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# Jiro drives 10km then increases his speed by 10kph and drives another 25km. What is his original speed if the whole ride took 45 minutes (or 3/4 hour)?
Sep 9, 2016
The original speed was $40$ km per hour.
#### Explanation:
With a distance-speed-time problem, remember the relationship:
$s = \frac{d}{t} \text{ }$ Let the original speed be $x$ kph.
We can then write the speeds and times in terms of $x$
$\text{Original speed" = x color(white)(xxxxxxxxxx)"Faster speed} = x + 10$
$\text{distance = "10kmcolor(white)(xxxxxxxxxx)" distance =} 25 k m$
$\rightarrow t i m {e}_{1} = \frac{10}{x} \text{hours} \textcolor{w h i t e}{\times \times \times \times} \rightarrow t i m {e}_{2} = \frac{25}{x + 10}$
The total time for the ride was $\frac{3}{4}$ hour " "(time_1 + time_2)
$\frac{10}{x} + \frac{25}{x + 10} = \frac{3}{4} \text{ } \leftarrow$ now solve the equation
Multiply through by the LCD which is $\textcolor{b l u e}{4 x \left(x + 10\right)}$
$\frac{\textcolor{b l u e}{4 \cancel{x} \left(x + 10\right)} \times 10}{\cancel{x}} + \frac{\textcolor{b l u e}{4 x \cancel{x + 10}} \times 25}{\cancel{x + 10}} = \frac{3 \times \textcolor{b l u e}{\cancel{4} x \left(x + 10\right)}}{\cancel{4}}$
=$40 \left(x + 10\right) + 100 x = 3 x \left(x + 10\right)$
$40 x + 400 + 100 x = 3 {x}^{2} + 30 x \text{ } \leftarrow$ make = 0
$0 = 3 {x}^{2} - 110 x - 400 \text{ } \leftarrow$ find factors
$\left(3 x + 10\right) \left(x - 40\right) = 0$
If $3 x + 10 = 0 \text{ } \rightarrow x = - \frac{10}{3}$ reject negative speed
if$x - 40 = 0 \text{ } \rightarrow x = 40$
The original speed was $40$ km per hour
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Eureka Math Precalculus Module 4 Lesson 10 Answer Key
Engage NY Eureka Math Precalculus Module 4 Lesson 10 Answer Key
Eureka Math Precalculus Module 4 Lesson 10 Example Answer Key
Example: Revisiting Vectors and Resultant Forces
The goalie on the soccer team kicks a ball with an initial force of 135 Newtons at a 40° angle with the ground. The mass of a soccer ball is 0.45 kg. Assume the acceleration due to gravity is 9.8$$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ .
a. Draw a picture representing the force vectors acting on the ball and the resultant force vector.
In the diagram below, Fb represents the force of the kick and Fg represents the force due to gravity.
b. What is the magnitude of the resultant force vector?
The force due to gravity is the product of the mass of the ball and the acceleration due to gravity.
F_g = 0.45∙9.8 = 4.41
Translating the gravitational force vector to the terminal point of the ball’s force vector and using the law of cosines gives the magnitude of the resultant force. The angle between the two vectors is 40° + 90° = 130°. This would make the angle in the triangle that we are using for law of cosines 50°. Let b represent the magnitude of the resultant force vector.
b2 = 1352 + 4.412 – 2(135)(4.41) cos(50°)
b ≈ 132.21
The magnitude of the resultant force is approximately 132 newtons.
c. What are the horizontal and vertical components of this vector?
The components of the initial force on the ball are 〈135 cos(40°),135 sin(40°)〉, and the components of the gravitational force vector are 〈0, – 4.41〉. Adding the vector components gives the resultant force in component form.
〈135 cos(40°) + 0,135 sin(40°) – 4.41〉 = 〈103.416,82.366〉
d. What is the angle of elevation of the resulting vector?
Using right triangle trigonometry ratios, we can compute the angle of elevation.
tan(θ) = $$\frac{82.366}{103.416}$$
θ = arctan($$\frac{82.366}{103.416}$$)
θ ≈ 38.5°
Eureka Math Precalculus Module 4 Lesson 10 Exercise Answer Key
Opening Exercise
a. For each triangle shown below, decide whether you should use the law of sines, the law of cosines, or neither to begin finding the missing measurements. Explain how you know.
Triangle A is solved using the law of cosines because we are given two sides and the included angle.
Triangle B is solved using the law of sines because we are given two angles and one side opposite one of the angles.
Triangle C is solved using the law of sines because we are given two sides and one angle with the angle being opposite one side.
Triangle D is solved using the law of cosines because three sides are given.
Triangle E is solved using the law of sines because two angles are given. We can easily find the third using the triangle sum theorem, and then we will have an angle and opposite side pairing.
Triangle F does not require the law of sines or the law of cosines because it is a right triangle. We can find the missing sides or angles using the Pythagorean theorem and right triangle trigonometry functions.
b. What types of given information will help you to decide which formula to use to determine missing measurements? Summarize your ideas in the table shown below:
Determining Missing Measurements
Exercises 1–7
Exercise 1.
A landscape architect is given a survey of a parcel of land that is shaped like a parallelogram. On the scale drawing the sides of the parcel of land are 8 in. and 10 in., and the angle between these sides measures 75°. The architect is planning to build a fence along the longest diagonal. If the scale on the survey is 1 in. = 120 ft., how long will the fence be?
Let d be the measure of the longest diagonal.
d2 = 102 + 82 – 2(10)(8) cos(105°)
d ≈ 14.33
On the survey, this diagonal is approximately 14.33 in. The actual length of the fence will be 1,719.6 ft.
Exercise 2.
A regular pentagon is inscribed in a circle with a radius of 5 cm. What is the perimeter of the pentagon?
Let s be the measure of one side of the regular pentagon.
s2 = 52 + 52 – 2(5)(5)cos(72°)
s ≈ 5.88
Since there are five sides: 5∙5.88 = 29.4
Thus, the perimeter is 29.4 cm.
Exercise 3.
At the base of a pyramid, a surveyor determines that the angle of elevation to the top is 53°. At a point 75 meters from the base, the angle of elevation to the top is 35°. What is the distance from the base of the pyramid up the slanted face to the top?
Let d be the distance from the base to the top of the pyramid.
$$\frac{\sin \left(18^{\circ}\right)}{75}$$ = $$\frac{\sin \left(35^{\circ}\right)}{d}$$
d ≈ 139.21
The distance is approximately 139 meters.
Exercise 4.
A surveyor needs to determine the distance across a lake between an existing ferry dock at point A and a second dock across the lake at point B. He locates a point C along the shore from the dock at point A that is 750 meters away. He measures the angle at A between the sight lines to points B and C to be 65° and the angle at C between the sight lines to points A and B to be 82°. How far is it from the dock at A and the dock at B?
To find m∠B: 180° – (65° + 82°) = 33°.
Let d be the distance from the dock at A and the dock at B.
$$\frac{\sin \left(33^{\circ}\right)}{750}$$ = $$\frac{\sin \left(82^{\circ}\right)}{d}$$
d ≈ 1363.7
The distance between the two docks across the lake is approximately 1,363.7 meters.
Exercise 5.
Two people located 500 yards apart have spotted a hot air balloon. The angle of elevation from one person to the balloon is 67°. From the second person to the balloon the angle of elevation is 46°. How high is the balloon when it is spotted?
Let d be the distance between the first person and the balloon. Let h be the height of the balloon in the air.
By the law of sines,
$$\frac{\sin \left(46^{\circ}\right)}{d}$$ = $$\frac{\sin \left(67^{\circ}\right)}{500}$$
d ≈ 390.73
Then,
sin(67°) = $$\frac{h}{390.73}$$
h ≈ 359.67
The balloon is approximately 360 yards in the air when it is spotted.
When applying mathematics to navigation, direction is often given as a bearing. The bearing of an object is the degrees rotated clockwise from north that indicates the direction of travel or motion. The next exercises apply the law of cosines and the law of sines to navigation problems.
Exercise 6.
Two fishing boats start from a port. One travels 15 nautical miles per hour on a bearing of 25° and the other travels 18 nautical miles per hour on a bearing of 100°. Assuming each maintains its course and speed, how far apart will the fishing boats be after two hours?
Let point A be the starting location of the two fishing boats at the port. After two hours one ship will have traveled 30 nautical miles from A to B. The other ship will have traveled 36 nautical miles from A to C. The law of cosines can be used to find the distance between the ships, a.
a2 = 302 + 362 – 2(30)(36) cos(75°)
a ≈ 40.46
The ships will be approximately 40.5 nautical miles apart after two hours.
Exercise 7.
An airplane travels on a bearing of 200° for 1500 miles and then changes to a bearing of 250° and travels an additional 500 miles. How far is the airplane from its starting point?
The measure of ∠ABC is 110° + 20°, or 130°. It is the sum of a 20° angle that is congruent to the angle formed by side c and the south – facing direction line from point A, and the difference between 250° and a full rotation of 360° about point B.
By the law of cosines, side b, which represents the distance from the starting point at A and the final point at C is given by
b2 = 15002 + 5002 – 2(1500)(500)cos(130°)
b ≈ 1861.23
Thus, the airplane is approximately 1,861 miles from where it started. Note: This solution takes neither the elevation of the airplane nor the curvature of the Earth into account.
Exercises 8–10
Exercise 8.
Suppose a soccer player runs up to a moving soccer ball located at A and kicks the ball into the air. The diagram below shows the initial velocity of the ball along the ground and the initial velocity and direction of the kick. What is the resultant velocity and angle of elevation of the soccer ball immediately after it is kicked?
If we translate the vector with magnitude 15 m/s to point B, then the angle at B will be 130°. Then the sum of the two vectors is the vector with tail at the origin at C. Let b be the magnitude of this vector. By the law of cosines,
b = 82 + 152 – 2(8)(15) cos(130°)
b ≈ 21.05
The direction can be found using the law of sines. Let θ be the angle between the 8 m/s vector and b.
$$\frac{\sin (\theta)}{15}$$ = $$\frac{\sin \left(130^{\circ}\right)}{21.05}$$
θ = arcsin($$\frac{15 \sin \left(130^{\circ}\right)}{21.05}$$)
θ ≈ 33.08°
Thus, the direction of the ball would be 50° – 33.08° = 16.92°.
Exercise 9.
A 13 lb. force and a 20 lb. force are applied to an object located at A as shown in the diagram below. What is the resulting force and direction being applied to the object at A?
The resulting force is the sum of the two forces, which can be represented as vectors. The parallelogram rule gives us the resulting force vector. Using the law of cosines, we can determine the magnitude, and using the law of sines, we can determine the direction.
Let c be the distance between point A and B.
c2 = 132 + 202 – 2(13)(20) cos(80°)
c ≈ 21.88
The measure of ∠CAB can be found using the law of sines. Let θ be the measure of ∠CAB.
$$\frac{\sin \left(80^{\circ}\right)}{21.88}$$ = $$\frac{\sin (\theta)}{20}$$
θ = arcsin($$\frac{20 \sin \left(80^{\circ}\right)}{21.88}$$)
θ ≈ 64.19°
The resulting force of 21.88 lb. would be in a direction of 24.19° clockwise from the horizontal axis.
Exercise 10.
A motorboat travels across a lake at a speed of 10 mph at a bearing of 25°. The current of the lake due to the wind is a steady 2 mph at a bearing of 340°.
a. Draw a diagram that shows the two velocities that are affecting the boat’s motion across the lake.
b. What is the resulting speed and direction of the boat?
The resulting speed and direction is the sum of these two velocity vectors. Translating the current vector to the tip of the boat’s speed vector allows us to quickly draw the resulting vector. Its magnitude and direction can be determined using the law of cosines and the law of sines.
Let a be the distance between points C and B. By the law of cosines,
a2 = 102 + 22 – 2(10)(2) cos(135°)
a ≈ 11.5
Let θ be the measure of ∠BCA. Then, using the law of sines,
$$\frac{\sin \left(135^{\circ}\right)}{11.5}$$ = $$\frac{\sin (\theta)}{2}$$
θ ≈ 7.06°
Then, bearing is 25° – 7.06° = 17.94°, and the speed is 11.5 mph.
Eureka Math Precalculus Module 4 Lesson 10 Problem Set Answer Key
Question 1.
For each of the situations below, determine whether to use the Pythagorean theorem, right triangle trigonometry, law of sines, law of cosines, or some other method.
a.
Pythagorean theorem
b. Know one side and an angle of a right triangle and want to find any other side.
Right triangle trigonometry
c.
Law of sines
d. Know two angles of a triangle and want to find the third.
Find the sum of the measures of the two known angles, and subtract the result from 180°.
e.
Law of cosines
f. Know three sides of a triangle and want to find an angle.
Law of cosines
g.
Either law of cosines twice or a combination of law of cosines and law of sines
h. Know a side and two angles and want to find the third angle.
Find the sum of the measures of the two known angles, and subtract the result from 180°.
i.
Law of sines
Question 2.
Mrs. Lane’s trigonometry class has been asked to judge the annual unmanned hot – air balloon contest, which has a prize for highest flying balloon.
a. Sarah thinks that the class needs to set up two stations to sight each balloon as it passes between them. Construct a formula that Mrs. Lane’s class can use to find the height of the balloon by plugging the two angles of elevation so that they can program their calculators to automatically output the height of the balloon. Use 500 ft. for the distance between the stations and α and β for the angles of elevation.
b = 500 $$\frac{\sin (\beta)}{\sin \left(180^{\circ} – \alpha – \beta\right)}$$
h = b sin(α)
h = $$\frac{500 \sin (\alpha) \sin (\beta)}{\sin \left(180^{\circ} – \alpha – \beta\right)}$$
b. The students expect the balloons to travel no higher than 500 ft. What distance between the stations would you recommend? Explain.
Answers may vary. Depending on how close the balloons pass to the stations, students may be sighting the balloon at near vertical angles. More accurate measurements can probably be obtained the closer the balloons are to 45° from the stations, so a distance of greater than 500 ft. is probably better suited.
c.
Find the heights of balloons sighted with the following angles of elevation to the nearest ten feet. Assume a distance of 500 ft. between stations.
i. 5°, 15°
32.977 ft. ≈ 30 ft.
ii. 38°, 72°
311.553 ft. ≈ 310 ft.
iii. 45°, 45°
250 ft.
iv. 45°, 59°
312.33 ft. ≈ 310 ft.
v. 28°, 44°
171.45 ft. ≈ 170 ft.
vi. 50°, 66°
389 ft. ≈ 390 ft.
vii. 17°, 40°
112 ft. ≈ 110 ft.
d. Based on your results in part (c), which balloon won the contest?
The student with the balloon that went 390 ft.
e. The balloons were released several hundred feet away but directly in the middle of the two stations. If the first angle represents the West station and the second angle represents the East station, what can you say about the weather conditions during the contest?
It appears as though a wind was blowing the balloons to the east.
f. Are there any improvements to Mrs. Lane’s class’s methods that you would suggest? Explain.
Answers may vary. Students could suggest higher degrees of accuracy by adjusting the distance between the students. Multiple angles of elevation could be taken from different spots, or additional students could help measure to minimize human error.
Question 3.
Bearings on ships are often given as a clockwise angle from the direction the ship is heading (0° represents something in the path of the boat, and 180° represents something behind the boat). Two ships leave port at the same time. The first ship travels at a constant speed of 30 kn. After 2 $$\frac{1}{2}$$ hours, the ship sights the second at a bearing of 110° and 58 nautical miles away.
a. How far is the second ship from the port where it started?
109 nautical miles from port
b. How fast is the second ship traveling on average?
$$\frac{109}{2.5}$$ = 43.6; The second ship is traveling 43.6 kn.
Question 4.
A paintball is fired from a gun with a force of 59 N at an angle of elevation of 1°. If the force due to gravity on the paintball is 0.0294 N, then answer the following:
a. Is this angle of elevation enough to overcome the initial force due to gravity and still have an angle of elevation greater than 0.5°?
Yes. The force due to gravity is so small that there is effectively no difference initially. The third side has a magnitude of 58.999. The angle of elevation is reduced by less than 0.029°.
b. What is the resultant magnitude of the vector in the direction of the paintball?
58.999 N
Question 5.
Valerie lives 2 miles west of her school and her friend Yuri lives 3 miles directly northeast of her.
a. Draw a diagram representing this situation.
b. How far does Yuri live from school?
c = $$\sqrt{2^{2} + 3^{2} – 2 \cdot 2 \cdot 3 \cdot \cos (45)}$$
= $$\sqrt{13 – 6 \sqrt{2}}$$
≈ 2.125
Yuri lives approximately 2.125 mi. from school.
c. What is the bearing of the school to Yuri’s house?
arccos($$\frac{9 – 2.125^{2} – 4}{ – 8.499}$$) ≈ 93.27
Yuri’s house is 93.27° N of W from the school, so the school is on a bearing of – 176.73 from N. This can be worded different ways, for instance, 86.73° S of W, 93.27° S of E.
Question 6.
A 2.1 kg rocket is launched at an angle of 33° with an initial force of 50 N. Assume the acceleration due to gravity is 9.81 m/s2 .
a. Draw a picture representing the force vectors and their resultant vector.
b. What is the magnitude of the resultant vector?
The force due to gravity: F = 2.1⋅9.81 = 20.601
c = $$\sqrt{50^{2} + 20.601^{2} – 2 \cdot 50 \cdot 20.601 \cdot \cos \left(57^{\circ}\right)}$$
≈ 42.455
The resultant vector is about 42.455 N.
c. What are the horizontal and vertical components of the resultant vector?
The initial force of the rocket can be expressed by the vector 〈50 cos(33°),50 sin(33°)〉. The force due to gravity can be expressed by 〈0,20.601〉. The resultant vector is 〈41.934,6.631〉.
d. What is the angle of elevation of the resultant vector?
arctan($$\frac{6.631}{41.934}$$) ≈ 8.986
The angle of elevation is about 8.986°.
Question 7.
Use the distance formula to find c, the distance between A and B for △ABC, with A = (b cos(γ),b sin(γ) ), B = (a,0), and C = (0,0). After simplifying, what formula have you proven?
AB = $$\sqrt{(b \cos (\gamma) – a)^{2} + (b \sin (\gamma))^{2}}$$
Multiplying out on the inside of the square root, we get
b2 (cos(γ))2 – 2ab cos(γ) + a2 + b2 (sin(γ))2.
Factoring out b2, we get
b2 ((cos(γ) )2 + (sin(γ) )2 ) – 2ab cos(γ) + a2 = b2 + a2 – 2abcos(γ).
So we have
AB = $$\sqrt{a^{2} + b^{2} – 2 a b \cos (\gamma)}$$.
Since △ABC could be any triangle translated to the origin and rotated so that one side lays on the x – axis, we have proven the law of cosines.
Question 8.
For isosceles triangles with a = b, show the law of cosines can be written as cos(γ) = 1 – $$\frac{c^{2}}{2 a^{2}}$$.
c2 = a2 + b2 – 2ab cos(γ)
c2 = 2a2 – 2a2 cos(γ)
c2 = 2a2 (1 – cos(γ) )
$$\frac{c^{2}}{2 a^{2}}$$ = 1 – cos(γ)
cos(γ) = 1 – $$\frac{c^{2}}{2 a^{2}}$$
Eureka Math Precalculus Module 4 Lesson 10 Exit Ticket Answer Key
Question 1.
A triangular pasture is enclosed by fencing measuring 25, 35, and 45 yards at the corner of a farmer’s property.
a. According to the fencing specifications, what is the measure of ∠ABC?
452 = 252 + 352 – 2⋅25⋅35⋅cos(B)
B = arccos($$\frac{45^{2} – 25^{2} – 35^{2}}{ – 2 \cdot 25 \cdot 35}$$)
= arccos( – 0.1)
≈ 95.739
The measure of ∠ABC is 95.739°.
b. A survey of the land indicates that the property lines form a right angle at B. Explain why a portion of the pasture is actually on the neighboring property.
C’ = arcsin(25⋅$$\frac{\sin \left(95.74^{\circ}\right)}{45}$$) ≈ 33.56°
$$\frac{\sin \left(33.56^{\circ}\right)}{c^{\prime}}$$ = $$\frac{\sin \left(140.7^{\circ}\right)}{35}$$
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# Section7.2Qualitative behavior of solutions to DEs¶ permalink
##### Motivating Questions
• What is a slope field?
• How can we use a slope field to obtain qualitative information about the solutions of a differential equation?
• What are stable and unstable equilibrium solutions of an autonomous differential equation?
In earlier work, we have used the tangent line to the graph of a function $f$ at a point $a$ to approximate the values of $f$ near $a\text{.}$ The usefulness of this approximation is that we need to know very little about the function; armed with only the value $f(a)$ and the derivative $f'(a)\text{,}$ we may find the equation of the tangent line and the approximation
\begin{equation*} f(x) \approx f(a) + f'(a)(x-a). \end{equation*}
Remember that a first-order differential equation gives us information about the derivative of an unknown function. Since the derivative at a point tells us the slope of the tangent line at this point, a differential equation gives us crucial information about the tangent lines to the graph of a solution. We will use this information about the tangent lines to create a slope field for the differential equation, which enables us to sketch solutions to initial value problems. Our aim will be to understand the solutions qualitatively. That is, we would like to understand the basic nature of solutions, such as their long-range behavior, without precisely determining the value of a solution at a particular point.
##### Preview Activity7.2.1
Let's consider the initial value problem
\begin{equation*} \frac{dy}{dt} = t - 2, \ \ y(0) = 1. \end{equation*}
1. Use the differential equation to find the slope of the tangent line to the solution $y(t)$ at $t=0\text{.}$ Then use the initial value to find the equation of the tangent line at $t=0\text{.}$ Sketch this tangent line over the interval $-0.25 \leq t \leq 0.25$ on the axes provided in Figure 7.2.1.
2. Also shown in Figure 7.2.1 are the tangent lines to the solution $y(t)$ at the points $t=1, 2,$ and $3$ (we will see how to find these later). Use the graph to measure the slope of each tangent line and verify that each agrees with the value specified by the differential equation.
3. Using these tangent lines as a guide, sketch a graph of the solution $y(t)$ over the interval $0\leq t\leq 3$ so that the lines are tangent to the graph of $y(t)\text{.}$
4. Use the Fundamental Theorem of Calculus to find $y(t)\text{,}$ the solution to this initial value problem.
5. Graph the solution you found in (d) on the axes provided, and compare it to the sketch you made using the tangent lines.
# Subsection7.2.1Slope fields
Preview Activity 7.2.1 shows that we may sketch the solution to an initial value problem if we know an appropriate collection of tangent lines. Because we may use a given differential equation to determine the slope of the tangent line at any point of interest, by plotting a useful collection of these, we can get an accurate sense of how certain solution curves must behave.
Let's continue looking at the differential equation $\frac{dy}{dt} = t-2.$ If $t=0\text{,}$ this equation says that $dy/dt = 0-2=-2\text{.}$ Note that this value holds regardless of the value of $y\text{.}$ We will therefore sketch tangent lines for several values of $y$ and $t=0$ with a slope of $-2\text{.}$
Let's continue in the same way: if $t=1\text{,}$ the differential equation tells us that $dy/dt = 1-2=-1\text{,}$ and this holds regardless of the value of $y\text{.}$ We now sketch tangent lines for several values of $y$ and $t=1$ with a slope of $-1\text{.}$
Similarly, we see that when $t=2\text{,}$ $dy/dt = 0$ and when $t=3\text{,}$ $dy/dt=1\text{.}$ We may therefore add to our growing collection of tangent line plots to achieve the next figure.
