely a discrete random variable is to occur, the closer the probability will be to 0, and the more likely a discrete random variable is to occur, the closer the probability will be to 1.
#### Example A
Which of the following can be represented by a discrete random variable?
a. The heights of the students in a high school
b. The number of sit-ups that you can do
c. The distances between stars in a galaxy
d. The number of wins by a professional hockey team
e. The speeds of the cars in a race
The heights of the students in a high school cannot be represented by a discrete random variable, since a height can take on any value within a certain range. For example, a height could be 64 inches, 64.5 inches, 64.55 inches, 64.555 inches, and so on. On the other hand, the number of sit-ups that you can do can be represented by a discrete random variable, because the number will always be an integer. The distances between stars in a galaxy are similar to the heights of the students in a high school in that there are an infinite number of possibilities, so these distances cannot be represented by a discrete random variable. However, the number of wins by a professional hockey team is similar to the number of sit-ups that you can do in that the number will always be an integer, so this number can be represented by a discrete random variable. Finally, the speeds of the cars in a race can take on any values, such as 100 MPH, 100.1 MPH, 100.12 MPH, 100.123 MPH, and so on, so these speeds cannot be represented by a discrete random variable. In summary, the answers to this question are as follows:
a. Cannot be represented by a discrete random variable
b. Can be represented by a discrete random variable
c. Cannot be represented by a discrete random variable
d. Can be represented by a discrete random variable
e. Cannot be represented by a discrete random variable
#### Example B
It is very likely, but not certain, that the high temperature will exceed $75^\circ \text{F}$ every day next week. Suppose that the discrete random variable $X$ represents the number of days next week that the high temperature will exceed $75^\circ \text{F}$ . Which of these could be $P(X=7)$ ?
a. $P(X=7)=0$
b. $P(X=7)=0.14$
c. $P(X=7)=0.5$
d. $P(X=7)=0.96$
e. $P(X=7)=1$
If the probability of an event is 0, it is impossible, and if the probability of the event is 1, it is certain. In this case, it is very likely, but not certain, that the high temperature will exceed $75^\circ \text{F}$ every day next week, so the probability will be close to 1, but not 1. Therefore, $P(X=7)$ could be equal to 0.96, so the correct answer is d.
#### Example C
Suppose that the discrete random variable $X$ represents the number of points out of 100 that Royce scores on a test. If $P(X=42)=0.88$ , which of these statements is most likely true?
a. Royce is likely to do well on the test.
b. Royce doesn't even know a single correct answer on the test.
c. Royce didn't study much for the test.
d. Royce knows all the correct answers on the test.
e. Royce studied a lot for the test.
Since $P(X=42)=0.88$ , the probability is high that Royce will get a low score on the test. However, if Royce gets 42 points out of 100, he at least knows some correct answers on the test. Therefore, the statement that is most likely true is that Royce didn't study much for the test, so the correct answer is C.
### Guided Practice
Put the following statements in order from least likely to most likely.
a. If a die is rolled 2 times, the same number will come up each time.
b. Christmas will be in December next year.
c. A letter chosen at random from the alphabet will be a consonant.
d. If a couple has a baby, it will be a girl.
e. The population of Wyoming will be greater than that of California in 10 years.
Let's look at each statement individually. The first statement is, "If a die is rolled 2 times, the same number will come up each time." The probability of this statement can actually be calculated to be $\frac{1}{6}$ . Next, we have, "Christmas will be in December next year." This probability of this statement is extremely high, as Christmas has traditionally been in December every year. After this, we have, "A letter chosen at random from the alphabet will be a consonant." Since 21 out of 26 letters in the alphabet are consonants, the probability of this statement is $\frac{21}{26}$ . The next statement is, "If a couple has a baby, it will be a girl." The probability of this statement will be about $\frac{1}{2}$ , since a boy and a girl are almost equally likely. Finally, we have, "The population of Wyoming will be greater than that of California in 10 years." The probability of this statement is extremely low, as the population of California is currently about 37,000,000, while the population of Wyoming is less than 600,000. In summary, here are our probabilities:
a. $\frac{1}{6}$
b. Extremely high
c. $\frac{21}{26}$
d. About $\frac{1}{2}$
e. Extremely low
Therefore, the order of the statements from least likely to most likely is e, a, d, c, b.
### Practice
1. Match the following statements from the first column with the probability values in the second column.
Probability Statement $P(X)$
a. The probability of this event will never occur. $\underline{\;\;\;\;\;} \ P(X) = 1.0$
b. The probability of this event is highly likely. $\underline{\;\;\;\;\;} \ P(X) = 0.33$
c. The probability of this event is very likely. $\underline{\;\;\;\;\;} \ P(X) = 0.67$
d. The probability of this event is somewhat likely. $\underline{\;\;\;\;\;} \ P(X) = 0.00$
e. The probability of this event is certain. $\underline{\;\;\;\;\;} \ P(X) = 0.95$
1. Match the following statements from the first column with the probability values in the second column.
Probability Statement $P(X)$
a. I bought a ticket for the State Lottery. The probability of a successful event (winning) is likely to be: $\underline{\;\;\;\;\;} \ P(X) = 0.80$
b. I have a bag of equal numbers of red and green jelly beans. The probability of reaching into the bag and picking out a red jelly bean is likely to be: $\underline{\;\;\;\;\;} \ P(X) = 0.50$
c. My dad teaches math, and my mom teaches chemistry. The probability that I will be expected to study science or math is likely to be: $\underline{\;\;\;\;\;} \ P(X) = 0.67$
d. Our class has the highest test scores in the State Math Exams. The probability that I have scored a great mark is likely to be: $\underline{\;\;\;\;\;} \ P(X) = 1.0$
e. The Chicago baseball team has won every game this season. The probability that the team will make it to the playoffs is likely to be: $\underline{\;\;\;\;\;} \ P(X) = 0.01$
1. Read each of the following statements and match the following words to each statement. You can put your answers directly into the table. Here is the list of terms you can add:
• certain or sure
• impossible
• likely or probable
• unlikely or improbable
• maybe
• uncertain or unsure
Statement Probability Term
Tomorrow is Friday.
I will be in New York on Friday.
It will be dark tonight.
It is snowing in August!
China is cold in January.
1. Read each of the following statements and match the following words to each statement. You can put your answers directly into the table. Here is the list of terms you can add:
• certain or sure
• impossible
• likely or probable
• unlikely or improbable
• maybe
• uncertain or unsure
Statement Probability Term
I am having a sandwich for lunch.
I have school tomorrow.
I will go to the movies tonight.
January is warm in New York.
My dog will bark.
1. Can the amount of lemonade in a pitcher be represented by a discrete random variable? Why or why not?
2. Can the number of musicians in an orchestra be represented by a discrete random variable? Why or why not?
3. Can the weights of the alligators in a swamp be represented by a discrete random variable? Why or why not?
4. Can the number of tickets sold to a movie be represented by a discrete random variable? Why or why not?
5. Give an example of something that can be represented with a discrete random variable. Explain your answer.
6. Give an example of something that cannot be represented with a discrete random variable. Explain your answer.
### Vocabulary Language: English Spanish
discrete random variables
discrete random variables
Discrete random variables represent the number of distinct values that can be counted of an event.
random variables
random variables
Random variables are quantities that take on different values depending on chance, or probability.
continuous random variable
continuous random variable
A continuous random variable is a random variable that can take on all values in an interval. |
## Precalculus (6th Edition) Blitzer
To find the x-intercepts of the function equate the function $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-x+4$ to zero: \begin{align} & {{x}^{3}}-4{{x}^{2}}-x+4=0 \\ & {{x}^{2}}\left( x-4 \right)-1\left( x-4 \right)=0 \\ & \left( {{x}^{2}}-1 \right)\left( x-4 \right)=0 \\ & \left( x+1 \right)\left( x-1 \right)\left( x-4 \right)=0 \end{align} $x=-1,1,4$ are the x-intercepts. To find the y-intercept of the function, find the value of $f\left( 0 \right)$. \begin{align} & f\left( 0 \right)=\left( 0+1 \right)\left( 0-1 \right)\left( 0-4 \right) \\ & f\left( 0 \right)=4 \\ \end{align} Thus, the y-intercept is 4. Substitute x with $-x$ to check the symmetry of the function: $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-x+4$: \begin{align} & f\left( -x \right)={{\left( -x \right)}^{3}}-4{{\left( -x \right)}^{2}}-\left( -x \right)+4 \\ & =-{{x}^{3}}-4{{x}^{2}}+x+4 \end{align} Since $f\left( x \right)\ne f\left( -x \right)$, the graph is not symmetric with respect to the y-axis and since $f\left( -x \right)\ne -f\left( x \right)$, the graph is not symmetric through the origin. |
Lesson Explainer: Arithmetic Series Mathematics
In this explainer, we will learn how to calculate the sum of the terms in an arithmetic sequence with a definite number of terms.
Sequences and series are commonly found in nature and can be used to model the spread of a virus or the decline of a population (these two things are not necessarily linked of course!). In the study of pure mathematics, we interest ourselves in finding the general, or th, term of such sequences, as well as the sum of a given number of terms.
We begin by recalling what we mean by an arithmetic sequence.
Definition: Arithmetic Sequences and Series
An arithmetic sequence is a sequence that has a common difference between consecutive terms. The general term, , of an arithmetic sequence with first term and common difference is given by
An arithmetic series is the sum of a given number of terms of an arithmetic sequence.
For instance, the sequence is an example of an arithmetic sequence.
It has a first term and a common difference of 3.
By using the general term with and , the term of this sequence is given by
The corresponding arithmetic series would be
Letβs look at a practical example before deriving a formula for the sum of a given number of terms in an arithmetic series.
Example 1: Finding the Sum of an Arithmetic Series given the First Three Terms
Find the sum of the first 17 terms of the arithmetic series .
This is an arithmetic series with first term 12. The common difference, , is found by subtracting a term by the term that precedes it:
The general term, , of an arithmetic sequence with first term and common difference is given by
This means the general term of our series is
The final two terms in the partial sum of this series are found by letting and .
When ,
When ,
This means the sum of the first 17 terms, , can be written as
Of course, we could reverse the series and still get the same result:
Notice how each of the 17 numbers in the list can be paired up with a number in the other list to give a sum of 168.
Adding these equations gives
The sum of the first 17 terms can now be found by dividing 2βββ856 by 2:
The sum of the first 17 terms of this arithmetic series is 1βββ428.
This method for finding the sum of a finite arithmetic series can be generalized to an arithmetic series with first term and common difference .
Example 2: Writing an Expression for the Sum of an Arithmetic Sequence
Find an expression for the sum of an arithmetic sequence whose first term is and whose common difference is .
We recall that the general term, , of an arithmetic sequence with first term and common difference is given by
We can use this formula to work out the first terms in this sequence.
When ,
When ,
When ,
The pattern continues in this manner.
The sum of the first terms, , is now given as
Of course, if we reverse the series, we will still get the same overall sum:
Notice how each number in the list can be paired up with a number in the other list to give a constant sum:
This means that when we add the two equations, we will have lots of this expression:
To find an expression for , we then divide through by 2:
Definition: The Sum of an Arithmetic Sequence
The sum of the first terms of an arithmetic series whose first term is and whose common difference is is given by , where
There will be occasions when we are given the first and last terms of an arithmetic series and asked to calculate its sum. We will now consider how to derive a formula for this sum using the formula for the sum of the first terms.
Example 3: Writing an Expression for the Sum of the First π Terms of an Arithmetic Sequence
Write an expression for the sum of the first terms of an arithmetic sequence with first term and last term .
We begin by recalling the formula that allows us to find the term of an arithmetic series with first term and common difference :
We also know that the sum of the first terms of an arithmetic series whose first term is and whose common difference is is given by , where
By writing as , we can substitute in, as shown:
In a sequence with terms, is the last term. This means we can replace with to find a formula for the sum of the first terms of an arithmetic sequence with first term and last term :
Definition: The Sum of an Arithmetic Sequence
The sum of the first terms of an arithmetic sequence with first term and last term is given by , where
We will now learn how to apply this formula to find the sum of the terms in a finite arithmetic series.
Example 4: Finding the Sum of an Arithmetic Series given the First and Last Terms
Find the sum of the terms of the 11-term arithmetic sequence whose first term is and last term is .
Recall that the sum of the first terms of an arithmetic sequence with first term and last term is given by , where
The first term in our sequence is and the last term is , so we will let and .
There are 11 terms in the sequence, so we will let .
Then, the sum of the first 11 terms is given by , where
The sum of the terms of this 11-term arithmetic sequence is .
We will now investigate how we can use the general term of an arithmetic sequence to calculate the sum of a given number of terms of the same arithmetic sequence.
Example 5: Finding the Sum of an Arithmetic Sequence given the General Term
Find the sum of the first 10 terms of the sequence , where .
We are given the general term of the sequence, . This is a formula that allows us to calculate any term given its position number.
For instance, the first term is found by substituting into the formula.
When ,
When ,
When ,
The first three terms of the sequence are 6, 8, and 10. We can therefore deduce the first term to be 6 and the common difference to be 2.
The sum of the first terms of an arithmetic series whose first term is and whose common difference is is given by , where
Since we are calculating the sum of the first 10 terms, we will let , , and :
The sum of the first 10 terms of this sequence is 150.
It is worth noting that the formulae for working with sequences and series can be adapted when a term or a number of terms are given as algebraic expressions. In our next example, we will see what that could look like.
Example 6: Finding the Sum of a Given Arithmetic Sequence in terms of π
Find, in terms of , the sum of the arithmetic sequence .
In order to find the sum of an arithmetic sequence, we need to know either the number of terms or the value of the last term. In this example, we are given the last term of the arithmetic sequence as an algebraic expression, .
In order to establish the number of the terms in this sequence, letβs look at the general term . When , , which is the first term in the sequence. When , , which is the second term in the sequence. This pattern continues, meaning that, as long as , is also the number of terms in the sequence.
This means we can use the formula to find the sum of the first terms of an arithmetic sequence with first term and last term :
Substituting and into our formula,
The sum of the arithmetic sequence is .
In our final example, we look at how we find the sum of a given number of terms of an arithmetic sequence given information about its terms. This process will involve some problem solving to βwork backwardβ to a solution.
Example 7: Finding the Sum of a Given Number of Terms of an Arithmetic Sequence under a Given Condition
Find the sum of the first 21 terms of an arithmetic sequence given and .
There are two formulae that we can use to find the sum of a given number of terms of an arithmetic sequence. The first requires us to know the value of the first term, , and the common difference, : while the second requires us to know the value of the first term, , and the last term, :
We have been given information about three of the terms in the sequence, so it follows that we might need to apply the formula for the term of an arithmetic sequence. Since this formula uses the value of the first term and the common difference, we might deduce that we will need the first version of the summation formula.
The general term, , of an arithmetic sequence with first term and common difference is given by
Letting ,
For ,
And for ,
Letβs substitute each of these expressions into the two equations given to us in the question.
The first gives us
Then, the equation becomes
Notice that we have a pair of linear simultaneous equations. These can be solved by subtracting one from the other:
Finally, we can substitute into either of our original equations:
The sum of the first terms of an arithmetic series whose first term is and whose common difference is is given by , where
Since we are finding the sum of the first 21 terms, we will substitute , , and into this formula:
Key Points
• The sum of the first terms of an arithmetic series whose first term is and whose common difference is is given by , where
• The sum of the first terms of an arithmetic sequence with first term and last term is given by , where |
# 5.3 - Inference for the Population Proportion
5.3 - Inference for the Population Proportion
Earlier in the lesson, we talked about two types of estimation, point, and interval. Let's now apply them to estimate a population proportion from sample data.
### Point Estimate for the Population Proportion
The point estimate of the population proportion, $p$, is:
Point Estimate of the Population Proportion
$\hat{p}=$ # of successes in the sample of size n
From our previous lesson on sampling distributions, we know the sampling distribution of the sample proportion under certain conditions. We can use this information to construct a confidence interval for the population proportion.
### Confidence Interval for the Population Proportion
Recall that:
If $np$ and $n(1-p)$ are greater than five, then $\hat{p}$ is approximately normal with mean, $p$, standard error $\sqrt{\frac{p(1-p)}{n}}$.
Under these conditions, the sampling distribution of the sample proportion, $\hat{p}$, is approximately Normal. The multiplier used in the confidence interval will come from the Standard Normal distribution.
# 5.3.1 - Construct and Interpret the CI
5.3.1 - Construct and Interpret the CI
## Constructing a Confidence Interval for the Population Proportion
To construct a confidence interval we're going to use the following 3 steps:
1. CHECK CONDITIONS
Check all conditions before using the sampling distribution of the sample proportion.
We previously used $np$ and $n(1-p)$. But $p$ is not known. Therefore, for the confidence interval, we will use
• $n\hat{p}>5$ and
• $n(1-\hat{p})>5$
##### What can one do if the conditions are NOT satisfied?
For a confidence interval for a proportion, there is a technique called exact methods. These methods can be used if the software offers it. These exact methods are more complicated and are based on the relationship between the binomial and another distribution we will later learn called the F-distribution. The Z-method is much simpler and fairly easy to compute. In fact if you ever come across a published random survey (e.g. a Gallup poll) you can use the methods in this lesson to construct a reliable proportion confidence interval rather quickly.
2. CONSTRUCT THE GENERAL FORM
The general form of the confidence interval is '$\text{point estimate }\pm M\times \hat{SE}(\text{estimate})$.' The point estimate is the sample proportion, $\hat{p}$, and the estimated standard error is $\hat{SE}(\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$. If the conditions are satisfied, then the sampling distribution is approximately normal. Therefore, the multiplier comes from the normal distribution. This interval is also known as the one-sample z-interval for $p$, or the Normal Approximation confidence interval for $p$.
$\boldsymbol{\left(1-\alpha \right) 100\%}$ confidence interval for the population proportion, $\boldsymbol{p}$
$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$
where $z_{\alpha/2}$ represents a z-value with $\alpha/2$ area to the right of it.
General notes about the confidence interval...
• The $\pm$ in the formula above means "plus or minus". It is a shorthand way of writing
$(\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}})$
• It is centered at the point estimate, $\hat{p}$.
• The width of the interval is determined by the margin of error.
• You must determine the multiplier.
3. INTERPRET THE CONFIDENCE INTERVAL
Applying the template from earlier in the lesson we can say we are $(1-\alpha)100\%$ confident that the population proportion is between $\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $\hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$. The examples will go into more detail regarding the interpretation of the confidence interval.
Think about it! What terms in the margin of error would change the width of the confidence interval? Do the changes make it narrower or wider?
## Derivation of the Confidence Interval
To calculate the confidence interval, we need to know how to find the z-multiplier. So where does this $z_{\alpha}$ come from?
The confidence interval can be derived from the following fact:
\begin{align} P\left(\left|\frac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\right|\le z_{\alpha/2}\right)=1-\alpha \\ P\left(-z_{\alpha/2}\le \dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\le z_{\alpha/2}\right)=1-\alpha \\ P\left(\hat{p}-z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\le p \le \hat{p}+z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\right)=1-\alpha \end{align}
The figure shows the general confidence interval on the normal curve.
## How to find the multiplier using the Standard Normal Distribution
$z_a$ is the z-value having a tail area of $a$ to its right. With some calculation, one can use the Standard Normal Cumulative Probability Table to find the value.
## Example 5-1: Finding $\boldsymbol{z_a}$
Find using the standard Normal table: $z_{0.15}$
$z_{0.15}$ means $P(Z>z_{0.15})=0.15$. This implies that $P(Z\le z_{0.15})=0.85$. The value from the table is 1.04.
For more detailed directions on reading the z-table or using Minitab refer to the examples on this page: 3.3.2 The Standard Normal Distribution.
## Try it!
Use the Standard Normal Table to find the following:
$z_{0.08}$
$z_{0.08}=1.40$
$z_{0.02}$
$z_{0.02}=2.05$
## Commonly Used Alpha Levels
The table is a list of frequently used alphas and their $z_{\alpha/2}$ multipliers.
Confidence level and corresponding multiplier.
Confidence Level $\boldsymbol{\alpha}$ $\boldsymbol{z_{\alpha/2}}$ $\boldsymbol{z_{\alpha/2}}$ Multiplier
90% .10 $z_{0.05}$ 1.645
95% .05 $z_{0.025}$ 1.960
98% .02 $z_{0.01}$ 2.326
99% .01 $z_{0.005}$ 2.576
The value of the multiplier increases as the confidence level increases. This leads to wider intervals for higher confidence levels. We are more confident of catching the population value when we use a wider interval.
## Example 5-2: Alpha Levels
For an 80% confidence interval find $\alpha$, $\alpha/2$, and $z_{\alpha/2}$.
Recall that $\alpha$ is used to find the confidence level by taking (1 - $\alpha)*100%$.
So for an 80% confidence we would take...
$(1 - \alpha)*100 = 80$ or...
$(1 - \alpha) = .8$
$\alpha = .2$
Therefore, $\alpha/2 = .2/2 = .1$
We would have $z_{0.10}$ which means $P(Z>z_{0.10})=0.10$.
This implies that $P(Z\le z_{0.10})=0.90$. The value from the table is 1.28.
Visually, you can see how these numbers relate to the normal distribution in the graph below.
## Example 5-3: Approval Ratings
A random sample of 1500 U.S. adults is taken. They are asked whether they approve or disapprove of the current president's performance so far (i.e. an approval rating). Of the 1500 surveyed, 660 respond with "approve". Calculate a 95% confidence interval for the overall approval rating of the the president.
Since we're dealing with a single proportion, we will examine the number of "successes" and the number of "failures". In this example there were 660 successes and 840 failures. With both successes and failures being at least 5, the condition to use the z-method to calculate the interval is acceptable.
For 95% confidence, the alpha value is 5% or 0.05 The multiplier would be a z-value with $\alpha/2$, or 0.025 area to the right of it. Examining the standard normal table, we find that this corresponds to a z-value of 1.96.
Important Note: Many students tend to use the multiplier of 2 instead of 1.96 due to the empirical rule. As a general rule, it is always best to use the exact values rather than the rounded value.
In this example, we have a sample proportion, $\hat{p}$, of 660/1500 = 0.44 and a sample size, $n$, of 1500.
\begin{align} \hat{p}& \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p}}{n}} &&\text{(General Form)} \\0.44 &\pm 1.96 \sqrt{\dfrac{0.44(1-0.44)}{1500}} &&\text{(Plug in the numbers)} \\ 0.44 &\pm 0.025 &&\text{(Simplify)}\end{align}
"We are 95% confident that the overall U.S. adult approval rating for the current president is from 41.5% to 46.5%." You could also see this written as, "The current U.S. approval rating for the president is 44% with a 95% margin of error of 2.5%." Commonly, the standard level of confidence is 95% so that reference is often left out as that is the assumed level of confidence unless otherwise stated. Also, the method calculates a proportion but often the reported values are converted to percentages. If you use the decimal formal (e.g. 0.415 and 0.465) then reference these as proportion and not percentage.
#### View the video explanation from Dr. Bulathsinhala
To construct a 1-proportion confidence interval...
1. In Minitab choose Stat > Basic Statistics > 1 proportion .
2. From the drop down box select the Summarized data option button. (If you have the raw data you would use the default drop down of One or more samples, each in a column.)
3. Enter the number of successes in the Number of Events text box, and the sample size in the Number of Trials text box.
4. Choose the Options button. The default confidence level is 95. If your desire another confidence level edit appropriately.
5. To use the z- interval method choose Normal Approximation from the Method text box. The exact interval is always appropriate and is the default. Under the conditions that: $n \hat{p} \ge 5, n(1− \hat{p}) \ge 5$, one can also use the z-interval to approximate the answers. The exact interval and the z-interval should be very similar when the conditions are satisfied.
6. Choose OK and OK again.
#### Using Minitab: Approval Ratings Example
We will now use Minitab to verify our by-hand results. Recall in that example a random sample of 1500 was taken from the population of U.S. adults, with 660 responding with a positive approval.
In Minitab and following the steps above, we would enter 660 for the Number of Events and 1500 for the Number of Trials. The confidence level was 95% and we satisfied the necessary conditions to use the Normal Approximation (or z-interval) method. The results are:
Test and CI for One Proportion
Sample X N Sample p 95% CI
1 660 1500 0.440000 (0.414880, 0.465120)
Using the normal approximation.
These results closely match our by-hand interval of 0.415 to 0.465
What if we had calculated the exact confidence interval (i.e. did not choose Normal Approximation as the method)? With the exact method the interval is (0.414685, 0.465550). Consistent to three decimal places in this case. You will notice that in the output Minitab does provide a notification that the normal approximation was used.
We want to know the proportion of graduate students at Penn State who are Democrats. To answer the question, we give out the following survey:
• Yes
• No
Suppose that we get 10 people that circled Yes and 20 people that circled No (that includes the case when people don't know whether they are Democrats!!)
• Let X = the number of successes (number of students who chose Yes) = 10
• n = number of trials = 30
Find a 90% confidence interval for the proportion of graduate students who are democrats.
You should first check the conditions. We know $\hat{p}=\frac{10}{30}=0.333$ and $n=30$ Therefore, $n\hat{p}=30(0.333)=10$ and $n(1-\hat{p})=20$. Since both values are greater than 5, we can use the Normal distribution.
The z multiplier will be $z_{0.1/2}=1.645$
$\hat{p}\pm 1.645\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=0.333\pm 1.645\sqrt{\dfrac{0.333(1-0.333)}{30}}=0.333\pm0.1415=(0.1915, 0.4745)$.
We are 90% confident that the population proportion of graduate students at Penn State who are democrats is between 19.15% and 47.45%.
The video demonstrates this same example using Minitab.
# 5.3.2 - Interpreting the CI
5.3.2 - Interpreting the CI
## More on the Interpretation of a Confidence Interval
In the graph below, we show 10 replications (for each replication, we sample 30 students and ask them whether they are Democrats) and compute an 80% Confidence Interval each time. We are lucky in this set of 10 replications and get exactly 8 out of 10 intervals that contain the parameter. Due to the small number of replications (only 10), it is quite possible that we get 9 out of 10 or 7 out of 10 that contain the true parameter. On the other hand, if we try it 10,000 (a large number of) times, the percentage that contains the true proportions will be very close to 80%.
If we repeatedly draw random samples of size n from the population where the proportion of success in the population is $p$ and calculate the confidence interval each time, we would expect that $100(1 - \alpha)$% of the intervals would contain the true parameter, $p$.
# 5.3.3 - Sample Size Computation
5.3.3 - Sample Size Computation
#### Sample Size Computation for the Population Proportion Confidence Interval
An important part of obtaining desired results is to get a large enough sample size. We can use what we know about the margin of error and the desired level of confidence to determine an appropriate sample size.
Recall that the margin of error, E, is half of the width of the confidence interval. Therefore for a one sample proportion,
$E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$
Precision
The wider the interval, the poorer the precision. Note that the higher the confidence level, the wider the width (or equivalently, half width) of the interval and thus the poorer the precision.
Since the confidence level reflects the success rate of the method we use to get the confidence interval, we like to have a narrower interval while keeping the confidence level at a reasonably higher level.
For most newspapers and magazine polls, it is understood that the margin of error is calculated for a 95% confidence interval (if not stated otherwise). A 3% margin of error is a popular choice also. For instance, you might see a television poll state that the "approval rating of the president is 72%; the margin of error of the poll is plus or minus 3%."
If we want the margin of error smaller (i.e., narrower intervals), we can increase the sample size. Or, if you calculate a 90% confidence interval instead of a 95% confidence interval, the margin of error will also be smaller. However, when one reports it, remember to state that the confidence interval is only 90% because otherwise, people will assume 95% confidence.
## Determining the Required Sample Size
If the desired margin of error E is specified and the desired confidence level is specified, the required sample size to meet the requirements can be calculated by two methods:
Educated Guess
$n=\dfrac{z^2_{\alpha/2}\hat{p}_g(1-\hat{p}_g)}{E^2}$
Where $\hat{p}_g$ is an educated guess for the parameter $p$.
*The educated guess method is used if it is relatively inexpensive to sample more elements when needed.
Conservative Method
$n=\dfrac{z^2_{\alpha/2}(\frac{1}{2})^2}{E^2}$
This formula can be obtained from part (a) using the fact that:
For $0 \le p \le 1, p (1 - p)$ achieves its largest value at $p=\frac{1}{2}$.
*The conservative method is used if the start-up cost of sampling is expensive and thus it is not economical to sample more elements later.
The sample size obtained from using the educated guess is usually smaller than the one obtained using the conservative method. This smaller sample size means there is some risk that the resulting confidence interval may be wider than desired. Using the sample size by the conservative method has no such risk.
## Example 5-4
Suppose a television poll states that the "approval rating of the president is 72%." For the next poll of the president's approval rating, we want to get a margin of error of 1% with 95% confidence. How many individuals should we sample?
##### Educated Guess:
$z_{0.025} = 1.96, E = 0.01$
Therefore,
$n=\dfrac{(1.96)^2(0.72)(1-0.72)}{(0.01)^2}=7744.67$
The sample size needed is 7745 people. We always need to round up to the next integer when the result is not a whole number. We discuss this in detail below.
##### Conservative Method:
$z_{0.025} = 1.96, E = 0.01$
Therefore,
$n=\dfrac{(1.96)^2(0.5)(1-0.5)}{(0.01)^2}=9604$
The sample size is 9604 people.
## Cautions About Sample Size Calculations
1. Why do we need to round up?
Because we are estimating the smallest sample size needed to produce the desired error. Since we cannot sample a portion of a subject (e.g. we cannot take 0.66 of a subject) we need to round up to guarantee a large enough sample.
2. Remember that this is the minimum sample size needed for our study.
If we encounter a situation where the response rate is not 100% then if we just sample the calculated size, in the end, we will end up with a less than desired sample size. To counter this, we can adjust the calculated sample size by dividing by an anticipated response rate. For instance, using the above example if we expected about 40% of the those contacted to actually participate in our survey (i.e. a 40% response rate) then we would need to sample 7745/0.4=19,362.5 or 19,363. In other words, our actual sample size would need to be 19,363 given the 40% response rate.
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FAQ
# Whats half of 250
One half written as a fraction is 1/2. You can also write it as a decimal by simply dividing 1 by 2 which is 0.50. If you multiply 0.50 with 250 you will see …
## What number is half of 250?
For, 2 × 250 = 500. That means that 250 is half of 500. 250 × 2 = 500.
## What is half of \$500?
Half of 500 is 250.
## How do you find half of a number?
This means that to find half of a number, we can divide it by 2. To divide by 2, split the total number into two equal parts. For example: To calculate half of 6 we can divide it by 2.
## What is a 150 half?
half of 150. = 1/2×150. =75.
## What are the multiples of 250?
What are the multiples of 250? Multiples of 250 are 250, 500, 750, 1000, 1250, 1500, 1750, 2000, 2250, 2500.
## How do you do halves?
To find one half of a number, divide it by 2. For larger numbers, split them into their tens and ones, halve these separately and then add them together. For example, to find one half of 8, we divide it by 2. Dividing by 2 means to split the total amount into two equal groups.
## What is half of \$90?
Half of 90 is 45.
## What is half of \$50?
Half of 50 equals 25.
## What is 7 in a half?
Half of 7 would be 3 1/2, or in decimal form, 3.5. Finding ‘half’ of a number is the same as dividing by 2.
## What is the half of the 25?
12 and a 1/2 or 12.5 is half of 25.
## How do you divide by half?
Technically, to divide a number by 1/2 is the same as to multiply it by 2. See also “multiply by double.”
## Whats is half of 35?
17.5
Answer: Half of 35 is 35/2 or 17½ as a fraction and 17.5 as a decimal.
## What can you divide 250 by?
The factors of 250 are integers that divide 250 without any remainder. Let’s begin dividing 250 with the natural numbers, starting with 1. All the divisors and the quotients thus obtained on division form the factors of 250. They are 1, 2, 5, 10, 25, 50, 125 and 250.
## What is the LCM of 250?
Free LCM Calculator determines the least common multiple (LCM) between 250 and 254 the smallest integer that is 31750 that is divisible by both numbers. Least Common Multiple (LCM) of 250 and 254 is 31750.
## How much is multiple of 250?
Multiples of 250: 250, 500, 750, 1000, 1250, 1500, 1750, 2000, 2250, 2500 and so on.
## What are halves math?
Splitting a whole thing into two equal parts gives a half.
## What is one half of a number?
One half is the irreducible fraction resulting from dividing one by two (2) or the fraction resulting from dividing any number by its double. Multiplication by one half is equivalent to division by two, or “halving”; conversely, division by one half is equivalent to multiplication by two, or “doubling”.
## What is half of of 5?
5/2
Answer: Half of 5 is 5/2 as a fraction and 2.5 as a decimal.
