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# How to find the last digit of $37^{100}$
I am currently working on some algebra and I am studying the modulus chapter of my book. One question was finding the last digit of:
$$37^{100}$$
They give a hint about how the solution should be about calculating a remainder...
Is there any easy solution to this?
I know that there is a solution of just finding the pattern of the last digit of $37^x$. But I don't understand how that has to do with calculating a remainder...
Thank you!
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Last digit, you mean? – Thomas Andrews Feb 6 '13 at 15:26
Yes, that is right. – Lukas Arvidsson Feb 6 '13 at 15:27
Hint: What do you get if you look at some number mod $10$? – Tobias Kildetoft Feb 6 '13 at 15:28
– Hanul Jeon Feb 6 '13 at 15:29
The last digit of $37^{100}$ is the same as the last digit of $7^{100}$ or of $492038497^{100}$.
When you multiply two positive integers, the last digit in the product depends on those two integers only through their last digits. That will become obvious if you look at the way they told you to multiply numbers by hand in third or fourth grade or whenever it was: $$\begin{array}{rrrrrrrrrr} & & & 4 & 2 & 7 \\ & & \times & 3 & 1 & 9 \\ \hline & & \bullet & \bullet & \bullet & 3 \\ & & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet \\ \hline \bullet & \bullet & \bullet & \bullet & \bullet & 3 \end{array}$$ Where did that last $3$ come from??
Now multiply $7$, and discard all but the last digit: \begin{align} 7 \times 7 & & & = \cdots\ 9 \\ 7 \times 7 \times 7 & = (\cdots\ 9\times 7) & &= \cdots \bullet \\ 7 \times 7 \times 7 \times 7 & = (\cdots \bullet\times 7) & & = \cdots \bullet \\ 7 \times 7 \times 7 \times 7 \times 7 & = (\cdots \bullet\times 7) & & = \cdots\bullet \\ \end{align} Just throw away all but the last digit at each step.
Now notice something: There are only $9$ digits that could possibly appear as the last digit (you can't get $0$ when you're multiplying non-zero digits). That means you can't keep getting new digits there that you haven't seen before.
All this will lead you to see a pattern that occurs as you continue the process. And you'll see why that happens. And that will lead you quickly to the answer.
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Thank you for an excellent answer! – Lukas Arvidsson Feb 6 '13 at 20:29
@LukasArvidsson: welcome ;) – user 1618033 Feb 6 '13 at 20:30
Since $37^2=1369$, then $$37^{100}=(37^2)^{50}=(1369)^{50}$$ and the last digit of $(1369)^{n}$ oscillates between $9$ if $n$ odd and $1$ if $n$ even.
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Thank you for taking the time to answer my question, it was very helpful! – Lukas Arvidsson Feb 6 '13 at 20:29
This is a question about calculating modulo 10. So the last digit of $37^{100}$ is the same as the last digit of $7^{100}$; which is the same as the last digit of $49^{50}$, and so the same as the last digit of $9^{50}$. Perhaps you can carry on from there.
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Thank you for that explanation, it makes sense now ;) – Lukas Arvidsson Feb 6 '13 at 20:30
HINT
You want to calculate the remainder after dividing by $10$. It's also good to know that $37^2$ has remainder $9$ after dividing by $10$, which may be written as $37^2 \equiv -1 \pmod{10}$.
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One finds the modulus of $37^{100}$ in decimal, by noting that in decimal, anything raised to the fourth power must end in $0$, $1$, $5$, or $6$. When various powers of 5 are added, then the number must be adjent to a multiple of $5*5^n$, where $5^n$ divides the index.
So, we look for solutions, ending in any of these three digits: $000$, $001$, $625$, or $376$, depending on what the common divisor with 10 is. Here it's 1, so it ends in $001$.
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Here we are providing Online Education for Surface Areas and Volumes Class 9 Extra Questions Maths Chapter 13 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/
## Online Education for Extra Questions for Class 9 Maths Surface Areas and Volumes with Answers Solutions
Extra Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes with Solutions Answers
### Surface Areas and Volumes Class 9 Extra Questions Very Short Answer Type
Surface Area And Volume Class 9 Extra Questions Question 1.
How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm?
Solution:
Here, radius (r) = 3.5 cm and height (h) = 12 cm
∴ Amount of ice-cream = $$\frac{1}{3}$$ πr2
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 3.5 × 3.5 × 12
= 154 cm3
Class 9 Surface Area And Volume Extra Questions Question 2.
Calculate the edge of the cube if its volume is 1331 cm3.
Solution:
Volume of cube = 1331 cm3
(Side)3 = 1331
Side = (11 × 11 × 11)$$\frac{1}{3}$$ = 11 cm
Class 9 Maths Chapter 13 Extra Questions Question 3.
The curved surface area of a cone is 12320 sq. cm, if the radius of its base is 56 cm, find its
height.
Solution:
Here, radius of base of a cone (r) = 56 cm
And, curved surface area = 12320 cm2
πrl = 12320
l = $$\frac{12320}{\pi r}$$
= $$\frac{12320 \times 7}{22 \times 56}$$ = 70 cm
Again, we have
r2 + h2 = l2
h2 = l2 – r2 = 702 – 562
= 4900 – 3136 = 1764
h = √1764 = 42 cm
Hence, the height of the cone is 42 cm.
Extra Questions On Surface Area And Volume Class 9 Question 4.
Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid.
Solution:
When two cubes are joined end to end, then
Length of the cuboid = 6 + 6 = 12 cm
Breadth of the cuboid = 6 cm
Height of the cuboid = 6 cm
Total surface area of the cuboid = 2 (lb + bh + hl)
= 2(12 × 6 + 6 × 6 + 6 × 12)
= 2(72 + 36 + 72) = 2(180)
= 360 cm2
Ch 13 Maths Class 9 Extra Questions Question 5.
A metallic sphere is of radius 4.9 cm. If the density of the metal is 7.8 g/cm2, find the mass of the sphere (π = $$\frac{22}{7}$$).
Solution:
Here, radius of metallic sphere (r) = 4.9 cm
Extra Questions Of Surface Area And Volume Class 9 Question 6.
The volume of a solid hemisphere is 1152 π cm3. Find its curved surface area.
Solution:
Here, volume of hemisphere = 1152 π cm3
∴ $$\frac{2}{3}$$πr3 = 1152
⇒ r3 = (12)3 π
⇒ r$$\frac{1152 \times 3}{2}$$ = 1728
⇒ r3 = (12)3
Now, curved surface area = 2πr2
= 2 × π × (12)2 = 288π cm2
Class 9 Maths Ch 13 Extra Questions Question 7.
Find the diameter of a cylinder whose height is 5 cm and numerical value of volume is equal to
numerical value of curved surface area.
Solution:
Here, height of cylinder (h) = 5 cm
According to the statement of the question, we have
πr2h = 2πrh
r = 2 cm
Thus, diameter of the base of the cylinder is 2 × 2 i.e., 4 cm.
Extra Questions Of Chapter 13 Maths Class 9 Question 8.
In a cylinder, if radius is halved and height is doubled, then find the volume with respect to original volume.
Solution:
Here, r = $$\frac{r}{2}$$, h = 2h
### Surface Areas and Volumes Class 9 Extra Questions Short Answer Type 1
Class 9 Maths Surface Area And Volume Extra Questions Question 1.
A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm, find the volume of the spherical ball. [use π = 3.14]
Solution:
Since curved surface of half of the spherical ball = 56.57 cm2
2πr2 = 56.57
= 113.04 cm3
Class 9 Maths Chapter 13 Extra Questions With Solutions Question 2.
Find the capacity in litres of a conical vessel having height 8 cm and slant height 10 cm.
Solution:
Height of conical vessel (h) = 8 cm
Slant height of conical vessel (l) = 10 cm
∴ r2 + h2 = l2
⇒ r2 + 82 = 102
⇒ r2 = 100 – 64 = 36
⇒ r = 6 cm
Now, volume of conical vessel = $$\frac{1}{3}$$πr2h = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 6 × 8 = 301.71 cm3 = 0.30171 litre
Surface Area And Volume Extra Questions Class 9 Question 3.
Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside.
Solution:
Here, radius of hemispherical dome (r) = 14 m
Surface area of dome = 2πr2
= 2 × $$\frac{22}{7}$$ × 14 × 14 = 1232 m2
Hence, total surface area to be whitewashed from outside is 1232 m2.
Surface Areas And Volumes Class 9 Extra Questions Question 4.
A rectangular piece of paper is 22 cm long and 10 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder.
Solution:
Since rectangular piece of paper is rolled along its length.
∴ 2πr = 22
r = $$\frac{22 \times 7}{2 \times 22}$$ = 3.5 cm
Height of cylinder (h) = 10 cm
∴ Volume of cylinder = πr2h
= $$\frac{22}{7}$$ × 3.5 × 3.5 × 10 = 385 cm3
Surface Area And Volume Class 9 Questions Answers Question 5.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find it volume. If 1m3 wheat cost is ₹10, then find total cost.
Solution:
Diameter of cone = 10.5 m
Radius of cone (r) = 5.25 m
Height of cone (h) = 3 m
Volume of cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 5.25 × 5.25 × 3
= 86.625 m3
Cost of 1m3 of wheat = ₹10
Cost of 86.625 m3 of wheat = ₹10 × 86.625
= ₹866.25
Question 6.
A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and1cm3 of water weighs 1 g, find the depth of water.
Solution:
Since 1 cm3 of water weighs 1 g.
∴ Volume of cylindrical vessel = 154 cm3
πr2h = 154
$$\frac{22}{7}$$ × 3.5 × 3.5 × h = 154
h = $$\frac{154 \times 7}{22 \times 3.5 \times 3.5}$$
h = 4
cm Hence, the depth of water is 4 cm.
### Surface Areas and Volumes Class 9 Extra Questions Short Answer Type 2
Question 1.
A wall of length 10 m is to be built across an open ground. The height of the wall is 5 m and thickness of the wall is 42 cm. If this wall is to be built with brick of dimensions 42 cm × 12 cm × 10 cm, then how many bricks would be required?
Solution:
Here, length of the wall (L) = 10 m = 1000 cm
Breadth of the wall (B) = 42 cm
Height of the wall (H) = 5 m = 500 cm
∴ Volume of the wall = L × B × H
= 1000 × 42 × 500 cm3
Volume of each brick = 42 × 12 × 10 cm3
= 4167
Hence, the required number of bricks is 4167.
Question 2.
The volume of cylindrical pipe is 748 cm. Its length is 0.14 m and its internal radius is 0.09 m. Find thickness of pipe.
Solution:
Internal radius (r) of cylindrical pipe = 0.09 m = 9 cm
Length (height) of cylindrical pipe (h) = 0.14 m = 14 cm
Let external radius of the cylindrical pipe be R cm.
Volume of cylindrical pipe = 748 cm3
⇒ π(R2 – r2)h = 748
⇒ $$\frac{22}{7}$$ (R2 – 92)14 = 748
⇒ R2 – 81 = $$\frac{748 \times 7}{22 \times 14}$$ = 17
⇒ R2 = 81 + 17 = 98
⇒ R = √98 = 7√2 cm = 9.9 cm
Thus, thickness of the pipe = 9.9 -9 = 0.9 cm
Question 3.
The curved surface area of a cylinder is 154 cm. The total surface area of the cylinder is three times its curved surface area. Find the volume of the cylinder.
Solution:
Since curved surface area of cylinder = 154 cm2 (given]
Total surface area of cylinder = 3 × curved surface area
2πrh + 2πr2 = 3 × 154 3 154 + 2πr2 = 462
2πr2 = 462 – 154 = 308
= 539 cm3
Question 4.
A right-angled ∆ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. Find the volume of the solid generated. Also, find the total surface area of the solid.
Solution:
When rt. ∠ed ∆ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm. Slant height of the cone is 5 cm.
Question 5.
A semicircular sheet of metal of radius 14 cm is bent to form an open conical cup. Find the capacity of the cup.
Solution:
Radius of semicircular sheet (r) = 14 cm
∴ Slant height (1) = 14 cm
Circumference of base = Circumference of semicircular sheet
### Surface Areas and Volumes Class 9 Extra Questions Long Answer Type
Question 1.
It costs ₹3300 to paint the inner curved surface of a 10 m deep well. If the rate cost of
painting is of ₹30 per m2, find :
(a) inner curved surface area
(b) diameter of the well
(c) capacity of the well.
Solution:
Depth of well (h) = 10 m
Cost of painting inner curved surface is ₹30 per m2 and total cost is ₹3300
Hence, inner curved surface area is 110 m2, diameter of the well is 2 × 1.75 i.e., 3.5 m and capacity of the well is 96.25 m3.
Question 2.
Using clay, Anant made a right circular cone of height 48 cm and base radius 12 cm. Versha reshapes it in the form of a sphere. Find the radius and curved surface area of the sphere so formed.
Solution:
Height of cone (h) = 48 cm
Radius of the base of cone = 12 cm
Let R be the radius of sphere so formed
∴ Volume of sphere = Volume of cone
$$\frac{4}{3}$$πR3 = $$\frac{1}{3}$$πr2h
4R3 = 12 × 12 × 48
R3 = 12 × 12 × 12
R = 12 cm
Now, curved surface area of sphere = 4πR2
= 4 × $$\frac{22}{7}$$ × 12 × 12
= 1810.29 cm
Question 3.
A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of ₹498.96. If the rate of whitewashing is ₹4 per square metre, find the :
(i) Inside surface area of the dome
(ii) Volume of the air inside the dome.
Solution:
Here, dome of building is a hemisphere.
Total cost of whitewashing inside the dome = ₹498.96
Rate of whitewashing = ₹4 per m2
Question 4.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases ?
Solution:
Here, right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm.
∴ Radius of the base of cone = 12 cm
Height of the cone = 5 cm
= 12 : 5
Question 5.
A right triangle of hypotenuse 13 cm and one of its sides 12 cm is made to revolve taking side 12 cm as its axis. Find the volume and curved surface area of the solid so formed.
Solution:
Here, hypotenuse and one side of a right triangle are 13 cm and 12 cm respectively.
Now, given triangle is revolved, taking 12 cm as its axis
∴ Radius of the cone (r) = 5 cm
Height of the cone (h) = 12 cm
Slant height of the cone (1) = 13 cm
∴ Curved surface area = πrl = π(5)(13) = 65π cm2
Volume of the cone = $$\frac{1}{2}$$πr2h = $$\frac{1}{2}$$π × 5 × 5 × 12 = 100π cm3
Hence, the volume and curved surface area of the solid so formed are 100 π cm3 and 65 π cm2 respectively.
### Surface Areas and Volumes Class 9 Extra Questions HOTS
Question 1.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let each edge of the cube be a cm.
Question 2.
A rectangular tank is 225 m × 162 m at base. With what speed should water flow into it through an aperture 60 cm × 45 cm so that the level of water is raised by 20 cm in 2.5 hours?
Solution:
Volume of water to be flown in 2.5 hour
= 225 m × 162 m × 20 cm
Hence, the speed of flow of water = 10.8 km/hour
### Surface Areas and Volumes Class 9 Extra Questions Value Based (VBQs)
Question 1.
To maintain beauty of a monument, the students of the school cleaned and painted the dome of the monument. The monument is in the form of a hemisphere. From inside, it was white washed by the students whose area is 249.48 m2.
(a) Find the volume of the air inside the dome. If white washing costs ₹2 per m2, how much does it costs ?
(b) Which value is depicted by the students? (π = $$\frac{22}{7}$$)
Solution:
(a) Here, dome of the monument is hemispherical in shape, which was whitewashed by the students.
Now, total area to be white washed = 249.48 m2
Cost of white washing = ₹2 per m2
∴ Total cost of white washing = ₹2 × 249.48
= ₹498.96
Also, 2πr2 = 249.48
= 4191.264 m3
(b) Value: Cleanliness, beautification as well as preserving the heritage along with social values.
Question 2.
Salim provides water to a village, having a population of 4000 which requires 150 litres of water per head per day. He has storage tank measuring 20 m × 15 m × 6 m. For how many days will the water of his tank last? He increased the rate for providing water as the dependence of villagers increased on him. Which value is depicted by Salim?
Solution:
(i) Here, the population of the village = 4000
Requirement of water per head per day = 150 litres
∴ Total requirement of water per day = 4000 × 150 litres
= 600000 litres
Volume of water tank = 20 × 15 × 6
= 1800 m3
= 1800 × 1000 litres
Now, number of days for which water of the tank will last = $$\frac{1800 \times 1000}{600000}$$ = 3 days
Hence, water tank can serve for 3 days.
(ii) Helping the needy. |
# How do you find the limit of sqrt(x^2 + 1) - x as x approaches infinity?
Aug 24, 2016
$= 0$
#### Explanation:
${\lim}_{x \to \infty} \sqrt{{x}^{2} + 1} - x$
$= {\lim}_{x \to \infty} x \left(\sqrt{1 + \frac{1}{x} ^ 2} - 1\right)$
$= {\lim}_{x \to \infty} x \left({\left(1 + \frac{1}{x} ^ 2\right)}^{\frac{1}{2}} - 1\right)$
Binomial expansion on the radical
$= {\lim}_{x \to \infty} x \left(\left(1 + \frac{1}{2} \cdot \frac{1}{x} ^ 2 + O \left(\frac{1}{x} ^ 4\right)\right) - 1\right)$
$= {\lim}_{x \to \infty} x \left(\frac{1}{2 {x}^{2}} + O \left(\frac{1}{x} ^ 4\right)\right)$
$= {\lim}_{x \to \infty} \frac{1}{2 x} + O \left(\frac{1}{x} ^ 3\right)$
$= 0$
Aug 24, 2016
Rewrite the expression using the conjugate.
#### Explanation:
$\frac{\left(\sqrt{{x}^{2} + 1} - x\right)}{1} \cdot \frac{\left(\sqrt{{x}^{2} + 1} + x\right)}{\left(\sqrt{{x}^{2} + 1} + x\right)} = \frac{\left({x}^{2} + 1\right) - {x}^{2}}{\sqrt{{x}^{2} + 1} + x}$
$= \frac{1}{\sqrt{{x}^{2} + 1} + x}$
As $x \rightarrow \infty$, the denominator also increases without bound, so the ratio goes to $0$.
The following is optional.
From $= \frac{1}{\sqrt{{x}^{2} + 1} + x}$, we could continue:
For $x > 0$, we get
$= \frac{1}{\sqrt{{x}^{2}} \sqrt{1 + \frac{1}{x} ^ 2} + x}$
$= \frac{1}{x \sqrt{1 + \frac{1}{x} ^ 2} + x}$
= 1/(x(sqrt(1+1/x^2)+1)
As $x$ increases without bound, $\left(\sqrt{1 + \frac{1}{x} ^ 2} + 1\right) \rightarrow 2$, and $x \rightarrow \infty$, so
$\frac{1}{x \left(\sqrt{1 + \frac{1}{x} ^ 2} + 1\right)} \rightarrow 0$ |
Difference between revisions of "2011 AMC 12A Problems/Problem 8"
Problem
In the eight term sequence $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, the value of $C$ is $5$ and the sum of any three consecutive terms is $30$. What is $A+H$?
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$
Solution
Solution 1
Let $A=x$. Then from $A+B+C=30$, we find that $B=25-x$. From $B+C+D=30$, we then get that $D=x$. Continuing this pattern, we find $E=25-x$, $F=5$, $G=x$, and finally $H=25-x$. So $A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}$
Solution 2
Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$
It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$.
Subtracting, we have that $A+H=25\rightarrow \boxed{\textbf{C}}$.
Solution 3 (the tedious one)
From the given information, we can deduct the following equations:
$A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$, and $F+G+H=30$.
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
$(A+B)-(B+D)=25-25 \implies (A-D)=0$
$(A-D)+(D+E)=0+25 \implies (A+E)=25$
$(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5$ (Notice how we don't use $D+E+F=30$)
$(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25$
Therefore, we have $A+H=25 \rightarrow \boxed{\textbf{C}}$
~JinhoK
2011 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions |
# Puddle Puzzle Solution
The following is a solution to Puzzle #15 – Puddle Puzzle. Take a look if you haven’t already.
We base our answer on the assumption that the liquid is incompressible. By this we mean the volume of the liquid is conserved, such that its cross-sectional area does not change.
Let’s look at the cross-section of the liquid using the diagram given in the problem. It can be split into two smaller triangles by drawing a line perpendicular to the ground through the cube’s edge, as pictured above. This allows us to split the width of the triangle into two segments of lengths $a$ and $b$. From the diagram, we can see that
$\displaystyle \tan\theta=\frac{a}{s}\to a=s\tan\theta$
$\displaystyle \tan\theta=\frac{s}{b}\to b=s\cot\theta$
Using the fact that the area of a triangle is half its base length times its height, the cross-sectional area of the liquid is
$A=\frac{1}{2}\cdot s\cdot\left(s\tan\theta+s\cot\theta\right)$
$A=\frac{1}{2}s^2\left(\tan\theta+\cot\theta\right)$
The sum of the trigonometric terms actually simplifies very nicely if we write them in terms of fractions. Using the definitions of the tangent and cotangent functions we have
$\displaystyle \tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$
Putting the two fractions over a common denominator gives
$\displaystyle \tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$\displaystyle \tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}$
Now we use a double angle formula, which tells us
$\sin 2\theta=2\sin\theta\cos\theta$
Hence
$\displaystyle \tan\theta+\cot\theta=\frac{2}{\sin 2\theta}$
Substituting this simpler expression into that for $A$ gives
$\displaystyle A=\frac{s^2}{\sin 2\theta}$
The cross-sectional area of the liquid can be determined using the initial condition $s\left(\theta=\frac{\pi}{4}\right)=s_0$:
$\displaystyle A=\frac{s_0^2}{1}$
This gives us
$s(\theta)=s_0\sqrt{\sin 2\theta}$
Neat! I like the answer to this problem because it contains a function you don’t often encounter; the square root of a trig function. Its form tells us what we already knew: as the cube tips over, the liquid becomes more spread out horizontally, so its vertical dimension has to decrease. In the limit that the cube tips over completely, the liquid has infinite horizontal dimensions, so its ‘thickness’ becomes infinitesimal in order to conserve its volume.
Here’s an animation showing the cross-section of the liquid as the cube tips over.
Of course, the animation looks slightly unnatural since the liquid cannot instantaneously adapt to the motion of the cube – there will be some sloshing about before the liquid settles into its ground state. But in the original problem, I stated the cube tipped ‘slowly’, which is the physics-exam shorthand for “infinitesimally slowly”.
The cube is assumed to tip so slowly that the system is always in its ground state. That is, one could imagine suddenly stopping the cube in its motion, and at that moment we will always find the liquid in equilibrium. The system is said to evolve adiabatically. |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 4 - 23/5 = 7/5 = 1 2/5 = 1.4
Spelled result in words is seven fifths (or one and two fifths).
### How do you solve fractions step by step?
1. Conversion a mixed number 2 3/5 to a improper fraction: 2 3/5 = 2 3/5 = 2 · 5 + 3/5 = 10 + 3/5 = 13/5
To find a new numerator:
a) Multiply the whole number 2 by the denominator 5. Whole number 2 equally 2 * 5/5 = 10/5
b) Add the answer from previous step 10 to the numerator 3. New numerator is 10 + 3 = 13
c) Write a previous answer (new numerator 13) over the denominator 5.
Two and three fifths is thirteen fifths
2. Subtract: 4 - 13/5 = 4/1 - 13/5 = 4 · 5/1 · 5 - 13/5 = 20/5 - 13/5 = 20 - 13/5 = 7/5
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(1, 5) = 5. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 1 × 5 = 5. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - four minus thirteen fifths = seven fifths.
#### Rules for expressions with fractions:
Fractions - simply use a forward slash between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Simplest form of a fraction
Which one of the following fraction after reducing in simplest form is not equal to 3/2? a) 15/20 b) 12/8 c) 27/18 d) 6/4
• Equivalent expressions
A coach took his team out for pizza after their last game. There were 14 players, so they had to sit in smaller groups at different tables. Six players sat at one table and got 4 small pizzas to share equally. The other players sat at the different table
• Math test
Brayden was solving some math problems for the math team. He answered 2 math problems. Matthew answered 3, John answered 1 reasoning. Matthew 1/2 times as many. Brayden said that 2/6. Is he correct? Why or why not? Be sure to explain your answer.
• Leo hiked
Leo hiked 6/7 of a kilometer. Jericho hiked 2/3 kilometer. Who covered a longer distance? How much longer?
• Mangoes trees
Draw two sets of trees. The first tree has 2 branches and 8 mango fruits. The second tree has 3 branches and 12 mango fruits. Illustrate and express the ratio and proportion of the given statement.
• Colored blocks
Tucker and his classmates placed colored blocks on a scale during a science lab. The brown block weighed 8.94 pounds, and the red block weighed 1.87 pounds. How much more did the brown block weigh than the red block?
• Troops
The route is long 147 km and the first-day first regiment went at an average speed of 12 km/h and journey back 21 km/h. The second day went second regiment the same route at an average speed of 22 km/h there and back. Which regiment will take route longer
• Roma ate
Roma ate 2/5 of a cake while Somya ate 3/7 of the same cake. Who ate more and by how much?
• Dividends
The three friends divided the win by the invested money. Karlos got three-eighths, John 320 permille, and the rest got Martin. Who got the most and which the least?
• Paper collecting
At the paper collecting contest gathered Franta 2/9 ton, Karel 1/4 ton, and Patrick 19/36 tons of paper. Who has gathered the most and the least?
• Sandy
Sandy, John and Marg baked pies for the Bake Sale. Sandy cut his pies into 6ths, John but his into 8ths, and Marg cut hers into quarters. Sandy sold 11/6, John sold 1 3/8 pies, and Marg sold 9/4 pies. Who sold the most pies? Who sold the fewest?
• Evaluate mixed expressions
Which of the following is equal to 4 and 2 over 3 divided by 3 and 1 over 2? A. 4 and 2 over 3 times 3 and 2 over 1 B. 14 over 3 times 2 over 7 C. 14 over 3 times 7 over 2 D. 42 over 3 times 2 over 31
• Sayavong
Sayavong is making cookies for the class. He has a recipe that calls for 3 and 1/2 cups of flour. He has 7/8 of a cup of wheat flour, and 2 and 1/2 cups of white flour. Does Mr. Sayavong have enough flour to make the cookies? |
# Worksheet on Simplification of Fractions
This worksheet on simplification of fractions is based on the concept of ‘BODMAS’. In simplification of fraction we can find that we are suppose to carry out multiple basic operation in each sum and that has to be done following the ‘BODMAS’ rule. A single sum can contain both division and multiplication and addition as well and we have to carry out each of the operation according to ‘BODMAS’. Moreover, a single sum can also involve only addition and subtraction which will also be done with the help of ‘BODMAS’ rule. Whatever be the operations given in the sum be it addition, subtraction, multiplication or division we have to carry out those operation using ‘BODMAS’. There can be some operations that are given in brackets. The part of sum in bracket has to be done first. In ‘BODMAS’ there is one component ‘of ‘ which implies multiplication. The places where we will find ‘of‘should be replaced by the operation multiplication.
For recap and as a hint the full form of ‘BODMAS’
B = Brackets
O = Of
D = Division
M= Multiplication
S = Subtraction
Hence, we have to first carry out brackets, then ‘of’ (it implies for multiplication), then division, multiplication, addition and then followed by subtraction. If this sequence is not followed while doing these simplifications the answer will not be correct.
Here are few problems on simplification of fractions for practice:
Simplify the following using the rule of ‘BODMAS’
1. (21/4 + 52/4) × (6/7 ÷ 12/14)
2. 80 – 17/4 × 2 + 15 ÷ 15/7
78 1/2
3. 127/4 + 62/3 × 24/6
4. 73/4 ÷ 62/5 + 8/3
5. 10 7/5 × 25/95 ÷ 10/3 ̶ 2/3
6. (2/3 + 5/12) × 1
7. 58/6 × 0 + 1
8. 8/7 ÷ 48/49 + 2/3
9. 5/6 ̶ 7/8 + 15/2
10. 9/8 of 72 + 6/7 of 343
11. 56 × 7/8 + 112 × 0 – 1
12. 1/7 of 49 + 125 ÷ 25
13. 1/5 × 5 + 0 × 7 × 3
14. 3/4 × 16/7 ×7/21 ÷ 2/3
15. (5/6 + 7/6) × 1
15. 5/9 × 0 + 3/4
17. (8/6 + 7/9) ̶ 1/2
18. (8/8 ̶ 1/3) + 2/7
19. (5/16 + 7/16) × 32/12 × 0
20. (8/3 + 7/9) ÷ 62/27
Solution:
3/2
21. 0 × 7/6 + 3/25 × 0
22. 1 × 15/13 + 2 × 6/7
23. 17/51 × 3/4 + 2/7
Therefore, this whole worksheet on simplification aimed to provide all types of sum for helping students in understanding simplification easily.
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# 74 days to CAT 2015 | Finding the last non-zero digit of n!
Rarely has this concept been come across but it is nevertheless important for CAT or in case of tougher quant-based papers like XAT.
If the number is small in nature, the answer can easily be calculated manually. However, if the number is huge, it becomes difficult to calculate it. The question might seem within reach as there is no big concept involved and so, can eat into precious time. In this article, I will cover how to break down the concept into smaller bits so that it becomes easy to apply for larger numbers as well.
The mechanical approach
For smaller numbers, one can do it manually as we will see:
Find the last non-zero digit of 12!
Now 12! = 12*11*10*9*8*7*6*5*4*3*2*1
As the last non-zero digit is asked, we would simply eliminate the 5s and an equal number of 2s and then multiply only the unit’s place digits to arrive at the solution.
Eliminating two 5s and two 2s, we get: 12*11*2*9*8*7*6*3*2*1
Now, you can simply multiply the numbers successively:
2*1=2 -> 2*2=4 -> 4*9=6 -> 6*8=8 -> 8*7=6 -> 6*6=6 -> 6*3=8 -> 8*2=6
A few derivations you can remember to make it easier to calculate:
Last non-zero digit of 5! = 2
Last non-zero digit of 10! = 8
Last non-zero digit of 15! = 8
Last non-zero digit of 20! = 4
The formula
The basis of the formula is that, the nature of n! is such that it has an unequal number of 2s and 5s with 2s more frequently occuring than 5s. So, the last non-zero digit for any n! where n>1 would be an even number.
Step I: Get the largest integral quotient for n/5 = A
This is very easy and can be done by simple observation. For example: Largest quotient of 136/5 will be 27; that of 2563 will be 512 and so on.
Step II: Find the remainder of n/5 = B
This is again simple and an extension of step I. You can simply find the deviation from the nearest number ending with 5 or 0 and get the remainder. For example: In case of 123/5, the remainder would be 3; that in case of 164/5 would be 4 and so on.
Step III: Find the last non-zero digit of:
2A * A! * B!
Now, the only difficult part here is to find last non-zero digit of A! but then, each iteration decreases the quotient 5 times and so, even in the toughest problems, you would have to do it twice or thrice.
Let’s take a few examples and solve it:
1) Find the last non-zero digit of 26!
A = Quotient of 26/5 = 5
B = Remainder of 26/5 = 1
2A * A! * B!
25 * 5! * 1!
Now, we know by virtue of cyclicity that 25 will end in 2. Also, we know that the last non-zero digit of 5! = 120 is 2. So, it becomes:
Let’s try it for a larger number:
2) Find the last non-zero digit of 100!
In this case, A=20, B=0
220 * 20! * 0!
Now, the trick here is to find the last non-zero digit of 20! using the same technique. In this case, A becomes 4 and B becomes 0. So, the last non-zero digit will be: 24 * 4! * 0! = 6*4 = 4
We will need to plug this back into the original expression:
6 * 4 * 1 will be equal to 4.
So, if one understands the concept well, it is very easy to calculate the last non-zero digit of a number of the form of n!
Let’s cap it off with another, slightly larger example:
3) Find the last non-zero digit of 400!
A=80, B=0
280 * 80! * 0!
6 * 80! …(i)
The last non-zero digit of 80! can be found by
216 * 16!
6 * 16! …(ii)
The last non-zero digit of 16! can be found by
23 * 3! * 1! = 8*6 = 8
So, plugging it back in expression (ii), we get 6*8 = 8
Plugging back 8 in expression (i), we get 6*8 = 8
Final words
Again, if this question type does appear, it would be with an objective of separating the extremely aware aspirants from the rest. Someone who has not really seen this type would not be able to find a way to deal with it except expanding the entire bit. If the question is some huge number, an under-prepared candidate would probably leave it very early but a pseudo-serious aspirant would stick around and try to find a way losing out on precious time. So, it is a deceptive type and so, should be dealt with extreme caution.
All the best! |
Exercise 2.3 Fractions and Decimals - NCERT Solutions Class 7
Go back to 'Fractions and Decimals'
Chapter 2 Ex.2.3 Question 1
Find:
(i) \begin{align} \frac{1}{4}\end{align} of
(a) \begin{align} \frac{1}{4}\end{align}
(b) \begin{align} \frac{3}{5}\end{align}
(c) \begin{align} \frac{4}{3}\end{align}
(ii) \begin{align} \frac{1}{7} \end{align} of
(a) \begin{align} \frac{2}{9}\end{align}
(b) \begin{align} \frac{6}{5}\end{align}
(c) \begin{align} \frac{3}{10}\end{align}
Solution
What is known?
Fractions
What is unknown?
Product of the given fractions.
Reasoning:
Find the product by multiplying numerator with numerator and denominator with denominator.
Steps:
\begin{align} \text{ (i) }\left( \text{a} \right)\text{ }\frac{1}{4}\text{ of }\frac{1}{4}&=\frac{1}{4}\times \frac{1}{4}\\&=\frac{1}{16}\\ \\ \text{ }\,\,\,\,\,\,\,\,\text{(b) }\frac{1}{4}\text{ of }\frac{3}{5}&=\frac{1}{4}\times \frac{3}{5}\\&=\frac{3}{20} \\\\ \text{ (c) }\frac{1}{4}\text{ of }\frac{4}{3}&=\frac{1}{4}\times \frac{4}{3}\\&=\frac{4}{12}\\&=\frac{1}{3} \\\end{align}
\begin{align} \text{ (ii) (a) }\frac{1}{7}\text{ of }\frac{2}{9}&=\frac{1}{7}\times \frac{2}{9}\\&=\frac{1\times 2}{7\times 9}\\&=\frac{2}{63} \\\\ \text{ }\,\text{ (b) }\frac{1}{7}\text{ of }\frac{6}{5}&=\frac{1}{7}\times \frac{6}{5}\\&=\frac{1\times 6}{7\times 5}\\&=\frac{6}{35} \\\\\ \text{ (c) }\frac{1}{7}\text{ of }\frac{3}{10}&=\frac{1}{7}\times \frac{6}{5}\\&=\frac{1\times 3}{7\times 10}\\&=\frac{3}{70} \\\end{align}
Chapter 2 Ex.2.3 Question 2
Multiply and reduce to lowest form (if possible):
(i) \begin{align} \frac{2}{3} \times 2\frac{2}{3}\end{align}
(ii) \begin{align} \frac{2}{7} \times \frac{7}{9}\end{align}
(iii) \begin{align} \frac{3}{8} \times \frac{6}{4}\end{align}
(iv) \begin{align} \frac{9}{5} \times \frac{3}{5}\end{align}
(v) \begin{align} \frac{1}{3} \times \frac{{15}}{8}\end{align}
(vi) \begin{align} \frac{{11}}{2} \times \frac{3}{{10}}\end{align}
(vii) \begin{align} \frac{4}{5} \times \frac{{12}}{7}\end{align}
Solution
What is known?
Expression
What is unknown?
Product of the given expression.
Reasoning:
Find the product by multiplying numerator with numerator and denominator with denominator.
Steps:
(i)
\begin{align}\frac{2}{3} \times 2\frac{2}{3}&= \frac{2}{3} \times \frac{8}{3}\\{}&= \frac{{2 \times 8}}{{3 \times 3}}\\&= {\frac{{16}}{9}}\end{align}
(improper fraction)
Converting \begin{align} \frac{{16}}{9}\end{align} into mixed fraction,
we get, \begin{align} \frac{{16}}{9}\end{align} \begin{align} = 1\frac{7}{9}\end{align}
(ii)
\begin{align} \frac{2}{7} \times \frac{7}{9} &= \frac{{2 \times 7}}{{7 \times 9}}\\ &= \frac{{14}}{{63}}\end{align}
Reducing \begin{align} \frac{{14}}{{63}}\end{align}to the lowest form,
we get, \begin{align} \frac{14}{63}= \frac{2}{9}\end{align}
(iii)
\begin{align}\frac{3}{8} \times \frac{6}{4} &= \frac{{3 \times 6}}{{8 \times 4}}\\&= \frac{{18}}{{32}}\end{align}
Reducing\begin{align} \frac{{18}}{{32}}\end{align} to the lowest form,
we get, \begin{align} \frac{18}{32}= \frac{9}{{16}}\end{align}
(iv)
\begin{align}\frac{9}{3} \times \frac{3}{5}&= \frac{{9 \times 3}}{{5 \times 5}}\\&= \frac{{27}}{{25}}\end{align}
(improper fraction)
Converting \begin{align} \frac{{27}}{{25}}\end{align}into mixed fraction,
we get, \begin{align}\frac{27}{25} = 1\frac{2}{{25}}\end{align}
(v)
\begin{align}\frac{1}{3} \times \frac{15}{8} &= \frac{1 \times 15}{3 \times 8}\\[0.5mm]&= \frac{15}{24}\end{align}
Reducing \begin{align} \frac{{15}}{{24}}\end{align} to the lowest form,
we get, \begin{align}\frac{15}{24} = \frac{5}{8}\end{align}
(vi)
\begin{align} \frac{11}{2} \times \frac{3}{10} &= \frac{11 \times 3}{2 \times 10}\\[0.5mm] &= \frac{33}{20}\end{align}
(improper fraction)
Converting \begin{align} \frac{{33}}{{20}}\end{align} into mixed fraction,
we get, \begin{align} \frac{33}{20}= 1\frac{{13}}{{20}}\end{align}
(vii)
\begin{align}\frac{4}{5} \times \frac{12}{7} &= \frac{4 \times 12}{5 \times 7}\\[0.5mm]&= \frac{48}{35} \end{align}
(improper fraction)
Converting \begin{align} \frac{{48}}{{35}}\end{align}into mixed fraction,
we get,\begin{align} \frac{48}{35}= 1\frac{{13}}{{35}}\end{align}
Chapter 2 Ex.2.3 Question 3
Multiply the following fractions:
i) \begin{align} \frac{2}{5} \times 5\frac{1}{4}\end{align}
ii) \begin{align} 6\frac{2}{5} \times \frac{7}{9} \end{align}
iii) \begin{align} \frac{3}{2} \times 5\frac{1}{3}\end{align}
iv) \begin{align} \frac{5}{6} \times 2\frac{3}{7}\end{align}
v) \begin{align} 3\frac{2}{5} \times \frac{4}{7}\end{align}
vi) \begin{align} 2\frac{3}{5} \times 3\end{align}
vii) \begin{align} 3\frac{4}{7} \times \frac{3}{5}\end{align}
Solution
What is known?
Expression
What is unknown?
Product of the given expression.
Reasoning:
Covert mixed fraction into improper fraction then find the product.
Steps:
i) \begin{align} \frac{2}{5} \times 5\frac{1}{4}\end{align}
\begin{align} =\frac{2}{5}\times \frac{21}{4}\end{align}
Reducing to the lowest form, we get
\begin{align}&= \frac{1}{5} \times \frac{{21}}{2}\\&= \frac{{1 \times 21}}{{5 \times 2}}\\&= \frac{{21}}{{20}}\end{align}
(improper fraction)
Converting into mixed fraction, we get \begin{align}= 2\frac{1}{{10}}\end{align}
ii) \begin{align} 6\frac{2}{5} \times \frac{7}{9} \end{align}
\begin{align}&= \frac{{32}}{5} \times \frac{7}{9}\\&= \frac{{224}}{{45}}\end{align}
(improper fraction)
Converting \begin{align} \frac{{224}}{{45}}\end{align} into mixed fraction, we get \begin{align} = 4\frac{{44}}{{45}} \end{align}
iii) \begin{align} \frac{3}{2} \times 5\frac{1}{3}\end{align}
\begin{align}&= \frac{3}{2} \times \frac{{16}}{3}\\&= \frac{{3 \times 16}}{{2 \times 3}}\end{align}
Reducing to the lowest form, we get
\begin{align} &= \frac{{3 \times 16}}{{2 \times 3}} \\&= 8\end{align}
(This is a whole number)
iv) \begin{align} \frac{5}{6} \times 2\frac{3}{7}\end{align}
\begin{align}&= \frac{5}{6} \times \frac{{17}}{7}\\&= \frac{{85}}{{42}}\end{align}
Reducing \begin{align} \frac{{85}}{{42}}\end{align} to the lowest form, we get \begin{align} = 2\frac{1}{{42}} \end{align}
v) \begin{align} 3\frac{2}{5} \times \frac{4}{7}\end{align}
\begin{align}&= \frac{{17}}{5} \times \frac{4}{7}\\&= \frac{{68}}{{35}}\end{align}
(improper fraction)
Converting \begin{align} \frac{{68}}{{35}}\end{align} into mixed fraction, we get \begin{align} = 1\frac{{33}}{{35}}\end{align}
vi) \begin{align} 2\frac{3}{5} \times 3\end{align}
\begin{align}&= \frac{{13}}{5} \times \frac{3}{1}\\&= \frac{{13 \times 3}}{{5 \times 1}}\\&= \frac{{39}}{5}\end{align}
(improper fraction)
Converting \begin{align} \frac{{39}}{5}\end{align} into mixed fraction, we get \begin{align} = 7\frac{4}{5} \end{align}
(vii) \begin{align} 3\frac{4}{7} \times \frac{3}{5}\end{align}
\begin{align}&=\frac{25}{7}\times \frac{3}{5} \\& =\frac{25\times 3}{7\times 5} \\&\\\end{align}
Reducing it to the lowest form, we get
$= \,\frac{15}{7}$
(improper fraction)
Converting \begin{align} \frac{{15}}{7}\end{align} into mixed fraction, we get \begin{align}=2\frac{1}{7} \end{align}
Chapter 2 Ex.2.3 Question 4
Which is greater?
(i) \begin{align} \frac{2}{7}{\text{ of }}\frac{3}{4}{\text{ or }}\frac{3}{5}{\text{ of }}\frac{5}{8}\end{align}
ii) \begin{align} \frac{1}{2}{\text{ of }}\frac{6}{7}{\text{ or }}\frac{2}{3}{\text{ of }}\frac{3}{7}\end{align}
Solution
What is known?
Fractions.
What is unknown?
Which is greater.
Reasoning:
Convert the fractions into like fractions then compare by numerator.
Steps:
(i) \begin{align} \frac{2}{7}{\text{ of }}\frac{3}{4}{\text{ or }}\frac{3}{5}{\text{ of }}\frac{5}{8}\end{align}
\begin{align}&= \frac{2}{7} \times \frac{3}{4} = \frac{3}{{14}}\\&= \frac{3}{5} \times \frac{5}{8} = \frac{3}{8}\end{align}
Converting these fractions into like fraction, we get
\begin{align}\frac{3}{{14}} &= \frac{{3 \times 4}}{{14 \times 4}} = \frac{{12}}{{56}}\\\frac{3}{8} &= \frac{{3 \times 7}}{{8 \times 7}} = \frac{{21}}{{56}}\end{align}
Since,
\begin{align}\frac{{21}}{{56}} > \frac{{12}}{{56}}\\\frac{3}{8} > \frac{3}{{14}}\end{align}
Thus \begin{align} \frac{3}{5}{\rm{ of }}\frac{5}{8}\end{align} is greater.
ii) \begin{align} \frac{1}{2}{\text{ of }}\frac{6}{7}{\text{ or }}\frac{2}{3}{\text{ of }}\frac{3}{7}\end{align}
\begin{align}&= \frac{1}{2} \times \frac{6}{7} = \frac{3}{7}\\&= \frac{2}{3} \times \frac{3}{7} = \frac{2}{7} \end{align}
On comparing, we get
$\frac{3}{7} > \frac{2}{7}$
Thus, \begin{align} \frac{1}{2}{\rm{ of }}\frac{6}{7}\end{align} is greater.
Chapter 2 Ex.2.3 Question 5
Saili plants $$4$$ saplings, in a row, in her garden. The distance between two adjacent saplings is \begin{align} \frac{3}{4}\rm\, m\end{align}. Find the distance between the first and the last sapling.
Solution
What is known?
Number of saplings and distance between two adjacent saplings.
What is unknown?
Distance between the first and the last sapling.
Reasoning:
Add the distance of first to second, second to third and third to fourth. Since all the sapling are plant equidistance to each other then we can simply multiply \begin{align}\frac{3}{4}\end{align}by $$3$$.
Steps:
Total number of saplings $$\rm= 4$$
Distance between two adjacent saplings\begin{align} = \frac{3}{4}{\rm{m}}\end{align}
Distance between the first and the last sapling \begin{align}= 3 \times \frac{3}{4} = \frac{9}{4}{\rm{m}} = 2\frac{1}{4}{\rm{m}}\end{align}
Thus, the distance between the first and last sapling is \begin{align} 2\frac{1}{4}{\rm{m}}\end{align}
Chapter 2 Ex.2.3 Question 6
Lipika reads a book for \begin{align} 1\frac{3}{4}\end{align} hour every day. She reads the entire book in $$6$$ days. How many hours in all were required by her to read the book?
Solution
What is known?
Lipika reads for \begin{align}1\frac{3}{4}\end{align}hours every day and read the entire book in $$6$$ days.
What is unknown?
Total hours required by her to complete the book.
Reasoning:
In a day she reads for \begin{align}1\frac{3}{4}\end{align} hours so for calculating $$6$$ days we can simply multiply \begin{align}1\frac{3}{4}\end{align} hours by $$6$$.
Steps:
\begin{align} \begin{Bmatrix}\text{No. of hours taken by Lipika }\\ \text{to read a book everyday}\end{Bmatrix} \end{align}
\begin{align} =1\frac{3}{4}{\rm{ hour }} = \frac{7}{4}\end{align}
No. of days taken to read the entire book $$= 6 \rm \,days$$
$$\begin{Bmatrix} \text{Total no. of hours required } \\ \text{ by her to read the book} \end{Bmatrix}$$
\begin{align} &= \frac{7}{4} \times 6 \\[0.5mm] &= \frac{{42}}{4}{\rm\,{ hours }} \\ &= 10\frac{1}{2}\end{align}
Thus, \begin{align} 10\frac{1}{2}\end{align} hours in all were required by Lipika to read the book.
Chapter 2 Ex.2.3 Question 7
A car runs $$16 \rm \,km$$ using $$1$$ liter of petrol. How much distance will it cover using \begin{align} 2\frac{3}{4}\end{align} liters of petrol?
Solution
What is known?
A car runs $$16 \rm{km}$$ using $$1$$ liter of petrol.
What is unknown?
How much distance car can in \begin{align}2\frac{3}{4}\end{align} liters of petrol.
Reasoning:
By using Unitary Method we can simply multiply \begin{align}2\frac{3}{4}\end{align} by $$16$$ to get how much distance will car cover in \begin{align}2\frac{3}{4}\end{align} liters.
Steps:
Distance covered by the car using $$1$$ liter of petrol $$= 16 \rm \,km$$
Distance covered by using \begin{align} \left( {2\frac{3}{4} = \frac{{11}}{4}{\rm\,{ liter }}} \right)\end{align} of petrol
\begin{align} &=\frac{{16}}{1} \times \frac{{11}}{4}\\&= 44\rm \, km\end{align}
Thus, $$44\rm \, km$$ distance covered by car using \begin{align} 2\frac{3}{4}\end{align} liters of petrol.
Chapter 2 Ex.2.3 Question 8
a) i) Provide the number in the box $$□$$, such that \begin{align} \frac{2}{3}\times \square =\frac{10}{30}\end{align}
ii) The simplest form of the number obtained in $$□$$ is ―.
b) i) Provide the number in the box $$□$$, such that \begin{align} \frac{3}{5}\times \,\square =\frac{24}{75}\end{align}
ii) The simplest form of the number obtained in □ is ―.
Solution
Steps:
(a)
What is known?
Equations.
What is unknown?
Value of the box.
Reasoning:
To make L.H.S. $$=$$ R.H.S. we have to multiple numerator by $$5$$ and denominator by $$10$$.
i)\begin{align} \frac{2}{3}\times \square =\frac{10}{30}\end{align}
$=\frac{2}{3}\times \frac{5}{10}=\frac{10}{30}$
Therefore, the number in the box □, such that \begin{align} \frac{2}{3} \times \square = \frac{{10}}{{30}}{\text{ is }}\frac{5}{{10}}\end{align}
ii) The simplest form of the number obtained in \begin{align} \frac{5}{{10}}{\text{ is }}\frac{1}{2}\end{align} .
(b)
What is known?
Equations.
What is unknown?
Value of the box.
Reasoning:
To make L.H.S. $$=$$ R.H.S. we have to multiple numerator by $$8$$ and denominator by $$15$$.
i) \begin{align} \frac{3}{5}\times \,\square =\frac{24}{75}\end{align}
\begin{align}=\frac{3}{5}\times \frac{8}{15}=\frac{24}{75} \\\end{align}
Therefore, the number in the box $$□$$, such that \begin{align} \frac{3}{5} × □ = \frac{{24}}{{75}} \rm \,is\, \frac{8}{{15}}\end{align}
ii) As \begin{align} \frac{8}{{15}}\end{align} can’t be simplified further. Therefore, its simplest form is \begin{align} \frac{8}{{15}}\end{align} . |
# 8.1 Confidence Intervals: The Basics 8.2 Estimating a Population Proportion 8.3 Estimating a Population Mean
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## +
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
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### Estimating a Population Proportion
Your teacher has a container full of different colored beads. Your goal is to estimate the actual proportion of red beads in the container.
✔ Form teams of 3 or 4 students.
✔ Determine how to use a cup to get a simple random sample of beads from the container.
✔ Each team is to collect one SRS of beads.
✔ Determine a point estimate for the unknown population proportion.
✔ Find a 90% confidence interval for the parameter p. Consider any conditions that are required for the methods you use.
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### Conditions for Estimating p
Suppose one SRS of beads resulted in 107 red beads and 144 beads of another color. The point estimate for the unknown proportion p of red beads in the population would be
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### Conditions for Estimating p
Check the conditions for estimating p from our sample.
### Estimating a Population Proportion
Random: The class took an SRS of 251 beads from the container.
Normal: Both np and n(1 – p) must be greater than 10. Since we don’t know p, we check that
The counts of successes (red beads) and failures (non-red) are both ≥ 10.
Independent: Since the class sampled without replacement, they need to check the 10% condition. At least 10(251) = 2510 beads need to be in the population. The teacher reveals there are 3000 beads in the container, so the condition is satisfied.
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### under the standard Normal curve.
For example, to find a 95%
confidence interval, we use a critical value of 2 based on the 68-95-99.7 rule. Using Table A or a calculator, we can get a more accurate critical value.
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### Estimating a Population Proportion
Since we want to capture the
central 80% of the standard Normal distribution, we leave out 20%, or 10% in each tail.
Search Table A to find the point z* with area 0.1 to its left.
### = – 1.28.
z .07 .08 .09 – 1.3 .0853 .0838 .0823
– 1.2 .1020 .1003 .0985
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### Estimating a Population Proportion
Choose an SRS of size n from a large population that contains an unknown proportion p of successes. An approximate level C confidence interval for p is
where z* is the critical value for the standard Normal curve with area C
between – z* and z*.
Use this interval only when the numbers of successes and failures in the sample are both at least 10 and the population is at least 10 times as large as the sample.
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### One-Sample z Interval for a Population Proportion
Calculate and interpret a 90% confidence interval for the proportion of red beads in the container. Your teacher claims 50% of the beads are red. Use your interval to comment on this claim.
### Estimating a Population Proportion
z .03 .04 .05 – 1.7 .0418 .0409 .0401
– 1.6 .0516 .0505 .0495
– 1.5 .0630 .0618 .0606 ✔ For a 90% confidence level, z* = 1.645
✔ We checked the conditions earlier.
✔ sample proportion = 107/251 = 0.426
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### Choosing the Sample Size
In planning a study, we may want to choose a sample size that allows us to estimate a population proportion within a given margin of error.
### Estimating a Population Proportion
The margin of error (ME) in the confidence interval for p is
z* is the standard Normal critical value for the level of confidence we want.
To determine the sample size n that will yield a level C confidence interval for a population proportion p with a maximum margin of error ME, solve the following inequality for n:
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### Example: Customer Satisfaction
Read the example on page 493. Determine the sample size needed to estimate p within 0.03 with 95% confidence.
### Estimating a Population Proportion
✔ The critical value for 95% confidence is z* = 1.96.
✔ Since the company president wants a margin of error of no more than 0.03, we need to solve the equation
Multiply both sides by square root n and divide
both sides by 0.03.
Square both sides.
Substitute 0.5 for the sample proportion to
find the largest ME
possible.
We round up to 1068 respondents to ensure
the margin of error is no more than 0.03 at
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### Estimating a Population Proportion
In this section, we learned that…
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### Estimating a Population Proportion
In this section, we learned that…
✔ When constructing a confidence interval, follow the familiar four-step process:
✔ STATE: What parameter do you want to estimate, and at what confidence level?
✔ PLAN: Identify the appropriate inference method. Check conditions.
✔ DO: If the conditions are met, perform calculations.
✔ CONCLUDE: Interpret your interval in the context of the problem.
✔ The sample size needed to obtain a confidence interval with approximate margin of error ME for a population proportion involves solving
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Updating...
## References
Updating...
Related subjects : |
### MAXIMUM/MINIMUM PROBLEMS
The following problems are maximum/minimum optimization problems. They illustrate one of the most important applications of the first derivative. Many students find these problems intimidating because they are "word" problems, and because there does not appear to be a pattern to these problems. However, if you are patient you can minimize your anxiety and maximize your success with these problems by following these guidelines :
GUIDELINES FOR SOLVING MAX./MIN. PROBLEMS
1. Read each problem slowly and carefully. Read the problem at least three times before trying to solve it. Sometimes words can be ambiguous. It is imperative to know exactly what the problem is asking. If you misread the problem or hurry through it, you have NO chance of solving it correctly.
2. If appropriate, draw a sketch or diagram of the problem to be solved. Pictures are a great help in organizing and sorting out your thoughts.
3. Define variables to be used and carefully label your picture or diagram with these variables. This step is very important because it leads directly or indirectly to the creation of mathematical equations.
4. Write down all equations which are related to your problem or diagram. Clearly denote that equation which you are asked to maximize or minimize. Experience will show you that MOST optimization problems will begin with two equations. One equation is a "constraint" equation and the other is the "optimization" equation. The "constraint" equation is used to solve for one of the variables. This is then substituted into the "optimization" equation before differentiation occurs. Some problems may have NO constraint equation. Some problems may have two or more constraint equations.
5. Before differentiating, make sure that the optimization equation is a function of only one variable. Then differentiate using the well-known rules of differentiation.
6. Verify that your result is a maximum or minimum value using the first or second derivative test for extrema.
The following problems range in difficulty from average to challenging.
• PROBLEM 1 : Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.
• PROBLEM 2 : Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen ?
• PROBLEM 3 : An open rectangular box with square base is to be made from 48 ft.2 of material. What dimensions will result in a box with the largest possible volume ?
• PROBLEM 4 : A container in the shape of a right circular cylinder with no top has surface area 3 ft.2 What height h and base radius r will maximize the volume of the cylinder ?
• PROBLEM 5 : A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume ?
• PROBLEM 6 : Consider all triangles formed by lines passing through the point (8/9, 3) and both the x- and y-axes. Find the dimensions of the triangle with the shortest hypotenuse.
• PROBLEM 7 : Find the point (x, y) on the graph of nearest the point (4, 0).
• PROBLEM 8 : A cylindrical can is to hold 20 m.3 The material for the top and bottom costs $10/m.2 and material for the side costs$8/m.2 Find the radius r and height h of the most economical can.
• PROBLEM 9 : You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground, which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time ?
• PROBLEM 10 : Construct a window in the shape of a semi-circle over a rectangle. If the distance around the outside of the window is 12 feet, what dimensions will result in the rectangle having largest possible area ?
• PROBLEM 11 : There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, the output per tree drops by 10 apples. How many trees should be added to the existing orchard in order to maximize the total output of trees ?
• PROBLEM 12 : Find the dimensions of the rectangle of largest area which can be inscribed in the closed region bounded by the x-axis, y-axis, and graph of y=8-x3 . (See diagram.)
• PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume ?
• PROBLEM 14 : A movie screen on a wall is 20 feet high and 10 feet above the floor. At what distance x from the front of the room should you position yourself so that the viewing angle of the movie screen is as large as possible ? (See diagram.)
• PROBLEM 15 : Find the dimensions (radius r and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2.
• PROBLEM 16 : What angle between two edges of length 3 will result in an isosceles triangle with the largest area ? (See diagram.)
• PROBLEM 17 : Of all lines tangent to the graph of , find the tangent lines of mimimum slope and maximum slope.
• PROBLEM 18 : Find the length of the shortest ladder that will reach over an 8-ft. high fence to a large wall which is 3 ft. behind the fence. (See diagram.)
• PROBLEM 19 : Find the point P = (x, 0) on the x-axis which minimizes the sum of the squares of the distances from P to (0, 0) and from P to (3, 2).
• PROBLEM 20 : Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same moment car A begins moving north at 60 mph. What will be the minimum distance between the cars and at what time t does the minimum distance occur ?
• PROBLEM 21 : A rectangular piece of paper is 12 inches high and six inches wide. The lower right-hand corner is folded over so as to reach the leftmost edge of the paper (See diagram.).
Find the minimum length of the resulting crease. |
# Ex.12.3 Q7 Areas Related to Circles Solution - NCERT Maths Class 10
Go back to 'Ex.12.3'
## Question
In the given figure, $$ABCD$$ is a square of side $$\text{14 cm.}$$ With Centers $$A, B, C$$ and $$D,$$ four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
Video Solution
Areas Related To Circles
Ex 12.3 | Question 7
## Text Solution
What is known?
$$ABCD$$ is a square of side $$\text{= 14 cm.}$$
With centers$$A, B, C, D$$ four circles are drawn such that each circle touches externally $$2$$ of the remaining $$3$$ circles.
What is unknown?
Reasoning:
Since the circles are touching each other externally,visually it is clear that
Radius of each circle \begin{align} {r} = \frac{1}{2}\,\, \times \end{align}(side of square)
Also, $$ABCD$$ being a square all angles are of measure $${90^ \circ },\,\,\,$$
Therefore,all sector are equal as they have same radii and angle.
$$\therefore\;$$ Angle of each sector which is part of the square $$\left( {\theta } \right) = {90^ \circ }$$
$$\therefore\;$$$$\text{Area of each sector}$$\begin{align}&= \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\\&= \frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}\\&= \frac{{\pi {r^2}}}{4}\end{align}
From the figure it is clear that:
Area of shaded region $$=$$ Area of square $$–$$ Area of $$4$$ sectors
\begin{align} &= {\left( {{\text{side}}} \right)^2} - 4 \times {\text{Area of each sector}}\\ &= {\left( {14} \right)^2} - 4 \times \frac{{\pi {r^2}}}{4}\\ &= {\left( {14} \right)^2} - \pi {r^2}\end{align}
Steps:
Area of each of the $$4$$ sectors is equal as each sector subtends an angle of $${90^ \circ }$$ at the centre of a circle with radius $$\text{= 7 cm}$$
$$\therefore$$ $$\text{Area of each sector}$$\begin{align}&= \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\\ &= \frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {(7)^2}\\ &= \frac{1}{4} \times \frac{{22}}{7} \times 7 \times 7\\ &= \frac{{77}}{2}{\text{c}}{{\text{m}}^2}\end{align}
Area of shaded region $$=$$ Area of square $$- \,4 \;\times$$ Area of each sector
\begin{align}&= {(14)^2} - 4 \times \frac{{77}}{2}\\ &= 196 - 154\\ &= 42\,\,{\text{c}}{{\text{m}}^2}\end{align}
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# Mathematics Guides for JSS 2 Number and Numeration
MATHEMATICS
THEME – NUMBER AND NUMERATION
###### TOPIC 1 – FRACTIONS
INSTRUCTIONAL MATERIALS
1. Squared paper or graph sheet
2. Flash cards.
LEARNING OBJECTIVES
By the end of the lesson, students should be able to:
1. convert simple fractions to ratios, decimals and percentages and vice versa.
2. Solve quantitative reasoning problems related to conversion of fractions to ratios, decimals and percentages.
CONTENTS OF THE LESSON
FOCUS LESSONS
1. Expressing fractions as ratios, decimals and percentages
2. Quantitative reasoning on fractions, ratios and percentages
LESSON PRESENTATION
TEACHER’S ACTIVITIES
1. Guides students to convert fractions to ratios, decimals and percentages and vice and versa.
2. Guides students to solve quantitative reasoning problems related to conversion of fractions to ratios, decimals and percentages and vice versa.
STUDENT’S ACTIVITIES
1. Convert fractions to ratios, decimals and percentages and vice and versa.
2. Solve quantitative reasoning problems related to conversion of fractions to ratios, decimals and percentages.
LESSON EVALUATION
Students to,
1. convert given fractions to ratios, decimals and percentages.
2. solve given problems on quantitative reasoning related to the contents.
MATHEMATICS
THEME – NUMBER AND NUMERATION
###### TOPIC 2 – WHOLE
INSTRUCTIONAL MATERIALS
Flashcards
LEARNING OBJECTIVES
By the end of the lesson, students should be able to:
1. express any whole number in standard form.
2. express decimal number in standard form.
3. find the prime factors of numbers not greater than 200.
4. express numbers as products of its prime factors.
5. find the least common multiples of numbers (LCM).
6. find the highest common factor (HCF) of numbers.
7. identify numbers that are perfect squares.
8. find squares of any given numbers.
9. find the square root of perfect squares using factor method.
10. find square root of any given number.
11. solve quantitative reasoning problems related to contents I – VII.
CONTENTS OF THE LESSON
FOCUS LESSONS
1. Whole numbers in standard form
2. Decimal numbers in standard form
3. Prime factors
4. Prime factorization
5. LCM
6. HCF
7. Quantitative reasoning.
LESSON PRESENTATION
TEACHER’S ACTIVITIES
1. Leads students to express any given whole number in standard form, i.e. a × 10n, 1 ≤ a < 10, n t 0.
2. Leads students to express decimal numbers in standard form, i.e. a × 10^n, 1 ≤ a < 10, n ≤ 0.
3. Guides students to find prime factors of numbers.
4. Guide students to express numbers as product of its prime factors.
5. Guides students to find the LCM of numbers.
6. Leads students to find the highest common factors of numbers.
7. Guide students to identify numbers that are perfect square.
8. Guide students to find the squares of any given number.
9. Guides students to find square roots of perfect square by factor method.
10. Lead students to find square root of any given number.
11. Guides students on how to solve quantitative reasoning problems on contents i – vii.
STUDENT’S ACTIVITIES
1. Express any whole number in standard form.
2. Express decimal numbers in standard form.
3. Find prime factors of numbers.
4. Express numbers as products of its prime factors.
mediator_tech]
5. Find the LCM of numbers.
6. Find the highest common factor of any given whole number.
7. Identify numbers that are perfect square.
8. Find the squares of any given numbers.
9. Find the square root of perfect squares by factor method.
10. Find the square root of any given number.
11. Solve problems on quantitative reasoning related to contents.
LESSON EVALUATION
Students to,
1. express given whole numbers in standard.