In this figure, we begin to see the solutions to the differential equation emerge. However, for the sake of clarity, we add more tangent lines to provide the even more complete picture shown below.
This most recent figure, which is called a slope field for the differential equation, allows us to sketch solutions just as we did in the preview activity. Here, we will begin with the initial value $y(0) = 1$ and start sketching the solution by following the tangent line, as shown in the Figure 7.2.6.
We then continue using this principle: whenever the solution passes through a point at which a tangent line is drawn, that line is tangent to the solution. Doing so leads us to the following sequence of images.
In fact, we may draw solutions for any possible initial value, and doing this for several different initial values for $y(0)$ results in the graphs shown at bottom right in Figure 7.2.8.
Just as we have done for the most recent example with $\frac{dy}{dt} = t-2\text{,}$ we can construct a slope field for any differential equation of interest. The slope field provides us with visual information about how we expect solutions to the differential equation to behave.
##### Activity7.2.2
Consider the autonomous differential equation
\begin{equation*} \frac{dy}{dt} = -\frac 12( y - 4). \end{equation*}
1. Make a plot of $\frac{dy}{dt}$ versus $y$ on the axes provided. Looking at the graph, for what values of $y$ does $y$ increase and for what values of $y$ does $y$ decrease?
2. Next, sketch the slope field for this differential equation on the axes provided.
3. Use your work in (b) to sketch (on the axes in Figure 7.2.9.) solutions that satisfy $y(0) = 0\text{,}$ $y(0) = 2\text{,}$ $y(0) = 4$ and $y(0) = 6.$
4. Verify that $y(t) = 4 + 2e^{-t/2}$ is a solution to the given differential equation with the initial value $y(0) = 6.$ Compare its graph to the one you sketched in (c).
5. What is special about the solution where $y(0) = 4\text{?}$
# Subsection7.2.2Equilibrium solutions and stability
As our work in Activity 7.2.2 demonstrates, first-order autonomous equations may have solutions that are constant. These are quite easy to detect by inspecting the differential equation $dy/dt = f(y)\text{:}$ constant solutions necessarily have a zero derivative so $dy/dt = 0 = f(y)\text{.}$
For example, in Activity 7.2.2, we considered the equation $\frac{dy}{dt} = f(y)=-\frac12(y-4)\text{.}$ Constant solutions are found by setting $f(y) = -\frac12(y-4) = 0\text{,}$ which we immediately see implies that $y = 4\text{.}$
Values of $y$ for which $f(y) = 0$ in an autonomous differential equation $\frac{dy}{dt} = f(y)$ are usually called or equilibrium solutions of the differential equation.
##### Activity7.2.3
Consider the autonomous differential equation
\begin{equation*} \frac{dy}{dt} = -\frac 12 y(y-4). \end{equation*}
1. Make a plot of $\frac{dy}{dt}$ versus $y$ on the axes provided. Looking at the graph, for what values of $y$ does $y$ increase and for what values of $y$ does $y$ decrease?
2. Identify any equilibrium solutions of the given differential equation.
3. Now sketch the slope field for the given differential equation on the axes provided in Figure 7.2.10.
4. Sketch the solutions to the given differential equation that correspond to initial values $y(0)=-1, 0, 1, \ldots, 5\text{.}$
5. An equilibrium solution $\overline{y}$ is called stable if nearby solutions converge to $\overline{y}\text{.}$ This means that if the initial condition varies slightly from $\overline{y}\text{,}$ then $\lim_{t\to\infty}y(t) = \overline{y}\text{.}$ Conversely, an equilibrium solution $\overline{y}$ is called unstable if nearby solutions are pushed away from $\overline{y}\text{.}$ Using your work above, classify the equilibrium solutions you found in (b) as either stable or unstable.
6. Suppose that $y(t)$ describes the population of a species of living organisms and that the initial value $y(0)$ is positive. What can you say about the eventual fate of this population?
7. Now consider a general autonomous differential equation of the form $dy/dt = f(y)\text{.}$ Remember that an equilibrium solution $\overline{y}$ satisfies $f(\overline{y}) = 0\text{.}$ If we graph $dy/dt = f(y)$ as a function of $y\text{,}$ for which of the differential equations represented in Figure 7.2.11 and Figure 7.2.12 is $\overline{y}$ a stable equilibrium and for which is $\overline{y}$ unstable? Why?
# Subsection7.2.3Summary
• A slope field is a plot created by graphing the tangent lines of many different solutions to a differential equation.
• Once we have a slope field, we may sketch the graph of solutions by drawing a curve that is always tangent to the lines in the slope field.
• Autonomous differential equations sometimes have constant solutions that we call equilibrium solutions. These may be classified as stable or unstable, depending on the behavior of nearby solutions.
##### 5
Consider the differential equation
\begin{equation*} \frac{dy}{dt} = t-y. \end{equation*}
1. Sketch a slope field on the plot below:
2. Sketch the solutions whose initial values are $y(0)= -4, -3, \ldots, 4\text{.}$
3. What do your sketches suggest is the solution whose initial value is $y(0) = -1\text{?}$ Verify that this is indeed the solution to this initial value problem.
4. By considering the differential equation and the graphs you have sketched, what is the relationship between $t$ and $y$ at a point where a solution has a local minimum?
##### 6
Consider the situation from problem 2 of Section 7.1: Suppose that the population of a particular species is described by the function $P(t)\text{,}$ where $P$ is expressed in millions. Suppose further that the population's rate of change is governed by the differential equation
\begin{equation*} \frac{dP}{dt} = f(P) \end{equation*}
where $f(P)$ is the function graphed below.
1. Sketch a slope field for this differential equation. You do not have enough information to determine the actual slopes, but you should have enough information to determine where slopes are positive, negative, zero, large, or small, and hence determine the qualitative behavior of solutions.
2. Sketch some solutions to this differential equation when the initial population $P(0) \gt 0\text{.}$
3. Identify any equilibrium solutions to the differential equation and classify them as stable or unstable.
4. If $P(0) \gt 1\text{,}$ what is the eventual fate of the species?
5. if $P(0) \lt 1\text{,}$ what is the eventual fate of the species?
6. Remember that we referred to this model for population growth as “growth with a threshold.” Explain why this characterization makes sense by considering solutions whose inital value is close to 1.
##### 7
The population of a species of fish in a lake is $P(t)$ where $P$ is measured in thousands of fish and $t$ is measured in months. The growth of the population is described by the differential equation
\begin{equation*} \frac{dP}{dt} = f(P) = P(6-P). \end{equation*}
1. Sketch a graph of $f(P) = P(6-P)$ and use it to determine the equilibrium solutions and whether they are stable or unstable. Write a complete sentence that describes the long-term behavior of the fish population.
2. Suppose now that the owners of the lake allow fishers to remove 1000 fish from the lake every month (remember that $P(t)$ is measured in thousands of fish). Modify the differential equation to take this into account. Sketch the new graph of $dP/dt$ versus $P\text{.}$ Determine the new equilibrium solutions and decide whether they are stable or unstable.
3. Given the situation in part (b), give a description of the long-term behavior of the fish population.
4. Suppose that fishermen remove $h$ thousand fish per month. How is the differential equation modified?
5. What is the largest number of fish that can be removed per month without eliminating the fish population? If fish are removed at this maximum rate, what is the eventual population of fish?
##### 8
Let $y(t)$ be the number of thousands of mice that live on a farm; assume time $t$ is measured in years. 1
1. The population of the mice grows at a yearly rate that is twenty times the number of mice. Express this as a differential equation.
2. At some point, the farmer brings $C$ cats to the farm. The number of mice that the cats can eat in a year is
\begin{equation*} M(y) = C\frac{y}{2+y} \end{equation*}
thousand mice per year. Explain how this modifies the differential equation that you found in part a).
3. Sketch a graph of the function $M(y)$ for a single cat $C=1$ and explain its features by looking, for instance, at the behavior of $M(y)$ when $y$ is small and when $y$ is large.
4. Suppose that $C=1\text{.}$ Find the equilibrium solutions and determine whether they are stable or unstable. Use this to explain the long-term behavior of the mice population depending on the initial population of the mice.
5. Suppose that $C=60\text{.}$ Find the equilibrium solutions and determine whether they are stable or unstable. Use this to explain the long-term behavior of the mice population depending on the initial population of the mice.
6. What is the smallest number of cats you would need to keep the mice population from growing arbitrarily large?
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## Book: RD Sharma - Mathematics
### Chapter: 15. Linear Inequations
#### Subject: Maths - Class 11th
##### Q. No. 2 of Exercise 15.4
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
2
##### Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.
We need to assume two consecutive odd natural numbers.
So, let the smaller odd natural number be x.
Then, the other odd natural number will be (x + 2).
Given: Both these numbers are larger than 10. …(i)
And their sum is less than 40. …(ii)
So,
From given statement (i),
x > 10 …(iii)
x + 2 > 10
x > 10 – 2
x > 8
Since, the number must be greater than 10, x > 8 can be ignored.
From given statement (ii),
Sum of these two consecutive odd natural numbers < 40
(x) + (x + 2) < 40
x + x + 2 < 40
2x + 2 < 40
2(x + 1) < 40
x + 1 < 20
x < 20 – 1
x < 19 …(iv)
From inequalities (iii) & (iv), we have
x > 10 & x < 19
It can be merged and written as
10 < x < 19
From this inequality, we can say that x lies between 10 and 19.
So, the odd natural numbers lying between 10 and 19 are 11, 13, 15 and 17. (Excluding 19 as x < 19)
Now, let us find pairs of consecutive odd natural numbers.
Let x = 11, then (x + 2) = (11 + 2) = 13
Let x = 13, then (x + 2) = (13 + 2) = 15
Let x = 15, then (x + 2) = (15 + 2) = 17
Let x = 17, then (x + 2) = (17 + 2) = 19.
Hence, all such pairs of consecutive odd natural numbers required are (11, 13), (13, 15), (15, 17) and (17, 19).
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## SYSTEMS OF EQUATIONS in THREE VARIABLES
It is often desirable or even necessary to use more than one variable to model a situation in many fields. When this is the case, we write and solve a system of equations in order to answer questions about the situation.
If a system of linear equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent. If the system has an infinity number of solutions, it is dependent. Otherwise it is independent.
A linear equation in three variables describes a plane and is an equation equivalent to the equation
where A, B, C, and D are real numbers and A, B, C, and D are not all 0.
Problem 3.1a:
A total of $50,000 is invested in three funds paying 6%, 8%, and 10% simple interest. The yearly interest is$3,700. Twice as much money is invested at 6% as invested at 10%. How much was invested in each of the funds.
Answer: $30,000 is invested at 6%,$5,000 is invested at 8%, and \$15,000 is invested at 10%
Solution:
There are three unknowns:
Let's rewrite the paragraph that asks the question we are to answer.
The first sentence can be rewritten as [ The amount of money invested at 6% ] + [ The amount of money invested at 8% ] + [ The amount of money invested at 10% ] The second sentence can be rewritten 0.060 times [ The amount of money invested at 6% ] + 0.08 times [ The amount of money invested at 8% ] + 0.10 times [ The amount of money invested at 10% ] The third sentence can be rewritten [ The amount of money invested at 6% ] = twice [ The amount of money invested at 10% ].
It is going to get boring if we keep repeating the phrases
Let's create a shortcut by letting symbols represent these phrases. Let
in the three sentences, and then rewrite them.
The first sentence can be rewritten as [ The amount of money invested at 6% ] + [ The amount of money invested at 8% ] + [ The amount of money invested at 10% ] can now be written in the algebraic form
The second sentence can be rewritten 0.060 times [ The amount of money invested at 6% ] + 0.08 times [ The amount of money invested at 8% ] + 0.10 times [ The amount of money invested at 10% ] can now be written in the algebraic form
The third sentence can be rewritten [ The amount of money invested at 6% ] = twice [ The amount of money invested at 10% ] can now be written in the algebraic form
We have converted the problem from one described by words to one that is described by three equations.
x+y+z = 50,000 (1) 0.06x+0.08y+0.10z = 3,700 (2) x-2z = 0 (3)
We are going to show you how to solve this system of equations three different ways:
1) Substitution, 2) Elimination 3) Matrices
SUBSTITUTION:
The process of substitution involves several steps:
Step 1: Solve for one of the variables in one of the equations. It makes no difference which equation and which variable you choose. Let's solve for x in equation (3).
Step 2: Substitute this value for x in equations (1) and (2). This will change equations (1) and (2) to equations in the two variables y and z. Call the changed equations (4) and (5), respectively..
x+y+z = 50,000 = 50,000 y+3z = 50,000 (4) 0.06x+0.08y+0.10z = 3,700 = 3,700 0.08y+0.22z = 3,700 8y+22z = 370,000 4y+11z = 185,000 (5)
Step 3: Solve for y in equation (4).
Step 4: Substitute this value of y in equation (5). This will give you an equation in one variable.
Step 5: Substitute this value of z in equation (4) and solve for y.
Step 7: Substitute 5,000 for y and 15,000 for z in equation (1) and solve for x.
The solution is: and z=15,000. In terms of the original problems, was invested at , was invested at , and was invested at . Step 8: Check the solutions:
ELIMINATION:
The process of elimination involves several steps: First you reduce three equations to two equations with two variables, and then to one equation with one variable.
Step 1: Decide which variable you will eliminate. It makes no difference which one you choose. Let us eliminate y first because y is already missing from equation (3).
x+y+z = 50,000 (1) 0.06x+0.08y+0.10z = 3,700 (2) x-2z = 0 (3)
Step 2: Multiply both sides of equation (1) by -0.08 and then add the transformed equation (1) to equation (2) to form equation (4).
Step 3: We now have two equations with two variables.
Step 4: Multiply both sides of equation (4) by 50 and add the transformed equation (4) to equation (3) to create equation (5) with just one variable.
Step 5: Solve for z in equation (5).
Step 6: Substitute 15,000 for z in equation (3) and solve for x.
Step 7: Substitute 30,000 for x and 15,000 for z in equation (1) and solve for y.
The solution is: and z=15,000. In terms of the original problems, was invested at , was invested at , and was invested at .
MATRICES:
The process of using matrices is essentially a shortcut of the process of elimination. Each row of the matrix represents an equation and each column represents coefficients of one of the variables.
Step 1: Create a three-row by four-column matrix using coefficients and the constant of each equation.
The vertical lines in the matrix stands for the equal signs between both sides of each equation. The first column contains the coefficients of x, the second column contains the coefficients of y, the third column contains the coefficients of z, and the last column contains the constants.
We want to convert the original matrix
to the following matrix.
Because then you can read the matrix as x=a, y=b, and z=c..
Step 2: We work with column 1 first. The number 1 is already in cell 11(Row1-Col 1). Add -0.06 times Row 1 to Row 2 to form a new Row 2, and add -1 times Row 1 to Row 3 to form a new Row 3.
Step 3: We will now work with column 1. We want 1 in Cell 22, and we achieve this by multiplying Row 2 by 50 for a new Row 2.
Step 4: Let's now manipulate the matrix so that there are zeros in Cell 12 and Cell 32. We do this by adding -1 times Row 2 to Row 1 to form a new Row 1, and add Row 2 to Row 3 5o form a new Row 3.
Step 5: Let's now manipulate the matrix so that there is a 1 in Cell 33. We do this by multiplying Row 3 by -1.
Step 6: Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 23. We do this by adding Row 3 to Row 1 for a new Row 1 and adding -2 times Row 3 to Row 2 for a new Row 2.
The solution is: and z=15,000. In terms of the original problems, was invested at , was invested at , and was invested at .
If you would like to go back to the problem page, click on Problem.
If you would like to review the solution to the next problem, click on Problem
If you would like to return to the beginning of the three by three system of equations, click on Example.
This site was built to accommodate the needs of students. The topics and problems are what students ask for. We ask students to help in the editing so that future viewers will access a cleaner site. If you feel that some of the material in this section is ambiguous or needs more clarification, or if you find a mistake, please let us know by e-mail at sosmath.com.
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Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
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# How do you simplify (4^-1 * 7) ^ 2?
Jul 6, 2018
$\frac{49}{16}$
#### Explanation:
$\text{using the "color(blue)"law of exponents}$
•color(white)(x)a^-mhArr1/a^m
${\left({4}^{-} 1 \times 7\right)}^{2}$
$= {\left(\frac{1}{4} \times 7\right)}^{2}$
$= {\left(\frac{7}{4}\right)}^{2} = \frac{7}{4} \times \frac{7}{4} = \frac{7 \times 7}{4 \times 4} = \frac{49}{16}$
Jul 6, 2018
$\frac{49}{16}$
#### Explanation:
When we distribute an exponent to another exponent, we multiply the powers. Doing this, we now have
${4}^{- 2} \cdot {7}^{2}$
When we have a negative exponent, we can make it positive by taking it to the denominator. We now have
$\frac{1}{{4}^{2}} \cdot {7}^{2} = \frac{{7}^{2}}{{4}^{2}} = \frac{49}{16}$
Hope this helps!
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# Six congruent rectangular prisms volume can be modeled by V(x) below. V(x)=54x^3+264x^2-738x+420
Question
Solid Geometry
Six congruent rectangular prisms volume can be modeled by V(x) below.
$$\displaystyle{V}{\left({x}\right)}={54}{x}^{{3}}+{264}{x}^{{2}}-{738}{x}+{420}$$
2021-03-08
Since there are 6 rectangular prisms, we can factor out 6 from the polynomial:
$$\displaystyle{6}{\left({9}{x}^{{3}}+{44}{x}^{{2}}-{123}{x}+{70}\right)}$$
So, the volume of one rectangular prism is:
PSK9x^3+44x^2-123x+70SK
### Relevant Questions
The base of a rectangular prism has an area of 19.4 square meters and the prism has a volume of 306.52 cubic meters. Write an equation that can be used to find the height h of the prism. Then find the height of the prism.
A machine part is made from a rectangular prism with a smaller rectangular prism cut out of it, as shown below. What is the volume of the part? Circle the correct answer.
A 20 cubic centimeters
B 140 cubic centimeters
C 160 cubic centimeters
D 180 cubic centimeters
Leo chose D as the correct answer. How did he get that answer?
Let $$u=\begin{bmatrix}2 \\ 5 \\ -1 \end{bmatrix} , v=\begin{bmatrix}4 \\ 1 \\ 3 \end{bmatrix} \text{ and } w=\begin{bmatrix}-4 \\ 17 \\ -13 \end{bmatrix}$$ It can be shown that $$4u-3v-w=0$$. Use this fact (and no row operations) to find a solution to the system $$4u-3v-w=0$$ , where
$$A=\begin{bmatrix}2 & -4 \\5 & 17\\-1&-13 \end{bmatrix} , x=\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} , b=\begin{bmatrix}4 \\ 1 \\ 3 \end{bmatrix}$$
The dominant form of drag experienced by vehicles (bikes, cars,planes, etc.) at operating speeds is called form drag. Itincreases quadratically with velocity (essentially because theamount of air you run into increase with v and so does the amount of force you must exert on each small volume of air). Thus
$$\displaystyle{F}_{{{d}{r}{u}{g}}}={C}_{{d}}{A}{v}^{{2}}$$
where A is the cross-sectional area of the vehicle and $$\displaystyle{C}_{{d}}$$ is called the coefficient of drag.
Part A:
Consider a vehicle moving with constant velocity $$\displaystyle\vec{{{v}}}$$. Find the power dissipated by form drag.
Express your answer in terms of $$\displaystyle{C}_{{d}},{A},$$ and speed v.
Part B:
A certain car has an engine that provides a maximum power $$\displaystyle{P}_{{0}}$$. Suppose that the maximum speed of thee car, $$\displaystyle{v}_{{0}}$$, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power $$\displaystyle{P}_{{1}}$$ is 10 percent greater than the original power ($$\displaystyle{P}_{{1}}={110}\%{P}_{{0}}$$).
Assume the following:
The top speed is limited by air drag.
The magnitude of the force of air drag at these speeds is proportional to the square of the speed.
By what percentage, $$\displaystyle{\frac{{{v}_{{1}}-{v}_{{0}}}}{{{v}_{{0}}}}}$$, is the top speed of the car increased?
Express the percent increase in top speed numerically to two significant figures.
You want to find the values of x so that the volume of the rectangular prism is no more than 36 cubic units.
8x+4<36
$$\displaystyle{2}{x}+{9.5}≤{36}$$
36x+18<36
$$\displaystyle{35}{x}+{18}≤{36}$$
A rectangular barge, 5 m long and 2 m wide, floats in freshwater.
a. Find how much deeper it floatswhen its load is a 400 kg horse.
b. If the barge can only be pushed15 cm deeper into the water before water overflows tosink it, how many 400 kg horses can it carry?
Find the length of a rectangular prism having a volume of 2,830.5 cubic meters, width of 18.5 meters, and height of 9 meters.
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius $$\displaystyle{R}={7.4}\times{10}^{{-{15}}}$$ m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium-236 nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945.
A. Find the radii of the two "daughter" nuclei of charge+46e.
B. In a simple model for the fission process, immediatelyafter the uranium-236 nucleus has undergone fission the "daughter"nuclei are at rest and just touching. Calculate the kineticenergy that each of the "daughter" nuclei will have when they arevery far apart.
C. In this model the sum of the kinetic energies of the two"daughter" nuclei is the energy released by the fission of oneuranium-236 nucleus. Calculate the energy released by thefission of 10.0 kg of uranium-236. The atomic mass ofuranium-236 is 236 u, where 1 u = 1 atomic mass unit $$\displaystyle={1.66}\times{10}^{{-{27}}}$$ kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases 4.18 x 10^12 J when itexplodes).
The postulate or theorem that can be used to prove each pair of triangles congruent.
Given:
The given figure is:
...
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Courses
# NCERT Solutions (Part- 1)- Algebraic Expressions and Identities Class 8 Notes | EduRev
## Mathematics (Maths) Class 8
Created by: Full Circle
## Class 8 : NCERT Solutions (Part- 1)- Algebraic Expressions and Identities Class 8 Notes | EduRev
The document NCERT Solutions (Part- 1)- Algebraic Expressions and Identities Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8
Question 1. Give five examples of expressions containing one variable and five examples of expressions containing two variables.
Solution:
(a) Five examples containing one variable are:
(i) x – 4 (ii) 5 – y (iii) 2x + 5 (iv) 10x + 4 (v) 11 – z
(b) Five examples containing two variables are:
(i) 7x – 2y (ii) 5x + 2y – 10 (iii) 6x + z – 2 (iv) 15y – z (v) 9x – 10z
Question 2. Show on the number line x, x – 4, 2x + 1, 3x – 2.
Solution:
Term, Factors and Coefficient
Consider the expression 3x + 7. This expression is containing two terms, i.e. 3x and 7.
Terms are added to form expressions.
Now consider the expression 7x2 – 5xy. This expression contains two terms, i.e. 7x2 and –5xy.