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# Examples On Integration By Parts Set-4
Go back to 'Indefinite Integration'
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We now consider some miscenalleous examples of integration by parts:
Example – 47
Evaluate the following integrals
(a) $$\int {{{\tan }^{ - 1}}\sqrt x \,\,dx}$$ (b) \begin{align}\int {\frac{{x\;{{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }}\,\,dx} \end{align}
Solution: (a) We apply integration by parts, taking unity as the second function.
\begin{align}& I = \int {\mathop {{{\tan }^{ - 1}}}\limits_{{\rm{Ist}}} \sqrt x } \cdot \mathop {1\,\,}\limits_{{\rm{IInd}}} dx\\ &= x{\tan ^{ - 1}}\sqrt x - \int {\frac{1}{{1 + {{(\sqrt x )}^2}}} \cdot \frac{1}{{2\sqrt x }} \cdot x\,\,dx} \\& = x{\tan ^{ - 1}}\sqrt x - \frac{1}{2}\int {\frac{{\sqrt x }}{{1 + x}}\,\,dx} \end{align}
To evaluate the integral on the right side above (call it $${I_2}$$ ), we let $$x = {t^2}$$ . Thus,
\begin{align}& dx = 2tdt\\& {I_2} = \int {\frac{{{t^2}}}{{1 + {t^2}}}\,\,dt}\\ &\quad = \int {\left( {1 - \frac{1}{{1 + {t^2}}}} \right)\,\,dt} \\&\quad = t - {\tan ^{ - 1}}t + C\\&\quad= \sqrt x - {\tan ^{ - 1}}\sqrt x + C\end{align}
Thus, $$I = x{\tan ^{ - 1}}\sqrt x + \sqrt x - {\tan ^{ - 1}}\sqrt x + C$$
(b) We can divide the given expression into two parts, \begin{align}\frac{x}{{\sqrt {1 + {x^2}} }}{\rm{ \;and \;}}{\tan ^{ - 1}}x\end{align} . The integration of the first expression is easy to carry out:
\begin{align}\int {\frac{x}{{\sqrt {1 + {x^2}} }}} \,\,dx = \frac{1}{2}\int {\frac{{dt}}{{\sqrt t }}} \left( where\;t=\text{ }1\text{ }+{{x}^{2}} \right) \end{align}
\begin{align}&= \sqrt t \\& = \sqrt {1 + {x^2}} \end{align}
Now we integrate our original expression by parts:
\begin{align}& I = \int {{{\mathop {\tan }\limits_{{\rm{Ist}}} }^{ - 1}}x \cdot \frac{x}{{\mathop {\sqrt {1 + {x^2}} }\limits_{{\rm{IInd}}} }}dx} \\& = \sqrt {1 + {x^2}} {\tan ^{ - 1}}x - \int {\frac{1}{{1 + {x^2}}} \cdot \sqrt {1 + {x^2}} } \,\,dx\\& = \sqrt {1 + {x^2}} {\tan ^{ - 1}}x - \mathop {\int {\frac{1}{{\sqrt {1 + {x^2}} }}} \,\,dx}\limits_{{\text{(of the standard form (22))}}} \\& = \sqrt {1 + {x^2}} {\tan ^{ - 1}}x - \ln \left| {x + \sqrt {1 + {x^2}} } \right| + C\end{align}
Example - 48
Evaluate \begin{align}\int {\frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} \,\,dx\end{align}
Solution: Consider the expression \begin{align}\frac{1}{{x\sin x + \cos x}}\end{align} carefully. See what happens when we differentiate this expression:
${\left( {\frac{1}{{x\sin x + \cos x}}} \right)^′} = \frac{{ - x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}$
Thus, from this relation, we can say that
$\int {\frac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dx} = \frac{{ - 1}}{{x\sin x + \cos x}} + C\qquad\qquad...{\text { }}\left( 1 \right)$
(1) gives us the required clue as to how to solve our original integral; we need to split the numerator $$x{{~}^{2}}$$ of the original integral into two parts, \begin{align} x\cos x\,\,{\rm{and}}\frac{x}{{\cos x}}:\end{align}
$I=\int{\frac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx=}\int{\underset{\begin{smallmatrix} \\ \text{Ist} \end{smallmatrix}}{\mathop{\frac{x}{\cos x}}}\,\,\,\cdot \underset{\begin{smallmatrix} \\ \left( \begin{smallmatrix} \text{We have evaluated the integral } \\ \text{of this IInd function in}\left( 1 \right) \end{smallmatrix} \right) \end{smallmatrix}}{\mathop{\frac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}}}\,}\,\,dx$
\begin{align}& = \frac{{ - x}}{{\cos x\left( {x\sin x + \cos x} \right)}} - \int {\frac{{\cos x + x\sin x}}{{{{\cos }^2}x}}} \cdot \frac{{ - 1}}{{x\sin x + \cos x}} \cdot dx\\& = \frac{{ - x}}{{\cos x\left( {x\sin x + \cos x} \right)}} + \int {{{\sec }^2}x\,dx} \\& = \frac{{ - x}}{{\cos x\left( {x\sin x + \cos x} \right)}} + \tan x + C\end{align}
How to think of such non-trivial manipulations is the skill that you absolutely require if you want to master integration!
Indefinite Integration
grade 11 | Questions Set 1
Indefinite Integration
Indefinite Integration
grade 11 | Questions Set 2
Indefinite Integration |
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Chapter Eleven — Polynomials
The Humongous Book of Algebra Problems
238
Classifying Polynomials
Labeling them based on the exponent and total terms
11.1 Classify the polynomial 7x + 10 according to its degree and the number of
terms it contains.
The degree of a polynomial is the highest exponent to which the variable in that
polynomial is raised. The only variable present in this expression is x, so the
polynomial has degree one. Such polynomials are described as linear.
The polynomial contains two terms, 7x and 10, so it is a linear binomial.
11.2 Classify the polynomial 6x
2
according to its degree and the number of terms it
contains.
The polynomial 6x
2
consists of a single term, so it is a monomial. That term
contains a variable raised to the second power, which classifies the polynomial
11.3 Classify the polynomial –3x
2
+ 5x according to its degree and the number of
terms it contains.
The polynomial –3x
2
contains two terms: –3x
2
and 5x. Of those terms, the
highest power of x is 2, so the polynomial has degree two. A polynomial with
two terms and degree two is called a quadratic binomial.
11.4 Classify the polynomial 12x
3
according to its degree and the number of terms
it contains.
This polynomial consists of a single term, so it is a monomial. Furthermore, the
polynomial has degree three, which means it is cubic. In conclusion, 12x
3
is a
cubic monomial.
Note: Problems 11.5–11.6 refer to the polynomial 4x
2
– 7x + 6.
11.5 Classify the polynomial according to its degree and the number of terms it
contains.
This polynomial contains three terms (4x
3
, –7x, and 6) and has degree two
(because the highest power of x is two). Therefore, 4x
2
– 7x + 6 is a quadratic
trinomial.
It doesnt
have degree
one because
theres only one
variable. It has
degree one because
the only variable
there is raised
to the rst
POWER.
It’s a BInomial
because it has
TWO terms (bi =
two). A MONOmial
has ONE term, and
a TRInomial has
THREE terms.
RULE
OF THUMB:
The degree is
the highest power
of the variable.
Heres how polynomial
classications break
down by degree:
Degree 1: Linear |
# What are the two numbers that are the sum 50 difference 10 ? thank you
May 25, 2018
See below.
#### Explanation:
Firstly, assign the two numbers random variables $x$ and $y$
The sum of them is equal to $50$ therefore
$x + y = 50$
The difference is $10$
$x - y = 10$
Now we have a simultaneous equation.
$x + y = 50$
$x - y = 10$
Add them together to cancel out the $y$.
$2 x = 60$
Now solve for $x$ $\implies x = 30$
Now put the value back into one of the equations to find $y$
$y + 30 = 50$
$\implies y = 20$
The two numbers are $30$ and $20$
May 25, 2018
$30 \text{ and } 20$
#### Explanation:
"let the 2 numbers be x and y ";x>y
$x + y = 50 \leftarrow \textcolor{b l u e}{\text{sum of numbers}}$
$x - y = 10 \leftarrow \textcolor{b l u e}{\text{difference of numbers}}$
$\text{add the 2 equations term by term on both sides}$
$\left(x + x\right) + \left(y - y\right) = \left(50 + 10\right)$
$2 x = 60$
$\text{divide both sides by 2}$
$x = \frac{60}{2} = 30 \Rightarrow x = 30$
$\text{substitute " x=30" into } x + y = 50$
$30 + y = 50$
$\text{subtract 30 from both sides}$
$y = 50 - 30 = 20 \Rightarrow y = 20$
$\text{the 2 numbers are 30 and 20}$
May 25, 2018
30 and 20
#### Explanation:
Okay let's define a couple numbers, let's call one of them $x$ and the other $y$.
We are told that the sum (addition) is:
$x + y = 50$
And the difference (subtraction):
$x - y = 10$
We have a system of equations; two equations and two unknown variables so it is solvable; we will use the "substitution" method:
add $y$ to both sides of: $x - y = 10$
$x - y + y = 10 + y$
$x = 10 + y$
now substitute the value we solved for $x$ into the other equation:
$x + y = 50$
$\left(10 + y\right) + y = 50$
$10 + 2 y = 50$
$2 y = 40$
$y = 20$
So one of the numbers is $20$. to find the other use either of our original equations and insert $y$ to solve for $x$, this one is simplest:
$x + y = 50$
$x + 20 = 50$
$x = 30$
Solved! Our numbers are 30 and 20
To check your solutions insert them into the original equations:
$x + y = 50$
$30 + 20 = 50$
and
$x - y = 10$
$30 - 20 = 10$ |
Coordinate Geometry - ESSENTIAL GEOMETRY SKILLS - SAT Test Prep
## CHAPTER 10ESSENTIAL GEOMETRY SKILLS
### Lesson 4: Coordinate Geometry
Plotting Points
Some SAT questions may ask you to work with points on the x-y plane (also known as the coordinate plane or the Cartesian plane, after the mathematician and philosopher René Descartes). When plotting points, remember these four basic facts to avoid the common mistakes:
• The coordinates of a point are always given alphabetically: the x-coordinate first, then the y-coordinate.
• The x-axis is always horizontal and the y-axis is always vertical.
• On the x-axis, the positive direction is to the right of the origin (where the x and y axes meet, point (0,0)).
• On the y-axis, the positive direction is up from the origin (where the x and y axes meet, point (0,0)).
Working with Slopes
Every line has a slope, which is the distance you move up or down as you move one unit to the right along the line. Imagine walking between any two points on the line. As you move, you go a certain distance “up or down.” This distance is called the rise. You also go a certain distance “left or right.” This distance is called the run.
The slope is simply the rise divided by the run.
You should be able to tell at a glance whether a slope has a positive, negative, or zero slope. If the line goes up as it goes to the right, it has a positive slope. If it goes down as it goes to the right, it has a negative slope. If it’s horizontal, it has a zero slope.
If two lines are parallel, they must have the same slope.
Finding Midpoints
The midpoint of a line segment is the point that divides the segment into two equal parts.
Think of the midpoint as the average of the two endpoints.
Concept Review 4: Coordinate Geometry
Questions 1–10 refer to the figure below. Horizontal and vertical lines are spaced 1 unit apart.
6. Draw a line through B that is parallel to the x-axis and label it 2.
7. Draw the line and label it ℓ3.
8. If point A is reflected over ℓ2, what are the coordinates of its image? __________
9. If line segment is rotated 90 clockwise about point B, what are the coordinates of the image of point A? __________
10. If point B is the midpoint of line segment , what are the coordinates of point C? __________
Rectangle ABCD has an area of 108.
Note: Figure not drawn to scale.
Questions 11–15 pertain to the figure above.
Questions 16–18 pertain to the figure above.
16. What is the area of the triangle above? __________
17. If the triangle above were reflected over the line x = 3, what would be the least x-coordinate of any point on the triangle? __________
18. If the triangle above were reflected over the line , what would the area of the new triangle be? __________
SAT Practice 4: Coordinate Geometry
1. If point A has coordinates (3, 5), point B has coordinates (3, 2), and ABCD is a square, which of the following could be the coordinates of point C?
(A) (4, 2)
(B) (6, 2)
(C) (6, 6)
(D) (4, 6)
(E) (8, 2)
2. If ℓ1 is a horizontal line passing through (1, 8) and ℓ2 is a vertical line passing through (–3, 4), then at what point do ℓ1 and ℓ2 intersect?
(A) (–3, 8)
(B) (1, 4)
(C) (–1, 6)
(D) (–2, 12)
(E) (0,0)
3. The point (–3, 4) is on a circle with its center at the origin. Which of the following points must also be on the circle?
I. (0, –5)
II. (–4, –3)
III. (3, 4)
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
4. If the point (3, –7) is the center of a circle and the point (8, 5) is on the circle, what is the circumference of the circle?
(A) 13π
(B) 15π
(C) 18π
(D) 25π
(E) 26π
5. In the figure above, point E is to be drawn so that ΔCDE has the same area as ΔABC. Which of the following could be the coordinates of E?
(A) (16, 5)
(B) (3, 8)
(C) (5, 12)
(D) (2, 16)
(E) (4, 24)
6. What is the area, in square units, of the shaded region in the figure above?
(A) 32
(B) 33
(C) 34
(D) 35
(E) 36
Note: Figure not drawn to scale.
7. Points A and B lie on the line , as shown above. Lines ℓ1 and ℓ2 pass through the origin, and ℓ1 has a slope of ½. If the distance from A to B is 4, what is the slope of line ℓ2?
Concept Review 4
1. (–2, 1)
2. (0.5, 2)
3. 2/5
4. Your line should be vertical (straight up and down) and pass through B.
5. The x-coordinate of all the points is 3, so ℓ1 can be described by the equation .
6. ℓ2 should be horizontal (straight across) and pass through B.
7. ℓ3 should be a horizontal (straight across) line one unit below the x-axis.
8. (–2, 5)
9. (1, 8) (Notice that the new segment must be the same length as AB, but its slope is the negative reciprocal of AB’s slope, that is, –5/2.)
10. (8, 5)
11. The area of the rectangle is 108, and its length DC is . So its height is . Therefore, , and .
12. AC is the hypotenuse of a right triangle with legs of 9 and 12. This is a 3-4-5 triangle times 3—a 9-12-15 triangle—so AC is 15. The perimeter of the rectangle is . So the ratio of the diagonal to the perimeter is .
13. The slope of DB is 3/4, or .75.
14. AC and DB intersect at the midpoint of each segment. The midpoint of AC is .
15. (26, 19)
16. Use the left side of the triangle as the base. This way, the base is vertical and the height is horizontal, so the lengths are easier to find. The base is 4 units and the height is 7 units, so the area is .
17. The reflection of the triangle over the line is shown above. The “leftmost” point has an x-coordinate of 0.
18. No matter how the triangle is reflected, the area remains the same. The area is still 14.
SAT Practice 4
1. B The vertices of a square must always be listed in consecutive order, so point C must follow consecutively after B and can be in either position shown in the figure at right. Therefore, C can be at (0, 2) or (6, 2).
2. A The horizontal line passing through (1, 8) is the line , and the vertical line passing through (–3, 4) is the line . So they intersect at (–3, 8).
3. E The distance from (0, 0) to (–3, 4) is 5, which is the radius of the circle. Therefore, any point that is 5 units from the origin is also on the circle. Each of the given points is also 5 units from the origin.
4. E The distance from (3, –7) to (8, 5) is The circumference is .
5. D ΔABC has a base of 4 and height of 8, so its area is . Since the base of ΔCDE is 2, its height must be 16 if it is to have the same area as ΔABC. The y-coordinate of E, then, must be 16 or –16.
6. E Draw a rectangle around the quadrilateral as in the figure at right. The rectangle has an area of . If we “remove” the areas of the four right triangles from the corners, the area of the shaded region will remain.
7. 3/8 The y-coordinate of point A is 6, which means the “rise” from O to A is 6. Since the slope of ℓ1 is ½, the “run” must be 12. The “run” from O to B is , and the “rise” is 6, so the slope of ℓ2 is .
|
# The Quadratic Formula and the Shortcut
### Background
There are two common ways for finding roots of quadratic equations, that is, equations of the form $ax^2 + bx + c = 0$
The one that’s usually taught first is a shortcut that works best when $$a = 1$$ and two factors of $$c$$ have a sum of $$b$$: In fact, that’s exactly how it’s taught.
For instance, in $$x^2 + 5x + 4$$, 4 has the factors 1 and 4, which add to 5; the quadratic can then be factored: $$x^2 + 5x + 4 = (x + 4)(x + 1)$$, and so the expression has roots at $$x = -4$$ and $$x = -1$$. (Graphed on WolframAlpha)
Meanwhile, in $$x^2 + 4x + 4$$, 4 has the factors 2 and 2, which add to 4. The quadratic can be factors as $$(x + 2)(x + 2)$$, and its roots are $$x = -2$$. (Graphed on WolframAlpha)
However, this doesn’t work well or at all in cases where $$c$$ doesn’t have integer factors that sum to $$b$$, or where $$a$$ is not 1. It can be modified, but at some point it becomes easier to use the general solution, which is using the Quadratic Formula: $\text{Given } ax^2 + bx + c = 0\text{, then} \\ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
I believe mathematics teachers tend to delay the Quadratic Formula because it looks complicated. The “shortcut” method involves trial and error and only works with a very limited subset of quadratics, but mathematics books are filled with deliberately constructed quadratic equations where it does work. So fair enough.
However, once the Quadratic Formula is introduced, it may be confusing on another level: It’s not immediately clear why the relatively complex-looking Quadratic Formula and the simple “find two numbers that multiply to C and add to B” method should return the same values. It’s an excellent opportunity for a teacher to illustrate that two different methods can return the same result. In this post, I’d like to show why it works.
Naturally, we could show how the Quadratic Formula was originally derived, but we don’t actually need to go that route. Instead, let’s think about what the formula means. The purpose of the formula is to find two binomial factors of a quadratic expression. Each of these factors will involve $$x$$ multiplied by some value, with the product added to some other value. In mathematical terms, $ax^2 + bx + c = (dx + e)(fx + g)$
We’re going to rewrite the Quadratic Formula in terms of these new parameters: $$d$$, $$e$$, $$f$$, and $$g$$.
First, expand the right side: $(dx + e)(fx + g) = dxfx + dxg + efx + eg \\ = dfx^2 + (dg + ef)x + eg$
We can then match each of the up with coefficient parameters from the general quadratic expression: $a = df \\ b = dg + ef \\ c = eg$
Let’s take the original Quadratic Formula and use these new substitutions: $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\ = \frac{-(dg + ef) \pm \sqrt{(dg + ef)^2 – 4dfeg}}{2df} | \text{ Substitute parameters} \\ = \frac{-(dg + ef) \pm \sqrt{(dg^2 + 2dgef + ef^2) – 4dfeg}}{2df} | \text{ Expand} \\ = \frac{-(dg + ef) \pm \sqrt{dg^2 – 2dfeg + ef^2}}{2df} | \text{ Combine} \\ = \frac{-(dg + ef) \pm \sqrt{(dg – ef)^2}}{2df} | \text{ Contract} \\ = \frac{-(dg + ef) \pm (dg – ef)}{2df} | \text{ Simplify}$
At this point, we want to look at the two cases represented by the $$\pm$$ sign: $$x = \frac{-(dg + ef) + (dg – ef)}{2df}$$ and $$x = \frac{-(dg + ef) – (dg – ef)}{2df}$$. Take the first one: $x = \frac{-(dg + ef) + (dg – ef)}{2df} \\ = \frac{- dg – ef + dg – ef}{2df} \\ = \frac{-2ef}{2df} = -\frac{e}{d}$ The second one returns a similar result: $x = \frac{-(dg + ef) – (dg – ef)}{2df} \\ = \frac{- dg – ef – dg + ef}{2df} \\ = \frac{-2dg}{2df} = -\frac{g}{f}$
So the Quadratic Formula gives us two roots: $$x = -\frac{e}{d}$$ and $$x = -\frac{g}{f}$$. This is to say, either $$x + \frac{e}{d} = 0$$ or $$x + \frac{g}{f} = 0$$. These should look familiar, because if $$x + \frac{e}{d} = 0$$ then $$dx + e = 0$$, which is one of the two factors at the beginning of this section.
### The Quadratic Formula and the special case shortcut
So now that we’ve written the roots of a quadratic expression in terms of different parameters, how does this align with the special case?
The short cut works best when $$a = 1$$. Since $$a = df$$, $$d$$ is the multiplicative inverse of $$f$$; for the sake of simplicity and illustration, let us use $$d = f = 1$$. This leads to $$b = g +e$$ and $$c = eg$$, with the roots $$x = -e$$ and $$x = -g$$… which is the shortcut, precisely.
Clio Corvid
## 1 Comment
1. James McKeogh says:
Awesome explanation. Cheers.
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
### 2.4 Real 0*s of Polynomial Functions
2.4 Real 0’s of Polynomial Functions
Long Division
q(x) called the quotient, r(x) called the remainder
f(x) = dividend d(x) = divisor
Remember when doing long division you must fill in holes
with 0, including the constant
example: if you are given x3 – 4x + 5 you will have to
rewrite it as x3 + 0x2 – 4x + 5
Examples: f(x) = 4x3 – 8x2 + 2x - 1; d(x) = x – 1
f(x) = x3 – 1; d(x) = x + 1
Synthetic division
Examples: f(x) = 2x3 – 3x2 – 5x – 12, d(x)=x -3
1) rewrite without the x’s, dividend: the signs stay the same,
divisor: the sign changes
3 2 -3 -5 -12
multi. on the diagonal
6 9 12
2 3
4 0
last # is always the remainder
Rewrite answer: 2x2 + 3x + 4
Examples: 2x4 – 5x3 + 7x2 - 3x -1
x -3
5x4 – 3x + 1
x-4
Remainder Theorem
If a polynomial f(x) is divided by x – k, then the remainder is
r = f(x)
To use the remainder theorem:
either solve the problem by long or synthetic division and
find the remainder
Example: find the remainder when f(x) = 3x2 + 7x – 20 is
divided by a) x – 2, b) x + 4
Factor Theorem
A polynomial of function f(x) has a factor x – k if and only if
f(k) = 0
To do: again either use long or synthetic division
example: use the Factor theorem to determine whether the
1st polynomial is a factor of the second polynomial
Example: x -3; x3 – x2 - x - 15
Finding rational 0’s
To do:
1) find all factors of your constant (p)
3) find all values of p/q – list all of them
4) pick any p/q to determine if it’s a 0 by using synthetic
division, need a remainder of 0 for it to work
5) rewrite answer & then factor it (if you get it down to x2)
6) find the 0’s by setting each factor = 0 & solving for x
Example: f(x) = 3x3 + 4x2 – 5x - 2
Upper & Lower bounds
k is an upper bound for the real 0’s of f if f(x) is never 0 when
x>k
k is a lower bound for the real 0’s of f if f(x) is never 0 when
x<k
Another way to say it:
if k> 0 and every # in the last line is non-negative (positive
or 0), then k is an upper bound for the real 0’s of f.
if k <0 and the #’s in the last line are alternately nonnegative and non-positive, then k is a lower bound for all the
real 0’s of f.
Proving you have an upper or lower
bound
Prove that the number k is an upper bound
k = 5, f(x) = 2x3 – 5x2 - 5x - 1
Prove that the number k is a lower bound
k = -3, f(x) = x3 + 2x2 + 2x + 5
Finding real 0’s
Use the Rational 0’s Theorem (which is what we use to find
rational 0’s)
Find p/q (p is the factors of your constant, q is the factors of
Pick any one you want & use synthetic division to determine
if it’s a 0 (your remainder will be 0)
Rewrite your answer: if it’s x3 or higher, then you need to
repeat what you did above. If it’s x2, you can factor it and
then solve each factor
Determine if the 0 is rational or irrational
example
Find all real 0’s of the function, finding exact values
whenever possible. Identify each 0 as rational or irrational
f(x) = x3 + 3x2 – 3x – 9
f(x) = 3x4 – 2x3 + 3x2 + x - 2 |
# Algorithms: Generating Combinations #100DaysOfCode
In how many different ways can we select r objects from a collection of n objects? In mathematics, this is called combinations.
The formula for the number of combinations is:
where, n! denotes the factorial of a number that is the product of all numbers from 1 to n (inclusive).
### Prelude : A function for calculating factorial of a number
public static long factorial(int n) {
long res = 1L;
for (int i = 1; i <= n; i++) {
res *= i;
}
return res;
}
### Calculating Combinations
That was simple! Let us now move on to calculating the number of combinations given n and r
public static long combinations(int n, int r) {
if (r < 0) {
return 0;
}
long res = 1L;
if (r > n - r) {
r = n - r;
}
for (int i = 0; i < r; i++) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
What does this algorithm do? Recall that we need to find n!/r!(n-r)! which will be of the form n(n-1)...(n-r+1)/1.2...r. Similar to factorial, we initialize the result as 1 and multiply by n-i and divide by i+1. Will this result in a fractional number? No. This is because first, we multiply by n and divide by 1. Next, we multiply by n-1 and divide by 2. Now, either n or n-1 have to be even (as they are consecutive numbers). Similarly, next when we divide by 3, one of n,n-1 and n-2 must be divisible by 3.
In the above code, we also make use of the mathematical property that combinations(n,r) = combinations(n,n-r). This way, we can do less number of operations for calculating the combinations.
### Generating the combinations
Counting the number of combinations was not so hard! Now, let's generate all the combinations.
• Given n and r, we will print out all the combinations.
• For the n objects, we will use the numbers from 0 to (n-1).
• Additionally, we will generate them in a lexicographical order which is math speak for sorted order.
• Finally, in a combination containing a and b, if a < b, we will print a b instead of b a. Formally stated, if a[k] and a[k+1] are the kth and (k+1)th elements in a generated combination, a[k] < a[k+1] for all k
For example, given n = 4, r = 2, we have:
0 1
0 2
0 3
1 2
1 3
2 3
i.e 6 combinations.
Notice that we have 0 1 and not 1 0. This is because we are generating each combination in lexicographical order and we take the minimum for each combination.
#### Generating for a specific value of r (r = 2)
If we have a specific value of r say 2, the code will involve 2 for loops like:
for (int i = 0; i < n-1; i++) {
for (int j = i+1; j < n; j++) {
println(i + " " + j);
}
}
In the code above, our first loop variable i goes from 0 to n-2 and the next variable j goes from i+1 to n-1. Why so? This is because we have a requirement for taking the lexicographical minimum combination, so i < j from our constraints.
If value of r is fixed, we can simply create r for loops. But it is not fixed...
#### Generalizing!
Now, let's move on to the main goal - generate combinations of n numbers taken r at a time. This section will be a little verbose as I have outlined how I arrived at the correct code. If you are interested in just the algorithm, feel free to skip to the bottom of the article
If we notice our previous code for r = 2, our first combination is always 0 1 as i = 0, j = 1. Similarly, if r was 3, our first combination would be 0 1 2. There is a pattern!
By creating an array a of size r, we can generate the first combination as 0 1 2 .. r-1. We have the first combination ready. What about the rest? Somehow, if we increment elements in this array, we will generate the combinations...
Again, looking at the r = 2 case, notice that the last combination is n-2 n-1. Similarly, for r = 3, it is n-3 n-2 n-1. Thus, for r elements, it will be n-r+1 n-r+2 .. n-1. There is one more insight - there is exactly one combination which starts with n-r+1. If our array's first element reaches n-r+1, we are done!
We now have a termination condition for our function: a[0] == n-r+1
The code we have so far will look like:
int n; // Given
int r; // Given
int[] a = new int[r]; // Initialize array of size r
for (int i = 0; i < r; i++) {
a[i] = i; // Initialize array with first combination
}
while (a[0] < n-r+1) { // Our termination condition
// DO SOMETHING!
}
We have a while loop that checks for termination condition. For the loop to terminate, we need to steadily progress from our first combination to the last combination. As we are generating elements in lexicographical order, the last element of the array must be incremented first. Then the second from last element and so on.
In our earlier example of n = 4, r = 2, we had
0 1
0 2
0 3
1 2
1 3
2 3
After 0 3, we get 1 2. This means once the r-1 element (last element) reaches its maximum, we increment r-2 element from 0 to 1 and also reset the value of r-1 element to a[r-2]+1 as it must always be at least 1 greater than the r-2 element (from our constraints). Moving to our pseudo-code, let's add this to the while loop
int i = r-1; // variable i keeps track of which element in array we are incrementing
while (a[0] < n-r+1) { // Our termination condition
if (i > 0 && a[i] == n-r+1) {
i = i-1; //If a[i] has reached the max allowable value, decrement i and move to next element in array
}
printArray(a); // pseudo-code to print out the combination
a[i] = a[i]+1; // increment
if (i < r-1) {
a[i+1] = a[i]+1; // Reset i+1 element as previous element + 1, according to our constraints
i = i+1; // Once you have reset the i+1 element, it is no longer < n-r+i and hence, we can move it back to old value
}
}
We have an index variable i which we use to check which is the element in the array to be incremented. In the first if in above code, we check if the a[i] has reached its maximum value of n-r+i. If yes, we decrement i as a[i] can no longer be incremented. Moving out of if, we then print the combination and increment a[i]. Now, if i is no longer r-1 i.e it is no longer last element of a, we must reset it to r-1 and also set the value of a[r-1] as a[r-2]+1. This works for r=2. Hooray! We have abstracted out the for loop in the earlier section into a while loop with a few conditionals.
But does this work for r > 2? No... We need a minor change to make it work! Change the if statements inside the loop to while loops and we are done! In case of first loop, we need to find the maximum i which is less than n+r-i. For example n=5, r=3 we have:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
As we move from 0 3 4 to 1 2 3, both i = 2 (a[2] = 4) and i = 1 (a[1] = 3) are at their maximum. We need to move to i = 0. Similarly, the second if must be a while loop because once we have incremented the a[i] for minimum i, we need to reset the outer elements of array to maintain our constraints.
### Our final code:
int[] a = new int[r];
// initialize first combination
for (int i = 0; i < r; i++) {
a[i] = i;
}
int i = r - 1; // Index to keep track of maximum unsaturated element in array
// a[0] can only be n-r+1 exactly once - our termination condition!
while (a[0] < n - r + 1) {
// If outer elements are saturated, keep decrementing i till you find unsaturated element
while (i > 0 && a[i] == n - r + i) {
i--;
}
printArray(a); // pseudo-code to print array as space separated numbers
a[i]++;
// Reset each outer element to prev element + 1
while (i < r - 1) {
a[i + 1] = a[i] + 1;
i++;
}
}
#### Proof of termination
Now that we have our algorithm, how can we show that it terminates? In each iteration of our outer while loop, we increment the element of the array with maximum index i which has not reached value n-r+i while maintaining our constraints. Due to the lexicographical ordering, our previous combination is always lesser than our currently generated combination. As there are only a finite number of combinations till we reach our "last" combination, we can say that our algorithm will terminate.
### Meta:
I began my 100 days of code challenge today with this problem. I will create a separate post explaining my motivations and plans.
Regarding this problem statement of generating combinations, I had some trouble initially moving from r=2 case to the general one. It took me some time to find the correct termination condition. I am happy that the final algorithm is relatively compact. I also want to do a proof of correctness for this algorithm later.
Posted on by:
### Raunak Ramakrishnan
Passionate about databases, distributed systems and functional programming.
### Discussion
Hello, thnx for this information. Do you also have the code in Fortran? KR, Marc |
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# Divisibility Tests
In math, we often need to find out if large integers are divisible by certain smaller numbers. For example, when factoring a large integer to find its prime factorization using the division method, we need to start by dividing by smaller numbers like 2, 3, or other smaller numbers. We don't even necessarily have to start with only prime numbers but can use numbers such as 8, 9, or 10 just to make the integers we are dividing by more manageable.
This article discusses some tips on how to find if certain larger integers are divisible by the smaller numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10.
## Divisibility by 2, 3, or 4
### Divisibility by 2
An integer is divisible by 2 if it is even. If it is even, its last digit will be 0, 2, 4, 6, or 8.
Example 1
3458 is divisible by 2 because it ends in the digit 8.
7385 is not divisible by 2 because it ends in the digit 5.
### Divisibility by 3
You can tell that an integer is divisible by 3 if the sum of its digits is divisible by 3.
Example 2
3483 is divisible by 3 because
$3+4+8+3=18$ and
$18÷3=6$ .
8249 is not divisible by 3 because
$8+2+4+9=23$
which is not divisible by 3.
Example 3
You can use this rule more than once in an extremely large number.
For example, to test the number 8,597,938, you can add the digits.
$8+5+9+7+9+3+8=49$
$4+9=13$
Therefore, 8,597,938 is not divisible by 3 because 3 does not divide evenly into 13.
Example 4
Let's try that again with another very large number. Is 9,645,984 divisible by 3?
$9+6+4+5+9+8+4=45$
$4+5=9$
Since 9 is evenly divisible by 3, so is 9,654,984.
### Divisibility by 4
Because 4 goes evenly into 100, an integer is divisible by 4 if the last two digits are divisible by 4.
Example 5
Is 3,588 divisible by 4? The last two digits are 88, which is divisible by 4. Therefore, 3,588 is divisible by 4.
Is 8,427 divisible by 4? The last two digits are 27, which is not divisible by 4. Therefore, 8,427 is not divisible by 4.
One thing to note is that, while not every even number is divisible by 4, only even numbers are divisible by 4. No odd number is divisible by 4.
## Divisibility by 5, 6, or 7
### Divisibility by 5
An integer is divisible by 5 if the last digit is either a 5 or a 0.
Example 6
Is 3,984,180 divisible by 5? Yes, because the last digit is 0.
Is 8,349 divisible by 5? No, because the last digit is not 0 or 5.
### Divisibility by 6
An integer that is divisible by both 2 and 3 is divisible by 6. So to determine if an integer is divisible by 6, you must run both the test for divisibility by 2 and the test for divisibility by 3 on it.
Example 7
Is 3,756 divisible by 6?
First, is it divisible by 2? Yes, because the last digit is 6, an even number.
Then, is it divisible by 3? Let's check. $3+7+5+6=21$ , which is divisible by 3. So the integer is divisible by 3.
Therefore, 3,756 is divisible by 6.
Example 8
Is 283,124 divisible by 6?
First, is divisible by 2? Yes, because the last digit is 4, an even number.
Then, it is divisible by 3? Let's check. $2+8+3+1+2+4=20$ , which is not divisible by 3. So the integer is not divisible by 3.
Therefore, 283,124 is not divisible by 6.
### Divisibility by 7
Unfortunately, there are no good divisibility tests for the number 7. The easiest way to find out if a large number is divisible by 7 is to use a calculator or to quickly do a long-division problem using the number 7.
## Divisibility by 8, 9, or 10
### Divisibility by 8
Because 8 goes evenly into 1000, you can tell that an integer is divisible by 8 if the last three digits are divisible by 8.
Example 9
Is 835,136 divisible by 8? It is because $8×17=136$ .
Is 20,375 divisible by 8? You can use a shortcut here. Like the numbers 2, 4, and 6, no odd number is divisible by 8. So you can immediately see that since this number ends in a 5, it is not divisible by 8.
Is 20,374 divisible by 8? It is not, because 374 is not evenly divisible by 8.
### Divisibility by 9
Similar to the test for divisibility by 3, you can tell that an integer is divisible by 9 if the sum of the digits is divisible by 9.
Example 10
Is 379,481 divisible by 9? Let's check and see.
$3+7+9+4+8+1=32$
Since 32 is not divisible evenly by 9, 379,481 is not divisible by 9.
How about 848,376? Is this divisible by 9? Let's check and see.
$8+4+8+3+7+6=36$ . Since $\frac{36}{9}=4$ , so 848,376 is divisible by 9.
### Divisibility by 10
This is probably the easiest test of the bunch. If an integer ends in a 0, it is divisible by 10.
Example 11
Is 83,476,950 divisible by 10? Yes, it is, because it ends in a 0.
Is 4,569 divisible by 10? No, it is not, because it does not end in a 0.