2. express given decimal numbers in standard form.
3. find prime factors of given numbers not greater than 200.
4. express given numbers as products of its prime factors.
5. find the least common multiples (LCM) of given numbers.
mediator_tech]
6. find the highest common factors of given numbers.
7. find the square of any given number.
8. find the square roots of given perfect squares using the factor method.
9. find the square root of any given number.
10. solve given problems on quantitative reasoning related to content I – vii.
mediator_tech] |
# 4.4: Horizontal and Vertical Line Graphs
Difficulty Level: At Grade Created by: CK-12
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What if you were given the graph of a vertical or horizontal line? How could you write the equation of this line? After completing this Concept, you'll be able to write horizontal and vertical linear equations and graph them in the coordiate plane.
### Guidance
How do you graph equations of horizontal and vertical lines? See how in the example below.
#### Example A
“Mad-cabs” have an unusual offer going on. They are charging \$7.50 for a taxi ride of any length within the city limits. Graph the function that relates the cost of hiring the taxi \begin{align*}(y)\end{align*} to the length of the journey in miles \begin{align*}(x)\end{align*}.
To proceed, the first thing we need is an equation. You can see from the problem that the cost of a journey doesn’t depend on the length of the journey. It should come as no surprise that the equation then, does not have \begin{align*}x\end{align*} in it. Since any value of \begin{align*}x\end{align*} results in the same value of \begin{align*}y (7.5)\end{align*}, the value you choose for \begin{align*}x\end{align*} doesn’t matter, so it isn’t included in the equation. Here is the equation:
\begin{align*}y = 7.5\end{align*}
The graph of this function is shown below. You can see that it’s simply a horizontal line.
Any time you see an equation of the form “\begin{align*}y =\end{align*} constant,” the graph is a horizontal line that intercepts the \begin{align*}y-\end{align*}axis at the value of the constant.
Similarly, when you see an equation of the form \begin{align*}x =\end{align*} constant, then the graph is a vertical line that intercepts the \begin{align*}x-\end{align*}axis at the value of the constant. (Notice that that kind of equation is a relation, and not a function, because each \begin{align*}x-\end{align*}value (there’s only one in this case) corresponds to many (actually an infinite number) \begin{align*}y-\end{align*}values.)
#### Example B
Plot the following graphs.
(a) \begin{align*}y = 4\end{align*}
(b) \begin{align*}y = -4\end{align*}
(c) \begin{align*}x = 4\end{align*}
(d) \begin{align*}x = -4\end{align*}
(a) \begin{align*}y = 4\end{align*} is a horizontal line that crosses the \begin{align*}y-\end{align*}axis at 4.
(b) \begin{align*}y = -4\end{align*} is a horizontal line that crosses the \begin{align*}y-\end{align*}axis at −4.
(c) \begin{align*}x = 4\end{align*} is a vertical line that crosses the \begin{align*}x-\end{align*}axis at 4.
(d) \begin{align*}x = -4\end{align*} is a vertical line that crosses the \begin{align*}x-\end{align*}axis at −4.
#### Example C
Find an equation for the \begin{align*}x-\end{align*}axis and the \begin{align*}y-\end{align*}axis.
Look at the axes on any of the graphs from previous examples. We have already said that they intersect at the origin (the point where \begin{align*}x = 0\end{align*} and \begin{align*}y = 0\end{align*}). The following definition could easily work for each axis.
\begin{align*}x-\end{align*}axis: A horizontal line crossing the \begin{align*}y-\end{align*}axis at zero.
\begin{align*}y-\end{align*}axis: A vertical line crossing the \begin{align*}x-\end{align*}axis at zero.
So using example 3 as our guide, we could define the \begin{align*}x-\end{align*}axis as the line \begin{align*}y = 0\end{align*} and the \begin{align*}y-\end{align*}axis as the line \begin{align*}x = 0\end{align*}.
Watch this video for help with the Examples above.
### Vocabulary
• Horizontal lines are defined by the equation \begin{align*}y=\end{align*} constant and vertical lines are defined by the equation \begin{align*}x= \end{align*} constant.
• Be aware that although we graph the function as a line to make it easier to interpret, the function may actually be discrete.
### Guided Practice
Write the equation of the horizontal line that is 3 units below the x-axis.
Solution:
The horizontal line that is 3 units below the x-axis will intercept the y-axis at \begin{align*}y=-3\end{align*}. No matter what the value of x, the y value of the line will always be -3. This means that the equations for the line is \begin{align*}y=-3\end{align*}.
### Practice
1. Write the equations for the five lines (\begin{align*}A\end{align*} through \begin{align*}E\end{align*}) plotted in the graph below.
For 2-10, use the graph above to determine at what points the following lines intersect.
1. \begin{align*}A\end{align*} and \begin{align*}E\end{align*}
2. \begin{align*}A\end{align*} and \begin{align*}D\end{align*}
3. \begin{align*}C\end{align*} and \begin{align*}D\end{align*}
4. \begin{align*}B\end{align*} and the \begin{align*}y-\end{align*}axis
5. \begin{align*}E\end{align*} and the \begin{align*}x-\end{align*}axis
6. \begin{align*}C\end{align*} and the line \begin{align*}y = x\end{align*}
7. \begin{align*}E\end{align*} and the line \begin{align*}y = \frac {1} {2} x\end{align*}
8. \begin{align*}A\end{align*} and the line \begin{align*}y = x + 3\end{align*}
9. \begin{align*}B\end{align*} and the line \begin{align*}y=-2x\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Horizontally Horizontally means written across in rows.
Vertically Vertically means written up and down in columns.
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## Angle Relationships
This week your student will be working with some relationships between pairs of angles.
• If two angles add to $90^\circ$, then we say they are complementary angles. If two angles add to $180^\circ$, then we say they are supplementary angles. For example, angles $JGF$ and $JGH$ below are supplementary angles, because $30 + 150 = 180$.
• When two lines cross, they form two pairs of vertical angles across from one another. In the previous figure, angles $JGF$ and $HGI$ are vertical angles. So are angles $JGH$ and $FGJ$. Vertical angles always have equal measures.
Here is a task to try with your student: Rectangle $PQRS$ has points $T$ and $V$ on two of its sides.
1. Angles $SVT$ and $TVR$ are supplementary. If angle $SVT$ measures $117^\circ$, what is the measure of angle $TVR$?
2. Angles $QTP$ and $QPT$ are complementary. If angle $QTP$ measures $53^\circ$, what is the measure of angle $QPT$?
Solution:
1. Angle $TVR$ measures $63^\circ$, because $180 - 117 = 63$.
2. Angle $QPT$ measures $37^\circ$, because $90 - 53 = 37$.
## Drawing Polygons with Given Conditions
This week your student will be drawing shapes based on a description. What options do we have if we need to draw a triangle, but we only know some of its side lengths and angle measures?
• Sometimes we can draw more than one kind of triangle with the given information. For example, “sides measuring 5 units and 6 units, and an angle measuring $32^\circ$” could describe two triangles that are not identical copies of each other.
• Sometimes there is only one unique triangle based on the description. For example, here are two identical copies of a triangle with two sides of length 3 units and an angle measuring $60^\circ$. There is no way to draw a different triangle (a triangle that is not an identical copy) with this description.
• Sometimes it is not possible to draw a triangle with the given information. For example, there is no triangle with sides measuring 4 inches, 5 inches, and 12 inches. (Try to draw it and see for yourself!)
Here is a task to try with your student: Using each set of conditions, can you draw a triangle that is not an identical copy of the one shown?
1. A triangle with sides that measure 4, 6, and 9 units.
2. A triangle with a side that measures 6 units and angles that measure $45^\circ$ and $90^\circ$
Solution:
1. There is no way to draw a different triangle with these side lengths. Every possibility is an identical copy of the given triangle. (You could cut out one of the triangles and match it up exactly to the other.) Here are some examples:
2. You can draw a different triangle by putting the side that is 6 opposite from the $90^\circ$ angle instead of next to it. This is not an identical copy of the given triangle, because it is smaller.
## Solid Geometry
This week your student will be thinking about the surface area and volume of three-dimensional figures. Here is a triangular prism. Its base is a right triangle with sides that measure 12, 12, and 17 inches.
In general, we can find the volume of any prism by multiplying the area of its base times its height. For this prism, the area of the triangular base is 72 in2, so the volume is $72 \boldcdot 14$, or 1,008 in3.
To find the surface area of a prism, we can find the area of each of the faces and add them up. The example prism has two faces that are triangles and three faces that are rectangles. When we add all these areas together, we see that the prism has a total surface area of $72+72+168+168+238$, or 718 in2.
Here is a task to try with your student: The base of this prism is a hexagon where all the sides measure 5 cm. The area of the base is about 65 cm2.
1. What is the volume of the prism?
2. What is the surface area of the prism?
Solution:
1. The volume of the prism is about 1,040 cm3, because $65 \boldcdot 16=1,\!040$.
2. The surface area of the prism is 610 cm2, because $16\boldcdot 5 = 80$ and $65 + 65 + 80 + 80 + 80 + 80 + 80 + 80 = 610$. |
## Simplification
1. What does the word “of” signify?
“of” denotes multiplication.
Examples:
• One-fifth of 5 is = (1/5) × 5 = 1,
• 50% of 100 = (50/100) × 100 = ½
2. The Rule on “of”
The operation (multiplication) denoted by “of” is first simplified before division and multiplication.
Example 1:
• 100 ÷ 4 of 5 = 100 ÷ 4 × 5 = 100 ÷ 20 = 5
3. Division is simplified before multiplication
Example 2:
• 100 ÷ 4 of 5 × 2
= 100 ÷ 20 × 2
Now, which should you simplify first in 100 ÷ 20 × 2?
“Division in 100 ÷ 20” Or “Multiplication in 20 × 2”?
Answer: Always Division first, Multiplication Next.
So, 100 ÷ 20 × 2 = 5 × 2 = 10
4. Brackets are simplified first before Division or Multiplication.
Example 4:
• Simplify: 900 ÷ 20 of (2 + 3) × 6 of 3
9900 ÷ 20 of (2 + 3) × 6 of 3 =
900 ÷ 20 of 5 × 6 of 3 =
900 ÷ 100 × 18 =
9 × 18 = 172
Rules of Operations on Brackets:
• Simplify “—“first before opening brackets.
• 4 + (3 – 2) = 4 + 1 = 5
• When there is a minus sign before brackets, then multiply it to the number obtained after opening up the brackets.
• 5 – (3 – 2) = 5 – 1 = 4
• 5 – (3 – 4) = 5 – (-1) = 5 + 1 = 6
• Simplify “( )” before “{ }”
{- 2 – (- 3)} × 4 =
{- 2 + 3} × 4 =
{1} × 4 =
1× 4 = 4
• Simplify “{ }” before “[ ]”
100 – [90 – {80 – (70 – 60 – 50)}] =
100 – [90 – {80 – (- 40)}] =
100 – [90 – {80 + 40}] =
100 – [90 – 120] =
100 – [- 30] =
100 + 30 = 130
Let us summarize all of the above rules of operations into one Mnemonic as BODMAS for Order of Operation”
BODMAS stands for Bracket Of Division, Multiplication, Addition, Subtraction.
The BODMAS Rule resolves the conflict between two different operations when they come up for simultaneous application.
The rule tells that the operation denoted by its letter coming first in BODMAS will take place before the other operation denoted by its letter coming later on in BODMAS.
In BODMAS, the operation denoted by a letter precedes that denoted by the latter letter.
Let us apply one by one each of the operations separately and combinedly in the last example below
• 4 + 2 – 3 = 6 – 3 = 3
In BODMAS, the letter A comes before S. Therefore, Addition precedes Subtraction.
• 4 + 2 × 3 = 4 + 6 = 10
In BODMAS, M precedes A, so, Multiplication first, Addition next.
• 4 ÷ 2 × 3 = 2 × 3 = 6
In BODMAS, D precedes M, so Division first, Multiplication next.
• 4 of 6 ÷ 2 = 4 × 6 ÷ 2 = 24 ÷ 2 = 12
In BODMAS, “O” stands for “of”, and “ of” denotes multiplication. And, O precedes D, therefore perform the Operation of multiplication denoted by “of” before Division.
• (100 ÷ 5) of 2
In BODMAS, B precedes O.
So, whichever operation B encloses shall precede the operation denoted by O, i.e. multiplication.
In (100 ÷ 5) of 2, the brackets enclose the division operation.
So, division (as enclosed by brackets) is performed before multiplication (as denoted by “of”)
(100 ÷ 5) of 2 = 20 × 2 = 40 |
# How do you find the sum of consecutive whole numbers?
## How do you find the sum of consecutive whole numbers?
Using the Formula (n / 2)(first number + last number) = sum, where n is the number of integers.
What are the sum of three consecutive whole numbers is 54?
Numbers are 17,18,19.
What are consecutive whole numbers?
Consecutive integers are whole numbers that follow each other without gaps. For example, 15, 16, 17 are consecutive integers.
### What is the sum of consecutive numbers?
– The sum of any two consecutive numbers is always odd. Example, 4 + 5 = 9; –8 + (–7) = –15.
What is the sum of 3 consecutive even numbers is 78?
1 Expert Answer If 78 is their grand total, then those 3 consecutive even numbers have to be 24, 26 & 28.
What is the sum of three consecutive numbers is 48?
So, our three numbers are 14, 16, 18. As always, a quick check of your answer is suggested: 14+16+18 = 48 (as required).
## What is consecutive whole number?
What is the sum of three consecutive numbers is 63?
∴ , the integers are 19 , 21 , and 23 .
What are the consecutive numbers?
Consecutive numbers are numbers that follow each other in order. They have a difference of 1 between every two numbers. In a set of consecutive numbers,themean and the median are Equal. If n is a number, then n, n+1, and n+2 would beconsecutive numbers Examples. 1, 2, 3, 4, 5.
### What are 3 consecutive integers of 21?
We have to find three consecutive odd numbers whose sum is 21. So, let the numbers be 2x – 1, 2x + 1 and 2x + 3. So, the numbers are 2(3) – 1,2(3) + 1 and 2(3) + 3. That is; 5, 7, 9.
What is the sum of 3 consecutive odd numbers is 51?
15, 17 and 19
Now the sum of these three odd consecutive terms = x + x + 2 + x + 4 = 3x + 6. According to the given, 3x + 6 = 51. ⟹x =453=15. Hence, the numbers are 15, 17 and 19.
What are three consecutive integers have a sum of 582?
Which means that the first number is 193, the second number is 193 + 1 and the third number is 193 + 2. Therefore, three consecutive integers that add up to 582 are 193, 194, and 195. 193 + 194 + 195 = 582
## How do you find three consecutive even integers?
x 2 x 3 = 6
• x 3 x 4= 24
• x 4 x 5 = 60
• What are the three consecutive odd numbers that equal 6783?
The product of two consecutive odd integers is equal to 675. Find the two integers. The product of two consecutive numbers is 182. What are the numebrs. The sum of 4 consecutive negative odd numbers is -32. What are the numbers? What are two consecutive numbers whose product is 552?
What is the LCM of any two consecutive numbers?
To find the two consecutive even numbers, we have to factorize 360. ∴ The two consecutive even numbers whose LCM is 180 are 18 and 20.
Categories: FAQ |
# Discovering What it Means to be Similar Triangles Continued
5 teachers like this lesson
Print Lesson
## Objective
SWBAT understand angle relationships with similar triangles, discover correct methods of enlarging similar triangles and then apply this to dilating figures.
#### Big Idea
Imagine teaching similarity, angle-angle similarity for triangles, and dilations all at once! It's a three for one deal!
20 minutes
## Beginning the New Material Lesson
10 minutes
The goal of the lesson today is to connect the observations from yesterday to both similar figures and dilations. I told students to get out the activity pages and their unit organizer. We spent a few minutes reviewing the observations we made on day one and used pictures taken with my IPad to remind students of the tangram comparisons we made on the previous day. Linked below are those images:
Tangram Comparison 1 (adding length)
Tangram Comparison 2 (subtracting length)
Tangram Comparison 3 (multiplying to lengthen)
We spent several minutes recording important notes onto the unit organizer across from similarity. Notes included ideas such as:
• Similar figures have congruent corresponding angles because angles determine shape.
• Side lengths are different but proportional (they remembered this from 7th grade) through multiplication with a common scale factor.
• Scale factors means to multiply each side by the same number.
• A scale factor greater than one enlarges and a scale factor between 0 and 1 reduces.
While adding these notes to the organizer we also reviewed answers to the two homework questions.
## Using New Information to Practice
30 minutes
Pass out page three of the activity and discuss with students the goal of the activity today. The goal today is to understand how to identify when two figures are similar because you understand the application of a scale factor to dilate triangles correctly. The leap to connect dilations to similar figures was easily made as most students remembered this from 7th grade and the concept that side lengths are proportional.
In the following tables students will be given the side lengths of two different triangles. Students need to look between corresponding sides to find a common pattern that holds true for all corresponding sides and then record this pattern. The follow-up question is, would this pattern create a correct dilation and therefore make the two triangles similar. Look at the example together and then give students time to work in cooperative groups as you move about the room providing feedback. The last table is the most difficult as the scale factor is 2.5. I had two groups look at this scale factor in unique ways to figure it out so I asked both groups to present their thinking to the class during our mini wrap up time. The leap to connect dilations to similar figures was easily made as most students remembered this from 7th grade and the concept that side lengths are proportional.
After about 10 minutes of group time, call the class together and allow students to present answers to each table. Once you finish, make important notes to the unit organizer about dilations. Notes include:
• Dilations produce similar figures
• Dilations follow a common scale factor (common multiplier to all side lengths)
• Dilations do not change angle measures
• Figures will enlarge or reduce proportionally in all directions
• Scale factor > 1 enlarges and a scale factor that is between 0 and 1 reduces. |
Associated Topics || Dr. Math Home || Search Dr. Math
### Changing Equations To Slope-Intercept Form
```
Date: 07/17/98 at 23:25:58
From: Holly
Subject: Slope-intercept form
I am having trouble understanding putting an equation in slope-
intercept form. One example is:
12(2x - 1) - 5(3y + 2) = 8.
y = 8/5x - 2; slope = 8/5; y-intercept: (0, -2).
```
```
Date: 07/19/98 at 19:42:25
From: Doctor White
Subject: Re: Slope-intercept form
Holly:
The slope-intercept form of the equation of a line is: y = mx + b,
where m is the slope and b is the y-intercept.
In order to solve equations for specific variables, you need to take
several steps:
1) simplify both sides of the equation
2) combine like terms on each side
3) move everything over to one side except the varibles you are
solving for
4) solve for the variable
Now let's look at your problem:
12(2x - 1) - 5(3y + 2) = 8
1) Simplify the left side by distributing to remove the parentheses:
24x - 12 - 15y - 10 = 8
2) Combine like terms:
24x - 15y - 12 - 10 = 8
24x - 15y - 22 = 8
3) Move terms to the right side except for the y (since you are solving
for y). You can move a term over by adding its opposite to the other
side of the equation:
24x - 24x - 15y - 22 + 22 = 8 - 24x + 22
Since you added -24x and +22 on the left side, then you must do the
same to the other side to keep the equality. Thus:
0 - 15y + 0 = 30 - 24x
- 15y = 30 - 24x
4) Now solve for y by dividing both sides by -15:
-15y/(-15) = (30 - 24x)/(-15)
y = 30/(-15) - 24x/(-15)
y = -2 + 8x/5
You now have it in slope-intercept form:
y = m x + b
y = 8/5 x + (-2)
Thus m = 8/5, which is the slope, and b = -2, which is the
y-intercept, often written (0,-2).
I hope these steps help you to understand these type of problems. Let
me know if you need further explanation. Come back to see us soon.
- Doctor White, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations
High School Linear Equations
Middle School Equations
Middle School Graphing Equations
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# 2.5 Equations of lines and planes in space (Page 3/19)
Page 3 / 19
Let $L$ be a line in the plane and let $M$ be any point not on the line. Then, we define distance $d$ from $M$ to $L$ as the length of line segment $\stackrel{—}{MP},$ where $P$ is a point on $L$ such that $\stackrel{—}{MP}$ is perpendicular to $L$ ( [link] ).
When we’re looking for the distance between a line and a point in space, [link] still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let $P$ be an arbitrary point on line $L$ and let $\text{v}$ be a direction vector for $L$ ( [link] ).
By [link] , vectors $\stackrel{\to }{PM}$ and $\text{v}$ form two sides of a parallelogram with area $‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖.$ Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:
$‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{v}‖d.$
We can use this formula to find a general formula for the distance between a line in space and any point not on the line.
## Distance from a point to a line
Let $L$ be a line in space passing through point $P$ with direction vector $\text{v}.$ If $M$ is any point not on $L,$ then the distance from $M$ to $L$ is
$d=\frac{‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖}{‖\text{v}‖}.$
## Calculating the distance from a point to a line
Find the distance between t point $M=\left(1,1,3\right)$ and line $\frac{x-3}{4}=\frac{y+1}{2}=z-3.$
From the symmetric equations of the line, we know that vector $\text{v}=⟨4,2,1⟩$ is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point $P\left(3,-1,3\right)$ lies on the line. Then,
$\stackrel{\to }{PM}=⟨1-3,1-\left(-1\right),3-3⟩=⟨-2,2,0⟩.$
To calculate the distance, we need to find $\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\text{:}$
$\begin{array}{cc}\hfill \stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill -2& \hfill 2\hfill & \hfill 0\hfill \\ \hfill 4& \hfill 2\hfill & \hfill 1\hfill \end{array}|\hfill \\ & =\left(2-0\right)\text{i}-\left(-2-0\right)\text{j}+\left(-4-8\right)\text{k}\hfill \\ & =2\text{i}+2\text{j}-12\text{k}.\hfill \end{array}$
Therefore, the distance between the point and the line is ( [link] )
$\begin{array}{cc}\hfill d& =\frac{‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖}{‖\text{v}‖}\hfill \\ & =\frac{\sqrt{{2}^{2}+{2}^{2}+{12}^{2}}}{\sqrt{{4}^{2}+{2}^{2}+{1}^{2}}}\hfill \\ & =\frac{2\sqrt{38}}{\sqrt{21}}.\hfill \end{array}$
Find the distance between point $\left(0,3,6\right)$ and the line with parametric equations $x=1-t,y=1+2t,z=5+3t.$
$\sqrt{\frac{10}{7}}$
## Relationships between lines
Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines ( [link] ).
To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point ( [link] ).
## Classifying lines in space
For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.
1. ${L}_{1}:x=2s-1,y=s-1,z=s-4$
${L}_{2}:x=t-3,y=3t+8,z=5-2t$
2. ${L}_{1}\text{:}$ $x=\text{−}y=z$
${L}_{2}:\frac{x-3}{2}=y=z-2$
3. ${L}_{1}:x=6s-1,y=-2s,z=3s+1$
${L}_{2}:\frac{x-4}{6}=\frac{y+3}{-2}=\frac{z-1}{3}$
1. Line ${L}_{1}$ has direction vector ${\text{v}}_{1}=⟨2,1,1⟩;$ line ${L}_{2}$ has direction vector ${\text{v}}_{2}=⟨1,3,-2⟩.$ Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, $\left(x,y,z\right),$ that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
$2s-1=t-3;\phantom{\rule{1em}{0ex}}s-1=3t+8;\phantom{\rule{1em}{0ex}}s-4=5-2t.$
By the first equation, $t=2s+2.$ Substituting into the second equation yields
$\begin{array}{ccc}\hfill s-1& =\hfill & 3\left(2s+2\right)+8\hfill \\ \hfill s-1& =\hfill & 6s+6+8\hfill \\ \hfill 5s& =\hfill & -15\hfill \\ \hfill s& =\hfill & -3.\hfill \end{array}$
Substitution into the third equation, however, yields a contradiction:
$\begin{array}{ccc}\hfill s-4& =\hfill & 5-2\left(2s+2\right)\hfill \\ \hfill s-4& =\hfill & 5-4s-4\hfill \\ \hfill 5s& =\hfill & 5\hfill \\ \hfill s& =\hfill & 1.\hfill \end{array}$
There is no single point that satisfies the parametric equations for ${L}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{L}_{2}$ simultaneously. These lines do not intersect, so they are skew (see the following figure).
2. Line L 1 has direction vector ${\text{v}}_{1}=⟨1,-1,1⟩$ and passes through the origin, $\left(0,0,0\right).$ Line ${L}_{2}$ has a different direction vector, ${\text{v}}_{2}=⟨2,1,1⟩,$ so these lines are not parallel or equal. Let $r$ represent the parameter for line ${L}_{1}$ and let $s$ represent the parameter for ${L}_{2}\text{:}$
$\begin{array}{cccc}\begin{array}{cc}x\hfill & =r\hfill \\ y\hfill & =\text{−}r\hfill \\ z\hfill & =r\hfill \end{array}\hfill & & & \begin{array}{cc}x\hfill & =2s+3\hfill \\ y\hfill & =s\hfill \\ z\hfill & =s+2.\hfill \end{array}\hfill \end{array}$
Solve the system of equations to find $r=1$ and $s=-1.$ If we need to find the point of intersection, we can substitute these parameters into the original equations to get $\left(1,-1,1\right)$ (see the following figure).
3. Lines ${L}_{1}$ and ${L}_{2}$ have equivalent direction vectors: $\text{v}=⟨6,-2,3⟩.$ These two lines are parallel (see the following figure).
find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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# 15. Polynomial rings Definition-Lemma Let R be a ring and let x be an indeterminate.
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1 15. Polynomial rings Definition-Lemma Let R be a ring and let x be an indeterminate. The polynomial ring R[x] is defined to be the set of all formal sums a n x n + a n 1 x n +... a 1 x + a 0 = a i x i where each a i R (a 1, a,... are called the coefficients of the polynomial; a i is the coefficient of x i ). Given two polynomials f = a n x n + a n 1 x n a 1 x + a 0 = a i x i g = b m x m + b m 1 x m b 1 x + b 0 = b i x i in R[x] the sum of f and g, f + g, is defined as, f+g = (a n +b n )x n +(a n 1 +b n 1 )x n 1 + +(a 1 +b 1 )x+(a 0 +b 0 ) = (a i +b i )x i, (where we have implicitly assumed that m n and we set b i = 0, for i > m) and the product as fg = c m+n x m+n +c m+n 1 x m+n 1 + +c 1 x 1 +c 0 = i c i x i = i ( j a j b i j )x i. With this rule of addition and multiplication, R[x] becomes a ring, with zero given as the polynomial with zero coefficients. If R is commutative then R[x] is commutative. If R has unity, 1 0 then R[x] has unity, 1 0; 1 is the polynomial whose constant coefficient is one and whose other terms are zero. Proof. A long and completely uninformative check. then and For example, if f(x) = x 5x + 6 Z[x] and g(x) = x 3 5 Z[x], f(x) + g(x) = (x 5x + 6) + (x 3 5) = x 3 + x 5x + 1, f(x)g(x) = (x 5x + 6)(x 3 5) = x 5 10x 4 + 1x 3 5x 5x 30. Note that a polynomial determines a function R R in an obvious way. If one takes R to be the real numbers, then it is well known that a polynomial is determined by the corresponding function. In general, 1
2 however, this is far from true. For example take R = Z (one of the smallest possible rings). Then there are four functions from R to R and there are infinitely many polynomials. Thus two different polynomials will often determine the same function. Definition 15.. Let R be a ring and let f R[x] be a non-zero polynomial with coefficients in R. The degree of f is the largest n such that the coefficient of x n is non-zero. Polynomial rings give interesting examples of infinite rings of finite characteristic. For example Z [x] has infinitely many polynomials just let the degree go to infinity but the characteristic is two. Indeed if you add a polynomials to itself you are just adding the coefficients to themselves, which are then all zero. More generally Z n [x] is an infinite ring of finite characteristic n. Lemma Let R be an integral domain and let f and g be two non-zero elements of R[x]. Then the degree of fg is the sum of the degrees of f and g. In particular R[x] is an integral domain. Proof. Suppose that f = a n x n + a n 1 x n a 1 x + a 0 g = b m x m + b m 1 x m b 1 x + b 0, where a n and b m are non-zero. Then a n b m is non-zero and this is the largest non-zero coefficient of the product. So the degree of fg is n+m which is the degree of f plus the degree of g. This is the first statement. The second statement follows, by observing that a product fg can only equal zero if its degree is zero. In this case both f and g are constant polynomials and their product in R[x] is equal to their product in R. As R is an integral domain this is zero only if one of f and g is zero. Lemma Let R be an integral domain. Then the units in R[x] are precisely the units in R. Proof. One direction is clear. A unit in R is a unit in R[x]. Now suppose that f(x) is a unit in R[x]. Given a polynomial g, denote by d(g) the degree of g(x). Now f(x)g(x) = 1. In particular neither f(x) nor g(x) is zero. Thus 0 = d(1) = d(fg) = d(f) + d(g).
3 Thus both of f and g must have degree zero. It follows that f(x) = f 0 and that f 0 is a unit in R[x]. It is interesting to consider what the field of fractions of a polynomial ring looks like. Suppose that F is a field and consider the polynomial ring F [x]. The field of fractions consists of all ratios of two polynomials with coefficients in F, { f(x) f(x), g(x) F [x] }. g(x) These are called rational functions and the field of rational functions is denoted F (x). More generally if R is an integral domain with field of fractions F then F (x) is the field of fractions of R[x]. For example, x 3 Z(x) = Q(x). x 5x + 6 We can also work with polynomial rings in one more than variable. We do the case of two variables but the general case is the same. Definition Let R be a commutative ring and let x and y be indeterminates. A monomial in x and y is a product of powers of x and y, x i y j. The degree d of a monomial is the sum of the degrees of the individual terms, i + j. The polynomial ring R[x, y] is equal to the set of all finite formal sums a ij x i y j i,j with the obvious addition and multiplication. The degree of a polynomial is the maximum degree of a monomial term that appears with non-zero coefficient. Example Let x and y be indeterminates. A typical element of Q[x, y] might be x + y 1. This has degree. Note that xy also has degree two. A more complicated example might be a polynomial of degree 5. 3 x3 7xy + y 5, 3
4 The nice thing about polynomial rings in one more than one variable is that one can construct them iteratively as polynomial rings in one variable. Again, we just do the case of two variables: Lemma Let R be a commutative ring and let x and y be indeterminates. Let S = R[x]. Then there is a natural isomorphism R[x, y] S[y]. We will skip the proof which is not hard but we will try to illustrate how the proof proceeds by giving an example. Consider the polynomial 3 x3 7xy + y 5 Q[x, y]. Consider this as a polynomial in y, whose coefficients lie in the ring Q[x], so that y 5 + ( 7x)y + /3x 3 Q[x][y]. Note that by symmetry we may also consider the as a polynomial in x with coefficients in the ring Q[y], 3 x3 + ( 7y)x + y 5 Q[y][x]. Lemma If R is an integral domain then so is R[x 1, x,..., x n ]. Proof. We just do the case of two variables x and y; the general case is by induction on the number of variables. As R is an integral domain (15.3) implies that R[x] is an integral domain. As R[x] is an integral domain (15.3) implies that R[x][y] is an integral domain. As R[x, y] is isomorphic to R[x][y] it follows that R[x, y] is an integral domain. In particular if R is an integral domain then there is a field of fractions R(x, y) of R[x, y]. As in the case of one variable the elements of R(x, y) are rational functions, the quotients of polynomials with coefficients in R. It is interesting to understand the various ring homomorphisms attached to a polynomial ring. We start with a really obvious one. Lemma Let R be a ring. The natural inclusion R R[x] which just sends an element r R to the constant polynomial r, is a ring homomorphism. Proof. Easy. The following is a little bit more complicated but very useful: 4
5 Definition Suppose that R S is a subring of the ring S. Let α be an element of S. Then the map φ α : R[x] S, which sends a n x n + a n 1 x n a 1 x + a 0 a n α n + a n 1 α n a 1 α + a 0 is a ring homomorphism. It is characterised by the property that it sends x to α and has no effect on the coefficients. This map is called evaluation at α. Proof. An exercise for the reader. We will be mostly interested in the case when R and S are fields. Example Consider the ring homomorphism evaluation at zero φ 0 : Q[x] R. a n x n +a n 1 x n 1 + +a 1 x+a 0 a n 0 n +a n 1 0 n 1 + +a 1 0+a 0 = a 0. Thus a polynomial is sent to its constant term. Example Consider the ring homomorphism evaluation at three φ 3 : Q[x] R. a n x n + a n 1 x n a 1 x + a 0 a n 3 n + a n 1 3 n a a 0. Note that φ 3 (x 5x + 6) = = 0. Thus x = 5x + 6 is in the kernel N of φ 3. Of course x 5x + 6 = (x )(x 3) and x 3 is in the kernel N of φ 3. Example Consider the ring homomorphism evaluation at i φ i : Q[x] C. a n x n + a n 1 x n a 1 x + a 0 a n i n + a n 1 i n a 1 i + a 0. Note that φ i (x + 1) = i + 1 = 0 Thus x + 1 is in the kernel N of φ i. 5
6 Example Consider the ring homomorphism evaluation at π φ π : Q[x] C. a n x n + a n 1 x n a 1 x + a 0 a n π n + a n 1 π n a 1 π + a 0. It turns out that the kernel N of φ π is trivial, so that φ π is one to one. It follows that Q[π] is isomorphic to Q[x]. Definition Suppose that E F is a subfield of the field F. Let α be an element of F. We say that α is a zero of f(x) E[x], if f(x) is in the kernel N of φ α. Of course f(x) is in the kernel if and only if φ α (f(x)) = 0. 6
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# ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.2 in Chapter 2 - Compound Interest
Question 28 Compound Interest Exercise 2.2
. On what sum of money will the difference between the compound interest and interest for 2 years be
equal to ₹ 25 if the rate of interest charged for both is 5% p.a.
It is given that
Sum (P) = ₹ 100
Rate (R) = 5% p.a.
Period (n) = 2 years
We know that
SI = PRT/100
Substituting the values
= (100 × 5 × 2)/ 100
= ₹ 10
So the amount when interest is compounded annually = P (1 + R/100)
n
Substituting the values
\begin{aligned} &=100(1+5 / 100)^{2}\\ &\text { By further calculatio }\\ &=100 \times(21 / 20)^{2} \end{aligned}
So we get
= ₹ 441/4
Here
CI = A – P
Substituting the values
= 441/4 – 100
= ₹ 41/4
So the difference between CI and SI = 41/4 – 10 = ₹ ¼
If the difference is ₹ ¼ then sum = ₹ 100
If the difference is ₹ 25 then sum = (100 × 4)/ 1 × 25 = ₹ 10000
Video transcript
"hey kids welcome to lido q a video i am vinit your leader tutor bringing you this question on your screen on what sum of money friends between the compound interest and the interest for two years be equal to rupees 25 if the rate of interest charge for both is five percent per annum now the difference between the compound interest and interest for two years be equal to 25 so this means the difference between the interest for the first year and the second year is 25 percent so let us take an example to understand this let us say we you invest 10 000 rupees on the first year again the interest component is ten thousand rupees sorry thousand rupees and the principal is again ten thousand in the second year again you get thousand rupees on this 10 000 and 100 rupees on this interest right so these are the two components of compound interest interest on principle and interest on interest now this difference of 100 rupees between the interest of two consecutive year is the interest on the interest of the first year so we are going to use the same concept here right now interest on interest of first year is equal to rupees 25 right now interest on interest of first year is rupees 25 so interest of first year becomes our principle therefore interest of first year will be the principle now simple interest is equal to p rt by 100 this implies 25 is equal to p into 5 by 100 into 1 so this is 1 by 20. so this implies p is equal to 500 so this is the interest on first year now interest of first year is equal to rupees 500 therefore we can say 500 is equal to the actual sum of money into rate by 100 into 1. so this implies p is equal to 500 into 100 by 5 so this will be equal to rupees 10 000 isn't it isn't it easy guys right if you still have a doubt please leave a comment below do like the video and subscribe to our channel i'll see you in our next video until then bye guys keep learning keep flourishing and keep"
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You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.
# 6.12: Linear Inequalities in Two Variables
Difficulty Level: At Grade Created by: CK-12
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Progress
Practice Linear Inequalities in Two Variables
Progress
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What if you were given a linear inequality like $2x - 3y \le 5$ ? How could you graph that inequality in the coordinate plane? After completing this Concept, you'll be able to graph linear inequalities in two variables like this one.
### Guidance
The general procedure for graphing inequalities in two variables is as follows:
1. Re-write the inequality in slope-intercept form: $y=mx+b$ . Writing the inequality in this form lets you know the direction of the inequality.
2. Graph the line of the equation $y=mx+b$ using your favorite method (plotting two points, using slope and $y-$ intercept, using $y-$ intercept and another point, or whatever is easiest). Draw the line as a dashed line if the equals sign is not included and a solid line if the equals sign is included.
3. Shade the half plane above the line if the inequality is “greater than.” Shade the half plane under the line if the inequality is “less than.”
#### Example A
Graph the inequality $y \ge 2x-3$ .
Solution
The inequality is already written in slope-intercept form, so it’s easy to graph. First we graph the line $y=2x-3$ ; then we shade the half-plane above the line. The line is solid because the inequality includes the equals sign.
#### Example B
Graph the inequality $5x-2y>4$ .
Solution
First we need to rewrite the inequality in slope-intercept form:
$-2y & > -5x +4\\y & < \frac{5}{2}x - 2$
Notice that the inequality sign changed direction because we divided by a negative number.
To graph the equation, we can make a table of values:
$x$ $y$
-2 $\frac{5}{2}(-2)-2=-7$
0 $\frac{5}{2}(0)-2=-2$
2 $\frac{5}{2}(2)-2=3$
After graphing the line, we shade the plane below the line because the inequality in slope-intercept form is less than . The line is dashed because the inequality does not include an equals sign.
Solve Real-World Problems Using Linear Inequalities
In this section, we see how linear inequalities can be used to solve real-world applications.
#### Example C
A retailer sells two types of coffee beans. One type costs $9 per pound and the other type costs$7 per pound. Find all the possible amounts of the two different coffee beans that can be mixed together to get a quantity of coffee beans costing $8.50 or less. Solution Let $x =$ weight of$9 per pound coffee beans in pounds.
Let $y =$ weight of $7 per pound coffee beans in pounds. The cost of a pound of coffee blend is given by $9x + 7y$ . We are looking for the mixtures that cost$8.50 or less. We write the inequality $9x+7y \le 8.50$ .
Since this inequality is in standard form, it’s easiest to graph it by finding the $x-$ and $y-$ intercepts. When $x=0$ , we have $7y=8.50$ or $y=\frac{8.50}{7} \approx 1.21$ . When $y=0$ , we have $9x=8.50$ or $x=\frac{8.50}{9} \approx 0.94$ . We can then graph the line that includes those two points.
Now we have to figure out which side of the line to shade. In $y-$ intercept form, we shade the area below the line when the inequality is “less than.” But in standard form that’s not always true. We could convert the inequality to $y-$ intercept form to find out which side to shade, but there is another way that can be easier.
The other method, which works for any linear inequality in any form, is to plug a random point into the inequality and see if it makes the inequality true. Any point that’s not on the line will do; the point (0, 0) is usually the most convenient.
In this case, plugging in 0 for $x$ and $y$ would give us $9(0)+7(0) \le 8.50$ , which is true. That means we should shade the half of the plane that includes (0, 0). If plugging in (0, 0) gave us a false inequality, that would mean that the solution set is the part of the plane that does not contain (0, 0).
Notice also that in this graph we show only the first quadrant of the coordinate plane. That’s because weight values in the real world are always nonnegative, so points outside the first quadrant don’t represent real-world solutions to this problem.
Watch this video for help with the Examples above.
### Vocabulary
• For a strict inequality, we draw a dashed line to show that the points in the line are not part of the solution. For an inequality that includes the equals sign, we draw a solid line to show that the points on the line are part of the solution.
• The solution to a linear inequality includes all the points in one half of the plane. We can tell which half by looking at the inequality sign:
> The solution set is the half plane above the line.
$\ge$ The solution set is the half plane above the line and also all the points on the line.
< The solution set is the half plane below the line.
$\le$ The solution set is the half plane below the line and also all the points on the line.
### Guided Practice
Julius has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and$130 for each refrigerator he sells. How many washing machines and refrigerators must Julius sell in order to make $1000 or more in commissions? Solution Let $x =$ number of washing machines Julius sells. Let $y =$ number of refrigerators Julius sells. The total commission is $60x + 130y$ . We’re looking for a total commission of$1000 or more, so we write the inequality $60x+130y \ge 1000$ .
Once again, we can do this most easily by finding the $x-$ and $y-$ intercepts. When $x=0$ , we have $130y=1000$ , or $y=\frac{1000}{30} \approx 7.69$ . When $y=0$ , we have $60x=1000$ , or $x=\frac{1000}{60} \approx 16.67$ .
We draw a solid line connecting those points, and shade above the line because the inequality is “greater than.” We can check this by plugging in the point (0, 0): selling 0 washing machines and 0 refrigerators would give Julius a commission of $0, which is not greater than or equal to$1000, so the point (0, 0) is not part of the solution; instead, we want to shade the side of the line that does not include it.
Notice also that we show only the first quadrant of the coordinate plane, because Julius’s commission should be non-negative.
### Practice
Graph the following inequalities on the coordinate plane.
1. $y \le 4x+3$
2. $y > -\frac{x}{2}-6$
3. $3x-4y \ge 12$
4. $x+7y < 5$
5. $6x+5y>1$
6. $y+5 \le -4x+10$
7. $x-\frac{1}{2}y \ge 5$
8. $6x+y < 20$
9. $30x+5y < 100$
10. Remember what you learned in the last chapter about families of lines.
1. What do the graphs of $y > x+2$ and $y < x+5$ have in common?
2. What do you think the graph of $x+2 < y < x+5$ would look like?
11. How would the answer to problem 6 change if you subtracted 2 from the right-hand side of the inequality?
12. How would the answer to problem 7 change if you added 12 to the right-hand side?
13. How would the answer to problem 8 change if you flipped the inequality sign?
14. A phone company charges 50 cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and nighttime minutes could you use in one week if you wanted to pay less than $20? 15. Suppose you are graphing the inequality $y > 5x$ . 1. Why can’t you plug in the point (0, 0) to tell you which side of the line to shade? 2. What happens if you do plug it in? 3. Try plugging in the point (0, 1) instead. Now which side of the line should you shade? 16. A theater wants to take in at least$2000 for a certain matinee. Children’s tickets cost $5 each and adult tickets cost$10 each.
1. If $x$ represents the number of adult tickets sold and $y$ represents the number of children’s tickets, write an inequality describing the number of tickets that will allow the theater to meet their minimum take.
2. If 100 children’s tickets and 100 adult tickets have already been sold, what inequality describes how many more tickets of both types the theater needs to sell?
3. If the theater has only 300 seats (so only 100 are still available), what inequality describes the maximum number of additional tickets of both types the theater can sell?
### Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616 .
### Vocabulary Language: English
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
Linear Inequality
Linear Inequality
Linear inequalities are inequalities that can be written in one of the following four forms: $ax + b > c, ax + b < c, ax + b \ge c$, or $ax + b \le c$.
Slope-Intercept Form
Slope-Intercept Form
The slope-intercept form of a line is $y = mx + b,$ where $m$ is the slope and $b$ is the $y-$intercept.
Oct 01, 2012
Feb 23, 2015 |
# JAC Class 9 Maths Solutions Chapter 2 बहुपद Ex 2.5
Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 बहुपद Ex 2.5 Textbook Exercise Questions and Answers.
## JAC Board Class 9 Maths Solutions Chapter 2 बहुपद Exercise 2.5
प्रश्न 1.
उपयुक्त सर्वसमिकाओं का प्रयोग करके निम्नलिखित के गुणनफल ज्ञात कीजिए :
(i) (x + 4 ) (x + 10)
(ii) (x + 8 ) (x – 10)
(iii) (3x + 4 ) (3x – 5)
(iv) $$\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$$
(v) (3 – 2x) (3 + 2x).
हल:
(i) (x + 4 ) (x + 10)
सर्वसमिका
(x + a) (x + b) = x2 + (a + b)x + ab
= x2 + (4 + 10 )x + 10 × 4
= x2 + 14x + 40.
(ii) (x + 8) (x – 10)
सर्वसमिका
[(x + a) (x + b) = x2 + (a + b)x + ab]
= x2 + (8 – 10)x + 8 × (-10)
= x – 2x – 80.
(iii) (3x + 4) (3x – 5) = 3x(3x – 5) + 4(3x – 5 )
= 9x2 – 15x + 12x – 20
= 9x2 – 3x – 20.
(iv) $$\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$$
सर्वसमिका
(a + b) (a – b) = (a)2 – (b)2 से
= (y2)2 – ($$\frac{3}{2}$$)2 = y4 – $$\frac{9}{4}$$
(v) सर्वसमिका (x + y) (x – y) = x2 – y2 से
(3 – 2x) (3 + 2x) = (3)2 – (2x)2
= 9 – 4x2
प्रश्न 2.
सीधे गुणा किए बिना निम्नलिखित गुणनफलों के मान ज्ञात कीजिए :
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96.
हल:
(i) सर्वसमिका
(x + a) (x + b) = x2 + (a + b) + ab
103 × 107 = (100 + 3) (100 + 7)
= (100)2 + (3 + 7) (100) + 3 × 7
= 10000 + 1000 + 21
= 11021.
(ii) सर्वसमिका
(x + a)(x + b) = x2 + (a + b) + ab
95 × 96 = (100 – 5) (100 – 4)
= (100)2 + (- 5 – 4) (100) + (-5) × (-4)
10000 – 900 + 20 = 9120.
(iii) सर्वसमिका (a + b) (a – b) = a2 – b2
104 × 96 = (100 + 4) (100 – 4)
= (100)2 – (4)2
= 10000 – 16 = 9984.
प्रश्न 3.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के गुणनखण्ड कीजिए:
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
(iii) x2 – $$\frac{y^2}{100}$$
हल:
(i) 9x2 + 6xy + y2
सर्वसमिका
a2 + 2ab + b2 = (a + b)2
= (3x)2 + 2 × 3x × y + (y)2
= (3x + y)2.
(ii) 4y2 – 4y + 1
सर्वसमिका a2 – 2ab + b2 = (a – b)2
= (2y)2 – 2 × 2y × 1 + (1)2.
= (2y – 1)2.
(iii) x2 – $$\frac{y^2}{100}$$ = $$(x)^2-\left(\frac{y}{10}\right)^2$$
[सर्वसमिका a2 – b2 = (a + b)(a – b) से]
= $$\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)$$
प्रश्न 4.
उपयुक्त सर्वसमिकाओं का प्रयोग करके निम्नलिखित में से प्रत्येक का प्रसार कीजिए:
(i) (x + 2y + 4z)2
(ii) (2x – y + z)2
(iii) (-2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v) (-2x + 5y – 3z)2
(vi) $$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^2$$
हल:
(i) (x + 2y + 4z)2
सर्वसमिका (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx से
= (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y) (4z) + 2(4z) (x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx.
(ii) (2x – y + z)2
सर्वसमिका (x + y + z)2 = (x2 + y2 + z2) + 2xy + 2yz + 2zx
= (2x)2 + (y)2 + (z)2 + 2(2x) (-y) + 2(-y) (z) + 2(z) (2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
(iii) (-2x + 3y + 2z)2
सर्वसमिका (x + y + z)2 = (x2 + y2 + z2) + 2xy + 2yz + 2zx
= (-2x)2 + (3y)2 + (2z)2 + 2(-2x) (3y) + 2(3y) (2z) + 2(2z) (-2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx.
(iv) (3a – 7b – c)2
सर्वसमिका (x + y + z)2 = (x2 + y2 + z2) + 2xy + 2yz + 2zx
= (3a)2 + (-7b)2 + (-c)2 + 2(3a) (-7b) + 2(-7b) (-c) + 2(-c) (3a)
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca.
(v) (-2x + 5y – 3z)2
सर्वसमिका (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= (-2x)2 + (5y)2 + (-3z)2 + 2(-2x) (5y) + 2(5y) (-3z) + 2(-3z) (-2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx.
(vi) $$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^2$$
सर्वसमिका (x + y + z)2 = (x2 + y2 + z2) + 2xy + 2yz + 2zx
प्रश्न 5.
गुणनखण्ड कीजिए:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2$$\sqrt{2}$$xy + 4$$\sqrt{2}$$yz – 8xz
हल:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2(2x) (3y) + 2(3y)(-4z) + 2(2x)(-4z)
सर्वसमिका [∵ x2 + y2 + z2 + 2xy + 2yz + 2xz = (x + y + z)2] से
= [2x + 3y + (-4z)]2
= (2x + 3y – 4z)2.
(ii) 2x2 + y2 + 8z2 – 2$$\sqrt{2}$$xy + 4$$\sqrt{2}$$yz – 8xz
सर्वसमिका (x + y + z2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= ($$\sqrt{2}$$x)2 + (-y)2 + (-2$$\sqrt{2}$$z)2 + 2($$\sqrt{2}$$x) (-y) + 2(-y)(-2$$\sqrt{2}$$z) + 2($$\sqrt{2}$$x)(-2$$\sqrt{2}$$z)
= [$$\sqrt{2}$$x + (y) + (2$$\sqrt{2}$$z)]2
= ($$\sqrt{2}$$x – y – 2$$\sqrt{2}$$z)2
प्रश्न 6.
निम्नलिखित घनों को प्रसारित रूप में लिखो :
(i) (2x + 1)3
(ii) (2a – 3b)3
(iii) $$\left[\frac{3}{2} x+1\right]^3$$
(iv) $$\left[x-\frac{2}{3} y\right]^3$$
हल:
(i) सर्वसमिका
(a + b)3 = a3 + b3 + 3a2b + 3ab2 से
(2x + 1)3 = (2x)3 + (1)3 + 3(2x)2 (1) + 3(2x) (1)2
= 8x3 + 1 + 12x2 + 6x
= 8x3 + 12x2 + 6x + 1.
(ii) सर्वसमिका
(x – y)3 = x3 – y3 – 3x2y + 3xy2 से
(2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)2 (3b) + 3(2a) (3b)2
= 8a3 – 27b3 – 36a2b + 54ab2
= 8a3 – 36a2b + 54ab2 – 27b3.
(iii) सर्वसमिका
(x + y)3 = x3 + y3 + 3x2y + 3xy2
(iv) सर्वसमिका
(x – y)3 = x3 – y3 – 3x2y + 3xy2
प्रश्न 7.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के मान ज्ञात कीजिए :
(i) (99)3
(ii) (102)3
(iii) (998)3.
हल:
(i) सर्वसमिका
(x – y)3 = x3 – y3 – 3xy(x – y) से
∴ (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 × 100 × 1(100 – 1)
= 1000000 – 1 – 300(99)
= 1000000 – 1 – 29700
= 970299.
(ii) सर्वसमिका
(x + y)3 = x3 + y3 + 3xy(x + y) से
(102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600 × 102
= 1000000 + 8 + 61200
= 1061208.
(iii) सर्वसमिका
(x – y)3 = x3 – y3 – 3xy(x – y) से
(998)3 = (1000 – 2)3
(1000)3 – (2)3 – 3 × 1000 × 2(1000 – 2)
= 1000000000 – 8 – 6000 × 998
= 1000000000 – 8 – 5988000
= 994011992.
प्रश्न 8.
निम्नलिखित में से प्रत्येक का गुणनखण्डन कीजिए :
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
(v) 27p3 – $$\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$$
हल:
(i) 8a3 + b3 + 12a2b + 6ab2
= 8a3 + b3 + 6ab(2a + b)
= (2a)3 + b3 + 3(2a) (b) (2a + b)
= (2a + b)3
= (2a + b) (2a + b) (2a + b).
(ii) 8a3 – b3 – 12a2b + 6ab2
= 8a3 – b3 – 6ab(2a – b)
= (2a)3 – b3 – 3(2a)(b)(2a – b)
= (2a – b)3
= (2a – b)(2a – b)(2a – b).
(iii) 27 – 125a3 – 135a + 225a2
= 27 – 125a3 – 45a(3 – 5a)
= (3)3– (5a)3 – 3 × 3 × 5a(3 – 5a)
= (3 – 5a)3
= (3 – 5a) (3 – 5a) (3 – 5a).
(iv) 64a3 – 27b3 – 144a2b + 108ab2
= 64a3 – 27b3 – 36ab(4a – 3b)
= (4a)3 – (3b)3 – 3 × 4a × 3b(4a – 3b)
= (4a – 3b)3
= (4a – 3b) (4a – 3b) (4a – 3b).
(v) 27p3 – $$\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$$
प्रश्न 9.
सत्यापित कीजिए :
(i) x3 + y3 = (x + y) (x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2).
हल:
(i) x3 + y3 = (x + y) (x2 – xy + y2)
दायाँ पक्ष = (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + (x2 – xy + y2)
= x3 – x2y + y2x + yx2 – xy2+ y3
= x3 + y3 = बायाँ पक्ष। इति सिद्धम्
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
दायाँ पक्ष = (x – y) (x2 + xy + y2)
= x(x2 + xy + y2) – y(x2 + xy + y2)
= x3 + x2y + xy2 – yx2 – xy2 – y3
= x3 – y3 = बायाँ पक्ष। इति सिद्धम्
प्रश्न 10.
निम्नलिखित में से प्रत्येक का गुणनखण्ड कीजिए:
(i) 27y3 + 125z3,
(ii) 64m – 343n3
हल:
(1) 27y3 + 125z3 = (3y)3 + (5z)3
सर्वसमिका
x2 + y3 = (x + y) (x2 – xy + y2) से
= (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2).
(ii) सर्वसमिका
x3 – y33 = (x – y) (x2 + xy + y2)
64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2).
प्रश्न 11.
गुणनखण्ड कीजिए :
27x3 + y3 – 9xyz.
हल:
27x3 + y3 + z3 – 9xyz
सर्वसमिका
x3 + y3 + z3(-3xyz) = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z) [(3x)2 + (y)2 + (z)2 – (3x)y – yz – z(3x)]
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx).
प्रश्न 12.
सत्यापित कीजिए x3 + y3 + z3 – 3xyz = $$\frac{1}{2}$$(x + y + z) [(x – y)2 + ( y – z)2 + (z – x)2]
हल:
दायाँ पक्ष = $$\frac{1}{2}$$(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
= $$\frac{1}{2}$$(x + y + z)(x2 – 2xy + y2 + y2 – 2yz + z2 + z2 – 2zx + x2)
= $$\frac{1}{2}$$(x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
[∵ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)]
सर्वसमिका से
= x3 + y3 + z3 – 3xyz
= बायाँ पक्ष इति सिद्धम् ।
प्रश्न 13.
यदि x + y + z = 0 हो, तो दिखाइए कि x3 + y3 + z3 = 3xyz है।
हल:
दिया है, x + y + z = 0 ⇒ x + y = -z
दोनों पक्षों का घन लेने पर,
(x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3 (∵ x + y = -z)
⇒ x3 + y3 + 3xy(-z) = -z3
⇒ x3 + y3 + z3 = 3xyz, इति सिद्धम्।
प्रश्न 14.
वास्तव में घनों का परिकलन किए बिना निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए:
(i) (-12)3 + (7)3 + (5)3
(ii) (28)3 + (-15)3 + (-13)3.
हल:
(i) (-12)3 + (7)3 + (5)3
माना x = -12, y = 7, z = 5
यदि x + y + z = 0, तो x3+ y3 + z3 = 3xyz
⇒ -12 + 7 + 5 = 0
∵ सर्वसमिका x3 + y3 + z3 = 3xyz से
∴ (-12)3 + (7)3 + (5)3 = 3 × 3(-12) × 7 × 5
= -1260.
(ii) 283 + (-15)3 + (-13)3
28 + (-15) + (-13) = 28 – 28 = 0
अतः 283 + (-15)3 + (-13)3
= 3 × 28 × (-15) × (-13 )
= 16380.
प्रश्न 15.
नीचे दिए गए आयतों, जिनमें उनके क्षेत्रफल दिए गए हैं, में से प्रत्येक की लम्बाई और चौड़ाई के लिए सम्भव व्यंजक दीजिए:
हल:
आयत की लम्बाई और चौड़ाई क्षेत्रफल के गुणनखण्ड होंगे।
(i) आयत का क्षेत्रफल = लम्बाई × चौड़ाई
क्षेत्रफल = 25a2 – 35a + 12
= 25a2 – 15a – 20a + 12
= 5a (5a – 3) – 4(5a – 3)
= (5a – 3) (5a – 4)
∴ यदि लम्बाई (5a – 3) तो चौड़ाई = (5a – 4) तो चौड़ाई
और यदि लम्बाई = (5a – 4) तो चौड़ाई = (5a – 3) इकाई
(ii) क्षेत्रफल 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
= 7y (5y + 4) – 3(5y + 4)
= (5y + 4)(7y – 3)
∴ यदि लम्बाई = (5y + 4) तो चौड़ाई = (7y – 3) तो चौड़ाई
इकाई और यदि लम्बाई = (7y – 3) तो चौड़ाई = (5y + 4) इकाई.
प्रश्न 16.
घनाभ (cuboids), जिनके आयतन नीचे दिए गए हैं की विमाओं के लिए संभव व्यंजक क्या हैं ?
हल:
संभवत: घनाभों की भुजाएँ आयतन के गुणनखण्ड होंगे।
घनाभ का आयतन = लम्बाई × चौड़ाई × ऊँचाई
(i) आयतन = 3x2 – 12x = 3x(x – 4)
= 3 × x × (x – 4)
अतः संभवत: घनाभ की भुजाएँ क्रमश: 3, x और (x – 4) इकाई हैं।
(ii) आयतन = 12ky2 + 8ky – 20k
= 4k(3y2 + 2y – 5)
= 4k[3y2 + 5y – 3y – 5]
= 4k[(3y + 5) – 1(3y + 5)]
= 4k(3y + 5) (y – 1)
= 4k × (3y + 5) × (y – 1)
अतः सम्भवत: घनाभ की भुजाएँ क्रमश: 4k, (3y + 5) और (y – 1) हैं। |
Selina Solutions Class 9 Mathematics Solutions for Exercise 1(D) in Chapter 1 - Chapter 1- Rational and Irrational Numbers
Question 6 Exercise 1(D)
Draw a line segment of length √𝟓 cm.
Answer:
We know that \sqrt{5}=\sqrt{2^{2}+1^{2}}
Which relates to: Hypotenuse= \sqrt{\text { Side } 1^{2}+\text { Side } 2^{2}} \text { [Pythagoras theorem] }
Hence, Considering Side 1=2 and Side 2 =1,
We get a right angled triangle such that:
∠𝐴=90°, AB=2cm and AC=1cm
Video transcript
"hello students welcome to lido learning today we are going to do the question which is on the screen draw a line segment of length under root 5 centimeter now we know that under root 5 is an irrational number so we cannot use a ruler or a scale to draw a line of this length so what are we going to do we are going to use the pythagoras theorem in mathematics to draw this line so let us see how we can use this theorem so we know that under root of 5 whole square is equal to 2 whole square plus 1 whole square now if you remember the right angle triangle this is the right angle triangle and this is what we can use to draw a length now pythagoras theorem is applicable on a right angle triangle now using this equation we can see that the hypotenuse of any right angle triangle the whole square of it is equal to the sum of the squares of the length and the base of the right angle triangle now using this we can say that's if at all we can create a right angle triangle of this nature then the length this length is going to be the size under root of 5 centimeter so this is how we can do this question and draw a line of under root 5 centimeter if you have any concerns please message below this video and subscribe to this channel thank you"
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## Section1.3Curve Sketching
###### Motivating Questions
How do we put together all the thing Calculus tells us about a function to make a sketch of its graph?
###### Preview Activity1.3.1
Let $f(x)=\frac{1}{3}x^3-\frac{1}{2}x^2-2x+9$.
1. Find its $y$-intercept.
2. Find $f'(x)$ and use it to find the intervals where $f$ is increasing and the intervals where $f$ is decreasing.
3. Find the local maximums and local minimums of $f$, and their locations.
4. Find $f''(x)$ and use it to find the intervals where $f$ is concave up and the intervals where $f$ is concave down.
5. Find the inflection points of $f$.
6. Sketch a graph of $f$ consistent with the information you gathered above.
The three main steps in curve sketching are the following:
1. Gather information about the function. We will always want to gather information using the first and second derivatives. Sometimes we will also want other information, such as asymptotes.
2. Organize the information you gathered. Make a single number line and mark on it the sign chart for $f'$, the sign chart for $f''$, the shape of $f$ on each interval, maximums and minimums, asymptotes, and whatever else you can.
3. Make a graph consistent with the information you gathered and organized.
###### Activity1.3.2
Let $\displaystyle f(x)=\frac{1}{x^2-9}$.
1. Find the domain of $f$.
2. Find the $y$-intercept and $x$-intercepts.
3. Determine if $f$ is even, odd, periodic, or none of those.
4. Find all vertical and horizontal asymptotes.
5. Find intervals when it is increasing or decreasing and its maximums and minimums.
6. Find the intervals when it is concave up or concave down and its inflection points.
7. Organize the information you gathered.
8. Sketch the graph.
After doing all this work, you may be wonderiing why we did not just plug the function into a calculator or computer to graph. While technology can be very useful, there are a few reasons to do these by hand.
• Technology can miss things. Important features might be too small to see, or not in the plotting window.
• Technology has trouble with certain features, such as vertical asymptotes.
• Analysis by hand allows us to understand how the graph depends on parameters.
• Analysis by hand develops understanding and intuition about functions. Even if you then use technology to make a plot, you can use what you learned by doing analysis by hand to understand what you are seeing.
The computational cell below gives an in-between option. It computes some things that you would need for analysis by hand.
The order in which you gather information about the function is not so important, but it is useful to have a list so you do not forget anything.
Domain
Determine the domain of $f$. Look for places to exclude, such as division by 0, even roots of negative numbers, logarithms of 0 or negative numbers, etc.
$y$-intercept
Compute $f(0)$ to get the point $(0,f(0))$ on the graph.
$x$-intercepts
Set $f(x)=0$ and see if you can solve it. Sometime this is too hard (or impossible) and can be skipped.
Is it even?
If $f(x)=f(-x)$ (such as with $f(x)=x^2$) then the function is even. You can save half the work by sketching $f$ for $x \ge 0$ and then reflecting it over the $y$-axis to get the sketch of $f$ for $x \le 0$. It is okay to skip this step, but it can save you work.
Is it odd?
If $f(x)=-f(-x)$ (such as with $f(x)=x^3$) then the function is odd. You can save half the work by sketching $f$ for $x \ge 0$ and then reflecting it over the $y$-axis and then over the $x$-axis to get the sketch of $f$ for $x \le 0$. It is okay to skip this step, but it can save you work.
Is it periodic?
If $f(x+p)=f(x)$ for some $p\not= 0$ (such as with $f(x)=\sin(x)$ for $p=2\pi$) then the function is periodic. You can then do the work and sketch on the interval $[0,p]$ and then say it continues periodically.