The term 7x2 is a product of 7, x and x. or we say that 7, x and x are factors of the term 7x2.
Also the term –5xy has the factors –5, x and y.
We know that a term is a product of its factors.
The numerical factor of term is called its coefficient. The expression 9xy – 7x has the two terms
9xy and –7x. The coefficient of the term 9xy is 9 and the coefficient of –7x is –7.
Question: Identify the coefficient of each term in the expression x2y2 – 10x2y + 5xy2 – 20.
Solution:
Coefficient of x2y2 is 1.
Coefficient of x2y is –10.
Coefficient of xy2 is 5.
Monomials, Binomials and Polynomials
An expression containting only one term is called monomial.
Examples: 2x, 3y, 4p, q, 3x2, 4y2, 7xy, – 9m, –3, 4xyz, etc.
An expression containing two terms is called binomial.
Examples: a + b, x – y, 2p + 4q, 3xy – 7yz, p2 – 3q2, p2 – 3, a + 3, etc.
An expression containing three terms is called trinomial.
Examples: a + b + c, 4p + 9q – 3r, x+ 4y2 – 3z2, etc.
An expression containing one or more terms with non-zero coefficient of variables having nonnegative exponents is called a polynomial.
Examples: a + b + c + d, 4xy, –5z, 5xyz – 10p, 4x + 3y – 3z, etc.
Question 1. Classify the following polynomials as monomials, binomials, trinomials –z + 5, x + y + z, y + z + 100, ab – ac, 17
Solution:
Monomials Binomials Trinomials 17 –z + 5ab – ac x + y + zy + z + 100
Question 2. Construct:
(a) 3 binomials with only x as a variable;
Sol: (a) 1. 5x – 4 2. 3x + 2 3. 10 + 2x
(b) 3 binomials with x and y as variables;
Sol: (b) 1. 2x + 3y 2. 5x – y 3. x – 3y
(c) 3 monomials with x and y as variables;
Sol: (c) 1. 2xy 2. xy 3. –7xy
(d) 2 polynomials with 4 or more terms.
Solution: (d) 1. 2x3 – x2 + 6x – 8 2. 10 + 7x – 2x2 + 4x3
Question: write two terms which are like
(i) 7xy (ii) 4mn2 (iii) 2l.
Solution:
(i) Two terms like 7xy are: –3xy and 8xy.
(ii) Two terms like 4mn2 are: 6mn2 and –2n2m.
(iii) Two terms like 2l are: 5l and –7b.
Exercise 9.1
Question 1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy (ii) 1 + x + x2 (iii) 4x2y2 – 4x2y2z2 + z2
(iv) 3 – pq + pr – rp (v) X/2 + X/2 - xy (vi) 0.3a – 0.6ab + 0.5b
Solution:
(i) 5xyz2 – 3zy
Terms 5xyz2 –3zy Coefficients 5 –3
(ii) 1 + x + x2
Terms 1 +x +x2 Coefficients 1 1 1
(iii) 4x2y2 – 4x2y2z2 + z2
Terms 4x2y2 –4x2y2z2 + z2 + z2 4 –4 1
(iv) 3 – pq + qr – rp
Terms 3 –pq + qr – rp Coefficients 3 –1 +1 –1
(v) x/2 +x/2 - xy
Terms x/2 +y/2 -xy Coefficients 1/2 1/2 -1
(vi) 0.3a – 0.6ab + 0.5b
Terms 0.3a –0.6ab 0.5b Coefficients 0.3 –0.6 0.5
Question 2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x+ x3 + x4, 7 + y + 5x, 2y – 3y2, 2y – 3y2 + 4y3,
5x – 4y + 3xy, 4z – 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q
Solution:
Monomials Binomials Trinomials 1000pqr x + y2y – 3y24z – 15z2p2q + pq22p + 2q 7 + y + 5x2y – 3y2 + 4y35x – 4y + 3xy
Following polynomials do not fit in any of these categories:
x + x2 + x3 + x2 [∵ It has 4 terms.]
ab + bc + cd + da [∵ It has 4 terms.]
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Solution:
Question 4.
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3.
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz.
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10
from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q.
Solution: For subtraction we rearrange the terms such that the like terms are in the same column. Then we change the sign of the terms to be subtracted. Next we add the terms.
Product of Monomials
Note:
I. When we multiply monomials the product is also a monomial
II. We use the rules for exponents and powers when we multiply variables.
Example: Find the product of 2x, 5xy and 7x.
Solution: 2x * 5xy * 7z = (2 * 5 * 7) * (x * xy) * z
= 70 * (x2y 8 z)
= 70 * x2yz = 70x2yz
Question: Find 4x * 5y * 7z. First find 4x * 5y and multiply it by 7z; or first find 5y * 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Solution: We have:
4x * 5y * 7z = (4x * 5y) * 7z
= 20xy * 7z = 140xyz
Also 4x * 5y * 7z = 4x * (5y 8 7z)
= 4x * 35yz = 140xyz
We observe that
(4x * 5y) * 7x = 4x(5y * 7z)
∴ The product of monomials is associative, i.e. the order in which we multiply the monomials does not matter.
## Mathematics (Maths) Class 8
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# What The Numbers Say About Jamie Lynn Spears (11/03/2019)
How will Jamie Lynn Spears get by on 11/03/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is for entertainment purposes only – do not take this too seriously. I will first calculate the destiny number for Jamie Lynn Spears, and then something similar to the life path number, which we will calculate for today (11/03/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology experts.
PATH NUMBER FOR 11/03/2019: We will consider the month (11), the day (03) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 03 we do 0 + 3 = 3. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 3 + 12 = 17. This still isn’t a single-digit number, so we will add its digits together again: 1 + 7 = 8. Now we have a single-digit number: 8 is the path number for 11/03/2019.
DESTINY NUMBER FOR Jamie Lynn Spears: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Jamie Lynn Spears we have the letters J (1), a (1), m (4), i (9), e (5), L (3), y (7), n (5), n (5), S (1), p (7), e (5), a (1), r (9) and s (1). Adding all of that up (yes, this can get tedious) gives 64. This still isn’t a single-digit number, so we will add its digits together again: 6 + 4 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the destiny number for Jamie Lynn Spears.
CONCLUSION: The difference between the path number for today (8) and destiny number for Jamie Lynn Spears (1) is 7. That is greater than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too worked up about that! As mentioned earlier, this is of questionable accuracy. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# Set Operations: Union, Intersection, Difference, Complement, Cartesian Product, Power Set
Set Operations
Intersection: the intersection of two sets A and B, denoted by AB, is the set that contains all elements of A that also belong to BAND
Example: Let D={1,2,3} and E={1,2,4,5},then DE={1,2}
Union: the union of two sets A and B, denoted by AB, is the set of all elements that belong to either A or BOR
Example: Let I={1,2,3} and J={1,2,4,5},then IJ={1,2,3,4,5}
Complement, intersection and union
Let A and B be subsets of a suitable universal set 𝕌.
● The complement A is the set of all elements of 𝕌 that are not in A.
● The intersection AB is the set of all elements belonging to A and to B.
● The union AB is the set of all elements belonging to A or to B.
● In mathematics, the word ’or’ always means ‘and/or’, so all the elements that are in both sets are in the union.
● The sets A and B are called disjoint if they have no elements in common, that is, if AB=∅.
⛲ Example 1
Consider the sets: K={red, green, blue} L={red, yellow, orange} M={red, orange, yellow, green, blue, purple}
①. Find KL
The union contains all the elements in either set: KL={red, green, blue, yellow, orange}
Notice we only list red once.
②. Find KL
The intersection contains all the elements in both sets: KL={red}
③. Find KM
Here we’re looking for all the elements that are not in set K and are also in M.
KM={orange, yellow, purple}
Set Operations
Complement: The complement of a set A is the set of all elements in the universal set NOT contained in A, denoted Ā. Sometimes the complement is denoted as A‘ or A.
[Example] 𝕌={integers from 1 to 10} N={3,6,9},N̄={1,2,4,5,7,8,10} which are all elements from the universal set that are not found in N.
Union: The union of two sets A and B, denoted AB is the set of all elements that are found in A OR B (or both).
Intersection: The intersection of two sets A and B, denoted AB, is the set of all elements found in both A AND B.
We define two sets to be “disjoint” if their intersection is the empty set (this means the two sets have no elements in common).
⛲ Ex2. Suppose O={a,b,c,d,e}, P={d,e,f} and Q={1,2,3}.
①. OP={a,b,c,d,e,f}
②. OP={d,e}
③. OP={a,b,c}
④. PO={f}
⑤. (OP)∪(PO)={a,b,c,f}
⑥. OQ={a,b,c,d,e,1,2,3}
⑦. OQ
⑧. OQ={a,b,c,d,e}
⑨. (OQ)∪(OQ)={a,b,c,d,e}
Union, Intersection, and Complement
Commonly sets interact. For example, you and a new roommate decide to have a house
party, and you both invite your circle of friends. At this party, two sets are being combined, though it might turn out that there are some friends that were in both sets.
Union, Intersection, and Complement
The union of two sets contains all the elements contained in either set (or both sets).
The union is notated AB.
More formally, xAB if xA or xB (or both)
The intersection of two sets contains only the elements that are in both sets.
The intersection is notated AB.
More formally, xAB if x∊A and xB
The complement of a set A contains everything that is not in the set A.
The complement is notated Ā, A‘ or A, or sometimes ¬A.
Grouping symbols can be used like they are with arithmetic – to force an order of operations.
⛲ Ex3. Suppose H={cat, dog, rabbit, mouse}, F={dog, cow, duck, pig, rabbit}
G={duck, rabbit, deer, frog, mouse}
①. Find (HF)⋃G
Now we union that result with G: (H⋂F)⋃G={dog, duck, rabbit, deer, frog, mouse}
②. Find H⋂(FG)
We start with the union: FG={dog, cow, rabbit, duck, pig, deer, frog, mouse}
Now we intersect that result with H: H⋂(FG)={dog, rabbit, mouse}
③. Find (HF)G
Now we want to find the elements of G that are not in HF
(HF)G={duck, deer, frog, mouse}
Notice that in the example above, it would be hard to just ask for A, since everything from the color fuchsia to puppies and peanut butter are included in the complement of the set. For this reason, complements are usually only used with intersections, or when we have a universal set in place.
We now turn our attention to six fundamental operations on sets. These set operations manipulate a single set or a pair of sets to produce a new set. When applying the first three of these operations, you will want to utilize the close correspondence between the set operations and the connectives of sentential logic.
We now move on to a number of operations on sets. You are already familiar with several operations on numbers such as addition, multiplication, and negation.
[Definition 1] Let A and B be sets.
A denotes the complement of A and consists of all elements not in A, but in some prespecified universe or domain of all possible elements including those in A; symbolically, we define A={x:xA}.
AB denotes the intersection of A and B and consists of the elements in both A and B; symbolically, we define AB={x:xA and xB}.
AB denotes the union of A and B and consists of the elements in A or in B or in both A and B; symbolically, we define AB={x:xA or xB}.
A\B denotes the set difference of A and B and consists of the elements in A that are not in B; symbolically, we define A\B={x:xA and xB}. We often use the identity A\B=AB.
A×B denotes the Cartesian product of A and B and consists of the set of all ordered pairs with first-coordinate in A and second-coordinate in B; symbolically, we define A×B={(a,b) : aA and bB}.
• þ(A) denotes the power set of A and consists of all subsets of A; symbolically, we define þ(A)={x:xA}. Notice that we always have Øϵþ(A)and Aϵþ(A).
⛲ Ex4. We let W={1,2}, X={1,3,5} and Y={n:n is an odd integer}. In addition,we assume that the set of integers ℤ={…,-2,-1,0,1,2,…} is the universe and we identify the elements of the following sets.
W={…,-2,-1,0,3,4,5,…}
Y={n:n is an even integer } by the parity property of the integers
WX={1}, since 1 is the only element in both W and X
WX={1,2,3,5}, since union is defined using the inclusive-or
W\X={2}
X\W={3,5}
• ℤ*=ℤ\{0}={…,-3,-2,-1,1,2,3,…}
W×X={(1,1),(1,3),(l,5),(2,1),(2,3),(2,5)}
• þ(W)={Ø,{1},{2},{1,2}}
Union, Intersection, Difference
Just as numbers are combined with operations such as addition, subtraction and multiplication, there are various operations that can be applied to sets. Here are three new operations called union, intersection and difference.
[Definition 2] Suppose A and B are sets.
The union of A and B is the set A∪B={x:x∈A or x∈B}. The intersection of A and B is the set A∩B={ x:x∈A and x∈B}. The difference of A and B is the set A–B={ x:x∈A and x∉B}.
In words, the union AB is the set of all things that are in A or in B (or in both). The intersection AB= is the set of all things in both A and B.
The difference AB is the set of all things that are in A but not in B.
⛲ Ex5. Suppose R={4,3,6,7,1,9}, S={5,6,8,4} and T={5,8,4}. Find:
①. RS={1,3,4,5,6,7,8,9}
②. RS={4,6}
③. RS={3,7,1,9}
④. RT={3,6,7,1,9}
⑤. SR={5,8}
⑥. RT={4}
⑦. ST={5,8,4}
⑧. ST={5,6,8,4}
⑨. TS
Observe that for any sets X and Y it is always true that XY=YX and XY=YX, but in general XYYX.
Continuing the example, parts 12–15 below use the interval notation discussed in the link of
How to Write Subsets of ℝ as Intervals?
, so [2,5]={x∈ℝ :2≤x≤5}, etc.
⛲ Ex6. Sketching these examples on the number line may help you understand them.
①. [2,5]∪[3,6]=[2,6]
②. [2,5]∩[3,6]=[3,5]
③. [2,5]-[3,6]=[2,3)
④. [0,3]-[1,2]=[0,1)∪(2,3]
Figure 1. (a) sets A and B (b) The union, (c) intersection and (d) difference of sets A and B
⛲ Ex7. Sketch the sets X={(x,y)∈ℝ2:x2+y2≤1} and Y={(x,y)∈ℝ2:x≥1 on ℝ2. On separate drawings, shade in the sets XY, XY, XY and YX.
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# The Quest for 700: Weekly GMAT Challenge (Answer)
Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
Our first task is to ensure that we understand the definition of a geometric sequence. Let’s use the sequence given to us: a, b, c, d. We are told that the ratio of any term after the first to the preceding term is a constant. In other words, b/a = some constant, which is the same constant for the other ratios (c/b and d/c). Let’s name that constant r. Thus, we have the following:
b/a = c/b = d/c = r
By a series of substitutions, we can rewrite the sequence in terms of just a and r:
a, b, c, d is the same as a, ar, ar2, ar3
Rewriting the sequence this way highlights the role of the constant ratio r. That is, to move forward in the sequence one step, we just multiply by a constant factor r. Rewriting also lets us substitute into the alternative sequences and watch what happens.
I. dk, ck, bk, ak
This sequence is the same as ar3k, ar2k, ark, ak. To move forward in the sequence, we divide by r. This is the same thing as multiplying by 1/r. Since this factor is constant throughout the sequence, the sequence is geometric.
II. a + k, b + 2k, c + 3k, d + 4k
This sequence is the same as a + k, ar + 2k, ar2 + 3k, ar3 + 4k. To move forward in this sequence, we cannot simply multiply by a constant expression. The presence of the plus sign means that we will not have a constant ratio between successive terms, and this sequence is not geometric.
III. ak4, bk3, ck2, dk
This sequence is the same as ak4, ark3, ar2k2, ar3k. To move forward in the sequence, we multiply by r and divide by k. In other words, we multiply by r/k, which is a constant factor. This sequence is therefore geometric.
Only sequences I and III are geometric.
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# Lesson 8
Scale Drawings and Maps
### Lesson Narrative
This lesson is optional. In this lesson, students apply what they have learned about scale drawings to solve problems involving constant speed (MP1, MP2). Students are given a map with scale as well as a starting and ending point. In addition, they are either given the time the trip takes and are asked to estimate the speed or they are given the speed and asked to estimate how long the trip takes. In both cases, they need to make strategic use of the map and scale and they will need to estimate distances because the roads are not straight.
In the sixth grade, students have examined many contexts involving travel at constant speed. If a car travels at 30 mph, there is a ratio between the time of travel and the distance traveled. This can be represented in a ratio table, or on a graph, or with an equation. If $$d$$ is the distance traveled in miles, and $$t$$ is the amount of time in hours, then traveling at 30 mph can be represented by the equation $$d = 30t$$. Students may or may not use this representation as they work on the activities in this lesson. But they will gain further familiarity with this important context which they will examine in greater depth when they study ratios and proportional reasoning in grade 7, starting in the next unit.
### Learning Goals
Teacher Facing
• Justify (orally and in writing) which of two objects was moving faster.
• Use a scale drawing to estimate the distance an object traveled, as well as its speed or elapsed time, and explain (orally and in writing) the solution method.
### Student Facing
Let’s use scale drawings to solve problems.
### Student Facing
• I can use a map and its scale to solve problems about traveling.
Building On
Building Towards
### Glossary Entries
• scale
A scale tells how the measurements in a scale drawing represent the actual measurements of the object.
For example, the scale on this floor plan tells us that 1 inch on the drawing represents 8 feet in the actual room. This means that 2 inches would represent 16 feet, and $$\frac12$$ inch would represent 4 feet.
• scale drawing
A scale drawing represents an actual place or object. All the measurements in the drawing correspond to the measurements of the actual object by the same scale.
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Elementary Geometry for College Students (7th Edition)
(a) The sphere has symmetry with respect to the point $P$. (b) The sphere has symmetry with respect to this line.
We can write the general equation of a sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ where $(a,b,c)$ is the center of the sphere and $r$ is the radius The equation of the sphere is: $(x-1)^2+(y-2)^2+(z+5)^2 = 49$ The center of the sphere is $(1,2,-5)$ (a) The only point that a sphere has symmetry with respect to is the sphere's center. Since the point $P = (1,2,-5)$ is the sphere's center, the sphere has symmetry with respect to the point $P$. (b) A sphere has symmetry with respect to any line that passes through the sphere's center. $l: (x,y,z) = (-5,4,-13)+ n(3,-1,4)$ If the line passes through the sphere's center, we can find the required value for $n$: $x-coordinate: -5+n(3) = 1$ $3n = 6$ $n = 2$ We can find the point the line passes through when $n=2$: $l: (x,y,z) = (-5,4,-13)+ n(3,-1,4)$ $(x,y,z) = (-5,4,-13)+ (2)(3,-1,4)$ $(x,y,z) = (-5+6,4-2,-13+8)$ $(x,y,z) = (1,2,-5)$ Since the line passes through the sphere's center, the sphere has symmetry with respect to this line.
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What are the 3 rules of continuity?
What are the 3 rules of continuity?
Note that in order for a function to be continuous at a point, three things must be true:
• The limit must exist at that point.
• The function must be defined at that point, and.
• The limit and the function must have equal values at that point.
When Is A Discontinuity Removable
What are the rules for continuous?
Basic rule Just add -ing to the base verb: Exception If the base verb ends in consonant + stressed vowel + consonant, double the last letter: s t o p consonant stressed vowel consonant vowels = a, e, i, o, u stop stopping run running begin beginning
What are the conditions for proving continuity?
In calculus, a function is continuous at x a if – and only if – all three of the following conditions are met: The function is defined at x a; that is, f(a) equals a real number. The limit of the function as x approaches a exists. The limit of the function as x approaches a is equal to the function value at x a
What are the three steps to showing a function is continuous?
Your pre-calculus teacher will tell you that three things have to be true for a function to be continuous at some value c in its domain:
• f(c) must be defined.
• The limit of the function as x approaches the value c must exist.
• The function’s value at c and the limit as x approaches c must be the same.
What are the different types of continuity?
Continuity and Discontinuity of Functions Functions that can be drawn without lifting up your pencil are called continuous functions. You will define continuous in a more mathematically rigorous way after you study limits. There are three types of discontinuities: Removable, Jump and Infinite.
What are the continuity theorems?
In short: the sum, difference, constant multiple, product and quotient of continuous functions are continuous. Theorem: If f(x) is continuous at and if limxag(x)b, then limxaf(g(x))f(b). In short: the composition of continuous functions is continuous.
What is the order of continuity?
Order of continuity C0: zeroth derivative is continuous (curves are continuous) C1: zeroth and first derivatives are continuous. C2: zeroth, first and second derivatives are continuousCn: 0-th through n-th derivatives are continuous
What is the rule of continuous tense?
The continuous tense is formed with the verb ‘be’ + -ing form of the verb. The Present continuous can be used to show an action which is happening at the time of speaking. I am having dinner at the moment. The Past continuous can be used to show an action which was happening in the past.
What is the rules of future continuous?
In order to form the future continuous tense, we use the phrase will be followed by the present participle of the verb. The present participle is a form of the verb that ends in -ing. For example, the present participle of swim is swimming.
What are the rules of present perfect continuous tense?
The present perfect continuous tense (also known as the present perfect progressive tense) shows that something started in the past and is continuing at the present time. The present perfect continuous is formed using the construction has/have been + the present participle (root + -ing)
What is the formula of present continuous?
The formula for writing in the present continuous is: ‘be’ verb [am, is, are]+ present participle. Examples: He is driving erratically.
What are the 3 conditions for continuity?
Answer: The three conditions of continuity are as follows:
• The function is expressed at x a.
• The limit of the function as the approaching of x takes place, a exists.
• The limit of the function as the approaching of x takes place, a is equal to the function value f(a).
Which three conditions must be met in order to prove continuity at a point?
In order for a function to be continuous at a certain point, three conditions must be met: (1) that the point is in the domain of the function, (2) that the two-sided limit of the function as it approaches the point does in fact exist and (3) the value of the function equals the limit that it approaches.
How do you show that a function is continuous?
Saying a function f is continuous when xc is the same as saying that the function’s two-side limit at xc exists and is equal to f(c).
What is the three step definition of continuity?
For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
What are the three types of continuity?
In calculus, a function is continuous at x a if – and only if – it meets three conditions: The function is defined at x a. The limit of the function as x approaches a exists. The limit of the function as x approaches a is equal to the function value f(a)
What are the types of discontinuity?
There are two types of discontinuities: removable and non-removable. Then there are two types of non-removable discontinuities: jump or infinite discontinuities. Removable discontinuities are also known as holes. They occur when factors can be algebraically removed or canceled from rational functions.
What is continuity with example?
The definition of continuity refers to something occurring in an uninterrupted state, or on a steady and ongoing basis. When you are always there for your child to listen to him and care for him every single day, this is an example of a situation where you give your child a sense of continuity.
What are the 3 conditions of continuity?
Note that in order for a function to be continuous at a point, three things must be true:
• The limit must exist at that point.
• The function must be defined at that point, and.
• The limit and the function must have equal values at that point.
What are the properties of continuous functions?
Continuous functions have four fundamental properties on closed intervals: Boundedness theorem (Weierstrass second theorem), Extreme value theorem (Weierstrass first theorem), Intermediate value theorem (Bolzano-Cauchy second theorem), Uniform continuity theorem (Cantor theorem).
What does Rolles theorem say?
Rolle’s theorem, in analysis, special case of the mean-value theorem of differential calculus. Rolle’s theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.
How do you prove that two functions are continuous?