## Flashcards covering the Divisibility Tests
Algebra 1 Flashcards
## Get help learning about divisibility tests
Tutoring is an excellent way to help your student learn and remember the divisibility tests for the numbers 2 through 10. These shortcuts can save them a lot of time in their future years as a math student, but first, they must memorize the various rules of divisibility. A tutor can help your student memorize these by discerning their learning style and teaching them memory tips that make the best use of their learning style. A tutor will work with your student at their pace, taking extra time on the rules that they find challenging and skimming through the rules that they pick up quickly and easily. We'd be happy to set your student up with a qualified math tutor who can help them learn the divisibility tests and more math tips and tricks. To get started, contact Varsity Tutors today and speak with one of our helpful Educational Directors. We look forward to helping your student.
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# Eureka Math Algebra 2 Module 3 Lesson 20 Answer Key
## Engage NY Eureka Math Algebra 2 Module 3 Lesson 20 Answer Key
### Eureka Math Algebra 2 Module 3 Lesson 20 Opening Exercise Answer Key
Opening Exercise:
a. Sketch the graphs of the three functions f(x) = x2, g(x) = (2x)2 + 1, and h(x) = 4x2 + 1.
i. Describe the sequence of transformations that takes the graph of f(x) = x2 to the graph of g(x) = (2x)2 + 1.
The graph of g is a horizontal scaling by a factor of $$\frac{1}{2}$$ and a vertical translation up 1 unit of the graph of f.
ii. Describe the sequence of transformations that takes the graph of f(x) = x2 to the graph of h(x) = 4x2 + 1.
The graph of h is a vertical scaling by a factor of 4 and a vertical translation up 1 unit of the graph of f.
iii. Explain why g and h from parts (I) and (ii) are equivalent functions.
These functions are equivalent and have the same graph because the expressions (2x)2 + 1 and 4x2 + 1 are equivalent. The blue graph shown is the graph off, and the green graph is the graph of g and h.
b. Describe the sequence of transformations that takes the graph of f(x) = sin(x) to the graph of g(x) = sin(2x) – 3.
The graph of g is a horizontal scaling by a factor of $$\frac{1}{2}$$ and a vertical translation down 3 units of the graph of f.
c. Describe the sequence of transformations that takes the graph of f(x) = sin(x) to the graph of h(x) = 4 sin(x) – 3.
The graph of h is a vertical scaling by a factor of 4 and a vertical translation down 3 units of the graph of f.
d. Explain why g and h from parts (b) – (c) are not equivalent functions.
These functions are not equivalent because they do not have the same graphs, and the two expressions are not equivalent.
### Eureka Math Algebra 2 Module 3 Lesson 20 Exploratory Challenge Answer Key
Exploratory Challenge:
a. Sketch the graph of f(x) = log2(x) by identifying and plotting at least five key points. Use the table below to get started.
x Log2(x) $$\frac{1}{4}$$ -2 $$\frac{1}{2}$$ -1 1 2 4 8
The graph of f is shown below.
x Log2(x) $$\frac{1}{4}$$ -2 $$\frac{1}{2}$$ -1 1 0 2 1 4 2 8 3
b. Describe a sequence of transformations that takes the graph off to the graph of g(x) = log2(8x).
The graph of g is a horizontal scaling by a factor of $$\frac{1}{8}$$ of the graph of f.
c. Describe a sequence of transformations that takes the graph off to the graph of h(x) = 3 + log2(x).
The graph of h is a vertical translation up 3 units of the graph of f.
d. Complete the table below for f, g, and h and describe any noticeable patterns.
The functions g and h have the same range values at each domain value in the table.
e. Graph the three functions on the same coordinate axes and describe any noticeable patterns.
The graphs of g and h are identical.
f. Use a property of logarithms to show that g and h are equivalent.
By the product property and the definition of logarithm, log2(8x) = log2(8) + log2(x) = 3 + log2(x), so g(x) = h(x) for all positive real numbers x and g and h are equivalent functions.
g. Describe the graph of p(x) = log2($$\frac{x}{4}$$) as a vertical translation of the graph of f(x) = log2(x). Justify your response.
The graph of p isa vertical translation down 2 units of the graph off because log2($$\frac{x}{4}$$) = log2(x) – 2.
h. Describe the graph of h(x) = 3 + log2(x) as a horizontal scaling of the graph of f(x) = log2(x). Justify your response.
The graph of h is a horizontal scaling by a factor of $$\frac{1}{8}$$ of the graph of f because
3 + log2(x)= log2(8) + log2(x)= log2(8x).
i. Do the functions h(x) = log2(8) + log2(x) and k(x) = log2(x + 8) have the same graphs? Justify your reasoning.
No, they do not. By substituting 1 for x in both f and g, you can see that the graphs of the two functions do not have the same y-coordinate at this point. Therefore, the graphs cannot be the same if at least one point is different.
j. Use properties of exponents to explain why graphs of f(x) = 4x and g(x) = (22)x are identical.
Using the power property of exponents, 22x = (22)x = 4x Since the expressions are equal, the graphs of the functions would be the same.
k. Use the properties of exponents to predict what the graphs of g(x) = 4 . 2x and h(x) = 2x + 2 look like compared to one another. Describe the graphs of g and h as transformations of the graph of f(x) = 2x. Confirm your prediction by graphing f, g, and h on the same coordinate axes.
The graphs of the two functions g and h are the same since 2x + 2 = 2x . 22 = 4. 2x by the multiplication property of exponents and the commutative property. The graph of g is the graph of f(x) = 2x scaled vertically by a factor of 4. The graph of h is the graph of f(x) = 2x translated horizontally 2 units to the left.
l. Graph f(x) = 2x, g(x) = 2-x and h(x) = ($$\frac{1}{2}$$)x on the same coordinate axes. Describe the graphs of g and h as transformations of the graph off. Use the properties of exponents to explain why g and h are equivalent.
The graph of g and the graph of h are both reflections about the vertical axis of the graph of f. They are equivalent because (($$\frac{1}{2}$$))x = (2-1)-x = 2-x by the definition of a negative exponent and the power property of exponents.
### Eureka Math Algebra 2 Module 3 Lesson 20 Example Answer Key
Example 1: Graphing Transformations of the Logarithm Functions
The general form of a logarithm function is given by f(x) = k + a log(x – h), where a, b, k, and h are real numbers such that b is a positive number not equal to 1, and x – h > 0.
a. Given g(x) = 3 + 2 log(x – 2), describe the graph of g as a transformation of the common logarithm function.
The graph of g is a horizontal translation 2 units to the right, a vertical scaling by a factor of 2, and a vertical translation up 3 units of the graph of the common logarithm function.
b. Graph the common logarithm function and g on the same coordinate axes.
Example 2: Graphing Transformations of Exponential Functions
The general form of the exponential function is given by f(x) = a . bx + k, where a, b, and k are real numbers such that b is a positive number not equal to 1.
a. Use the properties of exponents to transform the function g(x) = 32x + 1 – 2 to the general form, and then graph it. What are the values of a, b, and k?
Using the properties of exponents, 32x + 1 – 2 = 32x . 31 – 2 = 3 . 9x – 2. Thus, g(x) = 3(9)x – 2 so a = 3, b = 9, and k = -2.
b. Describe the graph of g as a transformation of the graph of f(x) = 9x.
The graph of g is a vertical scaling by a factor of 3 and a vertical translation down 2 units of the graph of f.
c. Describe the graph of g as a transformation of the graph of f(x) = 3x.
The graph of g is a horizontal scaling by a factor of $$\frac{1}{2}$$, a vertical scaling by a factor of 3, and a vertical translation down 2 units of the graph of f.
d. Sketch the graph of g using transformations.
The graph of f(x) = 9x is shown in blue, and the graph of g is shown in green.
### Eureka Math Algebra 2 Module 3 Lesson 20 Exercise Answer Key
Exercises:
Graph each pair of functions by first graphing f and then graphing g by applying transformations of the graph of f. Describe the graph of g as a transformation of the graph of f.
Exercise 1.
f(x) = log3(x) and g(x) = 2 log3(x – 1)
The graph of g is the graph of f translated 1 unit to the right and stretched vertically by a factor of 2.
Exercise 2.
f(x) = log(x) and g(x) = log(100x)
Because of the product property of logarithms, g(x) = 2 + log(x). The graph of g is the graph of f translated vertically 2 units.
Exercise 3.
f(x) = log5x and g(x) = – log5(5(x + 2))
Since -log5(5(x + 2)) = -1 – log5(x + 2) by the product property of logarithms and the distributive property, the graph of gis the graph off translated 2 units to the left, reflected across the horizontal axis, and translated down 1 unit.
Exercise 4.
f(x) = 3x and g(x) = -2 . 3x – 1
Since -2 . 3x – 1 = -2 . 3x. 3x – 1 = –$$\frac{2}{3}$$ . 3x by the properties of exponents and the commutative property, the graph of g is the graph of f reflected across the horizontal axis and compressed by afactor of $$\frac{2}{3}$$.
There are multiple ways to obtain the graph of g from the graph of f through a sequence of transformations. One choice is to use the structure of g(x) = -2 . 3x – 1to translate the graph off horizontally one unit left, reflect across the horizontal axis, and then scale vertically by a factor of 2. Another choice is to rewrite g(x) as g(x) = $$\frac{2}{3}$$(3x). Then the graph of g is obtained from the graph of f by reflecting the graph of f across the horizontal axis and then vertically scaling by a factor of $$\frac{2}{3}$$.
### Eureka Math Algebra 2 Module 3 Lesson 20 Problem Set Answer Key
Question 1.
Describe each function as a transformation of the graph of a function in the form f(x) = log(x). Sketch the graph of f and the graph of g by hand. Label key features such as intercepts, intervals where g is increasing or decreasing, and the equation of the vertical asymptote.
a. g(x) = log2(x – 3)
The graph of g is the graph of f(x) = log2(x) translated horizontally 3 units to the right. The function g is increasing on (3, ∞). The x-intercept is 4, and the vertical asymptote is x = 3.
b. g(x) = log2(16x)
The graph of g is the graph of f(x) = log2(x) translated vertically up 4 units. The function g is increasing on
(0, ∞). The x-intercept is 2-4. The vertical asymptote is x = 0.
c. g(x) = log2($$\frac{8}{x}$$)
The graph of g is the graph of f(x) = log2(x) reflected about the horizontal axis and translated vertically up 3 units. The function g is decreasing on (0, ∞). The x-intercept is 23. The vertical asymptote is x = 0. The reflected graph and the final graph are both shown in the solution.
d. g(x) = log2((x – 3)2)for x > 3
The graph of g is the graph of f(x) = log2(x) stretched vertically by a factor of 2 and translated horizontally 3 units to the right. The function g is increasing on (3, ∞). The x-intercept is 4, and the vertical asymptote is x = 3.
Question 2.
Each function graphed below can be expressed as a transformation of the graph of f(x) = log(x). Write an algebraic function for g and h, and state the domain and range.
In Figure 1, g(x) = -log(x – 2) for x > 2. The domain of g is x > 2, and the range of g is all real numbers.
In Figure 2, h(x) = 2 + log(x) for x > 0. The domain of h is x > 0, and the range of h is all real numbers.
Figure 1: Graphs of f(x) = log(x) and the function g
Figure 2: Graphs of f(x) = log(x) and the function h
Question 3.
Describe each function as a transformation of the graph of a function in the form f(x) = bx. Sketch the graph off and the graph of g by hand. Label key features such as intercepts, intervals where g is increasing or decreasing, and the horizontal asymptote.
a. g(x) = 2 . 3x – 1
The graph of g is the graph of f(x) = 3x scaled vertically by a factor of 2 and translated vertically down 1 unit. The equation of the horizontal asymptote is y = -1. The y-intercept is 1, and the x-intercept is approximately -0.631. The function g is increasing for all real numbers.
b. g(x) = 22x + 3
The graph of g is the graph of f(x) = 4x translated vertically up 3 units. The equation of the horizontal asymptote is y = 3. The y-intercept is 4. There is no x-intercept. The function g is increasing for all real numbers.
c. g(x) = 3x – 2
The graph of g is the graph of f(x) = 3x translated horizontally 2 units to the right OR the graph of f scaled
vertically by a factor of $$\frac{1}{9}$$. The equation of the horizontal asymptote is y = 0. The y-intercept is $$\frac{1}{9}$$. There is no x-intercept, and the function g is increasing for all real numbers.
d. g(x) = -9$$\frac{x}{2}$$ + 1
The graph of g is the graph of f(x) = 3x reflected about the horizontal axis and then translated vertically up 1 unit. The equation of the horizontal asymptote is y = 1. The y-intercept is 0, and the x-intercept is also 0. The function g is decreasing for all real numbers.
Question 4.
Using the function f(x) = 2x, create a new function g whose graph is a series of transformations of the graph of f with the following characteristics:
→ The function g is decreasing for all real numbers.
→ The equation for the horizontal asymptote is y = 5.
→ The y-intercept is 7.
One possible solution is g(x) = 2 . 2-x + 5.
Question 5.
Using the function f(x) = 2x, create a new function g whose graph is a series of transformations of the graph of f with the following characteristics:
→ The function g is increasing for all real numbers.
→ The equation for the horizontal asymptote is y = 5.
→ The y-intercept is 4.
One possible solution is g(x) = -(2-x) + 5.
Question 6.
Consider the function g(x) = ($$\frac{1}{4}$$)x – 3
a. Write the function g as an exponential function with base 4. Describe the transformations that would take the graph of f(x) = 4x to the graph of g.
($$\frac{1}{4}$$)x – 3 = (4-1)x – 3 = 4-x + 3 = 43 . 4-x
Thus, g(x) = 64 . 4-x. The graph of g is the graph of f reflected about the vertical axis and scaled vertically
by a factor of 64.
b. Write the function g as an exponential function with base 2. Describe two different series of transformations that would take the graph of f(x) = 2’ to the graph of g.
($$\frac{1}{4}$$)x – 3 = (2)-2x – 3 = 2-2(x – 3) = 2-2x + 6 = 64 . 2-2x
Thus, g(x) = 64 . 2-2x or g(x) = 2-2(x – 3). To obtain the graph of g from the graph of f, you can scale the graph horizontally by a factor of $$\frac{1}{2}$$, reflect the graph about the vertical axis, and scale it vertically by a factor of 64. Or, you can scale the graph horizontally by a factor of $$\frac{1}{2}$$, reflect the graph about the vertical axis, and translate the resulting graph horizontally 3 units to the right.
Question 7.
Explore the graphs of functions in the form f(x) = log(xn) for n > 1. Explain how the graphs of these functions change as the values of n increase. Use a property of logarithms to support your reasoning.
The graphs appear to be a vertical scaling of the common logarithm function by a factor of n. This is true because of the property of logarithms that states log(xn) = n log(x).
Question 8.
Use a graphical approach to so’ve each equation. If the equation has no solution, explain why.
a. log(x) = log(x – 2)
This equation has no solution because the graphs of y = log(x) and y = log(x – 2) are horizontal translations of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution.
b. log(x) = log(2x)
This equation has no solution because log(2x) = log(2) + log(x), which means that the graphs of y = log(x) andy = log(2x) are a vertical translation of each other. Thus, their graphs do not intersect, and the corresponding equation has no solution.
c. log(x) = log ($$\frac{2}{x}$$)
Ans:
The solution is the x-coordinate of the intersection point of the graphs of y = log(x) and y = log(2) – log(x). Since the graph of the function defined by the right side of the equation is a reflection across the horizontal axis and a vertical translation of the graph of the function defined by the left side of the equation, the graphs of these functions intersect in exactly one point with an x-coordinate approximately 1.5.
d. Show algebraically that the exact solution to the equation in part (c) is √2.
log(x) = log(2) – log(x)
2 log(x) = log(2)
log(x) = $$\frac{1}{2}$$log(2)
log(x) = log(2$$\frac{1}{2}$$)
x = 2$$\frac{1}{2}$$
Since 2$$\frac{1}{2}$$ = √2 the exact solution is √2.
Question 9.
Make a table of values forf(x) = $$x^{\frac{1}{\log (x)}}$$ for x > 1. Graph the function f for x > 1. Use properties of logarithms to explain what you see in the graph and the table of values.
We see that $$x^{\frac{1}{\log (x)}}$$ = 10 for all x > 1 because $$\frac{1}{\log (x)}=\frac{\log (10)}{\log (x)}$$ = logx(10). Therefore, when we substitute logx(10) into the expression $$x^{\frac{1}{\log (x)}}$$, we get xlogx(10), which is equal to 10 according to the definition of a logarithm.
### Eureka Math Algebra 2 Module 3 Lesson 20 Exit Ticket Answer Key
Question 1.
Express g(x) = -log4(2x) in the general form of a logarithmic function, f(x) = k + a logb(x – h). Identify a, b, h, and k.
Since -log4(2x) = -log4(2) + log4(x) = –$$\frac{1}{2}$$ – log4(x), the function is g(x) = –$$\frac{1}{2}$$ – log(x), and a = -1, b = 4, h = 0, and k = –$$\frac{1}{2}$$.
The graph of g is the graph of h reflected about the horizontal axis and translated down $$\frac{1}{2}$$ unit. |
#### Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (iii)
$\frac{-5}{12}$
Hint:
Eliminate the negative sign by using formula sec-1(-x)=$\pi$-sec-1(x) and then convert the sec-1 term into a cot-1 term to simplify the sum
Concept:
Inverse Trigonometric
Solution:
$cot(sec^{-1}(\frac{-13}{5}))=cot(\pi -sec^{-1}(\frac{13}{5}))$ $[sec^{-1}(-x)=\pi -sec^{-1}(x)]$
$=-cot(sec^{-1}(\frac{13}{5}))$ $[cot(\pi -\theta) =-cot\, \theta ]$
Let
$sec^{-1}(\frac{13}{5})=x$ ..............(1)
$sec\, x=\frac{13}{5}$
$\therefore tan\, x=\sqrt{sec^{2}x-1}$
$=\sqrt{(\frac{13}{5})^{2}-1}\: \: \: =\frac{12}{5}$
$cot\, x=\frac{5}{12}\Rightarrow x=cot^{-1}(\frac{5}{12})$ ............(2)
$sec^{-1}(\frac{13}{5})=cot^{-1}(\frac{5}{12})$
From (1) and (2)
$=cot(sec^{-1}(\frac{-13}{5}))$
$=-cot(sec^{-1}(\frac{13}{5}))$
$=-cot(cot(\frac{5}{12}))$
$=\frac{-5}{12}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities
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10.5 Graphing quadratic equations (Page 9/15)
Page 9 / 15
Find two consecutive even numbers whose product is 624.
A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.
The height of the banner is 13 cm and the length of the side is 54 cm.
Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is 70 square inches. Find the height and width of the case.
A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.
The lengths of the sides of the mosaic are 2.2 and 4.4 feet.
A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.
The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.
The width of the front walk is 8.1 feet and its length is 30.8 feet.
For Sophia’s graduation party, several tables of the same width will be arranged end to end to give a serving table with a total area of 75 square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth. Round answer to the nearest tenth.
A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula $h=-16{t}^{2}+{v}_{0}t$ to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.
The ball will reach 384 feet on its way up in 4 seconds and on the way down in 6 seconds.
A bullet is fired straight up from the ground at a velocity of 320 ft/sec. Use the formula $h=-16{t}^{2}+{v}_{0}t$ to determine when the bullet will reach 800 feet. Round to the nearest tenth.
10.5 Graphing Quadratic Equations in Two Variables
In the following exercises, graph by plotting point.
Graph $y={x}^{2}-2$
Graph $y=\text{−}{x}^{2}+3$
In the following exercises, determine if the following parabolas open up or down.
$y=-3{x}^{2}+3x-1$
down
$y=5{x}^{2}+6x+3$
$y={x}^{2}+8x-1$
up
$y=-4{x}^{2}-7x+1$
In the following exercises, find the axis of symmetry and the vertex.
$y=\text{−}{x}^{2}+6x+8$
$x=3$ $\left(3,17\right)$
$y=2{x}^{2}-8x+1$
In the following exercises, find the x - and y -intercepts.
$y={x}^{2}-4x+5$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,5\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(5,0\right),\left(-1,0\right)$
$y={x}^{2}-8x+15$
$y={x}^{2}-4x+10$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,10\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$
$y=-5{x}^{2}-30x-46$
$y=16{x}^{2}-8x+1$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,1\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{4},0\right)$
$y={x}^{2}+16x+64$
In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.
$y={x}^{2}+8x+15$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,15\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right),\left(-5,0\right);$
axis: $x=-4;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4,-1\right)$
$y={x}^{2}-2x-3$
$y=\text{−}{x}^{2}+8x-16$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-16\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(4,0\right);$
axis: $x=4;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(4,0\right)$
$y=4{x}^{2}-4x+1$
$y={x}^{2}+6x+13$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,13\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=-3;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-3,4\right)$
$y=-2{x}^{2}-8x-12$
$y=-4{x}^{2}+16x-11$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-11\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(3.1,0\right),\left(0.9,0\right);$
axis: $x=2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(2,5\right)$
$y={x}^{2}+8x+10$
In the following exercises, find the minimum or maximum value.
$y=7{x}^{2}+14x+6$
The minimum value is $-1$ when $x=-1$ .
$y=-3{x}^{2}+12x-10$
In the following exercises, solve. Rounding answers to the nearest tenth.
A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation $h=-16{t}^{2}+112t$ to find how long it will take the ball to reach maximum height, and then find the maximum height.
In 3.5 seconds the ball is at its maximum height of 196 feet.
A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using 180 feet of fencing on three sides of the yard. The quadratic equation $A=-2{x}^{2}+180x$ gives the area, $A$ , of the yard for the length, $x$ , of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.
Practice test
Use the Square Root Property to solve the quadratic equation: $3{\left(w+5\right)}^{2}=27$ .
$w=-2,-8$
Use Completing the Square to solve the quadratic equation: ${a}^{2}-8a+7=23$ .
Use the Quadratic Formula to solve the quadratic equation: $2{m}^{2}-5m+3=0$ .
$m=1,\frac{3}{2}$
Solve the following quadratic equations. Use any method.
$8{v}^{2}+3=35$
$3{n}^{2}+8n+3=0$
$n=\frac{-4±\sqrt{7}}{3}$
$2{b}^{2}+6b-8=0$
$x\left(x+3\right)+12=0$
no real solution
$\frac{4}{3}{y}^{2}-4y+3=0$
Use the discriminant to determine the number of solutions of each quadratic equation.
$6{p}^{2}-13p+7=0$
2
$3{q}^{2}-10q+12=0$
Solve by factoring, the Square Root Property, or the Quadratic Formula.
Find two consecutive even numbers whose product is 360.
Two consecutive even number are $-20$ and $-18$ and 18 and 20.
The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)
For each parabola, find which ways it opens, the axis of symmetry, the vertex, the x - and y -intercepts, and the maximum or minimum value.
$y=3{x}^{2}+6x+8$
up $x=-1$ $\left(-1,5\right)$ $y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$ minimum value of 5 when $x=-1$
$y={x}^{2}-4$
$y={x}^{2}+10x+24$
up $x=-5$ $\left(-5,-1\right)$ $y;\left(0,24\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-6,0\right),\left(-4,0\right)$ minimum value of $-5$ when $x=-1$
$y=-3{x}^{2}+12x-8$
$y=\text{−}{x}^{2}-8x+16$
down $x=-4$
$\left(-4,32\right)$ $y;\left(0,16\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-9.7,0\right),\left(1.7,0\right)$
maximum value of $32$ when $x=-4$
Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry.
$y=2{x}^{2}+6x+2$
$y=16{x}^{2}+24x+9$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,9\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{3}{4},0\right);$
axis: $x=-\frac{3}{4};\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{3}{4},0\right)$
Solve.
A water balloon is launched upward at the rate of 86 ft/sec. Using the formula $h=-16{t}^{2}+86t$ , find how long it will take the balloon to reach the maximum height and then find the maximum height. Round to the nearest tenth.
What is the lcm of 340
How many numbers each equal to y must be taken to make 15xy
15x
Martin
15x
Asamoah
15x
Hugo
1y
Tom
1y x 15y
Tom
find the equation whose roots are 1 and 2
(x - 2)(x -1)=0 so equation is x^2-x+2=0
Ranu
I believe it's x^2-3x+2
NerdNamedGerg
because the X's multiply by the -2 and the -1 and than combine like terms
NerdNamedGerg
find the equation whose roots are -1 and 4
Ans = ×^2-3×+2
Gee
find the equation whose roots are -2 and -1
(×+1)(×-4) = x^2-3×-4
Gee
there's a chatting option in the app wow
Nana
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you too.
Joan
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Nana
Hey you all there are several Free Apps that can really help you to better solve type Equations.
Debra
Debra, which apps specifically. ..?
Nana
am having a course in elementary algebra ,any recommendations ?
samuel
Samuel Addai, me too at ucc elementary algebra as part of my core subjects in science
Nana
me too as part of my core subjects in R M E
Ken
at ABETIFI COLLEGE OF EDUCATION
Ken
ok great. Good to know.
Joan
5x + 1/3= 2x + 1/2
sanam
Plz solve this
sanam
5x - 3x = 1/2 - 1/3 2x = 1/6 x = 1/12
Ranu
Thks ranu
sanam
a trader gains 20 rupees loses 42 rupees and then gains ten rupees Express algebraically the result of his transactions
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
Kim is making eight gallons of punch from fruit juice and soda. The fruit juice costs $6.04 per gallon and the soda costs$4.28 per gallon. How much fruit juice and how much soda should she use so that the punch costs $5.71 per gallon? Mohamed Reply (a+b)(p+q+r)(b+c)(p+q+r)(c+a) (p+q+r) muhammad Reply 4x-7y=8 2x-7y=1 what is the answer? Ramil Reply x=7/2 & y=6/7 Pbp x=7/2 & y=6/7 use Elimination Debra true bismark factoriz e usman 4x-7y=8 X=7/4y+2 and 2x-7y=1 x=7/2y+1/2 Peggie Ok cool answer peggie Frank thanks Ramil copy and complete the table. x. 5. 8. 12. then 9x-5. to the 2nd power+4. then 2xto the second power +3x Sandra Reply What is c+4=8 Penny Reply 2 Letha 4 Lolita 4 Rich 4 thinking C+4=8 -4 -4 C =4 thinking I need to study Letha 4+4=8 William During two years in college, a student earned$9,500. The second year, she earned $500 more than twice the amount she earned the first year. Nicole Reply 9500=500+2x Debra 9500-500=9000 9000÷2×=4500 X=4500 Debra X + Y = 9500....... & Y = 500 + 2X so.... X + 500 + 2X = 9500, them X = 3000 & Y = 6500 Pbp Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles. Josh Reply Reiko needs to mail her Christmas cards and packages and wants to keep her mailing costs to no more than$500. The number of cards is at least 4 more than twice the number of packages. The cost of mailing a card (with pictures enclosed) is $3 and for a package the cost is$7.
hey
Juan
Sup
patrick
The sum of two numbers is 155. The difference is 23. Find the numbers
The sum of two numbers is 155. Their difference is 23. Find the numbers
Michelle
The difference between 89 and 66 is 23
Ciid
Joy is preparing 20 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?
hello
bismark
I need a math tutor BAD
Stacie
Me too
Letha
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Xavier
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Bishal
teet
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misty |
GreeneMath.com - Solving Linear Equations in one Variable Test
Solving Linear Equations Test
In order to solve any linear equation in one variable, we must understand the four step process. First we collect terms and simplify each side. Then, we move all variable terms to one side and all numbers to the other. Next, we isolate the variable and finish our process by checking our result with substitution.
Test Objectives:
•Demonstrate an understanding of the addition property of equality
•Demonstrate an understanding of the multiplication property of equality
•Demonstrate the ability to solve any linear equation using the four step procedure
Solving Linear Equations Test:
#1:
Instructions: Solve each equation.
a) -2(5k - 2) - 10k = 2(1 + 2k) + 2
Watch the Step by Step Video Solution
|
View the Written Solution
#2:
Instructions: Solve each equation.
a) 11(1 - 8n) = -11(11n - 7)
Watch the Step by Step Video Solution
|
View the Written Solution
#3:
Instructions: Solve each equation.
a) 12 - 11(m + 11) = -8(m + 11)
Watch the Step by Step Video Solution
|
View the Written Solution
#4:
Instructions: Solve each equation.
a) -5p + 6(p + 7) = -2(-5p + 9) - 12
Watch the Step by Step Video Solution
|
View the Written Solution
#5:
Instructions: Solve each equation.
a) 10v - 2 + 1 + 7v = 8(1 + 2v) + 3(1 + v)
Watch the Step by Step Video Solution
|
View the Written Solution
Written Solutions:
#1:
Solution:
a) k = 0
Watch the Step by Step Video Solution
#2:
Solution:
a) n = 2
Watch the Step by Step Video Solution
#3:
Solution:
a) m = -7
Watch the Step by Step Video Solution
#4:
Solution:
a) p = 8
Watch the Step by Step Video Solution
#5:
Solution:
a) v = -6
Watch the Step by Step Video Solution |
# A Motivating Problem for the Alternating Series Test
Previous: The Alternating Series Test
Next: The Alternating Series Test
## Example
Does the following series
converge? First let's explore this problem using the tools we already have to see why a new convergence test is needed. This will demonstrate of how, in practice, determining whether an infinite series converges is not a matter of applying a set of algorithms that can be memorized. Rather, we often need to try different approaches and sometimes develop new approaches to solve a given problem.
## Why Do We Need Another Convergence Test?
We've covered the divergence dest and the integral test. Does the divergence test tell us if the series converges? Applying the divergence test, we find that the limit
equals zero. The divergence test yields no information, and we must use another test.
Can we apply the integral test? The series have negative and positive values:
But the integral test can only applied on a sequence if all of its terms are positive. The integral test yields no information, and so we must use another test.
In this lesson we introduce a convergence test that can be used to establish that this series converges: the alternating series test.
## Investigating the Partial Sums
Remember that for a series to converge, we must have that its partial sums tend to zero. In our example, this means that
tends to zero as N tends to infinity. Calculating the first 10 partial sums of our series, for N from 1 to 10, we obtain the values presented in the table below (numbers rounded to the first two decimal places).
k ak=1/k sN
1 -1.00 -1.00
2 0.50 -0.50
3 -0.33 -0.83
4 0.25 -0.58
5 -0.20 -0.78
6 0.17 -0.62
7 -0.14 -0.76
8 0.13 -0.63
9 -0.11 -0.75
10 0.10 -0.65
11 -0.09 -0.74
12 0.08 -0.65
13 -0.08 -0.73
14 0.07 -0.66
It appears that the terms of the sequence are tending towards zero, but does it appear that the partial sums converging? They could be, but our table doesn't prove that the sequence converges to zero. We need a better way to determine whether the series converges. We need a new convergence test.
The alternating series test, applicable to series whose terms alternate between positive and negative, can be applied here. We will next give the alternating series test, and then apply it to show that this series does converge.
Previous: The Alternating Series Test
Next: The Alternating Series Test |
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Algebra is one of the most important branches of mathematics, and it finds its application in various fields like engineering, science, finance, and more. However, many students struggle with Algebra, especially when it comes to solving equations. Lack of understanding of basic Algebraic principles can lead to errors, confusion, and frustration. In this guide, we will cover all the essential topics of Algebra and help you overcome the challenges you face in your homework assignments.
Basic Algebraic Equations
Algebraic equations lay the foundation of Algebra. Understanding basic equations is crucial to solving advanced equations.
Understanding the Order of Operations
The order of operations is essential in solving equations. Misunderstanding the order can lead to incorrect answers. The order of operations is Parentheses, Exponents, Multiplication and Division (left to right), Addition, and Subtraction (left to right).
Solving One-Step Equations
One-step equations are the easiest to solve. It only involves one mathematical operation to get the solution.
Solving Two-Step Equations
Two-step equations are a bit more complex than one-step equations. It involves two mathematical operations to get the solution.
Solving Multi-Step Equations
Multi-step equations are the most complex among the three. It involves multiple mathematical operations to get the solution.
Common Mistakes to Avoid in Algebraic Equations
Understanding equations is one thing, avoiding common mistakes in solving them is another. Let’s go through some common mistakes and how to avoid them.
Misunderstanding the Order of Operations
As mentioned above, misunderstanding the order of operations can be catastrophic. Sometimes, brackets can be placed incorrectly or not factored in, and the solution can become entirely different from the expected result.
Distributing Incorrectly
Distributing is the process of passing a value outside the parenthesis to all the terms inside the parenthesis. When distributing, it is essential to make sure the operations are correct.
Combining Like Terms Inappropriately
Like terms are terms that have the same variable, but their coefficients are different. For example, 2x and 6x are like terms. Combining like terms requires that only the coefficients are combined, but the values are kept as they are.
Signs and Negative Numbers
Many students make mistakes when dealing with negative numbers. The rules for negative numbers should be understood to avoid getting the wrong answer.
Cancelling out Incorrectly
Cancelling out incorrectly is one of the most common mistakes students make in Algebra. Not all terms are cancellable. Cancelling out can change the equation or make the solution incorrect.
Unclear Equations
Clear presentation of equations is essential to avoid confusion and misunderstandings. Ambiguous equations or notations can cause errors, misinterpretations, and lead to incorrect solutions.
Quadratic equations are the second-degree equations, where the highest power of the variable is 2. They are essential in Algebra and are commonly found in mathematics and science.
To solve Quadratic equations, it is crucial to understand what they represent. Quadratic equations typically represent parabolic shapes of graphs.
Factoring is one of the most common methods in solving quadratic equations. It is essential to know how to correctly factorize expressions to solve the equations.
Solving Quadratic Equations by Completing the Square
Completing the square is another method of solving quadratic equations. This method allows us to transform the equation to a perfect square.
The quadratic formula is another method of solving quadratic equations. It allows us to derive the roots of the equation.
Common Mistakes When Solving Quadratic Equations
Quadratic equations can be tricky. Common mistakes are made when multiplying binomials, factoring, or misapplication of quadratic formulas.
Solving Systems of Equations
A system of equations is a set of two or more equations with multiple variables.
Understanding Systems of Equations
A system of equations requires solving multiple equations simultaneously. It is crucial to understand the relationships between the equations to solve the system.
Solving Systems of Equations by Graphing
Graphing is a visual method of solving systems of equations. It involves plotting the equations on a graph and identifying intersections.
Solving Systems of Equations by Substitution Method
The substitution method is a way of solving systems of equations by replacing variables in one equation and solving the other.
Solving Systems of Equations by Elimination Method
The elimination method is a way of solving systems of equations by multiplying one or both equations by factors that allow us to eliminate one variable.
Common Mistakes When Solving Systems of Equations
Misapplying either method, not checking the solution, or making a sign error can make the solution incorrect and invalidate the answer.