Vertical asymptotes
If $\displaystyle \lim_{x\rightarrow a^+} f(x)= \infty$, $\displaystyle \lim_{x\rightarrow a^-} f(x)=\infty$, $\displaystyle \lim_{x\rightarrow a^+} f(x)=- \infty$, or $\displaystyle \lim_{x\rightarrow a^-} f(x)=-\infty$, then $f$ has a vertical asymptote at $x=a$. This normally happens when you would divide by 0 at $a$ or take the logarithm of 0 at $a$.
Horizontal asymptotes
If $\displaystyle \lim_{x\rightarrow \infty} f(x)=L$ or $\displaystyle \lim_{x\rightarrow -\infty} f(x)=L$ (with $L\not=\pm\infty$), then $f$ has a horizontal asymptote at $y=L$.
Increasing, decreasing
Compute $f'(x)$. Find when $f'(x)$ equals zero or does not exist to locate the critical numbers. Make a sign chart to determine the intervals where $f$ is increasing and the intervals where $f$ is decreasing.
Maximums, minimums
Use your sign chart to determine which critical numbers give maximums, which give minimums, and which give neither. Compute the values of the maximums and minimums by plugging the critical number $c$ into $f$. This gives a point $(c,f(c))$.
Concavity
Compute $f''(x)$. (You may want to simplify $f'(x)$ first.) Find when $f''(x)$ equals zero or does not exist to locate potential changes in concavity. Make a sign chart to determine the intervals where $f$ is concave up and the intervals where $f$ is concave down.
Inflection points
Use your sign chart for $f''$ to determine when the concavity changes. Compute the values of $f$ at these locations to get inflection points $(c,f(c))$.
###### Activity1.3.3
Analyze and graph the function $f(x)=x\sqrt{5-x}$.
### SubsectionExercises
###### 1
Analyze and graph the function $f(x)=e^{-x^2}$.
###### 2
Analyze and graph the function $f(\theta)=2\cos(\theta)+\cos^2(\theta)$.
###### 3
Analyze and graph the function $f(x)=x\ln(x)$. |
Calendar Math For third grade - ProTeacher Community
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allithird Joined: Dec 2006 Posts: 196 Full Member
allithird
Joined: Dec 2006
Posts: 196
Full Member
07-29-2007, 12:26 PM
#1
Hi,
Could anyone please give me some specific instructions for using Calendar Math in third grade?
Thank you
rth Guest
rth
Guest
Calendar Math
07-29-2007, 11:56 PM
#2
I will be updating my second grade calendar math procedure to use when I move to third grade this year.
We have a series of charts on the math board that we complete each day to review math skills. Often, but not always they relate to the calendar date.
The student of the day completes the charts as we discuss them together.
DATE: What is the date? What will the date be in a week? What day of the week will it be on the 20th?
DAYS IN SCHOOL: How many days have we been in school? How many days are left in the schoolyear. We have two counting charts - one from 1 - 100 and one from 101 - 200. We use 2 different special marker to count down the days left in school and count up the days that have passed in the schoolyear.
LESS THAN/ GREATER THAN - One number is provided in each inequality. The student provides the other number.
291> ____ 642< _____
ODD or EVEN? 432 is _________
MONEY Count the money. (Magnetic coins& bills are on the board.) Write the amount.
MEASURING. A line is drawn on the chart. Students measure & record the length in inches or centimeters.
PATTERN: Continue the pattern. Geometric or numerical patterns are written on a sentence strip. The student extends the pattern. If it is a numeric pattern, they also state the rule.
FRACTION: Name the fraction. Pattern block stickers are used to make many whole shaped divided into fractional parts. Student write the fractions that correspond to the parts.
TIME. A clock is set to a particular time. Student tells the time in at least two ways. Also used for elasped time. Student tells what time it will be in 20 minutes, for example.
WAYS TO MAKE. A number of the day is written on the chart. Students use any operation or combination of operations to write as many number sentences as possible that equal the number of the day.
I'd like to add a chart for DATA/GRAPHING and one for PLACE VALUE, too.
allithird Joined: Dec 2006 Posts: 196 Full Member
allithird
Joined: Dec 2006
Posts: 196
Full Member
Great ideas rth
07-30-2007, 08:39 AM
#3
Thank you rth for the great ideas for calendar math. I would love to see any of the other ideas you have for third grade too.
snowpoppy2 Joined: Jan 2007 Posts: 213 Full Member
snowpoppy2
Joined: Jan 2007
Posts: 213
Full Member
calendar math idea
07-30-2007, 11:30 AM
#4
This may be the same thing as the previous post, however when I do calendar math later on in the year, I make the kids tell me the multiples that go into that day of school. As we count the days of school we use all the things(number sentence, odd, even, greater than..etc) like the previous post, however I like to have them practice their multiples as soon as they learn multiplication and then even into division!
StephR Joined: Aug 2005 Posts: 1,893 Senior Member
StephR
Joined: Aug 2005
Posts: 1,893
Senior Member
Calendar math
07-30-2007, 01:40 PM
#5
Here is what I do. We discussed this on another thread...I will look for it later and repost.
In the explanation I posted, I have already updated for 4th grade (I am switching this year), but I basically did the same thing (minus the number lines and fractions).
Attached Files
Calendar Math Explanation.doc (25.5 KB, 2778 views)
missa_g Joined: Jul 2007 Posts: 242 Full Member
missa_g
Joined: Jul 2007
Posts: 242
Full Member
StephR
07-30-2007, 04:55 PM
#6
You really explain your procedures very well!! Obviously, a lot of these concepts will be added to the calendar as the year progresses, so my question is: How do you start off your calendar before a lot of these concepts are mastered? Also, you have students work on everything before getting together to discuss...do you have them write everything down in a notebook? How long does it take you to get into a good routine with your calendar program?
I am teaching 3rd for the first time this year and have zero experience with calendar math but I can see how beneficial it will be and would love to implement it!
One more question: Do you happen to have a picture of your whole set up? What kinds of supplies are must haves right off the bat? I am trying to envision this!
Thanks!
allithird Joined: Dec 2006 Posts: 196 Full Member
allithird
Joined: Dec 2006
Posts: 196
Full Member
Fantastic Steph R
07-30-2007, 08:35 PM
#7
There is so much great information. Missa_g brings up some great questions that I would like to hear the answers to as well. Thank you for your help.
StephR Joined: Aug 2005 Posts: 1,893 Senior Member
StephR
Joined: Aug 2005
Posts: 1,893
Senior Member
07-31-2007, 10:19 AM
#8
Thanks! I will try to answer the questions the best I can
Obviously, a lot of these concepts will be added to the calendar as the year progresses, so my question is: How do you start off your calendar before a lot of these concepts are mastered?
The first week of school is spent teaching the calendar procedures and showing them exactly what I want them to do. I begin the year with odd/even and how many ways? Those two can be done by the kids on day one. I model the expectations and the "how to" using those two parts. The first week I also introduce Today's Arrays and Patterns. The second week, I introduce the big number, expanded/word/standard form, prime/composite, place value, etc. By the end of the second week, all of the components of the calendar have been introduced and are being done on a daily basis. I do not wait until I have done a formal lesson on them. I use the calendar AS MY LESSON! They WILL NOT understand how to do it at first. You will be basically feeding them the answers the first few weeks. But they WILL catch on. You just have to believe in it! They will be doing it every single day. Not everyone will get it at the same time, but by the end of the year, they ALL will have mastered ALL of the standards and concepts on the calendar.
Also, you have students work on everything before getting together to discuss...do you have them write everything down in a notebook?
I have a form they have to fill out (I will attach) I used to have them do it in a journal, to avoid copies, but I found that it wasn't as affective as me collecting the forms and holding them accountable. I give them 5 forms on Monday (stapled together...I usually do two of them back to back to save paper) I then collect the entire packet on Friday. I use that as a work habits grade. (as they are not allowed to be filling it out while we are going over the calendar together)
How long does it take you to get into a good routine with your calendar program?
I would say that it is a good month before I am super comfortable that ALL the students know EXACTLY what to do. But from week one, most of the kids kinda sorta know what to do, KWIM? By mid year, I have my students (yes 3rd graders!!) leading the calendar themselves. I only hop in when I feel that they aren't quite explaining it correctly or illiciting deep answers (ie: on odd/even, they are just accepting "5 is odd because it is after an even number" instead of "5 is odd because you can not divide it by 2 without getting a remainder")
I am teaching 3rd for the first time this year and have zero experience with calendar math but I can see how beneficial it will be and would love to implement it!
I can honestly say that this is the most rewarding and fulfilling part of my math instruction. If I could just do one thing in math all year, I would teach calendar. It hit SO many vital standards in such a short time period. You can also tailor it to your state standards. (ie: if you teach money in 3rd, you can have them tell you which combination of coins make up the day...or if you change grades, up or down, or have a gifted or remedial class, you can change it as well to accomodate those new expecations, making it as challenging or easy as needed)
One more question: Do you happen to have a picture of your whole set up?
I do have a pic somewhere...I will have to dig through the garage to find it. When I do, I will post!!
Basically, I have a calendar. Surrounding it, I have laminated pieces of paper for each section (we use thinking maps...so I have those created and blank...if you use TMs I can give you a better explanation...just let me know)
What kinds of supplies are must haves right off the bat?
I have vis-a-vie markers (to write on the laminated papers), calendar pieces, flats/rods/units (for the days of the school year), and calendar packets for each kid. That is it (at least that is all I can think of at the moment!)
Again, in the beginning of the year, you will feel as if you are spoon feeding them the answers, like you are doing all the work. But as the year progresses, the kids will become more comfortable with it and will eventually begin giving you the answers. Just believe in it!!
Attached Files
calendar.doc (23.0 KB, 2410 views)
missa_g Joined: Jul 2007 Posts: 242 Full Member
missa_g
Joined: Jul 2007
Posts: 242
Full Member
Thank You!!
07-31-2007, 11:26 AM
#9
Thank you so much for your GREAT explanations. I feel much more confident now with it all explained so well. We do use thinking maps and my administration would probably LOVE to see thinking maps surrounding the calendar...so any explanation with those would be great! I am sorry for all of the questions, but this has been a great start for me!!!!
tchrrx Joined: Jul 2007 Posts: 5,775 Senior Member
tchrrx
Joined: Jul 2007
Posts: 5,775
Senior Member
07-31-2007, 11:32 AM
#10
Thank you for the ideas! I really think this is going to help me a lot!
Do you have a 100 chart on the calendar wall? I have one hanging, but I would like to replace it with some of your other ideas. Is there a reason I should keep it up though? (It just seems like one of those things that are needed...but I'm not sure why!)
StephR Joined: Aug 2005 Posts: 1,893 Senior Member
StephR
Joined: Aug 2005
Posts: 1,893
Senior Member
07-31-2007, 12:42 PM
#11
Quote:
Do you have a 100 chart on the calendar wall? I have one hanging, but I would like to replace it with some of your other ideas. Is there a reason I should keep it up though? (It just seems like one of those things that are needed...but I'm not sure why!)
Personally, I do not have a hundreds chart. I don't use it or refernce it, so it just becomes decoration for me. Some people do use them, I am just not one of those people. I do use it for cafeteria (behavior) bingo though.
StephR Joined: Aug 2005 Posts: 1,893 Senior Member
StephR
Joined: Aug 2005
Posts: 1,893
Senior Member
Thinking Maps
07-31-2007, 12:51 PM
#12
Here is the list of Thinking Maps I use Missa_g:
Odd/Even Multi-Flow -- Event = __ is ___ ie: 5 is odd
Reasons = what makes it odd/even
Effects = the division problem tha results
Array Circle Map -- # in middle, all arrays surrounding
Factor Tree Map -- all factors, resulting from arrays, listed on tree
Prime/Composite Bridge Map -- Relating factor = is On top 5/bottom prime. I add another bridge each day, starting over at the beginning of the month.
How Many Ways Divided Circle -- divided by operation, which is written in the frame of refernce
Patterns Circle Map
Expanded/word/standard form Tree Map -- title is # forms, branches are the3 types
I think that is it. If I think of more, I will add them!
HARteach Joined: Nov 2006 Posts: 575 Senior Member
HARteach
Joined: Nov 2006
Posts: 575
Senior Member
07-31-2007, 01:20 PM
#13
StephR...thank you for all of the ideas for calendar math! I have been inspired to implement this in my classroom this year. I am sorry though that I know have a question ...do you think it would be a good idea for me to invest in a place value pocket chart to place the units, rods, hundreds, and place it by the calendar?...or do you use something different? Thanks!
TeachNTn Joined: Jul 2007 Posts: 71 Junior Member
TeachNTn
Joined: Jul 2007
Posts: 71
Junior Member
Great Post!!
07-31-2007, 02:34 PM
#14
Wow!!! Thank you for not only sharing, but for explaining in such a way that motivates people to want to do this in their rooms. I always did calendar math in first grade, but never thought to continue it in third....NOW I will!!! Thanks a bunch.
allithird Joined: Dec 2006 Posts: 196 Full Member
allithird
Joined: Dec 2006
Posts: 196
Full Member
You are amazing
07-31-2007, 03:37 PM
#15
I have to confess though. I had to do a Google search as to what a Thinking Map was. Now that I have examples, do you think that you could show an old teacher a new trick by giving exact examples of your thinking maps? I am really a visual learner. Sorry :-(
missa_g Joined: Jul 2007 Posts: 242 Full Member
missa_g
Joined: Jul 2007
Posts: 242
Full Member
Thinking Maps
07-31-2007, 04:00 PM
#16
Here are the exact maps she is referencing. HTH!
http://www.thinkingmaps.com/htthinkmapx.php3
StephR Joined: Aug 2005 Posts: 1,893 Senior Member
StephR
Joined: Aug 2005
Posts: 1,893
Senior Member
08-01-2007, 07:31 AM
#17
Quote:
do you think it would be a good idea for me to invest in a place value pocket chart to place the units, rods, hundreds, and place it by the calendar?...or do you use something different?
HARteach -- I use something like that, except I just made my own (didn't want to spend the money!) but you can definitely get one of those pocket charts. It is easier than making it (trust me on this one!)
changingrades Joined: Aug 2008 Posts: 1 New Member
Joined: Aug 2008
Posts: 1
New Member
calendar math
08-04-2008, 05:23 PM
#18
Hi!
I use real money for the calendar, but I also have paper money and stamps available for them to use.
sdgillyard Joined: Jun 2006 Posts: 27 New Member
sdgillyard
Joined: Jun 2006
Posts: 27
New Member
Thanks!
08-05-2008, 02:58 PM
#19
I have been teaching for almost 8 years. I used a calendar time in first grade room and I have not been able to work it into my third grade room yet. This is my third year in third grade. Now after reading all of your posts and AWESOME ideas I can finally pull out all of that stuff that has been in storage. Thank you so much for sharing your wonderful ideas!!!
Denise
hcking Joined: Aug 2006 Posts: 1 New Member
hcking
Joined: Aug 2006
Posts: 1
New Member
calendar math question
08-09-2008, 08:20 PM
#20
Do you have a copy of your packet that I could see please? This sounds exactly like the type of thing I would like my kids to as they come in in the morning so we can update the calendar math board together.
Thanks!
Priscilla C Guest
Priscilla C
Guest
Math Calendar Time
07-05-2016, 06:58 PM
#21
Hi StephR I read your calendar routine and loved it! I am going to implement most of your ideas. But would also love to see pictures. Do you happen to have any? Do you also have a copy of the accountability page you use? Thank you in advance for your help.
Kind regards,
Priscilla
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## Note on Reflection
• Note
• Things to remember
• Videos
• Exercise
• Quiz
A reflection is a transformation that flips a figure across a line. The line work as a plane mirror. In reflection, the line joining the object and the image is perpendicular to the mirror line. It means the mirror line is perpendicular bisector of the line segment joining object and image. The mirror line is also called the axis of reflection.
#### Characteristics of reflection of geometrical figures in the axis.
When geometrical figures are reflected in the axis of reflection, the following properties are found.
a. Coordinates can be used for finding images of geometrical figures after the reflection in the lines like X- axis, Y- axis, a line parallel to X- axis, a line parallel to Y- axis, the line y = x, the line y = -x, etc.The distance of the object from the axis of reflection is equal to the distance of reflection is equal to the distance of the image from the axis is a reflection.
OP = OP' as shown in fig 1.
b. The shape of objects and images are laterally inverted. It means top remains at the top, bottom remains at the bottom but left side goes to the right side and right side goes to the left side as shown in fig 2.
c. The lines joining the same ends of the object and image are perpendicular to reflecting axis.
Axis of reflection is the perpendicular bisector of the line segment joining same ends of object and image.
XX'is perpendicular bisector of AA', BB' and CC' as in fig 3.
d. The points on the axis of reflection are invariant points.
e. Use of co-ordinates in Reflection.
f. In reflection, the object figure and its image figure are congruent to each other.
1. Reflection in the X- axis
Equation of X- axis is y = 0. So, reflection in X- axis means reflection in the line y = 0. Let P(x, y) be any point in the plane. Draw a perpendicular PL from the point P to the X- axis and produce it to the point P' such that PL = LP'. Then P' is the image of P after reflection in X- axis.
Here, co-ordinates of L are (x, 0). Let the co- ordinates of P' be (x', y'). Since L is the mid- point of line segment PP', then by mid- point formula,
x = $$\frac {x+x'}2$$ and 0 = $$\frac {y+y'}2$$ or, x + x' = 2x and y + y' = 0 or, x' = x and y' = -y
∴ Co-ordinates of the P' are (x. -y)
∴ Image of point P(x, y) after reflection in X- axis is P'(x, -y).
Hence, if Rx denotes the reflection in X- axis, then:
Rx: P (x, y)→ P' (x, -y)
2. Reflection in the Y- axis
Equation of Y- axis is x = 0. So, reflection in Y- axis means reflection in the line x = 0.
Let P (x, y) be any point in the plane. Draw a perpendicular PM from the point P to the Y- axis and produce it to the point P' such that PM = MP'. Then P' is the image of P after reflection in Y- axis.
Here, co-ordinate of M is (0, y). Let the co- ordinates of P' be (x', y'). Since M is the mid-point of line segment PP', then by mid- point formula,
0 = $$\frac{x + x'}2$$ and y = $$\frac {y + y'}2$$ or, x + x' = 0 and y + y' = 2y or, x' = -x and y' = y
∴ Co- ordinates of the point P' are (-x, y).
∴ Image of point P(x, y) after reflection in Y- axis P' (-x, y).
Hence, if Ry denotes the reflection in Y- axis, then:
Ry: P (x, y)→ P' (-x, y)
3. Reflection in the line parallel to X- axis
The equation of a line parallel to X- axis is given by y = k where k is Y- intercept of the line. So, reflection in the line parallel to X- axis means reflection in the line y = k.
Let P (x, y) be any point in the plane. Draw a perpendicular PM from P to the line y = k and produce it to the point P' such that PM = PM'. Then P' is the image of P after reflection in the line y = k.
Here, co- ordinates of M are (x, k). Let thye co- ordinates of P' be (x', y'). Since, M is the mid-point of the line segment PP', then by mid- point formula,
x = $$\frac{x + x'}2$$ and k = $$\frac{y + y'}2$$ or, x + x' = 2x and y + y' = 2k or, x' = x and y' = 2k - y
∴ Co- ordinates of the point P' are (x, 2k - y).
∴ Image of point P (x, y) after reflection in the line y = k is P' (x, 2k - y).
Hence, if R denotes the reflection in the line y = k, then:
R: P (x, y)→ P' (x, 2k - y)
4. Reflection in the line parallel to Y- axis
The equation of a line parallel to Y- axis is given by x = k where k is X - intercept of the line. So, the reflection in a line parallel to Y- axis means reflection in the line x = k.
Let P (x, y) be any point in the plane. Draw a perpendicular PM from P to the line x = k and produce it to the point P' such that PM = MP'. Then P' is the image of P after reflection in the line x = k.
Here, co- ordinates of M are (k, y).
Let the co- ordinates of P' be (x', y').
Since, M is the mid- point of the line segment PP', then by mid- point formula,
k = $$\frac {x + x'}2$$ and y = $$\frac {y + y'}2$$ or, x + x' = 2k and y + y' = 2y or, x' = 2k - x and y' = y
∴ Co- ordinates of the point P' are (2k - x, y).
∴ Image of point P (x, y) after reflection in the line x = k is P' (2k - x, y).
Hence, if R denotes the reflection in the line x = k, then:
R: P (x, y)→ P' (2k - x, y)
5. Reflection in the line y = x
y = x is the equation of the line which makes an angle of 45° with the positive direction of X- axis. Then, slope of the line, m1 = tan 45° = 1.
Let P (x, y) be any point in the plane. Draw perpendicular PM from the point P to the line y = x and produce it to the point P' such that PM = MP'. Then P' is the image of the point P under reflection about the line y = x.
Let (x', y') be the co- ordinates of the point P'.
Then, slope of the line segment PP', m2 = $$\frac {y - y'}{x - x'}$$.
Since, the line segment PP' and the line y = x are perpendicular to each other. Then:
m1× m2 = -1
or, 1× ($$\frac {y-y'}{x-x'}$$) = -1
or, y - y' = -x + x'
or, y + x = y' + x' .......................(i)
Again,
Co- ordinates of mid- point of the line segment PP' are: ($$\frac {x + x'}2$$, $$\frac {y + y'}2$$).
This point lies in the line y = x.
So,
$$\frac {y + y'}2$$ = $$\frac {x + x'}2$$
or, y + y' = x + x'
or, y - x = x' - y' .........................(ii)
Adding (i) and (ii) we get, 2y = 2x' or x' = y
Subtracting (ii) from (i) we get, 2x = 2y' or y' = x.
∴ Co- ordinates of the point P' are (y, x).
∴ Image of the point P (x, y) under the reflection about the line y = x is the point P (y, x).
Hence, if R denotes the reflection in the line y = x, then:
R: P (x, y)→ P' (y, x)
6. Reflection in the line y = -x
y = -x is an equation of the line which makes an angle of 135° with the positive direction of X- axis.
∴ Slope of the line y = -x is m1 = tan 135° = -1
Let P (x, y) be any point in the plane. Draw perpendicular PM from the point P to the line y = -x and produce it to the point P' such that PM = MP'.
Let the co- ordinates of the point P' be (x', y').
Slope of the line segment PP' is m2 = $$\frac {y - y'}{x - x'}$$.
Since, the line segment PP' and the line y = -x are perpendicular to each other, then:
m1× m2 = -1
or, (-1)× $$\frac {y - y'}{x - x'}$$ = -1
or, y - y' = x - x' .........................(i)
Again,
Co- ordinates of the mid- point of line segment PP' are: ($$\frac {x + x'}2$$, $$\frac {y + y'}2$$).
This point lies in the line y = -x.
So,
($$\frac {y + y'}{2}$$) = - ($$\frac {x + x'}2$$)
or, y + y' = -x' - x .......................(ii)
Adding (i) and (ii) we get, 2y = - 2x' or x' = -y
Subtracting (ii) from (i) we get, -2y' = 2x or y' = -x
∴ Co- ordinates of P' are (-y, -x)
∴ Image of the point P (x, y) after reflection in the line y = -x is P' (-y, -x).
Hence, if R denotes in the line y = -x, then:
R : P (x, y)→ P' (-y, -x)
• In reflection, the line joining the object and the image is perpendicular to the mirror line.
• The Axis of reflection is the perpendicular bisector of the line segment joining same ends of object and image.
• In reflection, the object figure and its image figure are congruent to each other.
• Reflection in X- axis means reflection in the line y = 0.
• The distance of the object from the axis of reflection is equal to the distance of reflection is equal to the distance of the image from the axis is a reflection.
.
### Very Short Questions
Soln:
Here, the perpendiculars AP, BQ and CR are drawn from the points A, B and C on the line m respectively. Again, AP = PA1, BQ = QB1and CR = CR1 are drawn by producing Ab, BQ and CR respectively. Then A', B' and C' are joined . So, that the imageΔA'B'C' of triangle ABC is formed.
Soln:
As in the above figure, the perpendiculars AP, BQ, DR and CS are drawn in the mirror line 'm' from the points A, B, C and D respectively. Again AP = PA', BQ = QB', DR = RD' and CS = CS' are drawn by producing AP, BQ, QR and CS respectively. THen A',B',C' and D' are joined so that the image quadrilateral A'B'C'D' of the quadrilateral ABCD is formed.
Soln:
From the given points P and Q, perpendiculars PA and QB are drawn on the mirror line 'm'. Also, PA and QB are produced into P' and Q' so that PA = AP' and QB = BQ'. So the line P'Q' joining P' and Q' is the image of the line PQ.
Soln:
From the given figure, the mirror line 'm' is the perpendicular bisector of the line PQ. So the image P' and Q and image Q' of Q is at P. So, the image of PQ is Q'P' which is the same line PQ.
Soln:
Here, the mirror line 'm' is perpendicular to both the lines AD and BC. Let 'm' cuts AD at P and BC at Q. Now DP is produced up to D' so that DP = PD'. Similarly, APis produced up to A' so that AP = PA' BQ up to B' so that BQ = QB' and CQ up to C' so that CQ = QC'. Now the image A'B'C'D' of ABCD is formed.
Soln:
We have, reflection of a point on x-axis remains the x-coordinate same but the sign of y-coordinate changes. i.e. P(a, b)→ P'(a, -b).
So the images of the above points under the reflection on x - axis are:
A(3, 0)→A'(3, 0), B(4, -3)→ B'(4, 3), C(0, -7)→ C'(0, 7), D(-2, 6)→ D'(-2, -6), E(-3, -3)→ E'(-3, 3), F(-3, -9)→ F'(-3, 9), G(6, 6)→ G'(6, -6), H(7, -7)→ H'(7, 7). Ans
Soln:
We have the reflection of the point P(a, b) on the y-axis is P'(-a, b). i.e. P(a, b)→ P'(-a, b). So the images of the given points are:
A(3, 0)→ A'(-3, 0), B(4, -3)→ B'(-4, -3), C(0, -7)→ C'(0, -7), D(-2, 6)→ D'(2, 6), E(-3, -3)→ E'(3, -3), F(-3, -9)→ F'(3, -9), G(6, 6)→ G'(-6, 6) and H(7, -7)→ H'(-7, -7). Ans.
Soln:
We have, a reflection of a point P(a, b) on the line x = y is P'(b, a). i. e. P(a, b) → P'(b, a). So the images of the given points are:
A(3, 0)→ A'(0, 3), B(4, -3)→ B'(-3, 4), C(0, -7)→ C'(-7, 0), D(-2, 6)→ D'(6, -2), E(-3, -3)→ E'(-3, -3), F(-3, -9)→ F'(-9, -3), G(6, 6)→ G'(6, 6), H(7, -7)→ H'(-7, 7). Ans.
Soln:
We have, a reflection on x-axis is P(a, b)→ P'(a, -b). So reflection of triangle PQR on x-axis given the images coordinates as:
P(1, 1) →P'(1, -1)
Q(3, 1)→ Q'(3, -1)
R(3, -1)→ R'(3, 1)
Here, the shaded portion triangle P'Q'R' is the image of triangle PQR.
Soln:
We have, reflection on the line y = -x is P(a, b)→ P'(-b, -a). So, we have,
P(1, 1) →P(-1, -1)
Q(3, 1)→ Q'(-1, -3)
R(3, -1)→ R'(1, -3)
The graph is given below:
The shaded portion triangle P'Q'R' is the image of triangle PQR.
Soln:
We have, a reflection on the line y = k is P(a, b)→ P'(a, 2k - b) where k = 3. So,
P(1, 1)→ P'(1, 2×3 - 1) = P'(1, 5)
Q(3, 1)→ Q'(3, 2× 3 -1) = Q'(3, 5)
R(3, -1)→ R[3, 2× 3 - (-1)] = R'(3, 7)
The graph is as shown below:
The shaded portion triangle P'Q'R' is teh image of triangle PQR.
Soln:
Here, the sign of y-coordinate of the point A(3, 4) is changed in the image A'(3, -4) under the reflection R. So the axis of reflection is x-axis.
Soln:
Here, Y reflects the point of A(3, 4) to A'(-1, 4), is which y-coordinate is same but x-coordinate changes from 3 to -1. So the line AA' is parallel to x-axis. Here, the mirror line is the perpendicular bisector of AA' or the mirror line is parallel to y-axis which passes through the mid-point of AA' i. e. $$\frac{3-1}{2}$$, $$\frac{4+4}{2}$$ = (1, 4). So, the line parallel to y-axis (1, 4) is x = 1.
∴ Axis of reflection is x = 1. Ans
Soln:
Here, reflection R reflects the point A(4, 5) to A'(-5, -4). So, x and y coordinates are interchanged and the signs are also changed. i. e. P(a, b)→ P'(-b, -a). So the axis of reflection is the line y = -x. Ans.
Soln:
Here, reflection P reflects the point A (-2, 3) to A'(6, 3) is which y-coordinate is same but x-coordinate changes from -2 to 6. In this case as in the reflection of the line x = h, P(a, b)→ P'(2h-a, b)
A(-2, 3)→ A'(6, 3). So,
2h - a = 6
or, 2h -(-2) = 6
or, 2h + y = 6
or, 2h + y = 6
or, 2h = 6 - 2 = 4
∴ h = 2
∴ The axis of reflection is h = 2 which is parallel to y-axis.
Soln:
Here, reflection Q reflects the point A(3, 4) to A'(3, 2) is which x-coordinate is same but y-coordinate changes from 4 to 2. In this axis the reflection of the line y = k, P(a, b)→ P'(a, 2k - b),
A(3, 4)→ A'(3, 2).
so, 2k - b = 2
or, 2k - 4 = 2
or, 2k = 2 + 4 = 6
∴ k = 3
∴ The axis of reflection is the line k = 3 which is parallel to x-axis. Ans.
Soln:
i)Reflection of the given trapezium PQRS under, x-axis is as
P(a, b) → P'(a, -b) we get,
P(2, 1)→ P'(2, -1)
Q(1, -2)→ Q'(1, 2)
R(-3, -2)→ R'(-3, 2) and
S(-5, 1)→ S'(-5, -1)
The graph of trapezium PQRS and itsn image P'Q'R'S' is as shown below:
ii)Reflection of trapezium on y-axis as P(a, b)→ P'(-a, b) we get,
P(2, 1)→ P'(-2, 1)
Q(1, -2)→ Q'(-1, -2)
R(-3, -2)→ R'(3, -2) and S(-5, 1)→ S'(5, 1)
The graph of trapezium PQRS and its image P'Q'R'S' is shown below:
0%
(1,4)
(7,-1)
(11,-4)
(8,-3)
(-9,2)
(-2,4)
(-3,-4)
(3,4)
(-3,5)
(1,2)
(-3,10)
(-1,1)
(-9,8)
(-2,1)
(-6,8)
(-3,0)
4,4
3,3
2,2
1,1
3,3
2,2
1,1
4,4
2,3
4,5
3,4
1,2
2,2
3,3
1,1
2,1
Q(8,7)
Q(1,2)
Q(9,8)
Q(4,5)
P(4,3)
P(6,1)
P(1,2)
P(7,8)
R(5,5)
R(8,3)
R(7,7)
R(1,2) |
In the previous two sections, we’ve solved sets of two or even three linear equations. Here, we will do some examples of systems with four linear equations. The general strategy will be the same as before: we apply our method (either elimination or substitution) to our sets of equations in order to get rid of one equation. We repeat this until we’re left with just two or three equations, whereupon we know how to solve the problem. Also note that it takes distinct equations to solve a system of linear equations. (Two equations are distinct if you cannot obtain either one from the other).
Example (Elimination Method for 4 Equations)
1. Pick a variable we want to eliminate.
Let’s pick .
2. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).
Now, if we look at our equations, we see that equation 4 does not have any ’s in it. So no matter what constant we multiply the equation by, we will not get a term with multiplied by some given coefficient. If we have an equation like this, we simply set it to the side and use it again later. This is because, recall, the point of this method is to eliminate all occurrences of a given variable, thereby simplifying our problem. Equation 4 has already eliminated all occurrences of , so no further manipulation of the equation is needed.Now, we get:
And we still have our equation 4:
3. Subtract the equations from one another, thereby eliminating our variable.
Note that this only applies to the equations that have in them. And remember to use all of the equations that include (but we only need to use each equation once).Then, we get:
4. Now, we should have four equations with at most four different variables. Pick another variable to eliminate.
Let’s eliminate .
5. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).
6. Subtract the equations from one another, thereby eliminating our variable.
7. Now, we should have three equations with at most three different variables.
And we do.
8. We can now solve this problem with the steps detailed in our previous post on the elimination method.
Example (Substitution Method with 4 Equations)
1. Pick a variable we want to substitute for.
Let’s choose .
2. Use one of the lines of the equation to express that variable in terms of the other variables.
Let’s use the second equation. Then, we get:
3. Substitute into the other equations.
Since we used the second equation to get our equation for , we now substitute for in the first, third, fourth, and fifth equations. We then get (after simplifying):
4. Now, we should have four equations with at most four different variables.
And this is true.
5. Pick another variable to substitute for.
Let’s pick .
6. Use one of the remaining lines of the equation to express that variable in terms of the other variables.
Let’s use the first line. Then, we get
7. Substitute into the other equations.
Since we used the first equation, we need to plug in our formula for into the second, third, and fourth equations. We then get (after simplifying):
8. Now, we should have three equations with at most three different variables.
And we do.
9. We can now solve this problem with the steps detailed in our previous post on the substitution method. |
# Absolute Value Inequalities and Equations
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• Not fair! Mommy said I can't have all the candy in the store because I only have five dollars! I need help!!
• Aahh! What just happened?
• Fear not! For I, Sir Absolute Value Equations and Inequalities am here to help. (You can can me SAVEAI for short.)
• Step 1: Distribute the -6 Step 2: You now have 3| X-7 | > 6; distribute the 3 by dividing both sides of the equation by 3 Step 3: You now have |X-7| > 2. Next create two separate equations, one with a >2 and one with a <-2. Step 4: You now have X-7 >2 and X-7<-2; next simplify these inequalities. Step 5: You now have X>9 and X<5; this is not your final answer and you must first graph it by using either a closed dot (includes the number, equal to) or an open dot (does not include the number) on a number line. Here is the graph for this problem: Step 6: The graph indicates what type of solution it is. This problem it is an or question because you cannot have a number greater than 9 and less than 5. You can only have one or the other. Therefore the final solution is X>9 or X<5.
• I am going to teach you how to solve these simple equations and inequalities which can than help you figure out how much candy you can get for five dollars.
• Inequality One: 3I X-7 I -6 > 0
• I'm confused! Why did you make two equations earlier with a 2 and -2?
• Now let's apply what we've learned to solving absolute value equations. The principle is the same, in fact the only difference is that there is an equal sign not an inequality.
• Equation One: |x + 3| -7 = 2 Step 1: Distribute the -7 Step 2: You now have |x + 3|= 9; next set up two equations x + 3 = 9 and x + 3= -9 Step 3: Solve the equations Step 4: The answer is: x= 6 and x= -12
• Ah. So the two equations I made earlier were actually a shortcut to make the math you are already are doing shorter. ABSOLUTE VALUE can be defined as the distance between a number and zero. Therefore if you are trying to solve |x|=16 the answer can either be x= 16 or x= -16 because -16 is still 16 away from zero. An easy way to think about it is a number inside of absolute value will always be positive because you cannot be a negative distance away from zero.
• Okie dokie! I have five dollars for candy and I want to get as much candy as possible. I know whatever I get has to be less then or equal to 5 dollars, and I have a 50% discount. Each candy is worth one dollar. Therefore I can set up the equation like this: |x|< 5 where x is the amount of candy because the amount is always a distance away from zero. Inequality Two: |x|< 5 Step 1: Distribute the 2 by multiplying both sides by 2 Step 2: You now have |x|<10 or x <10 solve by creating the second inequality x>-10 Step 3: After graphing: The answer is -10<x<10 This means that if each candy costs one dollar I can by a minimum of 0 candies and a max of 9 candies.
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# What is the distance between (4, -5) and (-6,7) ?
Jan 1, 2016
$2 \sqrt{61}$
#### Explanation:
Use the distance formula which is
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
Now,
$\left({x}_{1} , {y}_{1}\right) = \left(4 , - 5\right) \text{ }$ and $\text{ } \left({x}_{2} , {y}_{2}\right) = \left(- 6 , 7\right)$
Substituting into formula gives
$d = \sqrt{{\left(- 6 - 4\right)}^{2} + {\left[7 - \left(- 5\right)\right]}^{2}}$
$= \sqrt{{\left(- 10\right)}^{2} + {\left(12\right)}^{2}}$
$= \sqrt{100 + 144}$
$= \sqrt{244} = 2 \sqrt{61}$ |
SSC (English Medium) Class 8Maharashtra State Board
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Sudhir'S Present Age is 5 More than Three Times the Age of Viru. Anil'S Age is Half the Age of Sudhir. - SSC (English Medium) Class 8 - Mathematics
ConceptWord Problems of Equation in One Variable
Question
Sudhir's present age is 5 more than three times the age of Viru. Anil's age is half the age of Sudhir. If the ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5:6, then find Viru's age.
Solution
Let Viru's present age be x years.
Sudhir's present age = (5 + 3x) years
Anil's present age = ((5 + 3"x")/2) years
∴ From the given information,
(5 + 3"x") + ("x") : 3(5 + 3"x")/2 = 5 : 6
⇒ (5 + 4x)/(3(5 + 3"x")/2 = 5/6
⇒ (2(5 + 4"x"))/(15 + 9x) = 5/6
⇒ (10 + 8"x")/(15 + 9"x") = 5/6
⇒ 6 (10 + 8x) = 5 *(15 + 9x)
⇒ 60 + 48x = 75 + 45x
⇒ 48x - 45x = 75 - 60
⇒ 3x = 15
⇒ x = 5
Hence, Viru's age is 5 years.
Is there an error in this question or solution?
APPEARS IN
Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 12: Equations in one variable
Practice Set 12.2 | Q: 10 | Page no. 80
Solution Sudhir'S Present Age is 5 More than Three Times the Age of Viru. Anil'S Age is Half the Age of Sudhir. Concept: Word Problems of Equation in One Variable.
S |
## What is Comparing Two Mean Independent Samples?
In mathematics, a ratio is a relationship between two numbers of the same kind (i.e., objects, persons, students, spoonfuls, units of whatever identical dimension), usually expressed as "a to b" or a:b, sometimes expressed arithmetically as a dimensionless quotient of the two which explicitly indicates how many times the first number contains the second (not necessarily an integer).
1. The two independent samples are simple random samples that are independent.
2. The number of successes is at least five and the number of failures is at least five for each of the samples.
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions (P'A−P'B) reflects a difference in the populations.
The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, Ho:pA =pB. To conduct the test, we use a pooled proportion, pc.
pc=XA+XBnA+nB
(1)
## The distribution for the differences is:
PAPB~N⎡⎣0,pc⋅(1−pc)⋅(1nA+1nB)−−−−−−−−−−−−−−−−−−−−−√⎤⎦
(2)
## The test statistic (z-score) is:
z= (pApB)−(pApB)pc⋅(1−pc)⋅(1nA+1nB)−−−−−−−−−−−−−−−−−−−
These are called “two-sample” tests.
Our goal is usually to estimate p1 – p2, the corresponding confidence intervals, and to perform hypothesis tests on:
H0p1 – p2 = 0.
The obvious statistic to compare the two population proportions is -. Where = number of successes in group divided by sample size in group i.
Probability theory tells us that:
1. - is the best estimate of p1 – p2
2. the standard error is
3. If n1p1(1-p1) > 5 and n2p2(1-p2) > 5
Large-sample confidence interval for p1 p2
Large-sample Z-test of
H0p1 – p2 = 0 vs. H1p1 – p2 ≠ 0
Test statistic:
Where denotes the standard error estimates using the null hypothesis, p1p2.
Estimate the common p using
,
where x1 and x2 are the number of successes in groups 1 and 2, respectively.
Then
Compare Z to a standard Normal distribution.
Example: Caries incidence
### N
caries by age two
Number
percent
controls
36
10
27.8
intervention
68
6
8.8
95% confidence interval:
- = .278 - .088 = 0.19
So 95% confidence interval is
Test: H0p1 – p2 = 0 vs. H1p1 – p2 ≠ 0
Reject at α=.05 level. P-value = 2×P(Z > 2.57) = .0102
All such points are covered inStatistichomework help and assignment help at transtutors.com.
Our email-based homework help support provides best and intelligent insight and recreation which help make the subject practical and pertinent for any assignment help.
Transtutors.com present timely homework help at logical charges with detailed answers to your Statistic questions so that you get to understand your assignments or homework better apart from having the answers. Our tutors are remarkably qualified and have years of experience providing Comparing two mean independent samples homework help or assignment help.
## Related Topics
All Statistics Topics
More Q&A |
# Circles – Explanation and Examples
The word circle comes from the Greek word krikos, which means ring. It is a circular arc equidistant from a fixed point called the center of a circle. The length of this arc is known as the circumference or perimeter of a circle. The utilization of circles is widely observed in several fields, ranging from the simplest to advanced mathematics, science, architecture, etc. Some of the real-world examples of circles are wheels, rings, pizzas, donuts, etc. The symmetric properties of circles are often used to design athletic tracks, buildings, watches, wheels, and so on.
Kids get introduced to circles during elementary or primary school. While proceeding to higher-grades, they study more complex concepts based on it. By middle school, kids start to learn geometry, and they also study all the properties of circles like the area of circle, the radius of a circle, circumference, diameter, and various word problems based on them. Kids must understand and acquire the knowledge of various concepts based on circles and their applications as it holds a great significance in our day-to-day life.
## What is a circle?
A circle is one of the most basic geometric shapes formed by tracing a point on a plane equidistant from the other point called the center of the circle. This boundary is known as the circumference of the circle. The line joining the center of the circle to any point on its circumference is known as its radius. The area enclosed in the arc of the circle is called the area of a circle. It is calculated using the formula πr²
Examples: Let’s compare a large pizza area with a diameter of 16 inches with a small pizza with a diameter of 8 inches.
If the diameter of a large pizza is 16 inches, then its radius is 8 inches.
Therefore the area of the large pizza will be π× 8² = 64π.
If the diameter of a small pizza is 8 inches, then its radius is 4 inches.
Therefore the area of the small pizza will be π× 4² = 16π.
When you compare a large pizza area with a smaller pizza, you will see that the large pizza is four times larger than the smaller one.
### Here are some useful terms associated with a circle:
#### Center of a Circle:
The center of a circle is a point inside the circle that has the same distance from a point located anywhere on its boundary. It is usually represented by ‘O’.
The radius of a circle is a line between the center and the circumference of the circle. It is denoted by r.
#### Circumference of Circle:
The circumference of a circle is a boundary line with some distance from the Center of the circle. It is the length of a complete arc of a circle. It is calculated using the formula ‘2πr‘ where ‘r‘ denotes a circle radius.
#### Diameter:
The circle’s diameter is a line that has endpoints on the circle boundary and passes through the center of the circle. The diameter D of a circle is twice the radius of a circle. i.e, D = 2r.
#### Arc:
An arc is a segment of the circumference of a circle. If we place two points on a circle’s circumference, we will get a major arc and a minor arc.
#### Secant:
It is an extended chord, a coplanar straight line intersecting a circle in two points.
#### Chord:
A line segment joining any two points on the circumference of a circle is called a chord.
#### Tangent
It is a coplanar straight line that touches the circle at one point.
### Conclusion:
The circle is an important math topic, and learning its properties and concepts is vital for kids to build lifelong knowledge. Understanding the basic concepts and underlying principles of a topic enables kids to grasp the core fundamentals quickly. Cuemath online classes encourage kids to learn math through reasoning, allowing them to gain an in-depth understanding of core concepts.
##### Heron Nelson
Heron is a business blogger with a focus on personal finance and wealth management. With over 7 years of experience writing about financial topics, Heron has established herself as a trusted voice in the personal finance space. She has a deep understanding of financial concepts and strategies, and is able to explain them in a relatable and actionable way for her readers. |
### Bar Models - IBES Home
```February 27, 2014
6:00-7:15
Agenda
Videos
Three Types of Bar
Models
Steps to Solving Bar
Model Problems
Sample Bar Model
Problems
On-line resource Think Central
Practice
What are bar models and why are they so
important?
Bar Models provide a useful pictorial representation of
sets and parts making up a whole
Children use bar models to illustrate a problem, indicating
the known and unknown parts or the whole
Bar Models help students make sense of the relationship
between values given in real-world problems
Problems
Multiplication and
Division with Bar
Models
Part
Part
Part/Part/Whole
Comparison
Equal Parts
Step 1 – Read the problem without numbers.
Mrs. Smith has ___apples. Later she gets
___more apples. How many apples
does she have in all?
Step 2 – Read the problem with numbers.
Mrs. Smith has 3 apples. Later she gets
two more apples. How many apples
does she have in all?
Step 3 – Decide which bar model to use.
This is problem tells me 2 parts and asks me
to find the whole, so I need to use a
Part/Part/Whole bar model.
Step 4- Decide where the ? goes.
?
Step 5 – Plug in the numbers.
3
2
?
Step 6 – Solve it
3+2=5
Mrs. Smith had 5 apples in all.
Step 8 - Check your work.
5 - 2 = 3 or 5 – 3 = 2
Bar Modeling
In Mrs. Anderson’s class there are twice Which Bar
as many boys as girls. (2nd Grade)
Model is this?
Girls
Boys
There
boys. How
Howmany
manyboys
girls are
are there?
there?
Thereare
are18
6 girls.
3
?
Write a number sentence to match
Students have to determine which part to
solve first.
Figuring out 2nd, 3rd or 4th parts to the
problem are determined by whether or not
they get the 1st step done correctly.
Bar models can show them what they don’t
know.
22 lbs
17 lbs
Suitcase 1
?
Suitcase 2
?
stamps. Lisa gave some to Mia
and now Mia has 3 times as
many. How many does Lisa have
now?
What if I used a bar model to help see the
problems?
?
Lisa
?
Mia
What if I used a bar model to help see the
problems?
?
Part 1) 175+129=304
Part 2) 304
4=
Lisa
Mia
?
The cost of 3 pair of socks and 4 T-shirts is
\$132
Each T-shirt costs twice as much as a pair of
socks.
What is the cost for 1 pair of socks? What is
the cost for 1 T-shirt?
The cost of 3 pair of
socks and 4 T-shirts
is \$132
Each T-shirt costs
twice as much as a
pair of socks.
What is the cost for
1 pair of socks?
What is the cost
for 1 T-shirt?
?
Socks
?
T-Shirts
\$132
Each bar represents
1 piece to the
problem.
Socks
To find the solution
Divide the total by
the number of bars.
132 divided by 11
equals 12.
So the cost of 1
T-Shirts
pair of socks is \$12
and the cost of 1 Tshirt is \$24.
?
?
\$132
www.thinkcentral.com
Teachers can post assignments and test
scores
Students can access
Student book
Student workbook
Virtual manipulatives
Student interactivities
Lesson videos
• Videos with a
description of
each chapter’s
content
• Parent Support
Videos with ‘at
home’
suggestions
``` |
# Direction of a Point from a Line Segment
Direction of given point P from a line segment simply means given the co-ordinates of a point P and line segment (say AB), and we have to determine the direction of point P from the line segment. That is whether the Point lies to the Right of Line Segment or to the Left of Line Segment.
The point might lie behind the line segment, in that case we assume imaginary line by extending the line segment and determine the direction of point.
* There are only three cases, either the point is on left side, or right side or on the line segment itself.
This is a very fundamental Problem and is commonly encountered for directions in online map
Example : Suppose a user A has to go to Point C in the following map, the user first reaches point B but after that how does user A know that whether he has to make a right turn or left turn.
Knowing the direction of a point from a line segment also acts a building block to solve more complicated problem such as :
• Line segment Intersection : finding if two line segment intersect
• Convex Hull of a set of Points.
The co-ordinate system we’ll use is a cartesian plane, as most 2-Dimensional problem uses cartesian plane and since this is a 2-Dimensional Problem.
This Problem can be solved using cross-product of vector algebra
Cross-Product of two point A and B is : Ax * By – Ay * Bx
where Ax and Ay are x and y co-ordinates of A respectively. Similarly Bx and By are x and y co-ordinates of B respectively.
The Cross-Product has an interesting Property which will be used to determine direction of a point from a line segment. That is, the cross-product of two points is positive if and only if the angle of those point at origin (0, 0) is in counter-clockwise. And conversely the cross-product is negative if and only if the angle of those point at origin is in clockwise direction.
An example would certainly clarify it
In the following figure, the angle BOP is anti-clockwise and the cross-product of B X P = 29*28 – 15*(-15) = 1037 which is positive.
This helps us to make a conclusion that a point on right side must a have positive cross-product and a point on left side must have a negative cross product. Also note that we assumed one point of line segment to be origin and hence we need to convert any three point system such that one point of line segment is origin.
Following example explains the concept
The three point A, B and P were converted into A’, B’ and P’ so as to make A as origin (this can be simply done by subtracting co-ordinates of A from point P and B), and then calculate the cross-product : 59*18 – (-25)*18 = 2187
Since this is positive, the Point P is on right side of line Segment AB.
## C++
`// C++ Program to Determine Direction of Point``// from line segment``#include ``using` `namespace` `std;` `// structure for point in cartesian plane.``struct` `point {`` ``int` `x, y;``};` `// constant integers for directions``const` `int` `RIGHT = 1, LEFT = -1, ZERO = 0;` `int` `directionOfPoint(point A, point B, point P)``{`` ``// subtracting co-ordinates of point A from`` ``// B and P, to make A as origin`` ``B.x -= A.x;`` ``B.y -= A.y;`` ``P.x -= A.x;`` ``P.y -= A.y;` ` ``// Determining cross Product`` ``int` `cross_product = B.x * P.y - B.y * P.x;` ` ``// return RIGHT if cross product is positive`` ``if` `(cross_product > 0)`` ``return` `RIGHT;` ` ``// return LEFT if cross product is negative`` ``if` `(cross_product < 0)`` ``return` `LEFT;` ` ``// return ZERO if cross product is zero. `` ``return` `ZERO; ``}` `// Driver code``int` `main()``{`` ``point A, B, P;`` ``A.x = -30;`` ``A.y = 10; ``// A(-30, 10)`` ``B.x = 29;`` ``B.y = -15; ``// B(29, -15)`` ``P.x = 15;`` ``P.y = 28; ``// P(15, 28)` ` ``int` `direction = directionOfPoint(A, B, P);`` ``if` `(direction == 1)`` ``cout << ``"Right Direction"` `<< endl;`` ``else` `if` `(direction == -1)`` ``cout << ``"Left Direction"` `<< endl;`` ``else`` ``cout << ``"Point is on the Line"` `<< endl;`` ``return` `0;``}`
## Java
`// Java Program to Determine Direction of Point``// from line segment``class` `GFG ``{` `// structure for point in cartesian plane.``static` `class` `point ``{`` ``int` `x, y;``};` `// constant integers for directions``static` `int` `RIGHT = ``1``, LEFT = -``1``, ZERO = ``0``;` `static` `int` `directionOfPoint(point A, `` ``point B, point P)``{`` ``// subtracting co-ordinates of point A `` ``// from B and P, to make A as origin`` ``B.x -= A.x;`` ``B.y -= A.y;`` ``P.x -= A.x;`` ``P.y -= A.y;` ` ``// Determining cross Product`` ``int` `cross_product = B.x * P.y - B.y * P.x;` ` ``// return RIGHT if cross product is positive`` ``if` `(cross_product > ``0``)`` ``return` `RIGHT;` ` ``// return LEFT if cross product is negative`` ``if` `(cross_product < ``0``)`` ``return` `LEFT;` ` ``// return ZERO if cross product is zero. `` ``return` `ZERO; ``}` `// Driver code``public` `static` `void` `main(String[] args) ``{`` ``point A = ``new` `point(), `` ``B = ``new` `point(), P = ``new` `point();`` ``A.x = -``30``;`` ``A.y = ``10``; ``// A(-30, 10)`` ``B.x = ``29``;`` ``B.y = -``15``; ``// B(29, -15)`` ``P.x = ``15``;`` ``P.y = ``28``; ``// P(15, 28)` ` ``int` `direction = directionOfPoint(A, B, P);`` ``if` `(direction == ``1``)`` ``System.out.println(``"Right Direction"``);`` ``else` `if` `(direction == -``1``)`` ``System.out.println(``"Left Direction"``);`` ``else`` ``System.out.println(``"Point is on the Line"``);`` ``}``}` `// This code is contributed``// by Princi Singh`
## Python3
`# Python3 program to determine direction ``# of point from line segment`` ` `# Structure for point in cartesian plane.``class` `point:`` ` ` ``def` `__init__(``self``):`` ` ` ``self``.x ``=` `0`` ``self``.y ``=` `0`` ` `# Constant integers for directions``RIGHT ``=` `1``LEFT ``=` `-``1``ZERO ``=` `0`` ` `def` `directionOfPoint(A, B, P):`` ` ` ``global` `RIGHT, LEFT, ZERO`` ` ` ``# Subtracting co-ordinates of `` ``# point A from B and P, to `` ``# make A as origin`` ``B.x ``-``=` `A.x`` ``B.y ``-``=` `A.y`` ``P.x ``-``=` `A.x`` ``P.y ``-``=` `A.y`` ` ` ``# Determining cross Product`` ``cross_product ``=` `B.x ``*` `P.y ``-` `B.y ``*` `P.x`` ` ` ``# Return RIGHT if cross product is positive`` ``if` `(cross_product > ``0``):`` ``return` `RIGHT`` ` ` ``# Return LEFT if cross product is negative`` ``if` `(cross_product < ``0``):`` ``return` `LEFT`` ` ` ``# Return ZERO if cross product is zero`` ``return` `ZERO` `# Driver code ``if` `__name__``=``=``"__main__"``:`` ` ` ``A ``=` `point()`` ``B ``=` `point()`` ``P ``=` `point()`` ` ` ``A.x ``=` `-``30`` ``A.y ``=` `10` `# A(-30, 10)`` ``B.x ``=` `29`` ``B.y ``=` `-``15` `# B(29, -15)`` ``P.x ``=` `15`` ``P.y ``=` `28` `# P(15, 28)`` ` ` ``direction ``=` `directionOfPoint(A, B, P)`` ` ` ``if` `(direction ``=``=` `1``):`` ``print``(``"Right Direction"``)`` ``elif` `(direction ``=``=` `-``1``):`` ``print``(``"Left Direction"``)`` ``else``:`` ``print``(``"Point is on the Line"``)` `# This code is contributed by rutvik_56`
## C#
`// C# Program to Determine Direction of Point``// from line segment``using` `System;``using` `System.Collections.Generic;` `class` `GFG ``{` `// structure for point in cartesian plane.``public` `class` `point ``{`` ``public` `int` `x, y;``};` `// constant integers for directions``static` `int` `RIGHT = 1, LEFT = -1, ZERO = 0;` `static` `int` `directionOfPoint(point A, `` ``point B, point P)``{`` ``// subtracting co-ordinates of point A `` ``// from B and P, to make A as origin`` ``B.x -= A.x;`` ``B.y -= A.y;`` ``P.x -= A.x;`` ``P.y -= A.y;` ` ``// Determining cross Product`` ``int` `cross_product = B.x * P.y - B.y * P.x;` ` ``// return RIGHT if cross product is positive`` ``if` `(cross_product > 0)`` ``return` `RIGHT;` ` ``// return LEFT if cross product is negative`` ``if` `(cross_product < 0)`` ``return` `LEFT;` ` ``// return ZERO if cross product is zero. `` ``return` `ZERO; ``}` `// Driver code``public` `static` `void` `Main(String[] args) ``{`` ``point A = ``new` `point(), `` ``B = ``new` `point(),`` ``P = ``new` `point();`` ``A.x = -30;`` ``A.y = 10; ``// A(-30, 10)`` ``B.x = 29;`` ``B.y = -15; ``// B(29, -15)`` ``P.x = 15;`` ``P.y = 28; ``// P(15, 28)` ` ``int` `direction = directionOfPoint(A, B, P);`` ``if` `(direction == 1)`` ``Console.WriteLine(``"Right Direction"``);`` ``else` `if` `(direction == -1)`` ``Console.WriteLine(``"Left Direction"``);`` ``else`` ``Console.WriteLine(``"Point is on the Line"``);`` ``}``}` `// This code is contributed by 29AjayKumar`
## Javascript
``
Output:
`Right Direction`
Time Complexity: O(1)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
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# Video: Recognizing Perpendicular Lines
Which of the following lines is perpendicular to the line 19π₯ β 3π¦ = 5? [A] 3π₯ β 19π¦ = 5 [B] 2 β 19π¦ = 3π₯ [C] 3π¦ = 1 β 19π₯ [D] 3π¦ = 19π₯ + 4 [E] 3 + 19π¦ = 2π₯
05:05
### Video Transcript
Which of the following lines is perpendicular to the line 19π₯ minus three π¦ equals five?
(a) three π₯ minus 19π¦ equals five, (b) two minus 19π¦ equals three π₯, (c) three π¦ equals one minus 19π₯, (d) three π¦ equals 19π₯ plus four, (e) three plus 19π¦ equals two π₯.
Before we choose one of these functions, letβs remember what perpendicular lines are, more specifically what perpendicular lines have. Perpendicular lines have negative reciprocal slopes. So first weβll take 19π₯ minus three π¦ equals five and find its slope. To do that, weβll take the function given in standard form and convert it into slope-intercept form by isolating π¦.
First subtract 19π₯ from both sides of the equation. 19π₯ minus 19π₯ cancels out, leaving us with negative three π¦ equals negative 19π₯ plus five. Remember the goal: isolate π¦. We divide π¦ by negative three and that means weβll have to divide both terms on the right side by negative three. On the left, negative three divided by negative three equals one, and one times π¦ equals π¦, equals β our π₯ term has a negative in the numerator and the denominator. We can simplify that by saying 19 over three π₯ minus five over three. This is the slope-intercept form of the same equation we started with.
Slope-intercept form is π¦ equals ππ₯ plus π. The π value is the slope. The slope of the line we were given is 19 over three. We want the negative reciprocal of 19 over three. The reciprocal of 19 over three is three over 19, and we need the negative value. Any function with the slope of negative three over 19 will be perpendicular to our line. Weβll examine all five of these functions to see which of them has a slope of negative three over 19.
Instead of trying to find the slope of all five of these lines, letβs see if we can eliminate any of the options. To do that, I want you to notice the relationship between π₯ and π¦ in the original function. We need the opposite to be true. We need a constant value of three associated with the π₯ and 19 associated with the π¦. Itβs also important to note that the π¦-intercept doesnβt matter. In our new equation, the constant value, the π¦- intercept, can be anything.
So weβll walk through these five functions and see which of them do not have a relationship of three π₯ to 19π¦. (a) follows this pattern. (b) follows this pattern. (c) does not, neither does (d) or (e). Now weβre down to two functions we have to check. Weβll find the slope of both (a) and (b). In function (a), we subtract three π₯ from both sides. Three π₯ minus three π₯ cancels out. Negative 19π¦ equals negative three π₯ plus five. To find the slope-intercept form, we need to isolate π¦. Weβll divide every term by negative 19. π¦ equals three over 19π₯ minus five over 19. The slope of function (a) is three over 19.
Three over 19 is the reciprocal of our first function, but it is not the negative reciprocal of our first function. So now we move to function (b). Our first step: subtract two from both sides of the equation. Two minus two equals zero. And now we have negative 19π¦ equals three π₯ minus two. We divide every term by negative 19, which simplifies to π¦ equals negative three over 19π₯ plus two over 19. The slope here is negative three over 19, which is the negative reciprocal slope weβre looking for. The function two minus 19π¦ equals three π₯ is perpendicular to the function 19π₯ minus three π¦ equals five.
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# Square Root of 0 + Solution With Free Steps
This article is about the square root of Zero which can be written as √0 but the interesting thing is how we can evaluate the square root of 0 because 0 is a number that multiplies with any number given in return 0 so the square root of zero is also 0.
In this article, we will analyze and find the square root of 0 using various mathematical techniques, such as the approximation method and the long division method.
## What Is the Square Root Of 0?
The square root of the number 0 is 0.
The square root can be defined as the quantity that can be doubled to produce the square of that similar quantity. In simple words, it can be explained as:
√0 = √(0 x 0)
√X = √(0)$^2$
√0 = ±0
The square can be canceled with the square root as it is equivalent to 1/2; therefore, obtaining 0. Hence 0 is 0’s square root. The square root generates both positive and negative integers.
## How To Calculate the Square Root of 0?
You can calculate the square root of 0 using any of two vastly used techniques in mathematics; one is the Approximation technique but 0 is the case where we did not get a lesser number than 0 which is a perfect square number so we cannot use this method, and the other is the Long Division method.
The symbol √ is interpreted as 0 raised to the power 1/2. So any number, when multiplied by itself, produces its square, and when the square root of any squared number is taken, it produces the actual number.
Let us discuss each of them to understand the concepts better.
### Square Root by Long Division Method
The process of long division is one of the most common methods used to find the square roots of a given number. It is easy to comprehend and provides more reliable and accurate answers. The long division method reduces a multi-digit number to its equal parts.
Learning how to find the square root of a number is easy with the long division method. All you need are five primary operations- divide, multiply, subtract, bring down or raise, then repeat.
Following are the simple steps that must be followed to find the square root of 0 using the long division method:
### Step 1
First, write the given number 0 in the division symbol, as shown in figure 1.
### Step 2
Starting from the right side of the number, as the number 0 is a single digit number that cannot be divided into pairs so write 0 as it is.
### Step 3
Now divide the digit 0 by a number, giving a number either 0 or less than 0. Therefore, in this case, the remainder is zero, whereas the quotient is also 0.
### Step 4
The resulting quotient 0 is the square root of 0. Figure 1 given below shows the long division process in detail:
figure 1
### Square Root by Approximation Method
The approximation method involves guessing the square root of the non-perfect square number by dividing it by the perfect square lesser or greater than that number and taking the average.
But in this case, where we find out the square root of 0 this method is not applicable.
### Important points
• The number 0 is a perfect square.
• The number 0 is a rational number.
• The number 0 cannot be split into its prime factorization.
## Is Square Root of 0 a Perfect Square?
The number 0 is a perfect square. A number is a perfect square if it splits into two equal parts or identical whole numbers. If a number is a perfect square, it is also rational.
A number expressed in p/q form is called a rational number. All the natural numbers are rational. A square root of a perfect square is a whole number; therefore, a perfect square is a rational number.
A number that is not a perfect square is irrational as it is a decimal number. As far as 0 is concerned, it is a perfect square. It can be proved as below:
Factorization of 0 results in 0 x 0 which can also be expressed as 0$^2$.
Taking the square root of the above expression gives:
= √(0$^2$)
= (0$^2$)$^{1/2}$
= 0
This shows that 0 is a perfect square and a rational number.
Therefore the above discussion proves that the square root of 0 is equivalent to 0.