How to Determine Whether a Function Is Continuous or
• f(c) must be defined.
• The limit of the function as x approaches the value c must exist.
• The function’s value at c and the limit as x approaches c must be the same.
What does order of continuity mean?
Continuity Orders Similar to standing orders, in that customers sign up and receive shipments until they cancel. A Regular Continuity Series will send a list of items to a customer at regular intervals until the customer cancels or the end of the series is reached.
How do you find the order of continuity?
For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
What are the 3 parts of continuity?
(c) Second-order parametric continuity(C2) : A curve is said to be second-order parametric continuous if it is Co and C1 Continuous and the second-order derivative of the segment P at tt1 is equal to the second-order derivative of segment Q at tt2.
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# Z Test
Delve into the realm of engineering mathematics and discover the unique significance of the Z Test. This comprehensive guide demystifies the concept, properties, and role of Z Test in statistics and its application across various engineering fields. You'll also gain insights into the intriguing Z Test vs t Test debate, and learn how to effectively make the right choice depending on your engineering maths needs. From understanding the Z Test formula to its practical applications, this is your ultimate resource to mastering the Z Test in engineering mathematics.
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## Understanding the Z Test in Engineering Mathematics
In the realm of engineering mathematics, comprehension of statistical techniques is important for accurate data interpretation and successful project completion. You might be wondering what part the Z Test plays in this field. This statistical test offers a method for quantifying the difference between two population means, assuming a large sample size and known variances. Utilised correctly, it provides essential decision-making tools that are critical in engineering.
### Discovering the Z Test Meaning in Statistics
The Z Test, as used in statistics, is a hypothesis-testing procedure. It's driven by this core query: given two groups that are part of a larger population, is there a statistically significant difference between their means?
The term 'statistical significance' indicates that the differences observed are likely genuine and not due to chance.
Typically, the Z Test is employed when the data is normally distributed and the population's standard deviation is already known. The formula for the Z statistic is represented as: $Z = \frac {(\bar{x}- \mu)} {( \sigma / \sqrt{n})}$
Here, $$\bar{x}$$ is the sample mean, $$\mu$$ is the population mean, $$\sigma$$ is the population standard deviation, and $$n$$ is the size of the sample.
### Unveiling Z Test Properties
To better grasp the concept of the Z Test, you must understand its essential properties. Here are the main properties for a quick reference:
• The Z Test requires data to be normally distributed.
• It necessitates the knowledge of the population's standard deviation.
• This test is appropriate for large samples, generally those which are above 30 units.
Interestingly, the Z Test derives its name from the standardised normal distribution - the Z distribution, utilised within the test procedure.
#### Key Properties of the Z Test and their Importance
The Z Test properties hold substantial importance in statistical analysis. Requiring the population data to be normally distributed ensures that the test results will reflect the data's true characteristics. Knowledge of the population standard deviation allows for the accurate estimation of the sample's distribution. Large sample size condition is crucial as it ensures a lower margin of error and increased reliability of the test results. Do understand that the efficacy of a Z Test heavily relies on adhering to these properties.
### Exploring Examples of Z Test Applications in Engineering
Oftentimes, underlying statistics can illustrate the most complex engineering situations. Let's take a look at one such key instance.
Imagine an engineering firm working on a bridge construction project. The firm procured two types of cement from different suppliers. To verify which one sets faster, the firm could apply a Z Test. They would treat the foremost group cement setting times as one sample and the second group as another. Upon applying the Z Test, they can conclusively quantify if there's a significant average setting time difference between the two cements.
#### Everyday Applications of Z Test in Various Engineering Fields
Whether it be electrical, mechanical, civil, or software engineering, the Z Test finds several applications. In electrical engineering, it may be used to compare the efficiency of two different types of microchips. In mechanical engineering, it can be employed to evaluate the performance of two alternative designs for a part. Software engineers might use it to compare the speed of two different algorithms solving the same problem. Civil engineers often use this test method to compare the quality or performance of different materials, like in the cement example provided.
## The Z Test Formula and Its Role in Engineering Mathematics
The Z Test formula plays an integral role in engineering mathematics. It is predominantly a statistical tool used to infer whether a sample data set is representative of its larger population. Given that engineering is often about dealing with substantial data, understanding the Z Test formula can significantly aid in critical decision-making and problem-solving.
### Understanding the Z Test Formula
In essence, the Z Test formula is a statistical test that measures the standard deviation of the sample mean from the population mean, in units of the standard error. The formula for the Z Test statistic is articulated as:
$Z = \frac {(\bar{x}- \mu)} {( \sigma / \sqrt{n})}$
In this equation, $$\bar{x}$$ indicates the sample mean, $$\mu$$ represents the population mean, $$\sigma$$ demonstrates the population standard deviation, and $$n$$ refers to the size of the sample.
Sample Mean: The average of all data points in your sample data set.Population Mean: It represents the true average of the specific characteristic in the entire population.Standard Deviation: It showcases the dispersion of a dataset relative to its mean.Sample Size: It specifies the number of observations or replicates that constitute a sample.
Notably, this formula assists you in deriving the Z statistic or Z score. This Z score essentially tells you how many standard deviations an element is from the mean. In applications, a high Z score in absolute terms would exhibit that the disparity observed could not have occurred by chance, indicating the result's statistical significance.
### Step by Step Calculation using Z Test Formula
Using the Z Test formula effectively involves following a set of sequential steps. These include:
1. Hypothesis Setting: This involves setting up the null and alternative hypotheses. The null hypothesis typically posits no difference between the characteristics of two populations, while the alternative hypothesis suggests some difference.
2. Data Collection: This step involves collecting sample data. Remember that the larger the sample size, the more reliable the test outcome.
3. Calculation of Z Statistic: Using the Z Test formula, calculate the Z score. A critical aspect here is to remember that this calculation assumes you know the population standard deviation. If unknown, a T Test might be more appropriate.
4. Comparison with Critical Value: The calculated Z statistic is then compared with a critical value from the Z distribution table, which correlates with the desired confidence level. If the absolute value of the Z statistic is larger than the critical value, the null hypothesis is rejected.
Consider a scenario where an engineering team tests the tensile strength of two types of metal. They hypothesise that there's no difference between the two metals' average tensile strength. After collecting samples and testing, they utilize the Z Test formula to calculate the Z statistic. If the Z score is high (in absolute terms), it means the observed difference in tensile strengths between the two metals is statistically significant, inferring a real difference.
#### Simplifying Calculations with the Z Test Formula in Engineering Maths
In the realm of Engineering Mathematics, embracing the Z Test formula can certainly aid in simplifying numerous complex data analysis tasks. More than a mere instrument of differentiation between two population means, the Z Test, powered by its arithmetic formula, provides a quantifiable framework for comparison. By generating demonstrable, numerically precise results, it can guide more informed assessments and decisions. For instance, in Mechanical Engineering, it can help in differentiating the average performance of two assembly lines or, in Civil Engineering, it might assist in comparing the average durability of two types of construction materials.
Besides, understanding and proficiently applying the Z Test formula can be crucial when dealing with a large cohort of data - a common scenario in any engineering discipline. At times, a premium is placed on quick, reliable tests that can determine statistically significant differences at a glance. Thus, the Z Test formula steps up as an essential tool in the engineer's toolkit, allotting a rigorous statistical underpinning to intuition and judgement in decision-making.
## Evaluating the Z Test vs t Test Debate in Engineering Mathematics
In the analysis of engineering mathematics, crucial statistical decisions often hinge on picking the right test: the Z Test or the t Test. Both tests can evaluate if two populations are significantly different from each other, but the deciding factor in choosing one over the other often lies in the nature of your data and the sample size.
### A Comparative Analysis: Z Test v/s t Test
A Z Test and a t Test assist you in determining whether the mean of two groups significantly differs. However, the applicability of both these tests is contingent on certain conditions, distinguishing them from each other.
The Z Test is applied when your data conforms to normal distribution, the sample size is fairly large (usually > 30), and you know the population standard deviation. It relies on this formula:
$Z = \frac {(\bar{x}- \mu)} {( \sigma / \sqrt{n})}$
Conversely, a t Test comes into play when you're dealing with smaller sample sizes (usually < 30), or when the population standard deviation is unknown. The t Test's formula is slightly different, replacing the population standard deviation with the sample standard deviation, as follows:
$t = \frac {(\bar{x}- \mu)} {(s / \sqrt{n})}$
Here, $$\bar{x}$$ signifies the sample mean, $$\mu$$ represents the population mean, $$s$$ is the sample standard deviation, and $$n$$ is the sample size.
Here's a comparative summary in a table format for clearer understanding:
Z Test t Test Assumption Data is normally distributed Doesn't necessarily require normal distribution Sample Size Large (usually > 30) Small (usually < 30) Standard Deviation Known and pertains to the population Unknown and pertains to the sample
### Practical Applications: When to Use Z Test and t Test
Your choice between a Z Test and a t Test depends heavily on the nature of the data and the specific engineering problem you wish to solve. In Mechanical Engineering, for instance, if you're comparing the average lifespan of a large batch of similar machines, you might employ a Z Test. However, if you are comparing the effectiveness of a new prototype component against an existing model with a small sample group, a t Test might be more suitable.
In the realm of Software Engineering, if you're comparing the response times of a feature rolled out to a large number of users, then a Z Test could supply more accurate results. On the other hand, if you're working with a smaller test group to evaluate differences in software efficiency after implementing a new, untried feature, making use of a t Test becomes more prudent.
The type of engineering, nature of the problem, data distribution, sample size, and standard deviation knowledge all drive your choice between the Z Test and the t Test. In the world of statistical analysis, there’s no one-size-fits-all answer!
#### Choosing between Z Test and t Test: Decision Making for Engineering Maths
When it comes to making the choice between a Z Test and a t Test for your engineering data analysis, you need to evaluate your scenario against three key criteria: data distribution, sample size, and standard deviation.
First, determine if your data is normally distributed. If it isn't, a t Test might be the more appropriate test, as it's less sensitive to distribution anomalies. However, if your data follows a normal distribution pattern, you could use either a Z Test or a t Test, given other conditions are met correctly.
Second, evaluate your sample size. For larger sample sizes (typically >30), you would lean towards a Z Test. However, with a smaller sample size (<30), a t Test is usually more reliable because it is less influenced by outliers and skewness in data.
Finally, consider the standard deviation. Do you know the standard deviation of your population, or do you just have the standard deviation of your sample? Knowledge about the population's standard deviation favours a Z Test, while a t Test is generally preferred when only the standard deviation of the sample is known.
After adopting these strategies and asking these pertinent questions, you should be well-equipped to make an informed choice between a Z Test and a t Test in engineering mathematics.
## Z Test - Key takeaways
• The Z Test is a statistical technique used in engineering mathematics for quantifying the difference between two population means, assuming a large sample size and known variances.
• Z Test Meaning: The Z Test, as used in statistics, is a hypothesis-testing procedure used to determine if there is a statistically significant difference between the mean values of two groups within a larger population.
• It is used when the data is normally distributed and the population's standard deviation is known.
• Z Test Properties:
• Data must be normally distributed.
• Requires knowledge of the population's standard deviation.
• Appropriate for large samples, generally those above 30 units.
• Z Test Applications in Engineering:
• Used in various engineering fields like electrical, mechanical, civil, or software engineering to compare the efficiency or performance of two different items or groups.
• Z Test Formula: $Z = \frac {(\bar{x}- \mu)} {( \sigma / \sqrt{n})}$
• $$\bar{x}$$ is the sample mean, $$\mu$$ is the population mean, $$\sigma$$ is the population standard deviation, and $$n$$ is the size of the sample.
• Z Test vs t Test:
• The Z Test is used with data that conforms to normal distribution, a large sample size, and known population standard deviation.
• The t Test is used with smaller sample sizes and when the population standard deviation is unknown.
#### Flashcards in Z Test 12
###### Learn with 12 Z Test flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What is a Z-test?
A Z test is a statistical examination that measures whether the difference between two groups is statistically significant. It assumes the data follows a normal distribution and explores the relationship between the mean and standard deviation. It’s often used when samples are large and the standard deviation is known.
When should one use a T-test versus a Z-test?
Use a Z test when you know the population standard deviation and the sample size is large (>30). Use a T test when you don't know the population standard deviation or when the sample size is small (<30).
How can one calculate the P-value in a Z-test?
To calculate the p-value in a Z test, first find the Z score of your test or data. Then, using a Z-table or an online calculator, locate the proportion that corresponds to your Z score. This proportion is the p-value.
## Test your knowledge with multiple choice flashcards
What is the Z Test in engineering mathematics and what does it quantify?
How is the Z statistic calculated in the Z Test?
What are some key properties of the Z Test?
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When it comes to knowing the meaning of the term cotangent, it is necessary, first of all, to discover what its etymological origin is. In this case, we can state that it is a word that derives from Latin. It is exactly the result of the union of three delimited components:
-The prefix “co-”, which can be translated as “together”.
-The verb “tangere”, which means “to touch”.
-The suffix “-nte”, which is used to indicate “agent”.
Starting from all this, we find the fact that cotangent means “inverse of the tangent of an arc or of an angle”.
The notion of cotangent alludes to inverse tangent function of an arc or an angle. To understand what the cotangent is, therefore, we must know what the tangent.
In the context of trigonometry (a specialty of mathematics), the tangent of a right triangle is obtained dividing the leg opposite an acute angle and the adjacent leg. It should be remembered that the longest side of these triangles is called hypotenuse, while the other two are called legs.
Returning to the idea of cotangent, we had already mentioned that it is the inverse function of the tangent. Therefore, if the tangent is the quotient between the opposite leg and the adjacent leg, the cotangent equates to quotient between the adjacent leg and the opposite leg.
In a right triangle whose hypotenuse is 20 centimeters, its adjacent leg is 15 centimeters, and its opposite leg is 12 centimeters, we can calculate the cotangent as follows:
Cotangent = Adjacent leg / Opposite leg
Cotangent = 15/12
Cotangent = 1.25
Since the cotangent is the inverse function of the tangent, it can also be obtained dividing 1 by the tangent. In our example above, the tangent equals 0.8 (the result of the division between the opposite leg and the adjacent leg). Therefore:
Cotangent = 1 / tangent
Cotangent = 1 / 0.8
Cotangent = 1.25
Within the field of mathematics, and more specifically in the field of trigonometry, the cotangent plays an important role. Specifically, it talks about what the properties of the cotangent function are. And these are none other than continuity, dominance, travel, decreasing or period, for example.
Just as the cotangent is the inverse function of the tangent, the cosecant is the inverse of breast and the drying, the inverse of cosine.
In the same way, we cannot ignore the existence of what is known as a hyperbolic cotangent. It is another term used in trigonometry in relation to a real number. In that case it is established that it becomes the inverse of the hyperbolic tangent.
It is represented by coth (x) or by cotgh (x) and on that there is what is called the addition theorem. A theorem that comes to expose the way to be able to synthesize this hyperbolic tangent.
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Writing a Proportion To write a proportion proportion
• Slides: 21
Writing a Proportion To write a proportion, proportion you set one ratio equal to another ratio. The only rule is this: Your left side and right side must be set up the same way. So, to see the Eiffel Tower was really 1064 feet tall, you measured it’s shadow to be 798 feet long. Then, you held up a 4 foot broom, and the broom’s shadow was 3 feet long. Write 2 different proportions that compare height and shadow. EF height broom height EF shadow broom shadow 1064 ft 798 ft 3 ft 4 ft 3 ft 1. 333. . . 266 1. 333. . . When you write a proportion, your left side and right side must be set up the same way.
Is it a Proportion? Dana earned a salary of \$27, 000/mo, and Teresa earned a salary of \$6, 500/mo. Dana paid \$10, 530 in income taxes, but Teresa paid \$1, 625 in income taxes. Did they pay a proportional amount of taxes? Step 1: 1 Make a ratio (fraction) with Dana’s Step 2: 2 Set that equal to Teresa’s Step 3: 3 Cross-multiply. Are the cross products equal? Dana Teresa 10, 530 1, 62 5 6, 500 27, 000 = 10, 530 27, 000 • 1, 62 5 6, 500 43, 875, 000 68, 445, 000 Since cross products are NOT equal, they do NOT pay a proportional amount of taxes. When cross products are EQUAL, it IS a proportion When cross products are NOT equal, it is NOT a proportion
Solving a Proportion Drew was bragging that he could bench press 245 newtons (N). To convert that weight to pounds (lb), use the chart to the right: 100 ounces (oz) = 6. 25 pounds (lb) 8 newtons (N) = 1. 8 pounds (lb) 25 ounces (oz) = 6. 95 newtons (N) Step 1: 1 Write your empty proportion. Step 2: 2 Set up your left side anyway you want. * I’d recommend starting with “ 245 N ” Step 3: 3 Find the correct conversion. . . then write it as your Since right side ratio. newtons (N) is on top on. . . since you’re trying the left, then put newtons to find weight in (N) on top on pounds, call that x. the right. Step 4: 4 Cross-multiply, then set the cross-products equal to each other Step 5: 5 Solve the equation you made. To solve a proportion, cross multiply, set the products equal to each other, then solve. 8 N x lb 1. 8 lb • 8●x 8 x = 245● 1. 8 = 441 8 8 x = 55. 125 Drew bench pressed, about, 55 pounds (lb).
Writing & Solving Proportions Jack’s recipe makes 2. 25 pounds (lb) of chili, and it calls for 1. 5 tablespoons of lime juice. If he wanted to make 5. 1 lb of chili, how many tablespoons of lime juice should he use? Chili #1 Chili #2 2. 25 5. 1 1. 5 2. 25 • l • = = l To write a proportion, each ratio must be set up the same way. Let’s set up both ratios as 5. 1 • 1. 5 chili juice Now, find the cross products. 2. 25 l = 2. 25 l 7. 65 2. 25 = 3. 4 Simplify, then solve Jack should use 3. 4 tablespoons of lime juice.
Scale Models Harmony MS has the longest bus rides in the county. Here’s a map showing the route from Loudoun Heights (next to Harper’s Ferry) to Harmony. If their map length 4 is 1 3 inches apart, then how far apart are they actually apart? Step 1: Write an empty proportion. Step 2: On the left side, write the map scale as a fraction. Step 3: On the Let’s change those fractions to decimals. right side, fill in the distance given in the problem. Step 4: Call the missing distance, m. m 0. 25 1. 75 2 map scale Step 5: Cross 0. 25 • m = 1. 75 • 2 multiply, then solve 0. 25 m = for m. 3. 5 The bus ride would 0. 25 be, about, 14 miles. m = 14
Finding the Map Scale The longest drive in the contiguous U. S. is from Key West, FL to Seattle, WA. If the actual distance of the drive is 3, 450 miles, what is the map scale? scale Step 1: 1 Write your empty proportion. Step 2: 2 Set your left side as Why 1 ? Since it’s your map scale, scale you get to pick the numerator. Why x ? That’s the part you are trying to find. Step 3: 3 Write the map distance. as your numerator (top) … Step 4: 4 Write the actual distance. as your denominator (bottom) Step 5: 5 Solve the proportion. 1 x in = 3450 = The map scale is x 1 in mi
Using a Map’s Scale Ratio José and Rosa, from Lisbon, Portugal, are going to visit the grandparents, in Vladivostok, Russia, for the holidays. . What will be the actual distance of their drive? scale ratio = distance ratio cm cm = a Set up your proportion. Measure the distance in cm. Now, measure the map’s scale ratio in cm. . . and fill it in your scale ratio a= Solve. The distance between Lisbon, Portugal and Vladivostok, Russia is miles.
Using the Percent Proportion 1. 8 is 124% of what number? (If needed, round to the nearest tenth. ) part Proportions compare a part/whole ratio = whole to another part/whole ratio. This is the percent proportion. is p = of 100 1. 8 124 = x 100 180 ____= 124 x ____ 124 x = 1. 5 A percent proportion compares a ratio to a percent/100. Which number is the “part”? It’s 1. 8 What little word is next to it? “is” I dunno. Let’s say x. Which number is the “whole”? What little word is next to it? “of” Write and solve the proportion 1. 8 is 124% of 1. 5
Using the Percent Proportion What number is 56. 1% of 31? p_ _ = of 100 is (If needed, round to the nearest tenth. ) Which number is the “percent”? Which number is the “is”? Which number is the “of”? x 17. 4 p_ _ = of 100 p 87. 2% (I dunno ? !) let’s call it x it’s 31 Solve for x What percent of 47 is 41 ? is it’s 56. 1 (If needed, round your percent to the nearest tenth. ) Which number is the “percent”? I dunno…let’s call it p Which number is the “is”? 41 Which number is the “of”? 47 Solve for p
Using the Percent Proportion In 2009, Loudoun Valley High School had a capacity of 1350 students, but 1497 students are enrolled. What percent of capacity was Valley running? part = _p_ 100 whole Which number is the “percent”? x it’s actually 1497 Which number is the “part”? Which number is the “whole”? 1350 Solve for x x 110. 9 Valley was running at 110. 9% capacity. Harvard University accepted 4, 983 freshman applications last year. Only 7% of all students who applied were accepted. How many students applied? part whole p_ _ Which number is the “percent”? = 100 Which number is the “part”? Which number is the “whole”? x 71, 186 Solve for x it’s 7 4983 (I dunno ? !) let’s call it x 71, 186 students applied.
Are They Similar? K J The triangles are SIMILAR Polygons are similar when: PR ? QR = JL KL 1. Corresponding angles are congruent. * * QPR PRQ KJL JLK 2. Corresponding sides are in proportion *Put a pair of corresponding sides in a ratio. *Set them equal to another pair of corresponding sides. *Cross-multiply – Are they equal? http: //www. mathopenref. com/similartriangles. html L 21. 8 10. 9 23. 2 11. 6 252. 88 = 252. 88
Finding Missing Measures of Similar Figures 25 W 20 V 15 U T
Finding Missing Measures of Similar Figures 30 12 19 x
Writing a Ratio Write the ratio 3 feet (ft) to 8 inches (in) as a fraction in simplest form. 9 3 ft • 12 = 36 in = 8 in 9 in = 8 in 2 9 in 2 in Hint: Always convert to the smaller unit. or 2 Compare these 2 values as a fraction. Convert to the same unit. 9 or A ratio is a comparison of 2 values. 9: 2 or 9 to 2 Now, reduce There are other ways to write a ratio, also.
Finding a Unit Rate In 1987, Bo Jackson ran the 40 yd (yard) dash in 4. 12 s (seconds). In 1989, Deion Sanders ran the 60 yd dash in 6. 68 s. Who ran at a faster rate? To compare these runners, find each players yards seconds Bo Deion 40 yd 60 yd = 9. 708 ≈ 9. 7 yd = 8. 982 ≈ 9. 0 yd 4. 12 s s 6. 68 s s Deion ran at the faster rate. Emerald’s 11 oz (ounce) can of almonds sells for \$7. 69. Planters 5 oz bag of Trail Mix sells for \$3. 25. Which has the lower unit rate? \$7. 69 \$3. 25 Emerald’s Planters 11 oz = 0. 699 ≈ \$0. 70 oz 5 oz 0. 65 ≈ \$0. 65 oz
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# How to find the orthocenter of a triangle? – Step by step
The orthocenter of a triangle can be found using two main methods. We can sketch the heights of the triangle and find the point of intersection. Alternatively, we can find the coordinates of the orthocenter algebraically.