Rational Equations
Understanding Rational Equations
Rational equations can be complex, and it is essential to understand them to solve them.
Solving Rational Equations
Solving Rational equations requires knowledge of factors, multiplication, addition, subtraction, and division.
Common Mistakes When Solving Rational Equations
Not factoring correctly, applying the wrong operation or canceling wrongly are common mistakes made by students.
Absolute Value Equations and Inequalities
Absolute value equations and inequalities deal with the magnitude or size of a number, expressed in absolute terms.
Understanding Absolute Value Equations and Inequalities
Absolute value equations and inequalities include both positive and negative values. It is essential to understand how to solve them correctly.
Solving Absolute Value Equations
Solving Absolute Value equations requires understanding the concept of absolute value and solving the equations based on that.
Solving Absolute Value Inequalities
Solving Absolute Value inequalities requires understanding and applying different rules of inequalities.
Common Mistakes to Avoid in Absolute Value Equations and Inequalities
Ignoring or omitting the absolute value symbol, confusing the direction of the inequality sign, and calculating incorrectly are common mistakes to avoid.
Complex Numbers and Polynomials
Understanding complex numbers and polynomials form the basis for solving advanced Algebraic equations.
Understanding Complex Numbers
Complex numbers are those numbers that have both real and imaginary components.
Adding, Subtracting, Multiplying and Dividing Complex Numbers
Adding, subtracting, multiplying, and dividing complex numbers are essential in solving more complex equations.
Introduction to Polynomials
Polynomials are expressions with more than one term, and they form the building block for solving higher order equations.
Factoring Polynomials
Factoring Polynomials is the process of breaking down complex expressions into simpler ones.
Common Mistakes to Avoid in Dealing with Complex Numbers and Polynomials
Misapplying the properties of operations, making sign errors, and rushing through the process are common mistakes to avoid.
Conclusion
Algebra is a powerful tool that finds application in many fields of science and technology. Understanding the basics of Algebra is crucial to solving complex equations and succeeding in the career of your choice. We have reviewed all the critical topics of Algebra, including basic equations, quadratic equations, systems of equations, rational equations, absolute value equations and inequalities, complex numbers, and polynomials. We have also provided tips to avoid common mistakes and overcome challenges.
FAQs
Q. What are some common mistakes students make when solving algebraic equations?
Some common mistakes include misunderstanding the order of operations, distributing incorrectly, combining like terms inappropriately, making errors with signs and negative numbers, cancelling out incorrectly, and unclear equations.
Q. How can I become better at understanding algebraic equations?
Practice is the key to becoming better at understanding algebraic equations. Review the basics regularly and solve different types of problems.
Q. What is the best way to solve tricky algebraic equations?
The best way to solve tricky algebraic equations is to understand the basics of Algebra and avoid common mistakes. Try to break down complex equations into smaller parts and use different techniques to arrive at the solution.
Q. Why do I need to learn algebraic equations?
Algebraic equations are essential in various fields like science, finance, and engineering. They are also crucial in developing problem-solving skills.
Q. What are some common algebraic formulas I need to know?
Some common algebraic formulas include Quadratic Formula, Pythagorean Theorem, Distance Formula, Slope Formula, and Midpoint Formula.
Q. What mistakes should I avoid when solving quadratic equations?
Common mistakes include multiplying binomials incorrectly, factoring wrongly, or misapplication of quadratic formulas. Also, double check the solution and check for sign errors.
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# pi is a constant of 3
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``` GNVQ KEY SKILLS AREA Page 1 of 4 MPS @ WESTMINSTER COLLEGE
AREA
Areas apply to two-dimensional shapes, i.e. those with a length and width but no depth. This
means that you can have the area of a sheet of paper or a wall, but not the area of a solid shape
since this has three dimensions. You can have the surface area of a three-dimensional object
since this is the area of the outside or inside surfaces, and the surfaces are two-dimensional.
Firstly we will cover regular shapes
Rectangles and squares
Parallelograms
Triangles
Circles
Then there will be a short section on calculating the areas of irregularly shaped objects
Area of Regular Shapes
Rectangles and Squares
A rectangle is a regular four sided shape with each of the four angles being a right angle, (900).
Squares are rectangles whose four sides are the same length.
The area of a rectangle is calculated by multiplying the length by the width, with the area being
measured in square units. To do this you must first make sure that the length and width are in
the same units of measurement.
Example 1
A room is 8 metres long by 5 metres wide, so its floor area is 8x5 = 40 m2
A drawing pad is 32cm long and 24cm wide, so the area of a sheet is 32x24 = 768 cm2
A roll of paper is 52cm wide and 6.4m long, so the area of the roll is 52/100 x 6.4 = 3.328m2
Exercise 1
Calculate the area of the following shapes
1. A wall of height 2.1metres and width 5.2 metres
2. A square 3.5 inch floppy disc
3. A roll of wallpaper 50cm wide and 15m long
4. A picture of width 3 feet and height 2 feet 6 inches
Parallelograms and Trapeziums
Some four-sided figures do not have angles of 900, and this makes their area more difficult to
calculate. If the shape has two sides which are parallel, (they are a constant distance apart), then
the shape is a parallelogram or trapezium and its area can be calculated by-
adding the two parallel sides together and dividing by two,(if it is a trapezium)
then multiplying the answer by the vertical height of the shape.
To calculate the area you must make the units of measurement the same, (i.e. they must all be in
either metres, cm, feet etc.).
GNVQ KEY SKILLS AREA Page 2 of 4 MPS @ WESTMINSTER COLLEGE
Example 2 The area of the trapezium is 12 + 24 = 36 / 2 =18m x 10m = 180m2
12m
The area of the parallelogram is 20cm x 12cm = 240cm2
20cm
10 12cm
m
24m
Exercise 2
1. A parallelogram has sides of 14cm, and a vertical height of 8cm. What
is its area?
2. A parallelogram has sides of 1m, and a vertical height of 50cm. What is
its area?
3. A trapezium has sides of 6 inches and 10inches, and a vertical height of
1 foot. What is its area?
Triangles
Triangles are three sided figures. If two of them are placed side by side you can see that they
will make a parallelogram or rectangle. This means that the area of the triangle is half the area of
the rectangle or parallelogram, which is half the base multiplied by the vertical height
VERTICAL
HEIGHT
BASE
Example 3
1. The area of a triangle shown on the right with a base of
27cm and a vertical height of 15cm is 27 x 15 / 2 =
27 x 7.5 = 202.5cm2.
1.5m
2. The area of a right angle
triangle shown to the left is 1.5m x 2.5 m / 2 = 1.5m x 1.25m = 1.875m2.
2.5m
GNVQ KEY SKILLS AREA Page 3 of 4 MPS @ WESTMINSTER COLLEGE
Exercise 3
1. What is the area of a triangle with a base of 20cm and a vertical height
of 30cm?
2. What is the area of a triangle with a base of 1.2m and a vertical height of
50cm?
3. What is the area of triangle A? 25m
80cm
4. What is the area of triangle B? A B
40m
1m
CIRCLES
The area of a circle is found by using the equation
AREA = r2
Where (pi) is a constant of 3.14 and r is the radius (the distance from the centre of the circle to
the edge of the circle.
The diameter is the distance across the circle and is twice the radius
in length. The perimeter is the distance around the outside of the
D circle.
i
a
Example 4
m 1. The area of a circle of radius 32cm is found using r2, so the area
e
is 3.14 x 32 x 32 = 3.14 x 1024 = 3215.4cm2.
t
e 2. The area of a circle of diameter 800m is found using r2, where
r the radius is half the diameter. Since the diameter is 800m, the
radius will be 800/2 = 400m. The area of the circle will be 3.14 x
400 x 400 = 3.14 x 160,000 = 502,400m2.
Exercise 4
1. What is the area of a circle of radius10cm?
2. What is the area of a circle of radius 160m?
3. What is the area of a circle of diameter 8 inches?
4. What is the area of a circle of diameter 12.8m?
Area of Irregular Shapes
It is more difficult to calculate the area of irregularly shaped objects since you cannot just use a
simple equation for the calculation. It is still possible to calculate, but you might need to think
about the problem before you start calculating.
The area of irregular shaped rectangular objects can be measured by dividing the shape up into
rectangles and calculating the area of each part. The areas of each of the parts is then added
together to give the area of the whole.
GNVQ KEY SKILLS AREA Page 4 of 4 MPS @ WESTMINSTER COLLEGE
10m 3m
2m The plan of a room shown on the right
B has been split into three rectangles. The
area of the rectangles can be calculated as
A shown.
3.5m A 10m x 5.5m = 55m2
B 3m x 2m = 6m2
C 3m x 4m = 8m2
6m TOTAL = 69m2
C 3m
4m
EXERCISE 5
Calculate the areas of the following shapes
1. 2.
20m
8m
5m
4m mm
10m
4m
2.5
m
5m
3.
10m
8m
10m
16m
```
Related docs
Other docs by 1v84q8Yp |
# Natural numbers
Here you will learn about natural numbers, including the types of natural numbers and properties of natural numbers.
Students will first learn about natural numbers in Kindergarten with counting and cardinality and extend that knowledge through elementary school into middle school with properties of addition and multiplication.
## What are natural numbers?
Natural numbers help you count and represent objects or quantities. They are also called “counting numbers”. The set of natural numbers, usually represented by N, can be defined as positive whole numbers beginning with 1.
N = \{1, 2, 3, …\}
This number line shows the first 7 natural numbers.
Let’s look at some types of natural numbers and properties of natural numbers.
Prime and Composite Natural Numbers \hspace{1cm}
A prime number is a number with exactly
two factors: itself and 1.
2 and 7 are examples of prime numbers.
2 has two factors: \hspace{1cm} 7 has two factors:
1, 2 \hspace{3cm} 1, 7
1 \times 2 = 2 \hspace{2.2cm} 1 \times 7 = 7
A composite number is a number with more
than two factors.
4 and 12 are examples of composite numbers.
4 has three factors: \hspace{1cm} 12 has six factors:
1, 2, 4 \hspace{2.9cm} 1, 2, 3, 4, 6, 12
1 \times 4 = 4 \hspace{2.5cm} 1 \times 12 = 12
2 \times 2 = 4 \hspace{2.7cm} 2 \times 6 = 12
\hspace{4cm} 3 \times 4 = 12
Step-by-step guide: Prime number
Even and Odd Natural Numbers \hspace{1cm}
Even numbers are divisible by 2 without
remainders; they end in 0, 2, 4, 6, or 8.
Every other number on the number line
is even.
Odd numbers are not divisible by 2 without
remainders and end in 1, 3, 5, 7, or 9.
Every other number on the number line
is odd.
### There are four properties of natural numbers.
\bf{1} . Closure Property of Natural Numbers
The sum and product of two natural numbers are also natural numbers.
For example,
The closure property does not work all the time for subtraction and division.
For example,
\bf{2} . Associative Property of Natural Numbers
The sum and product of three natural numbers are the same even if the numbers
are grouped differently. → (a + b) + c = a + (b + c) \; \& \; (a \times b) \times c = a \times (b \times c)
For example,
(1 + 3) + 5 = 9
(1 \times 3) \times 5 = 15
The associative property does not work all the time for subtraction and division.
Step by step guide: Associative Property
\bf{3} . Commutative Property of Natural Numbers
The sum and product of two natural numbers is the same even if the order is
different. → a + b = b + a \; \& \; a \times b = b \times a
For example,
The commutative property does not work all the time for subtraction and division.
Step-by-step guide: Commutative Property
\bf{4} . Distributive Property of Natural Numbers
For three natural numbers, multiplication is distributive over addition and subtraction
meaning that a \times (b + c) = a \times b + a \times c \; \& \; a \times (b-c) = a \times b-a \times c
For example,
Step-by-step guide: Distributive Property
## Common Core State Standards
How does this apply to kindergarten through 6th grade?
• Kindergarten – Counting and Cardinality (K.CC.B.4)
Understand the relationship between numbers and quantities; connect counting to cardinality.
• Grade 1 – Operations and Algebraic Thinking (1.0A.B.3 )
Apply properties of operations as strategies to add and subtract.
Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition).
To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
• Grade 2 – Operations and Algebraic Thinking (2.OA.C.3)
Determine whether a group of objects (up to 20 ) has an odd or even number of members, e.g., by pairing objects or counting them by 2 s; write an equation to express an even number as a sum of two equal addends.
• Grade 3 – Operations and Algebraic Thinking (3.OA.B.5)
Apply properties of operations as strategies to multiply and divide.
Examples: If 6 \times 4 = 24 is known, then 4 \times 6 = 24 is also known. (Commutative property of multiplication).
3 \times 5 \times 2 can be found by 3 \times 5 = 15, then 15 \times 2 = 30, or by 5 \times 2 = 10, then 3 \times 10 = 30. (Associative property of multiplication).
Knowing that 8 \times 5 = 40 and 8 \times 2 = 16, one can find 8 \times 7 as 8 \times (5 + 2) = (8 \times 5) + (8 \times 2) = 40 + 16 = 56. (Distributive property).
• Grade 4 – Number and Operations Base Ten (4.NBT.B.5)
Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
• Grade 6 – Number Systems (6.NS.B.4)
Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 \, (9 + 2).
## How to use natural numbers
In order to classify natural numbers:
1. Recall the definition of the type of number needed.
2. Show whether the number fits or does not fit the definition.
In order to apply a property of natural numbers:
1. Recall the property.
2. Use the property to get an answer.
## Identifying natural numbers examples
### Example 1: identifying natural numbers
Is the number identified on the number line a natural number?
1. Recall the definition of the type of number needed.
Natural numbers are whole positive numbers beginning with 1.
2Show whether the number fits or does not fit the definition.
The number identified on the number line is \, \cfrac{2}{5} \, which is a fraction.
So, the number identified on the number line is not a natural number.
### Example 2: identifying types of natural numbers
Is 13 an odd natural number?
Recall the definition of the type of number needed.
Show whether the number fits or does not fit the definition.
### Example 3: identifying types of natural numbers
Is 19 a prime number?
Recall the definition of the type of number needed.
Show whether the number fits or does not fit the definition.
### Example 4: apply a property of natural numbers
Find the missing number using the associative property of addition.
(5 + 2) + \, \rule{0.5cm}{0.15mm} \, = 5 + (2 + 9)
Recall the property.
Use the property to get an answer.
### Example 5: apply a property of natural numbers
Find the missing number using the commutative property of multiplication.
19 \times 12 = 12 \times \, \rule{0.5cm}{0.15mm}
Recall the property.
Use the property to get an answer.
### Example 6: properties of natural numbers
Find the missing number in the distributive property equation.
8 \times (5 + 3) = \, \rule{0.5cm}{0.15mm} \, + 24
Recall the property.
Use the property to get an answer.
### Teaching tips for natural numbers lessons
• Natural numbers are students’ first experience with numbers because they are the counting numbers. When first teaching the concept of numbers to young mathematicians, associate the actual numeral with the correct number of dots so that students can associate the numeral with a quantity.
• Use a number line so students can visually see a representation of the numbers.
• To build number sense, expose students to number patterns.
• Use a Venn diagram when teaching students in middle school the real number system. Using the diagram will help students visualize how whole numbers, integers, and rational numbers (all the number sets) build upon the natural numbers.
### Easy mistakes to make
• Thinking \bf{0} is a natural number
Natural numbers are counting numbers and begin with the number 1.
0 is not part of the natural number set.
• Thinking all integers are natural numbers
Only the non-negative integers are natural numbers not including 0.
• Thinking that natural numbers are not part of integers
Natural numbers are a subset of the integers because they represent the positive integers.
This natural numbers topic guide is part of our series on types of numbers. You may find it helpful to start with the main types of numbers topic guide for a summary of what to expect or use the step-by-step guides below for further detail on individual topics. Other topic guides in this series include:
### Practice natural numbers
1. Which of the following numbers is a natural number?
0
1.03
15
\cfrac{4}{7}
Natural numbers are the set of whole positive numbers that start at 1.
15 is a positive whole number so it is a natural number.
2. Which group of numbers are natural numbers?
0, 1, 2
1, 2, 3
-3, -2, -1
-2, -1, 0
Natural numbers are the set of whole positive numbers that start at 1.
So, 1, 2, 3 are natural numbers.
3. Which group of numbers represents prime numbers?
2, 3, 7
3, 7, 9
1, 5, 9
-2, -3, -9
A prime number is a number that has exactly two factors, 1 and itself.
2, 3, 7 represent prime numbers because:
The factors of 2 are 1 and 2.
The factors of 3 are 1 and 3.
The factors of 7 are 1 and 7.
4. Find the missing number using the associative property.
(6 + 5) + 10 = 6 + ( \, \rule{0.5cm}{0.15mm} \, + 10)
21
16
10
5
The associative property is (a + b) + c = a + (b + c).
So, (6 + 5) + 10 = 6 + (5 + 10)
(6 + 5) + 10 = 21
6 + (5 + 10) = 21
5. Find the missing number using the commutative property.
12 \times \, \rule{0.5cm}{0.15mm} \, = 8 \times 12
12
96
12
8
The commutative property is a \times b = b \times a.
So, 12 \times 8 = 8 \times 12.
12 \times 8 = 96
8 \times 12 = 96
6. Find the missing number using the distributive property.
4 \times (2 + 9) = 8 + \, \rule{0.5cm}{0.15mm}
9
36
13
4
The distributive property is a \times (b + c) = a \times b + a \times c.
So, 4 \times (2 + 9) = 8 + 36 = 44
## Natural numbers FAQs
Is \bf{0} a natural number?
Are all positive numbers natural numbers?
No, only positive whole numbers are natural numbers. Positive fractions and decimals are not natural numbers.
Are all integers natural numbers?
No, only the non-negative integers starting with 1 are natural numbers.
What are cardinal numbers?
Cardinal numbers are natural numbers used for counting. They are countable numbers.
What are ordinal numbers?
Ordinal numbers are natural numbers used for ordering objects such as 1 st, 2 nd, 3 rd, etc…
Are rational numbers natural numbers?
Only the positive integers are rational numbers, not including 0. Negative numbers are not natural numbers, and fractions and decimals are not natural numbers. Natural numbers are a subset of rational numbers.
Why is set theory important?
Set theory serves as a foundation for everything that is done in mathematics because it builds concepts of numbers.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. |
# Introduction to Function Notation
Want some basic practice with functions first? Introduction to Functions
A function is a rule that takes an input, does something to it, and gives a unique corresponding output.
There is a special notation (called ‘function notation’) that is used to represent this situation:
if the function name is $\,f\,,$
and the input name is $\,x\,,$
then the unique corresponding output is called $\,f(x)\,.$
The notation ‘$f(x)\,$’ is read aloud as: ‘ $f$ of $\,x\,$ ’.
So, what exactly is $\,f(x)\,$? Answer: It is the output from the function $\,f\,$ when the input is $\,x\,\,.$
As a second example, what exactly is $\,g(t)\,$? Answer: It is the output from the function $\,g\,$ when the input is $\,t\,.$
Note that $\,f\,$ and $\,f(x)\,$ are different: $f\,$ is the name of the function (the ‘rule’); $f(x)\,$ is the output from this rule when the input is $\,x\,.$
It is often helpful to think of a function as a ‘box’. You drop an input in the top, something happens to the input inside the box, and the output drops out the bottom. The box is labeled with the name of the function.
The letter $\,f\,$ is commonly used as the name of a function, since it is the first letter in the word function.
A function box:
input: $x$
name of function: $f$
output: $f(x)$
If $\,\,x\,\,$ is dropped in the top of the box labeled $\,f\,,$ then $\,f(x)\,$ comes out the bottom.
If $\,\,t\,\,$ is dropped in the top of the box labeled $\,g\,,$ then $\,g(t)\,$ comes out the bottom.
If $\,x+2t\,$ is dropped in the top of the box labeled $\,h\,,$ then $\,h(x+2t)\,$ (read as ‘ $h$ of $x+2t$ ’) comes out the bottom.
The equation ‘$\,f(x) = x + 2\,$’ is function notation that describes the following situation: a function named $\,f\,$ acts on an input (here, indicated by $\,x\,$), and gives the output $\,f(x)\,,$ which is equal to $\,x+2\,.$ Thus, ‘$\, f(x) = x + 2\,$’ describes the ‘ add $2$ ’ function.
This same function $f$ could also be described by any of these:
$$\begin{gather} \cssId{s42}{f(t) = t + 2}\cr\cr \cssId{s43}{f(w) = w + 2}\cr\cr \cssId{s44}{f(u) = u + 2} \end{gather}$$
The variable used locally to give a name to the input is called a dummy variable.
In the equation $\,f(t) = t + 2\,,$
the dummy variable is $\,t\,.$
In the equation $\,f(w) = w + 2\,,$
the dummy variable is $\,w\,.$
In the equation $\,f(u) = u + 2\,,$
the dummy variable is $\,u\,.$
## Examples
Question: What does the function notation $\,g(7)\,$ represent?
Answer: the output from the function $\,g\,$ when the input is $\,7$
Question: Suppose $\,f(x) = x + 2\,.$ What is $\,f(3)\,$?
Answer: $\,f(3) = 3 + 2 = 5$
Question: Suppose $\,f(x) = x + 2\,.$ What is $\,f(x+5)\,$?
Answer: $\,\cssId{s63}{f(x+5)} \cssId{s64}{= (x+5) + 2} \cssId{s65}{= x + 7}$
Question: Suppose $\,g(x) = 3x + 2\,.$ Describe, in words, what the function $\,g\,$ does.
Answer: the function $\,g\,$ takes an input, multiplies by $\,3\,,$ and then adds $\,2$
Question: Write function notation for the ‘multiply by $\,5\,$’ rule. Use $\,f\,$ as the name of the function, and $\,t\,$ as the name for the input.
Answer: $\,f(t) = 5t$
Question: Write the function rule $\,f(x) = 5x - 2\,$ using the dummy variable $\,t\,.$
Answer: $\,f(t) = 5t - 2$ |
Finding the Equation of a Tangent Line
Using the First Derivative Certain problems in Calculus I call for using the first derivative to find the equation of the tangent line to a curve at a specific point. The following diagram illustrates these problems.
There are certain things you must remember from College Algebra (or similar classes) when solving for the equation of a tangent line. Recall : • A Tangent Line is a line which locally touches a curve at one and only one point. • • The slope-intercept formula for a line is y = mx + b, where m is the slope of the line and b is the y-intercept. The point-slope formula for a line is y – y1 = m (x – x1). This formula uses a point on the line, denoted by (x1, y1), and the slope of the line, denoted by m, to calculate the slope-intercept formula for the line.
Also, there is some information from Calculus you must use: Recall: • The first derivative is an equation for the slope of a tangent line to a curve at an indicated point. • The first derivative may be found using:
A) The definition of a derivative : f (x + h ) − f ( x ) h
lim
h →0
B) Methods already known to you for derivation, such as: • Power Rule • Product Rule • Quotient Rule • Chain Rule (For a complete list and description of these rules see your text)
which is 4. All that you need now is a point on the tangent line to be able to formulate the equation. 4). Therefore. Having a graph is helpful when trying to visualize the tangent line. the slope of the tangent line is f '(2). f(x) = x2. the derivative of f(x). Using the power rule yields the following: f(x) = x2 f '(x) = 2x Therefore. Consider the following problem: Find the equation of the line tangent to f ( x ) = x 2 at x = 2. you know the x coordinate for one of the points on the tangent line. You know that the tangent line shares at least one point with the original equation. for the point shared by f(x) and the line tangent to f(x) at x = 2. you know the slope of the tangent line. Now you have a point on the tangent line and the slope of the tangent line from step (1).With these formulas and definitions in mind you can find the equation of a tangent line. consider the following graph of the problem: 8 6 4 2 -3 -2 -1 1 2 3 The equation for the slope of the tangent line to f(x) = x2 is f '(x). By plugging the x coordinate of the shared point into the original equation you have: f(x) = (2) =4 2 or y=4 (3) Therefore. f '(2) = 2(2) =4 (1) (2) Now . at x = 2. (2. 2 . you have found the coordinates. Since the line you are looking for is tangent to f(x) = x2 at x = 2.
Revised Spring 2004 Created by Jay Whitehead. KY: D.C. 8 6 4 2 -3 -2 -1 1 2 3 (5) The tangent line appears to have a slope of 4 and a y-intercept at –4. R. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line. R.The only step left is to use the point (2. Health and Co. Therefore. in the point-slope formula for a line. Plug x value into f(x) to find the y coordinate of the tangent point. 4. 5) Graph your results to see if they are reasonable.San Marcos 3 . Bibliography Larson.). check with the graph to see if your answer is reasonable. Therefore: y − y1 = m( x − x1 ) y − 4 = 4( x − 2 ) y − 4 = 4x − 8 y = 4x − 4 (4) This is the equation for the tangent line. Calculus: With Analytical Geometry (5th ed. (1994). Here is a summary of the steps you use to find the equation of a tangent line to a curve at an indicated point: 1) 2) 3) 4) Find the first derivative of f(x). 4) and slope. the line y = 4x – 4 is tangent to f(x) = x2 at x = 2.P. March 1999 Student Learning Assistance Center (SLAC) Texas State University. therefore the answer is quite reasonable. Lexington. Finally. and Hostetler.E. Plug x value of the indicated point into f '(x) to find the slope at x. |
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## Related Articles
• RD Sharma Class 12 Solutions for Maths
# Class 12 RD Sharma Solutions- Chapter 22 Differential Equations – Exercise 22.1 | Set 1
### Question 1.
Solution:
We have,
Order of function:
The Highest order of derivative of function is 3 i.e.,
So, the order of derivative is equal to 3.
Degree of function:
As the power of the highest order derivative of function is 1 (i.e., power of is 1)
So, degree of function is 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 2.
Solution:
We have,
Order of function:
As the highest order of derivative of function is 2.(i.e.,)
So, Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 1(i.e., power of is 1)
So, Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is linear.
### Question 3.
Solution:
We have,
Order of function:
As the highest order of derivative of function is 1 (i.e., )
So, Order of the function is equal to 1.
Degree of function
As the power of the highest order derivative of the function is 3 (i.e., power of dy/dx is 3)
So, the degree of the function is equal to 3.
Linear or Non-linear:
The given equation is non-linear.
### Question 4.
Solution:
We have,
On squaring both side, we get
On cubing both side, we get
Order of function:
As the highest order of derivative of function is 2 (i.e.,
So, Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 2. (i.e., power of is 2)
So, the Degree of the function is equal to 2.
Linear or Non-linear:
The given equation is non-linear.
### Question 5.
Solution:
We have,
Order of function:
As the highest order of derivative of function is 2
So, Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of function is 1 (i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 6.
Solution:
We have,
On cubing both side, we get
On squaring both side, we get
Order of function:
As the highest order of derivative of function is 2 (i.e., )
So, the Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 2(i.e., power of is 2)
So, the Degree of the function is equal to 2.
Linear or Non-linear:
The given equation is non-linear.
### Question 7.
Solution:
We have,
On squaring both side, we get
Order of function:
The highest order of derivative of function is 4 (i.e., )
So, the order of the derivative is equal to 4.
Degree of function:
As the power of the highest order derivative of the function is 2 (i.e., power of is 2)
So, the degree of function is 2.
Linear or Non-linear:
The given equation is non-linear.
### Question 8:
Solution:
We have,
On squaring both side, we have
Order of function:
As the highest order of derivative of function is 1.
So, the Order of the function is equal to 1.
Degree of function:
As the power of the highest order derivative of the function is 1.
So, the degree of the function is equal to 1.
Linear or Non-linear:
The given equation is linear.
### Question 9:
Solution:
We have,
Order of function:
As the highest order of derivative of function is 2 (i.e.,)
So, order of derivative is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 1 (i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is linear.
### Question 10:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 2.
So, the Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 1 (i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 11:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 2
So, the Order of the function is equal to 2.
Degree of function:
As the power of the highest order derivative of the function is 3. (i.e., power of is 3)
So, the degree of the function is equal to 3.
Linear or Non-linear:
The given equation is non-linear.
### Question 12:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 3
So, the Order of the function is equal to 3.
Degree of function:
As the power of the highest order derivative of the function is 1.(i.e., power of is 1)
So, the Degree of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
### Question 13:
Solution:
We have,
Order of function:
As the highest order of derivative of the function is 1
So, the Order of the function is equal to 1.
Degree of function:
As the power of the highest order derivative of the function is 1. (i.e., power of dy/dx is 1)
So, the Order of the function is equal to 1.
Linear or Non-linear:
The given equation is non-linear.
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# CBSE Class 10 Maths Chapter 15 Probability Objective Questions
According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 14 in NCERT Class 10 Maths Textbook.
CBSE Class 10 Maths Chapter 15 Probability Objective Questions cover Unit 6 – Statistics and Probability. Probability is one of the chapters in Class 10 Maths that contains the concepts which are most likely to be used in daily life. Some of the key concepts discussed in this chapter include the difference between experimental probability and theoretical probability, how the probability of an impossible event is 0, complementary events, finding the probability of different events, and much more. Keeping in mind the recent modification of the exam pattern, CBSE students have to be more thorough with answering the Class 10 Maths Chapter 15 Probability Objective Questions and MCQs.
Students should be thorough with these CBSE Class 10 Maths Objective Questions, which will help them score well in the board exams because MCQs make up a major part of the exam questions, amounting to good scores.
## Download CBSE Class 10 Maths Chapter 15 – Probability Objective Questions Free PDF
### Sub-topics Covered in Chapter 15 – Probability
Students will explore concepts like the multiplication rule of probability, Bayes’ theorem, and the independence of events in this chapter. Find below the list of topics that are covered in the objective questions from this chapter:
15.1 Introduction to Probability (1 MCQ)
15.2 Complementary Events (4 MCQs)
15.3 Experimental Probability (4 MCQs)
15.4 Theoretical Probability (9 MCQs)
### Introduction to Probability
1. What is the probability that the minute and hour hands of a clock will form an acute angle at any given time?
A. PÂ >Â 0.5
B. PÂ =Â 0.5
C. PÂ <Â 0.5
D. P ≤ 0.25
Solution: The amount of time in a period of 12 hours when the hands will form an acute angle will be the same as that for an obtuse angle. So, you would think that the probability is 12. But if you take into account the small amounts of time when the hands are aligned (0∘), hands are at right angles, and hands are facing in opposite directions (180∘), then the probability would be slightly less than 0.5.
### Complementary Events
2. Two dice are thrown at the same time. Find the probability of getting different values on both.
A. 5/6
B. ½
C. 1/6
D. 1/36
Solution: Let E be the event of getting different values on both dice. The complementary event is getting the same value on both, for which there are 6 favourable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6). Thus, P (not E) = 1/6
Thus, P (E) = 1 – P (not E) = 1 – (1/6) = 5/6.
3. If P (A) and P (not A) are complementary events, and P (A) = 0.15, then P (not A) =?
A. 0.35
B. Cannot be determined
C. 0.85
D. 0.3
Solution: Given, P (A) = 0.15
As P (A) and P (not A) are complementary events, P (A) + P (not A) = 1
P (not A) = 1 – P (A) = 1 – 0.15 = 0.85.
4. What is the probability of not picking a king if you choose randomly from a pack of 52 cards?
A. 1/13
B. 12/13
C. 51/52
D. 1/52
Solution: Since there are 4 kings in a deck of 52, the probability of drawing a king is 4/52 = 1/13.
Hence, the probability of not picking a king is 1 – 1/13 = 12/13.
5. What is the probability of not picking a face card when you draw a card at random from a pack of 52 cards?
A. 1/13
B. 4/13
C. 10/13
D. 12/13
Solution: Since there are 12 face cards in a deck of 52 cards, the probability of drawing a face card is 12/52 = 3/13
Hence, the probability of not picking a face card = 1 – 3/13 = 10/13
### Experimental Probability
6. 24 cards numbered 1, 2, 3, …., 23, 24 are put in a box and mixed thoroughly. One person draws a card from the box. The probability that the number on the card is divisible by 2 or 3, or both is
A. 5/6
B. 2/3
C. 1/3
D. 1/6
Solution: The total possible outcomes = 24
Numbers divisible by only 2 are 2, 4, 8, 10, 14, 16, 20, 22 (8 numbers) ——— (1)
Numbers divisible only by 3 are 3, 9, 15, 21 (4 numbers) ——— (2)
Numbers divisible by both 2 and 3 are 6, 12, 18, and 24 (4 numbers) ——— (3)
From (1), (2) and (3), we see that the number of favourable outcomes is 16 (i.e., 8 + 4 + 4).
We know that the probability of an event E, P (E) = number of favourable outcomes/ total number of outcomes
= 16/24
= 2/3
Hence, the probability that the number on the card is divisible by 2 or 3, or both is 2/3.
7. A bag contains 6 black, 7 red, and 2 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is black or white.
A. 8/15
B. 3/5
C. 2/3
D. 1/5
Solution: Total number of balls = 15
Number of balls that are either black or white = 8
Hence, the number of favourable outcomes of the ball drawn being black or white is 8.
We know that the probability of an event E, P (E) =number of favourable outcomes/total number of outcomes.
So, the required probability is 8/15.
8. A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing a black card which is neither a face card nor an ace.
A. 10/13
B. 9/13
C. 9/26
D. 9/52
Solution: In each suit, there are 9 cards that are not face cards or ace cards.
Hence, there will be a total of 18 cards in a deck that are black and are not face cards or ace cards.
We know that, Probability of an event E, P (E) = number of favourable outcomes/total number of outcomes.
Hence, the required probability is 18/52 = 9/26.
9. Each of the letters of the word PILOTS is on separate cards, face down on the table. If you pick a card at random, what is the probability that the letter will be a T or an L?Â
A. 1/6
B. 1/3
C. 1/2
D. 2/3
Solution: There are 6 outcomes, out of which 2 are favourable (which are getting T or L).
Probability of an event E, P (E) = number of favourable outcomes/ total number of outcomes
The required probability = 2/6 = 1/3.
### Theoretical Probability
10. A single die is rolled. The probability of getting 1 or an even number is
A. 1/6
B. 4/6
C. 5/6
D. 3/6
Solution: The favourable outcomes are 1, 2, 4 and 6.
We have 4 favourable outcomes out of the total outcomes of 6.
Thus, the required probability = 4/6.
11. A bucket contains 10 brown balls, 8 green balls, and 12 red balls, and you pick one randomly without looking. What is the probability that the ball will be brown?Â
A. 0.33
B. 0.61
C. 1/3
D. 4/15
Solution: There are a total of 10 + 8 + 12 = 30 balls, out of which 10 are brown.
Thus, the required probability is 10/30 = 1/3.