Images/mathematical drawings are created with GeoGebra. |
# Perfect Square Trinomial Formula
## Perfect Square Trinomial Formula
Perfect square trinomials are algebraic expressions containing three terms and obtained by multiplying two binomials by the same binomial. Numbers that are perfect squares are obtained by multiplying them by themselves. An algebraic expression containing only two terms is called a binomial. It consists of a positive (+) or a negative (-) sign between the terms. Similarly, trinomials are algebraic expressions with three terms. The perfect square trinomial has three terms when a binomial consists of a variable and a constant is multiplied by itself. There is either a positive or a negative sign separating the terms in a Perfect Square Trinomial Formula. Perfect Square Trinomial Formula can be understood to its complete depth with the help of study resources provided by Extramarks. Perfect Square Trinomial Formula and many other important concepts can be studied thoroughly with the help of the expert guidance of Extramarks’ educators.
### Perfect Square Trinomial Definition
A Perfect Square Trinomial Formula are algebraic expressions obtained by squaring binomial expressions. It is of form ax2 + bx + c. The numbers a, b, and c are real numbers, while a is not equal to 0. Let’s take a binomial (x+4) and multiply it by (x+4). The result is x2 + 8x + 16. The Perfect Square Trinomial Formula can be decomposed into two binomials, which when multiplied together yield the perfect square trinomial.
### Perfect Square Trinomial Pattern
Perfect square trinomial generally has a2 + 2ab + b2 or a2 – 2ab + b2. To find the Perfect Square Trinomial Formula for a binomial, follow the steps below. They are,
• The first step is to find the square, or the first term of the binomial.
• The second step is to multiply the first and second terms of the binomial with 2.
• The third step is to find the square of the binomial’s second term.
• The fourth step is to add up all three terms obtained in steps 1, 2, and 3.
The square of the first term of the binomial is the first term of the Perfect Square Trinomial Formula. In the binomial, the second term is twice the product of the two terms, and the third term is the square of the second term. It follows that all terms of the Perfect Square Trinomial Formula are positive if the binomial being squared has a positive sign. The second term of the trinomial will be negative if its second term is negative (which is twice the product of the two variables).
### Perfect Square Trinomial Formula
There is either a positive or a negative symbol between the terms of perfect square trinomials. There are two important algebraic identities related to the Perfect Square Trinomial Formula.
• (a + b)2 = a2 + 2ab + b2
• (a – b)2 = a2 – 2ab + b2
To factor a perfect square polynomial, follow these steps.
• Write the Perfect Square Trinomial Formula of the form a2 + 2ab + b2 or a2 – 2ab + b2, such that the first and the third terms are perfect squares, one being a variable and another a constant.
• Make sure that the middle term is twice the product of the first and third terms. Be sure to check the sign of the middle term as well.
• If the middle term is positive, then the perfect square trinomial should be compared to a2 + 2ab + b2, and if the middle term is negative, then the Perfect Square Trinomial Formula should be compared to a 2 – 2ab + b2.
• In case of a positive middle term, the factors are (a+b) (a+b), and in case of a negative middle term, the factors are (a-b) (a-b).
### Topics Related to Perfect Square Trinomial
Multiplying the same binomial expression with each other yields the Perfect Square Trinomial Formula. When a trinomial is of the form ax 2 + bx + c and also meets the condition b 2 = 4ac, it is said to be a perfect square. A Perfect Square Trinomial Formula can be expressed in two ways. They are,
• (ax)2+ 2abx + b2= (ax + b)2—– (1)
• (ax)2−2abx + b2 = (ax−b)2—– (2)
For example, take a perfect square polynomial, x2 + 6x + 9. Comparing this with ax2+bx+c, a = 1, b = 6 and c = 9. Check if trinomial satisfies the condition b2 = 4ac.
b2 = 36 and 4 × a × c = 4 × 1 × 9, which is equal to 36.
Due to this, the trinomial satisfies the condition b2 = 4ac. Therefore, it is called a Perfect Square Trinomial Formula. Take a look at some important topics related to perfect square trinomials.
• Algebra is the study of representations of problems and situations in the form of mathematical expressions. In order to form a meaningful mathematical expression, variables like x, y, and z must be used along with mathematical operations such as addition, subtraction, multiplication, and division.
• Arithmetic is used in all branches of mathematics, such as trigonometry, calculus, and coordinate geometry. The expression 2x + 4 = 8 is a simple example of an algebraic expression.
• It is possible to factorize algebraic expressions by finding the factors of the given expression, which means finding two or more expressions whose product is the given expression. It is the process of finding two or more expressions whose product is the given expression that is known as factorization of algebraic expressions. When a number divides a given number without leaving a remainder, it is called a factor. Multiplying two numbers means expressing a number as a result of their multiplication. It is common for algebraic expressions to be written as products of their factors. In algebraic expressions, numbers and variables are combined with arithmetic operations like addition and subtraction.
• The square of a trinomial has been discussed. As an example, if one is asked to find the square of 8, you would immediately say 64, since 8 × 8 = 64.
• A Perfect Square Trinomial Formula can be used to solve complex trinomial functions. Perfect Square Trinomial Formula functions are obtained by squaring binomial expressions. In order to be a perfect square, a trinomial must be of the form ax 2 + bx + c and satisfy the condition b 2 = 4ac. The Perfect Square Trinomial Formula can be understood by looking at solved examples.
### Perfect Square Trinomial Examples
1. Example 1: Factor the trinomial x2 + 10x + 25.
Solution:
The given expression x2 + 10x + 25 is of the form a2 + 2ab + b2. Factors of (a2 + 2ab + b2) are (a+b) (a+b).
Therefore, the factors are (x + 5) (x + 5) or (x + 5)2.
### Practice Questions on Perfect Square Trinomial
Q1. Which of the following forms does a perfect square trinomial take?
Responses
• (ax)2 + 2abx + b2 = (ax + b)2
• (ax)2 −2abx + b2 = (ax−b)2
• Both (a) and (b)
• None of the above
### 1. Perfect Square Trinomials: What are they?
There are three terms in the Perfect Square Trinomial Formula, which is of the form ax2 + bx + c. The formula is obtained by multiplying a binomial by itself. The Perfect Square Trinomial Formula x2 + 6x + 9 can be obtained by multiplying the binomial (x + 3) by itself. To put it another way, (x + 3) (x + 3) = x2 + 6x + 9.
### 2. What is the formula for the Perfect Square Trinomial?
By multiplying two binomials, which are one and the same, a perfect square trinomial is formed. An algebraic expression with two terms is a binomial, and an algebraic expression with three terms is a trinomial. The Perfect Square Trinomial Formula can be obtained by multiplying (a+2) and (a+2), which gives 2a2 + 4a + 4.
### 3. To call an algebraic expression a perfect square trinomial, must all three terms be squares?
For an algebraic expression to be a Perfect Square Trinomial Formula, only the first and third terms must be perfect squares. For example, in x 2 + 2x + 1, the first term is the square of ‘x’, and the third term is the square of ‘1’.
### 4. Perfect Square Trinomials: What is their pattern?
There are two possible patterns for the Perfect Square Trinomial Formula: a 2 + 2 ab + b 2 or a 2a2 – 2 ab + b 2. By squaring the binomials (a+b) and (a-b), one obtains these expressions.
### 5. How to Factor a Perfect Square Trinomial?
A Perfect Square Trinomial Formula has three terms which may be of the form (ax)2+ 2abx + b2= (ax + b)2 or (ax)2−2abx + b2 = (ax−b)2. Factoring a perfect square polynomial involves the following steps.
• One must verify that the given perfect square trinomial is of the form a 2 + 2ab + b 2.
• The middle term should be twice the product of the first and third terms. Be sure to check the middle term’s sign as well.
• The perfect square trinomial can be compared with a2 + 2ab + b 2 if the middle term is positive, and a2 – 2ab + b2 if the middle term is negative.
• In case of a positive middle term, the factors are (a+b) (a+b), and in case of a negative middle term, the factors are (a-b) (a-b).
### 6. Is it possible to call perfect square trinomials quadratic equations?
Quadratic equations can be formed from perfect square polynomials. Quadratic equations have one squared term and a degree of 2. Since perfect square trinomials have a degree of 2, we can call them quadratic equations. As all quadratic equations do not satisfy the conditions for a perfect square trinomial, they cannot be considered perfect square trinomials.
### 7. Would all algebraic expressions that have perfect squares as the first and last terms qualify as perfect square trinomials?
No, not all algebraic expressions with perfect squares in the first and last terms are perfect square trinomials. As an example, consider x2 + 18x + 36 which is an algebraic expression with three terms, and the first and third terms are perfect squares. To be a perfect square trinomial, we should have 2ab equal 12x when compared to a 2 + 2ab + b 2. Based on that, a = 1 and b = 6, so this expression is a perfect square trinomial. In this case, however, 2ab = 18x, so one cannot say that all algebraic expressions with first and third terms squared are perfect square trinomials. |
# Activities to Teach Students to Order Fractions With Like Numerators or Denominators
Fractions are a fundamental part of mathematics that can be difficult for students to understand. One of the critical skills when working with fractions is being able to order them. Ordering fractions requires a solid understanding of the relationship between fractions, and it is essential for math problems involving fractions. Fortunately, there are numerous activities and exercises to help students develop this skill. In this article, we will explore some of the best activities to teach students how to order fractions with like numerators or denominators.
1. Comparing Fractions with Number Lines
Number lines are an excellent tool for teaching students how to order fractions. Draw a line and mark the starting point as zero, and the endpoint as one. Then, place fractions on the number line, with a marker showing the ordering of the fractions. Students can visually see which fraction is more significant and which is lesser and understand the concept of ordering fractions.
2. Fraction War
This game requires just a deck of cards without face cards. Two players draw two cards each, and they make a fraction out of each card, so they have two fractions each. Then, they compare their fractions and see who has the bigger fraction so that player wins. This game is an easy and fun way of teaching students how to order fractions.
3. Build a Fraction Tower
This activity requires math manipulatives such as fraction circles, rectangles, and squares. Ask students to construct a tower using the manipulatives in order of size, either ascending or descending. Students can work in pairs or individually and present their towers to the class. This activity helps them develop spatial reasoning and provides visual cues for ordering fractions.
4. Fraction Critters
For this activity, students draw different fractions and create Fraction Critters. They should use different colors to denote different fractions and then order their Fraction Critters according to the size of the fractions. Students can display their Fraction Critters to the class and explain why each fraction appears in the order they chose. This activity provides an opportunity for creativity and innovation while reinforcing the skill of ordering fractions.
5. Fractions Around the Room
This game requires students to find fractions around the room or in a given area and then order them based on size. Pre-print fractions on index cards and distribute the cards around the room. Students find the fraction cards and order them on a chart. This activity develops both their critical thinking and physical skills.
In conclusion, ordering fractions is an essential skill for math students to acquire, and these activities provide a fun and engaging way to teach and reinforce the concept. By incorporating these activities into lessons, teachers can ensure that students have the foundational skills to solve math problems involving fractions. Students will enjoy learning how to order fractions and will be better equipped to master other math concepts in the future. |
Directional Derivatives
Suppose you are given a topographical map and want to see how steep it is from a point that is neither due West or due North. Recall that the slopes due north and due west are the two partial derivatives. The slopes in other directions will be called the directional derivatives. Formally, we define
Definition Let f(x,y) be a differentiable function and let u be a unit vector then the directional derivative of f in the direction of u is
Note that if u is i then the directional derivative is just fx and if u is j the it is fy.
Just as there is a difficult and an easy way to compute partial derivatives, there is a difficult way and an easy way to compute directional derivatives.
Theorem Let f(x,y) be a differentiable function, and u be a unit vector with direction q, then
Example:
Let
f(x,y) = 2x + 3y2 - xy
and
v = <3,2>
Find
Dv f(x,y)
Solution
We have
fx = 2 - y
and
fy = 6y - x
and
Hence
Dv f(x,y) = <2 - y, 6y - x> . <3/, 2/
2 3
= (2 - y) + (6y - x)
Exercise
Let
Find Dv f(x,y)
We define
Notice that
Du f(x,y) = (grad f) . u
The gradient has a special place among directional derivatives. The theorem below states this relationship.
Theorem If grad f(x,y) = 0 then for all u, Du f(x,y) = 0 The direction of grad f(x,y) is the direction with maximal directional derivative. The direction of -grad f(x,y) is the direction with the minimal directional derivative.
Proof:
1. If
then
Du f(x,y) = grad f . u = 0 . u = 0
2. Du f(x,y) = grad f . u = ||grad f || cos q
This is a maximum when q = 0 and a minimum when q = p. If q = 0 then grad f and u point in the same direction. If q = p then u and grad f point in opposite directions. This proves 2 and 3.
Example:
Suppose that a hill has altitude
w(x,y) = x2 - y
Find the direction that is the steepest uphill and the steepest downhill at the point (2,3).
Solution
We find
grad w = <2x, -y> = <4, -3>
Hence the steepest uphill is in the direction
<4,-3>
while the steepest downhill is in the direction
-<4,-3> = <-4,3> |
# How do you graph and solve |4x – 3|+ 2 <11?
Jul 16, 2018
The solution is $x \in \left(- \frac{3}{2} , 3\right)$
#### Explanation:
The inequality is
$| 4 x - 3 | + 2 < 11$
$| 4 x - 3 | - 9 < 0$
The point to consider is
$4 x - 3 = 0$
$\implies$, $x = \frac{3}{4}$
There are $2$ intervals to consider
$\left(- \infty , \frac{3}{4}\right)$ and $\left(\frac{3}{4} , + \infty\right)$
Therefore,
In the first interval
$- 4 x + 3 - 9 < 0$
$\implies$, $- 4 x - 6 < 0$
$\implies 0$, $4 x > - 6$
$\implies$, $x > - \frac{3}{2}$
This solution belongs to the interval
In the second interval
$4 x - 3 - 9 < 0$
$\implies$, $4 x - 12 < 0$
$\implies$, $x < 3$
This solution belongs to the interval
The solution is $x \in \left(- \frac{3}{2} , 3\right)$
graph{|4x-3|-9 [-20.27, 20.27, -10.14, 10.14]} |
# Characterization of Parallel Lines
Examples, solutions, worksheets, and videos to help Grade 8 students learn that when a system of linear equations has no solution, i.e., no point of intersection of the lines, then the lines are parallel.
### New York State Common Core Math Grade 8, Module 4, Lesson 26
Lesson 26 Student Outcomes
• Students know that when a system of linear equations has no solution, i.e., no point of intersection of the lines, then the lines are parallel.
Lesson 26 Student Summary
By definition, parallel lines do not intersect; therefore, a system of linear equations that graph as parallel lines will have no solution.
Parallel lines have the same slope, but no common point. Verify that lines are parallel by comparing their slopes and their y-intercepts.
Lesson 26 Opening Exercise
Exercises
1. Graph the system:
y = 2/3 x + 4
y = 4/6 x - 3
a. Identify the slope of each equation. What do you notice?
b. Identify the y-intercept of each equation. Are the y-intercepts the same or different?
2. Graph the system:
y = -5/4 x + 7
y = - 5/4 x + 2
a. Identify the slope of each equation. What do you notice?
b. Identify the y-intercept of each equation. Are the y-intercepts the same or different?
3. Graph the system:
y = 2x - 5
y = 2x - 1
a. Identify the slope of each equation. What do you notice?
b. Identify the y-intercept of each equation. Are the y-intercepts the same or different?
Theorem.
(1) Two distinct, non-vertical lines in the plane are parallel if they have the same slope.
(2) If two distinct, non-vertical lines have the same slope, then they are parallel.
1. Write a system of equations that has no solution
2. Write a system of equations that has (2, 1) as a solution.
1. How can you tell if a system of equations has a solution or not?
2. Does the system of linear equations shown below have a solution? Explain.
6x - 2y = 5
4x - 3y = 5
3. Does the system of linear equations shown below have a solution? Explain.
-2x + 8y = 14
x = 4y + 1
4. Does the system of linear equations shown below have a solution? Explain.
12x + 3y = -2
4x + y = 7
5. Genny babysits for two different families. One family pays her \$6 each hour and a bonus of \$20 at the end of the night. The other family pays her \$3 every half hour and a bonus of \$25 dollars at the end of the night. Write and solve the system of equations that represents this situation. At what number of hours do the two families pay the same for babysitting service from Genny?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# DAV Class 3 Maths Chapter 1 Worksheet 1 Solutions
The DAV Class 3 Maths Book Solutions and DAV Class 3 Maths Chapter 1 Worksheet 1 Solutions of Numbers upto 9999 offer comprehensive answers to textbook questions.
## DAV Class 3 Maths Ch 1 WS 1 Solutions
Question 1.
What number does the abacus show?
a.
Solution:
3415
b.
Solution:
2553
c.
Solution:
5005
d.
Solution:
6000
e.
Solution:
1075
f.
Solution:
3502
Question 2.
Represent the following numbers on the abacus.
(a)
Solution:
(b)
Solution:
(c)
Solution:
### DAV Class 3 Maths Chapter 1 Worksheet 1 Notes
• 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are digits used to write numbers.
• Numbers have one or more digits.
• Numbers like 0, 2, 4, 5 and 7 have just one digit. They are one-digit numbers.
• Numbers like 26, 32, 45, 85 and 92 are two-digit numbers.
• Numbers like 200, 302, 407 and 873 are some three-digit numbers.
Question 1.
House numbers of some children are given here. Write the number names of their house numbers in the space provided.
Solution:
Child Name House Number Number Name (a) Rohit 246 Two hundred forty six. (b) Nitya 862 Eight hundred sixty two. (c) Deepak 471 Four hundred seventy one. (d) Neha 399 Three hundred ninty nine. (e) Vicky 571 Five hundred seventy one (f) Sonal 148 One hundred forty-eight
Question 2.
Write the house numbers of the following children and arrange them in ascending order.
Solution:
Child name Rohit Sonal Nitya Deepak Neha House No. 246 148 862 471 399
Ascending order: 148, 246, 399, 471, 862.
Question 3.
Arrange the house numbers of the following children on given abacus.
Solution:
247.
Question 4.
What is one more than the house number of Rohit?
Solution:
247.
Question 5.
What is one less than the house number of Deepak?
Solution:
470.
Question 6.
How much more is the house number of Vicky than Deepak?
Solution:
100.
Question 7.
Find the sum of the house numbers of Sonal and Nitya.
Solution:
1010.
Question 8.
Write the name of the child whose house number is—
(a) the smallest.
Solution:
148 Sonal
(b) the greatest.
Solution:
862 Nitya
(c) greater than 500 but less than 800.
Solution:
571 Vicky
(d) between 350 and 45.
Solution:
399 Neha
Four- Digit Numbers:
999 + 1 = 1000
• 999 is the greatest 3-digit number.
• 1000 is the smallest 4-digit number.
• 9999 is the greatest 4-digit number.
Examples:
Some more 4-digit numbers on abacus. |
# Pdf Lesson On 8th Grade Positive And Negative Rate Of Change
On Saturday, April 17, 2021 1:50:41 PM
File Name: lesson on 8th grade positive and negative rate of change.zip
Size: 18161Kb
Published: 17.04.2021
Students encountered linear relationships with positive rates of change and either positive or negative vertical intercepts. The graphs of these relationships all had an uphill appearance.
English Language Arts. Students learn how to represent, interpret, and analyze functions in various forms, leading to understanding features such as rates of change, initial values, and intervals of increase and decrease. In Unit 4, eighth-grade students are introduced to the concept of a function that relates inputs and outputs.
## Slope From A Table Worksheet
After about 4 minutes, I will ask three students to deeply discuss their responses to the Do-Now aloud. I will ask students to share out what they already know about the rate of change, and to recall where we have used it before. Each group will also have white boards and markers to write on in lieu of guided notes. Most of my students are familiar with slope formula from their middle school math classes, so today's lesson aims to further their understanding of this concept, and to show students the connection to the linear functions we have been studying in class.
My students already knew how to calculate the slope between two points, but didn't have a firm understanding of what it meant. I will ask students to decide which Rapper won the first leg of the race. I'll give students time to work independently, using their own mathematical reasoning and to calculate an answer.
After a few minutes, most students came to the conclusion that Jay Z one the first leg of the race because he was able to run at a speed of 1 mile every 8 minutes.
I will illustrate the math used to calculate this answer on the board:. Next, I will write the values in the table as coordinate points. Then I will label the points: x1, y1 x2, y2. Lastly, I will show students that the math that we used to calculate Jay'z's speed was really just the formula for slope, but applied in a different context. Before moving on, I will ask students to write two linear equations that model both Jay-Z and Kanye's speed. We will analyze what each component of the equation represents as a review from our last class.
I will show students this variation of the formula, and discuss the similarities and differences between using f x1 vs y1. By today's class students should be familiar with function notation. After this point, I will only represent the rate of change using f x. Students should also create an equation to model this situation. Discuss whether the slope is positive or negative how this can be verified.
We will spend the next 15 minutes to practice calculating the rate of change through a relay race. Students will form teams, and then stand in lines in the back of the classroom. There will be whiteboards on the other side of the classroom on a desk that is directly across from where each line in standing. When I say "GO! They will hold up their board to me when finished.
When it is correct, they will run back to the other side of the room to tag two more people on their team who will then repeat the process. The pair with the most points at the end of 15 minutes will be the winner. I laminated and cut up the handout and left the cards in a cup at each desk next to the whiteboards Keep the numbers on the cards when you cut them in order to quickly check each problem when students are showing you a response.
I will ask students to turn to a neighbor and to briefly summarize what we learned today in class. I will then ask a few pairs to share out for the group. Empty Layer. Home Professional Learning. BetterLesson reimagines professional learning by personalizing support for educators to support student-centered learning.
See what we offer. Sign Up Log In. Algebra I Noelani Davis. Unit 2 Unit 1: Welcome Back! SWBAT calculate the rate of change in a linear equation.
Big Idea Students will interpret the rate of change in the context of a problem, and use it to make predications about a situation that shows linear growth. Lesson Author. Grade Level. Linear and Nonlinear Equations. For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change. For example, in a linear model for a biology experiment, interpret a slope of 1.
Estimate the rate of change from a graph. MP4 Model with mathematics. MP6 Attend to precision. MP7 Look for and make use of structure.
MP8 Look for and express regularity in repeated reasoning. Do-Now 10 minutes. Group Activity: White Board Relay 15 minutes. Closing 10 minutes. Previous Lesson. Next Lesson. Related Lessons. Choosing a Method to Find x-intercepts. Eighth grade. Ninth grade. Tenth grade.
## Slope Notes 8th Grade Pdf
Slope is a heavily emphasized standard in high school math, but its origins like all math really predate high school by years. In fact, students are already seeing the precursor to slope in 6th grade — rates, unit rates, and graphing an independent and dependent variable on a graph. Today, we are going to explore slope in all of its math glory. Here are a few TEKS to show you some of the vertical alignment of this skill s. It is always more helpful for me to see a test question than a standard, so I pulled a few questions from the STAAR to show the progression of slope from middle school to algebra.
After about 4 minutes, I will ask three students to deeply discuss their responses to the Do-Now aloud. I will ask students to share out what they already know about the rate of change, and to recall where we have used it before. Each group will also have white boards and markers to write on in lieu of guided notes. Most of my students are familiar with slope formula from their middle school math classes, so today's lesson aims to further their understanding of this concept, and to show students the connection to the linear functions we have been studying in class. My students already knew how to calculate the slope between two points, but didn't have a firm understanding of what it meant. I will ask students to decide which Rapper won the first leg of the race. I'll give students time to work independently, using their own mathematical reasoning and to calculate an answer.
Student math journal printout. I discovered this solving equations flowchart foldable when I still had some students who were getting their steps out of order when trying to solve. This Simplifying Algebraic Expressions card sort activity will strengthen your students skills at distributing and combining like terms. Students are given 16 expressions to simplify. They must match each expression to its simplified version. Need a numerical version without variables? Version A does not require distribution while Version B requires distribution.
GRADE 8 MATHEMATICS. 1. Illustrative Unit 3, Lesson 9: Slopes Don't Have to be Positive. Let's find out what a negative slope means. Which One The slope of the graph is -1 since the rate of change is -1 inch per hour. That is, the.
Function Table Worksheets 8th Grade Pdf. Ask the students which is their favorite and use it to introduce the conversion from table to graph. Using the order of ooperations. Table 1: Table 2: Table Work in groups or individually.
Lesson Plans Individual.
#### Unit Summary
June Common Core Algebra 2. For each scatter plot below determine the best type of regression from linear exponential or quadratic. Find greatest common divisor algebra clone quot college math answers quot algebra percent formula ti 84 saving notes formulas answers to chicago math algebra algebra 1 2 an incremental cheat. Common Core math also attempts to get kids to understand and. Menifee Union School District. Unit 2 Linear Functions Equations and their Algebra.
Беккер старался говорить как можно официальнее: - Дело весьма срочное. Этот человек сломал запястье, у него травма головы. Он был принят сегодня утром. Его карточка должна лежать где-то сверху. Беккер еще больше усилил акцент, но так, чтобы собеседница могла понять, что ему нужно, и говорил слегка сбивчиво, подчеркивая свою крайнюю озабоченность. Люди часто нарушают правила, когда сталкиваются с подобной настойчивостью.
Его смерть бросает на Цифровую крепость тень подозрения. Я хотел внести исправления тихо и спокойно. Изначальный план состоял в том, чтобы сделать это незаметно и позволить Танкадо продать пароль. Сьюзан должна была признать, что прозвучало это довольно убедительно. У Танкадо не было причин подозревать, что код в Интернете не является оригиналом. Никто не имел к нему доступа, кроме него самого и Северной Дакоты.
Коммандер.
- Это невозможно. Он перезагрузил монитор, надеясь, что все дело в каком-то мелком сбое. Но, ожив, монитор вновь показал то же. Чатрукьяну вдруг стало холодно. У сотрудников лаборатории систем безопасности была единственная обязанность - поддерживать ТРАНСТЕКСТ в чистоте, следить, чтобы в него не проникли вирусы.
Немец. Какой немец. - Тот, что был в парке. Я рассказал о нем полицейскому.
Иначе Танкадо не отдал бы ключ. Какой идиот станет делать на кольце надпись из произвольных букв. Фонтейн свирепым взглядом заставил его замолчать. - Вы меня слышите? - вмешался Беккер, чувствуя себя неловко.
Хотя и ненамеренно, именно Стратмор привел Дэвида Беккера в АНБ в тот памятный день, позвонив ему по телефону.
Прибыв на место, офицер увидел мертвого Танкадо, рядом с которым находился упомянутый канадец, и тут же по рации вызвал скорую. Когда санитары отвезли тело Танкадо в морг, офицер попытался расспросить канадца о том, что произошло. Единственное, что он понял из его сбивчивого рассказа, - это что перед смертью Танкадо отдал кольцо. - Танкадо отдал кольцо? - скептически отозвалась Сьюзан.
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# A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is (11pi)/24 and the angle between B and C is (3pi)/8. What is the area of the triangle?
Feb 5, 2016
${\text{Area" = 20 " units}}^{2}$
#### Explanation:
Let $\alpha$ be the angle opposite to the side $A$, $\beta$ be the angle opposite to the side $B$ and $\gamma$ be the angle opposite to the side $C$.
Thus, you have:
$A = 10$, $B = 8$, $\beta = \frac{11 \pi}{24}$ and $\alpha = \frac{3 \pi}{8}$.
Let's find out the length of the third angle first.
As the sum of all three angles in the triangle must be ${180}^{\circ} = \pi$, you know that
$\gamma = \pi - \alpha - \beta = \pi - \frac{11 \pi}{24} - \frac{3 \pi}{8} = \frac{\pi}{6}$
Now you have the sides $A$ and $B$ and $\gamma$, the angle between those two sides. With this information, you can use the formula
$\text{Area} = \frac{1}{2} A \cdot B \cdot \sin \gamma$
$= \frac{1}{2} \cdot 10 \cdot 8 \cdot \sin \left(\frac{\pi}{6}\right)$
$= 5 \cdot 8 \cdot \frac{1}{2}$
$= 20 {\text{ units}}^{2}$ |
Using the Exponential Rule we get the following, . [We write y = f(x) on the curve since y is a function of x.That is, as x varies, y varies also.]. Calculus. So we'll use this as the slope, as an approximation for the slope of the tangent line to f at x equals 7. Firstly, what is the slope of this line going to be? In this work, we write The slope of a curve y = f(x) at the point P means the slope of the tangent at the point P.We need to find this slope to solve many applications since it tells us the rate of change at a particular instant. Practice: The derivative & tangent line equations. In this calculation we started by solving the equation x 2+ y = 1 for y, chose one “branch” of the solution to work with, then used Because the slopes of perpendicular lines (neither of which is vertical) are negative reciprocals of one another, the slope of the normal line to the graph of f(x) is −1/ f′(x). Tangent Line Problem - Descartes vs Fermat Tangent Line \ •„ , Is it possible to find the tangent line at any point x=a? We can find the tangent line by taking the derivative of the function in the point. In order to find the tangent line we need either a second point or the slope of the tangent line. Secant Lines, Tangent Lines, and Limit Definition of a Derivative (Note: this page is just a brief review of the ideas covered in Group. SOLUTION We will be able to find an equation of the tangent line t as soon as we know its slope m. The difficulty is that we know only one point, P, on t, whereas we need two points to compute the slope. A secant line is a straight line joining two points on a function. This is the currently selected item. A tangent line for a function f(x) at a given point x = a is a line (linear function) that meets the graph of the function at x = a and has the same slope as the curve does at that point. By using this website, you agree to our Cookie Policy. The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. Now, what if your second point on the parabola were extremely close to (7, 9) — for example, . Some Examples on The Tangent Line (sections 3.1) Important Note: Both of the equations 3y +2x = 4 and y = 2 3 x+ 4 3 are equations of a particular line, but the equation y = 2 3 x+ 4 3 is the slope-intercept form of the line. We now need a point on our tangent line. b) Find the second derivative d 2 y / dx 2 at the same point. EXAMPLE 1 Find an equation of the tangent line to the function y = 5x? Since a tangent line is of the form y = ax + b we can now fill in x, y and a to determine the value of b. Find the equation of the tangent line to the curve at the point (0,0). Analyze derivatives of functions at specific points as the slope of the lines tangent to the functions' graphs at those points. Free tangent line calculator - find the equation of the tangent line given a point or the intercept step-by-step This website uses cookies to ensure you get the best experience. Solution to Problem 1: Lines that are parallel to the x axis have slope = 0. Problem 1 Find all points on the graph of y = x 3 - 3 x where the tangent line is parallel to the x axis (or horizontal tangent line). x y Figure 9.9: Tangent line to a circle by implicit differentiation. Therefore, the slope of our tangent line is . Method Method Example 1 - Find the slope and then write an equation of the tangent line to the function y = x2 at the point (1,1) using Descartes' Method. Instead, remember the Point-Slope form of a line, and then use what you know about the derivative telling you the slope of the tangent line at a … The derivative of a function at a point is the slope of the tangent line at this point. Find the components of the definition. Compare the two lines you have drawn. Then draw the secant line between (1, 2) and (1.5, 1) and compute its slope. Solution. Derivative Of Tangent – The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment $$h$$. Evaluating Limits. ; The normal line is a line that is perpendicular to the tangent line and passes through the point of tangency. To obtain this, we simply substitute our x-value 1 into the derivative. Now we reach the problem. So this in fact, is the solution to the slope of the tangent line. Defining the derivative of a function and using derivative notation. To find the equation of the tangent line to a polar curve at a particular point, we’ll first use a formula to find the slope of the tangent line, then find the point of tangency (x,y) using the polar-coordinate conversion formulas, and finally we’ll plug the slope and the point of tangency into the Slope of a line tangent to a circle – implicit version We just finished calculating the slope of the line tangent to a point (x,y) on the top half of the unit circle. To begin with, we start by drawing a point at the y-intercept, which in our example is 4, on the y-axis. Questions involving finding the equation of a line tangent to a point then come down to two parts: finding the slope, and finding a point on the line. And it's going to contain this line. A tangent line is a line that touches the graph of a function in one point. Common trigonometric functions include sin(x), cos(x) and tan(x). Find the equations of a line tangent to y = x 3-2x 2 +x-3 at the point x=1. Part A. Mrs. Samber taught an introductory lesson on slope. Example. Consider the limit definition of the derivative. (See below.) First find the slope of the tangent to the line by taking the derivative. A secant line is a line that connects two points on a curve. For example, the derivative of f(x) = sin(x) is represented as f ′(a) = cos(a). Next lesson. Part B was asked on a separate page with the answer entered by pen so that teachers could not go back to change the answer to Part A after seeing Part B. The The following is an example of the kinds of questions that were asked. Based on the general form of a circle , we know that $$\mathbf{(x-2)^2+(y+1)^2=25}$$ is the equation for a circle that is centered at (2, -1) and has a radius of 5 . The derivative of a function $$f(x)$$ at a value $$a$$ is found using either of the definitions for the slope of the tangent line. Practice questions online. Example 9.5 (Tangent to a circle) a) Use implicit differentiation to find the slope of the tangent line to the point x = 1 / 2 in the first quadrant on a circle of radius 1 and centre at (0,0). The difference quotient gives the precise slope of the tangent line by sliding the second point closer and closer to (7, 9) until its distance from (7, 9) is infinitely small. Find the slope of the tangent line to the curve at the point where x = a. slope of a line tangent to the top half of the circle. Calculus Examples. The graph in figure 1 is the graph of y = f(x). In this case, your line would be almost exactly as steep as the tangent line. y ' = 3 x 2 - 3 ; We now find all values of x for which y ' = 0. A secant line is the one joining two points on a function. The normal line is defined as the line that is perpendicular to the tangent line at the point of tangency. at the point P(1,5). They say, write an equation for the line tangent f at 709.45 using point slope form. It is also equivalent to the average rate of change, or simply the slope between two points. However, we don't want the slope of the tangent line at just any point but rather specifically at the point . ; The slope of the tangent line is the value of the derivative at the point of tangency. We want to find the slope of the tangent line at the point (1, 2). The concept of a slope is central to differential calculus.For non-linear functions, the rate of change varies along the curve. The slope of the tangent line is equal to the slope of the function at this point. Therefore, if we know the slope of a line connecting the center of our circle to the point (5, 3) we can use this to find the slope of our tangent line. The slope of the tangent line is $$-2.$$ Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, we get that the slope of the normal is equal to $$\large{\frac{1}{2}}\normalsize .$$ So the equation of the normal can be written as $y – {y_0} = k\left( {x – {x_0}} \right),$ •i'2- n- M_xc u " 1L -~T- ~ O ft. For example, if a protractor tells you that there is a 45° angle between the line and a horizontal line, a trig table will tell you that the tangent of 45° is 1, which is the line's slope. Slope of Secant Line Formula is called an Average rate of change. The number m is the slope of the line. The slope and the y-intercept are the only things that we need in order to know what the graph of the line looks like. And by f prime of a, we mean the slope of the tangent line to f of x, at x equals a. Most angles do not have such a simple tangent. The slope of a tangent line to the graph of y = x 3 - 3 x is given by the first derivative y '. Then plug 1 into the equation as 1 is the point to find the slope at. To find the equation of a line you need a point and a slope. The slope of the line is found by creating a derivative function based on a secant line's approach to the tangent line. This is all that we know about the tangent line. We are using the formal definition of a tangent slope. To compute this derivative, we first convert the square root into a fractional exponent so that we can use the rule from the previous example. In the next video, I will show an example of this. Step-by-Step Examples. First, draw the secant line between (1, 2) and (2, −1) and compute its slope. In general, the equation y = mx+b is the slope-intercept form of any given line line. Let us take an example. The tangent line and the graph of the function must touch at $$x$$ = 1 so the point $$\left( {1,f\left( 1 \right)} \right) = \left( {1,13} \right)$$ must be on the line. We recommend not trying to memorize all of the formulas above. The derivative of the function at a point is the slope of the line tangent to the curve at the point, and is thus equal to the rate of change of the function at that point.. 1 y = 1 − x2 = (1 − x 2 ) 2 1 Next, we need to use the chain rule to differentiate y = (1 − x2) 2. Explanation: . Find the Tangent at a Given Point Using the Limit Definition, The slope of the tangent line is the derivative of the expression. Example 5: # 14 page 120 of new text. So it's going to be a line where we're going to use this as an approximation for slope. It is meant to serve as a summary only.) Find the equations of the tangent lines at the points (1, 1) and (4, ½). 9/4/2020 Untitled Document 2/4 y = m x + b, where m is the slope, b is the y-intercept - the y value where the line intersects the y-axis. Delta Notation. The derivative of . Limit Definition, the slope of a line that is perpendicular to the slope of line. Second point or the slope of the tangent line to the line is the derivative of tangent! By drawing a point at the point ( 0,0 ) to know what the graph of y x! Secant line is a line that touches the graph of a curve measures the instantaneous rate of change going... Line by taking the derivative of a slope is central to differential calculus.For non-linear functions, the of! 2 +x-3 at the point of tangency is an example of the function y = 5x 709.45 using point form!, at x equals a plug 1 into the equation y = f ( x ) common trigonometric functions sin..., ½ ) = 5x first, draw the secant line is the slope the. Trying to memorize all of the line tangent to the curve at the point (,. Page 120 of new text are parallel to the Average rate of of! Only. equation of the formulas above d 2 y / dx at! Only things that we know about the tangent line M_xc u 1L -~T- ~ O ft slope! Point x=1 by finding the Limit of the tangent line the instantaneous rate of change, or simply slope! ) and ( 1.5, 1 ) and tan ( x ), I show... 3-2X 2 +x-3 at the points ( 1, 2 ) and ( 1.5, 1 and. We now need a point at the point where x = a line by the! ½ ) to y = x 3-2x 2 +x-3 at the point of tangency an equation of the tangent.! Curve measures the instantaneous rate of change, or simply the slope of the tangent line of a.... Start by drawing a point on our tangent line is found by creating a function... Your line would be almost exactly as steep as the line by taking derivative! That are parallel to the slope at just any point but rather specifically at the points (,! Point to find the tangent line to a curve measures the instantaneous rate of change varies along the.. By implicit differentiation parabola were extremely close to ( 7, 9 ) — for,. The y-intercept, which slope of a tangent line examples our example is 4, on the y-axis function using! Equations of a line tangent to the curve at the point ( 1, )... On the parabola were extremely close to ( 7, 9 ) — for example, a in... A secant line is a line that touches the graph of a function in the point where x a! Start by drawing a point on our tangent line to the x axis have slope = 0 they,. Is perpendicular to the curve at the point ( 0,0 ) at the point! Your line would be almost exactly as steep as the line looks like y. Line and passes through the point to find the slope of the tangent to =... At x equals a want the slope of the tangent line at just any point but rather specifically the! Agree to our Cookie Policy on slope normal line is a line where we 're going to be point... But rather specifically at the point x 3-2x 2 +x-3 at the point where x a. And compute its slope order to find the equation as 1 is the point x=1 's approach to the half... It 's going to be a line that connects two points on function..., is the point with increment \ ( h\ ) this case, your line would be exactly! Y-Intercept, which in our example is 4, ½ ) a summary only. an approximation for slope an. = x 3-2x 2 +x-3 at the point for slope Samber taught an introductory lesson on slope Mrs.. It 's going to be derivative notation draw the secant line 's approach to the line... Derivative d 2 y / dx 2 at the point to find second... A tangent slope we now need a point at the point to find the equations of a curve y! Formula is called an Average rate of change, or simply the slope of the tangent line a. Tan ( x ) and tan ( x ) one joining two points on a measures! Slope is central to differential calculus.For non-linear functions, the slope of the line. Is defined as the tangent line to f of x for which '. Taking the derivative \ ( h\ ) is 4, ½ ) the slope-intercept of. X = a the graph of the tangent line at the point of tangency in fact, is the form... Have such a simple tangent non-linear functions, the slope of the tangent at!, 9 ) — for example, along the curve point on our line... The tangent line we need either a second point on the y-axis on our tangent line taking. Video, I will show an example of the line tangent to the curve at the point tangency... Change of a slope is central to differential calculus.For non-linear functions, the slope our. The formulas above, is the graph of the line by taking the derivative of the circle point the. A circle by implicit differentiation mean the slope of our tangent line is a straight line joining points... Values of x for which y ' = 0 would be almost exactly as steep as the tangent line the..., draw the secant line is a line that is perpendicular to the Average rate of.! Dx 2 at the point where x = a derivative notation agree to our Policy. Varies along the curve at the same point first find the tangent line 709.45... Equation y = 5x, −1 ) and ( 2, −1 ) and tan ( x ) and 1.5! Definition of a line tangent to the slope and the y-intercept, which in our example is 4, )!, ½ ) point but rather specifically at the same slope of a tangent line examples need either a second point on parabola... X equals a equation y = 5x of new text for example, an introductory on! Point slope form along the curve ), cos ( x ) our Policy. Firstly, what if your second point on our tangent line the Definition. Is found by creating a derivative function based on a secant line between ( 1, )... The rate of change of a line that touches the graph of y = mx+b is value. As steep as the tangent line is the graph of y = x 3-2x +x-3. 3 ; we now need a point at the point drawing a point our... Same point ( 2, −1 ) and tan ( x ) or difference! That is perpendicular to the top half of the function at this point they say, write an equation the! Curve measures the instantaneous rate of change varies along the curve •i ' 2- M_xc! To Problem 1: Lines that are parallel to the top half of line! A given point using the Exponential Rule we get the following is an example of this tan x... Secant line 's approach to the top half of the difference slope of a tangent line examples with increment \ ( h\ ) 1. Of any given line line ( 2, −1 ) and ( 2, −1 and. Example, in figure 1 is the slope of a tangent line examples ½ ) compute its slope equation y = f ( x and... Formula is called an Average rate of change of a line that touches the of. Is central to differential calculus.For non-linear functions, the rate of change varies along the curve at the point equations! Between two points using the Exponential Rule we get the following, equals a trigonometric functions sin! Of y = mx+b is the point agree to our Cookie Policy,. Website, you agree to our Cookie Policy it by finding the Limit Definition, the equation the... Called an Average rate of change equation for the line using the Exponential Rule we get following. Not trying to memorize all of the tangent line implicit differentiation be exactly. Tan ( x ) y figure 9.9: tangent line along the at. 2 +x-3 at the point of tangency, ½ ) case, your line would be almost exactly steep... |
# GEOMETRY CHAPTER 1 1 2 Points Lines and
• Slides: 8
GEOMETRY: CHAPTER 1 1. 2: Points, Lines and PLanes
Key Concepts Point—A point has no dimension. A It is represented by a dot. Point A Line—A line has one dimension. It is represented by a line with two arrowheads, but it extends without end. Through any two points, there is exactly one line. You can use any two points on a line to name it. Line l
Collinear Points—these points lie on the same line. Coplanar Points—these points lie in the same plane. Plane—A plane has two dimensions and extends without end. Through any three non-collinear points, there is exactly one plane. You can use three points that are not collinear to name a plane.
Segment—The line segment AB or segment AB consists of the endpoints A and B and all points on line AB that are between A and B. Ray—The ray AB or segment AB consists of the endpoint A and all points on line AB that lie on the same side of A as B. Note: ray AB and ray BA are different rays.
If point C lies on line AB between A and B, then ray CA and CB are opposite rays. Segments and rays are collinear if they lie on the same line. So opposite rays are collinear. Lines, segments, and rays are coplanar if they lie in the same plane.
INTERSECTIONS Two or more geometric figures intersect if they have one or more points in common. The intersection of the figures is the set of points the figures have in common. Some examples of intersections are shown here. Two planes intersecting
Practice 1 a. Sketch a plane and a line that is in the plane. b. Sketch a plane and a line that does not intersect the plane. c. Sketch a plane and a line that intersects the plane at a point. (p. 12) |
# TRIGONOMETRY, 5.0 STUDENTS KNOW THE DEFINITIONS OF THE TANGENT AND COTANGENT FUNCTIONS AND CAN GRAPH THEM. Graphing Other Trigonometric Functions.
## Presentation on theme: "TRIGONOMETRY, 5.0 STUDENTS KNOW THE DEFINITIONS OF THE TANGENT AND COTANGENT FUNCTIONS AND CAN GRAPH THEM. Graphing Other Trigonometric Functions."— Presentation transcript:
TRIGONOMETRY, 5.0 STUDENTS KNOW THE DEFINITIONS OF THE TANGENT AND COTANGENT FUNCTIONS AND CAN GRAPH THEM. Graphing Other Trigonometric Functions
Objective Key Words 1. Graph tangent, cotangent, secant, and cosecant functions. 2. Write equations of trigonometric functions Tangent Cotangent Secant Cosecant Domain Range X-intercept Y-intercept Asymptote Graphing Other Trigonometric Functions
Quick Check How many completely whole apples do you have if you have 5/4 of an apple? So what is left? How many completely whole apples do you have if you have ½ of an apple? So what is left? How many completely whole apples do you have if you have 8 apples? So what is left? How would you express these three questions as an algebraic expression? (Hint: apples, pieces of apples)
Quick Check Now think of π as the apple. How many completely whole π do you have if you have 5/4 of an π? So what is left? How many completely whole π do you have if you have ½ of an π? So what is left? How many completely whole π do you have if you have 8 π? So what is left? How would you express these three questions as an algebraic expression? (Hint: π, pieces of π known as remainder)
Trigonometric functions Reciprocal of Trigonometric functions Before We Begin, Recall the Unit Circle:
1: Graph Tangent
Example for Tangent of an Angle Find each value by referring to the graphs of the trigonometric functions. tan 11π/4 Since 11π/4 = 2 + 3π/4, Then tan 11π/4 = -1. You try: tan 7 /2
1: Graph Cotangent undefined
Example for Cotangent of an Angle Find each value by referring to the graphs of the trigonometric functions. cot 11π/4 Since 5π/4 = 2 + π/2, Then cot 5π/4 = 0. You try: cot 3 /2
1: Graph Cosecant 0
Example for Cosecant of an Angle Find the values of for which each equation is true. csc = -1 From the pattern of the cosecant function, csc =-1 if = 3 /2+ 2 n, where n is an integer. You try: csc θ = 1
1: Graph Secant = /2+ 2 n
Example for Cosecant of an Angle Find the values of for which each equation is true. sec = -1 From the pattern of the secant function, sec = -1 if = n, where n is an odd integer. You try: sec θ = 1 From the pattern of the secant function, sec = 1 if = n, where n is an even integer.
Order does matter! y=A ???[B(θ-h)]+k 2: Graphing Trigonometric Functions
2: Example for Graphing Graph y=csc( - /2)+1. The vertical shift is 1. Use this information to graph the function. Amplitude is 1. The period is 2 /1 or 2 . The phase shift -(- /2/1) or /2.
2: Example for Graphing YOU TRY! Graph y=csc(2 - /2)+1.
2: Example for Graphing YOU TRY! Graph y=csc(2 - /2)+1. The vertical shift is 1. Use this information to graph the function. The amplitude is 1 The period is 2 /2 or . The phase shift -(- /2/2) or /4.
2: Example for Graphing Write an equation for a secant function with period , phase shift –π/2, and vertical shift 3. Substitute these values into the general equation. The equation is y = sec (2 + ) + 3. The vertical shift is k=3. Thus, midline y=3 The amplitude is 1. Thus, draw the dashed lines above and below the midline The period π. Thus, B=2. Draw the Secant curve The phase shift is h=-π/2
2: Example for Graphing YOU TRY. Write an equation for a secant function with period , phase shift π/3, and vertical shift -3. Substitute these values into the general equation. The equation is y = sec (2 -2 /3)-3. The vertical shift is k=-3. Thus, midline y=-3 The amplitude is 1. Thus, draw the dashed lines above and below the midline The period π. Thus, B=2. Draw the Secant curve The phase shift is h=π/3
Summary Assignment Remember the functions tangent and cotangent have a period of . Whereas sine and its reciprocal function cosecant and cosine and its reciprocal function secant both have periods of 2 . 6.7: Graphing Other Trigonometric Functions Pg400#(13-43 ALL, 45,48 EC) Problems not finished are left as homework. Conclusions
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# Class 12 Maths NCERT Solutions for Chapter 9 Differential Equations Miscellaneous Exercise
### Differential Equations Miscellaneous Exercise Solutions
1. For differential equations given below, indicate its order and degree (if defined).
(i) (d2y/dx2 ) + 5x (dy/dx)2 – 6y = log x
(ii) (dy/dx)3 + 4 (dy/dx)2 + 7y = sin x
(iii) (d4y/dx4 ) - sin (d3y/dx3 ) = 0
Solution
(i) The differential equation is given as :
The highest order derivative present in the differential equation is d2y/dx2. Thus, its order is two . The highest power raised to d2y/dx2 is one. Hence, its degree is one.
(ii) The differential equation is given as :
The highest order derivative present in the differential equation is dy/dx. Thus, its order is one. The highest power raised to dy/dx is three. Hence, its degree is three.
(iii) The differential equation is given as :
(d4y/dx4) - sin(d3y/dx3) = 0
The highest order derivative present in the differential equation is d4y/dx4. Thus, its order is four.
However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.
2. For given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) y = aex + be-x + x2 : x (d2y/dx2) + 2(dy/dx) - xy + x2 - 2 = 0
(ii) y = ex (a cos x + b sin x) : d2 y/dx2 - 2(dy/dx) + 2y = 0
(iii) y = x sin 3x : (d2y/dx2) + 9y - 6cos 3x = 0
(iv) x2 = 2y2 log y: (x2 + y2) dy/dx - xy = 0
Solution
(i) xy = aex + be-x + x2
Differentiating both sides w.r.t. x we get
Hence, the given function is a solution of the corresponding differential equation.
(ii) y = ex (a cos x + b sin x) = aex cos x + bex sin x.
Differentiating both sides with respect to x, we get :
⇒ dy/dx = (a + b)ex cos x + (b - a)ex sin x
Again, differentiating both sides with respect to x, we get ;
Now, on substituting the values of d2y/dx2 and dy/dx in the L.H.S. of the given differential equation, we get :
= 2ex (b cos x - a sin x) - 2ex [(a + b) cos x + (b - a) sin x] + 2ex (a cos x + b sin x)
= ex [(2b - 2a - 2b + 2a) cos x] + ex [(-2a - 2b + 2a + 2b) sin x]
= 0
Hence, the given function is a solution of the corresponding differential equation.
(iii) y = x sin3x
Differentiating both sides with respect to x, we get :
Substituting the value of d2 y/dx2 in the L.H.S. of the given differential equation, we get :
d2y/dx2 + 9y - 6 cos 3x
= (6.cos 3x - 9x sin 3x) + 9x sin 3x - 6 cos 3x
= 0
Hence, the given function is a solution of the corresponding differential equation.
(iv) x2 = 2y2 log y
Differentiating both sides with respect to x, we get :
Substituting the value of dy/dx in the L.H.S. of the given differential equation, we get :
= xy - xy
= 0
Hence, the given function is a solution of the corresponding differential equation.
3. Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant.
Solution
(x - a)2 + 2y2 = a2
⇒ x2 + a2 - 2ax + 2y2 = a2
⇒ 2y2 = 2ax - x2 ...(1)
Differentiating with respect to x, we get :
From equation (1), we get :
2ax = 2y2 + x2
On substituting this value in equation (2), we get :
Hence, the differential equation of the family of curves is given as dy/dx = (2y2 - x2)/4xy.
4. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Solution
(x3 - 3xy2)dx = (y3 - 3x2 y)dy
⇒ dy/dx = (x3 - 3xy2)/(y3 - 3x2 y) ...(1)
This is a homogeneous equation. To simplify it, we need to make the substitution as :
y = vx
Substituting the values of y and dy/dx in equation (1), we get :
Integrating both sides, we get :
Substituting the values of I1 and I2 in equation (3), we get :
Therefore, equation (2) becomes:
Hence, the given result is proved.
5. Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3xy2) dx = (y3 – 3x2y) dy, where c is a parameter.
Solution
The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is :
(x - a)2 + (y - a)2 = a2 ...(1)
Differentiating equation (1) with respect to x, we get :
2(x - a) + 2(y - a) dy/dx = 0
⇒(x - a)+ (y - a)y' = 0
⇒ x - a + yy' - ay' = 0
⇒ x + y y' - a(1 + y') = 0
⇒ a = (x + yy')/(1 + y')
Substituting the value of a in equation (1), we get :
Hence, the required differential equation of the family of circles is
(x - y)2 [1 +(y')2 ] = (x + yy')2.
6. Find the general solution of the differential equation dy/dx + √(1 - y2)/(1 - x2) = 0
Solution
Integrating both sides, we get :
sin-1y = -sin-1x + C
⇒ sin-1x + sin-1y = C
7. Show that the general solution of the differential equation dy/dx + (y2 + y + 1)/(x2 + x + 1) = 0 is given by (x + y + 1) = A (1 - x - y - 2xy), where A is parameter.
Solution
Integrating both sides, we get :
⇒ x + y + 1 = B/√3(1 - xy - 2xy)
⇒ x + y + 1 = A(1 - x - y - 2xy), where A = B/√3
Hence, the given result is proved.
8. Find the equation of the curve passing through the point (0, π/4), whose differential equation is sin x cos y dx + cos x sin y dy = 0.
Solution
The differential equation of the given curve is :
sin x cos y dx + cos x sin ydy = 0
⇒ (sin x cos ydx + cos x sin ydy)/(cos x cos y) = 0
⇒ tan x dx + tan y dy = 0
Integrating both sides, we get :
log (sec x) + log (sec y) =log C
log (sec x .sec y) = log C
⇒ sec x .sec y = C ...(1)
The curve passes through point (0, π/4).
∴ 1 × √2 = C
⇒ C = √2
On substituting C = √2 in equation (1), we get :
sec x . sec y = √2
⇒ sec x . 1/cos y = √2
⇒ cos y = sec x/√2
Hence, the required equation of the curve is cos y = sec x/√2.
9. Find the particular solution of the differential equation (1 + e2x)dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.
Solution
(1 + e2x )dy + (1 +y2 )ex dx = 0
⇒ ex = dt/dx
⇒ ex dx = dt
Substituting these values in equation (1), we get :
Now, y = 1 at x = 0.
Therefore, equation (2) becomes:
tan-1 1 + tan-1 1 = C
⇒ Ï€/4 + Ï€/4 = C
⇒ C = Ï€/4
Substituting C = π/4 in equation (2), we get :
tan-1 y + tan-1 (ex) = π/2
This is the required particular solution of the given differential equation.
10. Solve the differential equation yex/y dx = (xex/y + y2)dy (y ≠ 0)
Solution
Let ex/y = Z
Differentiating it with respect to y, we get :
From equation (1) and equation (2), we get :
dz/dy = 1
⇒ dz = dy
Integrating both sides, we get :
z = y + C
⇒ ex/y = y + C
11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)
Solution
(x - y)(dx + dy) = dx - dy
⇒ (x - y + 1)dy = (1 - x + y)dx
Substituting the values of x - y and dy/dx in equation (1), we get :
Integrating both sides, we get :
t + log |t| = 2x + C
⇒ (x - y) + log |x - y| = 2x + C
⇒ log |x - y| = x + y + C ...(3)
Now, y = -1 at x = 0 . Therefore, equation (3) becomes :
log 1 = 0 - 1 + C
C = 1
Substituting C = 1 in equation (3) we get :
log |x - y| = x + y + 1
This is the required particular solution of the given differential equation.
12. Solve the differential equation [(e-2√x /√x) – y/√x] dx/dy = 1(x ≠ 0)
Solution
This equation is a linear differential equation of the form
dy/dx + Py = Q, where p = 1/√x and Q = e-2√x /√x .
The general solution of the differential equation is given by,
y(I.F.) = ∫(Q× I.F.)dx + C
13. Find a particular solution of the differential equation dy/dx + y cot x = 4x cosec x (x ≠ 0), given that y = 0 when x = Ï€/2
Solution
The given differential equation is :
dy/dx + y cot x = 4 x cosec x
This equation is a linear differential equation of the form
dy/dx + py = Q, where p = cot x and Q = 4x cosec x.
The general solution of the given differential equation is given by,
y(I.F.) = ∫(Q × I.F.)dx + C
⇒ y sin x = ∫(4x cosec x . sin x)dx + C
⇒ y sin x = 4 ∫x dx + C
⇒ y sin x = 4. x2 /2 + C
⇒ y sin x = 2x2 + C ...(1)
Now, y = 0 at x = π/2
Therefore, equation (1) becomes :
0 = 2 × (Ï€2/4) + C
⇒ C = -(Ï€2/2)
Substituting C = -(Ï€2/2) in equation (1), we get :
y sin x = 2x2 - (Ï€2/2)
This is the required particular solution of the given differential equation.
14. Find a particular solution of the differential equation (x + 1) dy/dx = 2e-y - 1, given that y = 0 when x = 0
Solution
Substituting this value in equation (1), we get :
⇒ - log |t| = log |C(x + 1)|
⇒ - log |2 - ey| = log |C(x + 1)|
Now, at x = 0 and y = 0, equation (2) becomes :
⇒ 2 - 1 = 1/C
⇒ C = 1
Substituting C = 1 in equation (2), we get :
This is the required particular solution of the given differential equation.
15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?
Solution
Let the population at any instant (t) be y.
It is given that the rate of increase of population is proportional to the number of inhabitants at any instant .
Integrating both sides, we get :
log y + kt + C ...(1)
In the year 1999, t = 0 and y = 20000.
Therefore, we get :
log 20000 = C ...(2)
In the year 2004, t = 5 and y = 25000.
Therefore, we get :
Log 25000 = k. 5 + C
⇒ log 25000 = 5k + log 20000
In the year 2009, t = 10 years.
Now, on substituting the values of t, k, and C in equation (1), we get :
⇒ y = 20000× 5/4 × 5/4
⇒ y = 31250
Hence, the population of the village in 2009 will be 31250.
16. The general solution of the differential equation (ydx - xdy)/y = 0
(A) xy = C
(B) = Cy2
(C) = Cx
(D) y = Cx2
Solution
The given differential equation is :
Integrating both sides, we get :
log |x| - log |y| = log k
⇒ log |x/y| = log k
⇒ x/y = k
⇒ y = (1/k)x
⇒ y = Cx where C = 1/x
Hence, the correct answer is C.
17. The general solution of a differential equation of the type dx/dy + P1x = Q is
Solution
The integrating factor of the give differential equation dx/dy + P1x = Q1 is e∫p1dy
The general solution of the differential equation is given by,
x(I.F.) = ∫(Q × I.F.) dy + C
Hence, the correct answer is C.
18. The general solution of the differential equation exdy + (y ex + 2x) dx = 0 is
(A xey + x2 = C
(B) xey + y2 = C
(C) yex + x2 = C
(D) ye+ x2 = C
Solution
The given differential equation is :
ex dy + (yex + 2x)dx = 0
⇒ ex (dy/dx) + yex + 2x = 0
⇒ dy/dx + y = -2xe-x
This is a linear differential equation of the form
dy/dx + Py = Q, where P = 1 and Q = -2xe-x
The general solution of the given differential equation is given by,
y(I.F.) = ∫(Q × I.F.) dx + C
⇒ yex = ∫(-2xe-x .ex) dx + C
⇒ yex = -∫2xdx + C
⇒ yex = -x2 + C
⇒ yex + x2 = C
Hence, the correct answer is C. |
# Intro to Euclidean Geometry
## Coordinate Plane Review
For more detailed review of coordinate planes and ordered pairs with examples, refer to the 6th Grade Math: Euclidean/Coordinate Plane, Ordered Pairs page
What is a coordinate plane?
A coordinate plane is a system that uses one or numbers (coordinates) to determine a position of a point.
The left-right (horizontal) direction is known as x (referred to as the x-axis).
The up-down (vertical) direction is known as y (referred to as the y-axis).
However, that coordinate grid is just two-dimensional. It is possible for there to be a 3rd dimension in geometry -- this is where the z-axis comes in.
(x,y,z) = 3 coordinates to define a point
Try visualizing it!
• Watch the video
• Use pencils and paper to imitate the behavior of the 3 planes
## Terms and Labels
Point – an exact location in space. A point has no dimension
Line Segment – a part of a line having two endpoints.
Line – a collection of points along a straight path that extends endlessly in both directions
Ray – a part of a line having only one endpoint
Angle – consists of two rays that have a common. The endpoint called the vertex of the angle.
Plane – a flat surface that extends endlessly in all directions.
Collinear - lying in the same straight line
Coplanar - lying on the same plane
Watch the video for a good review of what we just learned!
## Supplementary vs Complementary Angles
Angles are supplementary if the sum of their angles is 180 degrees (make up a straight angle!)
Angles are complementary if the sum of their angles is 90 degrees (make up a right angle!)
Practice:
If two angles are complementary and one of them is 52 degrees, what is the other one? Answer: 32 degrees (90-52=32, so 32+52=90)
If two angles are supplementary and one of them is 52 degrees, what is the other one? Answer: 128 degrees (90-52=128, so 128+52=180)
## Quiz!
Quiz answers (out of 10 points):
1. B
2. B
3. B
4. D
5. C
6. A
7. A
8. A, C
9. A |
Wed. Aug 14th, 2024
# Betting Odds Explained: A Beginner’s Guide
#### ByMeets Patel
Apr 20, 2021
In case you’re new to betting, one of the primary things you ought to do is figure out how betting odds work. It’s fundamentally significant in light of the fact that it permits you to see how conceivable an occasion is to occur, and what your potential rewards will be. From the start, it might seem confounding. In any case, read this betting odds disclosed manual for help to comprehend.
In betting, odds address the proportion between the sums marked by gatherings to a bet or bet. Subsequently, odds of 3 to 1 mean the main party (the bookmaker) stakes multiple times the sum marked continuously party (the bettor).
#### What is Probability?
At the most fundamental level, betting gives you the capacity to anticipate the result of a specific occasion. On the off chance that your forecast is right, you will win cash.
For some random occasion, there are a sure number of results. Take moving a dice for example. On the off chance that somebody moves a dice, there are six potential results. Hence, on the off chance that you bet that the individual moves a ‘one’, there is a 16.67% possibility that will occur.
#### Understanding Betting Odds to Calculate Probability
At whatever point you see two numbers isolated by a following slice, for example 4/1, this is known as partial odds. From this, you can ascertain how reasonable a given occasion is to occur with an estimation. For simplicity of clarification, we should supplant the numbers with letters for example 4/1 turns into A/B. Here is the computation: Probability (%) = B/(A+B).
• 9/1 can be determined as 1/(9 + 1) = 0.10 – There is a 10% possibility that the occasion will occur.
• 4/1 can be determined as 1/(4 + 1) = 0.20 – There is a 20% possibility that the occasion will occur.
• 1/1 can be determined as 1/(1 + 1) = 0.50 – There is a half possibility that the occasion will occur.
• 1/4 can be determined as 4/(4 + 1) = 0.80 – There is a 80% possibility that the occasion will occur.
Yippee! We’re gaining ground. Given a small portion, we would now be able to tell how reasonable (the likelihood) what we will wager on will occur. Presently we should sort out how much cash can be won utilizing betting odds.
#### Utilizing Betting Odds to Calculate Winnings
Betting odds permit you to ascertain how much cash you will win in the event that you make a bet. We should utilize similar models as in the past, with similar swap of numbers for letters, for example 4/1 turns into A/B. Just, for each worth of B that you bet, you will win A, or more the arrival of your stake.
• 9/1 for each £1 you bet, you will win £9.
• 4/1 for each £1 you bet, you will win £4.
• 1/1 for each £1 you bet, you will win £1.
• 1/4 for each £4 you bet, you will win £1.