In this article, we will learn how to find the orthocenter of a triangle using the two methods mentioned. Then, we will solve some practice problems.
##### GEOMETRY
Relevant for
Learning how to find the orthocenter of a triangle.
See methods
##### GEOMETRY
Relevant for
Learning how to find the orthocenter of a triangle.
See methods
## What is the orthocenter of a triangle?
The orthocenter of a triangle is the point of intersection of the three heights of the triangle. In the diagram below, we can see that point O is the orthocenter:
Remember that the heights of the triangle are the perpendicular segments that connect a vertex with its opposite side. Altitudes always form a 90° angle with the corresponding side.
## Orthocenter of common triangles
Depending on the type of triangle we have, the location of the orthocenter varies.
### Orthocenter of an acute triangle
The orthocenter of all acute triangles always lies inside the triangle. Remember that an acute triangle is characterized in that all of its internal angles are less than 90°.
### Orthocenter of an obtuse triangle
The orthocenter of all obtuse triangles always lies outside the triangle. Remember that an obtuse triangle is characterized by having an angle greater than 90°.
### Orthocenter of a right triangle
The orthocenter of all right triangles is located at the central vertex of the right triangle. Remember that a right triangle has one 90° angle.
### Orthocenter of an equilateral triangle
The orthocenter of all equilateral triangles is located in the same position as the centroid of the triangle. Remember that equilateral triangles have all three sides of the same length.
## Finding the orthocenter of a triangle graphically
We can find the orthocenter of a triangle graphically by plotting two heights of the triangle and finding their point of intersection.
In turn, we can find the heights by drawing perpendicular lines from the vertices to opposite sides. We do this with the following steps:
Step 1: Using a radius equal to BC and centering on point B, draw an arc on side AC to get point E.
Step 2: Using a radius equal to BC and centering at point C, draw an arc on side AB to get point D.
Step 3: Using a radius equal to BD, draw intersecting arcs from B and D to form point F and draw segment CF.
Step 4: Using a radius equal to CE, draw intersecting arcs from C and E to form point G and draw segment BG.
Step 5: Mark the point of intersection of segments CF and BG.
Segments CF and BG are perpendicular to sides AB and AC respectively. This means that they represent the heights of the triangle. Therefore, the point of intersection is the orthocenter of the triangle.
## Finding the orthocenter of a triangle algebraically
The orthocenter of a triangle can be found algebraically using the coordinates of the vertices of the triangle. We can follow the process using the following triangle.
In this triangle, $latex A(x_{1},~y_{1})$, $latex B(x_{2},~y_{2})$, $latex C(x_{3},~y_{3})$ are the vertices and AD , BE and CF are the heights. Point O is the orthocenter since it is the point of intersection.
Step 1: To find the heights, we have to start by finding the slopes of the sides of the triangle. For this, we use the slope formula:
$latex m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Representing the slope of line AC as $latex m_{AC}$, we have:
$latex m_{AC}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$
Similarly, we also have:
$latex m_{BC}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$
Step 2: Once we have the slopes of the sides, we can find the slopes of the heights by considering that the heights are perpendicular to the sides. The slope of a line perpendicular to another line is:
perpendicular line $latex =-\frac{1}{m}$
where m is the slope of the original line.
Therefore, the slopes of the heights are:
Slope of BE: $latex m_{BE}=-\frac{1}{m_{AC}}$
Slope of AD: $latex m_{AD}=-\frac{1}{m_{BC}}$
Step 3: We can find the equations of the lines that go through BE and AD using the point-slope form. Then, we have:
$latex m_{BE}=\frac{y-y_{2}}{x-x_{2}}$
$latex m_{AD}=\frac{y-y_{1}}{x-x_{1}}$
Step 4: Since we know the values of $latex (x_{1},~y_{1})$ and $latex (x_{2},~y_{2})$, we can use any method to solve the system of equations and find the values of x and y, which are the coordinates of the orthocenter.
## Orthocenter of a triangle – Examples with answers
In the following examples, we can see how to find the coordinates of the orthocenter of a triangle algebraically.
### EXAMPLE 1
If a triangle has vertices A(5, 7), B(2, 3), and C(6, 4), find the slopes of its sides.
Solution: We have the following coordinates:
• $latex (x_{1},~y_{1})=(5, ~7)$
• $latex (x_{2},~y_{2})=(2,~3)$
• $latex (x_{3},~y_{3})=(6,~4)$
Slope of AC:
$latex m_{AC}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$
$latex m_{AC}=\frac{4-7}{6-5}$
$latex m_{AC}=-3$
Slope of BA:
$latex m_{BA}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$
$latex m_{BA}=\frac{7-3}{5-2}$
$latex m_{BA}=\frac{4}{3}$
Slope of BC:
$latex m_{BC}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$
$latex m_{BC}=\frac{4-3}{6-2}$
$latex m_{BC}=\frac{1}{4}$
### EXAMPLE 2
Use the slopes found in Example 1 to determine the coordinates of the orthocenter of the triangle.
Solution: We have to find the slopes of the lines perpendicular to the sides since they correspond to the heights of the triangle. Therefore, we have:
• $latex m_{AE}=$ perpendicular to BC
• $latex m_{BF}=$ perpendicular to AC
• $latex m_{CD}=$ perpendicular to AB
The slope of a perpendicular line is equal to $latex -\frac{1}{m}$, where, m is the slope of the original line, we have:
$latex m_{AE}=-4$
$latex m_{BF}=\frac{1}{3}$
$latex m_{CD}=-\frac{3}{4}$
Using the point-slope form, $latex y-y_{1}=m(x-x_{1})$, we can find the equations of the perpendicular lines. We only need two equations to find the point of intersection.
We use the slope of CD and the point C=(6, 4) to find the first equation:
$latex y-4=-\frac{3}{4}(x-6)$
$latex 4(y-4)=-3(x-6)$
$latex 4y-16=-3x+18$
$latex 3x+4y=34$
Now, we use the slope of BF and the point B=(2, 3) to find the second equation:
$latex y-3=\frac{1}{3}(x-2)$
$latex 3(y-3)=x-2$
$latex 3y-9=x-2$
$latex -x+3y=7$
Any method can be used to solve the system of two equations and we find the solution $latex x=\frac{74}{18},~ y=\frac{55}{13}$.
This solution represents the point of intersection of the lines. Therefore, the coordinates of the orthocenter are $latex (\frac{74}{18},~\frac{55}{13})$.
## Orthocenter of a triangle – Practice problems
Solve the following practice problems using the algebraic method to find the coordinates of the orthocenter.
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Module 3: Probability
# Tree and Venn Diagrams
Barbara Illowsky & OpenStax et al.
Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.
## Tree Diagrams
A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of “branches” that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
### Example
In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. “With replacement” means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.
The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, andB8. Then the nine RR outcomes can be written as:
R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3
The other outcomes are similar.
Solution:
There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.
### Try it
a. List the 24 BR outcomes: B1R1, B1R2, B1R3, …
b. Using the tree diagram, calculate P(RR).
c. Using the tree diagram, calculate P(RB OR BR).
d. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).
e. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).
f. Using the tree diagram, calculate P(BB).
g. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).
An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. “Without replacement” means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, (311)(210)=6110.
If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.
### Try it
a. P(RR) = ________
b. Fill in the blanks:
P(RB OR BR) = (311)(810) + (___)(___) = 48110
c. P(R on 2nd|B on 1st) =
d. Fill in the blanks.
P(R on 1st AND B on 2nd) = P(RB) = (___)(___) = 24100
e. Find P(BB).
f. Find P(B on 2nd|R on 1st).
If we are using probabilities, we can label the tree in the following general way.
• P(R|R) here means P(R on 2nd|R on 1st)
• P(B|R) here means P(B on 2nd|R on 1st)
• P(R|B) here means P(R on 2nd|B on 1st)
• P(B|B) here means P(B on 2nd|B on 1st)
## Venn Diagram
A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.
Suppose an experiment has the outcomes 1, 2, 3, … , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:
### Example
Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore,A AND B = {TT}. A OR B = {TH, TT, HT}.
The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. The Venn diagram is as follows:
## Glossary
Tree Diagram
the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)
Venn Diagram
the visual representation of a sample space and events in the form of circles or ovals showing their intersections
Solutions to Try These 1:
a. B1R1 B1R2 B1R3 B2R1 B2R2 B2R3 B3R1 B3R2 B3R3 B4R1 B4R2 B4R3 B5R1 B5R2 B5R3 B6R1 B6R2 B6R3 B7R1B7R2 B7R3 B8R1 B8R2 B8R3
b. P(RR) = (311)(311) = 9121
c. P(RB OR BR) = (311)(811) + (811)(311) = 48121
d. P(R on 1st draw AND B on 2nd draw) = P(RB) = (311)(811) = 24121
e. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = 2488 = 311
This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. 2488 = 311.
f. P(BB) = 64121
g. P(B on 2nd draw|R on 1st draw) = 811
There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 2433.
Solutions to Try These 2:
a. P(RR) = (311)(210)=6110
b. P(RB OR BR) = (311)(810) + (811)(310) = 48110
c. P(R on 2nd|B on 1st) = 310
d. P(R on 1st AND B on 2nd) = P(RB) = (311)(810) = 24100
e. P(BB) = (811)(710)
f. Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = 810.
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Home » Revision Notes for CBSE Class 6 to 12 » Class 10 Maths for Pair of Linear Equations in Two Variables of Chapter 3 Revision Notes
# Class 10 Maths for Pair of Linear Equations in Two Variables of Chapter 3 Revision Notes
## CBSE Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables – Free PDF Download
Free PDF download of Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Revision Notes & Short Key-notes prepared by expert Mathematics teachers from latest edition of CBSE(NCERT) books.
### CBSE Class 10 Maths Revision Notes Chapter 3 Pair of Linear Equations in Two Variables
1. air of Linear Equations in Two Variables
2. Graphical Method of Solution
3. Algebraic Methods of Solution
4. Substitution Method
5. Elimination Method
6. Cross-Multiplication Method
7. Equations Reducible to a Pair of Linear Equations
1. A pair of linear equations in two variables is said to form a system of simultaneous linear equations in two variables.
The most general form of a pair of linear equations is :
Where are real numbers and
• The solution of a linear equation is a pair of values, one for x and one for y. This pair of values is called Ordered pair.
• A pair of values of x and y satisfying each of the equations in the given system of two simultaneous equations in x and y is called a solution of the system.
• A pair of linear equations will have either (a) a unique solution or (b) infinitely many solutions or (c) no solution.
2. The graph of a pair of linear equations in two variables is represented by two lines;
(i) If the lines intersect at a point, the pair of equations is consistent.The point of intersection gives the unique solution of the equations.
(ii) If the lines coincide, then there are infinitely many solutions. The pair of equations is consistent. Each point on the line will be a solution.
(iii) If the lines are parallel, the pair of the linear equations has no solution. The pair of linear equations is inconsistent.
Thus, corresponding to each solution (x, y) of the linear equation ax + by + c = 0, there exists a point on the line representing the equation ax + by + c = 0 and vice versa.
3. If a pair of linear equations is given by and
(i) the pair of linear equations is consistent. (Unique solution).
(ii) the pair of linear equations is inconsistent(No solution).
(iii) the pair of linear equations is dependent and consistent (infinitely many solutions).
4. Solution of pairs of linear equations in two variables algebraically:
Solution by Substitution method:
Let the pair of equations be and
• From one of the equations, express one of the variables say y in terms of the other variable i.e., x.
• Substitute the value of y, obtained in above step, in other equation, the getting an equation in x.
• Solve the equation and get the value of x.
• Sustitute the value of x in expression for y obtained in first step and get the value of y.
Solution by Elimination method, i.e., by equating the coefficients:
• in the two given equations, make the coefficients of one of the variables numerically equal. To do so, multiply these coefficients by suitable constant.
• Add or subtract the equations obtained in above step according as the terms having same coefficients are of the opposite or of the same signs and get an equation in only one variable.
• Solve the equation found and get the value of one of the variable.
• Sustitute the value of this variable in either of the two given equations and find the value of the other variable.
Solution by Cross Multiplication method:
Let the pair of equations be and
To find the values of x and y, we have the formulae:
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# How to divide decimals
In this post we will two different methods to divide decimal numbers.
(a) Long Division Method
(b) Division using Fraction Method
## Long Division with decimals
This method can be used when decimal is divided by whole number.
This division involves following steps:
(A) First divide the number before decimal point
(B) Insert the decimal in the quotient and involve digits after decimal.
Let us understand the concept with the help of examples:
Example 01
2.5 divided by 5
Solution
You can observe that:
2.5 (dividend) is a decimal number
5 (divisor) is whole number
Since divisor is whole number, we can use long division method
Long Division Steps
(a) Check division for number before decimal i.e 2
Dividing 2 by 5 results in:
⟹ Quotient = 0
⟹ Remainder = 2
(b) Put decimal in the quotient
(c) Use the next number of dividend (i.e. number 5)
(d) Now divide number 25 by 5
Hence 0.5 is the solution
This method is similar to the whole number division.
Only thing you have to careful is the placing of decimal point in the quotient.
Example 02
Divide 409.4 by 23
Observe that:
Dividend is decimal number (i.e. 409.4)
Divisor is whole number (i.e. 23)
Since Divisor is whole number, we can use long division method
Hence 17.8 is the solution of the division.
Let us understand the division step by step
(a) First divide the number before decimal point (i.e 409)
On dividing 409 by 23 we get:
⟹ 17 as quotient
⟹ 18 as remainder
Its now time to involve number after decimal point.
(b) Do the following Steps
⟹ Put decimal in the quotient
⟹ Involve number after decimal point (i.e. number 4) is the division
Hence, 17.8 is the final solution
Example 03
Divide 1.95 by 13
Notice that:
Divided (1.95) is a decimal number
Divisor (13) is a whole number
Since divisor is a whole number. Its possible to use Long Division Method.
Hence 0.15 is the solution of the division.
Let us understand the division process step by step
(a) First divide the number before decimal point
You can see that we got:
⟹ 0 as a quotient
⟹ 1 as remainder
Now involve the number after decimal point and do the rest of division
(b) Do the following steps
⟹ First put decimal point in the quotient
⟹ Divide rest of the number
From the above division, we found that 0.15 is the solution
## Division using Fraction Method
Step 01: Write the division in form of fraction
Step 02: Convert decimal into whole number by multiplying with 10, 100, 1000 etc..
Step 03: Divide the whole numbers to get final answer
Let us understand the process with examples
Example 01
18 divided by 0.2
Step 01: Writing in form of fraction
\Longrightarrow \ \frac{18}{0.2}\
Step 02: Convert decimal into whole number
18 ⟹ is already a decimal
0.2 ⟹ decimal with one decimal place value
To convert 0.2 into whole number, we multiply the fraction with 10
(Multiplication of new number is always done both on numerator & denominator)
\Longrightarrow \ \frac{18\times 10}{0.2\times 10}\\\ \\ \Longrightarrow \ \frac{180}{2}\\\ \\
Step 03: Simply divide the number
From the above division, we found 90 is the solution
Hence, \frac{18}{0.2} \ \Longrightarrow \ 90\
Let us see another example for this method:
Example 02
Divide 9.6 by 0.08
Step 01: Write the division in fraction form
\Longrightarrow \ \frac{9.6}{0.08}\
Step 02: Convert decimals into whole numbers
9.6 ⟹ contains one decimal place value
0.08 ⟹ contains two decimal place value
By Multiplying numerator & denominator by 100, both numbers become whole number
\Longrightarrow \ \frac{9.6\times 100}{0.08\times 100}\\\ \\ \Longrightarrow \ \frac{960}{8}\\\ \\
Step 03: Divide the whole numbers
Hence, 120 is the solution of the above division.
i.e. \frac{9.6}{0.08} \ \Longrightarrow \ 120\
Example 03
Divide 5.83 by 0.011
Step 01: Write the division in fraction form
\Longrightarrow \ \frac{5.83}{0.011}\
Step 02: Convert decimals into whole numbers
5.83 ⟹ contains two decimal place value
0.011 ⟹ contains three decimal place value
We multiply numerator and denominator by 1000 so that both numbers become whole numbers
\Longrightarrow \frac{5.83\times 1000}{0.011\times 1000} \\\ \\ \Longrightarrow \ \frac{5830}{11}
Step 03: Divide the whole numbers
Hence 530 is the solution of the problem
i.e. \frac{5.83}{0.011} \ \Longrightarrow \ 530\
I hope you have understood both the methods.
It is now time to solve some problems related to division of decimals
## Decimal Division Questions
(01) Divide 6.4 by 8
(a) 0.6
(b) 0.8
(c) 0.06
(d) 0.08
\Longrightarrow \frac{6.4}{8} \\\ \\ \Longrightarrow \ \frac{6.4\times 10}{8\times 10}\\\ \\ \Longrightarrow \ \frac{64}{8\times 10}\\\ \\
⟹ First divide 64 by 8
The equation can be written as :
⟹ Now divide 8 by 10
\Longrightarrow \frac{8}{10} \\\ \\ \Longrightarrow \ 0.8\
Hence 0.8 is the solution
Option (b) is the right answer
(02) Divide 1.2 by 3
(a) 0.08
(b) 0.04
(c) 0.8
(d) 0.4
\Longrightarrow \frac{1.2}{3} \\\ \\ \Longrightarrow \ \frac{1.2\times 10}{3\times 10}\\\ \\ \Longrightarrow \ \frac{12}{3\times 10}\\\ \\
⟹ First Divide 12 by 3
The equation can be expressed as
⟹ Now divide 4 by 10
\Longrightarrow \frac{4}{10} \\\ \\ \Longrightarrow \ 0.4\
Hence 0.4 is the solution
Option (d) is the right answer
(03) Divide 5.2 by 0.4
(a) 12
(b) 11
(c) 13
(d) 14
\Longrightarrow \frac{5.2}{0.4} \\\ \\ \Longrightarrow \ \frac{5.2\times 10}{0.4\times 10}\\\ \\ \Longrightarrow \ \frac{52}{4}\\\ \\
On dividing 52 by 4 we get:
Hence 13 is the solution
Option (c) is the right answer
(04) Divide 4.9 by 7
(a) 0.7
(b) 0.8
(c) 0.9
(d) 0.6
\Longrightarrow \frac{4.9}{7} \\\ \\ \Longrightarrow \ \frac{4.9\times 10}{7\times 10}\\\ \\ \Longrightarrow \ \frac{49}{7\times 10}\\\ \\
Dividing 49 and 7, we get number 7 as quotient
The equation can be written as:
⟹ Now divide 7 by 10
\Longrightarrow \frac{7}{10} \\\ \\ \Longrightarrow \ 0.7\
Hence 0.7 is the solution
Option (a) is the right answer
(05) Divide 2.4 by 0.6
(a) 6
(b) 8
(c) 4
(d) 2
\Longrightarrow \frac{2.4}{0.6} \\\ \\ \Longrightarrow \ \frac{2.4\times 10}{0.6\times 10}\\\ \\ \Longrightarrow \ \frac{24}{6}\\\ \\
Hence, 4 is the solution of the division
Option (c) is the right answer
(06) Divide 8.64 by 1.8
(a) 4.2
(b) 4.8
(c) 3.6
(d) 3.8
\Longrightarrow \frac{8.64}{1.8} \\\ \\ \Longrightarrow \ \frac{8.64\times 100}{1.8\times 100}\\\ \\ \Longrightarrow \ \frac{864}{180}\\\ \\
Dividing 864 by 180
Hence, 4.8 is the solution
Option (b) is the right answer
(07) Divide 13.11 by 0.19
(a) 9.6
(b) 6.9
(c) 96
(d) 69
\Longrightarrow \frac{13.11}{0.19} \\\ \\ \Longrightarrow \ \frac{13.11\times 100}{0.19\times 100}\\\ \\ \Longrightarrow \ \frac{1311}{19}\\\ \\
Dividing 1311 by 19, we get:
Hence 69 is the solution
Option (d) is the right answer
(08) Divide 36.75 by 3.5
(a) 10.5
(b) 9.7
(c) 10.8
(d) 9.5
\Longrightarrow \frac{36.75}{3.5} \\\ \\ \Longrightarrow \ \frac{36.75\times 100}{3.5\times 100}\\\ \\ \Longrightarrow \ \frac{3675}{350}\\\ \\
Dividing 3675 by 350, we get:
Hence, 10.5 is the solution
Option (a) is the right answer
(09) Divide 60.2 by 70
(a) 8.6
(b) 0.89
(c) 0.95
(d) 0.86
\Longrightarrow \frac{60.2}{70} \\\ \\ \Longrightarrow \ \frac{60.2\times 10}{70\times 10}\\\ \\ \Longrightarrow \ \frac{602}{700}\\\ \\
Dividing 602 by 700 using Long Division Method
0.86 is the solution
Option (d) is the right answer
(10) Divide 3012.54 by 35.4
(a) 85.1
(b) 85.2
(c) 85.3
(d) 85.4
\Longrightarrow \frac{3012.54}{35.4} \\\ \\ \Longrightarrow \ \frac{3012.54\times 100}{35.4\times 100}\\\ \\ \Longrightarrow \ \frac{301254}{3540}\\\ \\
Dividing 301254 by 3540
Hence 85.1 is the solution
option (a) is the right answer
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# Coordinate Geometry: Understanding the Cartesian Coordinate System and Plotting Points, Lines, and Distances.
Using the Cartesian coordinate system, this area of mathematics provides an effective method for representing and examining shapes, patterns, and relationships. This voyage into coordinate geometry aims to be both educational and useful, whether you’re a student, an aspiring mathematician, or someone wishing to brush up on your skills.
A coordinate system is used to represent and analyze geometric forms, points, lines, and equations in the field of mathematics known as coordinate geometry, also known as Cartesian geometry. The basic elements of coordinate geometry are as follows:
• Coordinate system: The horizontal x-axis and the vertical y-axis are two perpendicular number lines that make up a cartesian coordinate system.
• Coordinates: The ordered pairs (x, y) that make up this system’s coordinates serve to represent points. While the y-coordinate denotes the vertical position, the x-coordinate indicates the horizontal position.
• Plotting Points: You may see and deal with geometric figures and data by placing a point on the plane depending on its coordinates.
• Equations of lines: Linear equations can be used to describe lines. Two popular types are the slope-intercept form (y = mx + b) and the point-slope form (y – y1 = m(x – x1)). These equations assist inline analysis and graphing.
• Formulas for calculating distance and midpoint: Coordinate geometry offers formulas for determining the separation of two points and the midpoint of a line segment, allowing for a variety of applications in mathematics and science.
• Slope: Slope quantifies a line’s steepness and is crucial to comprehending lines and their characteristics.
## The Ideas Behind and Uses for Cartesian Coordinate Geometry
Analytical geometry: Using mathematics and algebra, one can explore geometric shapes using Cartesian coordinates. You can examine characteristics such as symmetry, crossings, and tangents by expressing figures as equations.
• Function Graphs: Plotting functions on a Cartesian plane is a fundamental algebraic and calculus tool. This enables you to comprehend how several functions interact with one another, visualize the behaviour of functions, and pinpoint crucial spots.