12. A number is chosen randomly among the first 100 natural numbers. Find the probability that the number chosen is prime.
A. 1/4
B. 3/10
C. 29/100
D. 27/100
Solution: There are 25 prime numbers in the set of the first 100 natural numbers, and they are as follows:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
We know that the Probability of an event is E, P (E) =number of favourable outcomes/total number of outcomes.
Hence, the required probability =Â 25/100Â =Â 1/4.
13. From a well-shuffled pack of 52 cards, a card is drawn at random; find the probability that it is a spade.
A. 1/2
B. 1/4
C. 1/51
D. 1/52
Solution: There are 13 spades in a deck of 52.
Hence, the probability of drawing a spade is 13/52 = ¼.
14. A die is thrown once; the probability of getting a composite number on the die is
A. 1/3
B. 1/2
C. 2/3
D. 1/6
Solution: The composite numbers among the numbers on a die are 4 and 6. Thus, we have 2 favourable outcomes out of a total of 6 outcomes. Hence, the required probability is 2/6 = 1/3.
15. The probability of an event of a trial
A. is greater than 1
B. 0
C. lies between 0 and 1 (both inclusive)
D. 1
Answer: (C) lies between 0 and 1 (both inclusive)
Solution: The probability of any event will lie between 0 and 1, both included
16. What is the probability of getting all heads or all tails when three coins are tossed simultaneously?
A. 3/4
B. 1/2
C. 1/4
D. 1/8
Solution: When three coins are tossed simultaneously, there are 8 possible outcomes, which are (HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT), where H represents the head, and T represents the tail.
The favourable outcomes of getting all heads or all tails are HHH and TTT.
We know that the probability of an event E, P (E) = Number of favourable outcomes /Total number of outcomes.
Hence, the required probability =Â 2/8Â =Â 1/4
17. There are 5 green, 6 black, and 7 white balls in a bag. A ball is drawn at random from the bag. Find the probability that it is not white.
A. 5/18
B. 2/13
C. 7/18
D. 11/18
Solution: Given,
Number of green balls = 5
Number of black balls = 6
Number of white balls = 7
Total number of outcomes = 5 + 6 + 7 = 18
There are 18 balls, out of which 11 are not white.
⇒ Number of favourable outcomes = 11
Probability of an event, P (E) = Number of favourable outcomes / Total number of outcomes
⇒ P (ball drawn is not white) = 11/18
Therefore, the probability that the ball drawn is not white is 11/18.
Alternate Method:
P (ball drawn is white) = 7/18
We know that,
P (ball drawn is white) + P (ball drawn is not white) = 1
Because the sum of the probability of an event and its complementary event is always 1,
⇒ P (ball drawn is not white) = 1 – P (ball drawn is white) = 1−7/18 = 11/18
18. From a set of 17 cards, numbered 1, 2, …. 17, one card is drawn. What is the probability that the number is a multiple of 3 or 7?
A. 5/17
B. 7/17
C. 8/17
D. 6/17
Solution: The total number of possible outcomes is 17
The number of favourable outcomes is 3, 6, 7, 9, 12, 14 and 15
= 7
Thus, the required probability = number of favourable outcomes/total number of outcomes
=Â 7/17.
Solving these CBSE Class 10 Maths Chapter 15 – Probability Objective Questions gives an idea to the students about the MCQs asked in the board exam from this chapter. This will help them prepare well for the board exams and score good marks.
Also, find below the extra MCQs from the CBSE Class 10 Maths Chapter 15 for the students to practise.
### CBSE Class 10 Maths Chapter 15 Extra Questions
1.  When a dice is thrown, what is the probability of getting an even number?
(a) 5/6
(b) 1
(c) 2/3
(d) 1/2
2. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, then how many blue balls are there in a bag?
(a) 5
(b) 20
(c) 10
(d) 15
3. What is the probability of getting a king of a red suit when a card is drawn from a well-shuffled deck of 52 cards?
(a) 1/13
(b) 3/26
(c) 7/52
(d) 1/26
Keep learning and stay tuned to BYJU’S for further updates on CBSE and other competitive exams. Download BYJU’S – The Learning App and subscribe to the YouTube channel to access interactive Maths and Science videos. |
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# What percent of 500 is 120?
Last updated date: 22nd Jul 2024
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Hint: In this problem, we have to find the percent of 500 which gives 120. Here we have to find the percent for the given number and the total number. We know that, we can find the number by dividing the given number by the total number and multiplying it with 100. Here we have the total number 500 and the given number 120, we can substitute in the formula for the final answer.
Complete step by step answer:
Here we have to find the percent of 35 is 14.
Here we can see that the given number is 120 and the total number is 500.
We know that,
$Percent=\dfrac{\text{Number}}{\text{Total number}}\times 100$
Here we can substitute the given values and find the percentage for it,
$\Rightarrow Percent=\dfrac{120}{500}\times 100$
We can now find the value of Percent by simplifying the above step.
We can first divide 120 and 500 and we can multiply it with 100, we get
$\Rightarrow Percent=0.24\times 100$
We can now multiply 100 on the above step, we get
$\Rightarrow Percent=24.0\%$
Here we have moved the decimal one point to the left.
Therefore, the 120 percent of 500 is $24\%$.
Note: Students make mistakes while finding the number from the given percentage. We should always remember that we can find the number by multiplying the given percentage and the total number and then divide by hundred, as we should remember that $Percent=\dfrac{\text{Number}}{\text{Total number}}\times 100$, we can also cancel the 100 and 500 and then simplify this problem, to get the same final answer. |
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# Sequences and Series
LEARNING
OBJECTIVES
## Find the nth term of a Sequence.
Find the Sum of a Sequence.
Find the Sum to Infinity of a GP.
Solve Quadratic Equation under EQUA mode.
Evaluate .
Find the Root(s) of an Equation under GRAPH mode.
Determine the Behaviour of a Sequence.
Example 4.1
A geometric series has common ratio r, and an arithmetic series has first term a
and common difference d, where a and d are non-zero. The first three terms of
the geometric series are equal to the first, sixth and tenth terms respectively of
the arithmetic series.
(i) Show that 5r2 9r + 4 = 0.
(ii) Deduce that the geometric series is convergent and find, in terms of a, the
sum to infinity.
(iii) The sum of the first n terms of the geometric series is denoted by S. Given
that a > 0, find the least value of n for which S exceeds 99% of the sum to
infinity.
28
Solution
(i)
r=
## a + (6 1)d a + (10 1)d
=
a
a + (6 1)d
If x, y an
d
1 + 9
d
a
r = 1 + 5 =
a 1 + 5 d
a
d r 1
=
r=
a
5
d z are in
y z
=
x y =r
G.P.,
r 1
1 + 9
5 r 2 = 1 + 9 r 1 5r 2 = 5 + 9 r 9
r 1
5
1 + 5
5r 2 9r + 4 = 0 (shown)
4
5
Since d 0, the three terms of GP are not the same, thus r 1.
4
Hence, r = Since r < 1, the geometric series is convergent.
5
a
And the sum to infinity =
= 5a.
4
1
5
(ii) 5r 2 9r + 4 = 0 (r 1)(5r 4) = 0 r = 1 or r =
S n > 0.99 S
(iii)
a (1 0.8n )
a
> 0.99
1 0.8
1 0.8
1 0.8n > 0.99
UNIT 4
## 0.8n < 0.01
n lg 0.8 < lg 0.01
lg 0.01
n>
lg 0.8
n > 20.6
Hence, the least value of n is 21.
## Practical Guide To GCE ALevel H2 Mathematics GC Approach
29
p1
iq
uuq
uy
0.8^
f\$,f
,1,2
00,1
kl
Select RUN.MAT
mode.
Select LIST.
Select SUM.
Select Seq.
Enter the parameters
in the following
format: Seq
(Expression, Variable,
Start, End, Increment)
is S where a = 0.8
and r = 0.8. (Here,
we assume a takes a
positive value.)
p8
eq
an = An + B.
0.8^
fl
30
0.8n.
Select SET.
1l50
ll
50 for End.
Increase the End
value if found
necessary later.
Lp
ql
Set On for
Display.
## GLM PTE LTD
Select TABL.
Press N until an
exceeds 0.99S, i.e.
0.99 4 = 3.96.
You may also
consider 0.8n falling
below 0.01.
Both approaches
arrive at the same
value of n is 21.
Example 4.2
(i)
Patrick saves \$20 on 1 January 2008. On the first day of each subsequent
month he saves \$4 more than in the previous month, so that he saves \$24
on 1 February 2008, \$28 on 1 March 2009, and so on. On what date will he
first have saved over \$5000 in total?
## Sequences and Series
(ii) Kenny puts \$20 on 1 January 2008 into a bank account which pays
compound interest at a rate of 3% per month on the last day of each month.
He puts a further \$20 into the account on the first day of each subsequent
month.
(a) How much compound interest has his original \$20 earned at the end
of 3 years?
(b) How much in total, correct to the nearest dollar, is in the account at
the end of 3 years?
## Practical Guide To GCE ALevel H2 Mathematics GC Approach
UNIT 4
(c) After how many complete months will the total in the account first
exceed \$5000?
31
Solution
(i)
T1 = 20, d = 4
n
[2(20) + 4(n 1)] > 5000 n2 + 9n 2500 > 0 n < 54.7 or n > 45.7
2
Hence, Patrick will first have saved over \$5000 in total on 1 October 2011.
pf
## Select EQUA mode.
Select F2:Polynomial.
Select Degree 2.
Enter the values of a,
b and c.
1l9l
n250
0l
Select SOLV.
q
45.7 or 54.7
## (ii) (a) The required compound interest = 20(1.03)36 20 = \$38.0 (3 s.f.)
(b) The required amount = 20(1.03) + 20(1.03) 2 + ... + 20(1.03)36
=
20(1.03) (1.03)36 1
1.03 1
= \$1303
p8
e
w
32
Select RECUR
mode.
Select TYPE.
Select
F2: an+1 = Aan + Bn
+ C.
1.03
jw+2
0kl
Enter the
recurrence
formula.
Press w for an.
Select SET.
N100
l
## Enter the start
and end value of
n as well as the
value of an.
du
Select TABL.
Keep pressing
N until the
value on the left
column reaches
36.
Record the
corresponding
value on the
right column:
1303.4.
## (c) 20(1.03) + 20(1.03) 2 + ... + 20(1.03) n > 5000
20(1.03)(1.03n 1)
> 5000
1.03 1
1.03n > 8.282
lg8.282
lg1.03
UNIT 4
n >
n > 71.5
The total in the account will first exceed \$5000 after 72 months.
## Practical Guide To GCE ALevel H2 Mathematics GC Approach
33
Keep pressing N
until the value on
the right column
first reaches 5000
or above.
Record the
corresponding
value on the right
column: 72.
Example 4.3
A sequence u1, u2, u3, is such that u1 = 1 and
un +1 = un
3n 2 + 3n + 1
, for all n 1.
n3 (n + 1)3
## (i) Use the method of mathematical induction to prove that un =
(ii) Hence find
1
.
n3
3n 2 + 3n + 1
.
3
3
n =1 n ( n + 1)
N
(iii) Give a reason why the series in part (ii) is convergent and state the sum to
infinity.
3n 2 3n + 1
.
3
3
n = 2 n ( n 1)
N
34
## GLM PTE LTD
Solution
(i)
1
3n 2 + 3n + 1
where
u
=
u
and u1 = 1 for n + .
+
n
n
1
n3
n3 (n + 1)3
1
When n = 1, L.H.S. = u1 = 1 ; R.H.S. = 2 = 1.
1
L.H.S. = R.H.S.
P1 is true.
Let Pn be the statement un =
## Assume Pk is true for some k + , i.e. uk =
To prove Pk +1 is true, i.e. uk +1 =
L.H.S. = uk +1 = uk
1
(k + 1)3
1
.
k3
3k 2 + 3k + 1
k 3 (k + 1)3
1 3k 2 + 3k + 1
3
k3
k (k + 1)3
(k + 1)3 3k 2 3k 1
k 3 (k + 1)3
k 3 + 3k 2 + 3k + 1 3k 2 3k 1
k 3 (k + 1)3
1
=
(k + 1)3
= R.H.S.
is true whenever Pk is true.
=
Pk +1
(ii)
3n 2 + 3n + 1 N
= (un un +1 )
3
3
n =1 n ( n + 1)
n =1
= u1 u2
N
+ u2 u3
UNIT 4
...
+ u N 1 u N
+ u N u N +1
= u1 u N +1
=1
## GLM PTE LTD
1
( N + 1)3
Practical Guide To GCE ALevel H2 Mathematics GC Approach
35
p1
Select RUN.MAT
mode.
ruw
Select MATH, ( .
N!f\$
1B8
## Enter the initial
value 1 and the end
value 8 (randomly
chosen).
N!!!
!z3f
s+3f
+1Nf
^3\$j
f+1k
^3l
## Enter the expression.
1-z1
N9^3
l
in part (ii) when N = 8.
fraction form, press
x.
(iii) When N ,
36
N
1
1
3n 2 + 3n + 1
3
3
( N + 1)3
( N + 1)3
n =1 n ( n + 1)
Hence, the series in part (ii) is convergent and the sum to infinity is 1.
## GLM PTE LTD
(iv)
3n 2 3n + 1 N 1 3(r + 1) 2 3(r + 1) + 1
=
when n = r + 1
3
3
(r + 1)3 r 3
n = 2 n ( n 1)
r =1
N
3(r + 1) 2 3(r + 1) + 1
r 3 (r + 1)3
r =1
N 1
3r 2 + 6r + 3 3r 3 + 1
r 3 ( r + 1)3
r =1
N 1
3r 2 + 3r + 1
3
3
r =1 r ( r + 1)
1
=1
( N 1 + 1)3
1
=1 3
N
N 1
[Apply similar
keystrokes found
in part (ii).]
UNIT 4
## Practical Guide To GCE ALevel H2 Mathematics GC Approach
37
Example 4.4
The diagram shows the graph of y = 2 x e 2 . The two roots of the equation are
denoted by and . where < .
(i)
## Find the values of and , each correct to 3 decimal places.
A sequence of real numbers x1, x2, x3, satisfies the recurrence relation
1 xn
xn+1 = e 2
2
for n 1.
(ii) Prove algebraically that, if the sequence converges, then it converges to
either or .
(iii) Use a calculator to determine the behaviours of the sequence for each of
the cases x1= 0, x1= 3, x1= 6.
(iv) By considering xn+1 xn, prove that
xn +1 < xn if < xn < ,
xn +1 > xn if xn < or xn > ,
(v) State briefly how the results in part (iv) relate to the behaviours determined in
part (iii).
38
Solution
p5
2f-L
GjfM
2kl
the graph as Y1.
Select DRAW.
yq
## Find the first root .
Find the second root .
\$
UNIT 4
39
(i)
1 xn
(ii) xn +1 = e 2
2
xn
2 xn +1 = e 2
xn
2 xn +1 e 2 = 0
L
2L e 2 = 0
x
## Since and are the roots of 2 x e 2 = 0, hence xn converges to or if the
sequence converges.
(iii)
p8
## Enter RECUR mode.
Select TYPE.
Select F2:
an+1 = Aan + Bn + C .
0.5L
GjwM
2kl
y
w
1l20
l0l
du
40
## Enter the recurrence
equation.
w corresponds to an.
Enter Table Setting.
Change a0 to a1.
Apply the settings
as shown. Use a
larger value for
End if later found
insufficient.
Use a1 = 0 for the
first case.
Select TABL.
## GLM PTE LTD
While pressing N,
observe the change
in an+1.
It is noted that the
sequence converges
to 0.7148 which is
when x1= 0.
[Set a1= 3 and
apply similar
key strokes.]
UNIT 4
## It is noted that the
sequence converges
to 0.7148 also which
is when x1= 3.
41
apply similar
key strokes.]
## It is noted that the
sequence diverges
when x1= 6 as shown
by the 'ERROR'
message.
xn
1 x
1 xn
1
(iv) xn +1 xn = e 2n xn = e 2 xn = (2 xn e 2 )
2
2
2
xn
xn
1
2
If < xn < , 2xn e > 0 (2 xn e 2 ) < 0 xn +1 xn < 0 xn +1 < xn .
2
xn
xn
1
If xn < or xn > , 2xn e 2 < 0 (2 xn e 2 ) > 0 xn +1 xn > 0 xn +1 > xn .
2
(v)
42
1 x
1
For x1 = 0 where x1 < xn +1 = e 2n xn < e 2 = , hence x1 < x2 < x3 < ... < .
2
2
1 x 1
For x1 = 3 where < x1 < xn > xn +1 = e 2n > e 2 = , hence x1 > x2 > x3 > ... > .
2
2
If x1 = 6 where x1 > xn < xn +1 , hence < x1 < x2 < x3 ... .
Checklist
TECHNIQUES
4.1
4.2
mode.
Evaluate .
4.3
4.4
UNIT 4
GRAPH mode.
43 |
## 1. Introduction and the three basic trigonometric functions
Triangles are basic geometric shapes comprising three straight lines, each of which is joined to the two others at its ends. A triangle has three included angles (angles defined by two of the sides) and together these angles sum to 180°. We will start with a special sort of triangle, where one of the included angles is a right-angle (90°). The lengths of the sides of a right-angled triangle can be calculated using Pythagoras' Theorem, and this is described graphically in the Appendix to this document.
Here, the main aim is to introduce a set of relationships between the sides and angles of triangles which form the basis of trigonometry. Most people encounter trigonometric functions in the context of geometry. However, some trigonometric functions are used in other applications, including the construction of regular cycles in models of physical and biological systems.
The three basic trigonometric functions relate an angle to the ratio between pairs of sides of a right-angled triangle. In this diagram, the angle α (Greek 'alpha') is at the junction between the hypotenuse, the longest side of the right-angled triangle, and a side designated the 'adjacent'. The side opposite angle α is designated the 'opposite'. Three ratios between pairs of sides are the sine, cosine and tangent:
You will notice that for the third angle of the triangle, which we can designate β:
β = 90 – α
sine β = cosine α
cosine β = sine α
tangent β = 1/tangent α
Given a value for one of the two non-right angles in a right-angled triangle, and the length of any side, it is possible to calculate the length of either of the other two sides. Conversely, given the length of any two sides, it is possible to calculate the value of either non-right angle from one of the trigonometric functions shown in the figure.
The three functions are usually written in their abbreviated forms, which are 'sin', 'cos' and 'tan'. These terms are used in spreadsheet calculations, for instance as SIN(A12), where the value contained in cell A12 (in brackets) is termed the 'argument' of the function. |
# Eureka Math Grade 1 Module 6 Mid Module Assessment Answer Key
## Engage NY Eureka Math 1st Grade Module 6 Mid Module Assessment Answer Key
Question 1.
Use the RDW process to solve the following problems. Write your statement on the line.
a. Lucy has 5 pencils. Kim has 7 pencils. How many more pencils does Kim have than Lucy?
__________________________
Explanation:
Kim have 2 pencils more than lucy.
b. Ben has 18 pencils. Anton has 9 pencils. How many fewer pencils does Anton have than Ben?
__________________________
Explanation:
Anton have 9 pencils fewer than Ben.
c. Julio has 5 more pencils than Fran. Fran has 6 pencils. How many pencils does Julio have?
__________________________
Explanation:
Julio have 11 pencils.
Question 2.
Fill in the missing numbers in the sequence.
a. 97, 98, ____, ____, ____, _____
97, 98, 99, 100, 101, 102.
b. 116, 117, ____, ____, ____
116, 117, 118, 119, 120.
c. ____, 14, ____, ____, 11, ____
15, 14, 13, 12, 11, 10.
d. 112, 111, ____, 109, ____, ____
112, 111, 110, 109, 108, 107.
Question 3.
Write the number as tens and ones in the place value chart, or use the place value chart to write the number.
a. 82
Explanation:
The number is 8 tens 2 ones.
b. 99
Explanation:
The number is 9 tens 9 ones.
c. _______
Explanation:
The number is 9 tens 6 ones.
d. ______
Explanation:
The number is 10 tens 5 ones.
Question 4.
Match the equal amounts.
Explanation:
I matched the numbers onthe left with their equal amounts on the right.
Question 5.
Use <, =, or > to compare the pairs of numbers.
a. 69 79
69 79
b. 15 50
15 50
c. 99 101
99 101
d. 110 108
110 108
e. 61 5 tens 11 ones
61 5 tens 11 ones
Question 6.
Ben thinks 92 ones is greater than 9 tens 2 ones. Is he correct? Explain your thinking using words, pictures, or numbers. Draw and write about tens and ones to explain your thinking.
No, Ben is not correct.
Explanation:
92 ones is the same as 9 tens 2 ones.
90 ones is 9 tens.So, 90+2 is the same as 9 tens 2 ones.
10+10+10+10+10+10+10+10+10=90
92=92
Therefore 92 ones is same as 9 tens 2 ones.
Question 7.
Find the mystery numbers. Explain how you know the answers.
a. 10 more than 90 is ______.
Explanation:
10 more than 90 is 100.
b. 10 less than 90 is _______.
Explanation:
10 less than 90 is 80.
c. 1 more than 90 is _______.
Explanation:
1 more than 90 is 91.
d. 1 less than 90 is ________.
Explanation:
1 less than 90 is 89.
Question 8.
Solve for each unknown number. Use the space provided to show your work.
a. 80 + 6 = _____
Explanation:
The sum of 80 and 6 is 86.
b. 20 + _____ = 80
Explanation:
The sum of 20 and 60 is 80.
c. 7 tens – _______________ = 4 tens
Explanation:
If we subtract 3 tens from 7 tens we get 4 tens
7 tens – 3 tens = 4 tens.
d. 90 – 40 = _____
Explanation:
If we subtract 4 tens from 9 tens we get 5 tens.
e. 68 + 7 = _____
Explanation:
The sum of 68 and 7 is 75.
f. 51 + 20 = _____
Explanation:
The sum of 51 and 20 is 71.
g. 46 + 31 = _____ |
# A new number system: Remainder numbers
Way back in elementary school, we used remainders in division problems. For example, the equation $5\div2=2$ r $1$ means "$5$ divided by $2$ is $2$ with a remainder of $1$." It's commonly understood that, in this context, $2$ r $1$ is a pair of numbers, with $2$ being the quotient and $1$ being the remainder. However, the way the equation is written makes it look like $2$ r $1$ is one number. This annoyed me for the longest time, but now, I've been able to come up with a new number system that actually makes sense of quantities like $2$ r $1$ as numbers in their own right. I call them, "remainder numbers."
Remainder Numbers 1.0: A prototype
A remainder number is of the form $a_1$ r $a_2$, where $a_1$ and $a_2$ are both integers. Addition and subtraction of remainder numbers are done component-wise, and multiplication and division are defined as follows:
For any three integers $a$, $b_1$, and $b_2$: $a\times(b_1$ r $b_2)=ab_1+b_2$.
For any two integers $a$ and $b$: $a\div b=\lfloor\frac{a}{b}\rfloor$ r $(a$ mod $b)$.
(In this convention, fractions represent standard division, and $\div$ represents remainder number division. Also, $a$ mod $b$ is defined to be between $0$ and $b$, including $0$ but not $b$, allowing $a$ mod $b$ to be negative when $b$ is negative.)
Let's look at some examples of multiplication and division:
$5\div2=\lfloor\frac52\rfloor$ r $(5$ mod $2)=2$ r $1$
$2\times(2$ r $1)=2\cdot2+1=5$
$5\div-3=\lfloor-\frac53\rfloor$ r $(5$ mod $-3)=-2$ r $-1$
$-3\times(-2$ r $-1)=(-3\cdot-2)-1=5$
As you can see, multiplication "undoes" division, which is a very nice property to have. Unfortunately, this number system has a glaring problem: multiplication and division aren't closed over the remainder numbers. Let's fix that.
Remainder Numbers 2.0: Improving Multiplication and Division
Fixing multiplication is pretty simple:
$(a_1$ r $a_2)\times(b_1$ r $b_2)=(a_1$ r $a_2)\times b_1+b_2=a_1b_1+a_2+b_2$
This derivation isn't exactly rigorous, but it's perfectly consistent with what multiplication represents in the remainder numbers, and it's a pretty reasonable definition. To multiply two remainder numbers, simply multiply the integer parts and add the remainders.
Fixing division is more difficult, so I'll build up a comprehensive definition case by case.
The first case is $a\div(b_1$ r $b_2)$. Since I want multiplication to undo division, this is essentially asking, "What number, when multiplied by $b_1$ r $b_2$, results in $a$?" Because of the way multiplication is defined, this is also equivalent to asking, "What number, after multiplying by $b_1$ and adding $b_2$, results in $a$?" This effectively reduces the problem to $(a-b_2)\div b_1$, which is already defined.
Since the divisor's remainder can easily be removed, the only real case left to consider is when the dividend has a non-zero remainder: $(a_1$ r $a_2)\div b$. Unfortunately, this reveals another glaring problem, which is that multiplication only returns integers.
Remainder Numbers 3.0: Final Version
To make division closed, I'll have to drastically update the definition of remainder numbers. From now on, a remainder number can be defined recursively as $a_1$ r $a_2$, where $a_1$ is an integer and $a_2$ is a remainder number. Alternatively, remainder numbers can be embedded as sequences of integers. Basically, think of a remainder number as a decimal expansion, but without a specified number base. Addition and subtraction are still done component-wise, and the definitions of multiplication and division are extended as follows:
Recursive notation: $(a_1$ r $a_2$ r $a_3)\times(b_1$ r $b_2$ r $b_3)=(a_1b_1+a_2+b_2)$ r $(a_3+b_3)$, where $a_3$ and $b_3$ are remainder numbers.
Sequence notation: $(a_1,a_2,a_3,a_4,\ldots)\times(b_1,b_2,b_3,b_4,\ldots)=(a_1b_1+a_2+b_2,a_3+b_3,a_4+b_4,\ldots)$
I'll skip a few steps in deriving the division formula, but here it is:
Recursive notation: $(a_1$ r $a_2)\div(b_1$ r $b_2$ r $b_3)=\lfloor\frac{a_1-b_2}{b_1}\rfloor$ r $[(a_1-b_2)$ mod $b_1]$ r $(a_2-b_3)$
Sequence notation:
$(a_1,a_2,a_3,\ldots)\div(b_1,b_2,b_3,b_4,\ldots)=(\lfloor\frac{a_1-b_2}{b_1}\rfloor,(a_1-b_2)$ mod $b_1,a_2-b_3,a_3-b_4,\ldots)$
Analysis and Final Thoughts
Admittedly, this number system looks incredibly ugly. Multiplication isn't associative, it doesn't distribute over addition, it's not one-to-one, and it doesn't even have an identity element. Also, the number system has zero divisors. (But hey, at least multiplication undoes division and is commutative, so I guess it has that going for it.) So, yeah, it's pretty clear why we use rational numbers instead of remainder numbers. Having said that, though, do you see any interesting properties or potential applications of this number system, or is it just the useless mess that it appears to be? Also, if you can find a way to modify this number system to make it more mathematically pleasing, that would be nice.
• Note representing an integer $n$ by $\{ n \bmod p^k,p \text{ prime }, k \ge 1\}$ is the idea behind the adele ring $\mathbb{A}_\mathbb{Q}$ which is useful to observe the unique factorization of integers and the local-global principle. – reuns Sep 4 '17 at 0:44
• Multiplication undoes division but not conversely: $((1\text{ r }1)\times1)\div1=2\div1=2$. – stewbasic Sep 4 '17 at 1:06
• This is going to take a lot of effort to rewrite as something understandable. I've read it several times and don't understand what you mean. Also there is some weird stuff like "[multiplication] isn't one-to-one". Can you just express your v3 in terms of how to multiply and divide two elements of some appropriate set? – rschwieb Sep 4 '17 at 2:28
• It would also help if there was some motivating construction. Right now it's just a mess of stuff that looks possibly made up, it is hard to see how (or even if) it works as advertised. – rschwieb Sep 4 '17 at 2:32
• When I said that multiplication isn't one-to-one, I meant that there are instances where $a\times b=a\times c$, but $b\neq c$. This is an instance of multiplication by $a$ not being one-to-one. For example, $1\times(1$ r $1)=1\times2=2$. It's an important thing to point out, as it implies that multiplication can't be undone by division, as stewbasic pointed out. – Tyler Borgard Sep 4 '17 at 2:45
According to me we can't say this is a new number system. We are seeing $\mathbb Z$ just from a different point, you are representing the integers (not a new number) on basis of three parameters divisor, quotient and remainder and making the known operations from that point of view. Formally, we can say a number new if it's totally different from the previous number systems(if you consider boolean algebra from abstract point of view) or a number system that contains a operation which was not in previous one(subtraction can be done in $\mathbb Z$ but not in $\mathbb N$) and contains the previous number system as a isomorphic copy. For this reason we can say $\mathbb C$ is a new number system than $\mathbb R$, but if we consider the elements of $\mathbb R$ in decimal representation rather than normal format($1.414....$ in place of $\sqrt 2$) and perform operations, it's not a new system, but new point of view of reals.......However if some system can allow division by zero then that's will be new.
• I'm not sure you understand how remainder numbers work. Every sequence of integers represents a unique remainder number, whereas every digit in the decimal expansion of a real number is limited to one of finitely many values (depending on the base). For example, $(1,2,3,4,\ldots)$ is a remainder number, but it can't be the decimal expansion of any real number regardless of base since the integer sequence is unbounded. Also, $\mathbb{Z}$ is a proper subset of the remainder numbers, so it's not just "seeing $\mathbb{Z}$ from a different point." – Tyler Borgard Sep 5 '17 at 16:56 |
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• Level: GCSE
• Subject: Maths
• Word count: 1668
# Investigate the relationship between the T-Total and the T-Number. Use grids of different sizes. Translate the T-Shapes to the different positions. Investigate
Extracts from this document...
Introduction
09th January 2006
GCSE Mathematics Coursework
T-Totals
• Investigate the relationship between the T-Total and the T-Number.
• Use grids of different sizes. Translate the T-Shapes to the different positions. Investigate relationships between the T-Total, the T-Number and the grid sizes.
• Use grids of different sizes again. Try other transformations and combinations of transformations. Investigate the relationships between the T-Total, the T-Number, the grid size and the transformations.
T-Total Grid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81
T-Numbers (T) 20 21 22 23 24 T-Total 37 42 47 52 57
+5 +5 +5 +5
5 x 20 = 100
100 – 37 = 63
The difference between the T-Totals is 5 this makes the first part of the formula: 5T
To work out the second part takes the T-Number and multiplies by the difference.
#### 5T - 63
The 50th Term is:
Middle
5 x 16 = 80
80 – 31 = 49
## 5T – 49
The 50th Term is:
5 x 50 = 250
250 - 49 = 201
To Test:
## T Number 40
5 x 40 = 200
200 – 49 = 151
T-Total Grid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
T-Number 14 15 16 17 18 T-Total 28 33 38 43 48
+5 +5 +5 +5
5 x 14 = 70
70 – 28 = 42
## 5T – 42
The 50th Term is:
5 x 50 = 250
250 - 42 = 208
To Test:
## T Number 30
5 x 30 = 150
150 – 42 = 108
### Relationships and Master Formula Working Out
Formulae’s
9 x 9 Grids: 5T - 63
8 x 8 Grids: 5T - 56
7 x 7 Grids: 5T – 49
6 x 6 Grids: 5T – 42
• Smaller the Grid Size the less you have to take away from the T-Total
• The Larger the grid size the more you have to take away from the T-Total
• All The Formulas have gaps of 7
(Ex. 63 – 56 = 7. 56 – 49 = 7. 49 – 42 = 7.)
I found out that the master formula is 5T-(7xG)
I got the 5 by looking at the differences of the number.
The T stands for the T-Number (which is always 5; as there are 5 squares in the T; and it is current throughout.)
Conclusion
## T-Number 30
5 x 30 = 150
150 – -63 = 213
8 x 8 Grids: Transformation
T-Number 11 12 13 14 15 T-Total 48 53 58 63 68
5 x 11 = 55
55 – 48 = 7
5T - 7
The 50th Term is:
5 x 50 = 250
250 – 7 = 243
To Test:
## T-Number 30
5 x 30 = 150
150 – 7 = 143
7 x 7 Grids: Transformation
T-Number 8 9 10 11 12 T-Total 47 52 57 62 67
5 x 8 = 40
40 - 47 = -7
5T - -7
The 50th Term is:
5 x 50 = 250
250 – -7 = 257
To Test:
## T-Number 30
5 x 30 = 150
150 – -7 = 157
7 x 7 Grids: Transformation
T-Number 2 3 4 5 6 T-Total 59 64 69 74 79
5 x 2 = 10
10 - 59 = -49
5T - -49
The 50th Term is:
5 x 50 = 250
250 – -49 = 299
To Test:
## T-Number 30
5 x 30 = 150
150 – -49 = 199
7 x 7 Grids: Transformation
T-Number 2 3 4 5 6 T-Total 59 64 69 74 79
5 x 2 = 10
10 - 59 = -49
5T - -49
The 50th Term is:
5 x 50 = 250
250 – -49 = 299
To Test:
## T-Number 30
5 x 30 = 150
150 – -49 = 199
6 x 6 Grids: Transformation
T-Number 7 8 9 10 11 T-Total 42 47 52 57 62
5 x 7 = 35
35 - 42 = -7
5T - -7
The 50th Term is:
5 x 50 = 250
250 – -7 = 257
To Test:
## T-Number 30
5 x 30 = 150
150 – -7 = 157
6 x 6 Grids: Transformation
T-Number 2 3 4 5 6 T-Total 52 57 62 67 72
5 x 2 = 10
10 - 52 = -42
5T - -42
The 50th Term is:
5 x 50 = 250
250 – -42 = 292
To Test:
## T-Number 30
5 x 30 = 150
150 – -42 = 192
6 x 6 Grids: Transformation
T-Number 9 10 11 12 13 T-Total 38 43 48 53 58
5 x 9 = 45
45 - 38 = 7
5T - 7
The 50th Term is:
5 x 50 = 250
250 – 7 = 243
To Test:
## T-Number 30
5 x 30 = 150
150 – 7 = 143
##### Relationships
This student written piece of work is one of many that can be found in our GCSE T-Total section.
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# Related GCSE T-Total essays
1. ## T Total and T Number Coursework
Vector T-Total of New Shape Change In T-Totals from original shape to new one. (0,0) (0,-1) (0,-2) (0,1) (0,2) (1,0) (2,0) (-1,0) (-2,0) 187 232 277 142 97 192 197 182 177 0 +45y +90y -45y -90y +5x +10x -5x -10x After looking at this table I have found the Formula for the 9x9 grid transformations.