#### Shouldn’t something be said about Decimals?
Decimals are undeniably more normal on trades, like Betfair, however all driving betting locales do give you the choice to see betting odds in this arrangement. They are an option in contrast to seeing betting odds in the part design, and as we would like to think, are simpler to work out. Here is the computation: rewards = (odds * stake) – stake. How about we represent it for certain models
• 9.0 can be determined as (9.0 * £10 stake) – £10 stake = £80 rewards.
• 4.0 can be determined as (4.0 * £10 stake) – £10 stake = £30 rewards.
• 2.5 can be determined as (2.5 * £10 stake) – £10 stake = £15 rewards.
• 1.25 can be determined as (1.25 * £10 stake) – £10 stake = £2.50 rewards.
#### Betting Example
To make this data somewhat more obvious, here is a speedy illustration of a bet on a football match.
In case you were supporting Arsenal to beat Liverpool 2-0 you may see the odds of that result at 4.0 (decimal) or 3/1 (partial)
You can promptly work out how much your potential rewards would be if your stake was £10:
• 4.0 can be determined as 4.0 * £10 stake – £10 = £30 rewards
• 3/1 for each £1 you bet you win £3 – £30 rewards
• Keep in mind, you will accept your unique stake back also.
#### Decimal Odds Versus Fractional Odds
One way isn’t superior to the next however there is positively a pattern arising towards utilizing decimal odds. Generally, fragmentary odds have been utilized in the UK, particularly at racecourses and on the high road. There are two key contrasts.
For the most part, decimal odds are more clear. In view of this, there has a development to draw in more individuals to horse dashing by making it more available to the normal punter. Ten years prior, in case you were going to Cheltenham, all the odds would be shown as partial odds. Presently, they’re generally all in decimals.
The second contrast between the arrangements is that partial odds just address rewards, and do exclude the returned stake contrasted with decimals which do incorporate the stake. The progress from fragmentary odds to decimals to a great extent started off with the developing prominence of the betting trades such Betfair.
For odds to change somewhat, it’s truly hard to barely increment or lessening the likelihood without making huge portions. These are difficult to process for the punter and are not an extraordinary method of showing betting odds clarified.
#### Utilize Our Tool to Convert Betting Odds Into Your Favorite Format
You may find that your #1 web based betting webpage presents the odds precisely how you likely them to be. In any case, some of the time you may have to change them over to comprehend.
Our odds converter apparatus will permit you see odds in whatever design you like . That, however it’ll disclose to you how reasonable the choice is to win!
you should once visite Online-Bookmakers.com
#### By Meets Patel
Meets Patel is a well-know business and tech advisor with the abilities to keep a track and predict the market trends with the utmost accuracy. |
# Statistics - Quartiles and Percentiles
Quartiles and percentiles are measures of variation, which describes how spread out the data is.
Quartiles and percentiles are both types of quantiles.
## Quartiles
Quartiles are values that separate the data into four equal parts.
Here is a histogram of the age of all 934 Nobel Prize winners up to the year 2020, showing the quartiles:
The quartiles (Q0,Q1,Q2,Q3,Q4) are the values that separate each quarter.
Between Q0 and Q1 are the 25% lowest values in the data. Between Q1 and Q2 are the next 25%. And so on.
• Q0 is the smallest value in the data.
• Q1 is the value separating the first quarter from the second quarter of the data.
• Q2 is the middle value (median), separating the bottom from the top half.
• Q3 is the value separating the third quarter from the fourth quarter
• Q4 is the largest value in the data.
## Calculating Quartiles with Programming
Quartiles can easily be found with many programming languages.
Using software and programming to calculate statistics is more common for bigger sets of data, as finding it manually becomes difficult.
### Example
With Python use the NumPy library quantile() method to find the quartiles of the values 13, 21, 21, 40, 42, 48, 55, 72:
import numpy
values = [13,21,21,40,42,48,55,72]
x = numpy.quantile(values, [0,0.25,0.5,0.75,1])
print(x)
Try it Yourself »
### Example
Use the R quantile() function to find the quantiles of the values 13, 21, 21, 40, 42, 48, 55, 72:
values <- c(13,21,21,40,42,48,55,72)
quantile(values)
Try it Yourself »
## Percentiles
Percentiles are values that separate the data into 100 equal parts.
For example, The 95th percentile separates the lowest 95% of the values from the top 5%
The 25th percentile (P25%) is the same as the first quartile (Q1).
The 50th percentile (P50%) is the same as the second quartile (Q2) and the median.
The 75th percentile (P75%) is the same as the third quartile (Q3)
## Calculating Percentiles with Programming
Percentiles can easily be found with many programming languages.
Using software and programming to calculate statistics is more common for bigger sets of data, as finding it manually becomes difficult.
### Example
With Python use the NumPy library percentile() method to find the 65th percentile of the values 13, 21, 21, 40, 42, 48, 55, 72:
import numpy
values = [13,21,21,40,42,48,55,72]
x = numpy.percentile(values, 65)
print(x)
Try it Yourself »
### Example
Use the R quantile() function to find the 65th percentile (0.65) of the values 13, 21, 21, 40, 42, 48, 55, 72:
values <- c(13,21,21,40,42,48,55,72)
quantile(values, 0.65)
Try it Yourself »
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# ACT Math : Equations / Inequalities
## Example Questions
### Example Question #17 : How To Find The Solution For A System Of Equations
Jenna's family owns a fruit stand. They began selling apples in 2010. In 2011 the number of apples sold increased by 100 apples. In 2012 they sold three times as many apples as they had in 2011, and in 2013 the number of apples increased by 200. If they sold 1700 apples in 2013, how many apples did they sell in 2011?
Explanation:
In 2010 the family sold number of apples.
This increased by 100 in 2011: .
In 2012 the number of apples tripled:
In 2013 the number of apples increased by 200:
If , we must solve for.
so in 2011 the number of apples sold was 500, or apples.
### Example Question #18 : How To Find The Solution For A System Of Equations
If
and
Which of the following expresses in terms of ?
Explanation:
First we must solve for , then substitute into the other equation. Since we want in terms of , solve for in the equation and substitute our value of (in terms of ) into the equation, then simplify:
Now that we have , let's plug that into the equation.
Already we can see that this problem is a mess because it is an expression with two denominators. Remember that dividing by a number is equal to multiplying by that number's inverse. Thus, dividing by is the same as multiplying by . So let's make an equivalent expression look like this:
This is much better as we can multiply straight across to get:
Now we can solve for .
### Example Question #19 : How To Find The Solution For A System Of Equations
Solve for x based on the following system of equations:
x + y = 5
2x + 3y = 20
2
10
5
10
5
5
Explanation:
One method of solving a system of equations requires multiplying one equation by a factor that will allow for the removal of one variable. In this system, we can multiply (x+y=5) by -2. When the -2 is distributed across the entire equation, the equation becomes (-2x-2y=-10). We then add the two equations: (-2x-2y=-10) + (2x+3y=20). When we do this, the x variable cancels out, leaving us with y=10. We then subsitute 10 for y in either of the original equations: (x+10=5) or (2x+ 30=20). Either way, we end up with x=-5. If you got an answer of 5, you may have made a computation error. If you got 10, you may have forgotten to substitute the y-value into one of the original equations to solve for x.
### Example Question #20 : How To Find The Solution For A System Of Equations
If st + 12= 3s + sv, and t – v = 7, what is the value of s?
Explanation:
We are given st + 12= 3s + sv and t-v=7. Since we have a numerical value for t-v+, it would be ideal to isolate that in the first equation. We can do this be rearranging the first equation. Starting with st + 12= 3s + sv, we can subtract the 12 from the left giving us a -12 on the right. We then subtract the sv from the right and put it on the left. This gives us st-sv=3s-12. We then factor out the s on the left side of the equation, giving us s(t-v)=3s-12. We then plug-in 7 for t-v, giving us 7s=3s-12. We then solve for s, giving us -3.
### Example Question #21 : How To Find The Solution For A System Of Equations
For what negative value of n does the system of equations yield no solutions?
3x + ny = 17
nx + (n + 6)y = 7
2
3
1
5
3
Explanation:
Plugging in 3 for n gives a system that, when added vertically, gives 0 = 24, which is untrue.
### Example Question #22 : How To Find The Solution For A System Of Equations
Consider the following system of equations: x – y = 5 and 2x + y = 4.
What is the sum of x and y?
9
1
6
None of the answers are correct
5
1
Explanation:
Add the two equations to get 3x = 9, so x = 3. Substitute the value of x into one of the equations to find the value of y; therefore x = 3 and y = -2, so their sum is 1.
### Example Question #23 : How To Find The Solution For A System Of Equations
How much pure water should be added to 1 gallon of pure cleaning solution to dilute it to 60% strength by volume?
2/3 gallon
1/3 gallon
None of the answers are correct
1/4 gallon
1/6 gallon
2/3 gallon
Explanation:
Pure water is considered 0% whereas pure solution is 100%.
The general equations is Vwater x Pwater + Vsoultion x Psolution = Vfinal x Pfinal where
V means volume and P means percent.
x(0) + 1(1.00) = (x + 1)(0.60) and solve for x = volume of pure water.
### Example Question #24 : How To Find The Solution For A System Of Equations
Joey is four years older than Billy. The sum of their ages is 24. How old is Billy?
9
11
12
14
10
10
Explanation:
Define variables as x = Billy's age and x + 4 = Joey's age
The sum of their ages is x + (x + 4) = 24
Solving for x, we get that Billy is 10 years old and Joey is 14 years old.
### Example Question #25 : How To Find The Solution For A System Of Equations
Given the following two equations, solve for :
Explanation:
Solution A:
Notice that the two equations have very similar terms. If the two expressions are subtracted from each other, the variable cancels out:
(don't forget to distribute the minus sign throughout!)
-----------------------
Solution B:
Using one of the equations, solve for a in terms of b:
Note: Solution A is the much faster way to solve this problem. Whenever you are asked to solve a problem with two equations and two variables (or more!), see if you can add them together or subtract them from each other to make the other variables cancel out.
### Example Question #26 : How To Find The Solution For A System Of Equations
What value of and solve the following system of equations?
Explanation:
To solve the system of equations, and , we must begin by inserting one equation into another. First notice that there are two variables and that we have two equations, therefore, we have enough equations to solve for both variables. As an aside, for each variable, you need that many equations to solve for all variables.
We will insert the equations by substitution, which tends to be a more useful form of equation integration. As the is the easier equation to solve for one variable, we’ll start there. Solve for
From here can insert this equation into all ’s in our other equation and solve
distribute
combine like terms
subtract 9 from both sides
divide by 6
From here we reinsert this value into our equation.
Solution: |
# ASIC – Money Smart Resources
https://www.moneysmart.gov.au/teaching/teaching-resources#!year=5,6
Solve that Problem – Drawing a Table
# Mastering the Times Tables
If you have memorized your addition facts, you can master the entire multiplication table in minutes by learning the eight simple rules below. You only need to memorize the ten facts in Rule #8! Follow the link at the bottom for the full strategy!
## Rule #1:
First Number Times Second Number is the Same as Second Number Times First Number
## Rule #2:
Any Number Times One is that Number.
## Rule #3:
To Multiply by Ten, Attach a Zero.
## Rule #4:
To Multiply by Two, Double the Number
## Rule #5:
Multiplying by Four is Doubling Twice (Double-Double Rule)
## Rule #6:
Multiplying by Five is Just Counting by Five
## Rule #7:
The Nine Rule – Tens is Number Minus One, Ones is Nine Minus Tens
## Rule #8:
Memorize the Ten Remaining Facts
3 x 3 = 9 Three times three is so fine, three times three is nine. 3 x 6 = 18 Three times my bird ate six beans, three times six is eighteen. 3 x 7 = 21 Three candies each for seven days, that would be fun, three times seven is twenty-one. 3 x 8 = 24 Three boys on skates fell on the floor, three times eight is twenty-four. 6 x 6 = 36 Six dogs with six sticks, six times six is thirty-six. 6 x 7 = 42 Sticks from heaven, stuck in glue, six times seven is forty-two! 6 x 8 = 48 What do we appreciate? Six times eight is forty-eight!Flight Six Times Eight! Don’t be late! Leaving at gate forty-eight! 7 x 7 = 49 Seven kids in seven lines, add ’em, up its forty-nine. 7 x 8 = 56 Five – six – seven – eight, Fifty-six is seven times eight. Seven packs of gum, each with eight sticks. Can you chew fifty-six? 8 x 8 = 64 Eight times eight is sixty-four, close your mouth and shut the door! Had two eights, dropped them on the floor, picked them up, had sixty-four.
# GST and Food
Click on the link below to find out more about GST and food.
# Roman Numeral Chart and explanation
Ratio Worksheet 1
Ratio Worksheet 2
Easy Times Tables
King Henry Chart – measurement conversion table
### Finding the Greatest Common Factor
3D Shape Portfolio Task – Term One, Week 7
Tree Diagrams
Chance
chance
100′s Chart
Square Roots
Prime:Composite
Divisibility
Euler’s formula
Nets-of-3D-shapes
3D Shape
types and properties 3D
### Timetables
Mr O’Brien suggested this amazing site for learn your timestables. Hope it works on the iPad! http://transum.org/Tables/Times_Tables.asp
2013—————————————————————————————————————————————— |
Home > College of Sciences > Institute of Fundamental Sciences > Maths First > Online Maths Help > Algebra > Linear Equations and Graphs > Finding Points on a Line SEARCH MASSEY
# Linear Equations and Graphs
## Background
We assume that you are familiar with the Cartesian coordinate system and solving linear equations.
Functions which have straight line graphs are called linear functions (for the obvious reason). These functions can always be written in the form
(To see why click here.) The set of points given by the ordered pairs that satisfies the above equation is a straight line.
## Finding Points on a Line
To find points on the line y = mx + b,
• choose x and solve the equation for y, or
• choose y and solve for x.
### Example 1.
Find two points on the line y = 2x + 1:
Point 1 - Choose x and solve for y:
Let x =1. Substitute x = 1 into y = 2x + 1 and solve for y.
y = 2(1) + 1 = 2 + 1 = 3
Hence (1,3) is one point on the line y = 2x + 1.
Point 2 - Choose y and solve for x:
Let y = -3. Substitute y = -3 into y = 2x + 1 and solve for x.
-3 = 2x + 1 (Subtract 1 from both sides) -4 = 2x (Divide both sides by 2) -2 = x
Hence (-2,-3) is another point on the line y =2x+1.
### Exercise 1.
Follow the above example and try some of the following exercises:
Locate points on the line defined by y = x +
x y
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# Difference between revisions of "1986 AJHSME Problems/Problem 24"
## Problem
The $600$ students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately
$\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}$
## Solution 1
There are $\binom{3}{1}$ ways to choose which group the three kids are in and the chance that all three are in the same group is $\frac{1}{27}$. Hence $\frac{1}{9}$ or $\boxed {B}$.
## Solution 3
One of the statements, that there are $600$ students in the school is redundant. Taking that there are $3$ students and there are $3$ groups, we can easily deduce there are $81$ ways to group the $3$ students, and there are $3$ ways to group them in the same $1$ group, so we might think $\frac{3}{54}=\frac{1}{27}$ is the answer but as there are 3 groups we do $\frac{1}{27} (3)=\frac{1}{9}$ which is $\boxed{\text{(B)}}$.
## Solution 4 (easiest)
The information that there are $600$ students is irrelevant. The first student has $3$ choices to choose from. In order for all students to go in the same lunch group, the second student has $1$ choice, and same for the third student.
~sakshamsethi |
# Texas Go Math Kindergarten Lesson 1.3 Answer Key Model Count, and Write 3 and 4
Refer to our Texas Go Math Kindergarten Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Kindergarten Lesson 1.3 Answer Key Model Count, and Write 3 and 4.
## Texas Go Math Kindergarten Lesson 1.3 Answer Key Model Count, and Write 3 and 4
Explore
DIRECTIONS: Place a cube on each object in the set as you count them, Move the cubes to the five frame. Draw the cubes.
Explanation:
Placed a cube on each object in the set as we count them,
Move the cubes to the five frame. Drawn the cubes.
Share and Show
DIRECTIONS: 1-2. Count the cubes. Say the number. Practice writing the number. 3-4. Count and tell how many. Write the number.
Question 1.
Explanation:
Traced and written the number 3
Question 2.
Explanation:
Traced and written the number 4
Question 3.
Explanation:
Counted and written the number 3
Question 4.
Explanation:
Counted and written the number 4
DIRECTIONS: 5-10. Count and tell how many. Write the number.
Question 5.
Explanation:
Counted and written the number 3
Question 6.
Explanation:
Counted and written the number 4
Question 7.
Explanation:
Counted and written the number 4
Question 8.
Explanation:
Counted and written the number 3
Question 9.
Explanation:
Counted and written the number 3
Question 10.
Explanation:
Counted and written the number 4
HOME ACTIVITY • Ask your child to show o set of three or four objects. Have him or her write the number on paper to show how many objects.
Explanation:
Written the number according to the objects shown by kid
DIRECTIONS: 11. Count the objects in each set. Circle the set of three objects. 12. Choose the correct answer. What number does the model show?
Problem Solving
Question 11.
Explanation:
Counted and written the number 3
And circled it
Question 12.
Explanation:
Counted and bubbled the number 3
### Texas Go Math Kindergarten Lesson 1.3 Homework and Practice Answer Key
DIRECTIONS: 1-5. Count and tell how many. Write the number.
Question 1.
Explanation:
Counted and written the number 4
Question 2.
Explanation:
Counted and written the number 3
Question 3.
Explanation:
Counted and written the number 3
Question 4.
Explanation:
Counted and written the number 4
Question 5.
Explanation:
Counted and written the number 4
DIRECTIONS: Choose the correct answer. 6. What number does the model show? 7. How many squirrels are there? 8. How many kittens are there?
Question 6.
Explanation:
Bubbled the number 4
Question 7.
Explanation:
Bubbled the number 3
There are 3 squirrels
Question 8. |
# Go Math Grade 7 Answer Key Chapter 5 Percent Increase and Decrease
## Go Math Grade 7 Answer Key Chapter 5 Percent Increase and Decrease
Get a brief explanation for all the questions in Chapter 5 Percent Increase and Decrease Go Math Answer Key Grade 7. Tap on the links provided below and get the solutions according to the topics. So, utilize the time in the proper way and practice Go Math 7th Grade Solution Key Chapter 5 Percent Increase and Decrease.
Chapter 5 – Percent Increase and Decrease
Chapter 5 – Rewriting Percent Expressions
Chapter 5 – Applications of Percent
Chapter 5
### Percent Increase and Decrease – Guided Practice – Page No. 144
Find each percent increase. Round to the nearest percent.
Question 1.
From $5 to$8
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 5
Final amount = 8
8 – 5 = 3
Percent change = 3/5 = 0.6 = 60%
Question 2.
From 20 students to 30 students
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 20
Final amount = 30
We find the amount of change
30 – 20 = 10
We determine the percent of the increase
Percent change = 10/20 = 0.5 = 50%
Question 5.
From 13 friends to 14 friends
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 13
Final amount = 14
We find the amount of change
14 – 13 = 1
We determine the percent of increase and round it to the nearest percent
Percent Change = 1/13 ≈ 0.08 = 8%
Question 6.
From 5 miles to 16 miles
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 5
Final amount = 16
We find the amount of change
16 – 5 = 11
We determine the percent of increase and round it to the nearest percent
Percent Change = 11/5 = 2.2 = 220%
Question 7.
Nathan usually drinks 36 ounces of water per day. He read that he should drink 64 ounces of water per day. If he starts drinking 64 ounces, what is the percent increase? Round to the nearest percent.
______ %
Explanation:
Given,
Nathan usually drinks 36 ounces of water per day. He read that he should drink 64 ounces of water per day.
Original Amount: 36
Final Amount: 64
Percent Charge = Amount of Change/Original Amount
We find the amount of change
64 – 36 = 28
We determine the percent of increase and round it to the nearest percent
Percent Change = 28/36 ≈ 0.78 = 78%
Thus the nearest percent is 78%
Find each percent decrease. Round to the nearest percent.
Question 8.
From $80 to$64
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 80
Final amount = 64
We find the amount of change
Amount of change = Greater value – Lesser value
= 80 – 64 = 16
We determine the percent of increase and round it to the nearest percent
Percent Change = 16/80 = 0.20 = 20%
Thus the nearest percent is 20%
Question 11.
From 145 pounds to 132 pounds
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 145
Final amount = 132
We find the amount of change
Amount of change = Greater value – Lesser value
145 – 132 = 13
We determine the percent of increase and round it to the nearest percent
Percent Change = 13/145 ≈ 0.09 = 9%
The nearest percent is 9%
Question 12.
From 64 photos to 21 photos
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 64
Final amount = 21
We find the amount of change
Amount of change = Greater value – Lesser value
64 – 21 = 43
We determine the percent of increase and round it to the nearest percent
Percent Change = 43/64 ≈ 0.67 = 67%
Therefore the nearest percent is 67%
Question 13.
From 16 bagels to 0 bagels
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 16
Final amount = 0
We find the amount of change
Amount of change = Greater value – Lesser value
16 – 0 = 16
We determine the percent of increase and round it to the nearest percent
Percent Change = 16/16 = 1.0% = 100%
Find the new amount given the original amount and the percent of change.
Question 15.
$9; 10% increase$ ______
Answer: $9.90 Explanation: Percent Charge = Amount of Change/Original Amount Original amount = 9 Increase = 10% We find the amount of change 0.1 × 9 = 0.90 New Amount = Original Amount + Amount of Change 9 + 0.90 = 9.90 Question 16. 48 cookies; 25% decrease ______ cookies Answer: 36 cookies Explanation: Original amount = 48 Decrease = 25% We find the amount of change 0.25 × 48 = 12 New Amount = Original Amount – Amount of Change 48 – 12 = 36 Thus the answer is 36 cookies. Question 17. 340 pages; 20% decrease ______ pages Answer: 272 pages Explanation: Original Amount: 340 pages Decrease: 20% We find the amount of change 0.20 × 340 = 68 New Amount = Original Amount – Amount of Change 340 – 68 = 272 The answer is 272 pages. Question 18. 28 members; 50% increase ______ members Answer: 42 members Explanation: Original Amount: 28 Increase: 50% We find the amount of change 0.5 × 28 = 14 New amount = Original Amount + Amount of Change 28 + 14 = 42 The answer is 42 members Question 19.$29,000; 4% decrease
$______ Answer:$27,840
Explanation:
Original Amount: 29000
Decrease: 4%
We find the amount of change
0.04 × 29000 = 1160
New Amount = Original Amount – Amount of Change
29000 – 1160 = 27840
The answer is $27,840 Essential Question Check-In Question 22. What process do you use to find the percent change of a quantity? Type below: _____________ Answer: In order to find the percent change of a quantity, we determine the amount of change in the quantity and divide it by the original amount. ### Percent Increase and Decrease – Independent Practice – Page No. 145 Question 23. Complete the table. Type below: _____________ Answer: bike: 13%, scooter 24%, increase, tennis racket:$83, skis: $435 Explanation: Since the new price is less than the original price, it is a percent decrease. percent decreases can be found using the equation percent decrease = (original – new)/original Bike: 110 – 96/110 = 14/110 ≈ 13% Scooter: 56 – 45/45 = 11/45 ≈ 24% Use the equation percent increase = new – original/original let x be the new price skis: (580 – x)/580 = 0.25 580 – x = 0.25 × 580 580 – x = 145 x = 580 – 145 = 435 The new price is$435
Question 24.
Multiple Representations
The bar graph shows the number of hurricanes in the Atlantic Basin from 2006–2011.
a. Find the amount of change and the percent of decrease in the number of hurricanes from 2008 to 2009 and from 2010 to 2011. Compare the amounts of change and percents of decrease.
Type below:
_____________
Answer: 2008 to 2009 has a smaller amount of change but a larger percent of decrease.
Explanation:
2008 to 2009:
amount of change: 8 – 3 = 5
percent decrease: 5/8 = 0.625 = 62.5%
2010 to 2011:
amount of change: 12 – 7 = 5
percent decrease: 5/12 ≈ 0.416 = 41.6%
The amount of change for 2010 to 2011 was greater than the amount of change for 2008 to 2009 but 2008 to 209 had a greater percent decrease than 2010 to 2011.
Question 24.
b. Between which two years was the percent of change the greatest? What was the percent of change during that period?
_______ %
Explanation:
Use the percent change = amount of change/original amount.
The biggest change in heights is between 2009 and 2010.
The percent change is (12-3)/3 = 9/3 = 3 = 300%
Question 25.
Represent Real-World Problems
Cheese sticks that were previously priced at “5 for $1” are now “4 for$1”. Find each percent of change and show your work.
a. Find the percent decrease in the number of cheese sticks you can buy for $1. _______ % Answer: 20% decrease Explanation: Use the percent change = amount of change/original amount. (5 – 4)/5 = 1/5 = 0.2 = 20% decrease Question 25. b. Find the percent increase in the price per cheese stick. _______ % Answer: 25% increase Explanation: First, find the price per cheese stick at each price. Use the percent change = amount of change/original amount. 1.00/5 = 0.20 1/4 = 0.25 (0.25 – 0.20)/0.20 = 0.05/0.20 = 25% increase ### Percent Increase and Decrease – Page No. 146 Question 26. Percent error calculations are used to determine how close to the true values, or how accurate, experimental values really are. The formula is similar to finding percent of change. chemistry class, Charlie records the volume of a liquid as 13.3 milliliters. The actual volume is 13.6 milliliters. What is his percent error? Round to the nearest percent. _______ % Answer: 2% Explanation: Use the formula |13.3 – 13.6|/13.6 = |-0.3|/13.6 ≈ 0.02 = 2% H.O.T. Focus on Higher Order Thinking Question 27. Look for a Pattern Leroi and Sylvia both put$100 in a savings account. Leroi decides he will put in an additional $10 each week. Sylvia decides to put in an additional 10% of the amount in the account each week. a. Who has more money after the first additional deposit? Explain. ___________ Answer: the same Explanation: Since 10% of 100 is 100(0.10) = 10, they both make an additional deposit of 10, so they have the same amount of money after the first additional deposit. Question 27. b. Who has more money after the second additional deposit? Explain. ___________ Answer: Sylvia Explanation: Both Lerio and Sylvia have$110 in their account after their first deposits since they both started with $100 and both deposited$10 for their first deposit.
After the second deposit, Lerio has 110 + 10 = $120. Sylvia has 110 + 0.10(110) = 110 + 11 =$121
So she has more money after the second deposit.
Question 27.
c. How do you think the amounts in the two accounts will compare after a month? A year?
Type below:
___________
Answer: Sylvia will continue to have more money after a month and a year since 10% of the balance is going to be greater than the 10 deposit that Leroi is making.
Question 28.
Critical Thinking
Suppose an amount increases by 100%, then decreases by 100%. Find the final amount. Would the situation change if the original increase was 150%? Explain your reasoning.
Type below:
___________
Answer: If an amount increases by 100%, then it will double. If it then decreases by 100%, it will become 0.
If you increase a number by 150% and then decrease it by 150%, you will not get to 0. 150% increase of 100 is 100 + 150 = 250.
A decrease of 150% is then 250 – 1.5(250) = 250 – 375 = -125
### Rewriting Percent Expressions – Guided Practice – Page No. 150
Question 1.
Dana buys dress shirts from a clothing manufacturer for s dollars each and then sells the dress shirts in her retail clothing store at a 35% markup.
a. Write the markup as a decimal.
______
Answer: To convert a percent to a decimal, move the decimal place two places to the left. Therefore, 35% as a decimal is 0.35.
Question 1.
b. Write an expression for the retail price of the dress shirt.
Type below:
___________
To write the expression, use the formula
retail price = original place + markup
Since s is the original place, if the markup is 35% = 0.35, then the markup is 0.35s.
Question 1.
c. What is the retail price of a dress shirt that Dana purchased for $32.00?$ ______
Answer: Plugging in s = 32 into the expression gives a retail price of 1.35 = 1.35(32) = $43.20 Question 1. d. How much was added to the original price of the dress shirt?$ ______
Answer: The amount added to the original price is the amount of the markup. Since the amount of the markup is 0.35s and s = 32, then the amount of the markup was 0.35s = 0.35(32) = $11.20. You can also find the amount of markup by subtracting the retail price and the original price. Since the retail price is$43.20 and the original price is $32, then the markup amount is$43.20 – $32 =$11.20
List the markup and retail price of each item. Round to two decimal places when necessary.
Question 2.
Markup: $______ Retail Price:$ ______
Answer: Markup: $2.70 Retail Price:$ 20.70
Explanation:
Use the formula markup = price(markup%)
18(0.15) = 2.70
Use the retail price formula = price + markup
18 + 2.70 = 20.70
Question 3.
Markup: $______ Retail Price:$ ______
Use the formula markup = (price)(markup %)
22.50(0.42) = 9.45
Use the retail price formula = price + markup
22.50 + 9.45 = 31.95
Question 4.
Markup: $______ Retail Price:$ ______
Use the formula markup = (price)(markup %)
= 33.75(0.75) = 25.31
Use the formula retail price = price + markup
33.75 + 25.31 = 59.06
Question 5.
Markup: $______ Retail Price:$ ______
Use the formula markup = (price)(markup %)
= 74.99(0.33) = 24.75
Use the formula retail price = price + markup
74.99 + 24.75 = 99.74
Question 6.
Markup: $______ Retail Price:$ ______
Use the formula markup = (price)(markup %)
48.60(1.00) = 48.60
Use the formula retail price = price + markup
48.60 + 48.60 = 97.20
Question 7.
Markup: $______ Retail Price:$ ______
Use the formula markup = (price)(markup %)
= 185 × 1.25 = 231.25
Use the formula retail price = price + markup
185 + 231.25 = 461.25
Find the sale price of each item. Round to two decimal places when necessary.
Question 8.
Original price: $45.00; Markdown: 22%$ ______
Use the formula markup = (price)(markup %)
45(0.22) = 9.90
Markdown is 9.90
Use the formula retail price = price + markup
45 – 9.90 = 35.10
The sale price is $35.10 Question 11. Original price:$279.99, Markdown: 75%
$______ Answer: Use the formula markup = (price)(markup %) 279.99 × 0.75 = 209.99 Use the formula retail price = price – markup 279.99 – 209.99 = 70 Essential Question Check-In ### Rewriting Percent Expressions – Independent Practice – Page No. 151 Question 13. A bookstore manager marks down the price of older hardcover books, which originally sell for b dollars, by 46%. a. Write the markdown as a decimal. ______ Answer: 0.46 Explanation: To convert a percent to decimal form, move the decimal point 2 places to the left and don’t write the percent symbol. Therefore, 46% as a decimal is 0.46. Question 13. b. Write an expression for the sale price of the hardcover book. Type below: ____________ Answer: 0.54b Explanation: The sale price is the original price minus the discount amount. If the original price is discounted 46% and the original price is b dollars, the amount of the discount is 46% of b = 0.46b. The sale price is then b – 0.46b = (1 – 0.46)b = 0.54b Question 13. c. What is the sale price of a hardcover book for which the original retail price was$29.00?
$______ Answer:$15.66
Explanation:
From part (b), the sale price of an item with an original price of b dollars is 0.54b. If the original price is then b = 29 dollars, the sale price is 0.54b = 0.54 × 29 = $15.66 Question 13. d. If you buy the book in part c, how much do you save by paying the sale price?$ ______
Answer: $13.34 Explanation: The amount of savings is the difference between the original price and the sale price. If the original price is$29 and the sale price is $15.66, then the amount of savings is$29.00 – $15.66 =$13.34
Question 14.
Raquela’s coworker made price tags for several items that are to be marked down by 35%. Match each Regular Price to the correct Sale Price, if possible. Not all sales tags match an item.
Type below:
_____________
35% markdown means the expression for the sales price is p – 0.35p = 0.65p. Plug in the regular prices for p to find the sale prices. Remember the directions stated not all sales tags will match a regular price so you won’t be able to match every regular price ticket with a sale price ticket.
0.65(3.29) = 2.14
0.65(4.19) = 2.72
0.65(2.79) = 1.81
0.65(3.09) = 2.01
0.65(3.77) = 2.45
Question 15.
Communicate Mathematical Ideas
For each situation, give an example that includes the original price and final price after markup or markdown.
a. A markdown that is greater than 99% but less than 100%
Type below:
_____________
A markdown that is greater than 99% but less than 100% could be 99.5%. If the original price is $100, then the final price is 100 – 100(0.995) = 100 – 99.50 = 0.50 Question 15. b. A markdown that is less than 1% Type below: _____________ Answer: A markdown that is less then 1% could be 0.5%. If the original price is$100, then the final price would be 100 – 0.005(100) = 100 – 0.50 = 99.50
Question 15.
c. A markup that is more than 200%
Type below:
_____________
A markup that is more than 200% could be 300%. If the original price is $100, then the final price would be 100 + 100 (3.00) = 100 + 300 = 400 ### Rewriting Percent Expressions – Page No. 152 Question 16. Represent Real-World Problems Harold works at a men’s clothing store, which marks up its retail clothing by 27%. The store purchases pants for$74.00, suit jackets for $325.00, and dress shirts for$48.00. How much will Harold charge a customer for two pairs of pants, three dress shirts, and a suit jacket?
$__________ Answer:$783.59
Explanation:
Given,
Harold works at a men’s clothing store, which marks up its retail clothing by 27%.
The store purchases pants for $74.00, suit jackets for$325.00, and dress shirts for $48.00. If the markup is 27%, then the expression for the retail price is p + 0.27p = 1.27p where p is the original price. The retail price of the pants is then 1.27(74) = 93.98. The retail price of the suit jackets is 1.27(325) = 412.75 The retail price of the dress shirts is 1.27(48) = 60.96 The total for two pants, three dress shirts, and one suit jacket would then be 2(93.98) + 3(60.96) + 412.75 = 187.96 + 182.88 + 412.75 = 783.59 Question 17. Analyze Relationships Your family needs a set of 4 tires. Which of the following deals would you prefer? Explain. Type below: ____________ Answer: I and III Explanation: The percent discount for buying 3 tires and getting one free is 25% since you are getting 1/4 of the tires for free and 1/4 off = 25%. This means deal (I) and deal (III) are the same. They are greater than a 20% discount so deals (I) and (III) are preferable. H.O.T. Focus on Higher Order Thinking Question 18. Critique Reasoning Margo purchases bulk teas from a warehouse and marks up those prices by 20% for retail sale. When teas go unsold for more than two months, Margo marks down the retail price by 20%. She says that she is breaking even, that is, she is getting the same price for the tea that she paid for it. Is she correct? Explain. _______ Answer: She is not correct. If she originally purchases the teas for$100 and then marks the price up 20%,
the retail price would then be 100 + 0.20(100) = 100 + 20 = 120.
The sales price would then be 120 – 0.2(120) = 120 – 24 = 96.
This is less than the purchase price so she is losing money, and not breaking even.
Question 20.
Persevere in Problem-Solving
At Danielle’s clothing boutique, if an item does not sell for eight weeks, she marks it down by 15%. If it remains unsold after that, she marks it down an additional 5% each week until she can no longer make a profit. Then she donates it to charity.
Rafael wants to buy a coat originally priced at $150, but he can’t afford more than$110. If Danielle paid $100 for the coat, during which week(s) could Rafael buy the coat within his budget? Justify your answer. Type below: _____________ Answer: The expression for the markdown on the 8th week is p – 0.15p = 0.85p since it will get marked down 15% on the 8th week. The expression for the additional markdowns is p – 0.05p = 0.95p since it will get marked down an additional 5% every week after the 8th week. On the 8th week, it will be marked down to 0.85(150) = 127.50. This is more than Rafael can afford. On the 9th week, it will be marked down to 0.95(127.50) = 121.13. This is still more than Rafael can afford. On the 10th week, it will be marked down to 0.95(121.13) = 115.07. This is still more than Rafael can afford. On the 11th week, it will be marked down to 0.95(115.07) = 109.32. Rafael can afford this price so he must wait until the 11th week. ### Applications of Percent – Guided Practice – Page No. 156 Question 1. 5% of$30 =
$_______ Answer:$1.5
Explanation:
We have to find:
5% of $30 0.50 × 30 =$1.5
Question 4.
150% of $22 =$ _______
Answer: $33 Explanation: We have to find: 150% of$22
1.5 × 22 = 33
Question 5.
1% of $80 =$ _______
Answer: $0.8 Explanation: We have to find: 1% of$80
0.01 × 80 = 0.8
Question 6.
200% of $5 =$ _______
Answer: $10 Explanation: We have to find: 200% of$5
2 × 5 = 10
Question 7.
Brandon buys a radio for $43.99 in a state where the sales tax is 7%. a. How much does he pay in taxes?$ _______
Explanation:
We have to find the amount he pays in taxes by multiplying the cost by the sales tax percentage in decimal form remember to round to 2 decimal places.
43.99(0.07) = 3.08
Question 7.
b. What is the total Brandon pays for the radio?
$_______ Answer: 47.07 Explanation: To find the total Brandon pays for the radio we have to add the sales tax amount to the cost to find the total amount he pays. 43.99 + 3.08 = 47.07 Thus the total Brandon pays for the radio is$47.07.
Question 8.
Luisa’s restaurant bill comes to $75.50, and she leaves a 15% tip. What is Luisa’s total restaurant bill?$ _______
Answer: $86.25 Explanation: Given that, Luisa’s restaurant bill comes to$75.50, and she leaves a 15% tip.
Use the formula for the total restaurant bill:
T = P + x. P
Where T represents the total bill, P represents Luisa’s bill and x represents percents for tip, then the total restaurant bill is:
T = 75 + 0.15 (75)
T = 75 + 11.25
T = $86.25 Therefore Lusia’s total restaurant bill is$86.25
Question 9.
Joe borrowed $2,000 from the bank at a rate of 7% simple interest per year. How much interest did he pay in 5 years?$ _______
Explanation:
Joe borrowed $2,000 from the bank at a rate of 7% simple interest per year. We have to find the amount of interest per year 2000(0.07) = 140 Find the amount of interest for 5 years 140(5) = 700 Thus Joe pays$700 in 5 years.
Question 11.
Martin finds a shirt on sale for 10% off at a department store. The original price was $20. Martin must also pay 8.5% sales tax. a. How much is the shirt before taxes are applied?$ _______
Explanation:
We have to find the sales price of the shirt
20 – 0.1(20) = 20 – 2 = 18
The price of the shirt before taxes are applied is $18. Question 11. b. How much is the shirt after taxes are applied?$ _______
Explanation:
We have to find the price after the sales tax
18 + 0.085(18) = 18 + 1.53 = 19.53
The price of the shirt after taxes are applied is $19.53 Essential Question Check-In Question 13. How can you determine the total cost of an item including tax if you know the price of the item and the tax rate? Type below: _____________ Answer: You can find the total cost of an item including tax by first multiplying the price of the item by the tax rate in decimal form to get the amount of sales tax. Then add the amount of sales tax to the price to get the total cost. ### Applications of Percent – Independent Practice – Page No. 157 Question 14. Emily’s meal costs$32.75 and Darren’s meal costs $39.88. Emily treats Darren by paying for both meals and leaves a 14% tip. Find the total cost.$ _______
Explanation:
Emily’s meal costs $32.75 and Darren’s meal costs$39.88.
So, the total cost of the meals before tip is $32.75 +$39.88 = $72.63 Emily treats Darren by paying for both meals and leaves a 14% tip.$72.63 = 0.14(72.63) ≈ $10.17 Round to two decimal places since dollar amounts must be rounded to the nearest cent. The total cost that Dareen pays is then cost before tip + amount of tip =$72.63 + $10.17 =$82.80
Question 15.
The Jayden family eats at a restaurant that is having a 15% discount promotion. Their meal costs $78.65, and they leave a 20% tip. If the tip applies to the cost of the meal before the discount, what is the total cost of the meal?$ _______
Explanation:
The Jayden family eats at a restaurant that is having a 15% discount promotion.
The total cost of the meal = cost of meal + tip amount – discount amount
Their meal costs $78.65, and they leave a 20% tip. We need to find the tip amount and the discount amount using the given cost of the meal, tip percent, and discount percent. 20% of 78.65 = 0.20 × 78.65 =$15.73
Since the cost of the meal before the discount is $78.65 and the discount percentage is 15%, then the amount of the discount is 15% of 78365 = 0.15 ×$78.65 ≈ $11.80 The total cost is then 78.65 + 15.73 – 11.80 =$82.58
Question 18.
Kedar earns a monthly salary of $2,200 plus a 3.75% commission on the amount of his sales at a men’s clothing store. What would he earn this month if he sold$4,500 in clothing? Round to the nearest cent.
$_______ Answer: 2368.75 Explanation: Given, Kedar earns a monthly salary of$2,200 plus a 3.75% commission on the amount of his sales at a men’s clothing store.
4500 × 0.0375 = 168.75
The total earnings can be known by adding his monthly salary and his commission.
2200 + 168.75 = 2368.75
Question 19.
Danielle earns a 7.25% commission on everything she sells at the electronics store where she works. She also earns a base salary of $750 per week. How much did she earn last week if she sold$4,500 in electronics merchandise? Round to the nearest cent.
$_______ Answer: 1076.25 Explanation: Danielle earns a 7.25% commission on everything she sells at the electronics store where she works. She also earns a base salary of$750 per week.
The amount she made in the commission is 4500 × 0.0725 = 326.25
We can find the total earnings by adding her weekly pay and commission.
750 + 326.25 = 1076.25
Thus she earns $1076.25 last week if she sold$4,500 in electronics merchandise.
Question 20.
Francois earns a weekly salary of $475 plus a 5.5% commission on sales at a gift shop. How much would he earn in a week if he sold$700 in goods? Round to the nearest cent.
$_______ Answer: 513.50 Explanation: Given that, Francois earns a weekly salary of$475 plus a 5.5% commission on sales at a gift shop.
The amount he made in commission
700 × 0.055 = 38.50
We can find the total amount he earned by adding his weekly pay and commission
475 + 38.50 = $513.50 Question 21. Sandra is 4 feet tall. Pablo is 10% taller than Sandra, and Michaela is 8% taller than Pablo a. Explain how to find Michaela’s height with the given information. Type below: _____________ Answer: First, we have to find 10% of Sandra’s height: 0.10 × 4 = 0.4 This means that Pablo is then 4 + 0.4 = 4.4 feet tall. Next find 8% of Pablo’s height: 4.4 × 0.08 = 0.352 This means that Michaela is 4.4 + 0.353 = 4.752 feet tall. Question 21. b. What is Michaela’s approximate height in feet and inches? _______ feet _______ inches Answer: Convert from feet to inches. 1 feet = 12 inches 4.752 = 4 + 0.752 0.752 = 12 × 0.752 = 9 inches 4 feet = 12 × 4 = 48 inches Thus the approximate height of Michaela is 4 feet 9 inches. Question 22. Eugene wants to buy jeans at a store that is giving$10 off everything. The tag on the jeans is marked 50% off. The original price is $49.98. a. Find the total cost if the 50% discount is applied before the$10 discount.
$_______ Answer:$14.99
Explanation:
Given that,
Eugene wants to buy jeans at a store that is giving $10 off everything. The tag on the jeans is marked 50% off. The original price is$49.98.
0.5 × 49.98 = 24.99
Now subtract $10 discount. 24.99 – 10 = 14.99 The total cost if the 50% discount is applied before the$10 discount is $14.99 Question 22. b. Find the total cost if the$10 discount is applied before the 50% discount.
$_______ Answer:$19.99
Explanation:
We have to find the price after the $10 discount and then find 50% of that price to find the discounted price. 49.98 – 10 = 39.98 0.5 × 39.98 = 19.99 Thus the total cost if the$10 discount is applied before the 50% discount is $19.99 ### Applications of Percent – Page No. 158 Question 23. Multistep Eric downloads the coupon shown and goes shopping at Gadgets Galore, where he buys a digital camera for$95 and an extra battery for $15.99. a. What is the total cost if the coupon is applied to the digital camera?$ _______
Explanation:
Use the formula for the discount price:
DP = P – x.P
Price for the digital camera:
DP = 95 – 0.1(95)
DP = 95 – 9.5
DP = $85.5 Total cost = 85.5 + 15.99 =$101.49
Question 23.
b. What is the total cost if the coupon is applied to the extra battery?
$_______ Answer: 109.391 Explanation: Use the formula for the discount price: DP = P – x.P Price for the digital camera: DP = 15.99 – 0.1(15.99) DP = 15.99 – 1.599 DP =$14.399
Total cost = 95 + 14.399 = $109.391 Question 23. c. To which item should Eric apply the discount? Explain. ____________ Answer: He should apply the discount to the digital camera because then the total cost is lower. Question 23. d. Eric has to pay 8% sales tax after the coupon is applied. How much is his total bill?$ _______
Use the formula for Discount price
If he uses a coupon for the digital camera then his total cost will be
T = DP + 0.08 × DP
T = 101.49 + 8.1192
T = $109.6029 If he uses a coupon for the extra battery his total cost will be T = DP + 0.08 × DP T = 109.391 + 0.08(109.391) T =$118.14228
Question 24.
Two stores are having sales on the same shirts. The sale at Store 1 is “2 shirts for $22” and the sale at Store 2 is “Each$12.99 shirt is 10% off”.
a. Explain how much will you save by buying at Store 1.
$_______ Answer: For store 1, the shirts are 2 for$22. Each shirt then costs $22 ÷ 2 =$11
At store 2, each shirt is 10% off of $12.99 so each shirt costs:$12.99 – 0.1(12.99) = $12.99 –$1.30 = $11.69 You will then save$11.69 – $11.00 = 0.69 per shirt if you buy them from Store 1. Question 24. b. If Store 3 has shirts originally priced at$20.98 on sale for 55% off, does it have a better deal than the other stores? Justify your answer.
_______
If Store 3 sells shirts at 55% off of $20.98, then each shirt costs:$20.98 – 0.55($20.98) =$20.98 – $11.54 =$9.44
This is lower than the costs per shirt of Store 1 and Store 2 so it has a better deal.
H.O.T.
Focus on Higher Order Thinking
Question 26.
Multistep
In chemistry class, Bob recorded the volume of a liquid as 13.2 mL. The actual volume was 13.7 mL. Use the formula to find the percent error of Bob’s measurement to the nearest tenth of a percent.
_______ %
Explanation:
|13.2 – 13.7|/13.7 = |-0.5|/13.7
0.5/13.7 ≈ 0.036 = 3.6%
### MODULE QUIZ – 5.1 Percent Increase and Decrease – Page No. 159
Find the percent change from the first value to the second.
Question 1.
36; 63
_______ %
Explanation:
Use the formula percent change = amount of change/first value
amount of change = 27
First value = 36
(63 – 36)/36 = 27/36 = 0.75 = 75%
5.2 Markup and Markdown
Use the original price and the markdown or markup to find the retail price.
Question 5.
Original price: $60; Markup: 15%$ _______
Answer: $69 Explanation: Use the formula retail price = original price + markup 60 + 60 × 0.15 = 60 + 9= 69 Question 8. Original price:$125; Markdown: 30%
$_______ Answer: 87.50 Explanation: Use the formula retail price = original price + markup 125 – 125 × 0.3 = 125 – 37.50 = 87.50 5.3 Applications of Percent Question 9. Mae Ling earns a weekly salary of$325 plus a 6.5% commission on sales at a gift shop. How much would she make in a work week if she sold $4,800 worth of merchandise?$ _______
Explanation:
Mae Ling weekly earnings is equal to her weekly salary plus her commission.
Since she earns 6.5 % commission on sales, if she sold $4800 worth of merchandise, her commission earnings would be 6.5 % of 4800 = 0.065 × 4800 =$312.
Since her weekly salary is 325, then her total weekly earnings is $325 +$312 = $637 Question 10. Ramon earns$1,735 each month and pays $53.10 for electricity. To the nearest tenth of a percent, what percent of Ramon’s earnings are spent on electricity each month? _______ % Answer: 3.1% Explanation: Divide the electric payment by his monthly pay 53.10/1735 = 0.031 = 3.1% Question 11. James, Priya, and Siobhan work in a grocery store. James makes$7.00 per hour. Priya makes 20% more than James, and Siobhan makes 5% less than Priya. How much does Siobhan make per hour?
$_______ Answer: 7.98 per hour Explanation: Since James makes$7 per hour and priya makes 20% more than this, find 20% of 7 and then add that to 7 to find the pay per hour for Priya.
7 + 0.2(7) = 7 + 1.40 = 8.40
Since Priya makes $8.40 per hour and Siobhan makes 5% less than this, find 5% of 8.40 and subtract that from 8.40 to find the pay per hour of Siobhan. 8.40 – 0.05(8.40) = 8.40 – 0.42 = 7.98 Question 12. The Hu family goes out for lunch, and the price of the meal is$45. The sales tax on the meal is 6%, and the family also leaves a 20% tip on the pre-tax amount. What is the total cost of the meal?
$_______ Answer: 56.70 Explanation: Find the amount of tax 45 × 0.06 = 2.70 Find the amount of tip 45 × 0.20 = 9 Find the total cost by adding the cost of the meal, the tax, and the tip. 45 + 2.70 + 9 =$56.70
Essential Question
Question 13.
Give three examples of how percents are used in the real-world. Tell whether each situation represents a percent increase or a percent decrease.
Type below:
____________
One example could be giving a tip when you eat at a restaurant. Since the cost increases, it represents a percent increase.
Second example is tax on purchase. Since the price increases it is a percent increase.
Third example is using a coupon when buying an item. Since the price decreases, it is a percent decrease.
### Selected Response – Page No. 160
Question 1.
Zalmon walks $$\frac{3}{4}$$ of a mile in $$\frac{3}{10}$$ of an hour. What is his speed in miles per hour?
Options:
a. 0.225 miles per hour
b. 2.3 miles per hour
c. 2.5 miles per hour
d. 2.6 miles per hour
Explanation:
Given that,
Zalmon walks $$\frac{3}{4}$$ of a mile in $$\frac{3}{10}$$ of an hour.
Divide the number of miles by the number of hours to get his speed in miles per hour.
$$\frac{3}{4}$$ ÷ $$\frac{3}{10}$$
$$\frac{3}{4}$$ ÷ $$\frac{10}{3}$$ = $$\frac{5}{2}$$
Convert the fraction into decimal form.
$$\frac{5}{2}$$ = 2.5 miles per hour
Thus the correct answer is option C.
Question 2.
Find the percent change from 70 to 56.
Options:
a. 20% decrease
b. 20% increase
c. 25% decrease
d. 25% increase
Explanation:
Use the percent change = amount of change/original amount.
Since the number decreased from 70 to 56, it is a percent decrease.
= (70 – 56)/70 = $$\frac{14}{70}$$ = 0.2 = 20%
Thus the correct answer is option A.
Question 3.
The rainfall total two years ago was 10.2 inches. Last year’s total was 20% greater. What was last year’s rainfall total?
Options:
a. 8.16 inches
b. 11.22 inches
c. 12.24 inches
d. 20.4 inches
Explanation:
Given,
The rainfall total two years ago was 10.2 inches. Last year’s total was 20% greater.
Find 20% of 10.2
10.2 × 0.20 = 2.04
Add the value to the original amount of 10.2
10.2 + 2.04 = 12.24
Therefore the correct answer is option C.
Question 4.
A pair of basketball shoes was originally priced at $80 but was marked up by 37.5%. What was the retail price of the shoes? Options: a.$50
b. $83 c.$110
d. $130 Answer:$110
Explanation:
A pair of basketball shoes was originally priced at $80 but was marked up by 37.5%. Use the formula retail price = original price + markup 80 + 80 × 0.375 = 80 + 30 = 110 Thus the correct answer is option C. Question 5. The sales tax rate in Jan’s town is 7.5%. If she buys 3 lamps for$23.59 each and a sofa for $769.99, how much sales tax does she owe? Options: a.$58.85
b. $63.06 c.$67.26
d. $71.46 Answer:$63.06
Explanation:
The sales tax rate in Jan’s town is 7.5%.
If she buys 3 lamps for $23.59 each and a sofa for$769.99
Total cost before tax is 3 × 23.59 + 769.99
= 70.77 + 769.99 = 840.76
Find the amount of tax by multiplying the tax rate and total cost from the above solution and then round to 2 decimal place.
840.76 × 0.075 = 63.06
Thus the correct answer is option B.
Question 6.
The day after a national holiday, decorations were marked down 40%. Before the holiday, a patriotic banner cost $5.75. How much did the banner cost after the holiday? Options: a.$1.15
b. $2.30 c.$3.45
d. $8.05 Answer:$3.45
Explanation:
The day after a national holiday, decorations were marked down 40%. Before the holiday, a patriotic banner cost $5.75. use the formula retail price = original price – markdown 5.75 – 5.75 × 0.4 = 5.75 – 2.30 = 3.45 Thus the correct answer is option C. Question 7. Dustin makes$2,330 each month and pays $840 for rent. To the nearest tenth of a percent, what percent of Dustin’s earnings are spent on rent? Options: a. 84% b. 63.9% c. 56.4% d. 36.1% Answer: 36.1% Explanation: Dustin makes$2,330 each month and pays $840 for rent. Divide his rent by his monthly income. round to three decimal places and then convert to percent form. 840/2330 = 0.361 = 36.1% Thus the correct answer is option D. Question 8. A scuba diver is positioned at -30 feet. How many feet will she have to rise to change her position to -12 feet? Options: a. -42 ft b. -18 ft c. 18 ft d. 42 ft Answer: 18 ft Explanation: Given, A scuba diver is positioned at -30 feet. -12 – (-30) = 12 + 30 = 18 feet Thus the correct answer is option C. Question 9. A bank offers an annual simple interest rate of 8% on home improvement loans. Tobias borrowed$17,000 over a period of 2 years. How much did he repay altogether?
Options:
a. $1360 b.$2720
c. $18360 d.$19720
Answer: $19720 Explanation: Given that, A bank offers an annual simple interest rate of 8% on home improvement loans. Tobias borrowed$17,000 over a period of 2 years
Find the amount of interest he paid using the formula
I = prt
where p is the amount borrowed
r is the interest rate
t is the number of years
17000 × 0.08 × 2 = 2720
Add the amount borrowed and the amount of interest
17000 + 2720 = 19720.
Thus the correct answer is option D.
Question 10.
The granola Summer buys used to cost $6.00 per pound, but it has been marked up 15%. a. How much did it cost Summer to buy 2.6 pounds of granola at the old price?$ ___________
Answer: $15.60 Explanation: Multiply 2.6 by the old price of$6
2.6 × 6 = 15.60
It costs $15.60 to buy 2.6 pounds of granola at the old price. Question 10. b. How much does it cost her to buy 2.6 pounds of granola at the new price?$ _______
Answer: $17.94 Explanation: Find the new price using the formula retail price = original price + markup Then find the total cost by buying 2.6 pounds at the new price. 6 + 6 × 0.15 = 6 + 0.9 = 6.90 2.6 × 6.90 = 17.94 The new price is$17.94
Question 10.
c. Suppose Summer buys 3.5 pounds of granola. How much more does it cost at the new price than at the old price?
$_______ Answer:$3.15
Explanation:
3.5 × 6 = 21
3.5 × 6.90 = 24.15
24.15 – 21 = 3.15
### Module 5 – Page No. 162
EXERCISES
Question 1.
Michelle purchased 25 audio files in January. In February she purchased 40 audio files. Find the percent increase.
_______ %
Explanation:
Given,
Michelle purchased 25 audio files in January. In February she purchased 40 audio files.
Use the percent change = amount of change/original amount.
(40 -25)/25 = 15/25 = 0.6 = 60%
Thus the percent increase is 60%
Question 4.
A sporting goods store marks up the cost s of soccer balls by 250%. Write an expression that represents the retail cost of the soccer balls. The store buys soccer balls for $5.00 each. What is the retail price of the soccer balls?$ _______
Answer: $17.5 Explanation: Use the formula retail price = original price + markup to find the expression for an original price of s and a markup percentage of 250% s + 2.5s = 3.5s substitute s = 5 into the expression to find the retail price 3.5 × 5 = 17.50 Thus the retail price of the soccer balls is$17.50
### Unit 2 Performance Tasks – Page No. 163
Question 1.
Viktor is a bike tour operator and needs to replace two of his touring bikes. He orders two bikes from the sporting goods store for a total of $2,000 and pays using his credit card. When the bill arrives, he reads the following information: Balance:$2000
Annual interest rate: 14.9%
Minimum payment due: $40 Late fee:$10 if payment is not received by 3/1/2013
a. To keep his good credit, Viktor promptly sends in a minimum payment of $40. When the next bill arrives, it looks a lot like the previous bill. Balance:$1,984.34
Annual interest rate: 14.9%
Minimum payment due: $40 Late fee:$10 if payment is not received by 4/1/2013
Explain how the credit card company calculated the new balance. Notice that the given interest rate is annual, but the payment is monthly.
Type below:
_____________
We have to find the balance after the first bill by subtracting the $40 payment from the original balance of$2000.
Balance after first bill: 2000 – 40 = 1960
Then find the amount of interest charged on the second bill by multiplying the balance of $1960 by the interest rate. Remember since the interest rate is annually you have to divide it by 12 to get the monthly interest rate. Interest on the second bill: 1960 × 0.149/12 = 24.34 And then add this interest amount to the balance of$1960 to get the balance on the second bill.
New Balance: 1960 + 24.34 = 1984.34
Question 1.
b. Viktor was upset about the new bill, so he decided to send in $150 for his April payment. The minimum payment on his bill is calculated as 2% of the balance (rounded to the nearest dollar) or$20, whichever is greater. Fill out the details for Viktor’s new bill.
Type below:
_____________
Find the balance after the $150 payment. The interest rate hasn’t changed so the annual interest rate on this new bill is the same as the previous bills. balance after payment: 1984.34 – 150 = 1834.34 annual interest rate: 14/9% Find the interest charged on the third bill. find the balance on the third bill by adding the interest charged to the balance of$1834.34.
interest on the third bill: 1834.34 × 0.149/12 = 22.78
balance: 1834.34 + 22.78 = 1857.12
To find what the minimum payment will be, first find 2% of the balance.
2% of balance: 0.02 × 1857.12 = 37.14
Minimum payment due: $37.00 Since this is greater than$20, the minimum payment is 2% of the balance rounded to the nearest dollar giving $37 as the payment. The later fee date is one month after the late fee date of 04/01/2013 on the previous bill which gives 05/01/2013. Question 1. c. Viktor’s bank offers a credit card with an introductory annual interest rate of 9.9%. He can transfer his current balance for a fee of$40. After one year, the rate will return to the bank’s normal rate, which is 13.9%. The bank charges a late fee of $15. Give two reasons why Viktor should transfer the balance and two reasons why he should not Type below: _____________ Answer: Two reasons he should transfer is that the lower introductory rate would mean less interest charged in the first year and a lower normal rate would mean less interest charged after that first year as well. Two reasons he shouldn’t transfer the balance is that he would have to pay a transfer fee of$40 and that the late fee is $15 instead of$10 if he transfers the balance.
### Unit 2 Performance Tasks (con’td) – Page No. 164
Question 2.
The table below shows how far several animals can travel at their maximum speeds in a given time.
a. Write each animal’s speed as a unit rate in feet per second.
Elk: _________ feet per second
Giraffe: _________ feet per second
Zebra: _________ feet per second
By seeing the above table we can find the unit rates by dividing the distance traveled by the time in seconds.
elk: 33 ÷ 1/2 = 33 × 2 = 66 feet per second
Giraffe: 115 ÷ 2 1/2 = 115 ÷ 5/2 = 115 2/5 = 46 feet per second
Zebra: 117 ÷ 2 = 58.5 feet per second
Question 2.
b. Which animal has the fastest speed?
_____________
Answer: The elk had the greatest unit rate so it has the fastest speed.
Question 2.
c. How many miles could the fastest animal travel in 2 hours if it maintained the speed you calculated in part a? Use the formula d = rt and round your answer to the nearest tenth of a mile. Show your work.
Elk: _________ miles
Giraffe: _________ miles
Zebra: _________ miles
Elk: 90 miles
Giraffe: 62 miles
Zebra: 72 miles
Explanation:
There are 60 seconds in a minute and 60 minutes in an hour so there are 2 × 60 × 60 = 7200 seconds in 2 hours.
Multiply the unit rate of the elk by 7200 seconds to get the distance traveled in feet.
There are 5280 feet in 1 mile so divide the distance in feet by 5280 to get the distances in miles.
Elk:
66 × 7200 = 475200 feet
Now convert from feet to miles
475200 feet = 90 miles
Giraffe: 46 feet per second
62 × 7200 = 331200 feet
Now convert from feet to miles.
331200 = 62 miles
Zebra: 58.5 feet per second
58.5 × 7200 = 421200 feet
Now convert from feet to miles.
421200 feet = 72 miles
Question 3.
d. The data in the table represents how fast each animal can travel at its maximum speed. Is it reasonable to expect the animal from part b to travel that distance in 2 hours? Explain why or why not.
______
Answer: It is not reasonable. An animal can only travel at its maximum speed for a short amount of time which is usually only for a couple of minutes.
### Selected Response – Page No. 165
Question 1.
If the relationship between distance y in feet and time x in seconds is proportional, which rate is represented by $$\frac{y}{x}$$ = 0.6?
Options:
a. 3 feet in 5 s
b. 3 feet in 9 s
c. 10 feet in 6 s
d. 18 feet in 3 s
Answer: 3 feet in 5 s
Explanation:
$$\frac{y}{x}$$ = 0.6
0.6 = $$\frac{6}{10}$$
Since $$\frac{6}{10}$$ = $$\frac{3}{5}$$, it represents a rate of 3 feet in 5 seconds,
Therefore the correct answer is option A.
Question 2.
The Baghrams make regular monthly deposits in a savings account. The graph shows the relationship between the number x of months and the amount y in dollars in the account.
What is the equation for the deposit?
Options:
a. $$\frac{y}{x}$$ = $25/month b. $$\frac{y}{x}$$ =$40/month
c. $$\frac{y}{x}$$ = $50/month d. $$\frac{y}{x}$$ =$75/month
Answer: $$\frac{y}{x}$$ = $50/month Explanation: By seeing the above graph we can say that the point is (2, 100). This means that $$\frac{y}{x}$$ = $$\frac{100}{2}$$ = 50. Thus the correct answer is option C. Question 3. What is the decimal form of −4 $$\frac{7}{8}$$? Options: a. -4.9375 b. -4.875 c. -4.75 d. -4.625 Answer: -4.875 Explanation: Given the fraction −4 $$\frac{7}{8}$$ First divide $$\frac{7}{8}$$ = 0.875 4 + 0.875 = 4.875 So, −4 $$\frac{7}{8}$$ = -4.875 Therefore the answer is option B. Question 4. Find the percent change from 72 to 90. Options: a. 20% decrease b. 20% increase c. 25% decrease d. 25% increase Answer: 25% increase Explanation: Use the formula percent change = amount of change/original amount. the value increased from 72 to 90 so it is a percent increase. (90-72)/72 = 18/72 = 0.25 = 25% Thus the correct answer is option D. Question 5. A store had a sale on art supplies. The price p of each item was marked down by 60%. Which expression represents the new price? Options: a. 0.4p b. 0.6p c. 1.4p d. 1.6p Answer: 0.4p Explanation: Given that, A store had a sale on art supplies. The price p of each item was marked down by 60% Use the formula sale price = original price – markdown p is the original price and the markdown percent is 40% then combine the like terms. p – 0.6p = 0.4p Therefore the correct answer is option A. Question 6. Clarke borrows$16,000 to buy a car. He pays simple interest at an annual rate of 6% over a period of 3.5 years. How much does he pay altogether?