• Vector and Vector Operations: Cartesian coordinates are essential to the study of vectors and vector operations. Addition and subtraction are performed using coordinate components, and vectors are visualized as points in space.
• Parametric Equations: Cartesian coordinates can also be used with parametric equations. These equations enable more intricate and dynamic geometric representations by describing how a point moves as it follows a path through space.
• Applications in Science and Engineering: Cartesian coordinates are used in physics to simulate the motion of objects, in engineering for structural analysis and design, and in computer science for computer graphics, navigation systems, and other applications.
• Integration with Technology: Graphing calculators and computer software have made it easier to employ Cartesian coordinates in the current day. Plotting points, graphing functions, and visualizing complicated data sets are all made simpler by them.
## Plotting Points and Coordinates:
Ordered pairs (x, y) are used to define points in the Cartesian plane, where:
1. The letter “x” stands for the x-axis’ horizontal position.
2. On the y-axis, “y” denotes the vertical position.
3. Plot a point by doing the following:
4. Locate the origin (0, 0), which is where the axes meet.
5. Move the quantity of units denoted by “x” horizontally (to the right or left) from the origin.
6. From the origin, move y units vertically (up or down).
7. Make a note of the intersection of your horizontal and vertical movements.
## Important Points to Remember About the Cartesian Coordinate System
• The origin is defined as the point (0, 0) where the two axes converge.
• There are countless potential points on a Cartesian coordinate plane.
• There are no points in any quadrant that are on any of the number lines.
## Some Typical Applications of Cartesian Geometry:
The ideas of Cartesian geometry are vital in many disciplines and are applied in a wide range of contexts. Following are some typical applications of Cartesian geometry:
• Mathematics: Mathematical foundations include Cartesian geometry. Mathematicians use it to graph equations, functions, and geometric figures in order to study relationships between variables and address mathematical issues.
• Physics: The positions and motions of objects in space are described using Cartesian geometry. It is crucial for deciphering vectors, interpreting motion, and illustrating physical processes.
• Engineering: For design, analysis, and modelling, engineers use Cartesian geometry.
• Navigation: GPS and other navigational technologies rely heavily on Cartesian geometry.
• Architecture: To create stable and aesthetically beautiful structures, architects use Cartesian geometry while designing buildings. It is frequently utilized in blueprints and architectural designs.
• Surveying: To precisely map and measure land parcels, land surveyors use Cartesian coordinates.
• Economics and Business: To graph economic data, such as supply and demand curves, and to model financial data, economists and business analysts employ Cartesian geometry.
• Data Analysis: Scatter plots, data visualization, and the examination of interrelationships between variables are all done using Cartesian geometry in data science and statistics.
• Geometry and Trigonometry: Cartesian coordinates are essential in geometry because they help define forms and angles. They are also important in trigonometry. A branch of mathematics called trigonometry also studies triangles and angles using Cartesian geometry.
• Geometry-Based Software: Software developed on the ideas of Cartesian geometry includes CAD (Computer-Aided Design) and GIS (Geographic Information Systems) programs.
In essence, Cartesian geometry is used to establish, analyse, or visualise precise spatial connections. It is a fundamental tool in mathematics and has several practical applications since it offers an all-encompassing and incredibly versatile method for working with points, lines, and objects in two or three dimensions.
Cartesian geometry is introduced in EuroSchool using a kid-centered method. Through visual aids and practical exercises, kids learn the fundamentals of the Cartesian plane first. Using relevant examples, they learn how to plot points, draw lines, and solve basic equations. Concepts are made interesting by using real-world examples, such as route mapping.
Children learn about coordinates through interactive games and puzzles, and with age-appropriate software, they advance to increasingly difficult activities like graphing functions. To prepare young minds for complex mathematical ideas, the focus is on developing a strong foundation in geometry, promoting discovery, and making learning fun.
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Home » What is 0.75 as a Fraction? – Decimal to Fraction Conversion
# What is 0.75 as a Fraction? – Decimal to Fraction Conversion
What is 0.75 as a Fraction?
• A. 17/34
• B. 11/20
• C. 1/2
• D. 3/4
• E. 1/8
## Solution:
The fraction 75/100 is the same as the decimal 0.75.
But you can make this even easier by dividing the numerator and the denominator by their largest common factor. In this case, you can divide both 75 and 100 by 25.
When we do this, we get (75 ÷ 25) / (100 ÷ 25) = 3/4.
So, 0.75 as a fraction is 3/4
##### Explanation
Converting decimals to fractions can seem hard initially, but it’s not too hard once you know how to do it. Today, we’ll look at how to change 0.75 from its decimal to its fraction form.
## Understanding the Decimal
0.75 is a decimal number with two digits after the decimal point. The first number shows how many tenths, and the second number shows how many hundredths. This means that 0.75 is the same as 75 hundredths.
## Conversion to a Fraction
Once we know what each place in a decimal means, converting it into a fraction is easy. 0.75 can be written as the fraction 75/100, where 75 is the number of hundredths and 100 is the total number of parts that make up a whole.
## Simplifying the Fraction
We divide the numerator and denominator by their greatest common divisor to simplify this fraction. 25 is the greatest common divisor of the numbers 75 and 100. Therefore, both parts of the fraction are divided by 25, resulting in (75 ÷ 25) / (100 ÷ 25) = 3/4.
## The Bottom Line
In conclusion, the fraction 3/4 can be expressed as the decimal 0.75. This conversion process—understanding the place values of the decimal, writing the decimal as a fraction, and simplifying that fraction—enables us to convert any decimal to its equivalent fraction form easily.
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A series of numbers that follows a particular pattern is called a sequence. Sometimes, each new term is found by adding or subtracting a certain constant, sometimes by multiplying or dividing. So long as the pattern is the same for every new term, the numbers are said to lie in a sequence.
Sequence questions will have multiple moving parts and pieces, and you will always have several different options to choose from in order to solve the problem. We’ll walk through all the methods for solving sequence questions, as well as the pros and cons for each. You will likely see two sequence questions on any given SAT, so keep this in mind as you find your perfect balance between time strategies and memorization.
This will be your complete guide to SAT sequence problems--the types of sequences you’ll see, the typical sequence questions that appear on the SAT, and the best ways to solve these types of problems for your particular SAT test taking strategies.
What Are Sequences?
You will see two different types of sequences on the SAT--arithmetic and geometric.
An arithmetic sequence is a sequence wherein each successive term is found by adding or subtracting a constant value. The difference between each term--found by subtracting any two pairs of neighboring terms--is called \$d\$, the common difference.
14, 11, 8, 5… is an arithmetic sequence with a common difference of -3. We can find the \$d\$ by subtracting any two pairs of numbers in the sequence, so long as the numbers are next to one another.
\$11 - 14 = -3\$
\$8 - 11 = -3\$
\$5 - 8 = -3\$
14, 17, 20, 23... is an arithmetic sequence in which the common difference is +3. We can find this \$d\$ by again subtracting pairs of numbers in the sequence.
\$17 - 14 = 3\$
\$20 - 17 = 3\$
\$23 - 20 = 3\$
A geometric sequence is a sequence of numbers in which each new term is found by multiplying or dividing the previous term by a constant value. The difference between each term--found by dividing any neighboring pair of terms--is called \$r\$, the common ratio.
64, 16, 4, 1, … is a geometric sequence in which the common ratio is \$1/4\$. We can find the \$r\$ by dividing any pair of numbers in the sequence, so long as they are next to one another.
\$16/64 = 1/4\$
\$4/16 = 1/4\$
\$1/4 = 1/4\$
Sequence Formulas
Luckily for us, sequences are entirely regular. This means that we can use formulas to find any piece of them we choose, such as the first term, the nth term, or the sum of all our terms.
Do keep in mind, though, that there are pros and cons for memorizing formulas.
Pros--formulas provide you with a quick way to find your answers. You do not have to write out the full sequence by hand or spend your limited test-taking time tallying your numbers (and potentially entering them wrong into your calculator).
Cons--it can be easy to remember a formula incorrectly, which would be worse than not having a formula at all. It also is an expense of brainpower to memorize formulas.
If you are someone who prefers to work with formulas, definitely go ahead and learn them! But if you despise using formulas or worry that you will not remember them accurately, then you are still in luck. Most SAT sequence problems can be solved longhand if you have the time to spare, so you will not have to concern yourself with memorizing your formulas.
That all being said, it’s important to understand why the formulas work, even if you do not plan to memorize them. So let’s take a look.
Arithmetic Sequence Formulas
\$\$a_n = a_1 + (n - 1)d\$\$
\$\$\Sum \terms = (n/2)(a_1 + a_n)\$\$
These are our two important arithmetic sequence formulas. We’ll look at them one at a time to see why they work and when to use them on the test.
Terms Formula
\$a_n = a_1 + (n - 1)d\$
This formula allows you to find any individual piece of your arithmetic sequence--the 1st term, the nth term, or the common difference. First, we’ll look at why it works and then look at some problems in action.
\$a_1\$ is the first term in our sequence. Though the sequence can go on infinitely, we will always have a starting point at our first term. (Note: you can also assign any term to be your first term if you need to. We’ll look at how and why we can do this in one of our examples.)
\$a_n\$ represents any missing term we want to isolate. For instance, this could be the 4th term, the 58th, or the 202nd.
So why does this formula work?
Imagine that we wanted to find the 2nd term in a sequence. Well each new term is found by adding the common difference, or \$d\$. This means that the second term would be:
\$a_2 = a_1 + d\$
And we would then find the 3rd term in the sequence by adding another \$d\$ to our existing \$a_2\$. So our 3rd term would be:
\$a_3 = (a_1 + d) + d\$
Or, in other words:
\$a_3 = a_1 + 2d\$
If we keep going, the 4th term of the sequence--found by adding another \$d\$ to our existing third term--would continue this pattern:
\$a_4 = (a_1 + 2d) + d\$
\$a_4 = a_1 + 3d\$
We can see that each term in the sequence is found by adding the first term, \$a_1\$, to a \$d\$ that is multiplied by \$n - 1\$. (The 3rd term is \$2d\$, the 4th term is \$3d\$, etc.)
So now that we know why the formula works, let’s look at it in action.
Now, there are two ways to solve this problem--using the formula, or simply counting. Let’s look at both methods.
Method 1--arithmetic sequence formula
If we use our formula for arithmetic sequences, we can find our \$a_n\$ (in this case \$a_12\$). So let us simply plug in our numbers for \$a_1\$ and \$d\$.
\$a_n = a_1 + (n - 1)d\$
\$a_12 = 4 + (12 - 1)7\$
\$a_12 = 4 + (11)7\$
\$a_12 = 4 + 77\$
\$a_12 = 81\$
Our final answer is B, 81.
Method 2--counting
Because the difference between each term is regular, we can find that difference by simply adding our \$d\$ to each successive term until we reach our 12th term.
Of course, this method will take a little more time than simply using the formula, and it is easy to lose track of your place. The test makers know this and will provide answers that are one or two places off, so make sure to keep your work organized so that you do not fall for bait answers.
First, line up your twelve terms and then fill in the blanks by adding 7 to each new term.
4, 11, 18, ___, ___, ___, ___, ___, ___, ___, ___, ___
4, 11, 18, 25, ___, ___, ___, ___, ___, ___, ___, ___
4, 11, 18, 25, 32, ___, ___, ___, ___, ___, ___, ___
And so on, until you get:
4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81
Again, the 12th term is B, 81.
Sum Formula
\$\Sum \terms = (n/2)(a_1 + a_n)\$
Our second arithmetic sequence formula tells us the sum of a set of our terms in a sequence, from the first term (\$a_1\$) to the nth term (\$a_n\$).
Basically, we do this by multiplying the number of terms, \$n\$, by the average of the first term and the nth term.
Why does this formula work?
Well let’s look at an arithmetic sequence in action:
10, 16, 22, 28, 34, 40
This is an arithmetic sequence with a common difference, \$d\$, of 6.
A neat trick you can do with any arithmetic sequence is to take the sum of the pairs of terms, starting from the outsides in. Each pair will have the same exact sum.
So you can see that the sum of the sequence is \$50 * 3 = 150\$.
In other words, we are taking the sum of our first term and our nth term (in this case, 40 is our 6th term) and multiplying it by half of \$n\$ (in this case \$6/2 = 3\$).
Another way to think of it is to take the average of our first and nth terms--\${10 + 40}/2 = 25\$ and then multiply that value by the number of terms in the sequence--\$25 * 6 = 150\$.
Either way, you are using the same basic formula. How you like to think of the equation and whether or not you prefer \$(n/2)(a_1 + a_n)\$ or \$n({a_1 + a_n}/2)\$, is completely up to you.
Now let’s look at the formula in action.
Kyle started a new job as a telemarketer and, every day, he is supposed to make 3 more phone calls than the day previous. If he made 10 phone calls his first day, and he meets his goal, how many total phone calls does he make in his first two weeks, if he works every single day?
1. 413
2. 416
3. 426
4. 429
5. 489
As with almost all sequence questions on the SAT, we have the choice to use our formulas or do the problem longhand. Let’s try both ways.
Method 1--formulas
We know that our formula for arithmetic sequence sums is:
\$\Sum = (n/2)(a_1 + a_n)\$
But, we must first find the value of our \$a_n\$ in order to use this formula. Once again, we can do this via our first arithmetic sequence formula, or we can find it by hand. As we are already using formulas, let us use our first formula.
\$a_n = a_1 + (n - 1)d\$
We are told that Kyle makes 10 phone calls on his first day, so our \$a_1\$ is 10. We also know that he makes 3 more calls every day, for a total of 2 full weeks (14 days), which means our \$d\$ is 3 and our \$n\$ is 14. We have all our pieces to complete this first formula.
\$a_n = a_1 + (n - 1)d\$
\$a_14 = 10 + (14 - 1)3\$
\$a_14 = 10 + (13)3\$
\$a_14 = 10 + 39\$
\$a_14 = 49\$
And now that we have our value for \$a_n\$ (in this case \$a_14\$), we can complete our sum formula.
\$(n/2)(a_1 + a_n)\$
\$(14/2)(10 + 49)\$
\$7(59)\$
\$413\$
Our final answer is A, 413.
Method 2--longhand
Alternatively, we can solve this problem by doing it longhand. It will take a little longer, but this way also carries less risk of incorrectly remember our formulas. As always, how you choose to solve these problems is completely up to you.
First, let us write out our sequence, beginning with 10 and adding 3 to each subsequence number, until we find our nth (14th) term.
10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49
Now, we can either add them up all by hand--\$10 + 13 + 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 + 43 + 46 + 49 = 413\$
Or we can use our arithmetic sequence sum trick and divide the sequence into pairs.
We can see that there are 7 pairs of 59, so \$7 * 59 = 413\$.
Again, our final answer is A, 413.
Only one more formula to go. Almost there!
Geometric Sequence Formulas
\$\$a_n = a_1( r^{n - 1})\$\$
(Note: while there is a formula to find the sum of a geometric sequence, but you will never be asked to find this on the SAT, and so it is not included in this guide.)
As with the first arithmetic sequence formula, this formula will allow you to find any number of missing pieces, including your 1st term, your nth term, or your \$r\$.
And, as always with sequences, you have the choice of whether to solve your problem longhand or with a formula.
Method 1--formula
If you’re one for memorizing formulas, we can simply plug in our values into our equation in place of \$a_n\$, \$n\$, and \$r\$ in order to solve for \$a_1\$.
We are told that Mr. Smith has 1 dollar 5 days later, which would be the 6th day (meaning our \$n\$ is 6), and that the ratio between each term is \$1/4\$.
\$a_n = a_1( r^{n - 1})\$
\$1 = a_1({1/4}^{6 - 1})\$
\$1 = a_1({1/4}^5)\$
\$1 = a_1(0.00097656)\$
\$1/0.00097656 = a_1\$
\$1024 = a_1\$
So the 1st term in the sequence is 1024, which means that Mr. Smith starts with \$1024 on Monday morning.
Method 2--longhand
Alternatively, we can, as always, solve them problem by hand.
First, set out our number of terms in order to keep track of them, with our 7th term, \$1, last.
___, ___, ___, ___, ___, 1
Now, because our ratio is \$1/4\$ and we are working backwards, we must multiply each term by 4. (Why? Because \${1/{(1/4)} = 1 * 4\$, according to the rules of fractions).
___, ___, ___, ___, 4, 1
___, ___, ___, 16, 4, 1
And, if we keep going, we will eventually get:
1024, 256, 64, 16, 4, 1
Which means that we can see that our 1st term is 1024.
Again, our final answer is 1024.
As with all sequence solving methods, there are benefits and drawbacks to solving the question in each way. If you choose to use formulas, make very sure you can remember them exactly.
And if you solve the questions by hand, be very careful to find the exact number of terms in the sequence. It can be all too easy to accidentally find one term more or fewer if you’re not carefully labeling or otherwise keeping track of your terms.
I'm preeeeetty sure it's not a proper math formula unless mystery variables and exploding test tubes are involved somehow.
Typical SAT Sequences Questions
Because all sequence questions on the SAT can be solved without the use or knowledge of sequence formulas, the test-makers will only ever ask you for a limited number of terms or the sum of a small number of terms (usually 12 or fewer).
As we saw above, you may be asked to find the 1st term in a sequence, the nth term, the difference between your terms (whether a common difference, \$d\$, or a common ratio, \$r\$), or the sum of your terms (in arithmetic sequences only).
You also may be asked to find an unusual twist on a sequence question that combines your knowledge of sequences or your knowledge of sequences and other SAT math topics.
For example:
Again, let us look at both formulaic and longhand methods for how to solve a problem like this.
Method 1--formulas
We are told that the ratio between the terms in our sequence is 2:1, successive term to previous term. This means that our common ratio is 2, as each term is being multiplied by 2 in order to find the next term. (Note: if you are not familiar with ratios, check out our guide to SAT ratios.)
Now, we can find the ratio between our 8th and 5th terms in a few different ways, but the simplest way--while still using formulas--is simply to reassign our 5th term as our 1st term instead. This would then make our 8th term become our 4th term.
(Why the 4th term? The 5th and 8th terms are 3 spaces from each other--5th to 6th, 6th to 7th, and 7th to 8th--which means our 1st term must be 3 spaces from our new nth term--1st to 2nd, 2nd to 3rd, 3rd to 4th).
Once we’ve designated our 5th term as our 1st term, we can use the strategy of plugging in numbers and assign a random value for our \$a_1\$. Then we will plug in our known values of \$r\$ (2) and \$n\$ (3) so that we can find our \$a_n\$.
Let us call \$a_1\$ 4. (Why 4? Why not!)
\$a_n = a_1( r^{n - 1})\$
\$a_4 = 4(2^{4 - 1})\$
\$a_4 = 4(2^3)\$
\$a_4 = 4(8)\$
\$a_4 = 32\$
So the ratio between our 4th term and our 1st term (the equivalent of the ratio to our 8th term and our 5th term) is:
\$32:4\$
Or, when we reduce:
\$8:1\$
The ratio between our 8th term and our 5th term is \$8:1\$
Our final answer is C, \$8:1\$.
As you can see, this problem was tricky because we had to reassign our terms and use our own numbers before we even considered having to use our formulas. Let us look at this problem were we to solve it longhand instead.
Method 2--longhand
If we choose to solve this problem longhand, we will not have to concern ourselves with reassigning our terms, but we will still have to understand that there are 3 spaces between our 8th and our 5th terms (8th to 7th, 7th to 6th, and 6th to 5th).
Since we used the technique of plugging in our own numbers last time, let us use algebra for our longhand method.
We know that each term is found by doubling the previous term. So let us say that our 5th term is \$x\$.
___, ___, ___, ___, x, ___, ___, ___
This would make our 6th term \$2x\$.
___, ___, ___, ___, x, 2x, ___, ___
And we can continue down the line until we get:
___, ___, ___, ___, x, 2x, 4x, 8x
This means that our ratio between our 8th term and our 5th term is:
\$8x:x\$
Or, in other words:
\$8:1\$
Our final answer is, again, C, \$8:1\$.
Again, you always have the choice to use formulas or longhand to solve these questions and how you prioritize your time (and/or how careful you are with your calculations) will ultimately decide which method you use.
Now let's take a look at our SAT sequence question strategies.
Tips For Solving Sequence Questions
Sequence questions can be somewhat tricky and arduous to work through, so keep in mind these SAT math tips on sequences as you go through your studies:
1) Decide before test day whether or not you will use the sequence formulas
Before you go through the effort of committing your formulas to memory, think about the kind of test-taker you are. If you are someone who loves to use formulas, then go ahead and memorize them now. Most sequence questions will go much faster once you have gotten used to using your formula.
However, if you would rather dedicate your time and brainpower to other math topics or if you would simply rather solve sequence questions longhand, then don’t worry about your formulas! Don’t even bother to try to remember them--just decide here and now not to use them and save your mental energy for other pursuits.
Unless you can be sure to remember them correctly, formulas will hinder more than help you on test day. So make the decision now to either memorize your formulas or forget about them entirely.
Though many calculators can perform long strings of calculations, sequence questions by definition involve many different values and terms. Small errors in your work can cause a cascade effect and one mistyped digit in your calculator can throw off your work completely. Even worse, you won’t know where the error happened if you do not keep track of your values.
Always write down your values and label your terms in order to prevent a misstep somewhere down the line.
3) Keep careful track of your timing
No matter how you solve a sequence question, these types of problems will generally take you more time than other math questions on the SAT. For this reason, most sequence questions are located in the last third of any particular SAT math section, which means the test-makers think of sequences as a “high difficulty” level problem.
Time is your most valuable asset on the SAT, so always make sure you are using yours wisely. If you feel you can (accurately) answer two other math questions in the time it takes you to answer one sequence question, then maximize your point gain by focusing on the other two questions.
Always remember that each question on the SAT math section is worth the same amount of points and you will get dinged if you get a question wrong. Prioritize both your quantity of answered questions as well as your accuracy, and don’t let your time run out trying to solve one problem.
If you feel that you can answer a sequence problem quickly, go ahead! But if you feel it will take up too much time, move on and come back to it later (or skip it entirely, if you need to).
No matter which method you choose to use, trust that you'll find the one that best suits your needs and abilities.
Now let’s test your sequence knowledge with real SAT math problems.
1)
2) What is the sum of the first 10 terms in the arithmetic sequence that begins: 13, 21, 29,...
1. 450
2. 458
3. 474
4. 482
5. 490
3)
1) The number of squirrels triples every three years, so this is a geometric sequence. As always, we can either count longhand or use our formulas. Let’s look at each way.
We first need to count how many times three years has passed between 1990 and 1999.
Including the year 1990 and the year 1999, there are 4 terms for every 3 years between 1990 and 1999.
1990, 1993, 1996, 1999
This means that 1999 is our 4th term and 1990 is our 1st term. Now let’s plug in our values into our formula.
\$a_n = a_1( r^{n - 1})\$
\$5400 = a_1(3^{4-1})\$
\$5400 = a_1(3^3)\$
\$5400 = a_1(27)\$
\$200 = a_1\$
Our first term is 200.
There were 200 squirrels in 1990.
Alternatively, we can simply find the number of squirrels in 1990 by counting by hand.
Again, we need to find the number of groups of 3 years between 1990 and 1999, inclusive.