2. ## Maths GCSE Coursework &amp;#150; T-Total
7 5 35 5 x 7 6 42 6 x 7 7 49 7 x 7 8 56 8 x 7 9 63 9 x 7 10 70 10 x 7 11 77 11 x 7 From this a grid of a T-Number based on the relationships between x (T-Number)
1. ## Objectives Investigate the relationship between ...
n+n+8+n+15+n+16+n+17 =n+n+n+n+n+8+15+16+17=5n+56 Check 5n+56 Substitute the T-number into 'n' 5x23+56 = 171 As you can see this formula will find the T-total of any T-shape rotated by180� So let's have a recap of all the algebraic formulas we have uncovered so far, and what they solve: Formula(8x8)
2. ## T-shapes. In this project we have found out many ways in which to ...
This is 9*7=63. The nine in this comes from the size of the grid this one been nine. If the grid size were 10 by 10 then it would be 10*7. At the end of this piece of coursework when we but all the formulas together we realise that the number we minus or plus by is divisible b y seven.
1. ## T totals. In this investigation I aim to find out relationships between grid sizes ...
EXPLAIN WHAT THE LETTERS STAND FOR. Preliminary work To begin with, we shall use a 9x9 grid and keep v constant (highlighted); 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
2. ## In this investigation Im going to find out relationships between the grid sizes and ...
- 19 + 34 - 18 + 34 - 17 = 107 Thus proving this equation can be used to find the T-Total (t) by substituting n for the given T-Number. The equation can be simplified more: t = n + n - 9 + n - 19 + n
1. ## T- total T -number coursework
The formula starts with 5* the t-number this is because there is a rise in the t-total by 5 for every t-number. I then -63 which do by working out the difference between the t-number and another number in the t-shape.
2. ## My aim is to see if theres a relation between T total and ...
8 by 8 grid of 90� anticlockwise: Would be the same as: T + 6 T-2 T -10 T-1 T T + 6 + T-2 + T - 10 + T -1 + T = 5T - 7 I'll test this using a random position from the grid: 27 19
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# Chapter 5: Exponential and Logarithmic Functions - PowerPoint PPT Presentation
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Chapter 5: Exponential and Logarithmic Functions. 5.1 Inverse Functions 5.2 Exponential Functions 5.3 Logarithms and Their Properties 5.4 Logarithmic Functions 5.5 Exponential and Logarithmic Equations and Inequalities
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Chapter 5: Exponential and Logarithmic Functions
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### Chapter 5: Exponential and Logarithmic Functions
5.1 Inverse Functions
5.2 Exponential Functions
5.3 Logarithms and Their Properties
5.4 Logarithmic Functions
5.5 Exponential and Logarithmic Equations and Inequalities
5.6 Further Applications and Modeling with Exponential and Logarithmic Functions
### 5.1 Inverse Functions
Example
Also, f[g(12)] = 12. For these functions, it can be
shown that
for any value of x. These functions are inverse functions
of each other.
### 5.1 One-to-One Functions
A function f is a one-to-onefunction if, for elements a and b from the domain of f,
a b implies f(a) f(b).
• Only functions that are one-to-one have inverses.
### 5.1 One-to-One Functions
Example Decide whether the function is one-to-one.
(a) (b)
Solution
(a) For this function, two different x-values produce two different y-values.
(b)If we choose a = 3 and b = –3, then 3 –3, but
### 5.1 The Horizontal Line Test
If every horizontal line intersects the graph of a function at no more than one point, then the function is one-to-one.
Example Use the horizontal line test to determine
whether the graphs are graphs of one-to-one functions.
(a) (b)
One-to-one
Not one-to-one
### 5.1 Inverse Functions
Let f be a one-to-one function. Then, g is the inversefunction of f and f is the inverse of g if
Example
are inverse functions of each other.
### 5.1 Finding an Equation for the Inverse Function
• Notation for the inverse function f-1 is read
“f-inverse”
Finding the Equation of the Inverse of y = f(x)
1. Interchange x and y.
2. Solve for y.
3. Replace y with f-1(x).
Any restrictions on x and y should be considered.
### 5.1 Example of Finding f-1(x)
ExampleFind the inverse, if it exists, of
Solution
Write f(x) = y.
Interchange x and y.
Solve for y.
Replace y with f-1(x).
### 5.1 The Graph of f-1(x)
• f and f-1(x) are inverse functions, and f(a) = b for real numbers a and b. Then f-1(b) = a.
• If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f-1.
If a function is one-to-one, the graph of its inverse f -1(x) is a reflection of the graph of f across the line y = x.
### 5.1 Finding the Inverse of a Function with a Restricted Domain
ExampleLet
SolutionNotice that the domain of f is restricted
to [–5,), and its range is [0, ). It is one-to-one and
thus has an inverse.
The range of f is the domain of f-1, so its inverse is
### 5.1 Important Facts About Inverses
• If f is one-to-one, then f-1 exists.
• The domain of f is the range of f-1, and the range of f is the domain of f-1.
• If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f-1, so the graphs of f and f-1 are reflections of each other across the line y = x.
### 5.1 Application of Inverse Functions
Example Use the one-to-one function f(x) = 3x + 1 and the
numerical values in the table to code the message BE VERY CAREFUL.
A1F6K 11P 16U21
B 2G 7L 12Q 17V22
C 3H 8M 13R 18W23
D4I 9N 14S 19X24
E 5J 10O 15 T 20Y 25
Z 26
SolutionBE VERY CAREFUL would be encoded as
7 16 67 16 55 76 10 4 55 16 19 64 37
because B corresponds to 2, and f(2) = 3(2) + 1 = 7,
and so on. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 7.15: Domain and Range of a Function
Difficulty Level: Basic Created by: CK-12
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Remember Kara and Marc? Well, they are still in Boston with their grandparents.
Kara and Marc decided to spend one morning helping out at a car wash to benefit a club that their grandparents have participated in for year.
The car wash was a busy place. At the beginning there weren’t any cars, but between 9 am and 10 am the class washed 5 cars. From 10 to 11, the class washed 10 cars, from 11 to 12 the class washed 15 cars and from 12 – 1 the class washed 20 cars.
Marc kept track of the cars washed each hour.
Hour 1 - 5 cars
Hour 2 - 10 cars
Hour 3 - 15 cars
Hour 4 - 20 cars
Can you write this list as ordered pairs? Can you identify the domain and range of the function?
This Concept will teach you how to do this.
### Guidance
We use the word “function” all the time in everyday speech. We say things like “It’s a function of time” or “It’s a function of price.” This is a real life application of a mathematical concept called a function. You will learn how to apply functions to real-world examples, but first let’s look at what a function is and how we can understand it better.
What is a function?
A function is a set of ordered pairs in which the first element in any pair corresponds to exactly one second element.
For example, look at this set of ordered pairs. Notice that braces, {}, are used to surround the set of ordered pairs.
{(0,5),(1,6),(2,7),(3,8)} \begin{align*}& \{ (0, 5), (\underline{1}, \underline{6}), (2, 7), (3, 8)\}\\ & \quad \ \ \qquad \uparrow \uparrow\end{align*}
In (1, 6), 1 is the first element and
6 is the second element.
Each of the first elements––0, 1, 2, and 3––corresponds to exactly one second element. So, this set of ordered pairs represents a function.
Let's take a look at another set of ordered pairs.
{(2,4),(5,3),(6,7),(2,8)} \begin{align*}& \{ (\underline{2}, 4), (5, 3), (6, 7), (\underline{2}, 8)\}\\ & \ \ \Box \qquad \qquad \qquad \quad \ \Box\end{align*}
The first element, 2, corresponds
to two different second elements––4 and 8.
Since one of the first elements corresponds to two different second elements, the set of ordered pairs above does not represent a function.
We can use these criteria to determine whether or not a series of ordered pairs forms a function.
Now that you know how to identify a function, let’s look at some of the key words associated with functions.
1. Domain
2. Range
The domain of a function is the set of all the first elements in a function. The range is the set of all the second elements in a function.
Let’s look at a set of ordered pairs and identify the domain and range of the series.
The ordered pairs below represent a function
{(0, -10), (2, -8), (4, -6), (6, -4)}
a. Identify the domain of the function.
b. Identify the range of the function.
Consider part a\begin{align*}a\end{align*} first.
The domain is the set of all the first elements in the function. These first elements are underlined below.
{(0,10),(2,8),(4,6),(6,4)}\begin{align*}\{ (\underline{0}, -10), (\underline{2}, -8), (\underline{4}, -6), (\underline{6}, -4)\}\end{align*}
The domain of this function is {0, 2, 4, 6}.
Next, consider part b\begin{align*}b\end{align*}.
The range is the set of all the second elements in the function. These second elements are underlined below.
{(0,10),(2,8),(4,6),(6,4)}\begin{align*}\{ (0, \underline{-10}), (2, \underline{-8}), (4, \underline{-6}), (6, \underline{-4})\}\end{align*}
The range of this function is {-10, -8, -6, -4}.
Now it's your turn. Identify the domain and range of each function.
#### Example A
(1, 3) (2, 4) (5, 7) (9, 11)
Solution: Domain {1,2,5,9}\begin{align*}\{1, 2, 5, 9\}\end{align*}, Range {3,4,7,11}\begin{align*}\{3, 4, 7, 11\}\end{align*}
#### Example B
(8, 12) (9, 22) (4, 7) (2, 5)
Solution: Domain {8,9,4,2}\begin{align*}\{8, 9, 4, 2\}\end{align*}, Range {12,22,7,5}\begin{align*}\{12, 22, 7, 5\}\end{align*}
#### Example C
(8, 9) (3, 5) (7, 6) (10, 12)
Solution: Domain {8,3,7,10}\begin{align*}\{8, 3, 7, 10\}\end{align*}, Range {9,5,6,12}\begin{align*}\{9, 5, 6, 12\}\end{align*}
Here is the original problem once again.
Kara and Marc decided to spend one morning helping out at a car wash to benefit a club that their grandparents have participated in for year.
The car wash was a busy place. At the beginning there weren’t any cars, but between 9 am and 10 am the class washed 5 cars. From 10 to 11, the class washed 10 cars, from 11 to 12 the class washed 15 cars and from 12 – 1 the class washed 20 cars.
Marc kept track of the cars washed each hour.
Hour 1 - 5 cars
Hour 2 - 10 cars
Hour 3 - 15 cars
Hour 4 - 20 cars
Can you write this list as ordered pairs? Can you identify the domain and range of the function?
First, use the hours as the domain and the number of cars as the range.
(1, 5) (2, 10) (3, 15) (4, 20)
The domain is {1, 2, 3, 4}.
The range is {5, 10, 15, 20}.
### Vocabulary
Here are the vocabulary words in this Concept.
Function
A set of ordered pairs in which one element corresponds to exactly one other element. Functions can be expressed as a set of ordered pairs or in a table.
Domain
the x\begin{align*}x\end{align*} value of an ordered pair or the x\begin{align*}x\end{align*} values in a set of ordered pairs.
Range
the y\begin{align*}y\end{align*} value of an ordered pair or the y\begin{align*}y\end{align*} values in a set of ordered pairs.
### Guided Practice
Here is one for you to try on your own.
Write the domain and range of this set.
(1, 3) (3, 9) (4, 6) (5, 12)
The first value of each ordered pair represents the domain.
Domain {1, 3, 4, 5}
The second value of each ordered pair represents the range.
Range {3, 9, 6, 12}
### Video Review
Here is a video for review.
### Practice
Directions: Identify whether or not each series of ordered pairs forms a function.
1. (1, 3)(2, 6)(2, 5) (3, 7)
2. (2, 5) (3, 6) (4, 7) (5, 8)
3. (6, 1) (7, 2) (8, 3)
4. (5, 2) (5, 3) (5, 4) (5, 5)
5. (81, 19)(75, 18) (76, 18) (77, 19)
Directions: Name the domain in numbers 1 – 5.
6.
7.
8.
9.
10.
Directions: Name the range in numbers 1 – 5.
11.
12.
13.
14.
15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
domain
The domain of a function is the set of $x$-values for which the function is defined.
Range
The range of a function is the set of $y$ values for which the function is defined.
Continuous
Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.
dependent variable
The dependent variable is the output variable in an equation or function, commonly represented by $y$ or $f(x)$.
Discrete
A relation is said to be discrete if there are a finite number of data points on its graph. Graphs of discrete relations appear as dots.
Formula
A formula is a type of equation that shows the relationship between different variables.
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
independent variable
The independent variable is the input variable in an equation or function, commonly represented by $x$.
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Real Number
A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.
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Converting a binary number to decimal is much simpler than the other way around. You dont need a pencil and paper for this. Lets say you have the number (110001)2 and you wish to convert it.
So start from the right. Keep ticking off powers of two in you head. Every time you encounter a 1 add the power of two to the current answer. If you encounter a 0 just move on to the next place.
Here doing this we would get:
1+16+32=49
Quite simple!
If the binary numer has a decimal point(?) then its a little more tricky(Maybe you should get your pencil now). Anyway lets say you have (.101)2. The process is the same except that you start from the left and must use inverse powers of two. So here we have (.101)2 = (.5+.125)10 = (.625)10 .
For those of us who can't keep all those numbers straight in our heads, here's a chart to help.
---------------------------------------------------------------
| 256 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
---------------------------------------------------------------
| | | | | | | | | |
---------------------------------------------------------------
The numbers in the top row are powers of 2, carried out to the eighth power. You can expand the chart if you need to, just double the previous number. Fill it in with the binary number you've got, say 010010110:
---------------------------------------------------------------
| 256 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
---------------------------------------------------------------
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
---------------------------------------------------------------
In the binary numbering system the only numbers that you use are 0 and 1. If you have a 0 in a columb, then nothing exists there, if you have a 1, then you have 1 of whatever that columb is. (For example, if you have a 1 in the 64 columb, then you have 64). All you have to do to convert from binary to decimal is add up all the columbs that have a 1 in them. So...
2 + 4 + 16 + 128 = 150. Voila!
The E2 Offline Scratchpad rocks for making charts :)
I was very proud when I figured out how to do this.
This method starts from the left and only requires you to remember one number (instead of remembering the powers of two like some methods require). You may find it complicated to convert higher numbers, if you discover you aren't good at multiplying by two in your head. (If so, learn the powers of two and try another method mentioned in this node.)
(1) start at the left and move right
(3) for each digit, add the digit to the number in your head and then multiply the answer by two unless you're at the last digit.
Like So: 110001 |
## About "Evaluate the Indicated Value of Composition Function From the Table"
Evaluate the Indicated Value of Composition Function From the Table :
Here we are going to see, how to evaluate the indicated value of composition functions from the table.
Evaluating the Indicated Value of Composition Function From the Table Examples :
Question 1 :
Evaluate the indicated expression assuming that f, g, and h are the functions completely defined by the tables below:
(i) (f ◦ g)(1) (ii) (f ◦ g)(3) (iii) (g ◦ f)(1)
(iv) (g ◦ f)(3) (v) (f ◦ f)(2) (vi) (f ◦ g ◦ h)(2)
(vii) (h ◦ g ◦ f)(2)
Solution :
(i) (f ◦ g)(1)
(f ◦ g)(1) = f [g(1)]
From the second table, the value of g(1) is 2.By applying the value of g(1), we get
= f(2)
Now let us find the value of f(2) from 1st table.The value is 1.
(f ◦ g)(1) = 1
(ii) (f ◦ g)(3)
Solution :
(f ◦ g)(3) = f [g(3)]
From the second table, the value of g(3) is 1.By applying the value of g(3), we get
= f(1)
Now let us find the value of f(2) from 1st table.The value is 1.
(f ◦ g)(3) = 4
(iii) (g ◦ f)(1)
Solution :
(g ◦ f)(1) = g [f(1)]
From the first table, the value of f(3) is 4.By applying the value of f(1), we get
= g(4)
Now let us find the value of g(4) from 2nd table.The value is 3.
(g ◦ f)(1) = 3
(iv) (g ◦ f)(3)
Solution :
(g ◦ f)(3) = g [f(3)]
From the first table, the value of f(3) is 2.By applying the value of f(2), we get
= g(2)
Now let us find the value of g(2) from 2nd table.The value is 1.
(g ◦ f)(3) = 1
(v) (f ◦ f)(2)
Solution :
(f ◦ f)(2) = f [f(2)]
From the first table, the value of f(2) is 1.By applying the value of f(2), we get
= f(1)
The value of f(1), from 1st table.
(f ◦ f)(2) = 4
(vi) (f ◦ g ◦ h)(2)
Solution :
(f ◦ g ◦ h)(2) = (f ◦ g)[h(2)]
= (f ◦ g)(3)
= f [g(3)]
= f(1)
= 4
(vii) (h ◦ g ◦ f)(2)
(h ◦ g ◦ f)(2) = (h ◦ g)[f(2)]
= (h ◦ g)(1)
= h [g(1)]
= h(2)
= 3
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
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OTHER TOPICS
Profit and loss shortcuts
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Decimal representation of rational numbers
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Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
Sunday, September 8, 2024
# What Does Denominator Mean In Math
## What Is The Difference Between Numerator And Denominator
What’s a Numerator and What’s a Denominator?
In a fraction, the top number is called the numerator and the bottom number is called the denominator. For example, is a fraction. Here, 4 is the numerator and 5 is the denominator. Similarly, the numerator defines how many parts we have and the bottom number represents how many equal parts the object is divided into.
## Meaning Of Denominator In Correlation
I can’t quite grasp the meaning of the denominator in the correlation coefficient.
$$\frac}$$
What exactly am I dividing with, and why?
I understood dividing with the standard deviation in the Z distribution, that got me the difference from the mean in terms of standard deviations. But what does this give?
The covariance measured in….what, the standard deviation of X times the standard deviation of Y?
That would explain where the n’s in the denominator have gone, but what does that mean?
• $\begingroup$One can think of them as normalizing terms, which keep the expression invariant if $X$ or $Y$ is rescaled. The correlation coefficient is dimensionless.$\endgroup$
Do you know the scalar/inner product of vectors on $\mathbb$ or $\mathbb$?
$\hat \cdot \hat = \|\hat\|\|\hat\|\cos\theta\$ or
$$\cos\theta = \frac \cdot \hat}\|\|\hat\|} = \frac}\|}\cdot\frac}\|} = \frac\sqrt}$$
Correlation then is analogous to finding the angle between two vectors. The denominator is normalizing the vectors so that we are taking the scalar/inner product between two vectors of unit length.
A way to better understand the denominator is to note that the coefficient of correlation is the square root of the coefficient of determination $r^2=\frac} ss_}$. Here, $ss_$ is the $xy$ covariance, $ss_$ is the $x$ variance, and $ss_$ is the $y$ variance.
## Often Asked: What Does And Mean In Math
But what does a squiggly line mean in math? There are two major varieties of mathematics used in the simple geometry. A single kind that uses straight lines plus the other variety that makes use of curved lines. In each forms, a squiggle represents an interval Math, Math Problems and Solutions, Tests, Formulas, Algebra. Menu Home; Question-Answer; Riddles; Puzzle; Rebus; Contact Us; Posted on May 16, 2021 May 16, 2021 by John Newton. What does the word pudding mean in the U.K.? What does the word pudding mean in the U.K.? Dessert A fancy meal Whipped cream Breakfast. Answer. What does it mean in the context of the above quote? Mr. Trump’s campaign manager says Professor Hawking should try to talk in English. Is this a farfetched way of using common denominator from math to politics? Did Professor Hawking misuse the common denominator Mean score of whole class . Mean Word Problems. Example 1: Pedro’s luncheonette is open six days a week. His income for the first five days was $1,200,$1,100, $2,000,$1,400 and $3,000. How much money must she make on the sixth day to average$2,000 for the six days? Example 2: George’s scores on three math tests were 70, 90 and 75 A fractional exponent is an alternate notation for expressing powers and roots together. For example, the following are equivalent. We write the power in numerator and the index of the root in the.
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## Cultural Definitions Forcommon Denominator
A number that will allow fractions with different denominators to be converted into fractions with the same denominator, so that these fractions can be added or subtracted. The fractions can be expressed as whole numbers divided by the common denominator. Thus, 12 is a common denominator for 1/3 and 1/4, since they can be written as 4/12 and 3/12, respectively.
## What Does The Lowest Common Denominator Mean In The Context Other Than Math
In the New Yorkers article under the title, Stephen Hawking angers Trump supporters with baffling array of long words, Andy Borowitz wrote;
Speaking to a television interviewer in London, the theoretical physicist, Hawking called Trump a demagogue who seems to appeal tothe lowest common denominator. — For a so-called genius, this was an epic fail, Trumps campaign manager, Corey Lewandowski, said. If Professor Hawking wants to do some damage, maybe he should try talking in English next time.
Later in the day, Hawking attempted to clarify his remark about the presumptive Republican Presidential nominee, telling a reporter, Trump bad man. Real bad man.
From the context of Professor Hawkings remark, I take the lowest common denominator as referring to the social segment of low educated, unsophisticated people, but Im not sure.
I thought “common denominator” is a simple mathematic term. What does it mean in the context of the above quote?
Mr. Trumps campaign manager says Professor Hawking should try to talk in English.Is this a farfetched way of using common denominator from math to politics?
Did Professor Hawking misuse “the common denominator”? Or does Mr. Trump’s campaign manager not understand the meaning of “common denominator”, which some call an “everyday-use” English phrase?
P.S.
I found the following definition of ‘common denominator’ in Oxford Advanced Learners English Dictionary;
Here’s a definition from ODO:
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## Mean Median Mode And Range Purplemat
Accepted Answer. There is a 2D for height and width of the video, which shows the resolution, the 3rd D is for example is of size 3 for RGB or of size 1 for grayscale. This 3 dimensions would describe one image. Since you have a video, there are all those single frames in the 4th dimension one after eachother What Does Compactness Really Mean? A set does not have to be infinite in length or area to be non-compact. When Math and Language Collide. September 30, 2013 Evelyn Lamb. The Sciences
What does this mean? – /sci/ – Science & Math is 4chan’s board for the discussion of science and math 9 Comments on What does 20/20 vision mean? DEPR says: 28 Oct 2010 at 8:17 pm No definitely the eye enhancement by Tiger Woods isn’t cheating. Jemal Kemal Nigo says: 29 Oct 2010 at 2:14 pm I like the article 20/20 vision, as it describe sight/eye related problem with the aid of mathematics.It is also an article in which one realize that mathematics. We use statistics such as the mean, median and mode to obtain information about a population from our sample set of observed values. Mean. The mean of a set of data values is the sum of all of the data values divided by the number of data values. That is: Example 1. The marks of seven students in a mathematics test with a maximum possible mark of 20 are given below Tim and Moby in a practical math movie where you can learn how mean, median, mode, and range help you work with sets and data
## Rationalizing The Denominator Vs Rationalizing The Denominator With Conjugate Method
Example
Rationalize the denominator.
???\frac}???
If we multiply the denominator by ???\sqrt???, well get rid of the square root there. But in order to keep the value of the fraction the same, we have to multiply both the numerator and the denominator by ???\sqrt???.
???\frac}\left???
???\frac}???
Now that the square root is out of the denominator, weve rationalized the denominator.
But how do we rationalize the denominator when its not just a single square root? Sometimes were going to have a denominator with more than one term, like
???\frac}???
In a case like this one, where the denominator is the sum or difference of two terms, one or both of which is a square root, we can use the conjugate method to rationalize the denominator.
The conjugate of a binomial is the same two terms, but with the opposite sign in between. For ???5-\sqrt???, we keep the same two terms, ???5??? and ???\sqrt???, but we change the sign in the middle. Since the sign in this case is negative, well change it to a positive sign.
Original binomial: ???5-\sqrt???
Its conjugate: ???5+\sqrt???
To rationalize the denominator of a fraction where the denominator is a binomial, well multiply both the numerator and denominator by the conjugate.
As were doing these problems, lets also remember these facts:
Fact 1: You can multiply any number by one without changing its value.
???\frac – 5} \cdot 1??? is the same as ???\frac-5}???
???\frac+5}+5} = 1???
Example
Simplify the expression.
???\frac}-5}???
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## What Does The Mean In Math Formulas
What Does n Mean in Math? By Staff Writer Last Updated Mar 27, 2020 8:10:56 PM ET. In mathematical operations, n is a variable, and it is often found in equations for accounting, physics and arithmetic sequences What does Dcmam mean in math? The acronym DCMAM stands for Don’t Call Me A Monster. It also stands for the steps needed to whittle a Monster down to a 2-step equation . Click to see full answer. Similarly, what is equation with example
What does ^ mean in math? Logical conjunction is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both of its operands are true. Answer: In Mathematics AND is denoted by the symbol ^ Let’s understand this An outlier is a number in a data set that is much smaller or larger than the other numbers in the data set.TranscriptWelcome to MooMooMath where we upload a. What Does FOIL in Math Mean? Start today. Try it now Math / Math for Kids Math for Kids 23 chapters | 325 lessons }. What does Annex in math mean? by annexing zero:x 5 = 2, In this case annexing means multiplication. in an addition or multiplication number do not change the result of the equation. When you compare two numbers that DO NOT have all the same number of digits after the decimal point, we can add ZEROS
## What Are Numerators And Denominators
What do we mean by LCM? | Don’t Memorise
Learn a quick tip to help you understand exactly what the numerator and denominator of a fraction tell you.
The fraction 1/2 means one piece of a whole object divided into two equally sized parts. The denominator indicates that two parts make a whole, and the numerator counts off the fact that the fraction 1/2 contains one of those parts. The fraction X/Y means X pieces of a whole object that is divided into Y equally sized parts.
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## What Does The Denominator In The Second Derivative Mean
This occurred to me a few days ago.
We know that the derivative of a function $y=f$ is $\frac$.This is because it represents how $y$ changes with $x$, which is the rate of change of $y$, or more specifically, the gradient of a function.
Then the second derivative is the rate of change of rate of change, or the rate of change of gradient. Since a general rate of change is $\frac$, the second derivative is $$. Thus, the expanded form is \frac. My question is, is the denominator d^2 or is it ^2? Surely, it would be the latter, because when you expand$$, the would become $^2$. But then why is it never written with brackets? I’m sure that would confuse some people and I only realised it myself when I started thinking about the second derivative properly, in terms of what it actually means.
## Faq: What Does U Mean In Math
Tech. Is the Universe Made of Math? In this excerpt from his new book, Our Mathematical Universe, M.I.T. professor Max Tegmark explores the possibility that math does not just describe. In music what does allegro mean math worksheet. M025 m025 82. Kindergarten Math Worksheets Fact Families Kiddo Shelter Music Faster i have the same worksheet and i cant figure out the full answer sorry but yea thats what it means Checks and Balance After mathwhat does a return to normal mean for American leadership? Our weekly podcast on democracy in America. Economist Radio Podcasts. Jun 11th 2021 ON HIS.
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## What Does It Mean If The Numerator Is Bigger Than The Denominator
In all the examples so far, the numerator has always been smaller than the denominator. In other words, in 1/2 and 45/77, 1 and 45 are smaller than 2 and 77, respectively. But what would it mean if the numerator were bigger than the denominator? Something like 7/4?
Well, lets try interpreting this the same way as before. The denominator, 4, indicates that a whole is divided into four equally sized parts, and the numerator, 7, indicates that we have seven of those parts. So, if four parts make a whole, and we have seven, then we must have a whole object plus three more of the equally sized parts. So 7/4 is equivalent to 1 3/4also known as one and three-quartersand we now know that a fraction whose numerator is greater than its denominator represents a number that is greater than one. In case youre wondering, that type of fraction is called improper, whereas fractions like 1/2 with numerators less-than denominators are called proper.
## What Does The Denominator Tell You
Heres what I mean. The denominator of a fraction tells you how many parts a whole is broken into. It can be a whole pineapple, a whole song, or a whole anything. If the denominator of a fraction is, say, 4, then that indicates that the whole whatever is broken up into 4 equally-sized pieces.
Or, if the denominator is 12, that means the whole whatever is broken-up into 12 equally-sized pieces. But how exactly does that name the type of fraction? Well, that leads us to the meaning of the numerator.
Recommended Reading: What Does Kw Mean In Chemistry
## Multiply Both Top And Bottom By The Conjugate
There is another special way to move a square root from the bottom of a fraction to the top … we multiply both top and bottom by theconjugate of the denominator.
The conjugate is where we change the sign in the middle of two terms:
Example Expression
It works because when we multiply something by its conjugate we get squares like this:
= a2 b2
Here is how to do it:
## Webster Dictionaryrate This Definition:
• Denominatornoun
one who, or that which, gives a name; origin or source of a name
Etymology:
• Denominatornoun
that number placed below the line in vulgar fractions which shows into how many parts the integer or unit is divided
Etymology:
• Denominatornoun
that part of any expression under a fractional form which is situated below the horizontal line signifying division
Etymology:
• You May Like: What Is Function Notation In Math
## What Does < And/or > Mean In Math What Does That Mean
• g: it means that the item on the left hand side is being defined to be what is on the right hand side. For example, y := 7 x + 2. means that y is defined to be 7 x + 2. This is different from, say, writing. 1 = sin 2
• The e stands for Exponent, which means the number of tens you multiply a number by. For example, if I square 123456789, I get 1.524157875019e+16, which means that the answer is 1.524157875019 times 10 raised to the sixteenth power . Click to read further detail
• It means pretty much the same thing that it does in English. In other words, it varies depending on how it’s used. Suppose that you’ve got a square inscribed in a circle. You can ask: how much of the circle is covered by the square? There’s a prec..
• A comma is placed every third digit to the left of the decimal point and so is used in numbers with four or more digits. What does this symbol mean in math? This symbol > means greater than, for example 4 > 2. These symbols mean ‘less than or equal to’ and ‘greater than or equal to’ and are commonly used in algebra
• How to Find 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. In other words it is the sum divided by the count
• ## Reciprocals And The Invisible Denominator
What does “LCM” mean? Math lesson
The reciprocal of a fraction is another fraction with the numerator and denominator exchanged. The reciprocal of 3 3 }} . The product of a fraction and its reciprocal is 1, hence the reciprocal is the multiplicative inverse of a fraction. The reciprocal of a proper fraction is improper, and the reciprocal of an improper fraction not equal to 1 is a proper fraction.
When the numerator and denominator of a fraction are equal , its value is 1, and the fraction therefore is improper. Its reciprocal is identical and hence also equal to 1 and improper.
Any integer can be written as a fraction with the number one as denominator. For example, 17 can be written as 17 1 }} , where 1 is sometimes referred to as the invisible denominator. Therefore, every fraction or integer, except for zero, has a reciprocal. For example. the reciprocal of 17 is 1
Read Also: What Is Cultural Diffusion In Geography
## Difference Between Numerator And Denominator
The difference between denominator and numerator in a research study is the same as that of mathematical elements: One element, the denominator, influences another variable element, the numerator, either by division, reduction, measurement or similar change. The research study has a question — what is the effect of X on Y?–and it answers it by observing how denominator X changes numerator Y. The decisive element of the study is X, the mover and shaker, so to speak. Y receives the action or effect of X, which gives resulting data to the study.
## What Does * Mean In Math
• Thus 2xy means 2 times x times y or said algebraically 2xy = 2 \times x \times y. If you want the operation between two symbols to be something else, like plus, you have to say it explicitly. Thus for x plus y you need to write x + y. Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences
• 2. Everyone does math, every day. 3. Girls are great at math. 4. There’s more than one way to skin a math problem. 5. Math anxiety is real and detrimentalbut it can be overcome. 6. Kids learn math best when they are allowed to discover their own approachesand fail. 7
• Whatdoes != do/meanin python. Ask Question Asked 7 years, 3 months ago. Active 3 years, 1 month ago. Viewed 62k times -17. 0. I see this code in my python reference guide, but no description. I am asked Browse other questions tagged python math or ask your own question
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# 3.3: The Complement of an Event
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Know the definition of the complement of an event.
• Use the complement of an event to calculate the probability of an event.
• Understand the Complement Rule.
## Introduction
In this lesson, you will learn what is meant by the complement of an event, and you will be introduced to the Complement Rule. You will also learn how to calculate probabilities when the complement of an event is involved.
### The Complement of an Event
The complement \begin{align*}A'\end{align*} of the event \begin{align*}A\end{align*} consists of all elements of the sample space that are not in \begin{align*}A\end{align*}.
Example: Let us refer back to the experiment of throwing one die. As you know, the sample space of a fair die is \begin{align*}S=\left \{1,2,3,4,5,6\right \}\end{align*}. If we define the event \begin{align*}A\end{align*} as observing an odd number, then \begin{align*}A=\left\{1,3,5\right\}\end{align*}. The complement of \begin{align*}A\end{align*} will be all the elements of the sample space that are not in \begin{align*}A\end{align*}. Thus, \begin{align*}A' =\left \{2,4,6\right \} \end{align*}
A Venn diagram that illustrates the relationship between \begin{align*}A\end{align*} and \begin{align*}A'\end{align*} is shown below:
This leads us to say that the sum of the possible outcomes for event \begin{align*}A\end{align*} and the possible outcomes for its complement, \begin{align*}A'\end{align*}, is all the possible outcomes in the sample space of the experiment. Therefore, the probabilities of an event and its complement must sum to 1.
### The Complement Rule
The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1.
\begin{align*}P(A)+P(A')=1\end{align*}
As you will see in the following examples, it is sometimes easier to calculate the probability of the complement of an event than it is to calculate the probability of the event itself. Once this is done, the probability of the event, \begin{align*}P(A)\end{align*}, is calculated using the relationship \begin{align*}P(A)=1-P(A')\end{align*}.
Example: Suppose you know that the probability of getting the flu this winter is 0.43. What is the probability that you will not get the flu?
Let the event \begin{align*}A\end{align*} be getting the flu this winter. We are given \begin{align*}P(A)=0.43\end{align*}. The event not getting the flu is \begin{align*}A'\end{align*}. Thus, \begin{align*}P(A')=1-P(A)=1-0.43=0.57\end{align*}.
Example: Two coins are tossed simultaneously. Let the event \begin{align*}A\end{align*} be observing at least one head.
What is the complement of \begin{align*}A\end{align*}, and how would you calculate the probability of \begin{align*}A\end{align*} by using the Complement Rule?
Since the sample space of event \begin{align*}A=\left \{HT, TH, HH\right \}\end{align*}, the complement of \begin{align*}A\end{align*} will be all events in the sample space that are not in \begin{align*}A\end{align*}. In other words, the complement will be all the events in the sample space that do not involve heads. That is, \begin{align*}A'=\left \{TT\right \}\end{align*}.