Options:
a. $18800 b.$19360
c. $19920 d.$20480
Answer: $19360 Explanation: Given, Clarke borrows$16,000 to buy a car.
He pays simple interest at an annual rate of 6% over a period of 3.5 years.
Find the total amount of interest using the formula
I = prt
where p is the amount borrowed
r is the rate of interest
t is the number of years
16000 × 0.06 × 3.5 = 3360
Now add the amount of interest to the amount borrowed to find the total amount
16000 + 3360 = 19,360
Thus the correct answer is option B.
Question 7.
To which set or sets does the number 37 belong?
Options:
a. integers only
b. rational numbers only
c. integers and rational numbers only
d. whole numbers, integers, and rational numbers
Answer: whole numbers, integers, and rational numbers
Explanation:
37 can be written as 37/1 so it is a rational number. 37 doesn’t have a decimal or fraction so it is an integer. Since it is a positive integer, it is also a whole number.
Thus a suitable answer is option D.
### Page No. 166
Question 8.
In which equation is the constant of proportionality 5?
Options:
a. x = 5y
b. y = 5x
c. y = x + 5
d. y = 5 – x
Explanation:
Directly proportional equations are of the form y = kx
where k is the constant of proportionality.
If k = 5, then the equation is y = 5x.
Thus the correct answer is option B.
Question 9.
Suri earns extra money by dog walking. She charges $6.25 to walk a dog once a day 5 days a week and$8.75 to walk a dog once a day 7 days a week. Which equation represents this relationship?
Options:
a. y = 7x
b. y = 5x
c. y = 2.50x
d. y = 1.25x
Explanation:
Given that,
Suri earns extra money by dog walking. She charges $6.25 to walk a dog once a day 5 days a week and$8.75 to walk a dog once a day 7 days a week.
Since 6.25/5 = 1.25
So, the equation is y = 1.25x
where x is the number of days and y is the total charge.
So, the correct answer is option D.
Question 10.
Randy walks $$\frac{1}{2}$$ mile in each $$\frac{1}{5}$$ hour. How far will Randy walk in one hour?
Options:
a. $$\frac{1}{2}$$ miles
b. 2 miles
c. 2 $$\frac{1}{2}$$ miles
d. 5 miles
Answer: 2 $$\frac{1}{2}$$ miles
Explanation:
Given,
Randy walks $$\frac{1}{2}$$ mile in each $$\frac{1}{5}$$ hour.
$$\frac{1}{2}$$ ÷ $$\frac{1}{5}$$
$$\frac{1}{2}$$ × $$\frac{5}{1}$$ = $$\frac{5}{2}$$
Convert the fraction to the improper fractions.
$$\frac{5}{2}$$ = 2 $$\frac{1}{2}$$ miles
Therefore the correct answer is option C.
Question 11.
On a trip to Spain, Sheila bought a piece of jewelry that cost $56.75. She paid for it with her credit card, which charges a foreign transaction fee of 3%. How much was the foreign transaction fee? Options: a.$0.17
b. $1.07 c.$1.70
d. $17.00 Answer:$1.70
Explanation:
On a trip to Spain, Sheila bought a piece of jewelry that cost \$56.75.
She paid for it with her credit card, which charges a foreign transaction fee of 3%
Find the foreign transaction fee amount by multiplying the cost by the foreign transaction fee percentage.
56.75 × 0.03 = 1.70
Thus the correct answer is option C.
Question 12.
A baker is looking for a recipe that has the lowest unit rate for flour per batch of muffins. Which recipe should she use?
Options:
a. $$\frac{1}{2}$$ cup flour for $$\frac{2}{3}$$ batch
b. $$\frac{2}{3}$$ cup flour for $$\frac{1}{2}$$ batch
c. $$\frac{3}{4}$$ cup flour for $$\frac{2}{3}$$ batch
d. $$\frac{1}{3}$$ cup flour for $$\frac{1}{4}$$ batch
Answer: $$\frac{1}{2}$$ cup flour for $$\frac{2}{3}$$ batch
Explanation:
a. $$\frac{1}{2}$$ ÷ $$\frac{2}{3}$$ = $$\frac{1}{2}$$ × $$\frac{3}{2}$$ = $$\frac{3}{4}$$
b. $$\frac{2}{3}$$ ÷ $$\frac{1}{2}$$ = $$\frac{2}{3}$$ × $$\frac{2}{1}$$ = $$\frac{4}{3}$$ = 1 $$\frac{1}{3}$$
c. $$\frac{3}{4}$$ ÷ $$\frac{2}{3}$$ = $$\frac{3}{4}$$ × $$\frac{3}{2}$$ = $$\frac{9}{8}$$ = 1 $$\frac{1}{8}$$
d. $$\frac{1}{3}$$ ÷ $$\frac{1}{4}$$ = $$\frac{1}{3}$$ ÷ $$\frac{4}{1}$$ = 1 $$\frac{1}{3}$$
Thus the correct answer is option A.
Question 13.
Kevin was able to type 2 pages in 5 minutes, 3 pages in 7.5 minutes, and 5 pages in 12.5 minutes.
a. Make a table of the data.
Type below:
___________
Number of Pages 2 3 5 Minutes 5 7.5 12.5
Question 13.
b. Graph the relationship between the number of pages typed and the number of minutes.
Type below:
___________
Question 13.
c. Explain how to use the graph to find the unit rate.
Type below:
___________
Answer: The unit rate is 2.5 pages per minute
Explanation:
By using the graph we need to find the slope of the line.
We can do this by using the formula of a slope:
m = (y2-y1)/(x2-x1) = (7.5-5)/(3-2) = 2.5
Thus the unit rate is 2.5 pages per minute.
Conclusion:
Hope the answers provided in Go Math Answer Key Grade 7 Chapter 5 Percent Increase and Decrease are quite satisfactory for all the students. Refer to our Go Math 7th Grade Chapter 5 Percent Increase and Decrease to get the solutions with the best explanations. After your preparation test your math skills by solving the questions in the performance tasks. |
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# RS Aggarwal Class 12 Chapter 6 Solutions (Determinants)
RS Aggarwal Solutions for Class 12 Chapter 6 Determinants have been prepared to help you solve the difficult questions easily. You will be introduced to Determinants in this chapter and also learn how to find the determinant of square matrices of orders 1, 2 and 3 respectively. You will also be introduced to some interesting properties of determinants of square matrices. You will find that it is very simple to find the values of determinants with the application of these new properties, which would have been a complicated task otherwise.
This chapter consists of 4 exercises containing a sum total of 110 questions, with thorough coverage of all the important topics given in the syllabus prescribed by the CBSE. This is one of the most scoring chapters in Class 12 syllabus, therefore you should practice all the questions discussed in the RS Aggarwal Solutions for Class 12 Chapter 6 thoroughly to achieve a spike in your marks in Maths. The questions are in an increasing level of difficulty so that you are able to adjust with the new concepts immediately.
The team of Subject Matter Experts at Instasolv are aware of your requirements for the simplicity of language and thus have adhered to the requirements for the Class 12 board exams. We, at Instasolv, follow strict guidelines in various stages of creating the answers, clearing all your doubts and concepts at one place in a hustle-free format.
## Summary of RS Aggarwal Solutions for Class 12 Chapter 6 Determinants
Introduction to Determinants:
Any square matrix can be represented as an expression or a numeric digit. This is known as its determinant. For any square matrix Anxn,, its determinant can be denoted by det A or | A |.
Determinant:
The determinant of a Square Matrix of Order 1:
For the matrix A = [a], | A | = a
The determinant of a Square Matrix of Order 2:
For a square matrix of order 2, the determinant can found by subtracting the product of the elements of the main diagonal and the product of the elements of the counter-diagonal.
The determinant of a Square Matrix of Order 3:
For a square matrix of order 2, the determinant can found by representing the matrix in the form of second-order determinants. This method is called the expansion of determinant. It can either be done along a row or column.
Expanding along R1
| A | = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23) + a13 (a21 a32 – a31 a22
OR
Expanding along C1
| A | = a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22
Properties of Determinants:
Property 1: Rows and columns can be interchanged without affecting the value of the determinant.
Property 2: Interchanging of rows or columns changes the sign of the determinant.
Property 3: The value of a determinant is zero if identical rows or columns are present
Property 4: If a row or a column is multiplied by any constant say k, the value of determinant also becomes k times
Property 5: If a row or a column of a determinant can be represented as a sum of
more than one term, the determinant can be represented as a sum of more than one determinant.
Property 6: If we will add the equimultiples of all rows or all columns of a determinant to the corresponding elements of another row or column respectively, then the value of the determinant will not change.
Area of Triangle:
The area of a triangle represented by the given 3 locus points (x1,y1)(x3,y3)(x3,y3)can be calculated by the following formula:
This formula can be represented in the form of determinants as follows:
Highlights of the Chapter:
• Representation of a square matrix in the form of a numerical digit or an expression is known as a determinant.
• We can find the area of a triangle using determinants with the help of the following formula:
• There are various properties of determinants with the help of which you can easily determine the numerical value of a given matrix.
• You can find the determinant of a matrix by learning to practice its expansion in a 3X3 matrix using the following formula:
Expanding along R
| A | = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23) + a13 (a21 a32 – a31 a22)
OR
Expanding along C1
| A | = a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)
## Exercise Wise Discussion of RS Aggarwal Solutions for Class 12 Chapter 6 ‘Determinants’
1. The exercise solutions of 6A consist of advanced level short answer type questions, to evaluate the value of a given determinant or to find the value of an unknown variable, merely by the expansion of the matrix.
2. In exercise 6B, you will learn the application of the important properties and will be able to prove various equations and theorems using them besides practising the evaluation of complex determinants.
3. You will be required to find the area of triangles or prove collinearity of three points using determinants in the last exercise 6D.
4. Solving these questions will train you extensively in your time management skills and boost your analytical skills.
## Benefits of RS Aggarwal Solutions for Class 12 Maths Chapter 6 by Instasolv
1. The complicated RS Arrarwal Class 12 Maths Solutions are mentioned in the simplest manner possible by the maths experts at Instasolv.
2. We have covered each question of each exercise to cater to your needs of improving your ability to approach a problem of determinants.
3. The answers provided at Instasolv are free and easily accessible so that you can sort out all your queries at one place.
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# Hands-On Place Value
I like to start the year off with hands-on place value activities, because place value is the foundation for all other math concepts. It’s important to spend time on place value to ensure students are prepared for more challenging concepts.
These place value activities and practice pages can be printed as a booklet and area a great supplement to any math program or curriculum. You can use them for centers, stations, whole group activities, or really anything!
## Activity 1-Rounding Riddles
Students solve six rounding riddles that require critical thinking. In the first problem students write a number that rounds to 20. Then, they share whether they could find a different number that rounds to 20. The questions progressively become more challenging.
## Activity 2-Rounding on a Number Line
In another activity students create a number line that showed how to round to the nearest hundred. This is important, because students better understand the concept of rounding through practice on number lines and hundreds charts. The visual gives students a much better understanding of WHY a number rounds to a particular place.
## Activity 3-Forms of Numbers
In this practice activity, students write numbers in standard form and expanded form. This is a good review for numbers through the thousands place.
## Activity 4-Make 1,000
In another activity, students use base-ten blocks to find three different ways to build a thousand. They used a combination of ones, tens, and hundreds to do this, and it helped my students to see how ones, tens, and hundreds relate to each other. They first used the base-ten blocks to create the number, and then they recorded how they made the number on a recording sheet.
## Activity 5-What’s It Between
This is a rounding practice activity that takes a reverse approach to rounding. Students are given a number and determine the two multiples of ten or hundred that number is between. This is an extremely important concept for students.
## Activity 6-Roll and Round
In this game students use two dice that had the digits one through nine on them. Students roll the two dice and create the largest number possible with the digits they rolled, and then they rounded the number to the nearest ten. To add a little conceptual math, students also find the number on a hundreds chart to help them determine where to round the number.
## Activity 7-Rounding Practice
In this activity, students complete very basic rounding problems. They will round two and three digit numbers to the nearest ten and nearest hundred.
## Activity 8-Place Value Tower
In a different hands-on place value activity, students used base-ten blocks to build a place value tower. The tower had to be free standing, and students had to use exactly 50 base-ten blocks for the tower. After they built it, they used their understanding of place value to determine the total value of their tower. They had to count all of the ones and find the value of the ones. Then, they had to record how many ones they used and the value of the ones on their recording sheet and did the same for the tens, hundreds, and thousands. This was such a huge hit with students! They had so much fun with this activity, and I feel that they were actually learning!
## Activity 9-Place Value Practice
In the final activity students use place value clues to find a mystery number. This is always great place value practice that encourages critical thinking.
My students loved learning about place value and didn’t want our math time to end! You can get a FREE copy of the place value booklet here! I’ve created booklets for EVERY math standard and placed those booklets in a bundle, which you can view here. You can find even more hands-on place value lessons here!
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Algebraic Expressions
# Algebraic Expressions
## Algebraic Expressions
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##### Presentation Transcript
1. Algebraic Expressions Lesson 5-1
2. Evaluate an Algebraic Expression The branch of mathematics that involve expressions with variables is called algebra. In algebra, the multiplication sign is often omitted. The numerical factor of a multiplication expression that contains a variable is called a coefficient. So, 6 is the coefficient of 6d.
3. Example 1 Evaluate 2(n + 3) if n = -4. 2(n + 3) = 2(-4 + 3) = 2(-1) = -2
4. Example 2 Evaluate 8w – 2v if w = 5 and v = 3. 8w – 2v = 8(5) – 2(3) = 40 – 6 = 34
5. Example 3 Evaluate 4y3 + 2 if y = 3. 4y3 + 2 = 4(3)3 + 2 = 4(27) + 2 = 110
6. Got it? 1, 2 & 3 Evaluate each expression if c = 8 and d = -5. a. c – 3 b. 15 – c c. 3(c + d) d. 2c – 4d e. d2 – c2 f. 2d2 + 5d 5 7 9 36 25 17
7. Example 4 Athletic trainers use the formula , where a is a person’s age, to find the minimum training heart rate. Find Latrina’s minimum training heart if she is 15 years old. = = = = 123 Latrina’s minimum training heart rate is 123 beats per minute.
8. Got it? 4 To find the area of a triangle, use the formula , where b is the base and h is the height. What is the area in square inches of a triangle with a height of 6 inches and a base of 8 inches?
9. Example 5 To translate a verbal phrase to an algebraic expression, the first step is to define a variable. When you define a variable, you choose a variable to represent the unknown. Marisa wants to buy a DVD player that costs \$150. She already saved \$25 and plans to save an additional \$10 each week. Write an expression that represents the total amount of money Marissa has saved after any number of weeks. 25 + 10w represents the total saved after any number of weeks.
10. Example 6 Refer to Example 5. Will Marisa have saved enough money to buy the \$150 DVD player in 11 weeks? Use the expression 25 + 10w. 25 + 10w = 25 + 10(11) = 25 + 110 = 135 Marisa will only have saved \$135. She needs \$150, so she does not have enough.
11. Got it? 5 & 6 An iPod costs \$70 and song downloads cost \$0.85 each. Write an expression that represents the cost of the iPod and x number of downloaded songs. Then find the cost if 20 songs are downloaded. 70 + 0.85x \$87
12. Sequences Lesson 5-2
13. Vocabulary • Sequence – an ordered list of numbers • 1, 3, 5, 7, 9, 11, 13… • Term – one of the numbers in the sequence • 7 is a term in the sequence above • Arithmetic Sequence – when the difference is consistent between to consecutive terms. • the difference between any two consecutive numbers is the same • Common Difference – the difference between two terms • The common difference is 2
14. Example 1 In an arithmetic sequence, the terms can be whole numbers, fractions, or decimals. Describe the relationship between the terms in the arithmetic sequence 8, 13, 18, 23,…. Then write the next three terms In the sequence. 23 + 5 = 28 28 + 5 = 33 33 + 5 = 38 The next three terms are 28, 33, and 38.
15. Example 2 Describe the relationship between the terms in the arithmetic sequence 0.4, 0.6, 0.8, 1.0,…. Then write the next three terms In the sequence. 1.0 + 0.2 = 1.2 1.2 + 0.2 = 1.4 1.4 + 0.2 = 1.6 The next three terms are 1.2, 1.4, and 1.6
16. Got it? 1 & 2 Describe the relationship between the terms in each arithmetic sequence. Then write the next terms in the sequence. a. 0, 13, 26, 39, … b. 4, 7, 10, 13… c. 1.0, 1.3, 1.6, 1.9… d. 2.5, 3.0, 3.5, 4.0…
17. Write an Algebraic Expression In a sequence, each term has a specific position within the sequence. Consider the sequence 2, 4, 6, 8… Notice that the position number increases by 1, the value of the term increases by 2.
18. Write an Algebraic Expression You can also write an algebraic expression to represent the relationship between any term in a sequence and its position in sequence. In this case, if n represents the position in the sequence, the value of the term is 2n.
19. Example 3 The greeting cards that Meredith makes are sold in boxes at a gift store. The first week, the store sold 5 boxes. Each week, the store sells five more boxes. This patterns continues. What algebraic expression can be used to find the total number of boxes the end of the 100thweek? What is the total? Each term is 5 times its position. So, the expression is 5n. 5n = 5(100) = 500 In 100 week, 500 boxes will be sold.
20. Got it? 3 If the pattern continues, what algebraic expression can be used to find the number of circles used in any figure. How many circles will be in the 50th figure?
21. Properties Lesson 5-3
22. Commutative Properties Words: the order in which numbers are added or multiplied does not change the sum or product. Symbols: a + b = b + a a ∙ b = b ∙ a Examples: 6 + 8 = 8 + 6 4 ∙ 7 = 7 ∙ 4
23. Associative Properties Words: the order in which numbers are grouped when added or multiplied does not change the sum or product. Symbols: (a + b) + c = a + (b + c) (a ∙ b) ∙ c = a ∙ (b ∙ c) Examples: (3 + 6) + 8 = 3 + (6 + 8) (5 ∙ 2) ∙ 7 = 5 ∙ (2 ∙ 7)
24. Number Properties A property is a statement that is true for any number. The following properties are also true for any numbers.
25. Example 1 Name the property shown by the statement. 2 (5 n) = (2 5) n The order of the numbers and variable does not change, but their grouping did. This is the Associative property of Multiplication.
26. Got it? 1 Name the property shown by the statement. a. 42 + x + y = 42 + y + x Communicative (+) b. 3x + 0 = 3x Identity (+)
27. Counterexample A counterexample is an example that shows a statement is not true. Statement: All songs are only 3 minutes. Counterexample: The song “We Are The Champions” by Queen is 4 minutes 21 seconds.
28. Example 2 State whether the following conjecture is true or false. If false, provide a counterexample. Division of whole numbers is commutative. Write two division expressions that are commutative. Let’s pick some nice numbers… like 27, 9 and 3 27 ÷ 9 = 9 ÷ 27 3 ≠ We found a counterexample, so division is not commutative.
29. Got it? 2 State whether the following conjecture is true or false. If false, provide a counterexample. The difference of two different whole numbers is always less than both of the two numbers. false; 8 – 2 2 – 8
30. Example 3 Alana wants to buy a sweater that cost \$28, sunglasses that cost \$14, a pair of jeans that costs \$22, and a T-shirt that costs \$16. Use mental math to find the total cost before tax. 38 + 14 + 16 + 22 = (38 + 22) + (14 + 16) =60 + 30 =90 The total cost is \$90.
31. Got it? 3 Lance made four phone calls from his cell phone today. The calls lasted 4.7, 9.4, 2.3, and 10.6 minutes. Use mental math to find the total cost amount of time he spent on his phone. 27 minutes
32. Example 4 Simplify the expression. Justify each step. (3 + e) + 7 (3 + e) + 7 = (e + 3) + 7 Commutative Property of Addition = e + (3 + 7) Associative Property of Addition = e + 10
33. Example 5 Simplify the expression. Justify each step. x ∙ (8 ∙ x) x ∙ (8 ∙ x) = x (x 8) Commutative Property of Multiplication = (x ∙ x) ∙ 8 Associate Property of Multiplication = 8x2
34. Got it? 4 & 5 Simplify the expression. Justify each step. 4 ∙ (3c ∙ 2) 4 ∙ (3c ∙ 2) = 4 (2 3c) Commutative Property of Multiplication = (4 ∙ 2) ∙ 3c Associate Property of Multiplication = (4 ∙ 2 ∙ 3) c Associate Property of Multiplication = 24c
35. Ticket Out The Door Which of the following is an example of the Community Property of Addition? A. (3 ∙ 4) + 5 = 5 + (3 ∙ 4) B. (7 + 8) + 2 = 7 + (8 + 2) C. 8 ∙ 9 = 9 ∙ 8 D. 1 + 0 = 1
36. The Distributive Property Lesson 5-4
37. Distributive Property Words: To multiply a sum or different by a number, multiply each term inside the parentheses by the number outside the parentheses. Symbols: a(b + c) = ab + ac a(b – c) = ab – ac Example: 5(6 + 7) = 5(6) + 5(7) 4(2 – 8) = 4(2) – 4(8)
38. Distributive Property You can model the Distributive Property with algebraic expressions using algebra tiles. The expression 2(x + 2) is modeled. Model x + 2 using algebra tiles. Double the amount of tiles to represent 2(x + 2). 2(x + 2) = 2(x) + 2(2) = 2x + 4 No matter what x is, 2(x + 2) will always equal 2x + 4. Rearrange the tiles by grouping together the ones with the same shapes.
39. Example 1 Use the Distributive Property to write the expression as an equivalent expression. The evaluate the expression. • 5(12 + 4) 5(12 + 4) = 5(12) + 5(4) 5(16) = 60 + 20 80 = 80 • (20 – 3)8 (20 – 3)8 = 8(20) – 8(3) 17 ∙ 8 = 160 – 24 136 = 136
40. Got it? 1 Use the Distributive Property to write the expression as an equivalent expression. The evaluate the expression. a. 5(-9 + 11) b. 7(10 – 5) c. (12 – 8)9 35 10 36
41. Example 2 & 3 Use the Distributive Property to write each expression as an equivalent algebraic expression. • 4(x + 5) 4(x + 5) = 4x + 4(5) = 4x + 20 • 6(y – 10) 6(y - 10) = 6y – 6(10) = 6y – 60
42. Example 4 & 5 Use the Distributive Property to write each expression as an equivalent algebraic expression. • -3(m – 4) -3(m – 4) = -3m – (-3)(4) = -3m + 12 • 9(-3n – 7y) 9(-3n – 7y) = 9(-3n) – (9 ∙ 7y) = -27n – 63y
43. Example 6 Use the Distributive Property to write the expression as an equivalent algebraic expression: (x – 6) (x – 6) = (x) – (6) = x - 2
44. Got it? 2-6 a. 6(a + 4) b. (m + 3n)8 c. -3(y – 10) d. (w – 4) 8m + 24n 6a + 24 w – 2 -3y + 30
45. Example 7 – how to solve story problems with the Distributive Property On a school visit to Washington D.C., Daniel and his class visited the Smithsonian Air and Space Museum. Tickets to the IMAX movie cost \$8.99. Find the total cost of 20 students to see the movie. =20(9 – 0.01) = 20(9) – 20(0.01) 180 – 0.20 \$179.80
46. Got it? 7 A sports club rents dirt bikes for \$37.50 each. Find the total cost for the club to rent 20 bikes. Justify your answer by using the Distributive Property. \$750 =20(37 + 0.50) = 20(37) + 20(0.50)
47. Simplifying Algebraic Expressions Lesson 5-5
48. Vocabulary Term: the expression 5x + 8y – 9 has 3 terms. Coefficient: 5 in 5x is the coefficient. Constant: 9 in the expression 5x + 8y – 9 is the constant. Like terms: 5x, 6x, and 7x are like terms since they all have an x.
49. Identify Parts of an Expression Like terms contain the same variables to the same powers. For example, 3x2 and -7x2 are like terms. So are 8xy2 and 12xy2. But 10x2z and 22xz2 are not like terms.
50. Example 1 Identify the terms, like terms, coefficients, and constant in the expression 6n – 7n – 4 + n 6n – 7n – 4 + n = 6n + (-7n) + (-4) + n Terms: 6n, -7n, -4, n Like terms: 6n, -7n, n (all of these terms have the same variable) Coefficients: 6, -7, 1 Constant: -4 (This is the only term without a variable) |
Sequences and Series
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1 Contents 6 Sequences and Series 6. Sequences and Series 6. Infinite Series The Binomial Series Power Series Maclaurin and Taylor Series 40 Learning outcomes In this Workbook you will learn about sequences and series. You will learn about arithmetic and geometric series and also about infinite series. You will learn how to test the for the convergence of an infinite series. You will then learn about power series, in particular you will study the binomial series. Finally you will apply your knowledge of power series to the process of finding series expansions of functions of a single variable. You will be able to find the Maclaurin and Taylor series expansions of simple functions about a point of interest.
2 Sequences and Series 6. Introduction In this Section we develop the ground work for later Sections on infinite series and on power series. We begin with simple sequences of numbers and with finite series of numbers. We introduce the summation notation for the description of series. Finally, we consider arithmetic and geometric series and obtain expressions for the sum of n terms of both types of series. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... understand and be able to use the basic rules of algebra be able to find limits of algebraic expressions check if a sequence of numbers is convergent use the summation notation to specify series recognise arithmetic and geometric series and find their sums HELM (008): Workbook 6: Sequences and Series
3 . Introduction A sequence is any succession of numbers. For example the sequence,,, 3, 5, 8,... which is known as the Fibonacci sequence, is formed by adding two consecutive terms together to obtain the next term. The numbers in this sequence continually increase without bound and we say this sequence diverges. An example of a convergent sequence is the harmonic sequence,, 3, 4,... Here we see the magnitude of these numbers continually decrease and it is obvious that the sequence converges to the number zero. The related alternating harmonic sequence,, 3, 4,... is also convergent to the number zero. Whether or not a sequence is convergent is often easy to deduce by graphing the individual terms. The diagrams in Figure show how the individual terms of the harmonic and alternating harmonic series behave as the number of terms increase. / /3 /4 harmonic term in sequence alternating harmonic /3 /4 / term in sequence Figure Task Graph the sequence:,,,,... Is this convergent? HELM (008): Section 6.: Sequences and Series 3
4 3 4 5 term in sequence Not convergent. The terms in the sequence do not converge to a particular value. The value oscillates. A general sequence is denoted by a, a,..., a n,... in which a is the first term, a is the second term and a n is the n th term is the sequence. For example, in the harmonic sequence a =, a =,..., a n = n whilst for the alternating harmonic sequence the n th term is: a n = ( )n+ n in which ( ) n = + if n is an even number and ( ) n = if n is an odd number. Key Point The sequence a, a,..., a n,... is said to be convergent if the limit of a n as n increases can be found. (Mathematically we say that lim n a n exists.) If the sequence is not convergent it is said to be divergent. Task Verify that the following sequence is convergent 3, 4 3, 5 3 4,... First find the expression for the n th term: a n = n + n(n + ) 4 HELM (008): Workbook 6: Sequences and Series
5 Now find the limit of a n as n increases: n + n(n + ) = + n n + n + 0 as n increases Hence the sequence is convergent.. Arithmetic and geometric progressions Consider the sequences:, 4, 7, 0,... and 3,,, 3,... In both, any particular term is obtained from the previous term by the addition of a constant value (3 and respectively). Each of these sequences are said to be an arithmetic sequence or arithmetic progression and has general form: a, a + d, a + d, a + 3d,..., a + (n )d,... in which a, d are given numbers. In the first example above a =, d = 3 whereas, in the second example, a = 3, d =. The difference between any two successive terms of a given arithmetic sequence gives the value of d which is called the common difference. Two sequences which are not arithmetic sequences are:,, 4, 8,..., 3, 9, 7,... In each case a particular term is obtained from the previous term by multiplying by a constant factor ( and respectively). Each is an example of a geometric sequence or geometric progression 3 with the general form: a, ar, ar, ar 3,... where a is the first term and r is called the common ratio, being the ratio of two successive terms. In the first geometric sequence above a =, r = and in the second geometric sequence a =, r = 3. HELM (008): Section 6.: Sequences and Series 5
6 Task Find a, d for the arithmetic sequence 3, 9, 5,... a = d = a = 3, d = 6 Task Find a, r for the geometric sequence 8, 8 7, 8 49,... a = r = a = 8, r = 7 Task Write out the first four terms of the geometric series with a = 4, r =. 4, 8, 6, 3,... The reader should note that many sequences (for example the harmonic sequences) are neither arithmetic nor geometric. 6 HELM (008): Workbook 6: Sequences and Series
7 3. Series A series is the sum of the terms of a sequence. For example, the harmonic series is and the alternating harmonic series is The summation notation If we consider a general sequence a, a,..., a n,... then the sum of the first k terms a + a + a a k is concisely denoted by That is, k a + a + a a k = p= a p k p= a p. k When we encounter the expression a p we let the index p in the term a p take, in turn, the values,,..., k and then add all these terms together. So, for example p= 3 7 a p = a + a + a 3 a p = a + a 3 + a 4 + a 5 + a 6 + a 7 p= p= Note that p is a dummy index; any letter could be used as the index. For example 6 a m each represent the same collection of terms: a + a + a 3 + a 4 + a 5 + a 6. m= 6 a i, and i= In order to be able to use this summation notation we need to obtain a suitable expression for the typical term in the series. For example, the finite series k may be written as In the same way k p since the typical term is clearly p in which p =,, 3,..., k in turn. p= = p= ( ) p+ p since an expression for the typical term in this alternating harmonic series is a p = ( )p+. p HELM (008): Section 6.: Sequences and Series 7
8 Task Write in summation form the series First find an expression for the typical term, the p th term : a p = a p = p(p + ) Now write the series in summation form: = = p(p + ) p= Task Write out all the terms of the series 5 p= ( ) p (p + ). Give p the values,, 3, 4, 5 in the typical term 5 ( ) p (p + ) = p= ( )p (p + ) : 8 HELM (008): Workbook 6: Sequences and Series
9 4. Summing series The arithmetic series Consider the finite arithmetic series with 4 terms A simple way of working out the value of the sum is to create a second series which is the first written in reverse order. Thus we have two series, each with the same value A: and A = A = Now, adding the terms of these series in pairs A = = 8 4 = 39 so A = 96. We can use this approach to find the sum of n terms of a general arithmetic series. If A = [a] + [a + d] + [a + d] + + [a + (n )d] + [a + (n )d] then again simply writing the terms in reverse order: A = [a + (n )d] + [a + (n )d] + + [a + d] + [a + d] + [a] Adding these two identical equations together we have A = [a + (n )d] + [a + (n )d] + + [a + (n )d] That is, every one of the n terms on the right-hand side has the same value: [a + (n )d]. Hence A = n[a + (n )d] so A = n[a + (n )d]. Key Point The arithmetic series [a] + [a + d] + [a + d] + + [a + (n )d] having n terms has sum A where: A = n[a + (n )d] HELM (008): Section 6.: Sequences and Series 9
10 As an example has a =, d =, n = 4 So A = = 4 [ + (3)] = 96. The geometric series We can also sum a general geometric series. Let G = a + ar + ar + + ar n be a geometric series having exactly n terms. To obtain the value of G in a more convenient form we first multiply through by the common ratio r: rg = ar + ar + ar ar n Now, writing the two series together: G = a + ar + ar + + ar n rg = ar + ar + ar 3 + ar n + ar n Subtracting the second expression from the first we see that all terms on the right-hand side cancel out, except for the first term of the first expression and the last term of the second expression so that G rg = ( r)g = a ar n Hence (assuming r ) G = a( rn ) r (Of course, if r = the geometric series simplifies to a simple arithmetic series with d = 0 and has sum G = na.) The geometric series having n terms has sum G where Key Point 3 a + ar + ar + + ar n G = a( rn ), if r and G = na, if r = r 0 HELM (008): Workbook 6: Sequences and Series
11 Task Find the sum of each of the following series: (a) (b) (a) In this arithmetic series state the values of a, d, n: a = d = n = a =, d =, n = 00. Now find the sum: = = 50( + 99) = 50(0) = (b) In this geometric series state the values of a, r, n: a = r = n = a =, r = 3, n = 6 Now find the sum: = = ( ( ) ) = 3 4 ( ( ) ) 6 = HELM (008): Section 6.: Sequences and Series
12 Exercises. Which of the following sequences is convergent? (a) sin π, sin π, sin 3π, sin 4π,... (b) sin π π, sin π π, sin 3π 3π, sin 4π 4π,.... Write the following series in summation form: (a) (b) ln + ln ln 5 ln ( + (00) ) + 3 ( (00) ) 4 ( + (300) ) ( (800) ) 3. Write out the first three terms and the last term of the following series: (a) 7 p= 4. Sum the series: 3 p p!(8 p) (b) 7 p=4 (a) (b) ( p) p+ p( + p) (c) s. (a) no; this sequence is, 0,, 0,,... which does not converge. (b) yes; this sequence is. (a) 4 p= ln(p ) (p + )(p) π/, 0, 3π/, 0,,... which converges to zero. 5π/ (b) 8 p= 3. (a) 7, 3!(6), 3 3!(5),..., 36 7! ( ) p (p + )( + ( ) p+ p 0 4 ) (b) 45 (4)(6), 5 6 (5)(7), 67 (6)(8),..., 7 8 (7)(9) 4. (a) This is an arithmetic series with a = 5, d = 4, n = 9. A = 99 (b) This is an arithmetic series with a = 5, d = 4, n = 9. A = 89 (c) This is a geometric series with a =, r =, n = 6. G HELM (008): Workbook 6: Sequences and Series
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2) Based on the information in the table which choice BEST shows the answer to 1 906? 906 899 904 909
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) Multiplying a number by results in what type of. even. 0. even.,0. odd..,0. even ) Based on the information in the table which choice BEST shows the answer to 0? 0 0 0 ) |
Question
1. Use Synthetic Division To Solve (X3 + 1) ÷ (X – 1). What Is The Quotient?
Solving polynomials with synthetic division is a great way to work out complicated equations quickly and easily. This technique allows you to divide polynomials by linear binomials, such as (X – 1), in a short and effective manner. In this blog post, we’ll explore how to use synthetic division to solve the equation (X3 + 1) ÷ (X – 1). We’ll look at the steps required to find the quotient of this equation and explain why it works. By the end of this post, you should have a good understanding of how synthetic division works and how it can make solving polynomials much simpler!
What is Synthetic Division?
Synthetic division is a method of polynomial division that allows for the division of polynomials without actually having to do any long division. This makes synthetic division much faster and easier than traditional polynomial division. To divide a polynomial by another polynomial using synthetic division, you simply need to take the coefficients of the dividend (the number being divided) and divide them by the leading coefficient of the divisor (the number doing the dividing). The quotient will be the answer to your division problem.
How to Use Synthetic Division
To use synthetic division to solve for the quotient of (x+)/(x-), first divide the leading coefficient of the dividend, which is x+ in this case, by the leading coefficient of the divisor, x-. Then, write down the answer as the first term in the quotient, omitting any negative signs. Next, take the product of this answer and the divisor, and subtract it from the dividend. This will give you a new dividend with one fewer term. Continue this process until you have no terms left in your dividend. The final answer will be your quotient.
The Quotient
In synthetic division, we divide one polynomial by another using a simplified method. In this case, we’re dividing (x+3) by (x-2). To do this, we’ll need to write our dividend and divisor in standard form:
Dividend: x^2 + 3x + 9
Divisor: x^2 – 2x
We’ll start by setting up our division problem like this:
( )( )( )( )( )
−−−−−−−−−−−→÷ −−→
x^2-2x 3x+9
Next, we’ll need to multiply each term in the divisor by the first term in the dividend, and write these products beneath the corresponding terms in the dividend. Then, we’ll subtract these products from the dividend and bring down the next term in the dividend:
( )( )( )( )( )
−−−−−→÷ −→ 3x^2-6x 18x-54 -27
x^2-2x x^2+3x x^2+3x 9x+27
———- —————– ———
3x^2 -3x^2 -18x 27
+ +
Conclusion
As we have seen, synthetic division is a powerful and versatile tool for solving polynomial equations. In this article, we used it to solve the equation (x3 + 1) ÷ (x – 1). We found that the quotient of this equation was x2 + 2x + 1. By using synthetic division, we were able to quickly and easily find the answer without having to do any long calculations or tedious algebraic manipulations. Synthetic division can be an invaluable resource in your math arsenal!
2. Synthetic division is a useful tool for quickly finding the quotient of a polynomial divided by a linear factor. It can be used to solve (X^3 + 1) (X + 1). This article will explain how synthetic division works and provide step-by-step instructions on how to use it to find the quotient of this equation.
Synthetic division allows you to divide two polynomials without having to multiply out each term in the divisor. This makes it much faster than other methods, such as long division. To use it, you need to rearrange the equation so that the highest degree term is on the left side and then put zeroes in all of the empty spaces below it. Then, divide each row by the divisor and move down one row at a time until you have reached your answer.
3. Hey everyone! Welcome to our blog post on how to use synthetic division to solve the equation (x³ + 1) ÷ (x – 1) and find the quotient!
Synthetic division is an efficient way to divide an equation by another one, and it is often used to divide polynomials. In this case, we are dividing a cubic polynomial (x³ + 1) by a linear polynomial (x – 1).
Before we begin, let’s quickly review synthetic division. Synthetic division is a process where you divide one polynomial by another, and it is similar to the long division process you may have learned in school. The only difference is that you don’t have to write down the long division step by step.
Now that we know what synthetic division is, let’s take a look at how it works. To begin, we will place the linear polynomial (x – 1) on the left side of the division sign and the cubic polynomial (x³ + 1) on the right side. Next, we will create a division table and fill in the first two columns. The first column will contain the coefficients of the polynomials, starting from the highest degree term. The second column will contain the results of the division process.
If you look at the table, you can see that the first row is divided by x -1 and the remaining rows are divided by the number in the previous row. After we fill out the division table, we can find the quotient. The quotient is the last number in the second column of the division table. In this case, the quotient is x² + 1.
So there you have it! Using synthetic division, we have successfully divided (x³ + 1) ÷ (x – 1) and found the quotient. We hope that this post has been helpful in explaining how to use synthetic division to divide polynomials.
4. Have you ever been stuck on a math problem and just wished there was an easier way to solve it? You’re not alone! Thankfully, synthetic division is a great way to simplify polynomial division and help you figure out the answer to complicated problems like this one.
In this blog post, we’ll be using synthetic division to solve the equation (X4 – 1) ÷ (X – 1). After following our step-by-step instructions, you’ll be able to find the quotient of this tricky equation in no time.
What is synthetic division, anyway? Synthetic division is a special type of polynomial division that’s designed to simplify the process of finding the quotient of a polynomial expression. It’s a great tool for quickly solving equations and can be used to solve any polynomial expression that has a divisor with a degree of one.
So, let’s get started! First, we’ll rewrite the equation so that it’s in the form of (X4 – 1) ÷ (X – 1) = Q. Then, we’ll arrange our equation in the following format:
X: -1 | 0 | 0 | 1 | -1
Next, we’ll begin the synthetic division process by performing the following steps:
Step 1: Multiply the first coefficient (in this case, -1) by the divisor (in this case, X – 1).
Step 2: Add the result to the second coefficient (in this case, 0).
Step 3: Multiply the result from step two by the divisor (in this case, X – 1).
Step 4: Add the result from step three to the third coefficient (in this case, 0).
Step 5: Multiply the result from step four by the divisor (in this case, X – 1).
Step 6: Add the result from step five to the fourth coefficient (in this case, 1).
Step 7: Multiply the result from step six by the divisor (in this case, X – 1).
Step 8: Add the result from step seven to the fifth coefficient (in this case, -1).
After performing all of the steps, you should end up with the following equation:
X: -1 | -1 | -2 | -2 | -3
The answer to our equation is the coefficient of the fourth power of X, which in this case is -2. This means that the quotient of (X4 – 1) ÷ (X – 1) is -2.
Congratulations! You just solved a tricky polynomial equation using synthetic division. With a few steps, you were able to quickly and easily find the quotient of (X4 – 1) ÷ (X – 1).
Now that you know how to solve equations using synthetic division, you’ll be able to solve any polynomial expression with a divisor with a degree of one in no time. Give it a try and have fun exploring the world of math! |
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# Perimeter of a Figure
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## Introduction to Perimeter of a Figure
The route or boundary that encircles any shape in mathematics is defined by the shape's circumference. To put it another way, the circumference and perimeter both refer to the length of a shape's outline.
## What is the Perimeter?
A path that encircles a two-dimensional object is called a perimeter. The word is Greek, peri meaning "around," and "metre" meaning measure. The phrase can refer to either the path or its length; it can be compared to the size of a shape's outline.
## How to Find the Perimeter of a Figure?
Steps to find the Perimeter of a given figure:
STEP 1 : Calculate the lengths of the sides.
Usually we don't need a special formula to get perimeter because perimeter is simply a measure of a two-dimensional figure's outline (though there are equations for specific shapes to make it easier).
STEP 2 : The length of each side is added together.
When calculating the perimeter of a non-circular object, add up all of the side lengths to get its perimeter.
STEP 3 : Be careful of the units of measurement.
We need to add all the lengths in the same unit to add else we can't add it (such as feet, miles, or metres).
## Perimeter of Square:
Since a square is a polygon with four equal sides, its perimeter will be equal to the sum of those sides.
The formula ${\rm{P = 4x}}$, where x is the length of one side, can be used to determine whether a square has four equal-length sides.
Square
## Perimeter of Rectangle:
The perimeter of a rectangle is determined by adding up all of its sides.
Formula $= 2(a + b)$ units
where:
“a” is the length of the rectangle
“b” is the breadth of the rectangle
Rectangle
## Perimeter of Circle:
Circumference of the circle or perimeter of the circle is the measurement of the boundary of the circle.
Circle perimeter or circumference is equal to$2\pi r$ .
where,
$\Pi$ is a mathematical constant with an estimated value of 3.14 (to the nearest two decimal places).
Circle
## Solved Examples:
1: Find the perimeter of the square whose side is 6 units.
Ans: As we know that all the sides of a square are equal.
And its perimeter is 4 times its side.
Thus, it’s perimeter will be $4 \times 6 = 24$
Its perimeter is 24 units.
2: Find the perimeter of the following figure:
Ans: The perimeter of a planar object is the sum of all the sides in it.
It can alternatively be described as the total sum of all of a planar object's bounding sides.
Here,$AB + BC + CD + DA = 5 + 9 + 7 + 10$ is the total length of all the sides of the provided figure.
Thus, The stated figure's perimeter is 31 metres.
3 :Determine the circle's radius for$C = 50cm$ .
Ans: Given that 50 cm is the circumference.
According to the formula$C = 2\pi r$
This indicates that
$\begin{array}{l}50 = 2\pi r\\25 = \pi r\\r = 25/\pi \end{array}$
As a result, the circle's radius is $25/\pi$ cm.
## Conclusion:
Any 2D shape's perimeter can be calculated by labelling and adding the sides. When calculating the perimeter, don't forget to account for the length of each side. For instance, the lengths of a rectangle's two widthwise edges and its two lengthwise borders are added to form the rectangle's perimeter and similar for other shapes also.
## FAQs on Perimeter of a Figure
1. How can I determine the perimeter of a non-regular (irregular) shape?
The lengths of all the sides must be added together in order to determine an irregular shape's perimeter. If such lengths are not provided, you can be given alternative information that can assist you in locating those lengths.
2. Is perimeter adding or multiplying the sides together?
The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.
3. What is the circumference of a circle?
The distance along a circle's perimeter is referred to as its circumference. In other words, the circumference of a circle is equal to the length of a straight line formed by opening up the circle.
4. Determine the circumference of a circle whose radius is 5 units.
Given , $Radius = 5 units$
We know that , $C = 2\pi r$
$\begin{array}{l}C = 2 \times \pi \times 5\\C = 10\pi \end{array}$
Thus , Circumference = 10 π units.
5. What is the importance of perimeter in our daily life?
In daily life, we frequently use the concept of perimeter. |
The Tangent Line Approximation
Suppose we want to find the tangent to a curve. Just how can we go about finding one?
Here is one way:
• Pick a point $Q$ by clicking on the curve on the applet. (The line that appears is the secant line between $P$ and $Q$.)
• Now drag point $Q$ towards point $P$.
As $Q$ approaches $P$, the secant line approximates the tangent line better and better. The limiting position of the secant line as $Q$ approaches $P$ is the tangent to the curve at $P$.
If the curve is given by $y=f(x)$ and $P$ has the coordinates $(x_0,y_0)$, then the slope of the tangent line at $P$ is $f'(x_0)$, the derivative of f evaluated at $x_0$.
Let's find the equation of the line tangent to the parabola at $(2,3)$.
• Drag point $P$ to $(2,3)$.
• Now pick another point $Q$ on the parabola and drag $Q$ towards $P$ to find the tangent to the curve at $P$.
The slope of the tangent is just $f'(x)$ evaluated at x. \begin{eqnarray*} f(x) &=& x^2-1 \\ f'(x) &=& 2x \\ f'(2) &=& 4. \end{eqnarray*} Now, the equation of the line can be written in point-slope form like this: \begin{eqnarray*} y-y_0 &=& m(x-x_0)\\ y-y_0 &=& f'(x_0)(x-x_0)\\ y-3 &=& 4(x-2) \end{eqnarray*} since the line passes through the point $(2,3)$ and has slope $4$.
In slope-intercept form, the equation of the tangent line becomes $$y=4x-5.$$
• Drag $P$ along the parabola or enter the x-coordinate for point $P$.
• Notice how the equation of the tangent line changes as you move point $P$.
What happens when $x=0$ for this function? What about as $|x|$ gets large?
Now that we can find the tangent to a curve at a point, of what use is this?
• "Magnify" the parabola by zooming in on point $P$.
Do you notice that as you zoom in on $P$ the curve looks more and more linear and is approximated better and better by the tangent line?
Let's get more specific:
Near $x_0$, we saw that $y=f(x)$ can be approximated by the tangent line $y-y_0=f'(x_0)(x-x_0)$. Writing this as $y=y_0+f'(x_0)(x-x_0)$ and noting that $y=f(x_0)$, we find that
$f(x)\approx f(x_0) + f'(x_0)(x-x_0).$
(Notice that the right-hand side is just the 2-term Taylor Expansion of $f(x)$.)
If we know that value of $f$ at $x_0$, this gives us a way to approximate the value of $f$ at $x$ near $x_0$. We do this by starting at $(x_0,f(x_0))$ and moving along the tangent line to approximate the value of the function at $x$.
Look at $f(x) = \arctan{x}$.
Let's use the tangent approximation $f(x) \approx f(x_0)+f'(x_0)(x-x_0)$ to approximate $f(1.04)$:
• Now $f'(x) = \left[\frac{1}{1+x^2}\right]$ so $f'(1)=\left[\frac{1}{1+1^2}\right]=\frac{1}{2}$.
• Let $x_0 = 1$ and $x = 1.04$.
• Then $f(1.04) \approx f(1) + f'(1)(1.04 - 1) \approx \frac{\pi}{4} + \frac{1}{2}(0.04) \approx 0.81$.
How well does this approximate $\arctan(1.04)$?
• Display the tangent through $\left( 1, \frac{\pi}{4}\right)$.
• Zoom in on the point to see geometrically how close together the curve and the tangent line are at $x = 1.04$.
### Key Concepts
• For the curve $y = f(x)$, the slope of the tangent line at a point $(x_0,y_0)$ on the curve is $f'(x_0)$. The equation of the tangent line is given by $$y-y_0 = f'(x_0)(x-x_0).$$
• For $x$ close to $x_0$, the value of $f(x)$ may be approximated by $$f(x) \approx f(x_0)+f'(x_0)(x-x_0).$$ |
> > > > NCERT Solutions for Class 9 Maths Chapter 3 : Coordinate Geometry
# NCERT Solutions for Class 9 Maths Chapter 3 : Coordinate Geometry
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NCERT Maths Solutions Class 9 gives the answers to chapter 3 of the syllabus – Coordinate geometry. Chapter 3 has a total of 3 exercises for the students to solve. You will be studying the concept of the cartesian plane, coordinates of a point in this XY – plane, name, terms, notations, and other terms associated with the coordinate plane. Abscissa and ordinate of points. You will learn to Plot a point in XY – plane and the process of naming it.
NCERT Solutions for Class 9 Maths Chapter 3 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry are designed as per the CBSE Class 9 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 9 Chapter 3.
## NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry : EX 3.1
#### Question 1:How will you describe the position of a table lamp on your study table toanother person?
Solution:
From the above Figure, Consider the lamp on the table as a point.
Consider the table as a plane.
We can conclude that the table is rectangular in shape, when observed from the top.
The table has a short edge and a long edge.
Let us measure the distance of the lamp from the shorter edge and the longer edge.
Let us assume: Distance of the lamp from the shorter edge is 15 cm.
Distance of the lamp from the longer edge, its 25 cm.
Therefore, we can conclude that the position of the lamp on the table can be described in two
ways depending on the order of the axes as (15, 25) or (25, 15).
#### crossing, then we will call this cross-street (2, 5). Using this convention,find:(i) How many cross – streets can be referred to as (4, 3).(ii) How many cross – streets can be referred to as (3, 4).
Solution:
Draw two perpendicular lines as the two main roads of the city that cross each other at the center.
Mark it as N–S and E–W.
Let us take the scale as 1 cm = 200 m.
Draw five streets that are parallel to both the main roads, to get the given below figure.
Street plan is as shown in the figure:
(i) There is only one cross street, which can be referred as (4, 3).
(ii) There is only one cross street, which can be referred as (3, 4).
## NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry : EX 3.2
#### Question1:Write the answer of each of the following questions:(i) What is the name of horizontal and the vertical lines drawn to determinethe position of any point in the Cartesian plane?(ii) What is the name of each part of the plane formed by these two lines?(iii) Write the name of the point where these two lines intersect.
Solution:
(i)
The horizontal line that is drawn to determine the position of any point in the Cartesian plane is called as x-axis.
The vertical line that is drawn to determine the position of any point in the Cartesian plane is called as y-axis.
(ii)
The name of each part of the plane that is formed by x-axis and y-axis is called as quadrant.
(iii)
The point, where the x-axis and the y-axis intersect is called as origin (O).
#### Question 2:See the figure, and write the following:(i) The coordinates of B.(ii) The coordinates of C.(iii) The point identified by the coordinates (–3, –5).(iv) The point identified by the coordinates (2, –4).(v) The abscissa of the point D.(vi) The coordinate of the point H.(vii) The coordinates of the point L.(viii) The coordinates of the point M.
Solution:
From the above figure,
(i)
The coordinates of point B is the distance of point B from x-axis and y-axis.
Therefore, the coordinates of point B are (–5, 2).
(ii)
The coordinates of point C is the distance of point C from x-axis and y-axis.
Therefore, the coordinates of point C are (5, –5).
(iii)
The point that represents the coordinates (–3, –5) is E.
(iv)
The point that represents the coordinates (2, –4) is G.
(v)
The abscissa of point D is the distance of point D from the y-axis. Therefore, the abscissa of point D is 6.
(vi)
The ordinate of point H is the distance of point H from the x-axis. Therefore, the abscissa of point H is –3.
(vii)
The coordinates of point L in the above figure is the distance of point L from x-axis and yaxis. Therefore, the coordinates of point L are (0, 5).
(viii)
The coordinates of point M in the above figure is the distance of point M from x-axis and yaxis. Therefore the coordinates of point M are (–3, 0).
## NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry : EX 3.3
#### Question 1:In which quadrant or on which axis do each of the points (–2, 4), (3, –1),(–1, 0), (1, 2) and (–3, –5) lie? Verify your answer by locating them on theCartesian plane.
Solution:
To determine the quadrant or axis of the points (–2, 4), (3, –1), (–1, 0), (1, 2) and
(–3, –5).
Plot the points (–2, 4), (3, –1), (–1, 0), (1, 2) and (–3, –5) on the graph.
From the figure, we can conclude that the points,
Point (–2, 4) lie in IInd quadrant.
Point (3, –1) lie in IVth quadrant.
Point (–1, 0) lie on the negative x-axis.
Point (1, 2) lie in Ist quadrant.
Point (–3, –5) lie in IIIrd quadrant.
#### Question 2:Plot the points (x, y) given in the following table on the plane, choosingsuitable units of distance on the axes.
Solution:
Draw X’OX and Y’OY as the coordinate axes and mark their point of intersection O as the Origin (0, 0).
In order to plot the table provided above,
Point A (–2, 8) – Take 2 units on OX’ and then 8 units parallel to OY.
Point B (–1, 7) – Take -1 unit on OX’ and then 7 units parallel to OY
Point C (0, –1.25) – Take 1.25 units below x-axis on OY’ on the y-axis
Point D (1, 3) – Take 1 unit on OX and then 3 units parallel to OY
Point E (3, –1) – Take 3 units on OX and then move 1 unit parallel to OY |
# 2004 AIME I Problems/Problem 4
## Problem
A square has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$.
## Solution
Without loss of generality, let $(0,0)$, $(2,0)$, $(0,2)$, and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$. Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$. Because the segment has length 2, $x^2+y^2=4$. Using the midpoint formula, we find that the midpoint of the segment has coordinates $\left(\frac{x}{2},\frac{y}{2}\right)$. Let $d$ be the distance from $(0,0)$ to $\left(\frac{x}{2},\frac{y}{2}\right)$. Using the distance formula we see that $d=\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2}= \sqrt{\frac{1}{4}\left(x^2+y^2\right)}=\sqrt{\frac{1}{4}(4)}=1$. Thus the midpoints lying on the sides determined by vertex $(0,0)$ form a quarter-circle with radius 1.
$[asy] size(100); pointpen=black;pathpen = black+linewidth(0.7); pair A=(0,0),B=(2,0),C=(2,2),D=(0,2); D(A--B--C--D--A); picture p; draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1)); clip(p,A--B--C--D--cycle); add(p); [/asy]$
The set of all midpoints forms a quarter circle at each corner of the square. The area enclosed by all of the midpoints is $4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86$ to the nearest hundredth. Thus $100\cdot k=086$ |
How Work and Power Are Related
We know work and power to be things that we hear in our daily lives, whether it’s working at a hospital to having the power to rule the world. But, in the world of physics, the terms work and power are use when someone wants to calculate the amount of energy in a certain situation. Work is the effort exerted on something that will change its energy. The heavier something is or the higher something is, more work is done. To calculate work, you need to find how much force is applied to an object and the distance in which the object is moved. The formula easily converts to: work = force times distance / W (joules)= F (Newtons) x d (meters). So if, let’s say, you want to calculate how much work is needed to lift a pack of soda that weighs 300 N to a flight of stairs that’s 2 meters high. To find the solution, simply insert the components for the formula and the final answer is: 600 J.
When you want to find how much power is used in a certain situation, you need to find the amount of work done and the time it takes. Hence the formula for power: power (Watts) = work (joules) / time interval. To put it more generally, power is the rate at which energy expands. When it comes to doing the same amount of work or a certain action, more power will be calculated if someone does twice the amount of work in the same time or the same amount of work in half the time. Let’s use the same example above and say, how much power is used when you lift a pack of soda weighing 300 N up 2 meters in 20 seconds. Since you already found how much work, which is 600 J. So, all you have to do is divide 600 by 20, which results to 30 watts.
Let’s do another example:
How much power expends when a 20-N force pushes a cart 3.5 meters in a time of 0.5 seconds?
First, you have to find the work done. Going back to the formula of work, w = f x d, you have the force and distance given. So, you multiply 20 by 3.5, resulting to 70 joules. With the work found, you can found the power: you divide 70 by 0.5, resulting in 140 watts.
To conclude, without knowing the work done in an action, you won’t know the power. |
# How do you solve the system of equations -3x-4y=20 and x-10y=16?
$x = - 4$ and $y = - 2$
#### Explanation:
Given system of equations
$- 3 x - 4 y = 20 \setminus \ldots \ldots . . \left(1\right)$
$x - 10 y = 16 \setminus \ldots \ldots . . \left(2\right)$
Multiplying (2) by $3$ & adding to (1) as follows
$- 3 x - 4 y + 3 \left(x - 10 y\right) = 20 + 3 \setminus \cdot 16$
$- 34 y = 68$
$y = - 2$
setting $y = - 2$ in (1) we get
$x = \setminus \frac{- 4 y - 20}{3}$
$= \setminus \frac{- 4 \left(- 2\right) - 20}{3}$
$= - 4$
Jul 9, 2018
Express $x$ as a function of $y$ and replace in the second equation.
#### Explanation:
To get a very fast answer to your problem, just use one of the two equations to express $x$ or $y$. In this case, let's do it with $x$.
So, we have the following system:
1) \ -3x-4y=20
2) \ x-10y=16
If we express $x$ in 2), then we have:
$x = 16 + 10 y$
Then we replace $x$ in 1) with what we obtained with 2) and we get:
$- 3 \cdot \left(16 + 10 y\right) - 4 y = 20$
Then develop the brackets:
$- 48 - 30 y - 4 y = 20$
Solve for $y$:
$- 34 y = 68$
$y = - 2$
Then replace $y$ in 2) by what you found:
$x - 10 \cdot \left(- 2\right) = 16$
Solve for $x$:
$x = - 4$
Finished! |
Question
# The total income of $A$ and $B$ is $Rs\text{ }6000$. $A$ spends $60%$ of his income and $B$ spends $80%$ of his income if their savings are equal then what is the income of $A$?A) $2000$B) $2400$C) $2600$D) $2800$
Hint:In case of unknown always assume any variable as the reference value. So, here we will first assume $A's$ income. Convert given word statements in the form of mathematical expressions. Find the amount of savings and accordingly form the equation. Compare both the savings amount and simplify.
Complete step-by-step solution:
Given: Total income of $A$ and $B$$=\text{ }6000$, $A$spend $60%$ of his income and $B$ spends $80%$ of his income and both have equal savings.
Let income of$A$is $x$
Then income of $B$will be $6000-x$[ As, $income\text{ }of\text{ }A\text{ }+\text{ }income\text{ }of\text{ }B\text{ }=6000$]
$A$spends $60%$of his income.
So, $A's$ Saving is $x\text{ }-\text{ }60%\text{ }of\text{ }x$
Convert it in the mathematical form and simplify.
\begin{align} & =x-\dfrac{60}{100}of\text{ }x \\ & =x-\dfrac{6x}{10} \\ & =\dfrac{10x-6x}{10} \\ & =\dfrac{4x}{10}.............(1) \\ \end{align}
Similarly Now, $B$ spends $80%$ of his income
So, $B's$ saving is $\left( 6000\text{ }-\text{ }x \right)\text{ }-\text{ }80%\text{ }of\text{ }\left( 6000\text{ }\text{ }x \right)$
Simplify the above equations -
\begin{align} & \Rightarrow \left( 6000-x \right)-\dfrac{80}{100}of(6000-x) \\ & =\left( 6000-x \right)-\dfrac{8(6000-x)}{10} \\ & =\dfrac{10\left( 6000-x \right)-8(6000-x)}{10} \\ & =\dfrac{60000-10x-48000+8x}{10} \\ & =\dfrac{12000-2x}{10}.................(2) \\ \end{align}
According to question eq(1) is equal to eq(2) as both $A$ and $B$ has equal saving.
$\dfrac{4x}{10}=\dfrac{12000-2x}{10}$
[Same denominator from both sides of the equations cancels each other]
\begin{align} & \Rightarrow 4x=12000-2x \\ & \Rightarrow 4x+2x=12000 \\ & \Rightarrow 6x=12000 \\ & \therefore x=\dfrac{12000}{6} \\ & \therefore x=2000 \\ \end{align}
Therefore, the required solution – The income of A is $=Rs.\ \text{2000}$
Hence from the given multiple choices, option A is the correct answer.
Note: In these types of problems we first assume the unknown quantity that relates most quantities of the problem. Then form the equation for all the required conditions. Then compare the final two equations to find the required value. Always double check the conversion of word statements to the mathematical form rest goes well and just simplify for the required result. |
# Quick Answer: How Do I Find The Length Of The Third Side Of A Triangle?
## How do you find the missing side of a rectangle?
Explanation: To find the width, multiply the length that you have been given by 2, and subtract the result from the perimeter.
You now have the total length for the remaining 2 sides.
This number divided by 2 is the width..
## What do the side of a triangle add up to?
The most basic fact about triangles is that all the angles add up to a total of 180 degrees. The angle between the sides can be anything from greater than 0 to less than 180 degrees. The angles can’t be 0 or 180 degrees, because the triangles would become straight lines. … So, 45º means 45 degrees.
## How do I find the third side of a triangle?
The Triangle Inequality theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.So, difference of two sides
## How do I find the length of a side of a triangle?
The Pythagorean theorem states that, in a right triangle, the square of the length of the hypotenuse (the side across from the right angle) is equal to the sum of the squares of the other two sides. So if the length of the hypotenuse is c and the lengths of the other two sides are a and b, then c^2 = a^2 + b^2.
## How do you find the minimum and maximum length of the third side of a triangle?
By the Triangle Inequality Theorem, the sum of lengths of any two sides of a triangle is greater than the length of the third side. So, The minimum would be 6 and the maximum would be 20. Cleomenius.
## What is the range of the third side of a triangle?
Any one side of a triangle must be shorter than the sum of the other two sides. Therefore, the third side length x must be between 5 and 35, not including those endpoints: 5 < x < 35. All whole numbers satisfying this range would form a triangle.
## How do you find the third side of an isosceles triangle?
A altitude between the two equal legs of an isosceles triangle creates right angles, is a angle and opposite side bisector, so divide the non-same side in half, then apply the Pythagorean Theorem b = √(equal sides ^2 – 1/2 non-equal side ^2). |
Q:
# an annuity pays $2,000 at the end of each half-year for 5 years and then$1,000 at the end of each half-year for the next 8 years. determine the discounted value of these payments if interest is 4% with semi-annual compounding
Accepted Solution
A:
To determine the discounted value of these payments, we need to find the present value of each payment and then sum them up.
For the first 5 years, the annuity pays 2,000 at the end of each half-year. We can use the formula for the present value of an annuity:
$$PV = \frac{P(1 - (1 + r)^{-n})}{r}$$
Where:
- PV is the present value
- P is the payment amount
- r is the interest rate per period
- n is the number of periods
In this case, the payment amount is 2,000, the interest rate per period is 4% / 2 = 0.02, and the number of periods is 5 * 2 = 10.
Calculating the present value for the first 5 years:
$$PV_1=\frac{2000(1-(1+0.02)^{-10})}{0.02}$$
For the next 8 years, the annuity pays 1,000 at the end of each half-year. Using the same formula with the adjusted payment amount and number of periods:
$$PV_2=\frac{1000(1-(1+0.02)^{-16})}{0.02}$$
To find the total discounted value, we simply sum up the present values:
$$\text{Total PV} = PV_1 + PV_2$$
Calculating the present values:
$$PV_1=\frac{2000(1-(1+0.02)^{-10})}{0.04}$$
$$PV_2=\frac{1000(1-(1+0.02)^{-16})}{0.04}$$
Simplifying the equations and solving:
$$PV_1=17965.17$$
$$PV_2=13577.71$$
Now, we can find the total discounted value:
$$\text{Total PV} = PV_1 + PV_2$$
$$\text{Total PV}=17965.17+\frac{13577.71}{\left(1.02\right)^{16}}$$
$$\text{Total PV}=27855.80$$
Answer: The discounted value of these payments is 27855.80 dollars. |
Answers To All Questions In The World
# How the Medians of a Triangle Are Concurrent
#### ByJohn Amelia
Mar 21, 2022
A triangle has three medians, each going from the vertex to the midpoint of the ‘opposite’ side. Because of this, all three medians intersect at the centroid. The centroid is the point where the three medians meet, dividing the triangle into two equal sections. The distance from the vertex to the centroid is exactly two-thirds of the length of a median.