1990, 1993, 1996, 1999
Now, let us plug in our known value for 1999 and find the rest of our terms by dividing each term by 3.
___, ___, ___, 5400
___, ___, 1800, 5400
And so on, until you get:
200, 600, 1800, 5400
Again, our first term is 200.
There were 200 squirrels in 1990.
2) We are asked to find the sum of this arithmetic sequence, which means we can either use our formula or count our sequence by hand.
Method 1--formulas
First, we need to determine our common difference, \$d\$, in the sequence. To do so, let us subtract one of our neighboring pairs of numbers.
\$21 - 13 = 8\$
Before we can find our sum, however, we must find our \$a_10\$. This means we need to use our first arithmetic sequence formula:
\$a_n = a_1 + (n - 1)d\$
\$a_10 = 13 + (10 - 1)8\$
\$a_10 = 13 + 72\$
\$a_10 = 85\$
Now that we know our \$d\$ and our \$a_10\$, we can plug in our values to find our sum.
\$(n/2)(a_1 + a_n)\$
\$(10/5)(13 + 85)\$
\$(5)(98)\$
\$490\$
Our final answer is E, 490.
Method 2--counting
If you do not want to remember or use your formulas, you can always find your answer by counting.
First, we must still determine our \$d\$ by subtracting our neighboring terms:
\$29 - 21 = 8\$
Now, we can find the value of all our terms by continuing to add 8 to each new term until we reach our 10th term.
13, 21, 29, ___, ___, ___, ___, ___, ___, ___
13, 21, 29, 37, ___, ___, ___, ___, ___, ___
13, 21, 29, 37, 45, ___, ___, ___, ___, ___
And so on, until we finally get:
13, 21, 29, 37, 45, 53, 61, 69, 77, 85
Now, we can either add them up individually (\$13 + 21 + 29 + 37 + 45 + 53 + 61 + 69 + 77 + 85 = 490\$), or you can, find your pairs of numbers, beginning from the outside in.
We can see that there are 5 pairs of 98, so \$5 * 98 = 450\$
Our final answer is E, 490.
3) Because the price of our mystery item raises by \$2 every year, this is an arithmetic sequence. Again, we have multiple ways to solve this kind of problem--using formulas, or counting longhand.
Method 1--formulas
\$a_n = a_1 + (n - 1)d\$
\$100 = 10 + (n - 1)2\$
\$100 = 10 + 2n - 2\$
\$100 = 8 + 2n\$
\$92 = 2n\$
\$n = 46\$
Now, we know that 100 is the price at our 46th term, but this is not the same thing as 46 years from 1990. Remember: the number of terms from the 1st is always 1 fewer space than the actual count of the term.
For instance, the 1st term in a sequence is 4 spaces from the 5th term and 5 spaces from the 6th term. Why?
1st to 2nd, 2nd to 3rd, 3rd to 4th, 4th to 5th. We can see it takes 4 total spaces to go from the 1st term to the 5th.
For our price problem, our \$n\$ is 46, which means that the year will be \$46 - 1 = 45\$ actual spaces away from our starting term.
So:
\$1990 + (46 - 1)\$
\$1990 + 45\$
\$2035\$
The price will be \$100 in 2035.
Method 2--counting
Because each new term is determined by adding 2, it will take us a long time to get from 10 to 100. We can speed up this process by first finding the difference between the 1st and last term:
\$100 - 10 = 90\$
And then we can divide this difference by the common difference, \$d\$:
\$90/2 = 45\$
It will take 45 years to get to the price to raise to \$100. 45 years after 1990 is:
\$1990 + 45\$
\$2035\$
Again, the price will be \$100 in 2035.
Yeah! You toppled those sequence questions!
The Take Aways
Though sequence questions can take some little time to work through, they are usually made complicated by their number of terms and values rather than being actually difficult to solve.
So long as you remember to keep all your work organized and decide before test-day whether or not you want to spend your study efforts memorizing, and you’ll be able to tackle any number of sequence questions the SAT can throw your way. As long as you keep your values straight (and don’t get tricked by bait answers!), you will be able to grind through these problems without fail.
What’s Next?
Now that you've taken on sequences and dominated, it's time to make sure you have a solid handle on the rest of your SAT math topics. The SAT presents familiar concepts in unfamiliar ways, so check out our guides on all your individual SAT topic needs. We'll provide you with all the strategies and practice problems on any SAT math topic you could ask for.
Running out of time on SAT math? Not to worry! Our guide will show you how to maximize both your time and your score so that you can make the most of your time on test day.
Don't know what score to aim for? Follow our simple steps to figure out what score is best for you and your needs.
Looking to get a perfect score? Check out our guide to getting a perfect 800 on SAT math, written by a perfect-scorer!
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# Assignment 7
## Tangent Circles
Kate Berryman
cavaleri@uga.edu
*Please make sure your browser is maxiumized to view this write up;
In this write up, we will discuss different cases of tangent circles.
First let's look at how to construct a tanget circle.
First, construct two circles with center A and center B. (We will look at the case when B is inside of A, however, when completed, B can be moved.) Construct point P to be any point on the circle A Construct a line passing through AC. Construct a radius on circle B to any point. Let's call this point D Construct a circle with center C and radius of circle BD. Let's call this point E. Construct a line segment connecting E and B. Find its midpoint F. Construct a line that is perpendicular to BE. Where this line intersects line AC, call this point G. Construct line segments between GE and GB. Now, construct a circle with center G and radius of CB. As C moves around circle A, you can see that circle FC is our tangent circle. Click Here to have interaction with this tangent circle!!! Move C around the circle A and see what happens! Click Here to see what happens when circle with center A and circle with center B are disjoint! You can also see what happens when circle with center A intersects circle with center B! You can also look in Assignment #5 for the script tools.
## Exploration #1When Circle B is Inside Circle A
As in our construction, we see the example of when our Circle B is completely contained in Circle A. What happens when we animated our point C and trace our point G?
We see that this gives us an ellipse. Let's remind ourselves what an ellipse is. To be an ellipse, the two focii, in our case points A and B, and their distance to G must be a positive constant.
Let's look back at our construction.
In order to prove this is an ellipse, we must show that AG+BG=a postive contant no matter where C is on circle A. Since line GF is perpendicular to the midpoint of BE, then BG=EG. We also know that AE is equal to the length of the radius of circle B and circle A. Therefore:
AE = AG + GE = AG + BG
So no mattter where C is on circle A, we know that AG+BG is always a positive constant, so this is an ellipse with focii A and B.
## Exploration #2When Circle B and Circle A Intersect
As in Exploration #1, let's trace point G and see what happens as C moves around the perimeter of A.
Again, we see what appears to be an ellipse. Let's look at our original construction to see if we can prove that this is true.
As in Exploration #1, we have that BG = GE and that AE is equal to the radius of circle A and circle B. Therefore, again
AE = AG + GE = AG + BG
So AG + BG will always be a postive constant no matter the point of P on circle C.
## Exploration #3When Circle B is outside of Circle A
Now, let's animate point C when circle B is outside circle A.
Here we get a very different picture. Instead of an ellipse, we get a hyperbola. We still have to prove that our distance from AG+BG is always a constant positive. Let's look at our original construction.
So we know that GB = GE and that GE = GA + AE. Therefore GA = GE - AE or GA = GB - AE. Also, AE = AC + CE. So, we can conclude that AG + BG will always be a postive constant no matter where P is on the perimeter of circle A.
Now we will begin to look at Circle B is constructed inside the Circle G
## Exploration #4When Circle B is inside of Circle A
Again, we know that DG = AG. Also, BD = BG +DG. Therefore:
BD = BG + DG = BG + AG.
This again gives us a positive number for AG+BG. So no matter the point on P on circle A, these will remain positive constants. Therefore, A and B are focii of our ellipse.
## Exploration #5When Circle B and Circle A Intersect
So from construction, AG = DG. Notice that DG = BD-BG or DG = BG + BD. Therefore, again we have the result of:
BD = BG + DG = BG + AG.
## Exploration #6When Circle B is outside of Circle A
This is also a hyperbola. Again, we have AG=DG. Also, DG - BG = BD or DG = BG + BD. Therefore, again we have the result of:
BD = BG + DG = BG + AG.
To summarize we notice that the loci of the center of the two circles form either a hyperbola or an ellipse. When our circle B was not contained inside the tangent circle, then the trace of the center point of the tangent circle formed a hyperbola. When our circle B was contained or intersected our tangent circle, then the trace of the center point of our tangent circle formed an ellipse. In all explorations, we see that A and B are the focii of either an ellipse or a hyperbola.
Return the Kate's Homepage
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Lesson 24 of 47
In Progress
# Quick Sort
##### Yasin Cakal
Quicksort is a popular sorting algorithm that is widely used in computer science and programming. It is an efficient, in-place sorting algorithm that is based on the divide-and-conquer approach. Quicksort is a comparison-based sorting algorithm, meaning that it sorts elements by comparing them to each other. It is a recursive algorithm, meaning that it breaks down a problem into smaller problems and solves them. Quicksort is often used for sorting large datasets, due to its efficiency. In this article, we will discuss the basics of Quicksort, its time and space complexity, and provide some code examples in Python.
## What is Quicksort?
Quicksort is a sorting algorithm that works by partitioning a list of elements into two parts – one part with elements that are less than a given pivot element, and the other part with elements that are greater than the pivot element. The algorithm then recursively sorts the two parts. Quicksort has the following steps:
• Select a pivot element from the list.
• Partition the list around the pivot element.
• Recursively sort the two parts.
Let’s look at an example of Quicksort in action.
## Example of Quicksort
Suppose we have a list of numbers: [8, 4, 1, 6, 5, 7, 3, 2] and we want to sort them in ascending order. We can use Quicksort to do this.
First, we select a pivot element. For this example, let’s select the first element, 8. Next, we partition the list around the pivot element. This means that we move all elements that are less than 8 to the left of 8 and all elements greater than 8 to the right of 8. This results in the following partitioned list: [4, 1, 6, 5, 7, 3, 2, 8].
Finally, we recursively sort the two parts. We apply Quicksort to the left part [4, 1, 6, 5, 7, 3, 2] and the right part [8]. This results in the following sorted list: [1, 2, 3, 4, 5, 6, 7, 8].
## How does Quicksort Work?
Quicksort is an efficient sorting algorithm that uses a divide-and-conquer approach. It works by picking a pivot element from the array and then partitioning the array into two sub-arrays. The elements in the first sub-array are all less than the pivot element while the elements in the second sub-array are all greater than the pivot element. The two sub-arrays are then sorted recursively using quicksort until all elements in the array are sorted.
The steps of Quicksort are as follows:
• Pick a pivot element from the array. This is usually the first or the last element.
• Partition the array into two sub-arrays. All elements in the first sub-array should be less than the pivot element while all elements in the second sub-array should be greater than the pivot element.
• Recursively apply Quicksort to the two sub-arrays.
• When all elements in the sub-arrays have been sorted, combine the sorted sub-arrays and return the sorted array.
The advantages of Quicksort are that it is relatively easy to understand, fast, and can be implemented in place. The disadvantages are that it is not stable and can be vulnerable to worst-case scenarios.
## Time Complexity of Quicksort
The time complexity of Quicksort depends on whether it is implemented as an in-place algorithm or not. The best, average, and worst case time complexities of Quicksort are given below.
### Best Case: O(n log n)
The best case time complexity of Quicksort occurs when the partitioning process always results in two equal-sized subarrays. In this case, Quicksort has a time complexity of O(n log n).
### Average Case: O(n log n)
The average case time complexity of Quicksort is also O(n log n). This is because the partitioning process is random and the size of the two subarrays produced is mostly equal.
### Worst Case: O(n^2)
The worst case time complexity of Quicksort occurs when the partitioning process always produces two subarrays of unequal size. In this case, Quicksort has a time complexity of O(n^2).
### Space Complexity of Quicksort
The space complexity of Quicksort is O(log n). This is because Quicksort is an in-place sorting algorithm and it only requires a constant amount of extra space for the partitioning process.
## Example of Quicksort in Python
We can implement Quicksort in Python using the following code:
def quicksort(arr):
if len(arr) <= 1:
return arr
pivot = arr[len(arr) // 2]
left = [x for x in arr if x < pivot]
middle = [x for x in arr if x == pivot]
right = [x for x in arr if x > pivot]
return quicksort(left) + middle + quicksort(right)
arr = [8, 4, 1, 6, 5, 7, 3, 2]
sorted_arr = quicksort(arr)
print(sorted_arr)
# Output: [1, 2, 3, 4, 5, 6, 7, 8]
## Conclusion
Quicksort is a popular sorting algorithm that is based on the divide-and-conquer approach. It is an efficient, in-place sorting algorithm that has a time complexity of O(n log n) in the best, average, and worst cases and a space complexity of O(log n). Quicksort is often used for sorting large datasets due to its efficiency.
## Exercises
#### Write a Python function to implement Quicksort.
def quicksort(arr):
if len(arr) <= 1:
return arr
pivot = arr[len(arr) // 2]
left = [x for x in arr if x < pivot]
middle = [x for x in arr if x == pivot]
right = [x for x in arr if x > pivot]
return quicksort(left) + middle + quicksort(right)
#### Write a Python function to calculate the time complexity of Quicksort in the worst case.
def quicksort_time_complexity(arr):
if len(arr) <= 1:
return 0
else:
return len(arr) * quicksort_time_complexity(arr[len(arr) // 2:]) + quicksort_time_complexity(arr[:len(arr) // 2])
#### Write a Python function to calculate the space complexity of Quicksort.
def quicksort_space_complexity(arr):
if len(arr) <= 1:
return 0
else:
return len(arr) + quicksort_space_complexity(arr[len(arr) // 2:]) + quicksort_space_complexity(arr[:len(arr) // 2])
#### Write a Python function to sort a list of numbers using Quicksort.
def quicksort(arr):
if len(arr) <= 1:
return arr
pivot = arr[len(arr) // 2]
left = [x for x in arr if x < pivot]
middle = [x for x in arr if x == pivot]
right = [x for x in arr if x > pivot]
return quicksort(left) + middle + quicksort(right)
#### Write a Python function to implement the partitioning step of Quicksort.
def partition(arr, pivot):
left = [x for x in arr if x < pivot]
right = [x for x in arr if x > pivot]
return left, right
|
# Square formula
In this topic square formula, we are going to discuss about two formulas which are being used to expand the terms like in the form (a + b) ². This particular formula (a + b) ² plays a vital role in math from 8 th grade, 9 th grade, 10 th grade to higher studies.
A students who studies math in his school days can never miss this formula. Particularly in the topic Algebra, we can solve many problems by using this formula. We are going to see some of the example problem.After getting clear of using this you can try the worksheet also.
We are giving this worksheet for the purpose of making practice.If you practice this worksheets it will become easy to face problems in the topic algebra.We will use these formulas in most of the problem.
## Example problems of square formula:
Question 1 :
Expand (5x + 3) ²
Solution:
Here the question is in the form of (a+b)². Instead of a we have "5x" and instead of b we have "3" . So we need to apply the formula a² + 2ab + b ² and we need to apply those values instead of a and b
(5x + 3)² = (5x)² + 2 (5x) (3) + (3)²
= 25x² + 30 x + 9
= 25x² + 30 x + 9 square formula
Question 2 :
Expand (x + 2) ²
Solution:
Here the question is in the form of (a+b) ². Instead of a we have "x" and instead of b we have "2" . So we need to apply the formula a² + 2ab + b ² and we need to apply those values instead of a and b
(x + 2)² = (x)² + 2 (x) (2) + (2)²
= x² + 4 x + 4
Question 3 :
If a - b = 3 and a² + b² = 29,find the value of ab.
Solution:
(a - b)² = a² + b² - 2 a b
3² = 29 - 2ab
9 = 29 - 2 ab
2 a b = 29 - 9
2 a b = 20
ab = 20/2
ab = 10
Question 4 :
[√2 + (1/√ 2)]² is equal to
Solution:
(a + b)² = a² + b² + 2 a b
a = √2 b = 1/√2
[√2 + (1/√ 2)]² = ( √2 )² + (1/√2)² + 2 √2 (1/√2)
= 2 + (1/2) + 2
= 4 + (1/2)
= 9/2
r
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Solving $2x+8=6x-12$ by using the guess and check method
How do you solve this equation: $2x+8=6x-12$ by using the guess and check method?
I divide $2x+8$ and I get $4$ then I divide $6x-12$ and I get $-2$ but I don't know what to do next or is it wrong?
• Is it $2^x$, instead of $2x$? For $2x$ there isn't much to guess. – Karolis Juodelė Jun 24 '14 at 17:11
• Try guessing $x = 5$. – JimmyK4542 Jun 24 '14 at 17:12
The answer is $x = 5$. You can get this solution using normal analytical methods (algebraic manipulation), i.e., \begin{align} 2x + 8 = 6x - 12 &\iff -4x + 8 = -12 \\&\iff -4x = -20 \\&\iff x = 5 \,\,. \end{align} In terms of a "guess and check method", here's my strategy: factor $2$ out of the LHS and $6$ out of the RHS to obtain the equation $2(x + 4) = 6(x - 2)$. Now, we can divide $2$ from both sides to obtain $x + 4 = 3(x - 2)$. The solution is fairly easy to see in this form by guessing and checking. You will come to $x = 5$, as before.
Your instructor wants you to try plugging in a bunch of numbers to guess the correct value. Just by eyeballing this, we might try 4 or 5. If you plug in $5$ on the left you get $$2(5)+8=18$$ and on the right you get $$6(5)-12 = 30 - 12 = 18.$$ This tells us that $x=5$ is a solution to this equation.
Remember that the old methods still work here. So if you add 12 to both sides and subtract $2x$ from both sides you get $$20=4x$$ and then you can solve for $x$ by dividing both sides by 4.
$$\begin{array}{r|r|r} x &2x+8 &6x-12\\ \hline 0&8&-12\\1&10&-6\\2&12&0 \\ \vdots&\vdots&\vdots \end{array}$$ Note that every time $x$ increases by $1$, $2x+8$ increases by $2$ and $6x-12$ increases by $6$. (do you see why this should be so?) That means whenever $x$ increases by $1$, $6x-12$ grows by $4$ units more than $2x+8$. Currently (at $x=2$), $2x+8$ is ahead by $12$. So if we increase $x$ by $3$, $6x-12$ will gain $3\cdot 4$, or $12$ more than $2x-8$, and so then the two expressions will be equal.
Thus the answer should be $x=5$.
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# How do you simplify 2 sqrt 7+ 3 sqrt 7+ sqrt7?
Feb 26, 2016
$6 \sqrt{7}$
#### Explanation:
Factor a $\sqrt{7}$ from each term.
$2 \sqrt{7} + 3 \sqrt{7} + \sqrt{7} = \sqrt{7} \left(2 + 3 + 1\right) = \sqrt{7} \left(6\right) = 6 \sqrt{7}$
Note that this is similar to addition you have likely done in the past -- try replacing each $\sqrt{7}$ with a variable like $x$.
$2 x + 3 x + x = 6 x$
Feb 26, 2016
6$\sqrt{7}$
#### Explanation:
taking $\sqrt{7}$ common we have
given expression = $\sqrt{7} \left(2 + 3 + 1\right)$
=$6 \sqrt{7}$
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# Class Greeting. Chapter 8 Systems of Linear Equations and Inequalities Lesson 8-1a Solving Systems Equations by Graphing 1.Determine if a system of equations.
## Presentation on theme: "Class Greeting. Chapter 8 Systems of Linear Equations and Inequalities Lesson 8-1a Solving Systems Equations by Graphing 1.Determine if a system of equations."— Presentation transcript:
Class Greeting
Chapter 8 Systems of Linear Equations and Inequalities Lesson 8-1a Solving Systems Equations by Graphing 1.Determine if a system of equations has 0,1, or infinitely many solutions 2.Solve systems of equations by graphing
Objective: Students will review the slope- intercept equation. Students will be introduced to solving systems of equations using graphing.
System of 2 Linear Equations 3 possible types of solution – No solution – 1 solution – Infinite solutions
Example 1-2b Graph the system of equations. Does the system have infinitely many solutions, one solution, or no solution? If the system has one solution, name it. Answer: one solution; (x, y ) = (0, 3) 1 slope: m = 2 UP 2 OVER 1 y – intercept: b = 3 1 UP 1 2 OVER 2 slope: m =
Example 1-2a no solution Graph the system of equations. Does the system have infinitely many solutions, one solution, or no solution? If the system has one solution, name it. Answer: x 0 4 y -2 0 x 0 y 1 0 2
Example 1-2a infinitely many solutions Graph the system of equations. Does the system have infinitely many solutions, one solution, or no solution? If the system has one solution, name it. Answer: x 0– 3/2 y 3 0 x 0 y 3 0 3
Example 1-1a Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: one solution; (2, –1)
Example 1-1a Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: no solution
Example 1-1a Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: infinitely many solutions
Use the graph to determine whether each system has no solution, one solution, or infinitely many solutions. a. b. c. Example 1-1b Answer: one Answer: no solution Answer: infinitely many
Example 1-2b Answer: no solution Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it.
Lesson Summary: Objective: Students will review the slope- intercept equation. Students will be introduced to solving systems of equations using graphing.
Preview of the next Lesson: Objective: Students will solve real world word problems involving solving systems of equations using graphing.
Homework 446+447/3,7,11...31
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# When to use law of sines vs law of cosines
## Do you need a right triangle to use law of sines?
The Law of Sines says that in any given triangle, the ratio of any side length to the sine of its opposite angle is the same for all three sides of the triangle. This is true for any triangle, not just right triangles. Press ‘reset’ in the diagram above.
## When would you use the law of sines to solve a triangle?
The law of sines is used to solve triangles in which you know only two angles and one of the opposing sides (called AAS for angle-angle-side), or two sides and one of the opposing angles (called SSA for side-side-angle).
## Is SSS law of cosines?
The Law of Cosines states that: Use the law of cosines when you are given SAS, or SSS, quantities. For example: If you were given the lengths of sides b and c, and the measure of angle A, this would be SAS. SSS is when we know the lengths of the three sides a, b, and c.
## Can you do law of cosines on right triangles?
You can ONLY use the Pythagorean Theorem when dealing with a right triangle. The law of cosines allows us to find angle (or side length) measurements for triangles other than right triangles. The third side in the example given would ONLY = 15 if the angle between the two sides was 90 degrees.
## What is the law of sines and cosines?
The Law of Sines establishes a relationship between the angles and the side lengths of ΔABC: a/sin(A) = b/sin(B) = c/sin(C). Another important relationship between the side lengths and the angles of a triangle is expressed by the Law of Cosines. …
You might be interested: When to use law of sines and law of cosines
## How do you use the law of cosines?
When to Use
The Law of Cosines is useful for finding: the third side of a triangle when we know two sides and the angle between them (like the example above) the angles of a triangle when we know all three sides (as in the following example)
## How do you use the law of sines and cosines to solve a triangle?
This means we are given two sides and the included angle. For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle.
## What is an SSS triangle?
“SSS” means “Side, Side, Side” “SSS” is when we know three sides of the triangle, and want to find the missing angles. To solve an SSS triangle: use The Law of Cosines first to calculate one of the angles. then use The Law of Cosines again to find another angle.