We can draw a simple Venn diagram that shows \begin{align*}A\end{align*} and \begin{align*}A'\end{align*} when tossing two coins as follows:
The second part of the problem is to calculate the probability of \begin{align*}A\end{align*} using the Complement Rule. Recall that \begin{align*}P(A)=1-P(A')\end{align*}. This means that by calculating \begin{align*}P(A')\end{align*}, we can easily calculate \begin{align*}P(A)\end{align*} by subtracting \begin{align*}P(A')\end{align*} from 1.
\begin{align*}P(A')& =P(TT)=\frac{1}{4}\\ P(A)& =1-P(A')=1-\frac{1}{4}=\frac{3}{4}\end{align*}
Obviously, we would have gotten the same result if we had calculated the probability of event \begin{align*}A\end{align*} occurring directly. The next example, however, will show you that sometimes it is much easier to use the Complement Rule to find the answer that we are seeking.
Example: Consider the experiment of tossing a coin ten times. What is the probability that we will observe at least one head?
What are the simple events of this experiment? As you can imagine, there are many simple events, and it would take a very long time to list them. One simple event may be \begin{align*}HTTHTHHTTH\end{align*}, another may be \begin{align*}THTHHHTHTH\end{align*}, and so on. There are, in fact, \begin{align*}2^{10}=1024\end{align*} ways to observe at least one head in ten tosses of a coin.
To calculate the probability, it's necessary to keep in mind that each time we toss the coin, the chance is the same for heads as it is for tails. Therefore, we can say that each simple event among the 1024 possible events is equally likely to occur. Thus, the probability of any one of these events is \begin{align*}\frac{1}{1024}\end{align*}.
We are being asked to calculate the probability that we will observe at least one head. You will probably find it difficult to calculate, since heads will almost always occur at least once during 10 consecutive tosses. However, if we determine the probability of the complement of \begin{align*}A\end{align*} (i.e., the probability that no heads will be observed), our answer will become a lot easier to calculate. The complement of \begin{align*}A\end{align*} contains only one event: \begin{align*}A'=\left \{TTTTTTTTTT\right \}\end{align*}. This is the only event in which no heads appear, and since all simple events are equally likely, \begin{align*}P(A')=\frac{1}{1024}\end{align*}.
Using the Complement Rule, \begin{align*}P(A)=1-P(A')=1-\frac{1}{1024}=\frac{1023}{1024}=0.999\end{align*}.
That is a very high percentage chance of observing at least one head in ten tosses of a coin.
## Lesson Summary
The complement \begin{align*}A'\end{align*} of the event \begin{align*}A\end{align*} consists of all outcomes in the sample space that are not in event \begin{align*}A\end{align*}.
The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event \begin{align*}A, P(A)+P(A')=1\end{align*}.
For an explanation of complements and using them to calculate probabilities (1.0), see jsnider3675, An Event's Complement (9:40).
## Review Questions
1. A fair coin is tossed three times. Two events are defined as follows: \begin{align*}& A: {\text{at least one head is observed}}\end{align*} \begin{align*}& B: {\text{an odd number of heads is observed}}\end{align*}
1. List the sample space for tossing the coin three times.
2. List the outcomes of \begin{align*}A\end{align*}.
3. List the outcomes of \begin{align*}B\end{align*}.
4. List the outcomes of the following events: \begin{align*}A \cup B, A', \ A\cap B\end{align*}.
5. Find each of the following: \begin{align*}P(A), P(B), P(A \cup B), P(A'), P(A \cap B).\end{align*}
2. The Venn diagram below shows an experiment with five simple events. The two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are shown. The probabilities of the simple events are as follows: \begin{align*}P(1)=\frac{1}{10}, P(2)=\frac{2}{10}, P(3)=\frac{3}{10}, P(4)=\frac{1}{10}, P(5)=\frac{3}{10}\end{align*}. Find each of the following: \begin{align*}P(A'), P(B'), P(A' \cap B), P(A \cap B), P(A \cup B'), P(A \cup B), P(A \cap B'), P \left [(A \cup B)' \right ]\end{align*}.
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In two right triangles ΔABC and ΔDEF,if sinB = sin E then prove that angle B = angleE - Future Study Point
# In two right triangles ΔABC and ΔDEF,if sinB = sin E then prove that angle B = angleE
## In two right triangles ΔABC and ΔDEF,if sinB = sin E then prove that ∠ B = ∠E
NCERT Solutions Class 10 Science from chapter 1 to 16
In future study point, you can find free study material, you can ask any question by writing the questions in the comment box, you can communicate with us directly in our Whatsapp number 9891436286 or our facebook page future point coaching center, we are always with you to boost your progress in reaching your target. We can guide you for your carrier choice, competitive entrance exams, etc.
Question.In two right triangles ΔABC and ΔDEF,if sinB = sin E then prove that angle B = angleE
We are given two right triangles ΔABC and ΔDEF in which sinB =sinE
Let ∠A and ∠D are right triangles of ΔABC and ΔDEF respectively
GIVEN : sinB = sinE
TO PROVE: ∠B =∠E
Proof: In ΔABC and ΔDEF
sinB = sinE
BC = kEF & AC = kDF……(i)
Applying Pythagoras theorem in both triangles ΔABC and ΔDEF, we have
Putting the value of BC and AC in the equation (i)
Since we have to prove the angles of two triangles, so we are needed to prove both triangles similar to each other.
And proving both triangles similar to each other, here we have to prove all three sides in proportional to each other.
From (i),(ii) and (iii) we have
∴ΔABC ∼ DEF (SSS rule)
Hence ∠B =∠E
OR
you can understand it through our video
We have recently prepared a few sets of maths questions for the year 2020 CBSE board exams, studying these questions definitely boosts your preparations for the CBSE exams.
We have selected a few specific questions of mensuration which are generally asked in every board exam of CBSE, revise these questions and crack the board exam with ninety to hundred percent achievement in the exam.
You can study here the solutions of last years question papers asked in the exams,ncert solutions of maths and science, assignment of maths and science, download e-books of physics, chemistry,blog post related to science and maths in which every question explained in a systematic and detailed way with the help of proper diagram so that every student could understand the problems into its depth. We have determined to explain every fact and figure related to science and maths in simple words so that we could contribute to enhancing your interest and skill in science and maths subjects.
### NCERT Solutions for class 10 maths
Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry
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# When the slope is 0 What does that mean?
Now we see why a vertical line has an undefined slope. Because we can't divide by zero! Usually in math, when something is "undefined" it means that somewhere, something is being divided by 0. And that's a no-no. It has no meaning.
A.
### What happens when the slope has a 0 on top?
When the 0 is on the "top" of the fraction, that would mean that the two y-values are the same. Thus that line is horizontal (slope of 0). If the "bottom" of the fraction is 0 that means the two x-values are the same. Thus that line is vertical (undefined slope).
• #### How do you graph a slope?
Algebra 1 – How to Graph a Linear Equation Using Slope and y
1. Step 1: Put the equation in Slope Intercept Form.
2. Step 2: Graph the y-intercept point (the number in the b position) on the y-axis.
3. Step 3: From the point plotted on the y-axis, use the slope to find your second point.
4. Step 4: Draw your line using the two points you plotted (y-intercept (b) first, slope (m) second.
• #### How do you find the slope and y intercept of a line?
The coordinates of every point on the line will solve the equation if you substitute them in the equation for x and y. The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept.
• #### What is the horizontal line?
In geometry, a horizontal line is one which runs from left to right across the page. It comes from the word 'horizon', in the sense that horizontal lines are parallel to the horizon. The horizon is horizontal. Its cousin is the vertical line which runs up and down the page.
B.
### What do you do if the slope is 0?
Since we did not have a change in the x values, the denominator of our slope became 0. This means that we have an undefined slope. If you were to graph the line, it would be a vertical line, as shown above. This form can be handy if you need to find the slope of a line given the equation.
• #### How do you know when a slope is steep?
When you look at the two lines, you can see that the blue line is steeper than the red line. It makes sense the value of the slope of the blue line, 4, is greater than the value of the slope of the red line, . The greater the slope, the steeper the line. The next example shows a line with a negative slope.
• #### Is a horizontal line undefined?
It is a vertical line through 2 on the - axis. Vertical Lines A vertical line is a line whose equation is of the form =ℎ, where ℎ is a real number. All points that lie on the line have an -coordinate of ℎ. The slope of any vertical line is undefined.
• #### How does a graph fail the vertical line test?
To use the vertical line test, take a ruler or other straight edge and draw a line parallel to the y-axis for any chosen value of x. If the vertical line you drew intersects the graph more than once for any value of x then the graph is not the graph of a function.
Updated: 7th September 2018 |
# Number of combinations and permutations of letters
I am confused by the following exercise.
Exercise. Find the number of (a) combinations and (b) permutations of 4 letters each that can be made from the letters of
the word Tennessee.
Ideas:
We have:
T-1
E-4
N-2
S-2
If we would have unlimited number of letters, the number of arrangements is $4*4*4*4$, however the number of T’s, N’s, S’s doesn’t all us to do so, and I don’t see the easy way to eliminate impossible cases.
#### Solutions Collecting From Web of "Number of combinations and permutations of letters"
You have to be aware of the differences between combinations and permutations.
The first are all the posible agrupations that you can make with the elements of your set, here the order doesn’t matter, $(123)$ is the same as $(312)$, but it’s important that the element doesn’t repeat.
The latter are all the differente agrupations of the elements of your set where the order matter, it’s not the same to say $(123)$ than $(312)$, but you can’t repeat elements, say $(111)$.
First the permutations. You can take the set that cointains the letters from $Tenessee$ like $\{T_1\;E_2\;N_3\;E_4\;S_5\;S_6\;E_7\;E_8\}$, and the size of the sample as 4, imagine that you have 4 boxes, in each box you can put those elements, but with the restriction that you can’t repeat the elements, and the order matters,
$\mathbf (Box 1)$ you can put any of the 8 elements
$\mathbf (Box 2)$ since you already put one element in B1 and you can’t repeat, you have 7 choices
$\mathbf (Box 3)$ using the same thinking, now you have 6 choices
$\mathbf (Box 4)$ now you have 5 choices
So using the Rule of product, you have that the permutations that you can get are $8*7*6*5$
Now, with the combinations, you want to arrange the 8 elements in 4 boxes again, but you’ll consider cases like $\{T_1\;E_2\;N_3\;E_4\}$, $\{E_2\;T_1\;N_3\;E_4\}$, $\{E_2\;N_3\;T_1\;E_4\}$, $\{E_2\;N_3\;E_4\:T_1 \}$, etc. to be the same. In the case of permutations, these elements were different, so will have to discard this cases if we want to get combinations, so the number of extra elements is $4*3*2*1=4!$ (see the EDIT), so the combinations are $$8*7*6*5 \over 4*3*2*1$$
You could also use formulas. If $n$ is the number of elements in your set, and $m$ is the size of your sample, then the permutation is given by: $$n! \over (n-m)!$$ There are books that use the term variation for this formula, and the term permutation for $n!$, but you can see that it’s the same. And the combinations are given by: $$n! \over m!(n-m)!$$ wich is the definition of binomial coefficient
In your excersize $n=8$ and $m=4$.
$\mathbf {EDIT:}$ Why there are $4!$ extra cases?
I think it’ll be easier with a smaller example first.
There’s a group of 5 students that want to form a committee of 3, how many different committees can they form?
Using the same box method, we have 3 spots, we know that the first spot can be taken by 5 students, the second by 4 students and the third by 3 students, now we are not done yet, because by now we are considering that the committee of $(A,B,C)$ is different from $(B,A,C),(B,C,A),(A,C,B),(C,A,B),(C,B,A)$ (wich as you can see, is the permutations of 3 elements in 3 spots, hence $3*2*1=3!$), our interest is to know how many times are we counting the same committee, so let’s consider one fixed committee, like the one we had $(A,B,C)$, and count how many times we can order those elements, we know that: $3!$; now lets fixed another one: $(B,C,D)$ we are also counting $3!$ more times, another one $(C,D,E)$ the same, etc. etc. The same for the rest. Then for every 3 people we are counting $3!$ more times the product $5*4*3$. So the answer is $$5*4*3 \over 3!$$
Now with your problem, I took one fixed posiblity $\{T_1\;E_2\;N_3\;E_4\}$, and saw that you can arrange those elements $4!$, using the same thinking as the problem above, what we are doing is counting $4!$ times more the product $8*7*6*5$, hence what we are looking for is $$8*7*6*5 \over 4!$$
Hope this helps. You can always make drawings to see what happens, try it with the committee’s problem.
Multisets of size $4$ made from the letters in TENNESSEE can have frequencies of $T$, $N$ and $S$ as per the following table; all of the remaining letters must be E.
$$\begin{array}{c|ccc|c} \text{number} & \# \text{T} & \# \text{N} & \# \text{S} & \text{multiset} \\ \hline 1 & 0 & 0 & 0 & \{\text{E,E,E,E}\} \\ 2 & 0 & 0 & 1 & \{\text{S,E,E,E}\} \\ 3 & 0 & 0 & 2 & \{\text{S,S,E,E}\} \\ 4 & 0 & 1 & 0 & \{\text{N,E,E,E}\} \\ 5 & 0 & 1 & 1 & \{\text{N,S,E,E}\} \\ 6 & 0 & 1 & 2 & \{\text{N,S,S,E}\} \\ 7 & 0 & 2 & 0 & \{\text{N,N,E,E}\} \\ 8 & 0 & 2 & 1 & \{\text{N,N,S,E}\} \\ 9 & 0 & 2 & 2 & \{\text{N,N,S,S}\} \\ 10 & 1 & 0 & 0 & \{\text{T,E,E,E}\} \\ 11 & 1 & 0 & 1 & \{\text{T,S,E,E}\} \\ 12 & 1 & 0 & 2 & \{\text{T,S,S,E}\} \\ 13 & 1 & 1 & 0 & \{\text{T,N,E,E}\} \\ 14 & 1 & 1 & 1 & \{\text{T,N,S,E}\} \\ 15 & 1 & 1 & 2 & \{\text{T,N,S,S}\} \\ 16 & 1 & 2 & 0 & \{\text{T,N,N,E}\} \\ 17 & 1 & 2 & 1 & \{\text{T,N,N,S}\} \\ \end{array}$$
For the permutation problem, we just sum the corresponding multinomial coefficients for the multisets in the above table. |
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# In the Following Figure, Ad is a Straight Line. Op ⊥ Ad and O is the Centre of Both the Circles. If Oa = 34 Cm. Ob = 20 Cm and Op = 16cm; Find the Length of Ab. - ICSE Class 10 - Mathematics
ConceptChord Properties - a Straight Line Drawn from the Center of a Circle to Bisect a Chord Which is Not a Diameter is at Right Angles to the Chord
#### Question
In the following figure, AD is a straight line. OP ⊥ AD and O is the centre of both the circles. If OA = 34 cm. OB = 20 cm and OP = 16cm; find the length of AB.
#### Solution
For the inner circle, BC is a chord and OP⊥ BC.
We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ BP = PC
By Pythagoras theorem,
OA2 = OP2 = BP2
⇒ BP2 = (20)2 - (16)2 =144
∴ BP= 12cm
We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ AP = PD
By Pythagoras Theorem,
OA2 = OP2 + AP2
⇒ AP2 = (34)2 − (16)2 = 900
⇒ AP = 30 cm
AB = AP − BP = 30 − 12 = 18 cm
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution In the Following Figure, Ad is a Straight Line. Op ⊥ Ad and O is the Centre of Both the Circles. If Oa = 34 Cm. Ob = 20 Cm and Op = 16cm; Find the Length of Ab. Concept: Chord Properties - a Straight Line Drawn from the Center of a Circle to Bisect a Chord Which is Not a Diameter is at Right Angles to the Chord.
S |
### Factoring polynomials completely
Example 1 — Factor:. Example 2 — Factor:. Example 3 — Factor:. Example 4 — Factor:. Make sure that the trinomial is written in the correct order; the trinomial must be written in descending order from highest power to lowest power. Decide if the three terms have anything in common, called the greatest common factor or GCF. If so, factor out the GCF. Do not forget to include the GCF as part of your final answer. Multiply the leading coefficient and the constant, that is multiply the first and last numbers together.
List all of the factors from Step 3 and decide which combination of numbers will combine to get the number next to x. After choosing the correct pair of numbers, you must give each number a sign so that when they are combined they will equal the number next to x and also multiply to equal the number found in Step 3.
Opengl draw tube
Rewrite the original problem with four terms by splitting the middle term into the two numbers chosen in step 5. Step 1 : Make sure that the trinomial is written in the correct order; the trinomial must be written in descending order from highest power to lowest power. In this case, the problem is in the correct order. Step 2 : Decide if the three terms have anything in common, called the greatest common factor or GCF.
In this case, the three terms only have a 1 in common which is of no help. Step 3 : Multiply the leading coefficient and the constant, that is multiply the first and last numbers together.
In this case, you should multiply 6 and —2. Step 4 : List all of the factors from Step 3 and decide which combination of numbers will combine to get the number next to x. In this case, the numbers 3 and 4 can combine to equal 1.
Step 5 : After choosing the correct pair of numbers, you must give each number a sign so that when they are combined they will equal the number next to x and also multiply to equal the number found in Step 3. Step 6 : Rewrite the original problem with four terms by splitting the middle term into the two numbers chosen in step 5.
Step 7 : Now that the problem is written with four terms, you can factor by grouping.During these challenging times, we guarantee we will work tirelessly to support you. We will continue to give you accurate and timely information throughout the crisis, and we will deliver on our mission — to help everyone in the world learn how to do anything — no matter what.
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We will get through this together. Updated: September 6, References. A trinomial is an algebraic expression made up of three terms. There are several tricks to learn that apply to different types of quadratic trinomial, but you'll get better and faster at using them with practice. Higher degree polynomials, with terms like x 3 or x 4are not always solvable by the same methods, but you can often use simple factoring or substitution to turn them into problems that can be solved like any quadratic formula.
Next, use factoring to guess at the Last terms. To factor, find two numbers that multiply to form the Last term. Do this until you narrow the Last terms down to a few possibilities. Then, test which possibilities work with Outside and Inside multiplication. When you find the terms that match the original polynomial, you have the correct answer. If you want to learn how to factor trinomials when the variables have a coefficient, keep reading the article! Did this summary help you?
By using our site, you agree to our cookie policy. As the COVID situation develops, our hearts ache as we think about all the people around the world that are affected by the pandemic Read morebut we are also encouraged by the stories of our readers finding help through our site. Article Edit. Learn why people trust wikiHow. To create this article, 32 people, some anonymous, worked to edit and improve it over time. Together, they cited 5 references.
This article has also been viewedtimes. Learn moreA trinomial is a polynomial with 3 terms. This page will focus on quadratic trinomials. The degree of a quadratic trinomial must be '2'. In other words, there must be an exponent of '2' and that exponent must be the greatest exponent. In fact, this is not even a trinomial because there are 2 terms. It's always easier to understand a new concept by looking at a specific example so you might want scroll down and do that first. In other words, we will use this approach whenever the coefficient in front of x 2 is 1.
Remember a negative times a negative is a positive. If you'd like, you can check your work by multiplying the two binomials and verify that you get the original trinomial.
Substitute that factor pair into two binomials. Note: since c is negative, we only need to think about pairs that have 1 negative factor and 1 positive factor. Remember a negative times a positive is a negative.
### Trinomial Factoring Calculator?
Factor Trinomial Worksheet. Factor Trinomial Calculator.
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Answer: A trinomial is a polynomial with 3 terms. Note: For the rest of this page, 'factoring trinomials' will refer to factoring 'quadratic trinomials'. The only difference being that a quadratic trinomial has a degree of 2. Video Tutorial of Factoring a Trinomial. Formula Steps. Next step. Step 4 Substitute that factor pair into two binomials.
### How to Factor a Polynomial Expression
Step 5 If you'd like, you can check your work by multiplying the two binomials and verify that you get the original trinomial. Popular pages mathwarehouse. Surface area of a Cylinder.
Unit Circle Game. Pascal's Triangle demonstration. Create, save share charts. Interactive simulation the most controversial math riddle ever!Polynomials are mathematical equations that contain variables and constants. They may also have exponents. The constants and the variables are combined by addition, while each term with the constant and the variable is connected to the other terms by either addition or subtraction.
Factoring polynomials is the process of simplifying the expression by division. In order to factor polynomials, you must determine whether it is a binomial or a trinomial, understand the standard factoring formats, find the greatest common factor, find which numbers corresponds to the product and sum of the various parts of the polynomial and then check your answer. Determine whether the polynomial is a binomial or a trinomial. A binomial has two terms, and a trinomial has three terms.
Understand the difference between the difference of two perfect squares, the sum of two perfect cubes and the difference of two perfect cubes. These types of polynomials are binomials and have a special format for factoring.
Find the greatest common factor. The greatest common factor is the highest number that is divisible by all of the constants in the polynomial. For example, in 4x, the greatest common factor is 4. Four divided by four is one, and 12 divided by four is three. By factoring out the four, the expression simplifies to 4 x Find the numbers which correspond to the product and the sum of the second and third terms of the polynomial.
This is how you factor trinomials. Check your answer. In order to make sure you factored the polynomial correctly, multiply the contents of the answer.
For example, for the answer 4 x-3you would multiply four by x, and then subtract four times three, such as 4x Since 4x is the original polynomial, your answer is correct. Mara Pesacreta has been writing for over seven years. She has been published on various websites and currently attends the Polytechnic Institute of New York University. Things You'll Need. About the Author.
Photo Credits. Copyright Leaf Group Ltd.Learning Objectives. View a video of this example. Note that if we multiply our answer out, we should get the original polynomial. In this case, it does check out.
Factoring gives you another way to write the expression so it will be equivalent to the original problem. Note that this is not in factored form because of the plus sign we have before the 5 in the problem.
To be in factored form, it must be written as a product of factors.
## How To Factor Trinomials
Step 1: Group the first two terms together and then the last two terms together. Step 2: Factor out a GCF from each separate binomial. Step 3: Factor out the common binomial. Where the number in front of x squared is 1.
Step 1: Set up a product of two where each will hold two terms. Step 2: Find the factors that go in the first positions.
As you are finding these factors, you have to consider the sign of the expressions:. So we go right into factoring the trinomial of the form. Anytime you are factoring, you need to make sure that you factor everything that is factorable.
Sometimes you end up having to do several steps of factoring before you are done. The difference between this trinomial and the one discussed above, is there is a number other than 1 in front of the x squared.
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This means, that not only do you need to find factors of cbut also a. Step 2: Use trial and error to find the factors needed. The trick is to get the right combination of these factors.
You can check this by applying the FOIL method. If your product comes out to be the trinomial you started with, you have the right combination of factors. If the product does not come out to be the given trinomial, then you need to try again. In the second terms of the binomials, we need factors of 2. This would have to be 2 and 1.
I used positives here because the middle term is positive. Also, we need to make sure that we get the right combination of these factors so that when we multiply them out we get. Second try:. This is our original polynomial. So this is the correct combination of factors for this polynomial.
In the second terms of the binomials, we need factors of The process of factoring is essential to the simplification of many algebraic expressions and is a useful tool in solving higher degree equations. In fact, the process of factoring is so important that very little of algebra beyond this point can be accomplished without understanding it.
In earlier chapters the distinction between terms and factors has been stressed. You should remember that terms are added or subtracted and factors are multiplied. Three important definitions follow. Terms occur in an indicated sum or difference. Factors occur in an indicated product. Note in these examples that we must always regard the entire expression. Factors can be made up of terms and terms can contain factors, but factored form must conform to the definition above.
Factoring is a process of changing an expression from a sum or difference of terms to a product of factors. Note that in this definition it is implied that the value of the expression is not changed - only its form. Upon completing this section you should be able to: Determine which factors are common to all terms in an expression.
Factor common factors. In general, factoring will "undo" multiplication. Next look for factors that are common to all terms, and search out the greatest of these.
This is the greatest common factor. In this case, the greatest common factor is 3x. The terms within the parentheses are found by dividing each term of the original expression by 3x. Note that this is the distributive property. It is the reverse of the process that we have been using until now. The original expression is now changed to factored form.
To check the factoring keep in mind that factoring changes the form but not the value of an expression. If the answer is correct, it must be true that. Multiply to see that this is true. A second check is also necessary for factoring - we must be sure that the expression has been completely factored. In other words, "Did we remove all common factors? Can we factor further? Multiplying to check, we find the answer is actually equal to the original expression.
However, the factor x is still present in all terms. Hence, the expression is not completely factored. This expression is factored but not completely factored. For factoring to be correct the solution must meet two criteria: It must be possible to multiply the factored expression and get the original expression.
FThe expression must be completely factored. At this point it should not be necessary to list the factors of each term. You should be able to mentally determine the greatest common factor.Factoring a polynomial is the opposite process of multiplying polynomials.
Recall that when we factor a number, we are looking for prime factors that multiply together to give the number; for example. When we factor a polynomial, we are looking for simpler polynomials that can be multiplied together to give us the polynomial that we started with.
You might want to review multiplying polynomials if you are not completely clear on how that works. The simplest type of factoring is when there is a factor common to every term. In that case, you can factor out that common factor.
What you are doing is using the distributive law in reverse—you are sort of un-distributing the factor. Notice that each term has a factor of 2 xso we can rewrite it as:. If you see something of the form a 2 - b 2you should remember the formula. We are interested here in factoring quadratic trinomials with integer coefficients into factors that have integer coefficients.
Factor Polynomials - Understand In 10 min
Not all such quadratic polynomials can be factored over the real numbers, and even fewer into integers they all can be factored of we allow for imaginary numbers and rational coefficients, but we don't. Therefore, when we say a quadratic can be factored, we mean that we can write the factors with only integer coefficients. If a quadratic can be factored, it will be the product of two first-degree binomials, except for very simple cases that just involve monomials.
For example x 2 by itself is a quadratic expression where the coefficient a is equal to 1, and b and c are zero. Obviously, x 2 factors into x xbut this is not a very interesting case. A slightly more complicated case occurs when only the coefficient c is zero. Then you get something that looks like.
We look at two cases of this type. Now look at this and think about where the terms in the trinomial came from. Obviously the x 2 came from x times x. The last term in the trinomial, the 6 in this case, came from multiplying the 2 and the 3.
Where did the 5 x in the middle come from? We got the 5 x by adding the 2 x and the 3 x when we collected like terms.
## comments on “Factoring polynomials completely”
#### Vikazahn
Ich kann Ihnen empfehlen, die Webseite, mit der riesigen Zahl der Informationen nach dem Sie interessierenden Thema zu besuchen. |
6
Q:
# The average of 26,29,n,35 and 43 lies between 25 and 35. If n is always an integer and greater than the average of the given integers then the value of n is :
A) 33 B) 34 C) 33 D) None of these
Explanation:
Average of 26,29,35 and 43 is 33.25 . Also the average of 26 , 29, n, 35 and 43 lies between 25 and 35 i.e,
$25<26+29+n+35+435<35$
=> 125 < 26+29+n+35+43 < 175
=> 125 < 133 + n < 175
=> n < 42
Since the value of n is an integer and greater than 33.25 then 33 < n < 42 for every integer n.
Q:
The average score of boys in an examination in a school is 81 and that of the girls is 83 and the average score of the school is 81.8. Then what will be ratio between the number of girls to the number of boys, in the examination?
A) 3:2 B) 2:3 C) 4:3 D) 3:4
Explanation:
Let the number of boys = b
Let the number of girls = g
From the given data,
81b + 83g = 81.8(b + g)
81.8b - 81b = 83g - 81.8g
0.8b = 1.2g
b/g = 1.2/0.8 = 12/8 = 3/2
=> g : b = 2 : 3
Hence, ratio between the number of girls to the number of boys = 2 : 3.
2 228
Q:
Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, what will be the sum of the highest number and the lowest number?
A) 120 B) 160 C) 80 D) 60
Explanation:
Let the three numbers be x, y, z.
From the gien data,
x = 2y ....(1)
x = z/2 => z = 2(2y) = 4y .....(From 1) ...........(2)
Given average of three numbers = 56
Then,
Now,
x = 2y => x = 2 x 24 = 48
z = 4y = 4 x 24 = 96
Now, the highest number is z = 96 & smallest number is y = 24
Hence, required sum of highest number and smallest number
= z + y
= 96 + 24
= 120.
2 359
Q:
How many window coverings are necessary to span 50 windows, if each window covering is 15 windows long?
A) 4 B) 15 C) 3 D) 50
Explanation:
Given that,
Number of windows = 50
Each window covering covers 15 windows
=> 50 windows requires 50/15 window coverings
= 50/15 = 3.333
Hence, more than 3 window coverings are required. In the options 4 is more than 3.
Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.
1 1420
Q:
Ian has 14 boxes of paper and divides them evenly between 4 coworkers. How many whole boxes did each coworker get?
A) 2 B) 2.5 C) 3 D) 3.5
Explanation:
Given number of boxes = 14
Number of workers = 4
Now, number of whole boxes per worker = 14/4 = 3.5
Hence, number of whole boxes per each coworker = 3
1 4318
Q:
Five boxes of bananas sell for Rs. 30. how many boxes can you buy for Rs.9?
A) 2 B) 1.5 C) 1.25 D) 2.5
Explanation:
Given Five boxes of bananas sell for Rs. 30.
=> 1 Box of Bananas for = 30/5 = Rs. 6
Then, for Rs. 9
=> 9/6 = 3/2 = 1.5
Hence, for Rs. 9, 1.5 box of bananas can buy.
8 2777
Q:
Third proportion of 10 and 20 is
A) 30 B) 40 C) 25 D) 20
Explanation:
The third proportional of two numbers p and q is defined to be that number r such that
p : q = q : r.
Here, required third proportional of 10 & 20, and let it be 'a'
=> 10 : 20 = 20 : a
10a = 20 x 20
=> a = 40
Hence, third proportional of 10 & 20 is 40.
8 1741
Q:
The average weight of 45 passengers in a bus is 52 kg. 5 of them whose average weight is 48 kg leave the bus and other 5 passengers whose average weight is 54 kg join the bus at the same stop. What is the new average weight of the bus?
A) 54.21 kgs B) 51.07 kgs C) 52.66 kgs D) 53.45 kgs
Explanation:
Given total number of passengers in the bus = 45
First average weight of 45 passengers = 52 kgs
Average weight of 5 passengers who leave bus = 48
Average weight of passengers who joined the bus = 54
Therefore, the net average weight of the bus is given by
24 2733
Q:
12 boys decided to constribute Rs. 750 each to an Orphange. Suddenly few of them boys dropped out and consequently the rest had to pay Rs. 150 more. Then the number of boys who dropped out?
A) 4 B) 6 C) 2 D) 3
Explanation:
Total money decided to contribute = 750 x 12 = 9000
Let 'b' boys dropped
The rest paid 150/- more
=> (12 - b) x 900 = 9000
=> b = 2
Hence, the number of boys who dropped out is 2. |
Patrick Dempsey Under The Microscope (11/09/2019)
How will Patrick Dempsey fare on 11/09/2019 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is not scientifically verified – take it with a grain of salt. I will first calculate the destiny number for Patrick Dempsey, and then something similar to the life path number, which we will calculate for today (11/09/2019). 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 11/09/2019: We will analyze the month (11), the day (09) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. We’ll show you how it works now. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 09 we do 0 + 9 = 9. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 9 + 12 = 23. This still isn’t a single-digit number, so we will add its digits together again: 2 + 3 = 5. Now we have a single-digit number: 5 is the path number for 11/09/2019.
DESTINY NUMBER FOR Patrick Dempsey: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Patrick Dempsey we have the letters P (7), a (1), t (2), r (9), i (9), c (3), k (2), D (4), e (5), m (4), p (7), s (1), e (5) and y (7). Adding all of that up (yes, this can get tedious) gives 66. This still isn’t a single-digit number, so we will add its digits together again: 6 + 6 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the destiny number for Patrick Dempsey.
CONCLUSION: The difference between the path number for today (5) and destiny number for Patrick Dempsey (3) is 2. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is of questionable accuracy. If you want a forecast that we do recommend taking seriously, 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. |
# A soft drink is available in two packs - (i) a tin can with a rectangular base of length $5 \mathrm{~cm}$ and width $4 \mathrm{~cm}$, having a height of $15 \mathrm{~cm}$ and (ii) a plastic cylinder with circular base of diameter $7 \mathrm{~cm}$ and height $10 \mathrm{~cm}$. Which container has greater capacity and by how much?
AcademicMathematicsNCERTClass 9
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Given:
A soft drink is available in two packs -
(i) a tin can with a rectangular base of length $5\ cm$ and width $4\ cm$, having a height of $15\ cm$ and
(ii) a plastic cylinder with circular base of diameter $7\ cm$ and height $10\ cm$.
To do:
We have to find the container that has greater capacity and by how much.
Solution:
In the first case,
The length of the base of the tin can $= 5\ cm$
Width of the tin can $= 4\ cm$
Height of the tin can $= 15\ cm$
Therefore,
Volume of the soft drink $= lbh$
$= 5 \times 4 \times 15$
$= 300\ cm^3$
In the second case,
Diameter of the base of the plastic cylinder $= 7\ cm$
Radius of the plastic cylinder $=\frac{7}{2} \mathrm{~cm}$
Height of the plastic cylinder $=10 \mathrm{~cm}$
Therefore,
Volume of the soft drink $=\pi r^{2} h$
$=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 10$
$=385 \mathrm{~cm}^{2}$
The soft drink in the second container is greater by $385\ cm^3 - 380\ cm^3 = 85\ cm^3$.
Hence, the plastic cylinder has a greater capacity of $85\ cm^3$.
Updated on 10-Oct-2022 13:46:38
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## Wednesday, 9 January 2013
### PRIME, HCF, LCM - SIEVE
INTRODUCTION
Search the definition of the following using reliable websites (post as comments and remember to quote the sites in your comments)
1. PRIME NUMBER
2. COMPOSITE NUMBER
Question:
3. Are all odd numbers prime? Justify
4. Are all even numbers composite numbers? Justify
SIEVE OF ERASTOTHENES
View the following video before completing the exercise in the booklet.
Exercise Respect and Responsibility while watching
1. 1&2:
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example, 5 is prime because only 1 and 5 divide it, whereas 6 is composite because it has the divisors 2 and 3 in addition to 1 and 6.
http://en.wikipedia.org/wiki/Prime_number
3:Since a prime number is a number that that can be divided by 1or itself, then the rule does not apply to some of the odd numbers:9, 15, 21, and 27
4:No,not all even numbers are composite numbers. one example is the number two. It may be an even number, but it is a prime number, as it is only able to be divided by the number 1 and itself,2.