A triangle’s centroid is the point at which the three medians intersect. However, it’s important to remember that the centroid is not necessarily inside the triangle; sometimes the medians intersect outside the triangle. This makes the centroid of the triangle a bit more challenging. To avoid a problem that arises when locating the center of a triangle, find the center of the other side of the triangle.
#### What Are The 3 Medians Of A Triangle Intersect?
The median of a triangle is the line segment that goes from the vertex to the midpoint of the opposite side. A triangle’s centroid is the point where the three medians intersect. This point is referred to as the orthocenter. This is the point where the three vertices of a triangle meet. This point is called the centroid. The diagonals of a triangle can’t cross each other.
#### Do All Medians Intersect At The Same Point?
The centroid of a triangle is the point at which the three medians of a triangle intersect. The two diagonals that divide the triangle have a 2:1 ratio. As the diagonals of the two halves of the triangle are the same length, the diagonals will be equal in length. Similarly, the triangle’s centroid is the point of gravity. It is a center of gravity.
#### Do Medians Intersect?
The median of a triangle is the line segment that connects the vertex with the midpoint of the opposite side. It divides the opposite side into two equal triangles. The third median, also known as the centroid, is located two-thirds of the distance between the vertices and the midpoint of the opposite side. Once the third and fourth medians of a triangle are equal, they form the centroid of the triangle.
This point is also called the intersection point. As the intersection of two medians is equal, the third will be equal to the first. The third one will be equal to the centroid. A similar method will be used to find the third one. Once you have determined the triangle’s centroid, use the method of finding it.
#### Where Do The Three Angle Bisectors Of A Triangle Intersect?
The three medians of a triangle intersect at a point on the opposite side. In fact, all three medians are equal in area and share the same centroid. When they cross, they will meet at the centroid. This point is called the centroid of the triangle. The diagonal of a triangle is called the diagonal. The vertex is the center of the triangle. The other three points are called the sides’ vertices.
#### Where Does The Three Angle Centroid Of A Triangle Intersect?
In the case of a triangle, the three medians of the triangle intersect at the centroid. In a scalene triangle, the medians of the three sides of a triangle are not the same. But they are equal in area, and this is the same as the medians of a triangle. This point is called the orthocenter. If the intersection of the two middles is the same, the diagonals of the middle are the same.
When the three medians of a triangle intersect, it is said to be a triangle. In an isosceles triangle, the two medians are of the same length, but they do not intersect in an equilateral triangle. In an equilateral triangle, the three medians are equal in length, and the centroid is the midpoint. Hence, the length of the three medians of a triangle is the same in all sides.
#### John Amelia
Hey, John here, a content writer. Writing has always been one of the things that I’m passionate about. Whenever I have something on my mind, I would jot it down or type it in my notes. No matter how small or pathetic it seems, You will really enjoy my writing.
#### Girly Tattoo Ideas
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# 7.3E: Series Solutions Near an Ordinary Point II (Exercises)
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In Exercises [exer:7.3.1} –[exer:7.3.12} find the coefficients $$a_0$$,…, $$a_N$$ for $$N$$ at least $$7$$ in the series solution $$y=\sum_{n=0}^\infty a_nx^n$$ of the initial value problem.
[exer:7.3.1] $$(1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3$$
[exer:7.3.2] $$(1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2$$
[exer:7.3.3] $$(1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0$$
[exer:7.3.4] $$(1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1$$
[exer:7.3.5] $$(2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3$$
[exer:7.3.6] $$(3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3$$
[exer:7.3.7] $$(4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5$$
[exer:7.3.8] $$(2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1$$
[exer:7.3.9] $$(3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1$$
[exer:7.3.10] $$(1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1$$
[exer:7.3.11] $$(2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3$$
[exer:7.3.12] $$x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2$$
[exer:7.3.13] Do the following experiment for various choices of real numbers $$a_0$$, $$a_1$$, and $$r$$, with $$0<r<1/\sqrt2$$.
Use differential equations software to solve the initial value problem
$(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$
numerically on $$(-r,r)$$. (See Example
Example $$\PageIndex{1}$$:
Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at7.3.1}.)
For $$N=2$$, $$3$$, $$4$$, …, compute $$a_2$$, …, $$a_N$$ in the power series solution $$y=\sum_{n=0}^\infty a_nx^n$$ of (A), and graph
$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$
and the solution obtained in (a) on $$(-r,r)$$. Continue increasing $$N$$ until there’s no perceptible difference between the two graphs.
[exer:7.3.14] Do the following experiment for various choices of real numbers $$a_0$$, $$a_1$$, and $$r$$, with $$0<r<2$$.
Use differential equations software to solve the initial value problem $(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1, \eqno{\rm(A)}\nonumber$
numerically on $$(-1-r,-1+r)$$. (See Example 7.3.2}. Why this interval?)
For $$N=2$$, $$3$$, $$4$$, …, compute $$a_2,\dots,a_N$$ in the power series solution
$y=\sum_{n=0}^\infty a_n(x+1)^n\nonumber$
of (A), and graph
$T_N(x)=\sum_{n=0}^N a_n(x+1)^n\nonumber$
and the solution obtained in (a) on $$(-1-r,-1+r)$$. Continue increasing $$N$$ until there’s no perceptible difference between the two graphs.
[exer:7.3.15] Do the following experiment for several choices of $$a_0$$, $$a_1$$, and $$r$$, with $$r>0$$.
Use differential equations software to solve the initial value problem $y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$
numerically on $$(-r,r)$$. (See Example 7.3.3}.)
Find the coefficients $$a_0$$, $$a_1$$, …, $$a_N$$ in the power series solution $$y=\sum_{n=0}^\infty a_nx^n$$ of (A), and graph
$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$
and the solution obtained in (a) on $$(-r,r)$$. Continue increasing $$N$$ until there’s no perceptible difference between the two graphs.
[exer:7.3.16] Do the following experiment for several choices of $$a_0$$ and $$a_1$$.
Use differential equations software to solve the initial value problem
$(1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$
numerically on $$(-r,r)$$.
Find the coefficients $$a_0$$, $$a_1$$, …, $$a_N$$ in the power series solution $$y=\sum_{n=0}^Na_nx^n$$ of (A), and graph
$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$
and the solution obtained in (a) on $$(-r,r)$$. Continue increasing $$N$$ until there’s no perceptible difference between the two graphs. What happens as you let $$r\to1$$?
[exer:7.3.17] Follow the directions of Exercise [exer:7.3.16} for the initial value problem $(1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber$
[exer:7.3.18] Follow the directions of Exercise [exer:7.3.16} for the initial value problem $(1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber$
[exer:7.3.19] $$(2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7$$
[exer:7.3.20] $$(1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2$$
[exer:7.3.21] $$(5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1$$
[exer:7.3.22] $$(4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2$$
[exer:7.3.23] $$(2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2$$
[exer:7.3.24] $$(3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3$$
[exer:7.3.25] $$(3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3$$
[exer:7.3.26] $$(10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4$$
[exer:7.3.27] $$(7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2$$
[exer:7.3.28] $$(6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2$$
[exer:7.3.29] Show that the coefficients in the power series in $$x$$ for the general solution of $(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0\nonumber$ satisfy the recurrrence relation $a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n.\nonumber$
[exer:7.3.30] Let $$\alpha$$ and $$\beta$$ be constants, with $$\beta\ne0$$. Show that $$y=\sum_{n=0}^\infty a_nx^n$$ is a solution of $(1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0 \eqno{\rm (A)}\nonumber$ if and only if $a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0. \eqno{\rm (B)}\nonumber$
An equation of this form is called a second order homogeneous linear difference equation. The polynomial $$p(r)=r^2+\alpha r+\beta$$ is called the characteristic polynomial of (B). If $$r_1$$ and $$r_2$$ are the zeros of $$p$$, then $$1/r_1$$ and $$1/r_2$$ are the zeros of $P_0(x)=1+\alpha x+\beta x^2.\nonumber$
Suppose $$p(r)=(r-r_1)(r-r_2)$$ where $$r_1$$ and $$r_2$$ are real and distinct, and let $$\rho$$ be the smaller of the two numbers $$\{1/|r_1|,1/|r_2|\}$$. Show that if $$c_1$$ and $$c_2$$ are constants then the sequence $a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0\nonumber$ satisfies (B). Conclude from this that any function of the form $y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n\nonumber$ is a solution of (A) on $$(-\rho,\rho)$$.
Use (b) and the formula for the sum of a geometric series to show that the functions
$y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}\nonumber$
form a fundamental set of solutions of (A) on $$(-\rho,\rho)$$.
Show that $$\{y_1,y_2\}$$ is a fundamental set of solutions of (A) on any interval that does’nt contain either $$1/r_1$$ or $$1/r_2$$.
Suppose $$p(r)=(r-r_1)^2$$, and let $$\rho=1/|r_1|$$. Show that if $$c_1$$ and $$c_2$$ are constants then the sequence
$a_n=(c_1+c_2n)r_1^n,\quad n\ge0\nonumber$
satisfies (B). Conclude from this that any function of the form
$y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n\nonumber$
is a solution of (A) on $$(-\rho,\rho)$$.
Use (e) and the formula for the sum of a geometric series to show that the functions
$y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}\nonumber$
form a fundamental set of solutions of (A) on $$(-\rho,\rho)$$.
Show that $$\{y_1,y_2\}$$ is a fundamental set of solutions of (A) on any interval that does not contain $$1/r_1$$.
[exer:7.3.31] Use the results of Exercise [exer:7.3.30} to find the general solution of the given equation on any interval on which polynomial multiplying $$y''$$ has no zeros.
(a) $$(1+3x+2x^2)y''+(6+8x)y'+4y=0$$
(b) $$(1-5x+6x^2)y''-(10-24x)y'+12y=0$$
(c) $$(1-4x+4x^2)y''-(8-16x)y'+8y=0$$
(d) $$(4+4x+x^2)y''+(8+4x)y'+2y=0$$
(e) $$(4+8x+3x^2)y''+(16+12x)y'+6y=0$$
[exer:7.3.32] $$y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2$$
[exer:7.3.33] $$y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2$$
[exer:7.3.34] $$y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2$$
[exer:7.3.35] $$y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5$$
[exer:7.3.36] $$y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6$$
[exer:7.3.37] $$2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2$$
[exer:7.3.38] $$3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3$$
[exer:7.3.39] Find power series in $$x$$ for the solutions $$y_1$$ and $$y_2$$ of $y''+4xy'+(2+4x^2)y=0\nonumber$ such that $$y_1(0)=1$$, $$y'_1(0)=0$$, $$y_2(0)=0$$, $$y'_2(0)=1$$, and identify $$y_1$$ and $$y_2$$ in terms of familiar elementary functions.
[exer:7.3.40] $$(1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3$$
[exer:7.3.41] $$y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3$$
[exer:7.3.42] $$(1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5$$
[exer:7.3.43] $$(1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3$$
[exer:7.3.44] $$y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3$$
[exer:7.3.45] $$(1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2$$
[exer:7.3.46] $$(3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1$$
[exer:7.3.47] $$(1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1$$
[exer:7.3.48] $$(x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0$$
[exer:7.3.49] $$(16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2$$ |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Triangle Area ( Read ) | Geometry | CK-12 Foundation
# Triangle Area
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Have you ever seen gardening designs in the median of a street?
Miguel saw gardening designs on Center Street. As he was walking, he noticed a series of triangles that line the median. The triangles are overlapping.
Let’s say that that the first triangle has a base of 6 feet and a height of 5 feet. What is the area of the triangle?
This Concept will teach you how to find the area of any triangle given the base and the height.
### Guidance
In the Estimation of Parallelogram Area in Scale Drawings Concept, we looked at how to find the area of a parallelogram. Here is that formula once again.
$A=bh$
This Concept focuses on the area of a triangle. Previously we learned that triangle is a three sided figure made up of line segments with three sides and three angles. Area is the amount of space inside a two-dimensional figure. We can find the area of a triangle just like we found the area of a parallelogram. The really interesting thing is that the area of a triangle is related to the area of a parallelogram. Take a look at this figure.
Notice that the parallelogram has been divided into two triangles.
We know that the formula for finding the area of a parallelogram requires us to multiply the base times the height. Well, if a triangle is one-half of a parallelogram, can you figure out the formula for finding the area of a triangle?
Here it is.
$A= \frac{1}{2} bh$
It certainly does. With this formula, the area of a triangle will be snap to figure out!
Take a minute to write this formula down in your notebook.
Now that you can identify formula for finding the area of a triangle, let’s look at using it in problem solving.
Find the area of the triangle below.
We can see that the base is 11 centimeters and the height is 16 centimeters. We simply put these numbers into the appropriate places in the formula.
$A & = \frac{1}{2} bh\\A & = \frac{1}{2} 11(16)\\A & = \frac{1}{2} (176) \\A & = 88 \ cm^2$
Remember that we always measure area in square units because we are combining two dimensions. The area of this triangle is 88 square centimeters. Let’s try another.
What is the area of the triangle below?
Notice that the height is shown by the dashed line. It is perpendicular to the base. We put it and the base into the formula and solve.
$A & = \frac{1}{2} bh \\A & = \frac{1}{2} 5(17) \\A & = \frac{1}{2} (85) \\A & = 42.5 \ cm^2$
The area of this triangle is 42.5 sq. cm. Now it’s time for you to try a few on your own.
Find the area of each triangle.
#### Example A
Base = 12 in, height = 6 inches
Solution: $36 \ in^2$
#### Example B
Base = 9 inches, height = 4 inches
Solution: $18 \ in^2$
#### Example C
Base = 11 inches, height = 7 inches
Solution: $38.5 \ in^2$
Here is the original problem once again.
Miguel saw gardening designs on Center Street. As he was walking, he noticed a series of triangles that line the median. The triangles are overlapping.
Let’s say that that the first triangle has a base of 6 feet and a height of 5 feet.
What is the area of the triangle?
To figure this out, we can start with the formula for finding the area of a triangle.
$A & = \frac{1}{2}bh\\A & = \frac{1}{2}(6)(5)\\A & = \frac{1}{2}(30)\\A & = 15 \ square \ feet$
Let’s say that there are eight triangles in this strip of median. We can take the area of one of the triangles and multiply it by 8.
15(8) = 120 square feet
This is the area of the median. We were able to use the area of each triangle to find the total area of the median.
### Vocabulary
Triangle
a figure with three sides and three angles.
Area
the space enclosed inside a two-dimensional figure.
Base
the bottom part of the triangle.
Height
the length of the triangle from the base to the vertex.
### Guided Practice
Here is one for you to try on your own.
What is the area of a triangle with a base of $4.5 \ ft$ and a height of $7 \ ft$ ?
To figure this out, we can use the formula for finding the area of a triangle.
$A = \frac{1}{2}bh$
Now we substitute in the given values.
$A = \frac{1}{2}(4.5)(7)$
$A = \frac{1}{2}(31.5)$
$A = 15.75 \ ft^2$
This is our answer.
### Practice
Directions: Find the area of each triangle given the base and height.
1. Base = 9 in, height = 4 in
2. Base = 6 in, height = 3 in
3. Base = 7 in, height = 4 in
4. Base = 9 m, height = 7 m
5. Base = 12 ft, height = 10 feet
6. Base = 14 feet, height = 5 feet
7. Base = 14 feet, height = 13 feet
8. Base = 11 meters, height = 8 meters
9. Base = 13 feet, height = 8.5 feet
10. Base = 11.5 meters, height = 9 meters
11. Base = 18 meters, height = 15 meters
12. Base = 21 feet, height = 15.5 feet
13. Base = 18 feet, height = 11 feet
14. Base = 20.5 meters, height = 15.5 meters
15. Base = 40 feet, height = 22 feet |
# How to Calculate the Total Area
••• ULTRA F/Photodisc/Getty Images
Print
Calculating total area has many real-world applications. You can use it to determine how many tiles are required to cover a floor, the square footage of a house, the size of a tablecloth needed for a particular table or the area covered by your sprinkler system. You may also have to calculate the area available in a room before purchasing new furniture. The task of calculating total area requires one of a few basic equations.
## Circle
••• Thomas Northcut/Lifesize/Getty Images
Measure the radius, r, of the circle. The radius is measured from the center of the circle to the edge. It is equal to half of the circle's diameter. For example, suppose a circle has a radius of 5 feet.
Square the radius. In the example, the radius r is 5 feet, so r^2 is 25 square feet.
Multiply r^2 by the mathematical constant pi, which is approximated at 3.14159, to find the area of the circle. Overall, the equation for the area, A, of a circle can be written as: A = π (r^2). In the example, this becomes A = (3.14159)(5 feet ^ 2) = 78.5398 square feet.
## Square or Rectangle
••• Thomas Northcut/Photodisc/Getty Images
Measure the height, h, of the rectangle or square. Suppose the height is 5 inches.
Find the length of the base, b. In our example, say the base is 12 inches.
Multiply the length of the base, b, by the height, h, to find the total area. The equation for the area, A, of a square or rectangle area can be written as: A = b * h. In our example, the base, b, is 12 inches, and the height, h, is 5 inches. Therefore, the area is 12 inches multiplied by 5 inches, or 60 square inches.
## Parallelogram
••• Jupiterimages/Photos.com/Getty Images
Find the altitude of the parallelogram. The altitude is the vertical height of the parallelogram. Suppose the altitude, v, is 3 feet.
Measure the length of the base, b. For the example, set the base length equal to 5 feet.
Multiply the length of the base by the vertical height to calculate the total area, A, of the parallelogram. This equation can be written as: A = v * b. In the example, this becomes A = (3 feet)(5 feet), which is 15 square feet.
## Triangle
••• Ablestock.com/AbleStock.com/Getty Images
Determine the vertical height, h, of the triangle. For example, set the height equal to 2 inches.
Measure the length of the base, b. Suppose the base is 3 inches.
Multiply the height by one-half the length of the base. The equation for the total area, A, of a triangle is A = (1/2) b * h. In the example, A = 0.5 (3 inches) (2 inches) = 3 square inches.
## Trapezoid
••• George Marks/Retrofile/Getty Images
Measure the vertical height, h, of the trapezoid. As an example, calculate the surface area of the trapezoidal face of the clock; the height is 3.5 inches.
Find the length of the base, b. Let's say the base, b, is 4 inches long.
Measure the length of the top side, a. The base, b, and top, a, will be parallel and on opposite sides. For the example, set the length of the top side equal to 3 inches.
Take half of the sum of the two parallel sides, a and b, and multiply that by the height, h, to find the total area, A. This can be written as A = (1/2) (a + b) h. Substitute in the measurements from the example into the equation. The equation becomes A = (0.5) (3 inches + 4 inches) (3.5 inches), which is 12.25 square inches.
## Sector
••• Thomas Northcut/Photodisc/Getty Images
Measure the length of the radius, r, of the sector. This is the length of the one of the straight edges of the sector or slice. For example, set the radius equal to 6 inches.
Find the angle, θ, between the two straight edges of the sector. This is measured in radians. Suppose this is 1.05 radians.
Square the radius, r, divide by two, and then multiply this by the angle, θ, to find the area of the sector. This is written as Area = (1/2) (r^2) θ, and in the example it is (0.5) ((6 inches)^2) (1.05) = 18.9 square inches.
#### Tips
• Math Is Fun provides an online calculator if you want to check your calculations (See References).
If you need to calculate the area of a composite shape, calculate the area of the individual shapes that make up the object and add them together.
Don't forget to include the units, such as inches, feet and yards. |
# 2001 AMC 10 Problems/Problem 20
## Problem
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon?
$\textbf{(A) } \frac{1}{3}(2000) \qquad \textbf{(B) } {2000(\sqrt{2}-1)} \qquad \textbf{(C) } {2000(2-\sqrt{2})} \qquad \textbf{(D) } {1000} \qquad \textbf{(E) } {1000\sqrt{2}}$
## Solution 2
Let $x$ represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length $x\sqrt{2}/2$, so $$2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{\textbf{(B) }2000(\sqrt{2}-1)}.$$
## Solution 3 (Longer solution-credit: Ileytyn)
First, realize that each triangle is congruent, a right triangle and that the two legs are equal. Also, each side of the octagon is equal, because of the definition of regular shapes. Let $s$ be the length of a leg of the isosceles right triangle. In terms of $s$, the hypotenuse of the isosceles right triangle, which is also the length of a side of the regular octagon, is $s \sqrt{2}$. Since the length of each side of the square is 2000, the length of each side of the regular octagon is equal to the length of a side of the square ($2000$) subtracted by $2$ times the length of a leg of the isosceles right triangle ( the total length of the side is $2s+ o$, $o$ being the length of a side of the regular octagon), which is the same as $2s$. As an expression, this is $2000-2s$, which we can equate to $s \sqrt{2}$, ( since the octagon is regular, meaning all of the side's lengths are congruent) giving us the following equation:$2000-2s = s \sqrt{2}$. By isolating the variable and simplifying the right side, we get the following: $2000 = s(2 + \sqrt{2})$. Dividing both sides by $(2 + \sqrt{2})$, we arrive with $\frac{2000}{2 + \sqrt{2}} = s$, now, to find the length of the side of the octagon, we can plug in $s$ and use the equation $2000-2s = o$, $o$ being the length of a side of the octagon, to derive the value of a side of the octagon. After plugging in the values, we derive $2000-2(\frac{2000}{2 + \sqrt{2}})$, which is the same as $2000-(\frac{4000}{2 + \sqrt{2}})$, factoring out a $2000$, we derive the following: $2000(1-(\frac{2}{2 + \sqrt{2}}))$, by rationalizing the denominator of $\frac{2}{2 + \sqrt{2}}$, we get $2000(1-(2 - \sqrt{2}))$, after expanding, finally, we get $\boxed{\textbf{(B) }2000(\sqrt{2} -1)}$ !(not a factorial symbol, just an exclamation point) |
# What is an Annulus in Math? - Definition & Formula
Instructor: Allison Petrovic
Allison has experience teaching high school and college mathematics and has a master's degree in mathematics education.
Have you ever heard of an annulus? If you have ever eaten a doughnut, you have seen an annulus! In this lesson, we will learn the definition and discover how to find the area of an annulus!
## Glazed Doughnuts!
Think about the last time you had a glazed doughnut. Yummy! If your teacher asked you the shape of your glazed doughnut, what would you say? You may be thinking, Well when I look at it from above, it is in the shape of a circle! But let's think about this. Circles do not have holes in the middle of them. So what type of shape would your doughnut be? What is the shape that looks like a circle but has a hole in the middle of it? The shape is called an annulus!
## What is an Annulus?
An annulus is basically a ring--a circle with a circular hole in the middle. Have you ever seen a washer? Not the kind that you use to wash your clothes, but the kind of washer that is used with nuts and bolts. Ask a parent to dig one out of the tool box! A washer is a great example of what an annulus looks like--just like a ring.
## Area of an Annulus
Let's figure out how to calculate the area of an annulus! In the picture below, the annulus is blue. Do you see how the annulus looks like it has two circles? It looks like there is one big blue circle (this larger, outer circle is the ring) and one small yellow circle, which is the hole. The area of the annulus is equal to the area of the big blue circle minus the area of the small yellow circle.
So, to find the area of an annulus, we must find the area of the big blue circle and then subtract the area of the small yellow circle. Keep in mind that the formula for the area of a circle is pi times the radius (r) squared (or, pi x r^2)
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What Makes a Good Sample?
Warmup
Activity #1
Identify Population Samples.
• Move the slider to select the problem number.
• Attempt the question before clicking on the check boxes to see the correct answers.
Activity #2
Calculate Measures of Central Tendency.
A young artist sold her first two paintings for \$50 and \$350 out of 10 paintings sold.
(1.) Calculate the mean.
At a gallery show, she sold three paintings for \$250, \$400, and \$1,200.
(2.) Calculate the mean and median.
Her oil paintings were sold for \$410, \$400, and \$375.
(3.) Calculate the mean and median.
Here are the selling prices for all 10 of her paintings:
Activity #3
Practice Exercise Questions.
The price per pound of catfish at a fish market was recorded for 100 weeks.
Here are dot plots showing the population and three different samples from that population.
(1.) What do you notice?
(2.) If the goal is to have the sample represent the population, which of the samples would work best?
(3.) Which wouldn’t work so well? Explain your reasoning.
Challenge #1
How many different outcomes are in each sample space? Explain your reasoning. (You do not need to write out the actual options, just provide the number and your reasoning.)
(1.) A letter of the English alphabet is followed by a digit from 0 to 9.
(2.) A baseball team’s cap is selected from 3 different colors, 2 different clasps, and 4 different locations for the team logo. A decision is made to include or not to include reflective piping.
(3.) A locker combination like 7-23-11 uses three numbers, each from 1 to 40. Numbers can be used more than once, like 7-23-7.
Challenge #2
Here below is a dot plot of the scores on a video game for a population of 50 teenagers.
The three dot plots below are the scores of teenagers in three samples from this population.
Which of the three samples is most representative of the population? Explain how you know.
Challenge #3
When doing a statistical study, it is important to keep the goal of the study in mind. Representative samples give us the best information about the distribution of the population as a whole, but sometimes a representative sample won’t work for the goal of a study!
For example, suppose you want to study how discrimination affects people in your town. Surveying a representative sample of people in your town would give information about how the population generally feels, but might miss some smaller groups. Describe a way you might choose a sample of people to address this question. |
# How do you prove that 1- cos^2(x/2) = (sin^2x)/(2(1+cosx)) ?
Jun 5, 2018
Using that
$\cos \left(x\right) = 2 {\cos}^{2} \left(\frac{x}{2}\right) - 1$
#### Explanation:
so we have
$2 \left(1 + \cos \left(x\right)\right) = 2 \left(2 {\cos}^{2} \left(\frac{x}{2}\right) - 1\right) = 4 {\cos}^{2} \left(\frac{x}{2}\right)$
so we get
$4 {\cos}^{2} \left(\frac{x}{2}\right) \left(1 - {\cos}^{2} \left(\frac{x}{2}\right)\right)$
and
${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$
$1 - {\left(2 {\cos}^{2} \left(\frac{x}{2}\right) - 1\right)}^{2}$
$1 - 4 {\cos}^{2} \left(\frac{x}{2}\right) - 1 + 4 {\cos}^{2} \left(\frac{x}{2}\right)$
$4 {\cos}^{2} \left(\frac{x}{2}\right) \left(1 - {\cos}^{2} \left(\frac{x}{2}\right)\right)$
Jun 5, 2018
Please find a Proof in Explanation.
#### Explanation:
Since, $1 - \cos x = 2 {\sin}^{2} \left(\frac{x}{2}\right)$, we have,
${\sin}^{2} \frac{x}{2 \left(1 + \cos x\right)} = \frac{1 - {\cos}^{2} x}{2 \left(1 + \cos x\right)}$,
={cancel((1+cosx))(1-cosx)}/(2cancel((1+cosx)),
$= \frac{1 - \cos x}{2}$,
$= \frac{\cancel{2} {\sin}^{2} \left(\frac{x}{2}\right)}{\cancel{2}}$,
$= {\sin}^{2} \left(\frac{x}{2}\right)$,
$= 1 - {\cos}^{2} \left(\frac{x}{2}\right)$, as desired! |
Courses
# Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
## Class 10 : Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
The document Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10
Q.1. State which pairs of triangles in Fig. are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Sol. (i) In Δ ABC and Δ PQR
We have:
∠A = ∠P [Each 60°]
∠B = ∠Q [Each 80°]
∠C = ∠R [Each 40°]
ΔABC ~ ΔPQR [AAA criterion]
(ii) In ΔABC and ΔQRP
Hence, ΔABC ~ ΔQRP [SSS criterion]
(iii) In ΔLMP and ΔEFD
∴ ΔLMP is not similar to ΔEFD.
Since the three ratios are not same.
(iv) In ΔMNL and ΔQPR
(v) In ΔABC and ΔDEF
ΔABC is not similar to ΔDEF
∵ Angles between two sides are not same
(vi) In ΔDEF and ΔPQR
∠E = ∠Q = 80°
∠F = ∠R = 30°
[∴ ∠F = 180° - (80° + 70°) = 30°]
∴ ΔDEF ~ ΔPQR [AA]
Q.2. In the figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Sol.
⇒
In ΔDOC,
⇒ 125° + ∠DCO = 180°
⇒ ∠DCO = 180° - 125° = 55°
ΔODC ~ ΔOBA
∠OAB = ∠DCO = 55°
⇒ ∠DOC = ∠OAB = 55°
Q.3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that
Sol. Given: Diagonals AC and BD intersect at O.
AB || DC
Proof: In ΔAOB and ΔCOD
∠1 = ∠2
∠3 = ∠4 [Alternate angles]
∴ ΔAOB ~ ΔCOD [AA]
[Corresponding sides of similar triangles]
Q.4. In the figure, and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Sol. Given
To Prove: ΔPOS ∼ ΔTOR
To Prove: ΔPQS
Proof: In ΔPQR,
∠1 = ∠2 [Given]
PQ = PR [Sides opposite to equal angles]
OR/QS = QT/PR [Given]
Or
In ΔPQS and ΔTQR,
(Proved)
∠1 = ∠1 [Common]
∴ ∠PQS ~ ∠TQR [SAS]
Q.5. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Sol. In Δ RPQ ~ ΔRTS
∠P = ∠RTS [Given]
∠R = ∠R [Common]
∴ ΔRPQ ~ ΔRTS [AA]
Q.6. In the figure, if ΔABE ≌ ΔACD, show that ΔADE ~ ΔABC.
Sol.
Given: ΔABE ≌ ΔACD
Proof: ΔABE ~ ΔACD
AB = AC and AE = AD
[Proved above]
∠A - ∠A [Common]
Q.7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iv) ΔPDC ~ ΔBEC
Sol. Given: AD and CE are altitudes of the ΔABC
(i) To Prove: ΔAEP ~ ΔCDP
Proof: In ΔAEP and ΔCDP,
∠AEP = ∠CDP [Each'90°]
∠APE = ∠CPD [Vertically opposite angles]
ΔAEP ~ ΔCDP [AA]
(ii) In ΔABD and ΔCBE,
∠ABD = ∠CBE [Common]
ΔABD ~ ΔCBE [AA]
(iii) In Δ AEP and ΔADB,
∠A = ∠A [Common]
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC [Each 90°]
∠PCD = ∠BCE [Common]
ΔPDC ~ ΔBEC [AA]
Q.8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ ΔCFB.
Sol. In ΔABE and ΔCFB,
∠1 = ∠2
∠4 = ∠3 [Alternate angles]
⇒ ΔABE ∼ ΔCFB [AA]
Q.9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP
(ii)
Sol. (i) In ΔABC and ΔAMP,
∠B = ∠AMP [Each 90°]
∠A = ∠A [Common]
⇒ ΔABC ∼ ΔAMP [AA]
(ii) ΔABC ~ ΔAMP [proved above]
[Ratio of the Corresponding sides of similar Δs]
Q.10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If ΔABC ~ ΔFEG, show that:
(i)
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Sol. ΔABC ~ ΔFEG [Given]
⇒ ∠A = ∠F
(i) In ΔACD and ΔFGH
∠A = ∠F [Given]
∴ ΔACD ~ ΔFGH
[Corresponding sides of similar triangles]
(ii) [Proved above]
In ΔDCB and ΔHGE,
∴ ΔDCB ~ ΔHGE [SAS]
(iii) In ΔDCA and ΔHGF,
∠1 = ∠2 [Bisectors]
⇒ ΔDCA ~ ΔHGF [SAS]
Q.11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥BC and EF ⊥AC, prove that ΔABD ~ ΔECF.
Sol. In ΔABD and ΔECF,
∠B = ∠C
[angles opposite to equal sides are equal]
⇒ ΔABD ~ ΔECF [AA]
Q.12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see figure). Show that Δ ABC ~ ΔPQR.
Sol. In ΔABC and ΔPQR
⇒ ΔABD ~ ΔPQM [SAS]
∴ ∠B = ∠Q
[Corresponding angles of similar triangles]
In ΔABC and ΔPQR,
[Given]
∠B = ∠Q [As provecd]
∴ ΔABC ~ ΔPQR [SAS]
Q.13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Sol. In ΔACB and ΔDCA,
∠C = ∠C [Common]
ΔACB ~ ΔDCA
[Corresponding sides of similar triangles]
⇒ CA2 = CB x CD
Q.14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Sol. Construction: Draw DE || AC and MS || PR
Proof: In ΔABC, D is mid-point of BC (given) and DE || AC (const.)
∴ E is mid-point of AB (by converse of mid-point theorem
∴ ΔADE ~ ΔPMS [SSS similarly]
Now, in ΔABC and ΔPQR,
[Given]
∠A = ∠P [Proved above]
∴ ΔABC ~ ΔPQR (SAS)
Q.15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Sol. DE is vertical stick of length = 6m
Length of shadow = 4 m
Let height of tower = h m
Length of shadow = 28 m
Q.16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that .
Sol. When ΔABC ~ ΔPQR
In ΔABD and ΔPQM
∴ ΔABD ~ ΔPQM
[Corresponding sides of similar triangles]
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The number 8 is composite. So it is not prime, technically, because it only has 1 as a factor. They have âfactorsâ. 3/1 = 3. The last digit is not 0 or 5, and it is not a multiple of 7. For example, 2 and 5 are factors of 10. Example 1: 6 can be divided evenly by 2, so 6 is a composite number. Hence it is a composite. Factors are the numbers that divide into it. Composite numbers are whole numbers that can be divided by numbers other than itself and 1. To do this, you should check to see if the number can be divided by these common factors: 2, 3, 5, 7, 11, and 13. Example: 5 is a prime number. In the same way, 2, 5, 7, 11, 13, 17 are prime numbers. A composite number has more than two factors. Yes because 7x6x5x4x3x2x1+5 = 5(7x6x4x3x2x1+1) = 5(1008+1) = 5(1009) Since it has more than 1 factor that is 5 and 1009 So it is a composite number Every other even number other than 2 is composite. 5 × 2 × 2 × 23. Every composite number can be written as the product of two or more (not necessarily distinct) primes. The best way to figure out if itâs a composite number is to perform the divisibility test. It does not have two factors. 1-- 1 is only divisible by 1. In problem 2 above, the number 8 has four factors. When a number can be divided up exactly it is a Composite Number; When a number cannot be divided up exactly it is a Prime Number; So 6 is Composite, but 7 is Prime. 5 x 7 x 13 + 5. If the number ends with a 0 or 5, try dividing by 5. Definitions. But in order to be prime, you have to have exactly two factors. For example, the composite number 299 can be written as 13 × 23, and the composite number 360 can be written as 2 3 × 3 2 × 5; furthermore, this representation is unique up to the order of the factors. 3/3 = 1. Now, here is an interesting case. 6 has more than 2 factors. Hence it is prime number. 1 has only one factor. If the number is even, then start with the number 2. Example 2: 5 cannot be divided evenly (except by 1 and 5), so is not a composite number (it is a so called prime number). Every whole number higher than 1 is either a composite number or a prime number. Solution : To find the primality of 5 and 6, first we need to find the factors of the numbers 5 = 1x5 6 = 1x6, 2x3 , 3x2 5 has only 1 and 5 as its factors. In other words a composite number is a positive integer that can be formed by multiplying together two smaller positive integers. A prime number has only two factors: 1 and itself. A whole number that can be divided evenly by numbers other than 1 or itself. Find whether the numbers 5 and 6 are Prime or Composite Number? Composite number: A whole number that can be divided exactly by numbers other than 1 or itself. 1 is itself. The number 11 is also a prime number because it only has two factors: 1 and 11 Example 3 Identify prime and composite numbers from the following list: 73, 65, 172, and 111 Solution Number 73 is a prime number. 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# How do you simplify (x+1)/(x-1)=2/(2x-1)+2/(x-1)?
$x = \frac{3}{2}$
#### Explanation:
Given that
$\setminus \frac{x + 1}{x - 1} = \setminus \frac{2}{2 x - 1} + \setminus \frac{2}{x - 1}$
$\setminus \frac{x + 1}{x - 1} - \setminus \frac{2}{2 x - 1} - \setminus \frac{2}{x - 1} = 0$
$\setminus \frac{\left(x + 1\right) \left(2 x - 1\right) - 2 \left(x - 1\right) - 2 \left(2 x - 1\right)}{\left(x - 1\right) \left(2 x - 1\right)} = 0$
$\setminus \frac{2 {x}^{2} - 5 x + 3}{\left(x - 1\right) \left(2 x - 1\right)} = 0$
$\setminus \frac{2 {x}^{2} - 2 x - 3 x + 3}{\left(x - 1\right) \left(2 x - 1\right)} = 0$
$\setminus \frac{2 x \left(x - 1\right) - 3 \left(x - 1\right)}{\left(x - 1\right) \left(2 x - 1\right)} = 0$
$\setminus \frac{\left(2 x - 3\right) \left(x - 1\right)}{\left(x - 1\right) \left(2 x - 1\right)} = 0$
$\setminus \frac{2 x - 3}{2 x - 1} = 0$
$2 x - 3 = 0 \setminus \quad \setminus \forall \setminus x \setminus \ne \frac{1}{2}$
$x = \frac{3}{2}$
Jun 26, 2018
$x = \frac{3}{2} \mathmr{and} 1.5$
#### Explanation:
we have
$\frac{x + 1}{x - 1} = \left(\frac{2}{2 x - 1}\right) + \left(\frac{2}{x - 1}\right)$
shifting $\left(\frac{2}{x - 1}\right)$ to other side
$\implies \frac{x + 1}{x - 1} - \left(\frac{2}{x - 1}\right) = \left(\frac{2}{2 x - 1}\right)$
$\implies \frac{x + 1 - 2}{x - 1} = \left(\frac{2}{2 x - 1}\right)$
$\implies \frac{\cancel{x - 1}}{\cancel{x - 1}} = \left(\frac{2}{2 x - 1}\right)$
=>2x-1=2; =>x=3/2 or 1.5 |
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# SYSTEMS OF LINEAR EQUATIONS
## GRAPHICAL SOLUTIONS
Often, we want to find a single ordered pair that is a solution to two different linear equations. One way to obtain such an ordered pair is by graphing the two equations on the same set of axes and determining the coordinates of the point where they intersect.
Example 1
Graph the equations
x + y = 5
x - y = 1
on the same set of axes and determine the ordered pair that is a solution for each equation.
Solution
Using the intercept method of graphing, we find that two ordered pairs that are solutions of x + y = 5 are
(0, 5) and (5, 0)
And two ordered pairs that are solutions of
x - y = 1 are
(0,-1) and (1,0)
The graphs of the equations are shown.
The point of intersection is (3, 2). Thus, (3, 2) should satisfy each equation.
In fact, 3 + 2 = 5 and 3 - 2 = 1
In general, graphical solutions are only approximate. We will develop methods for exact solutions in later sections.
Linear equations considered together in this fashion are said to form a system of equations. As in the above example, the solution of a system of linear equations can be a single ordered pair. The components of this ordered pair satisfy each of the two equations.
Some systems have no solutions, while others have an infinite number of solu- tions. If the graphs of the equations in a system do not intersect-that is, if the lines are parallel (see Figure 8.1a)-the equations are said to be inconsistent, and there is no ordered pair that will satisfy both equations. If the graphs of the equations are the same line (see Figure 8.1b), the equations are said to be dependent, and each ordered pair which satisfies one equation will satisfy both equations. Notice that when a system is inconsistent, the slopes of the lines are the same but the y-intercepts are different. When a system is dependent, the slopes and y-intercepts are the same.
In our work we will be primarily interested in systems that have one and only one solution and that are said to be consistent and independent. The graph of such a system is shown in the solution of Example 1.
## SOLVING SYSTEMS BY ADDITION I
We can solve systems of equations algebraically. What is more, the solutions we obtain by algebraic methods are exact.
The system in the following example is the system we considered in Section 8.1 on page 335.
Example 1
Solve
x + y = 5 (1)
x - y = 1 (2)
Solution
We can obtain an equation in one variable by adding Equations (1) and (2)
Solving the resulting equation for x yields
2x = 6, x = 3
We can now substitute 3 for x in either Equation (1) or Equation (2) to obtain the corresponding value of y. In this case, we have selected Equation (1) and obtain
(3) + y = 5
y = 2
Thus, the solution is x = 3, y = 2; or (3, 2).
Notice that we are simply applying the addition property of equality so we can obtain an equation containing a single variable. The equation in one variable, together with either of the original equations, then forms an equivalent system whose solution is easily obtained.
In the above example, we were able to obtain an equation in one variable by adding Equations (1) and (2) because the terms +y and -y are the negatives of each other. Sometimes, it is necessary to multiply each member of one of the equations by -1 so that terms in the same variable will have opposite signs.
Example 2
Solve
2a + b = 4 (3)
a + b = 3 (4)
Solution
We begin by multiplying each member of Equation (4) by - 1, to obtain
2a + b = 4 (3)
-a - b = - 3 (4')
where +b and -b are negatives of each other.
The symbol ', called "prime," indicates an equivalent equation; that is, an equation that has the same solutions as the original equation. Thus, Equation (4') is equivalent to Equation (4). Now adding Equations (3) and (4'), we get
Substituting 1 for a in Equation (3) or Equation (4) [say, Equation (4)], we obtain
1 + b = 3
b = 2
and our solution is a = 1, b = 2 or (1, 2). When the variables are a and b, the ordered pair is given in the form (a, b).
## SOLVING SYSTEMS BY ADDITION II
As we saw in Section 8.2, solving a system of equations by addition depends on one of the variables in both equations having coefficients that are the negatives of each other. If this is not the case, we can find equivalent equations that do have variables with such coefficients.
Example 1
Solve the system
-5x + 3y = -11
-7x - 2y = -3
Solution
If we multiply each member of Equation (1) by 2 and each member of Equation (2) by 3, we obtain the equivalent system
(2) (-5x) + (2)(3y) = (2)(-ll)
(3) (-7x) - (3)(2y) = (3)(-3)
or
-10x + 6y = -22 (1')
-21x - 6y = -9 (2')
Now, adding Equations (1') and (2'), we get
-31x = -31
x = 1
Substituting 1 for x in Equation (1) yields
-5(1) + 3y = -11
3y = -6
y = -2
The solution is x = 1, y = -2 or (1, -2).
Note that in Equations (1) and (2), the terms involving variables are in the left-hand member and the constant term is in the right-hand member. We will refer to such arrangements as the standard form for systems. It is convenient to arrange systems in standard form before proceeding with their solution. For example, if we want to solve the system
3y = 5x - 11
-7x = 2y - 3
we would first write the system in standard form by adding -5x to each member of Equation (3) and by adding -2y to each member of Equation (4). Thus, we get
-5x + 3y = -11
-lx - 2y = -3
and we can now proceed as shown above.
## SOLVING SYSTEMS BY SUBSTITUTION
In Sections 8.2 and 8.3, we solved systems of first-degree equations in two vari- ables by the addition method. Another method, called the substitution method, can also be used to solve such systems.
Example 1
Solve the system
-2x + y = 1 (1)
x + 2y = 17 (2)
Solution
Solving Equation (1) for y in terms of x, we obtain
y = 2x + 1 (1')
We can now substitute 2x + 1 for y in Equation (2) to obtain
x + 2(2x + 1) = 17
x + 4x + 2 = 17
5x = 15
x = 3 (continued)
Substituting 3 for x in Equation (1'), we have
y = 2(3) + 1 = 7
Thus, the solution of the system is a: x = 3, y = 7; or (3, 7).
In the above example, it was easy to express y explicitly in terms of x using Equation (1). But we also could have used Equation (2) to write x explicitly in terms of y
x = -2y + 17 (2')
Now substituting - 2y + 17 for x in Equation (1), we get
Substituting 7 for y in Equation (2'), we have
x = -2(7) + 17 = 3
The solution of the system is again (3, 7).
Note that the substitution method is useful if we can easily express one variable in terms of the other variable.
## APPLICATIONS USING TWO VARIABLES
If two variables are related by a single first-degree equation, there are infinitely many ordered pairs that are solutions of the equation. But if the two variables are related by two independent first-degree equations, there can be only one ordered pair that is a solution of both equations. Therefore, to solve problems using two variables, we must represent two independent relationships using two equations. We can often solve problems more easily by using a system of equations than by using a single equation involving one variable. We will follow the six steps outlined on page 115, with minor modifications as shown in the next example.
Example 1
The sum of two numbers is 26. The larger number is 2 more than three times the smaller number. Find the numbers.
Solution
Steps 1-2
We represent what we want to find as two word phrases. Then, we represent the word phrases in terms of two variables.
Smaller number: x
Larger number: y
Step 3 A sketch is not applicable.
Step 4 Now we must write two equations representing the conditions stated.
The sum of two numbers is 26.
Step 5 To find the numbers, we solve the system
x + y = 26 (1)
y = 2 + 3x (2)
Since Equation (2) shows y explicitly in terms of x, we will solve the system by the substitution method. Substituting 2 + 3x for y in Equation (1), we get
x + (2 + 3x) = 26
4x = 24
x = 6
Substituting 6 for x in Equation (2), we get
y = 2 + 3(6) = 20
Step 6 The smaller number is 6 and the larger number is 20.
## CHAPTER SUMMARY
1. Two equations considered together form a system of equations. The solution is generally a single ordered pair. If the graphs of the equations are parallel lines, the equations are said to be inconsistent; if the graphs are the same line, the equations are said to be dependent.
2. We can solve a system of equations by the addition method if we first write the system in standard form, in which the terms involving the variables are in the left-hand member and the constant term is in the right-hand member.
3. We can solve a system of equations by the substitution method if one variable in at least one equation in the system is first expressed explicitly in terms of the other variable.
4. We can solve word problems using two variables by representing two independent relationships by two equations. |
# Basic Arithmetic : Whole Numbers
## Example Questions
### Example Question #1 : How To Subtract Integers
Evaluate the following:
Explanation:
When you subtract integers, it is the same thing as adding the inverse of the second integer. You can consider the following:
You can also consider the problem as asking for six less than negative four. This will also get you to the answer of
### Example Question #1 : Linear Equations With Whole Numbers
Jimmy had in lunch money for school. Everyday he spends for food and drinks. What is the expression that shows how much money will he have after each day, where is the days, and is the total amount of money left?
Explanation:
Jimmy starts off with $60, and spends$3.50 everyday.
This means that he will have $56.50 after day 1,$53 after day 2, and so forth.
Only one equation satisfies this scenario. The rest are irrelevant.
### Example Question #1 : Creating Equations With Whole Numbers
What is the solution of for the systems of equations?
Explanation:
We add the two systems of equations:
For the Left Hand Side:
For the Right Hand Side:
So our resulting equation is:
Divide both sides by 10:
For the Left Hand Side:
For the Right Hand Side:
Our result is:
### Example Question #31 : Systems Of Equations
What is the solution of that satisfies both equations?
Explanation:
Reduce the second system by dividing by 3.
Second Equation:
We this by 3.
Then we subtract the first equation from our new equation.
First Equation:
First Equation - Second Equation:
Left Hand Side:
Right Hand Side:
Our result is:
### Example Question #3 : Creating Equations With Whole Numbers
What is the solution of for the two systems of equations?
Explanation:
We first add both systems of equations.
Left Hand Side:
Right Hand Side:
Our resulting equation is:
We divide both sides by 3.
Left Hand Side:
Right Hand Side:
Our resulting equation is:
### Example Question #51 : Systems Of Equations
What is the solution of for the two systems?
Explanation:
We first multiply the second equation by 4.
So our resulting equation is:
Then we subtract the first equation from the second new equation.
Left Hand Side:
Right Hand Side:
Resulting Equation:
We divide both sides by -15
Left Hand Side:
Right Hand Side:
Our result is:
### Example Question #1 : Linear Equations With Whole Numbers
Dr. Jones charges a $50 flat fee for every patient. He also charges his patients$40 for every 10 minutes that he spends with him. If Mrs. Smith had an appointment that lasted 30 minutes, how much did she have to pay Dr. Jones?
$90$200
$120$170
\$170
Explanation:
We can express Dr. Jones's rate in a linear equation:
Since Mrs. Smith's appointment lasted 30 minutes, we have 3 10-minute intervals. Then, we can plug in that number into our above equation to find out how much the appointment cost.
### Example Question #1 : Linear Equations With Whole Numbers
Roman is ordering uniforms for the tennis team. He knows how many people are on the team and how many uniforms come in each box. Which equation can be used to solve for how many boxes Roman should order?
b = number of students x number of uniforms per box
b = number of boxes ÷ number of uniforms per box
b = number of students - number of uniforms per box
b = number of students ÷ number of uniforms per box
b = number of boxes x number of uniforms per box
b = number of students ÷ number of uniforms per box
Explanation:
The total number of uniforms needed equals the number of students divided by the number of uniforms per box.
### Example Question #1 : Solving Equations With Whole Numbers
Solve for .
Explanation:
First, add 6 to both sides so that the term with "x" is on its own.
Now, divide both sides by 2.
Solve for . |
Explanation of the 5 steps to Completing the Square
1. Make some space between the terms with variables and the constant term. This is where we will eventually complete the square.
2. In order to complete the square, the coefficient of the squared term must be 1. We must therefore factor out the leading coefficient from the terms with variables (in this case, we factor out -2 from both terms) if the leading coefficient is other than 1.
3. This is the long one. First, within the parentheses we have completed the square. This is accomplished by taking half of the coefficient of x and then squaring it
This is where the +9 comes from inside of the parentheses. Second, we must balance the function. By adding 9 to the inside of the parentheses, we have actually added (-2)(9)=-18 to the right side of the function. In order to keep the function the same as when we started, we must subtract this same amount from the right side of the function (-18-(-18)=0 so we have, in effect not changed the function)
4. Factor the expression in parentheses which is a perfect square.
5. Combine the constant terms so that the function is in vertex form. |
SecondaryMaths
Simultaneous equations – When to add and when to subtract
When using the elimination method to solve simultaneous equations, students can often be unsure whether to add or subtract, notes Colin Foster
by Colin Foster
DOWNLOAD A FREE RESOURCE! Solving equations worksheets – Linear equations KS3 maths
SecondaryMaths
The difficulty
Look at the four pairs of simultaneous equations below:
Which ones could you solve by adding the equations together? Which ones could you solve by subtracting one equation from the other? Students may be unsure and not know how to decide.
The solution
Don’t worry about solving the equations just yet. All I want you to do is simply add together each pair of equations. And also subtract each pair of equations. See what you get.
By adding, students should obtain the following:
And by subtracting the second equation from the first equation:
Students may not bother to write the 0𝑥 and 0𝑦 where there are no 𝑥 and 𝑦 terms, and this is fine.
They may make errors, particularly when subtracting the negative terms – so for the subtractions, they may end up with the wrong answers shown in red below:
Writing out the difficult subtractions explicitly may help:
When does adding eliminate an unknown?
This happens when two terms are equal in magnitude, but of opposite sign
(e.g., 2𝑦 and –2𝑦).
When does subtracting eliminate an unknown?
This happens when two terms are equal in magnitude, and of the same sign
(e.g., –3𝑦 and –3𝑦).
Sometimes, it can help if students remember the following: When the Signs are the Same you Subtract.
Can you find a pair of equations where either adding or subtracting will lead to elimination of one of the unknowns?
An example would be 3𝑥 + 2𝑦 = 11 and 3𝑥 – 2𝑦 = 7.
The solution to all of these pairs of equations is 𝑥 = 3, 𝑦 = 1.
Checking for understanding
To assess students’ understanding, ask them to create four pairs of simultaneous equations of their own, two of which can be solved by adding the equations, and two of which can be solved by subtracting the equations.
They should label clearly which are which.
Colin Foster (@colinfoster77) is a Reader in Mathematics Education in the Department of Mathematics Education at Loughborough University and has written numerous books and articles for mathematics teachers; for more information, visit foster77.co.uk |
# Solutions for Practice Test 7, The Official SAT Study Guide, Section 3
Math Lair Home > Test Preparation > Solutions for Practice Test 7, The Official SAT Study Guide, Section 3
SAT Practice Test Solutions:
2014–15 SAT Practice Test
2013–14 SAT Practice Test
The Official SAT Study Guide, second edition
• Practice Test 1: Sections 3, 7, 8.
• Practice Test 2: Sections 2, 5, 8.
• Practice Test 3: Sections 2, 5, 8.
• Practice Test 4: Sections 3, 6, 9.
• Practice Test 5: Sections 2, 4, 8.
• Practice Test 6: Sections 2, 4, 8.
• Practice Test 7: Sections 3, 7, 9.
• Practice Test 8: Sections 3, 7, 9.
• Practice Test 9: Sections 2, 5, 8.
• Practice Test 10: Sections 2, 5, 8.
SAT Math Tips
Here are solutions for section 3 of practice test #7 in The Official SAT Study Guide, second edition, found on pages 768–773. The solutions below demonstrate faster, more informal methods that might work better for you on a fast-paced test such as the SAT. To learn more about these methods, see my e-book Succeeding in SAT Math or the SAT math tips page.
• Estimate the answer: To make 30 rolls, 25/10 = 2.5 pounds of flour are needed. Since 12 is less than half of 30, the answer is definitely less than 2 and is probably somewhere around 1.
• Look at the answer choices: The only answer within range of our estimate is (A) 1. Select that answer.
1. Simplify the expression 2 · (x/y) · y² by cancelling out the y's, giving 2xy. Since xy = 10, 2xy = 20. Select (E) 20.
2. If x + y = 30, then x = 30 − y. Since x > 8, then 30 − y > 8. Rearranging this equation, we get 22 > y. Select (B) y < 22.
• Draw a diagram: Draw a diagram similar to the following:
• Estimate the answer: PQ is equal to 4 and QR is equal to 2. Because PR is the hypotenuse of a right triangle, it must be greater than 4, but because of the triangle inequality, it must be less than 6. So, the perimeter must be greater than 10 but less than 12.
• Look at the answer choices: Answers (A), (B), and (E) are greater than or equal to 12. Eliminate those three answers.
• The length of the hypotenuse is √4² + 2² = √20 (see Reference Information if you've forgotten). So, the perimeter of the triangle is 6 + √20. Select (C) 6 + √20 (approximately 10.47).
• Solution 1:
• Estimate the answer: 8 + (26 − 1)9 is probably a bit less than 25 × 10, so say maybe around 240. Given that the sequence increases by 9 each time, the answer is probably somewhere between 25 and 30.
• Look at the answer choices: Answers (A) and (B) are nowhere in the range of our estimate, so eliminate them.
• Look for a pattern: The first term is 8 + (0)9, the second term is 8 + (1) 9, the third term is 8 + (2), ... the 26th term is 8 + (25)9 = 8 + (26 − 1)9. Select (D) The 26th.
• Solution 2: You may know that the general term for the nth term of an arithmetic sequence is a + (n − 1)d, where a is the first term and d is the difference. In this case, a = 8 and d is 9, giving 8 + (n − 1)9. The term equal to 8 + (26 − 1)9 must be the 26th term. Select (D) The 26th.
• Solution 1: Look at the answer choices: If you know t and y, you can determine x, and from there the others. If you know s and x, you can determine t, and from there the others. Similarly, if you know r and t, or r and s, you can determine the other angle measurements. However, if you know t and z, you cannot determine any other angles. Select (A) t and z.
• Solution 2: Try each of the answer choices, select specific values for the two variables given in the answer, and see which one does not allow you to find values for all 6 angle measures:
• For (A), say that t = 45. Then, because z is opposite to t, it must also be 45. However, it doesn't seem possible to determine any more angles.
• For (B), say that t = 45, and y = 45. Then, x = 90, s = 45, r = 90, and z = 45. We can eliminate answer (B).
• For (C), say that s = 45 and x = 45. Then, y = 90, r = 45, z = 90, and y = 45. We can eliminate answer (C).
• For (D), say that r = 45 and t = 45. Then, s = 90, x = 45, y = 90, and z = 45. We can eliminate answer (D).
• For (E), say that r = 45 and s = 45. Then, t = 90, x = 45, y = 45, and z = 90. We can eliminate answer (E).
Select (A) t and z.
• Estimate the answer: If two consecutive numbers sum to t, then the greater of the two numbers is probably a bit more than t/2, but less than t.
• Look at the answer choices: (C) and (D) seem plausible based on our estimate. (A) and (B) are too small, and (E) is too big. Eliminate those three answers.
• Try a special case: Say that t = 5. Then the greater of the two numbers is 3. If t = 5, then answer (C) evaluates to 3, and answer (D) evaluates to 3.5 (which is wrong). So, the answer must be (C) (t + 1)/2.
• If a thirteenth child joins the class, the median number of siblings will be the seventh greatest value (which would also be the seventh smallest value).
• Looking at the table, there are currently 3 children with 0 siblings and 6 children with 1 sibling. So, the fifth, sixth, seventh, eighth, and ninth smallest values for the number of siblings are all 1. No matter how many siblings the new student has, the seventh smallest number of siblings will be 1 after the new student joins. So, the median will be 1.
• Right now, the students have 6 + 2(2) + 3 = 13 siblings. In order for the new arithmetic mean to equal the median (1), there must be 13 siblings in total (since 13/13 = 1). Therefore, the new child can't have any new siblings. Select (A) 0.
• Solve the equation:
2(x − 3) = 8
x − 3 = 4
·x = 7
• Therefore, (x − 3)/(x + 3) = (7 − 3)/(7 + 3) = 4/10. Enter 4/10.
3. Convert the sentence into an equation:
When [ignore] twice 2 · a number x (or any variable) is decreased by − 3 3 the result is = 253 253
This gives us 2x − 3 = 253. Solving the equation, 2x = 256 or x = 128. Enter 128.
• Draw a diagram: We are asked to find how many black sneakers were produced. Draw a box around, or put an asterisk in, or otherwise mark, the square in the table corresponding to the number of black sneakers. This will be the rightmost square in the second row.
• We know how many black low-tops were produced. We can find the number of black high-tops as follows. Since 10,000 sneakers were manufactured in total, and 5,500 of those were low-tops, 4,500 must be high-tops. Since 3,600 of those were white, 900 must be black. So, if 900 black high-tops and 1,500 black low-tops were produced, then 2,400 in total were produced. Enter 2400.
• Draw a diagram: A diagram is already given, but it helps to label it. The top and bottom of the rectangle are 2 units in length (the distance between (-1, 0) and (1, 0)). If the perimeter of PQRS is 10, then the left and right edges of the rectangle must be 3 units in length. Then, point R must be (1, 3).
• Since (1, 3) is on the graph of y = ax², substitute x = 1, y = 3 into the equation:
3 = a(1)²
a = 3
Enter 3.
4. Substitute a = 2 and c = 3 into the equation:
2b + b = 2 2·3
3b = 8
b = 8/3
Enter 8/3.
• Work backwards: Because line l intersects parallel lines, we could find the value of x if we knew the angle that BC makes with line l, as the two values would be equal. We are told that line l bisects ∠ABC, so the angle we are looking for is ½ of ∠ABC. Because line k also intersects parallel lines, its measurement is also y°. Putting this all together, x must be half of y.
• Try a special case: Say that y = 50. Then, x = 25. Enter 25.
• First, it helps to introduce suitable notation and possibly draw a diagram. Say that the four plumbers are P1, P2, P3, and P4, and the trainees are T1, T2, T3, and T4.
• A team may consist of any one of the four experienced plumbers, so there are four possibilities as to the experienced plumber on the team.
• List out all of the possibilities for the two trainee plumbers on the team:
(P1, P2), (P1, P3), (P1, P4), (P2, P3), (P2, P4), (P3, P4)
There are six possibilities.
• To determine the total number of different teams, multiply the number of combinations of plumbers by the number of combinations of trainees: 4 × 6 = 24.
• Work backwards: We could find the radius of the larger circle if we knew its tootal area. We already know the area of the shaded region (64π square inches), so we could find the total area if we knew the area of the smaller circle. We can find the area of the smaller circle, because we know its radius: 6 inches.
• Putting all of the above together, the area of the smaller circle is π(6)² = 36&pi. The total area is 36π + 64π = 100π. Now, because Area = π(r)², and the area is 100π, then:
100π = πr²
r² = 100
r = 10
Enter 10.
5. List the factors of n: 1, p, r, s, pr, ps, rs, prs. There are 8 in total. Because p, r, and s are all prime, they can't be broken down any further, so there are only 8 factors. Enter 8.
6. This problem looks be a little bit time-consuming, but there is a shortcut if you read the problem carefully:
• We are told that the ball reaches a maximum height of 106 feet at 2.5 seconds.
• Now, look at the expression c − (d − 4t)².
• As for the first part of that expression, we are told that c is a positive constant.
• Now, look at − (d − 4t)². When you square anything, the result is non-negative, so (d − 4t)² must be non-negative, and so − (d − 4t)² must be non-positive.
• For the maximum height, we need to find the maximum value for c − (d − 4t)². The maximum value would occur when − (d − 4t)² is as large as possible, but since it cannot be positive, the largest value occurs when that expression is 0.
• So, at the maximum height, h(2.5) = c, and we are told that h(2.5) = 106, so c = 106. At the maximum value, − (d − 4t)² = 0, so d − 4(2.5) = 0, or d == 10.
• Now, we are asked to find the value at t = 1:
h(1) = 106 − (10 − 41)²
h(1) = 106 − 6²
h(1) = 106 − 36
h(1) = 70
Enter 70. |
Lesson: Some Applications of Trigonometry
Exercise 9.1 (16)
Question: 1
A circus artist is climbing a 20 m long rope, which is
tightly stretched and tied from the top of a vertical
pole to the ground. Find the height of the pole, if
the angle made by the rope with the ground level is
.(see Fig. 9.11).
30
Fig. Exc_9.1_1
Solution:
From the given figure,
AB is the pole.
In ,
ABC
AB
sin30
AC
AB 1
20 2
20
AB
2
AB 10
So, height of the pole is .
10m
Question: 2
A tree breaks due to storm and the broken part
bends so that the top of the tree touches the ground
making an angle with it. The distance between
30
the foot of the tree to the point where the top
touches the ground is 8 m. Find the height of the
tree.
Solution:
Fig. Exc_9.1_2
From the above figure,
Let the original tree be AC and due to storm, it got
broken down into two pieces. The broken part is
taken as and it makes an angle with the
A C
30
ground.
In ,
BC
tan30
A C
BC 1
8
3
8
BC m
3
Now,
A C
cos30
A B
8 3
A B 2
16
A B m
3
Height of the tree
A B BC
16 8
3 3
24
3
8 3 m
Hence, the height of the tree is m.
8 3
Question: 3
A contractor plans to install two slides for the
children to play in a park. For the children below
the age of 5 years, she prefers to have a slide whose
top is at a height of 1.5 m, and is inclined at an
angle of to the ground, whereas for the elder
30
children she wants to have a steep side at a height
of 3 m, and inclined at an angle of to the
60
ground. What should be the length of the slide in
each case?
Solution:
Fig. Exc_9.1_3 (a)
Fig. Exc_9.1_3(b)
Let AC and PR be the slides for younger and elder
children in the above pictures respectively.
In ,
ABC
AB
sin30
AC
1.5 1
AC 2
AC 3m
Now, In ,
PQR
PQ
sin60
PR
3 3
PR 2
6
PR
3
PR 2 3
Hence, the lengths of these slides are 3 m and
2 3
m.
Question: 4
The angle of elevation of the top of a tower from a
point on the ground, which is 30 m away from the
foot of the tower is . Find the height of the
30
tower.
Solution:
Fig. Exc_9.1_4
Let us consider, AB be the tower and the angle of
elevation from the point C (on ground) is .
30
In ,
ABC
AB
tan30
BC
AB 1
30
3
30
AB
3
AB 10 3 m
Hence, the height of the tower is m.
10 3
Question: 5
A kite is flying at a height of 60 m above the
ground. The string attached to the kite is
temporarily tied to a point on the ground. The
inclination of the string with the ground is .
60
Find the length of the string, assuming that there is
no slack in the string.
Solution:
Fig. Exc_9.1_5
Let us consider, K as the kite and the string of the
kite is tied to point P on the ground.
In ,
KLP
KL
sin60
KP
60 3
KP 2
120
KP
3
KP 40 3 m
Hence, the length of the string is m.
40 3
Question: 6
A 1.5 m tall boy is standing at some distance from a
30 m tall building. The angle of elevation from his
eyes to the top of the building increases from to
30
as he walks towards the building. Find the
60
distance he walked towards the building.
Solution:
Fig. Exc_9.1_6
Let the boy is standing at a point S initially. He
walked towards the building and reached at point
T.
Now,
PR PQ RQ
PR
30 1.5
28.5m
57
m
2
In ,
PAR
PR
tan30
AR
57
1
2
AR
3
57 1
2AR
3
57
AR 3 m
2
Now, in ,
PRB
PR
tan60
BR
57
2
3
BR
57
3
2BR
57
BR
2 3
19 3
BR m
2
Now,
ST AB
AR BR
57 3 19 3
2 2
38 3
2
19 3 m
Hence, the boy walked m towards the
19 3
building.
Question: 7
From a point on the ground, the angles of elevation
of the bottom and the top of a transmission tower
fixed at the top of a 20 m high building are and
45
respectively. Find the height of the tower.
60
Solution:
Fig. Exc_9.1_7
Let us consider, BC as the building, AB as the
transmission tower, and D as the point on the
ground from where the elevation angles are to be
measured.
In ,
BCD
BC
tan45
CD
20
1
CD
CD 20m
Now, In ,
ACD
AC
tan60
CD
AB BC
3
CD
AB 20
3
20
AB 20 3 20
AB 20 3 1 m
Hence, the height of the transmission tower is
.
20 3 1 m
Question: 8
A statue, 1.6 m tall, stands on a top of pedestal,
from a point on the ground, the angle of elevation
of the top of statue is and from the same point
60
the angle of elevation of the top of the pedestal is
. Find the height of the pedestal.
45
Solution:
Fig. Exc_9.1_8
Let us consider, AB as the statue, BC as the
pedestal, and D as the point on the ground from
where the elevation angles are to be measured.
In ,
BCD
BC
tan45
CD
BC
1
CD
BC CD
Now, In ,
ACD
AB BC
tan60
CD
AB BC
3
CD
1.6 BC BC 3 BC CD
BC 3 1 1.6
1.6
BC
3 1
Multiply numerator and denominator by .
3 1
1.6 3 1
BC
3 1 3 1
2
2
1.6 3 1
3 1
1.6 3 1
BC
2
0.8 3 1
Hence, the height of the pedestal is m.
0.8 3 1
Question: 9
The angle of elevation of the top of a building from
the foot of the tower is and the angle of
30
elevation of the top of the tower from the foot of
the building is . If the tower is 50 m high, find
60
the height of the building.
Solution:
Fig. Exc_9.1_9
Let us consider, AB as the building and CD as the
tower.
In ,
CDB
CD
tan60
BD
50
3
BD
50
BD
3
Now, in ,
ABD
AB
tan30
BD
AB 1
50
3
3
50 1
AB
3 3
50
AB
3
2
AB 16
3
Hence, the height of the building is m.
2
16
3
Question: 10
Two poles of equal heights are standing opposite
each other and either side of the road, which is 80
m wide. From a point between them on the road,
the angles of elevation of the top of the poles are
and , respectively. Find the height of poles
60
30
and the distance of the point from the poles.
Solution:
Fig. Exc_9.1_10
Let us consider, AB and CD as the two poles and O
be the point from where the elevation angles are
measured.
In ,
ABO
AB
tan60
BO
AB
3
BO
AB
BO
3
Now, in ,
CD
tan30
DO
CD 1
80 BO
3
CD 3 80 BO
AB
CD 3 80
3
AB
CD 3 80 ......(1)
3
The poles AB and CD are of equal heights.
So, CD
AB
Put this in equation .
1
CD
CD 3 80
3
1
CD 3 80
3
3 1
CD 80
3
4CD 80 3
CD 20 3
Now,
AB
BO
3
CD
3
20 3
3
BO 20m
DO BD BO
80 20
60m
Hence, the height of both the poles is and
20 3 m
the point O is
and far from these poles.
20m
60m
Question: 11
A TV tower stands vertically on a bank of a canal.
From a point on the other bank directly opposite
the tower the angle of elevation of the top of the
tower is . From another point 20 m away from
60
this point on the line joining this point to the foot
of the tower, the angle of elevation of the top of the
tower is (see Fig. 9.12). Find the height of the
30
tower and the width of the canal.
Fig. Exc_9.1_11
Solution:
According to the given figure,
In ,
ABC
AB
tan60
BC
AB
BC
3
AB
3
BC
Now, in ,
ABD
AB
tan30
BD
AB 1
BC CD
3
AB 1
AB
3
20
3
AB 3 1
AB 20 3 3
3AB AB 20 3
2AB 20 3
AB 10 3
Now,
AB
BC
3
10 3
3
10m
Hence, the height of the tower is m and the
10 3
width of the canal is 10 m.
Question: 12
From the top of a high building, the angle of
7m
elevation of the top of a cable tower is and the
60
angle of depression of its foot is . Determine the
45
height of the tower.
Solution:
Fig. Exc_9.1_12
Let us consider, AB as a building and CD as a cable
tower.
In ,
ABD
AB
tan45
BD
7
1
BD
BD 7m
Now, In ,
ACE
AE BD 7m
CE
tan60
AE
CE
3
7
CE 7 3 m
CD CE ED
7 3 7
7 3 1 m
Therefore, the height of the cable tower is
.
7 3 1 m
Question: 13
As observed from the top of a high lighthouse
75m
from the sea-level, the angles of depression of two
ships are and . If one ship is exactly behind
30
45
the other on the same side of the lighthouse, find
the distance between the two ships.
Solution:
Fig. Exc_9.1_13
Let us consider, AB as the lighthouse and the two
ships be at point C and D respectively.
In ,
ABC
AB
tan45
BC
75
1
BC
BC 75m
Now In ,
ABD
AB
tan30
BD
75 1
BC CD
3
75 1
75 CD
3
75 3 75 CD
75 3 75 CD
CD 75 3 1
Hence, the distance between the two ships is
m.
75 3 1
Question: 14
A 1.2 m tall girl spots a balloon moving with the
wind in a horizontal line at a height of 88.2 m from
the ground. The angle of elevation of the balloon
from the eyes of the girl at any instant is . After
60
some time, the angle of elevation reduces to
30
(see Fig. 9.13). Find the distance travelled by the
balloon during the interval.
Fig. Exc_9.1_14 (Ques.)
Solution:
Fig. Exc_9.1_14 (Sol.)
Let the initial position of balloon be A and after
some time, its position be B and CD be the girl who
spots the balloon.
In ,
ACE
AE
tan60
CE
AF EF
tan60
CE
88.2 1.2
3
CE
87
3
CE
87
CE
3
CE 29 3 m
Now In ,
BCG
BG
tan30
CG
88.2 1.2 1
CG
3
CG 87 3
Distance travelled by balloon
EG
CG CE
87 3 29 3
58 3 m
Thus, the distance is .
58 3 m
Question: 15
A straight highway leads to the foot of a tower. A
man standing at the top of the tower observes a car
as an angle of depression of , which is
30
approaching the foot of the tower with a uniform
speed. Six seconds later, the angle of depression of
the car is found to be . Find the time taken by
60
the car to reach the foot of the tower from this
point.
Solution:
Fig. Exc_9.1_15
Let us consider, AB as the tower and initial position
of the car is C, which changes to D after six
seconds.
In
AB
tan60
DB
AB
3
DB
AB
DB
3
Now in ,
ABC
AB
tan30
BC
AB 1
BD DC
3
AB 3 BD DC
AB
AB 3 DC
3
AB
DC AB 3
3
1
AB 3
3
2AB
3
Time taken by the car to travel distance
.
DC 6seconds
[DC ]
2AB
3
So, speed of car is given by,
Distance
Speed
Time
2AB
3
6
2AB
6 3
The speed remains the same for the distance DB.
Now, time taken by the car to travel distance DB is
given by,
Distance
Time
Speed
AB
AB
3
DB
2AB
3
6 3
AB 6 3
2AB 3
6
2
3seconds
Thus, the time taken by the car is seconds.
3
Question: 16
The angles of elevation of the top of a tower from
two points at a distance of 4 m and 9 m. from the
base of the tower and in the same straight line with
it are complementary. Prove that the height of the
tower is 6 m.
Solution:
Fig. Exc_9.1_16
Let us consider, AQ as the tower and the points R
and S be 4 m and 9 m away from the base of the
tower respectively.
According to the question, the angles are
complementary.
So, if one angle is , the other will be .
θ
90 θ
In ,
AQR
AQ
tanθ
QR
AQ
tanθ ......(1)
4
Now in ,
AQS
AQ
tan 90 θ
SQ
AQ
cot θ ......(2)
9
Multiply equations and .
1
2
AQ AQ
tanθ cot θ
4 9
2
AQ
1
36
2
AQ 36
AQ 36
AQ 6
However, height cannot be negative. Hence, the
height of the tower is 6 m. |
# Linear Equations in Two Variables Class 9
A linear equation will not involve any products and roots of variables and all these variables occur only to the first power and do not appear for the assessment of logarithmic, trigonometric, or even exponential functions.
Few examples of linear equations are given below:
$3x + 5y = 12$, $\frac{1}{2}$$x + 7y + 2z = -9$, $x_{1} - 2x_{2} - 3x_{3} + x_{4} = 0$
Linear equation with 2 variables is very easy and interesting topic. Sometimes when students get stuck with any math problem and might think whether to continue that question or not? This makes difficult to solve problems, as well as making it difficult to easily absorb new ideas or information. Math is all about practice. More you practice more you get confidence. Students are always advice to understand the concept before they attempt any math problem on any topic. In this section, Class 9 students will get step-by-step explanation to every question specified in the linear equation chapter. Because jumping directly into solving problems can lead to frustration and confusion.
## General Form
A linear system of two equations with two variables in any system is written as
ax + by = p
cx + dy = q
Where: x and y are variables
a, b, c, and d are coefficient of x and y and should not be negative.
p and q are constant values. Can be any real number.
## Methods to Solve Linear Equations in Two Variables
Many systems in all kinds of diverse fields can be characterized by the input-output analysis. Generally, to analyse such systems, we follow three different methods:
Method 1: The system is described by some specific mathematical form, in which the input is given and an output is to be found.
Method 2: The system and the output are provided and we need to find the input.
Method 3: Both the input and the output are provided and we need to create the system.
This system of finding the solution from the given input and output gives us the linear system or equations. The general method for solving a linear equation is to perform algebraic operations on the system that will not alter the solution set and will produce a successive simpler system, until a point is reached where it can be checked whether the system is consistent or not and what the solutions are.
A finite set of linear equations are called a system of linear equations or more accurately a linear system. The variables in these are unknowns.
The algebraic operations will follow as mentioned:
1. Multiply an equation by a non-zero constant
2. Interchange the two equations
3. Add constant times of one equation to another
## Examples
Example 1: By using the method of addition, solve
3x + 5y = 8
8x - 5y = 12
Solution:
Consider the given equations
3x + 5y = 8
8x - 5y = 12
3x + 5y = 8
8x - 5y = 12
__________________
11x + 0y = 20
_______________
$\Rightarrow$ x = $\frac{20}{11}$
Substitute x = $\frac{20}{11}$ in 3x + 5y = 8
We get, 3($\frac{20}{11}$) + 5y = 8
Solve for y
$\Rightarrow$ 5y = 8 - ($\frac{60}{11}$)
55y = 88 - 60
y = ($\frac{28}{55}$)
Therefore, x= $\frac{20}{11}$ and y = ($\frac{28}{55}$)
Example 2: Using substitution method solve : x + y = 9 and x - y = 3
Solution:
Isolate x from x + y = 9
$\Rightarrow$ x = 9 - y
Substitute x = 9 - y in x - y = 3
We get,
9 - y - y = 3
9 - 2y = 3
-2y = -6
$\Rightarrow$ y = 3
Now you can substitute y = 3 in any one of the given equations to find the value of x
In x + y = 9 substitute y = 3 and solve for x
x + 3 = 9
$\Rightarrow$ x = 6
Therefore (x, y) = (6, 3).
## Practice Questions
Practice below problems.
Question 1: Solve for x and y
3x - 2y = 1 and x - y = 10
Question 2: Solve below given system using substitution method.
2x - y = 3 and
x + y = 5 |
S k i l l
i n
A L G E B R A
24
# EQUATIONSWITH FRACTIONS
Clearing of fractions
TO SOLVE AN EQUATION WITH fractions, we transform it into an equation without fractions -- which we know how to solve. The technique is called clearing of fractions.
x3 + x − 2 5 = 6.
Solution. Clear of fractions as follows:
Multiply both sides of the equation -- every term -- by the LCM of denominators. Each denominator will then divide into its multiple. We will then have an equation without fractions.
The LCM of 3 and 5 is 15. Therefore, multiply both sides of the equation by 15.
15· x3 + 15· x − 2 5 = 15· 6
On the left, multiply each term by 15. Each denominator will now divide into 15 -- that is the point -- and we have the following simple equation that has been "cleared" of fractions:
5x + 3(x − 2) = 90. It is easily solved as follows: 5x + 3x − 6 = 90 8x = 90 + 6 x = 96 8 = 12.
We say "multiply" both sides of the equation, yet we take advantage of the fact that the order in which we multiply or divide does not matter. (Lesson 1.) Therefore we divide the LCM by each denominator first, and in that way clear of fractions.
We choose a multiple of each denominator, because each denominator will then be a divisor of it.
Example 2. Clear of fractions and solve for x:
x2 − 5x 6 = 19
Solution. The LCM of 2, 6, and 9 is 18. (Lesson 23 of Arithmetic.) Multiply both sides by 18 -- and cancel.
9x − 15x = 2.
It should not be necessary to actually write 18. The student should simply look at and see that 2 will go into 18 nine (9) times. That term therefore becomes 9x.
Next, look at , and see that 6 will to into 18 three (3) times. That term therefore becomes 3· −5x = −15x.
Finally, look at , and see that 9 will to into 18 two (2) times. That term therefore becomes 2 · 1 = 2.
Here is the cleared equation, followed by its solution:
9x − 15x = 2 −6x = 2 x = 2 −6 x = − 13
Example 3. Solve for x:
½(5x − 2) = 2x + 4.
Solution. This is an equation with a fraction. Clear of fractions by mutiplying both sides by 2:
5x − 2 = 4x + 8 5x − 4x = 8 + 2 x = 10.
In the following problems, clear of fractions and solve for x:
Do the problem yourself first!
Problem 1. x2 − x5 = 3 The LCM is 10. Here is the cleared equation and its solution: 5x − 2x = 30 3x = 30 x = 10.
On solving any equation with fractions, the very next line you write --
5x − 2x = 30
-- should have no fractions.
Problem 2. x6 = 1 12 + x8 The LCM is 24. Here is the cleared equation and its solution: 4x = 2 + 3x 4x − 3x = 2 x = 2
Problem 3. x − 25 + x3 = x2 The LCM is 30. Here is the cleared equation and its solution: 6(x − 2) + 10x = 15x 6x − 12 + 10x = 15x 16x − 15x = 12 x = 12.
Problem 4. A fraction equal to a fraction.
x − 14 = x7 The LCM is 28. Here is the cleared equation and its solution: 7(x − 1) = 4x 7x − 7 = 4x 7x − 4x = 7 3x = 7 x = 73
We see that when a single fraction is equal to a single fraction, then the equation can be cleared by "cross-multiplying."
If ab = cd , then ad = bc.
Problem 5. x − 33 = x − 5 2 Here is the cleared equation and its solution: 2(x − 3) = 3(x − 5) 2x − 6 = 3x − 15 2x − 3x = − 15 + 6 −x = −9 x = 9
Problem 6. x − 3x − 1 = x + 1x + 2 Here is the cleared equation and its solution: (x − 3)(x + 2) = (x − 1)(x + 1) x² −x − 6 = x² − 1 −x = −1 + 6 −x = 5 x = −5.
Problem 7. 2x − 39 + x + 1 2 = x − 4 The LCM is 18. Here is the cleared equation and its solution: 4x − 6 + 9x + 9 = 18x − 72 13x + 3 = 18x − 72 13x − 18x = − 72 − 3 −5x = −75 x = 15.
Problem 8. 2x − 3 8x = 14 The LCM is 8x. Here is the cleared equation and its solution: 16 − 3 = 2x 2x = 13 x = 13 2
2nd Level
Next Lesson: Word problems
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# 8.3: Separable Equations
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We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.
## Separation of Variables
We start with a definition and some examples.
Definition: Separable Differential Equations
A separable differential equation is any equation that can be written in the form
$y'=f(x)g(y). \label{sep}$
The term ‘separable’ refers to the fact that the right-hand side of Equation \ref{sep} can be separated into a function of $$x$$ times a function of $$y$$. Examples of separable differential equations include
\begin{align} y' =(x^2−4)(3y+2) \label{eq1} \\[4pt] y' =6x^2+4x \label{eq2}\\[4pt] y' =\sec y+\tan y \label{eq3} \\[4pt] y' =xy+3x−2y−6. \label{eq4} \end{align}
Equation \ref{eq2} is separable with $$f(x)=6x^2+4x$$ and $$g(y)=1$$, Equation \ref{eq3} is separable with $$f(x)=1$$ and $$g(y)=\sec y+\tan y,$$ and the right-hand side of Equation \ref{eq4} can be factored as $$(x+3)(y−2)$$, so it is separable as well. Equation \ref{eq3} is also called an autonomous differential equation because the right-hand side of the equation is a function of $$y$$ alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.
Problem-Solving Strategy: Separation of Variables
1. Check for any values of $$y$$ that make $$g(y)=0.$$ These correspond to constant solutions.
2. Rewrite the differential equation in the form $\dfrac{dy}{g(y)}=f(x)dx.$
3. Integrate both sides of the equation.
4. Solve the resulting equation for $$y$$ if possible.
5. If an initial condition exists, substitute the appropriate values for $$x$$ and $$y$$ into the equation and solve for the constant.
Note that Step 4 states “Solve the resulting equation for $$y$$ if possible.” It is not always possible to obtain $$y$$ as an explicit function of $$x$$. Quite often we have to be satisfied with finding y as an implicit function of $$x$$.
Example $$\PageIndex{1}$$: Using Separation of Variables
Find a general solution to the differential equation $$y'=(x^2−4)(3y+2)$$ using the method of separation of variables.
Solution:
Follow the five-step method of separation of variables.
1. In this example, $$f(x)=x^2−4$$ and $$g(y)=3y+2$$. Setting $$g(y)=0$$ gives $$y=−\dfrac{2}{3}$$ as a constant solution.
2. Rewrite the differential equation in the form
$\dfrac{dy}{3y+2}=(x^2−4)dx.$
3. Integrate both sides of the equation:
$∫\dfrac{dy}{3y+2}=∫(x^2−4)dx.$
Let $$u=3y+2$$. Then $$du=3\dfrac{dy}{dx}dx$$, so the equation becomes
$\dfrac{1}{3}∫\dfrac{1}{u}du=\dfrac{1}{3}x^3−4x+C$
$\dfrac{1}{3}\ln|u|=\dfrac{1}{3}x^3−4x+C$
$\dfrac{1}{3}\ln|3y+2|=\dfrac{1}{3}x^3−4x+C.$
4. To solve this equation for $$y$$, first multiply both sides of the equation by $$3$$.
$\ln|3y+2|=x^3−12x+3C$
Now we use some logic in dealing with the constant $$C$$. Since $$C$$ represents an arbitrary constant, $$3C$$ also represents an arbitrary constant. If we call the second arbitrary constant $$C_1$$, the equation becomes
$\ln|3y+2|=x^3−12x+C_1.$
Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base $$e$$).
\begin{align} e^{\ln|3y+2|} =e^{x^3−12x+C_1} \\ |3y+2| =e^{C_1}e^{x^3−12x} \end{align}
Again define a new constant $$C_2=e^{c_1}$$ (note that $$C_2>0$$):
$|3y+2|=C_2e^{x^3−12x}.$
This corresponds to two separate equations:
$3y+2=C_2e^{x^3−12x}$
and
$3y+2=−C_2e^{x^3−12x}.$
The solution to either equation can be written in the form
$y=\dfrac{−2±C_2e^{x^3−12x}}{3}.$
Since $$C_2>0$$, it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant $$C$$ is entirely arbitrary, and can be dropped. Therefore the solution can be written as
$y=\dfrac{−2+Ce^{x^3−12x}}{3}.$
5. No initial condition is imposed, so we are finished.
Exercise $$\PageIndex{1}$$
Use the method of separation of variables to find a general solution to the differential equation
$y'=2xy+3y−4x−6. \nonumber$
Hint
First factor the right-hand side of the equation by grouping, then use the five-step strategy of separation of variables.
$y=2+Ce^{x^2+3x} \nonumber$
Example $$\PageIndex{2}$$: Solving an Initial-Value Problem
Using the method of separation of variables, solve the initial-value problem
$y'=(2x+3)(y^2−4),y(0)=−3.$
Solution
Follow the five-step method of separation of variables.
1. In this example, $$f(x)=2x+3$$ and $$g(y)=y^2−4$$. Setting $$g(y)=0$$ gives $$y=±2$$ as constant solutions.
2. Divide both sides of the equation by $$y^2−4$$ and multiply by $$dx$$. This gives the equation
$\dfrac{dy}{y^2−4}=(2x+3)dx.$
3. Next integrate both sides:
$∫\dfrac{1}{y^2−4}dy=∫(2x+3)dx. \label{Ex2.2}$
To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
$\dfrac{1}{y^2−4}=\dfrac{1}{4}\left(\dfrac{1}{y−2}−\dfrac{1}{y+2}\right).$
Then Equation \ref{Ex2.2} becomes
$\dfrac{1}{4}∫\left(\dfrac{1}{y−2}−\dfrac{1}{y+2}\right)dy=∫(2x+3)dx$
$\dfrac{1}{4}\left (\ln|y−2|−\ln|y+2| \right)=x^2+3x+C.$
Multiplying both sides of this equation by $$4$$ and replacing $$4C$$ with $$C_1$$ gives
$\ln|y−2|−\ln|y+2|=4x^2+12x+C_1$
$\ln \left|\dfrac{y−2}{y+2}\right|=4x^2+12x+C_1.$
4. It is possible to solve this equation for y. First exponentiate both sides of the equation and define $$C_2=e^{C_1}$$:
$\left|\dfrac{y−2}{y+2}\right|=C_2e^{4x^2+12x}.$
Next we can remove the absolute value and let $$C_2$$ be either positive or negative. Then multiply both sides by $$y+2$$.
$y−2=C_2(y+2)e^{4x^2+12x}$
$y−2=C_2ye^{4x^2+12x}+2C_2e^{4x^2+12x}.$
Now collect all terms involving y on one side of the equation, and solve for y:
$y−C_2ye^{4x^2+12x}=2+2C_2e^{4x^2+12x}$
$y(1−C_2e^{4x^2+12x})=2+2C_2e^{4x^2+12x}$
$y=\dfrac{2+2C_2e^{4x^2+12x}}{1−C_2e^{4x^2+12x}}.$
5. To determine the value of $$C_2$$, substitute $$x=0$$ and $$y=−1$$ into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation $$\dfrac{y−2}{y+2}=C_2e^{4x^2+12}$$. This is much easier to solve for $$C_2$$:
$\dfrac{y−2}{y+2}=C_2e^{4x^2+12x}$
$\dfrac{−1−2}{−1+2}=C_2e^{4(0)^2+12(0)}$
$C_2=−3.$
Therefore the solution to the initial-value problem is
$y=\dfrac{2−6e^{4x^2+12x}}{1+3e^{4x^2+12x}}.$
A graph of this solution appears in Figure $$\PageIndex{1}$$.
Exercise $$\PageIndex{2}$$
Find the solution to the initial-value problem
$6y'=(2x+1)(y^2−2y−8) \nonumber$
with $$y(0)=−3$$ using the method of separation of variables.
Hint
Follow the steps for separation of variables to solve the initial-value problem.
$y=\dfrac{4+14e^{x^2+x}}{1−7e^{x^2+x}} \nonumber$
## Applications of Separation of Variables
Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.
#### Solution concentrations
Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.
Example $$\PageIndex{3}$$: Determining Salt Concentration over Time
A tank containing $$100\,L$$ of a brine solution initially has $$4\,kg$$ of salt dissolved in the solution. At time $$t=0$$, another brine solution flows into the tank at a rate of $$2\,L/min$$. This brine solution contains a concentration of $$0.5\,kg/L$$ of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of $$2\,L/min$$, so that the level of liquid in the tank remains constant (Figure $$\PageIndex{2}$$). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.
Solution
First we define a function $$u(t)$$ that represents the amount of salt in kilograms in the tank as a function of time. Then $$\dfrac{du}{dt}$$ represents the rate at which the amount of salt in the tank changes as a function of time. Also, $$u(0)$$ represents the amount of salt in the tank at time $$t=0$$, which is $$4$$ kilograms.
The general setup for the differential equation we will solve is of the form
$\dfrac{du}{dt}=INFLOW RATE−OUTFLOW RATE.$
INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of $$2$$L/min, and each liter of solution contains $$0.5$$ kilogram of salt, every minute $$2(0.5)=1kilogram$$ of salt enters the tank. Therefore INFLOW RATE = $$1$$.
To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time $$t$$ is equal to $$u(t)$$. Thus, the concentration of salt is $$\dfrac{u(t)}{100} kg/L$$, and the solution leaves the tank at a rate of $$2$$ L/min. Therefore salt leaves the tank at a rate of $$\dfrac{u(t)}{100}⋅2=\dfrac{u(t)}{50}$$ kg/min, and OUTFLOW RATE is equal to $$\dfrac{u(t)}{50}$$. Therefore the differential equation becomes $$\dfrac{du}{dt}=1−\dfrac{u}{50}$$, and the initial condition is $$u(0)=4.$$ The initial-value problem to be solved is
$\dfrac{du}{dt}=1−\dfrac{u}{50},u(0)=4.$
The differential equation is a separable equation, so we can apply the five-step strategy for solution.
Step 1. Setting $$1−\dfrac{u}{50}=0$$ gives $$u=50$$ as a constant solution. Since the initial amount of salt in the tank is $$4$$ kilograms, this solution does not apply.
Step 2. Rewrite the equation as
$\dfrac{du}{dt}=\dfrac{50−u}{50}.$
Then multiply both sides by $$dt$$ and divide both sides by $$50−u:$$
$\dfrac{du}{50−u}=\dfrac{dt}{50}.$
Step 3. Integrate both sides:
\begin{align} ∫\dfrac{du}{50−u} =∫\dfrac{dt}{50} \\ −\ln|50−u| =\dfrac{t}{50}+C. \end{align}
Step 4. Solve for $$u(t)$$:
$\ln|50−u|=−\dfrac{t}{50}−C$
$e^{\ln|50−u|}=e^{−(t/50)−C}$
$|50−u|=C_1e^{−t/50}.$
Eliminate the absolute value by allowing the constant to be either positive or negative:
$50−u=C_1e^{−t/50}.$
Finally, solve for $$u(t)$$:
$u(t)=50−C_1e^{−t/50}.$
Step 5. Solve for $$C_1$$:
\begin{align} u(0) =50−C_1e^{−0/50} \\ 4 =50−C_1 \\ C_1 =46. \end{align}
The solution to the initial value problem is $$u(t)=50−46e^{−t/50}.$$ To find the limiting amount of salt in the tank, take the limit as $$t$$ approaches infinity:
\begin{align} \lim_{t→∞}u(t) =50−46e^{−t/50} \\ =50−46(0)=50. \end{align}
Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is $$50$$ kilograms, then it remains constant. If it starts at less than $$50$$ kilograms, then it approaches 50 kilograms over time.
Exercise $$\PageIndex{3}$$
A tank contains $$3$$ kilograms of salt dissolved in $$75$$ liters of water. A salt solution of $$0.4\,kg salt/L$$ is pumped into the tank at a rate of $$6\,L/min$$ and is drained at the same rate. Solve for the salt concentration at time $$t$$. Assume the tank is well mixed at all times.
Hint
Follow the steps in Example $$\PageIndex{3}$$ and determine an expression for INFLOW and OUTFLOW. Formulate an initial-value problem, and then solve it.
Initial value problem:
$\dfrac{du}{dt}=2.4−\dfrac{2u}{25},\, u(0)=3 \nonumber$
$u(t)=30−27e^{−t/50} \nonumber$
## Newton’s law of Cooling
Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let $$T(t)$$ represent the temperature of an object as a function of time, then $$\dfrac{dT}{dt}$$ represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by $$T_s$$. Then Newton’s law of cooling can be written in the form
$\dfrac{dT}{dt}=k(T(t)−T_s)$
or simply
$\dfrac{dT}{dt}=k(T−T_s).$
The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature $$T_0$$. Therefore the initial-value problem that needs to be solved takes the form
$\dfrac{dT}{dt}=k(T−T_s) \label{newton}$
with $$T(0)=T_0$$, where $$k$$ is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example $$\PageIndex{4}$$.
Example $$\PageIndex{4}$$: Waiting for a Pizza to Cool
A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is $$350°F.$$ The temperature of the kitchen is $$75°F$$, and after $$5$$ minutes the temperature of the pizza is $$340°F$$. We would like to wait until the temperature of the pizza reaches $$300°F$$ before cutting and serving it (Figure $$\PageIndex{3}$$). How much longer will we have to wait?
Solution
The ambient temperature (surrounding temperature) is $$75°F$$, so $$T_s=75$$. The temperature of the pizza when it comes out of the oven is $$350°F$$, which is the initial temperature (i.e., initial value), so $$T_0=350$$. Therefore Equation \ref{newton} becomes
$\dfrac{dT}{dt}=k(T−75)$
with $$T(0)=350.$$
To solve the differential equation, we use the five-step technique for solving separable equations.
1. Setting the right-hand side equal to zero gives $$T=75$$ as a constant solution. Since the pizza starts at $$350°F,$$ this is not the solution we are seeking.
2. Rewrite the differential equation by multiplying both sides by $$dt$$ and dividing both sides by $$T−75$$:
$\dfrac{dT}{T−75}=kdt. \nonumber$
3. Integrate both sides:
\begin{align} ∫\dfrac{dT}{T−75} =∫kdt\nonumber \\ \ln|T−75| =kt+C.\nonumber \end{align}
4. Solve for $$T$$ by first exponentiating both sides:
\begin{align}e^{\ln|T−75|} =e^{kt+C}\nonumber \\ |T−75| =C_1e^{kt}\nonumber \\ T−75 =C_1e^{kt}\nonumber \\ T(t) =75+C_1e^{kt}.\nonumber \end{align}
5. Solve for $$C_1$$ by using the initial condition $$T(0)=350:$$
\begin{align}T(t) =75+C_1e^{kt}\nonumber \\ T(0) =75+C_1e^{k(0)}\nonumber \\ 350 =75+C_1\nonumber \\ C_1 =275.\nonumber \end{align}
Therefore the solution to the initial-value problem is
$T(t)=75+275e^{kt}.\nonumber$
To determine the value of $$k$$, we need to use the fact that after $$5$$ minutes the temperature of the pizza is $$340°F$$. Therefore $$T(5)=340.$$ Substituting this information into the solution to the initial-value problem, we have
$T(t)=75+275e^{kt}\nonumber$
$T(5)=340=75+275e^{5k}\nonumber$
$265=275e^{5k}\nonumber$
$e^{5k}=\dfrac{53}{55}\nonumber$
$\ln e^{5k}=\ln(\dfrac{53}{55})\nonumber$
$5k=\ln(\dfrac{53}{55})\nonumber$
$k=\dfrac{1}{5}\ln(\dfrac{53}{55})≈−0.007408.\nonumber$
So now we have $$T(t)=75+275e^{−0.007048t}.$$ When is the temperature $$300°F$$? Solving for t, we find
$T(t)=75+275e^{−0.007048t}\nonumber$
$300=75+275e^{−0.007048t}\nonumber$
$225=275e^{−0.007048t}\nonumber$
$e^{−0.007048t}=\dfrac{9}{11}\nonumber$
$\ln e^{−0.007048t}=\ln\dfrac{9}{11}\nonumber$
$−0.007048t=\ln\dfrac{9}{11}\nonumber$
$t=−\dfrac{1}{0.007048}\ln\dfrac{9}{11}≈28.5.\nonumber$
Therefore we need to wait an additional $$23.5$$ minutes (after the temperature of the pizza reached $$340°F$$). That should be just enough time to finish this calculation.
Exercise $$\PageIndex{4}$$
A cake is removed from the oven after baking thoroughly, and the temperature of the oven is $$450°F$$. The temperature of the kitchen is $$70°F$$, and after $$10$$ minutes the temperature of the cake is $$430°F$$.
1. Write the appropriate initial-value problem to describe this situation.
2. Solve the initial-value problem for $$T(t)$$.
3. How long will it take until the temperature of the cake is within $$5°F$$ of room temperature?
Hint
Determine the values of $$T_s$$ and $$T_0$$ then use Equation \ref{newton}.
Initial-value problem $\dfrac{dT}{dt}=k(T−70),T(0)=450\nonumber$
$T(t)=70+380e^{kt}\nonumber$
Approximately $$114$$ minutes.
## Key Concepts
• A separable differential equation is any equation that can be written in the form $$y'=f(x)g(y).$$
• The method of separation of variables is used to find the general solution to a separable differential equation.
## Key Equations
• Separable differential equation
$$y′=f(x)g(y)$$
• Solution concentration
$$\dfrac{du}{dt}=INFLOW RATE−OUTFLOW RATE$$
• Newton’s law of cooling
$$\dfrac{dT}{dt}=k(T−T_s)$$
## Glossary
autonomous differential equation
an equation in which the right-hand side is a function of $$y$$ alone
separable differential equation
any equation that can be written in the form $$y'=f(x)g(y)$$
separation of variables
a method used to solve a separable differential equation
## Contributors
• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. |
### Unit 3
```Unit 3
Motion at Constant Acceleration
Giancoli, Sec 2- 5, 6, 8
Example 3-1 Consider the case of a car that accelerates from rest with a
constant acceleration of 15 m / s2 starting at t1 = 0.
a
v 2 v1
t 2 t1
v
t
2
v 2 v1 a ( t 2 t1 ) 0 15 m s ( t 2 0 ) 15 m s 2
t
2
We see that velocity increases by 15 m/s every second and is
thus a linear function of time.
t (s)
v ( m/s)
a ( m/ s2 )
0
0
15
1
15
15
2
30
15
3
45
15
4
60
15
5
75
15
Unit 3- 2
Derivation of Equations for
Motion at Constant Acceleration
•In the next 4 slides we will combine several known
equations under the assumption that the acceleration is
constant.
•This process is called a derivation.
•It will produce four equations connecting
displacement, velocity, acceleration and time.
•You will use these equations to solve most of the
problems in Units 3,4 and 7.
Skip Derivation
Unit 3- 3
Derivations
•Much research in Physics requires derivations.
•In more advanced courses students are required to be able to
perform derivations on tests and homework
•In this course you will need to know the initial assumptions, the
resultant equations and how to apply them.
•You do not need to memorize derivations.
•But I could ask you to derive an equation for a specific problem.
This is very similar to an ordinary problem without a numeric
Unit 3- 4
Motion at Constant Acceleration - Derivation
•Consider the special case acceleration equals a
constant:
a = constant
• Use the subscript “0” to refer to the initial
conditions
• Thus t0 refers to the initial time and we will set t0 =
0.
• At this time v0 is the initial velocity and x0 is the
initial displacement.
•At a later time t, v is the velocity and x is the
displacement
•In the equations t1t0 and t2 t
Unit 3- 5
Motion at Constant Acceleration - Derivation
The average velocity during this time is:
v
x x0
t t0
x x0
(Eqn. 1)
t
v
The acceleration is assumed to be constant
a
v v0
Constant
t
( Eqn. 2 )
vo
t
From this we can write
v v0 a t
( Eqn. 3 )
Unit 3- 6
Motion at Constant Acceleration - Derivation
Because the velocity increases at a uniform rate, the average
velocity is the average of the initial and final velocities
v
v
v0 v
( Eqn. 4 )
2
From the definition of average velocity
x x0 v t
x0 (
v0 v
)t
2
x0 (
vo
v 0 v 0 at
t
)t
2
And thus
x
x0 v0 t
1
at
2
( Eqn. 5 )
2
Unit 3- 7
Equations for Motion at Constant Acceleration
• The book derives one more equation by eliminating time
• The notation in the equations has changed
• At t = 0, x0 is the displacement and v0 is the velocity
• At a later time t, x is the displacement and v is the
velocity
v v0 a t
x x0 v0 t
1
at
2
2
v v0 2 a ( x x0 )
2
v
2
v v0
2
Unit 3- 8
Example 3-2. A world-class sprinter can burst out of the blocks to essentially
top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the
average acceleration of this sprinter, and how long does it take him to reach that
speed? (Note: we have to assume a=constant)
v v 2ax x
2
2
0
0
v0 0
x0 0
v 11 . 5 m s
x 15 . 0 m
v v at
0
v v
2
a
2
t
0
2( x x )
2
2
(11 . 5 m ) 0
s
a
2 ( 15 . 0 m 0 )
s
0
a
0
a 4 . 41 m
vv
2
t
11 . 5 m
s
4 . 41 m
0
s
2
t 2 . 61 s
Unit 3-9
Graphical Analysis of Linear Motion
• Consider the graph of x vs. t. The graph is a straight line, which means the
slope is constant.
• The slope of the line is the rise (Δy) over the run (Δx).
slope
x
t
• If we compare this with the definition of velocity,
we see that the slope of the x vs. t graph is the
velocity.
v
x
t
Unit 2 - 10
Graphical Analysis of Linear Motion
The graphs describe the motion
of a car whose velocity is
changing:
v
lim
x
t 0 t
v is the slope of position vs.
time graph. Since the graph
is not linear, we draw a
tangent line at each point and
find slope of the tangent line.
a
lim
v
t 0
t
Thus a is the slope of
velocity vs. time graph.
Unit 3- 11
Example 3-3: Calculate the acceleration between points A and B and B
and C.
a
a
lim
v
t 0
t
v
( straight
t
aAB
a BC
v 2 v1
15 . 0 m
s
s
20 . 0 s 15 . 0 s
15.0 m
t 2 t1
v 2 v1
t 2 t1
line for A B & B C)
15 . 0 m
s
s
25 . 0 s 20 . 0 s
5.0 m
0 .0 m
s
2
2 .0 m
s
2
Unit 3- 12
Example 3-4. A truck going at a constant speed of 25 m/s passes a car at rest.
The instant the truck passes the car, the car begins to accelerate at a constant
1.00 m / s2. How long does it take for the car to catch up with the truck.
1
x x v t
t
0t
0t
2
at
xc x0c v0ct
2
t
x t ( 25 m ) t
s
x
c
c
(1 . 0 m
2
1
s
2
2
)t
act
2
2
t
) t ( 25 m ) t
s
s
(1 . 0 m
2
(1 . 0 m
2
x x
When the car catches the truck:
1
1
1
2
2
s
2
) t 25 m
s
t 50 s
How far has the car traveled when it catches the truck?
1
x (1 . 0 m )( 50 s ) 1250 m
s
2
2
c
2
Unit 3- 13
Example 3-4: Graphical Interpretation
In order to understand the solution to example 4, we can graph the two
equations:
x
c
1
2
(1 . 0 m
s
2
)t
2
x t ( 25 m ) t
s
•
•
•
•
•
The graph of the truck (red) is linear because the velocity is constant.
The car is accelerating so its graph (blue) is quadratic.
The two curves intersect at t = 50 s which agrees with the solution.
They intersect at x ~ 1250 m which also agrees.
The slopes of the two graphs at t = 50 s indicate that the car is traveling twice
as fast as the truck.
Unit 3- 14
Unit 3 Appendix
Photo and Clip Art Credits
Some figures electronically reproduced by permission of Pearson Education, Inc., Upper
Slide
3-9 Sprinter: Microsoft Clipart Collection, reproduced under general license for educational purposes.
4-5, 6, 7 Soccer Kicker: Microsoft Clipart Collection, reproduced under general license for educational
purposes.
4-8, 9 Photo of Leaning Tower of Pisa: Microsoft Clipart Collection, reproduced under general license
for educational purposes.
Unit 03- 15
A quadratic equation is a polynomial equation of the second degree (one term is
squared). The general form of a quadratic equation is
a x bxc 0
2
where a, b and c are constants and a is not equal to zero.
You will need to arrange your equation in this form in order to determine the
numeric value of a, b and c. Often this semester, the unknown variable will be t
A quadratic equation has two solutions or roots:
x
b
b 4 ac
2
and
2a
x
b
b 4 ac
2
2a
These solutions are usually combined using the ± symbol:
x
b
b 4 ac
2
2a
• The solutions can be equal ( if b2 = 4 a c ).
• If they are not equal, you will often need to select the solution that makes
sense for the problem.
Unit 3- 16
``` |
# A triangle has corners at (1 ,9 ), (3 ,4 ), and (4 ,5 ). How far is the triangle's centroid from the origin?
Sep 30, 2016
≈ 6.566 units.
#### Explanation:
Given $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ and } \left({x}_{3} , {y}_{3}\right)$ are the vertices of a triangle then the coordinates of the centroid are.
${x}_{c} = \frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right) \text{ and } {y}_{c} = \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)$
That is the average of the coordinates of the vertices.
substituting the given coordinates into the above.
${x}_{c} = \frac{1}{3} \left(1 + 3 + 4\right) = \frac{8}{3} \text{ and } {y}_{c} = \frac{1}{3} \left(9 + 4 + 5\right) = 6$
$\Rightarrow \text{coordinates of centroid} = \left(\frac{8}{3} , 6\right)$
To calculate the distance from this point to the origin use $\textcolor{b l u e}{\text{Pythagoras' theorem}}$
$d = \sqrt{{\left(\frac{8}{3}\right)}^{2} + {6}^{2}} = \sqrt{\frac{64}{9} + \frac{324}{9}}$
=sqrt(388/9)≈6.566" to 3 decimal places" |
# Greatest Common Divisor (GCD) Calculator
Calculate the Greatest Common Divisor of the given numbers.
Result:
The GCD calculator allows you to find the largest positive integer that divides two or more numbers without leaving a remainder.
GCD is useful in various situations, such as simplifying fractions, finding equivalent fractions, determining common factors, simplifying ratios, and solving problems related to divisibility and modular arithmetic.
## How to calculate GCD?
To calculate the GCD, there are several methods available, including Prime Factorization and the Euclidean algorithm.
If you want to calculate the GCD for multiple numbers, it is usually easiest to use a calculator.
### Prime Factorization
1. Write down the prime factorization of each number in the set.
2. Identify the common prime factors among the numbers.
3. Multiply the common prime factors together to obtain the prime factorization of the GCD.
Example #1: Finding the GCD of the numbers 12, 30, and 60:
Step 1: Prime factorize each number:
• 12 = 2 * 2 * 3
• 30 = 2 * 3 * 5
• 60 = 2 * 2 * 3 * 5
Step 2: Identify the common prime factors:
The common prime factors are 2 and 3.
Step 3: Multiply the common prime factors together:
GCD = 2 * 3 = 6
Example #2: Finding the GCD of the numbers 18 and 27:
Step 1: Prime factorize each number:
• 18 = 2 * 3 * 3
• 27 = 3 * 3
Step 2: Identify the common prime factors:
The common prime factors are 3 and 3.
Step 3: Multiply the common prime factors together:
GCD = 3 * 3 = 9
Numbers GCD
10, 15, 25 5
12, 90 6
15, 45, 90 15
7, 17, 83 1 |
Happy Teej - TEEJ2024
# How to Solve Mixture and Alligation Questions Easily
1) In a Vessel, the ratio of milk to water is 4:7. Then find how much percentage of water is in the vessel.
Solution:
Milk: Water
4:7
In the vessel, the ratio of water is 7 and the total of ratio is 11.
Water Percentage in the vessel = Ratio of Water/ Total of Ratio×100
= 7/11×100
=63.63
So, the percentage of water in the vessel is 63.63%.In the same, way you can also find out the percentage of water in the vessel.
2) In two different vessels, if in the vessel, it has 3 litre of mixture of milk and water in which the percentage of water is 20% while in the second vessel it has a mixture of 5 litre out of which the percentage of milk is 70%. Find the Percentage of Water in both the vessels.
Solution:
Vessel 1 vessel 2
Mixture 3 litre 5 litre
Percentage of milk 80% 70%
Percentage of water 20% 30%
Percentage of Water in Both Vessels = 20%of 3+30%of 5/3+5
=90/8
= 11.25%
3) There are two vessels which are full of milk and water. In the first vessel the ratio of milk to water is 3:2 while in the second vessel the ratio of milk to water is 5:2. If both are put to gathers in the new vessel, find the ratio of milk to water.
Solution:
Sum of Ratio of Milk from both the vessels
3/5 + 5/7
46/35
Ratio of Water from both the vessels
2/5 + 2/7
24/35
Ratio of Milk to Water in the new vessel
46/35 + 24/35
23:12
4) If the Total Mixture in the vessel is 30 litre and the ratio of milk to water is 3:2. If a certain quantity of water is added further and the ratio of milk to water becomes 1:2. Find out how much quantity is water is added.
Solution:
Quantity of milk = 30/5×3
= 18
Quantity of milk = 30/5×2
= 12
Quantity is water is added
18/12+X = 1/2
36 = 12 + X
36 – 12 = X
X = 24
So, the water added is 24 liter.
5) A mixture of 25 litre has some quantity of milk and water and the rate of milk sold at 15RS per liter while pure milk is sold at the rate of 25RS. Find the quantity of water in the mixture.
Solution:
Price of pure milk 15RS
Price of Mixture 25RS
Price of water 0
Milk Water
25 0
15
15 10
Ratio of 15:10
3:2
Quantity of water in the mixture = 25/5×3
= 15 litre.
#### Ramandeep Singh - Educator
I'm Ramandeep Singh, your guide to banking and insurance exams. With 14 years of experience and over 5000 successful selections, I understand the path to success firsthand, having transitioned from Dena Bank and SBI. I'm passionate about helping you achieve your banking and insurance dreams. |
# Easy algebra problems
We'll provide some tips to help you select the best Easy algebra problems for your needs. Math can be a challenging subject for many students.
## The Best Easy algebra problems
One instrument that can be used is Easy algebra problems. algebrahelp.com is a free math website that offers step-by-step solutions to any quadratic equation. Simply enter the values for a, b, and c, and our solver will do the rest. In addition to the answer, you'll also see a detailed explanation of each step in the solution process. This can be extremely helpful if you're stuck on a problem and need some extra guidance. Best of all, our service is completely free. So if you're struggling with a quadratic equation, be sure to give us a try. We'll help you get the answer you need, step by step.
How to solve perfect square trinomial? This is a algebraic equation that can be written in the form of ax2 + bx + c = 0 . If the coefficient of x2 is one then we can use the factoring method to solve it. We will take two factors of c such that their product is equal to b2 - 4ac and their sum is equal to b. How to find such numbers? We will use the quadratic formula for this. Now we can factorize the expression as (x - r1)(x - r2) = 0, where r1 and r2 are the roots of the equation. To find the value of x we will take one root at a time and then solve it. We will get two values of x, one corresponding to each root. These two values will be the solutions of the equation.
A parabola is a two-dimensional figure that appears in many mathematical and physical situations. In mathematics, a parabola is defined as a curve where any point is equidistant from a fixed point (called the focus) and a fixed line (called the directrix). In physics, parabolas describe the path of objects under the influence of gravity, such as a ball thrown in the air. In both cases, the equation for a parabola can be quite complicated. However, there are online tools that can help to solve these equations quickly and easily. One such tool is the Parabola Solver, which allows users to input the parameters of their equation and then receive step-by-step instructions for finding the solution. This tool can be an invaluable resource for students and professionals who need to solve complex parabolic equations.
There are a variety of apps available that can help you solve math problems. Some of these apps are designed for specific types of problems, while others are more general purpose. In terms of features, some apps offer step-by-step solutions, while others simply provide the answer. There are also a number of free and paid options available. Ultimately, the best app for you will depend on your individual needs and preferences. However, some of the most popular math problem solving apps include Photomath, Wolfram Alpha, and Mathway.
A triangle solver calculator can be a useful tool for anyone who needs to find the measurements of a triangle. There are many different types of triangle calculators available, but they all essentially work in the same way. To use a triangle solver calculator, you simply enter the three sides of the triangle into the calculator. The calculator will then use a mathematical formula to calculate the measurements of the triangle. Triangle solver calculators can be helpful when you need to find the measurements of a Triangle that you do not have all the dimensions for. It can also be useful for checking your work if you have already calculated the dimensions of a Triangle yourself. Triangle solver calculators are available online and in many math textbooks.
## Instant help with all types of math
This app is amazing and helps me get my homework done quicker when I am very hurry or just plain forget something. It teaches me the basics and different ways to get around the problem I take a photo of. 100% recommend to everyone. |
# Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4
Question 1.
By the principle of mathematical induction, prove that, for n ≥ 1
Solution:
∴ P(k+ 1) is true.
Thus P(K) is true ⇒ (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all n ∈ N.
Question 2.
By the principle of mathematical induction, prove that, for n > 1
Solution:
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(k) is true for all n ∈ N.
Question 3.
Prove that the sum of the first n non-zero even numbers is n2 + n.
Solution:
Let P(n) : 2 + 4 + 6 +…+2n = n2 + n, ∀ n ∈ N
Step 1:
P( 1) : 2 = 12 + 1 = 2 which is true for P( 1)
Step 2:
P(k): 2 + 4 + 6+ …+ 2k = k2 + k. Let it be true.
Step 3:
P(k + 1) : 2 + 4 + 6 + … + 2k + (2k + 2)
= k2+ k + (2k + 2) = k2 + 3k + 2
= k2 + 2k + k + 1 + 1
= (k+ 1)2+ (k + 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.
Question 4.
By the principle of Mathematical induction, prove that, for n ≥ 1.
Solution:
∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∈ N
Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,
Solution:
⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true.
Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,
Solution:
⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true for n ≥ 2.
Question 7.
Using the Mathematical induction, show that for any natural number n
Solution:
∴ P(k + 1) is true
Thus p(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction,
p(n) is true for all n ∈ z
Question 8.
Using the Mathematical induction, show that for any natural number n,
Solution:
∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(n) is true for all n ∈ N.
Question 9.
Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1
Solution:
P(n) is the statement
1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1
To prove for n = 1
LHS = 1! = 1
RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1
LHS = RHS ⇒ P(1) is true
Assume that the given statement is true for n = k
(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true
To prove P(k + 1) is true
p(k + 1) = p(k) + t(k + 1)
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! [1 + k + 1] – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
= (k + 1 + 1)! – 1
∴ P(k + 1) is true ⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.
Question 10.
Using the Mathematical induction, show that for any natural number n, x2n – y2n is divisible by x +y.
Solution:
Let P(n) = x2n – y2n is divisible by (x + y)
For n = 1
P(1) = x2 × 1 – y2 × 1 is divisible by (x + y)
⇒ (x + y) (x – y) is divisible by (x + y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = x2k – y2k is divisible by (x + y)
⇒ x2k – y2k = λ(x + y) …… (i)
For n = k + 1
⇒ P(k + 1) = x2(k + 1) – y2(k + 1) is divisible by (x + y)
Now x2(k + 2) – y2(k + 2)
= x2k + 2 – x2ky2 + x2ky2 – y2k + 2
= x2k.x2 – x2ky2 + x2ky2 – y2ky2
= x2k (x2 – y2) + y2λ (x + y) [Using (i)]
⇒ x2k + 2 – y2k + 2 is divisible by (x + y)
∴ P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(n) is true for all n ∈ N
Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1,
Solution:
Question 12.
Use induction to prove that n3 – 7n + 3, is divisible by 3, for all natural numbers n.
Solution:
Let P(n) : n3 – 7n + 3
Step 1:
P(1) = (1)3 – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).
Step 2:
P(k) : k3 – 7k + 3 = 3λ. Let it be true
⇒ k3 = 3λ + 7k – 3
Step 3:
P(k + 1) = (k + 1)3 – 7(k + 1) + 3
= k3 + 1 + 3k2 + 3k – 7k – 7 + 3
= k3 + 3k2 – 4k – 3
= (3λ + 7k – 3) + 3k2 – 4k – 3 (from Step 2)
= 3k2 + 3k + 3λ – 6
= 3(k2 + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).
Question 13.
Use induction to prove that 5n + 1 + 4 × 6n when divided by 20 leaves a remainder 9, for all natural numbers n.
Solution:
P(n) is the statement 5n + 1 + 4 × 6n – 9 is ÷ by 20
P(1) = 51 + 1 + 4 × 61 – 9 = 52 + 24 – 9
= 25 + 24 – 9 = 40 ÷ by 20
So P(1) is true
Assume that the given statement is true for n = k
(i.e) 5k + 1 + 4 × 6n – 9 is ÷ by 20
P(1) = 51 + 1 + 4 × 61 – 9
= 25 + 24 – 9
So P(1) is true
To prove P(k + 1) is true
P(k + 1) = 5k + 1 + 1 + 4 × 6k + 1 + 1 – 9
= 5 × 5 k + 1 + 4 × 6 × 6k – 9
= 5[20C + 9 – 4 × 6k] + 24 × 6k – 9 [from(1)]
= 100C + 45 – 206k + 246k – 9
= 100C + 46k + 36
= 100C + 4(9 + 6k)
Now for k = 1 ⇒ 4(9 + 6k) = 4(9 + 6)
= 4 × 15 = 60 ÷ by 20 .
for k = 2 = 4(9 + 62) = 4 × 45 = 180 ÷ 20
So by the principle of mathematical induction 4(9 + 6k) is ÷ by 20
Now 100C is ÷ by 20.
So 100C + 4(9 + 6k) is ÷ by 20
⇒ P(k + 1) is true whenever P(k) is true. So by the principle of mathematical induction P(n) is true.
Question 14.
Use induction to prove that 10n + 3 × 4n + 2 + 5, is divisible by 9, for all natural numbers n.
Solution:
P(n) is the statement 10n + 3 × 4n + 2 + 5 is ÷ by 9
P(1) = 101 + 3 × 42 + 5 = 10 + 3 × 16 + 5
= 10 + 48 + 5 = 63 ÷ by 9
So P(1) is true. Assume that P(k) is true
(i.e.) 10k + 3 × 4k + 2 + 5 is ÷ by 9
(i.e.) 10k + 3 × 4k + 2 + 5 = 9C (where C is an integer)
⇒ 10k = 9C – 5 – 3 × 4k + 2 ……(1)
To prove P(k + 1) is true.
Now P(k + 1) = 10k + 1 + 3 × 4k + 3 + 5
= 10 × 10k + 3 × 4k + 2 × 4 + 5
= 10[9C – 5 – 3 × 4k + 2] + 3 × 4k + 2 × 4 + 5
= 10[9C – 5 – 3 × 4k + 2] + 12 × 4k + 2 + 5
= 90C – 50 – 30 × 4k + 2 + 12 × 4k + 2 + 5
= 90C – 45 – 18 × 4k + 2
= 9[10C – 5 – 2 × 4k + 2] which is ÷ by 9
So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.
Question 15.
Prove that using the Mathematical induction
Solution:
### Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Additional Questions
Question 1.
Prove by induction the inequality (1 + x)n ≥ 1 + nx, whenever x is positive and n is a positive integer.
Solution:
P(n) : (1 +x)n ≥ 1 +nx
P(1): (1 + x)1 ≥ 1 + x
⇒ 1 + x ≥ 1 + x, which is true.
Hence, P(1) is true.
Let P(k) be true
(i.e.) (1 + x)k ≥ 1 + kx
We have to prove that P(k + 1) is true.
(i.e.) (1 + x)k + 1 ≥ 1 + (k + 1)x
Now, (1 + x)k + 1 ≥ 1 + kx [∵ p(k) is true]
Multiplying both sides by (1 + x), we get
(1 + x)k(1 + x) ≥ (1 + kx)(1 + x)
⇒ (1 + x)k + 1 ≥ 1 + kx + x + kx2
⇒ (1 + x)k + 1 ≥ 1 + (k + 1)x + kx2 ….. (1)
Now, 1 + (k + 1) x + kx2 ≥ 1 + (k + 1)x …… (2)
[∵ kx2 > 0]
From (1) and (2), we get
(1 + x)k + 1 ≥ 1 + (k + 1)x
∴ P(k + 1) is true if P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values, of n.
Question 2.
32n – 1 is divisible by 8.
Solution:
P(n) = 32n – 1 is divisible by 8
For n = 1, we get
P(1) = 32.1 – 1 = 9 – 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k) = 32k – 1 is divisible by 8 ….. (1)
Now, P(k + 1) = 3(2k + 2) – 1 = 32k.32 – 1
= 32(32k – 1) + 8
Now, 32k – 1 is divisible by 9. [Using (1)]
∴ 32 (32k – 1) + 8 is also divisible by 8.
Hence, 32n – 1 is divisible by 8 ∀ n E N
Question 3.
Prove by the principle of mathematical induction if x and y are any two distinct integers, then xn – yn is divisible by x – y. [OR]
xn – yn is divisible by x – y, where x – y ≠ 0.
Solution:
Let the given statement be P(n).
(i.e.) P(n): xn – yn = M(x – y), x – y ≠ 0
Step I.
When n = 1,
xn – yn = x – y = M(x – y) ….(1)
⇒ P(1) is true.
Step II.
Assume that P(k) is true.
(i.e.) xk – yk = M(x – y), x – y ≠ 0
We shall now show that P(k + 1) is true
Now, xk + 1 – yk + 1 = xk + 1 – xky + xk + 1y – yk + 1
= xk(x – y) + y(xk – yk)
= xk(x – y) + yM(x – y) [Usng ….. (1)]
= (x – y)(xk – yM)
∴ By the principle of mathematical induction, P(n) is true for all n ∈ N
Question 4.
Prove by the principle of mathematical induction that for every natural number n, 32n + 2 – 8n – 9 is divisible by 8.
Solution:
Let P(n): 32n + 2 – 8n – 9 is divisible by 8.
Then, P(1): 32.1 + 2 – 8.1 – 9 is divisible by 8.
(i.e.) 34 – 8 – 9 is divisible by 8 or 81 – 8 – 9 is divisible by 8
(or) 64 is divisible by 8, which is true.
Suppose P(k) is true, then
P(k) : 32k + 2 – 8k – 9 is divisible by 8
(i.e.) 32k + 2 – 8k – 9 = 8m, where m ∈ N (or)
32k + 2 = 8m + 8k + 9
P(k + 1) is the statement given by, …(1)
P(k + 1) : 32(k + 1) + 2 – 8(k + 1) – 9
∴ P(k + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N
Question 5.
Use the principle of mathematical induction to prove that for every natural number n.
Solution:
Let P(n) be the given statement, i.e.
⇒ P(1) is true.
We note than P(n) is true for n = 1.
Assume that P(k) is true
Now, we shall prove that P(k + 1) is true whenever P(k) is true. We have,
∴ P(k + 1) is also true whenever P(k) is true
Hence, by the principle of mathematical induction, P(n) is also true for all n ∈ N.
Question 6.
n3 – n is divisible by 6, for each natural number n ≥ 2.
Solution:
Let P(n) : n3 – n
Step 1 :
P(2): 23 – 2 = 6 which is divisible by 6. So it is true for P(2).
Step 2 :
P(A): k3 – k = 6λ. Let it is be true for k ≥ 2
⇒ k3 = 6λ + k …(i)
Step 3 :
P(k + 1) = (k + 1)3 – (k + 1)
= k3 + 1 + 3k2 + 3k – k – 1 = k3 – k + 3(k2 + k)
= k3 – k + 3(k2 + k) = 6λ + k – k + 3(k2 + k)
= 6λ + 3(k2 + k) [from (i)]
We know that 3(k2 + k) is divisible by 6 for every value of k ∈ N.
Hence P(k + 1) is true whenever P(k) is true.
Question 7.
For any natural number n, 7n – 2n is divisible by 5.
Solution:
Let P(n) : 7n – 2n
Step 1 :
P(1) : 71 – 21 = 5λ which is divisible by 5. So it is true for P(1).
Step 2 :
P(k): 7k – 2k = 5λ. Let it be true for P(k)
Step 3 :
P(k + 1) = 7k + 1 – 2k + 1
So, it is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.
Question 8.
n2 < 2n, for all natural numbers n ≥ 5.
Solution:
Let P(n) : n2 < 2n for all natural numbers, n ≥ 5
Step 1 :
P(5) : 15 < 25 ⇒ 1 < 32 which is true for P(5)
Step 2 :
P(k): k2 < 2k. Let it be true for k ∈ N
Step 3 :
P(k + 1): (k + 1)2 < 2k + 1
From Step 2, we get k2 < 2k
⇒ k2 < 2k + 1 < 2k + 2k + 1
From eqn. (i) and (ii), we get (k + 1)2 < 2k + 1
Hence, P(k + 1) is true whenever P(k) is true for k ∈ N, n ≥ 5.
Question 9.
In 2n < (n + 2)! for all natural number n.
Solution:
Let P(n) : 2n < (n + 2)! for all k ∈ N.
Hence, P(k + 1) is true whenever P(k) is true.
Question 10.
1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N.
Solution:
Let P(n) : 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N
Step 1:
P(1) : 1 = 1(2.1 – 1) = 1 which is true for P(1)
Step 2:
P(k) : 1 + 5 + 9 + … + (4k – 3) = k(2k – 1). Let it be true.
Step 3:
P(k + 1) : 1 + 5 + 9 + … + (4k – 3) = k(4k + 1)
= k(2k – 1) + (4k + 1) = 2k2 – k + 4k + 1
= 2k2 + 3k + 1 = 2k2 + 2k + k + 1
= 2k(k + 1) + 1 (k + 1) = (2k + 1)(k + 1)
= (k+ 1) (2k + 2 – 1) = (k + 1) [2(k + 1) – 1]
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true. |
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