2. Definition of a prime number : A prime number is a number is a number whose only factors are 1 and itself. That means there is no whole number that evenly divides the prime number. (source:http://www.mathwarehouse.com/arithmetic/numbers/prime-number/)
Definition of a composite number : A Composite Number can be divided evenly by numbers other than 1 or itself.
Example: 9 can be divided evenly by 1, 3 and 9, so 9 is a composite number.
But 7 can only be divided evenly by 1 and 7, so 7 is NOT a composite number (it is a Prime Number).
(source:http://www.mathsisfun.com/definitions/composite-number.html)
---------------------------------------------------------------------------------------------------------------------------
Are all odd numbers prime ? : Its a yes and now i guess . Considering that all even numbers are divisible by 2 , it means that the only even prime number is 2 itself as a prime number must only have two factors - one and itself . All the other prime numbers are odd numbers . So , yes and no to the question but mainly yes , all prime numbers are odd except for one (two)
Are all even numbers composite numbers ? : Again , yes and no . Most even numbers are composite numbers except for numbers like 27 and 49 which are composite numbers but are odd numbers , not even numbers . In addition , 2 is an even number but it is not a composite number - it is a prime number .
3. A prime number is a natural number greater than 1 that has no other divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number.
http://en.wikipedia.org/wiki/Prime_number
Not all odd numbers are prime numbers, eg. 9, 15, 21 can be divided by more than one number. http://wiki.answers.com/Q/Are_all_odd_numbers_prime_numbers
No. 2 is even, but it's a prime number, and not a composite number. http://answers.yahoo.com/question/index?qid=20091012175114AAN2ikf
4. 1. A Prime Number can be divided evenly only by 1, or itself and it must be a whole number greater than 1. (www.mathsisfun.com/definitions/prime-number.html)
2. A Composite Number can be divided evenly by numbers other than 1 or itself. (http://www.mathsisfun.com/definitions/composite-number.html)
3. Not all odd numbers are prime numbers, eg. 9, 15, 21 can be divided by more than one number. (9 can be divided by 1, 3 and 9). (http://wiki.answers.com/Q/Are_all_odd_numbers_prime_numbers)
4. Not all even numbers are composite. Take 2 for example; its an even number but it only has 2 factors; 1 and itself. (http://wiki.answers.com/Q/Are_all_even_numbers_composite_why)
5. 1.A prime number is a number that is a whole number and have exactly 2 factors and that is the number 1 and the number itself.
2.Composite numbers are numbers that are whole numbers other than 1 and prime numbers ,composite numbers have more than 2 factors.(source: person longman maths insights express 1A)
3.No not all odd numbers are prime 9,27 have more than 2 factors 9: 1,3,9 27:1,3,9,27.so not all odd numbers are prime.
4.No not all even numbers are primes.2 has 2 factors so it is a prime number,not a composite number
6. 1.A prime number is a number that can be divided by 1 and itself.
2.A composite number have more than only two factors.
3.No,21 can be divided by 7and 3, 15 can be divided by 5 and 3.
4.No,Two only can be divided by itself and one.
7. 1. A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example, 5 is prime because only 1 and 5 divide it, whereas 6 is composite because it has the divisors 2 and 3 in addition to 1 and 6.
http://en.wikipedia.org/wiki/Prime_number
2. A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words a composite number is any positive integer greater than one that is not a prime number.
http://en.wikipedia.org/wiki/Composite_number
3. Not all odd numbers are prime. E.g. the number 49 is odd, but it is not a prime.
4. Not all even numbers are composite. The number 2 is an even number, but it is not a composite.
8. 1. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.or example,
2. A natural number greater than 1 that is not a prime number is called a composite number. Ex. (6 is composite because it has the divisors 2 and 3 in addition to 1 and 6.)
Source : http://en.wikipedia.org/wiki/Prime_number
3. Not all odd numbers are prime numbers, eg. 9, 15, 21 can be divided by more than one number.
4.No. A composite number has factors in addition to one and itself while 2 has only 2 factors, 1 and itself.
9. 1. A prime number is an integer that cannot be factorized into other integers but is only divisible by itself or 1, such as 2, 3, 5, 7, and 11.
2. A composite number is an integer that can be divided by at least two other integer besides itself and 1 without leaving a remainder.
3. Not all odd numbers are prime. For instance, 9, 21, and 27 are multiples of 3 and though they are odd, they are not prime numbers.
4. No, 2 is the only even number that is prime as it can be divided by 1 and itself.
(http://www.thefreedictionary.com/composite+number)
10. 1.A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
2. A natural number greater than 1 that is not a prime number is called a composite number
3.no, because some odd numbers can be divided by another number other than 1 and itself.
4. no because 2 is an even number and it is a prime, though the rest are composites.
1. http://en.wikipedia.org/wiki/Prime_numbers
11. 1:A positive integer that is not divisible without remainder by any integer except itself and 1, with 1 often excluded
2:A number that is a multiple of at least two whole numbers other than itself and 1.
3:No, some odd numbers can still be divided by a number other than itself and 1, e.g. 9 (9/3=3)
4:No, the only non-composite number which is larger than 1, an even number but not able to be divided by at least two whole numbers other than itself and 1, is 2. The factor of 2 is 2 and 1, therefore 2 is a prime and not a composite number.
Source:http://dictionary.reference.com/
12. A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words a composite number is any positive integer greater than one that is not a prime number.
(Source: http://en.wikipedia.org/wiki/Prime_number,http://en.wikipedia.org/wiki/Composite_number)
13. 3. A prime number is a number that can only be divided by 1 or by itself. Not all odd numbers are prime numbers, eg. 9, 15, 21 can be divided by more than one number.
4. No. 2 is prime. All the rest are divisible by 2,so they are composite.
14. 1. an integer that has no integral factors but itself and 1. (wordnetweb.princeton.edu/perl/webwn)
2. A Composite Number can be divided evenly by numbers other than 1 or itself. (http://www.mathsisfun.com/definitions/composite- number.html)
3. No. 9 is an odd number but it is not a prime number.
4. No. 2 is an even number but it is not a composite number.
15. 1. A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
2. A natural number greater than 1 that is not a prime number is called a composite number.
(1&2= http://en.wikipedia.org/wiki/Prime_number )
3. No, some odd numbers may be multiples of other numbers.
4. No, 2 is not a composite number because it can be divided by only 1 and itself.
16. 1. A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example, 5 is prime because only 1 and 5 divide it, whereas 6 is composite because it has the divisors 2 and 3 in addition to 1 and 6. The fundamental theorem of arithmetic establishes the central role of primes in number theory: any integer greater than 1 can be expressed as a product of primes that is unique up to ordering. The uniqueness in this theorem requires excluding 1 as a prime because it is the multiplicative identity.
2. composite number is a positive integer that has at least one positive divisor other than one or itself. In other words a composite number is any positive integer greater than one that is not a prime number.
(http://en.wikipedia.org/wiki/)
3. No. 9 is an odd number but it is not a prime number.
4. No, 2 is not a composite number because it can be divided by only 1 and itself.
17. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. (e.g. 2, 3, 5, 7, 13...)
A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words a composite number is any positive integer greater than one that is not a prime number. ( e.g. 4, 6, 8, 9, 10...)
Source: Wikipedia |
Scientific Notation and Significant Figures Explained Discover the power of scientific notation and significant figures in math and science. Learn to express extreme values efficiently and report results accurately for success in various scientific fields.
Now Playing:Introduction scientific notation and significant figures – Example 0a
Intros
1. Introduction to significant figures
2. Significant figures and scientific notation.
3. How to use very small or very large numbers.
Examples
1. Apply significant figures and scientific notation/standard form to very large and small numbers.
Boltzmann's constant and Avogadro's number are two important constants in chemistry.
The approximate value of Boltzmann's constant is 0.0000000000000000000000138064852
The approximate value of Avogadro's constant is 602214085700000000000000
1. Round both of these values to 2 decimal places.
Afterward, round both of these values to 3 significant figures.
2. Round both of these values to 3 significant figures using standard form.
Introduction to the periodic table
Notes
In this lesson, we will learn:
• Reasons for using significant figures and scientific notation.
• Situations where rounding, scientific notation and significant figures are appropriate.
• Rules for approximating and reporting values in science.
Notes:
• When doing any calculations in science or math, using standard form and significant figures is useful when dealing with extremely large or extremely small numbers.
• These techniques are used to avoid time wasting writing long meaningless numbers when dealing with extreme numerical values.
• A significant figure is any non-zero number in a value, or any zeros between non-zero numbers in a value.
• Standard form, AKA scientific notation, is written in the format a x 10$^b$. Here, a is your original number set between 1 and 10 and b is the number of places the decimal point got moved. To change a number to standard form:
• Step one: Write your original number with the decimal place in, for example 150 written as 150.00.
• Step two: Set your number as between 1 and 10 by moving the decimal place, for example 150.00 becomes 1.5000.
• Step three: Count the number of decimal places your decimal point had to move for step 2. Moving to the left should be counted as 1, moving to the right should be counted as –1. For example 1.5000 from 150.00 was two moves to the left, so 2.
• Working this way means numbers larger than 1-10 have a positive number of decimal places moved and numbers smaller than 1-10 should have a negative number of decimal places moved.
• Step four: Multiply your number from step 2 by 10 to the power of your number in step 3. For example: 1.5000 x 102
• Further examples: 1,650,000 could be written 1.65 x 106 or 0.0000592 could be written 5.92 x 10$^{-5}$.
• If nothing is specifically requested then some general rules are:
• Try not to round your values until the final stage or answer of a calculation.
• Do not round to a large degree at an early stage calculation, then to a small degree (for example, don't round to 3 significant figures then later to 5 S.F.)
• Use scientific notation (standard form) for any values larger than 1 x 10$^3$ or smaller than 1 x 10$^{-3}$ or on those orders of magnitude.
Concept
Introduction to Scientific Notation and Significant Figures
Scientific notation and significant figures are fundamental concepts in scientific calculations and measurements. Scientific notation is a standardized way of expressing very large or very small numbers using powers of 10, making them easier to read and manipulate. Significant figures, on the other hand, indicate the precision of a measurement by showing which digits are reliable. Our introduction video provides a comprehensive overview of these essential topics, helping students grasp their importance in scientific work. Understanding scientific notation allows for efficient representation of extreme values, while mastering significant figures ensures accurate reporting of experimental results. These concepts are crucial in various scientific fields, from physics and chemistry to astronomy and engineering. By watching the introduction video, learners will gain a solid foundation in these principles, enabling them to perform calculations with confidence and interpret scientific data accurately. Mastering scientific notation and significant figures is a key step in developing strong scientific literacy and critical thinking skills.
Example
In this guide, we will explore the concepts of significant figures and scientific notation, which are essential for accurately reporting numbers in scientific and mathematical calculations. These methods help maintain the meaning of values, especially when dealing with very large or very small numbers.
Step 1: Understanding the Importance of Significant Figures
Significant figures are crucial in scientific measurements because they help convey the precision of a value. When we measure something in science, we need to ensure that the reported value reflects the accuracy of the measurement. Significant figures include all non-zero numbers, any zeros between non-zero numbers, and any trailing zeros in a decimal number.
For example, in the number 105, the digits 1 and 5 are significant, and the zero between them is also significant. In the number 100.5, the digits 1, 0, 0, and 5 are all significant. However, in a number like 10,000, the trailing zeros are not considered significant unless specified otherwise.
Step 2: The Problem with Decimal Places
Using decimal places to report very large or very small numbers can be problematic. For very large numbers, adding extra decimal places often adds no value. For instance, the difference between 10 billion and 10 billion point five is negligible compared to the overall magnitude of the number. Similarly, for very small numbers, rounding to a few decimal places can cause a loss of meaning. For example, rounding the mass of an atom to two decimal places would lose the precision needed for scientific accuracy.
Step 3: Defining Significant Figures
A significant figure is any non-zero number or a zero between non-zero numbers. For example, in the number 105, the digits 1 and 5 are significant, and the zero between them is also significant. In the number 100.5, the digits 1, 0, 0, and 5 are all significant. However, in a number like 10,000, the trailing zeros are not considered significant unless specified otherwise.
To illustrate, consider the numbers 105, 100.5, and 10 million. In 105, the digits 1 and 5 are significant, and the zero between them is also significant. In 100.5, the digits 1, 0, 0, and 5 are all significant. In 10 million, the digit 1 is significant, but the trailing zeros are not unless specified otherwise.
Step 4: Introduction to Scientific Notation
Scientific notation is a method of writing very large or very small numbers in a more concise form. It is especially useful in science, where such numbers are common. Scientific notation expresses numbers as a product of a number between 1 and 10 and a power of 10. For example, the number 1,000,000 can be written as 1.0 x 10^6.
To convert a number to scientific notation, you need to move the decimal point to create a number between 1 and 10. For example, to convert 10,000,000 to scientific notation, you would move the decimal point 7 places to the left, resulting in 1.0 x 10^7. Similarly, to convert 0.0000003 to scientific notation, you would move the decimal point 7 places to the right, resulting in 3.0 x 10^-7.
Step 5: Applying Scientific Notation
Let's apply scientific notation to a few examples. Consider the number 10,000,000. To convert this to scientific notation, we move the decimal point 7 places to the left, resulting in 1.0 x 10^7. For the number 0.0000003, we move the decimal point 7 places to the right, resulting in 3.0 x 10^-7.
Another example is the number 105. To convert this to scientific notation, we move the decimal point 2 places to the left, resulting in 1.05 x 10^2. For the number 100.5, we move the decimal point 2 places to the left, resulting in 1.005 x 10^2.
Step 6: Benefits of Using Significant Figures and Scientific Notation
Using significant figures and scientific notation helps maintain the precision and meaning of values in scientific calculations. It also saves time and reduces errors when dealing with very large or very small numbers. By following the rules for significant figures and scientific notation, you can ensure that your reported values are accurate and meaningful.
In summary, significant figures and scientific notation are essential tools for accurately reporting numbers in science and mathematics. They help maintain the precision and meaning of values, especially when dealing with very large or very small numbers. By understanding and applying these concepts, you can improve the accuracy and clarity of your scientific calculations.
FAQs
Here are some frequently asked questions about scientific notation and significant figures:
1. How do you write significant figures in scientific notation?
To write significant figures in scientific notation, express the number as a coefficient between 1 and 10, multiplied by a power of 10. Include all significant digits in the coefficient. For example, 0.00345 with three significant figures becomes 3.45 × 10^-3.
2. How many significant figures does 10.0 have?
10.0 has three significant figures. The decimal point indicates that the zero is significant, as it provides information about the precision of the measurement.
3. What is 0.00186 written in three significant figures with scientific notation?
0.00186 written in three significant figures with scientific notation is 1.86 × 10^-3. The coefficient 1.86 contains the three significant figures, and the exponent -3 represents the position of the decimal point.
4. How many significant figures does 100.0 have?
100.0 has four significant figures. The decimal point indicates that all digits, including the trailing zero, are significant.
5. What is the standard notation for Avogadro's number?
The standard notation for Avogadro's number is 6.022 × 10^23 mol^-1. This represents approximately 602,200,000,000,000,000,000,000 particles per mole of a substance.
Prerequisites
Understanding scientific notation and significant figures is crucial in various scientific and mathematical fields. To fully grasp these concepts, it's essential to have a solid foundation in several prerequisite topics. These fundamental skills will enhance your ability to work with large numbers, perform precise calculations, and communicate scientific data effectively.
One of the key prerequisites is scientific notation. This topic is directly related to our main subject, as it forms the basis for expressing very large or very small numbers in a concise and standardized format. Mastering scientific notation is crucial for converting between standard and scientific notation, which is a fundamental skill when dealing with significant figures.
Another important prerequisite is understanding how to multiply fractions and whole numbers. This skill is particularly relevant when working with significant figures in whole numbers and fractions. Being able to manipulate these numbers accurately is essential for maintaining the correct number of significant figures in calculations and measurements.
The quotient rule of exponents is also a crucial prerequisite topic. This rule is particularly important when dealing with exponents and powers of 10, which are fundamental to scientific notation. Understanding how to simplify and manipulate exponents is essential for working efficiently with scientific notation and maintaining the correct number of significant figures.
By mastering these prerequisite topics, you'll be better equipped to handle the intricacies of scientific notation and significant figures. For instance, when converting from scientific notation to standard form or vice versa, you'll need to apply your knowledge of exponents and powers of 10. Similarly, when performing calculations with significant figures, your understanding of multiplying fractions and whole numbers will ensure you maintain the appropriate level of precision.
Moreover, these prerequisite skills will help you avoid common errors and misconceptions. For example, a solid grasp of scientific notation will prevent mistakes in placing decimal points when dealing with very large or small numbers. Understanding the quotient rule of exponents will help you simplify complex expressions in scientific notation more efficiently.
In conclusion, a strong foundation in these prerequisite topics is essential for mastering scientific notation and significant figures. By investing time in understanding these fundamental concepts, you'll be better prepared to tackle more advanced problems and applications in scientific and mathematical fields. Remember, each of these topics builds upon the others, creating a comprehensive skill set that will serve you well in your studies and future career. |
Important MCQ questions Class 10 Maths Chapter 6 - Triangles - GMS - Learning Simply
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# Important MCQ questions Class 10 Maths Chapter 6 - Triangles
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# Important MCQ questions Class 10 Maths Chapter 6 - Triangles
Class 10 Maths Chapter 6 (Triangles) MCQs are provided here online for students, who are preparing for the board exam, along with their answers. The objective questions are as per the latest CBSE syllabus and NCERT curriculum. These multiple-choice questions for triangles are presented here with detailed explanations, by which students can easily score good marks.
## Class 10 Maths MCQs for Triangles
Multiple choice questions for triangles are given below for students to enhance their problem-solving abilities and confidence. Triangles are the most common concept whose applications are seen frequently in day to day life. Solve the objective questions here and learn more about triangles. Get important questions for class 10 Maths here as well.
We have provided some important questions of this chapter, along with the detailed solutions. After that, we have also provided some questions for students practice which does not have solutions. They must solve all of them to gain command over Triangles topic.
Below are the MCQs for Triangles
1. Which of the following triangles have the same side lengths?
(a)Scalene
(b)Isosceles
(c)Equilateral
(d)None of these
Explanation: Equilateral triangles have all its sides and all angles equal.
2. Area of an equilateral triangle with side length a is equal to:
(a)√3/2a
(b)√3/2a2
(c)√3/4 a2
(d)√3/4 a
3.D and E are the midpoints of side AB and AC of a triangle ABC, respectively and BC=6cm. If DE || BC, then the length of DE is:
(a)2.5
(b)3
(c)5
(d)6
Explanation: By midpoint theorem,
DE=½ BC
DE = ½ of 6
DE=3cm
4. The diagonals of a rhombus are 16cm and 12cm, in length. The side of rhombus in length is:
(a)20cm
(b)8cm
(c)10cm
(d)9cm
Explanation: Here, half of the diagonals of a rhombus are the sides of the triangle and side of the rhombus is the hypotenuse.
By Pythagoras theorem,
(16/2)2+(12/2)2=side2
82+62=side2
64+36=side2
side=10cm
5. Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of small triangle is 48 sq.cm, then the area of large triangle is:
(a)230 sq.cm.
(b)106 sq.cm
(c)107 sq.cm.
(d)108 sq.cm
Solution: Let A1 and A2 are areas of the small and large triangle.
Then,
A2/A1=(side of large triangle/side of small triangle)
A2/48=(3/2)2
A2=108 sq.cm.
6. If perimeter of a triangle is 100cm and the length of two sides are 30cm and 40cm, the length of third side will be:
(a)30cm
(b)40cm
(c)50cm
(d)60cm
Solution: Perimeter of triangle = sum of all its sides
P = 30+40+x
100=70+x
x=30cm
7. If triangles ABC and DEF are similar and AB=4cm, DE=6cm, EF=9cm and FD=12cm, the perimeter of triangle is:
(a)22cm
(b)20cm
(c)21cm
(d)18cm
Explanation: ABC ~ DEF
AB=4cm, DE=6cm, EF=9cm and FD=12cm
AB/DE = BC/EF = AC/DF
4/6 = BC/9 = AC/12
BC = (4.9)/6 = 6cm
AC = (12.4)/6 = 8cm
Perimeter = AB+BC+AC
= 4+6+8
=18cm
8. The height of an equilateral triangle of side 5cm is:
(a)4.33
(b)3.9
(c)5
(d)4
Explanation:The height of the equilateral triangle ABC divides the base into two equal parts at point D.
Therefore,
BD=DC= 2.5cm
In triangle ABD, using Pythagoras theorem,
9. If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if
(a)∠A=∠F
(b)∠B=∠D
(c)∠A=∠D
(d)∠B=∠E
10. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio
(a)2: 3
(b)4: 9
(c)81: 16
(d)16: 81
Explanation: Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16: 81
11. If △ ABC ~ △ DEF such that AB = 12 cm and DE = 14 cm. Find the ratio of areas of △ ABC and △ DEF.
1. 49/9
2. 36/49
3. 49/16
4. 25/49
Solution: We know that the ratio of areas of two similar triangles is equal
to the ratio of the squares.
Of any two corresponding sides,
area of △ ABC / area of △ DEF = (AB/DE) 2= (12/14) 2= 36/49
12. D and E are points on the sides AB and AC respectively of a △ABC such that DE || BC. Which of the following statement is true?
1. △ ADE ~ △ ABC
1. only (iii)
2. only (i)
3. only (i) and (ii)
4. all (i) , (ii) and (iii)
Solution:
In △ ADE and △ ABC, we have
[Since DE || BC ∠ ADE = ∠ B (Corresponding angles)]
and, ∠ A = △ A [Common]
Therefore, (area of △ ADE / area of △ ABC) = (AD2/AB2)
13.
In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm and area (ΔQOA) = 150 cm2, find the area of ΔPOB.
1. 233 cm2
2. 294 cm2
3. 300 cm2
4. 420 cm2
Solution: Consider Δ~QOA and Δ POB
QA || PB,
Therefore, ∠ AQO = ∠ PBO [Alternate angles]
∠ QAO = ∠ BPO [Alternate angles]
and
∠ QOA = ∠ BOP [Vertically opposite angles]
Δs QOA ~ BOP [by AAA similarity]
Therefore, (OQ/ OB) = (OA/OP)
Now, area (POB)/ area (QOA) = (OP) 2/ (OA) 2= 72/ 52
Since area (QOA) =150cm2
⇒area (POB) =294cm2
14. Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The ratio of corresponding heights is:
1. 4:5
2. 5:4
3. 3:2
4. 5:7
Solution: For similar isosceles triangles,
Area (Δ1) / Area (Δ2) = (h1)2 / (h2)2
(h1 / h2) = 4/5
15. In △ABC, AB = 3 and, AC = 4 cm and AD is the bisector of ∠A. Then, BD : DC is —
1. 9: 16
2. 4:3
3. 3:4
4. 16:9
Solution:
The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides (It may be similar or may not depending on type of triangle it divides)
In △ABC
as per the statement AB/ AC= BD/DC i.e. a/b= c/d
So, BD/ DC= AB/AC= ¾
So, BD: DC = 3: 4
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Chapter 1.1 (Mathematics 4th)
# Place value of numbers up to tens of thousands
• Place value up to thousands
• Place value of numbers up to tens of thousands
## Place value up to thousands
### Example 1
What is the place value of each digit in 368?
##### Working
Place value Digit Hundreds 3 Tens 6 Ones 8
• The place value of digit 3 is hundreds.
• The place value of digit 6 is tens.
• The place value of digit 8 is ones.
### Example 2
Write the place value of each digit in the number 2 341.
##### Working
Place value Digit Thousands 2 Hundreds 3 Tens 4 Ones 1
• The place value of digit 2 is thousands.
• The place value of digit 3 is hundreds.
• The place value of digit 4 is tens.
• The place value of digit 1 is ones.
## Exercise A
### Question 1
Show the place value of the digits in the following number.
1. 529
2. 613
3. 802
### Question 2
Use an abacus to represent the place value of the digit in each of the following numbers.
1. 2 434
2. 5 086
3. 9 034
Number Th H T O a) 735 b) 900 c) 1 948 d) 2 018 e) 5 314 f) 8 059
1. 826
2. 1 408
3. 2 178
4. 9 634
A shopkeeper had sh 2 785 in a cash box. What is the place value of digit 7?
∴ The place value of digit 7 is
## Place value of numbers up to tens of thousands
Complete the pattern.
9 + 1 = 10 99 + 1 = 100 999 + 1 = 1 000 9 999 + 1 = 10 000
1. The next number after 9 999 is 10 000.
2. The place value of digit 1 in 10 000 is ten thousands.
### Example 3
What is the place value of digit 4 in 43 905?
##### Working
TTh Th H T O 4 3 9 0 5
The place value of digit 4 is ten thousands.
## Exercise B
Number TTh Th H T O a) 2 063 b) 1 975 c) 8 462 d) 2 011 e) 3 689 f) 9 709
1. 842
2. 2 374
3. 49 621
4. 84 005
Digit Place value 8 Ten ThousandsThousandsHundredsTensOnes 9 Ten ThousandsThousandsHundredsTensOnes 7 Ten ThousandsThousandsHundredsTensOnes 5 Ten ThousandsThousandsHundredsTensOnes 6 Ten ThousandsThousandsHundredsTensOnes
There were 68 920 people in a location. What is the place value of digit 6 in the number?
∴ The place value of digit 6 is
Use IT device to learn more on place value of numbers up to tens of thousands.
What did you learn?
∴ |
# Difference of squares Rule
The difference of the squares of any two quantities is equal to the product of their sum and difference, is called the difference of squares rule.
## Introduction
Let the literals $a$ and $b$ denote two variables.
1. The addition of them forms a binomial $a+b$.
2. The subtraction of them forms another binomial $a-b$.
3. The difference of their squares also forms another binomial $a^2-b^2$.
According to a mathematical property, the difference of the squares of any two quantities is exactly equal to the product of their sum and difference. It can be written in mathematical form as follows.
$\implies$ $a^2-b^2$ $\,=\,$ $(a+b) \times (a-b)$
$\,\,\,\therefore\,\,\,\,\,\,$ $a^2-b^2$ $\,=\,$ $(a+b)(a-b)$
It is an algebraic identity and it is called the difference of squares law.
### Other form
The difference of squares identity is also expressed in terms of variables $x$ and $y$ alternatively.
$x^2-y^2$ $\,=\,$ $(x+y)(x-y)$
#### Uses
The difference of squares algebraic identity is mainly used as a formula in two cases in mathematics.
1. It is used to find the difference of squares of two quantities by calculating the product of sum and difference of the quantities.
2. It is also used to convert the difference of squares of two quantities into the factor form for simplifying expressions and solving the equations.
List of the understandable examples to learn how to use the difference of squares rule in mathematics.
#### Verification
Take $a \,=\, 4$ and $b \,=\, 2$. Now, evaluate both sides of the expressions by substituting the values.
$(1).\,\,$ $a^2-b^2$ $\,=\,$ $4^2-2^2$ $\,=\,$ $16-4$ $\,=\,$ $12$
$(2).\,\,$ $(a+b)(a-b)$ $\,=\,$ $(4+2)(4-2)$ $\,=\,$ $6 \times 2$ $\,=\,$ $12$
Similarly, take $x \,=\, 7$ and $y \,=\, 3$. Now, calculate values of both sides of the equation.
$(1).\,\,$ $x^2-y^2$ $\,=\,$ $7^2-3^2$ $\,=\,$ $49-9$ $\,=\,$ $40$
$(2).\,\,$ $(x+y)(x-y)$ $\,=\,$ $(7+3)(7-3)$ $\,=\,$ $10 \times 4$ $\,=\,$ $40$
The above two examples have proved that the difference of squares of any two numbers is exactly equal to the product of their sum and difference. In this way, the difference of squares identity can be verified arithmetically in mathematics.
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# All Formulas of Limits Formulas of Useful Limits 1) If
• •
and
, then
, Where
, Where , Where , Where , Where is a real number. is a real number. is measured in radians.
2) 3) 4) 5) 6) 7)
think "approaching" It is a mathematical way of saying "we are not talking about when x=∞.. but you can use a limit. the answer gets closer and closer to 0". sometimes Infinity cannot be used directly. .8) nd write it like this: In other words: As x approaches infinity... Summary So. but we know as x gets bigger.. What happens at ∞ is undefined . then 1/x approaches 0 When you see "limit". but we do know that 1/x approaches 0as x approaches infinity 1/∞ .
but in "limit" language the limit is infinity (which is really saying the function is limitless).. so does "2x": x 1 2 4 10 100 . So as "x" approaches infinity.Limits Approaching Infinity What is the limit of this function? y = 2x Obviously as "x" gets larger. y=2x 2 4 8 20 200 . We write this: But don't be fooled by the "=". then "2x" also approaches infinity.. . You cannot actually get to infinity...
In fact. This is also true for 1/x2 etc A function such as x will approach infinity.Infinity and Degree We have seen two examples. so you have to look at the signs of x. Likewise functions with x2 or x3 etc will also approach infinity But be careful. In fact many infinite limits are actually quite easy to work out. the other went to infinity. or x/9 and so on. if you can figure out "which way it is going". a function like "-x" will approach "infinity". one went to 0. as well as 2x. like this Functions like 1/x approach 0 as x approaches infinity. if we look at the Degree of the function (the highest exponent in the function) we can tell what is going to happen: If the Degree of the function is: • greater than 0. the limit is infinity (or -infinity) .
and Q(x)=6x2: Following on from our idea of the Degree of the Equation. the first step to find the limit is to .... the limit is 0. . Compare the Degree of P(x) to the Degree of Q(x): If the Degree of P is less than the Degree of Q . divide the coefficients of the terms with the largest exponent. the limit is 0 But if the Degree is 0 or unknown then we need to work a bit harder to find a limit Rational Functions A Rational Function is one that is the ratio of two polynomials: For example. If the Degree of P and Q are the same . like this: .• less than 0.. .. here P(x)=x3+2x-1......
or maybe negative infinity... • x3 (the term with the largest exponent in the top) and 6x2 (the term with the largest exponent in the bottom) • ..If the Degree of P is greater than the Degree of Q . . But this will head for negative infinity. then the limit is positive infinity . because both . .... are positive.... because -2/5 is negative.. just like how we found the coefficients above: For example this will go to positive infinity. You need to look at the signs! You can work out the sign (positive or negative) by looking at the signs of the terms with the largest exponent... .
48832 2.71815 2.71692 2..70481 2.000 100.25000 2.59374 2..71827 It settles down to a value (2.71828. which is the magic number e) So again we have an odd situation: . let's try larger and larger values of n: n 1 2 5 10 100 1. we don't know! So instead of trying to work it out for infinity (because we can't get a sensible answer).A Harder Example: Working Out "e" There is a formula for the value of e (Euler's number) based on infinity and this formula: (1+ 1/n)n At Infinity: (1+1/∞)∞ = ??? ..00000 2.000 10.000 (1 + 1/n)n 2..
..71828... ! You can see by the graph and the table that as n get larger the function approaches 2. But trying to use infinity as a "very large real number" (it isn't!) would give this: (1+1/∞)∞ = (1+0)∞ = (1)∞ = 1 So. but we know as n gets bigger. So. the answer gets closer and closer to the value of e". Don't Do It The Wrong Way .• • We don't know what the value is when n=infinity But we can see that it settles towards 2.71828.. you will get wrong answers! Limits are the right way to go. we use limits to write the answer like this: It is a mathematical way of saying "we are not talking about when n=∞. Evaluating Limits I have taken a gentle approach to limits so far. and shown tables and graphs to illustrate the points.. don't try to use Infinity as a real number.. .
I will show you how in Evaluating Limits. Limits (Evaluating) You should read Limits (An Introduction) first Quick Summary of Limits Sometimes you can't work something out directly . but you can see what it should be as you get closer and closer! For example: At x=1: (x2-1)/(x-1) (12-1)/(1-1) = (1-1)/(1-1) = 0/0 But 0/0 is "indeterminate".50000 1.99999 ..9999 0.99000 1. But instead of trying to work it out for x=1 let's try approaching it closer and closer: x 0. meaning we can't determine its value.90000 1.99 0.But to "evaluate" (in other words calculate) the value of a limit can take a bit more effort.... ..99999 .99990 1.9 0.. (x2-1)/(x-1) 1.999 0.5 0.99900 1.
so instead mathematicians say exactly what is going on by using the special word "limit" The limit of (x2-1)/(x-1) as x approaches 1 is 2 And it is written in symbols as: So it is a special way of saying. in truth. "ignoring what happens when you get there. the limit is 2. Evaluating Limits "Evaluating" means to find the value of (think e-"value"-ating) . But you can say that as you approach 1. you cannot say what the value at x=1 is. then (x2-1)/(x-1) gets close to 2 We are now faced with an interesting situation: • • When x=1 we don't know the answer (it is indeterminate) But we can see that it is going to be 2 We want to give the answer "2" but can't.Now we can see that as x gets close to 1. but as you get closer and closer the answer gets closer and closer to 2" As a graph it looks like this: So.
but the second example gave us a quick and easy answer. Factors You can try factoring. Just Put The Value In The first thing to try is just putting the value of the limit in. 2. Example: By factoring (x2-1) into (x-1)(x+1) we get: .In the example above we said the limit was 2 because it looked like it was going to be. Let's look at some: 1. But that is not really good enough! In fact there are many ways to get an accurate answer. Let's try some examples: Example Substitute Value Works? (1-1)/(1-1) = 0/0 10/2 = 5 It didn't work with the first one (we knew that!). and see if it works (in other wordssubstitution).
Conjugate If it's a fraction. then multiplying top and bottom by a conjugate might help. The conjugate is where you change the sign in the middle of 2 terms like this: Here is an example where it will help you to find a limit: Evaluating this at x=4 gives 0/0. which is not a good answer! So.Now we can just substitiute x=1 to get the limit: 3. let's try some rearranging: Multiply top and bottom by the conjugate of the top: .
. now we have: Done! 4. Infinite Limits and Rational Functions A Rational Function is one that is the ratio of two polynomials: For example. Infinity. or easily calculated from the coefficients. here P(x)=x3+2x-1.Simplify top using : Simplify top further: Eliminate (4-x) from top and bottom: So. and Q(x)=6x2: By finding the overall Degree of the Function we can find out whether the function's limit is 0. -Infinity. Read more at Limits To Infinity.
. Formal Method The formal method sets about proving that you can get as close as you want to the answer by making "x" close to "a".5. |
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