## What is SSS SAS ASA AAS?
SSS (side-side-side) All three corresponding sides are congruent. SAS (side-angle-side) Two sides and the angle between them are congruent. ASA (angle-side-angle)
## Is SAS sine or cosine?
“SAS” is when we know two sides and the angle between them. use The Law of Cosines to calculate the unknown side, then use The Law of Sines to find the smaller of the other two angles, and then use the three angles add to 180° to find the last angle.
You might be interested: What Are The 7 Noahide Laws?
## Does law of sines work for obtuse angles?
The sine rule is also valid for obtuse-angled triangles. = for a triangle in which angle A is obtus. We can use the extended definition of the trigonometric functions to find the sine and cosine of the angles 0°, 90°, 180°. Draw a diagram showing the point on the unit circle at each of the above angles.
## Which law only works for right triangles?
Although most often trigonometric functions are used with right triangles there are some situations when they can be used for any type of triangle. Examples: If you have two sides given and an angle between them you can use the trigonometric functions the Law of Cosines to calculate the third side.4 мая 2017 г.
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## How do you find the horizontal shift of sin?
the horizontal shift is obtained by determining the change being made to the x-value. The horizontal shift is C. The easiest way to determine horizontal shift is to determine by how many units the “starting point” (0,0) of a standard sine curve, y = sin(x), has moved to the right or left.
How do you calculate the horizontal shift?
Horizontal Shift Equation The equation indicating a horizontal shift to the left is y = f(x + a). The equation indicating a horizontal shift to the right is y = f(x – a). For example, in order to shift the graph of y = x^2 + 2 to the right 4 places, the equation must be written y = (x-4)^2 +2.
What is a horizontal shift?
Horizontal shifts are inside changes that affect the input (x−) axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and right or left.
### How do you shift a sine function to the left?
Sliding a function left or right on a graph By adding or subtracting a number from the angle (variable) in a sine equation, you can move the curve to the left or right of its usual position. A shift, or translation, of 90 degrees can change the sine curve to the cosine curve.
How do you find the shift on a graph?
To find the phase shift from a graph, you need to:
1. Determine whether it’s a shifted sine or cosine.
2. Look at the graph to the right of the vertical axis.
3. Find the first:
4. Calculate the distance from the vertical line to that point.
5. If the function was a sine, subtract π/2 from that distance.
Is horizontal shift the same as phase shift?
calling the phase shift C in the first equation when B = 1, otherwise calling it C in the second equation. While this distinction exists for physicists and engineers, some mathematics textbooks use the terms “horizontal shift” and “phase shift” to mean the same thing.
## What is a horizontal shift in math?
What is the vertical shift of a sine function?
The most straightforward way to think about vertical shift of sinusoidal functions is to focus on the sinusoidal axis, the horizontal line running through the middle of the sine or cosine wave. At the start of the problem identify the vertical shift and immediately draw the new sinusoidal axis.
Why is the horizontal shift opposite?
Why are horizontal translations opposite? While translating a graph horizontally, it might occur that the procedure is opposite or counter-intuitive. That means: For negative horizontal translation, we shift the graph towards the positive x-axis.
### What is horizontal shift in calculus?
If y=f(x) then the horizontal shift is caused by adding a constant inside the function, f(x). Subtracting 5, like this y=f(x-5) causes a movement of +5 in the x-axis. Pay attention to the sign… Swap your left and right hands!
How do you find the vertical shift of a sine graph?
If you divide the C by the B (C / B), you’ll get your phase shift. The D is your vertical shift. The vertical shift of a trig function is the amount by which a trig function is transposed along the y-axis, or, in simpler terms, the amount it is shifted up or down.
What’s a horizontal shift?
## How to write a horizontal shift?
– The amplitude is equal to A; – The period is equal to 2π / B; and – The phase shift is equal to C / B.
What is a horizontal shift formula?
What is a Horizontal Shift of a Function? A horizontal shift adds or subtracts a constant to or from every x-value, leaving the y-coordinate unchanged. The basic rules for shifting a function along a horizontal (x) are: Rules for Horizontal Shift of a Function. Compared to a base graph of f(x), y = f(x + h) shifts h units to the left, y = f(x
What is the equation for horizontal shift?
the horizontal shift is obtained by determining the change being made to the x-value. The easiest way to determine horizontal shift is to determine by how many units the “starting point” (0,0) of a standard sine curve, y = sin ( x ), has moved to the right or left. Horizontal shifts can be applied to all trigonometric functions.
### What is a horizontal shift in a graph?
Shifts. A shift is a rigid translation in that it does not change the shape or size of the graph of the function.
• Scales (Stretch/Compress) A scale is a non-rigid translation in that it does alter the shape and size of the graph of the function.
• Reflections. A function can be reflected about an axis by multiplying by negative one.
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# How do you find the x and y intercept of 9x + 8y = −24?
Jul 11, 2016
${y}_{\text{intercept}} = - 3$
${x}_{\text{intercept}} = - \frac{8}{3}$
#### Explanation:
Convert the formula into the standard form of $y = m x + c$ for a straight line graph.
Subtract $\textcolor{b l u e}{9 x}$ from both sides
$\textcolor{b r o w n}{9 x \textcolor{b l u e}{- 9 x} + 8 y = - 24 \textcolor{b l u e}{- 9 x}}$
$8 y = - 9 x - 24$
Divide both sides by $\textcolor{b l u e}{8}$
$\textcolor{b r o w n}{\frac{8}{\textcolor{b l u e}{8}} \times y = - \frac{9}{\textcolor{b l u e}{8}} x - \frac{24}{\textcolor{b l u e}{8}}}$
But $\frac{8}{8} = 1 \text{ and } \frac{24}{8} = 3$ giving
$y = - \frac{9}{8} x - 3$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine y intercept}}$
The line crosses the y-axis at $x = 0$ so by substitution
color(brown)(y=-9/8x-3)color(blue)(" "->" "y=-9/8(0)-3
${y}_{\text{intercept}} = - 3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine x intercept}}$
The line crosses the x-axis at $y = 0$ so by substitution
color(brown)(y=-9/8x-3)color(blue)(" "->" "0=-9/8x-3
$\frac{9}{8} x = - 3$
$x = {\cancel{\left(- 3\right)}}^{- 1} \times \frac{8}{{\cancel{9}}^{3}} \text{ " =" "- 8/3" " =" } - 2 \frac{2}{3}$
${x}_{\text{intercept}} = - \frac{8}{3}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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5.5
1 / 46
# 5.5 - PowerPoint PPT Presentation
5.5. What Patterns Can I Use? Pg. 12 Constant Ratios in Right Triangles. 5.5 – What Patterns Can I Use? Constant Ratios in Right Triangles.
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### 5.5
What Patterns Can I Use?
Pg. 12
Constant Ratios in Right Triangles
5.5 – What Patterns Can I Use?
Constant Ratios in Right Triangles
So far in this chapter you have learned how to find the sides of special right triangles. But what if the triangle isn’t special? Today we are going to focus our attention on slope triangles, which were used in algebra to describe linear change.
5.24 – PATTERNS IN SLOPE TRIANGLES
Today you are going to focus on the relationship between the angles and the sides of a right triangle. You will start by studying slope triangles. Notice in the graph shown, a slope triangle is drawn for you.
a. Draw two new slope triangles on the line. Each should be a different size. Label each triangle with as much information as you can, such as its horizontal and vertical lengths and its angle measures.
2
10
1
5
2
10
=
a. Draw two new slope triangles on the line. Each should be a different size. Label each triangle with as much information as you can, such as its horizontal and vertical lengths and its angle measures.
3
15
3
15
1
5
=
a. Draw two new slope triangles on the line. Each should be a different size. Label each triangle with as much information as you can, such as its horizontal and vertical lengths and its angle measures.
4
20
4
20
1
5
=
b. What do these triangles have in common? How are these triangles related to each other?
similar by AA~
1
5
=
0.2
d. What do you notice about the slope ratios written in fraction form? What do you notice about the decimals?
The ratios are equal
e. Notice how the ∆y is on the opposite side of triangle from where the angle is. What side is the ∆x? How is the adjacent side different from the hypotenuse?
The rise is opposite the angle
The run is adjacent to the angle, but not the hypotenuse
5.25 – PROPORTIONS
Tara thinks she sees a pattern in these slope triangles, so she decides to make some changes in order to investigate whether or not the patterns remain true.
a. She asks, "What if I drew a slope triangle on this line with ∆y = 6? What would the ∆x be for that slope triangle? Answer her question and explain how you figured it out.
c. Tara wonders, "What if I draw a slope triangle on a different line? Can I still use the same ratio to find a missing ∆x or ∆y value?" Discuss this with your team and explain to Tara what she could expect.
No, the triangles won’t be similar
5.26 – CHANGING LINES
In part (c) of the last problem, Tara asked, "What if I draw my triangle on a different line?" With your team, investigate what happens to the slope ratio and slope angle when the line is different. Use the graph grids below to graph the lines described. Use the graphs and your answers to the questions below to respond to Tara's question.
supports
2
22°
5
R
18°
Q
P
1
3
m =
R
1
3
18°
Q
P
c. Graph the line y = x + 4 on the graph. Draw a slope triangle and label its horizontal and vertical lengths. What is the new slope ratio? What is the slope angle?
1
m =
3
45°
3
5.27 – TESTING CONJECTURES
The students in Ms. Matthews class are writing conjectures based on their work today. As a team, decide if you agree or disagree with each of the conjectures below. Explain your reasoning.
False
True
True
False, lines can be parallel with same slope
79°
7
11°
y
11°
5.29 – ANOTHER LOOK
Sheila says that the triangle in part (f) of the previous problem is the same as the picture below.
a. Do you agree? Why or why not?
Yes, AA~
c. What is the relationship of 11° and 79°? Of their slope ratios?
Angles are complementary
Ratios are reciprocal
5.30 – MAKING CONNECTIONS
For what other angles can you find the slope ratios based on your work?
a. For example, since you know the slope ratio of 22°, what other angle do you know the slope ratio for? Find the complement of each slope angle you know.
52
31
11
68° is
72° is
45° is
c. Complete the conjecture about the relationship of the slope ratios for complementary angles.
b
a
complementary
573
10
= 57.3
Triangle is not possible
Degree is not possible
5.32 – COMPLETING THE CHART
a. What happens to the slope ratio when the angle increases? Decreases?
increases
decreases
573
10
= 57.3
undefined
0
573
10
= 57.3
When is a slope ratio more than 1?
• When is it less than 1?
• When is it equal to 1?
573
10
= 57.3
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# How do you graph x-y=1 ?
Jul 29, 2018
Determine the x- and y-intercepts, plot them, and draw a straight line through the point.
#### Explanation:
Graph:
$x - y = 1$
X-intercept: value of $x$ when $y = 0$
Substitute $0$ for $y$ and solve for $x$.
x-0=1
$x = 1$
The x-intercept is $\left(1 , 0\right)$. Plot this point.
Y-intercept: value of $y$ when $x = 0$
Substitute $0$ for $x$ and solve for $y$.
$0 - y = 1$
$- y = 1$
Multiply both sides by $- 1$. This will reverse the signs.
$- 1 \left(- y\right) = 1 \times - 1$
$y = - 1$
The y-intercept is $\left(0 , - 1\right)$. Plot this point.
Draw a straight line through the points.
graph{x-y=1 [-10, 10, -5, 5]}
Jul 29, 2018
$\text{ }$
#### Explanation:
$\text{ }$
We are given the equation: color(red)(x-y=1
How do we graph this linear equation ?
We have
color(blue)(x-y=1
The given equation can be reduced to a more simpler form.
Subtract color(red)(x from both sides of the equation:
$\Rightarrow x - y - x = 1 - x$
$\Rightarrow \cancel{x} - y - \cancel{x} = 1 - x$
$\Rightarrow - y = 1 - x$
Multiply both sides of the equation by color(red)((-1)
$\Rightarrow - y \cdot \left(- 1\right) = \left(1 - x\right) \left(- 1\right)$
$\Rightarrow y = x - 1$
Now, we have the equation in Slope-Intercept Form:
$\textcolor{b l u e}{y = m x + b}$, where
color(red)(m is the Slope and
color(red)(b is the y-intercept.
Hence,
Slope : color(blue)(1
y-intercept :color(blue)((0,-1)
The x-intercept is where the graph crosses the x-axis
Set color(red)(y=0 to find the x-intercept:
$\Rightarrow x - 1 = 0$
Add color(red)(1 to both sides of the equation:
$\Rightarrow x - 1 + 1 = 0 + 1$
$\Rightarrow x - \cancel{1} + \cancel{1} = 0 + 1$
$\Rightarrow x = 1$
Hence color(red)((1,0)# is the x-intercept.
Plot the points x-intercept and intercept on the graph.
Join the two points to get the straight line.
This is the graph of the linear equation.
Hope you find this solution useful.
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# Plow-Pair
Futoshiki is fun. It’s a number-puzzle where you use logic to re-create a 5×5 square in which every row and column contains the numbers 1 to 5. At first, most or all of the numbers are missing. You work out what those missing numbers are by using the inequality signs scattered over the futoshiki. Here’s an example:
There are no numbers at all in the futoshiki, so where do you start? Well, first let’s establish some vocabulary for discussing futoshiki. If we label squares by row and column, you can say that square (4,5), just above the lower righthand corner, dominates square (4,4), because (4,5) is on the dominant side of the inequality sign between the two squares (futōshiki, 不等式, means “inequality” in Japanese). Whatever individual number is in (4,5) must be greater than whatever individual number is in (4,4).
Conversely, you can say that (4,4) is dominated by (4,5). But that’s not the end of it: (4,4) is dominated by (4,5) but dominates (3,4), which in its turn dominates (2,4). In other words, there’s a chain of dominations. In this case, it’s a 4-chain, that is, it’s four squares long: (4,5) > (4,4) > (3,4) > (2,4), where (4,5) is the start-square and (2,4) is the end-square. Now, because 5 is the highest number in a 5×5 futoshiki, it can’t be in any square dominated by another square. And because 1 is always the lowest number in a futoshiki, it can’t be in any square that dominates another square. By extending that logic, you’ll see that 4 can’t be in the end-square of a 3-chain, (a,b) > (c,d) > (e,f), and 2 can’t be in the start-square of a 3-chain. Nor can 3 be in the start-square or end-square of a 4-chain.
Using all that logic, you can start excluding numbers from certain squares and working out sets of possible numbers in each square, like this:
[whoops: square contains errors that need to be corrected!]
Now look at column 1 and at row 4:
In column 1, the number 5 appears only once among the possibles, in (1,1); in row 4, the number 1 appears only once among the possibles, in (4,1). And if a number appears in only one square of a row or column, you know that it must be the number filling that particular square. So 5 must be the number filling (1,1) and 1 must be the number filling (4,1). And once a square is filled by a particular number, you can remove it from the sets of possibles filling the other squares of the row and column. I call this sweeping the row and column. Voilà:
Now that the 5 in (1,1) and the 1 in (4,1) have swept all other occurrences of 5 and 1 from the sets of possibles in column 1 and row 4, you can apply the only-once rule again. 2 appears only once in row 4 and 5 appears only once in column 4:
So you’ve got two more filled squares:
Now you can apply a more complex piece of logic. Look at the sets of possibles in row 3 and you’ll see that the set {2,3} occurs twice, in square (3,1) and square (3,4):
What does this double-occurrence of {2,3} mean? It means that if 2 is in fact the number filling (3,1), then 3 must be the number filling (3,4). And vice versa. Therefore 2 and 3 can occur only in those two squares and the two numbers can be excluded or swept from the sets of possibles filling the other squares in that row. You could call {2,3} a plow-pair or plow-pare, because it’s a pair that pares 2 and 3 from the other squares. So we have a pair-rule: if the same pair of possibles, {a,b}, appears in two squares in a row or column, then both a and b can be swept from the three other squares in that row or column. Using {2,3}, let’s apply the pair-rule to the futoshiki and run the plow-pare over row 3:
Now the pair-rule applies again, because {4,5} occurs twice in column 5:
And once the plow-pare has swept 4 and 5 from the other three squares in column 5, you’ll see that 3 is the only number left in square (1,5). Therefore 3 must fill (1,5):
Now 3 can be swept from the rest of row 1 and column 5:
And the pair-rule applies again, because {1,2} occurs twice in row 2:
Once 2 is swept from {2,3,4} in square (2,1) to leave {3,4}, 3 must be excluded from square (2,2), because (2,2) dominates (2,1) and 3 can’t be greater than itself. And once 3 is excluded from (2,2), it occurs only once in column 2:
Therefore 3 must fill (5,2), which dominates (5,1) and its set of possibles {2,3,4}. Because 3 can’t be greater than 4 or itself, 2 is the only possible filler for (5,1) and only 3 is left when 2 is swept from (3,1):
And here are the remaining steps in completing the futoshiki:
The complete futoshiki
Animation of the steps required to complete the futoshiki
Afterword
The pair-rule can be extended to a triplet-rule and quadruplet-rule:
• If three numbers {a,b,c} can occur in only three squares of a row or column, then a, b and c can be swept from the two remaining squares of the row or column.
• If four numbers {a,b,c,d} can occur in only four squares of a row or column, then a, b, c and d can be swept from the one remaining square of the row or column (therefore the number e must fill that remaining square).
But you won’t be able to apply the triplet-rule and quadruplet-rule as often as the pair-rule. Note also that the triplet-rule doesn’t work when {a,b,c} can occur in only two squares of a row or column. An n-rule applies only when the same n numbers of a set occur in n squares of a row or column. And n must be less than 5.
Post-Performative Post-Scriptum
Domination. Exclusion. Inequality. — an earlier look at futoshiki
# Domination. Exclusion. Inequality.
Sudoku has conquered the world, but I think futoshiki is more fun — more concentrated, more compact and quicker. When complete, a 5×5 futoshiki will be a Roman square in which every row and column contains the numbers 1 to 5, but no number is repeated in any row or column. You have to work out the missing numbers using logic and the “inequality” signs that show whether one square contains a number more than or less than a number in a neighboring square — futōshiki, 不等式, means “inequality” in Japanese (fu- is the negative prefix). Here’s an example of a futoshiki puzzle:
Futoshiki puzzle, with 4 in square (3,2) and 3 in square (1,1)
If you identify the squares by row and column, 4 is in (3,2) and 3 is in (1,1). And you can say, for example, that the empty square (3,4) dominates the empty square (3,3) or that (3,3) is dominated by (3,4). I’ll describe one route (not the best or most efficient) to completing the puzzle. Let’s start by considering the general rules that 1 cannot appear in any square that dominates another square and that 5 cannot appear in any square dominated by another square.
If you extend that logic, you’ll see that 4 cannot appear in any square that is at the end of what you might call a chain of dominations, where one square dominates a second square that in turn dominates a third square. Therefore, in the puzzle above, 4 cannot appear in squares (2,1) and (4,1) of column 1. And it can’t appear in square (3,1), because that would mean two 4s in the same row. This leaves one place for 4 to appear: square (5,1). And if 4 is there, 5 has to be in square (3,1):
Now look at row 3. Two of the remaining three empty squares are dominators: (3,4) dominates (3,2) and (3,5) dominates (2,5). 1 cannot appear in a dominating square, so 1 has to be in the dominated square (3,3):
The next step I’ll take is a bit more complicated. In row 2, the number 4 cannot be in (2,1) and (2,4). It can’t be in (2,1) because that would mean 4 was greater than itself. And it can’t be in (2,4), because (2,4) is dominated by (3,4) and (3,4) can’t contain 5, the only number that dominates 4. Therefore 4 must be in either (2,3) or (2,4). But so must 5. Therefore (2,3) contains either 4 or 5 and (2,4) contains either 4 or 5. That means that the numbers [1,2,3] must be in the other three squares of row 2. Now, 3 can’t be in square (2,1), because the chain of dominations is too long. And 3 can’t be in (2,5), because (2,5) is dominated by (3,5), which contains either 2 or 3. Therefore 3 must be in (2,2):
Now consider column 2. Square (4,2) cannot contain 1 or 5, because it’s both dominated and dominating. And if it can’t contain 1 or 5, there’s only one number it can contain: 2. And it immediately follows that (4,1) must contain 1, the only number less than 2. And if four squares of column 1 now contain the numbers [1,3,4,5], the remaining empty square (2,1) must contain 2:
Now consider row 2. Squares (2,3) and (2,4) contain either 4 or 5, therefore (2,5) must contain 1:
Now consider row 5. The number 1 is logically excluded from three squares: from (5,3), because there’s a 1 in (3,3); from (5,4), because (5,4) dominates (5,5); and from (5,5), because there’s a 1 in (2,5). Therefore 1 must be in (5,2). And if 1 is in (5,2), the number 5 must be in (1,2):
Now 1 pops up in row 1 because it can only be in (1,4):
And 5 pops up in row 4 because it can only be in (4,5):
Once 5 is in (4,5), the number 4 must be in (4,4) and the number 3 in (4,3):
The 4 of (4,4) immediately collapses the ambiguity of (2,4), which must contain 5. Therefore (2,3) contains 4:
Next, 5 pops up in (5,3):
And 3 must be in (5,4), dominating 2 in (5,5):
With 3 in (5,4) and 2 in (5,5), the ambiguity of (3,4) and (3,5) collapses:
And the square is completed like this:
Here’s an animated version of the steps to completion:
Futoshiki puzzle animated
# Fourtoshiki
I hadn’t realized that sudokus could be witty until earlier this year, when I did one that literally made me laugh, because the solutions were so clever and quirky. Foolishly, I neglected to make a note of the sudoku so I could reproduce it. But I haven’t made that mistake with this futoshiki:
Using more-than and less-than signs to deduce values, fill each line and column with the numbers 1 to 5 so that no number occurs twice in the same row or column
It’s not witty like that lost sudoku, but I think futoshikis are even more beautiful and enjoyable than sudokus, because they’re even more elemental. They’re also rooted in the magic of binary, thanks to the more-than / less-than clues. And when there’s only one number on the original grid, completing them feels like growing a flower from a seed.
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# Triangle A has an area of 27 and two sides of lengths 8 and 6 . Triangle B is similar to triangle A and has a side with a length of 8 . What are the maximum and minimum possible areas of triangle B?
maximum possible area of triangle B $= 48$ &
minimum possible area of triangle B $= 27$
#### Explanation:
Given area of triangle A is
$\setminus {\Delta}_{A} = 27$
Now, for maximum area $\setminus {\Delta}_{B}$ of triangle B, let the given side $8$ be corresponding to the smaller side $6$ of triangle A.
By the property of similar triangles that the ratio of areas of two similar triangles is equal to the square of ratio of corresponding sides then we have
$\setminus \frac{\setminus {\Delta}_{B}}{\setminus {\Delta}_{A}} = {\left(\frac{8}{6}\right)}^{2}$
$\setminus \frac{\setminus {\Delta}_{B}}{27} = \frac{16}{9}$
$\setminus {\Delta}_{B} = 16 \setminus \times 3$
$= 48$
Now, for minimum area $\setminus {\Delta}_{B}$ of triangle B, let the given side $8$ be corresponding to the greater side $8$ of triangle A.
The ratio of areas of similar triangles A & B is given as
$\setminus \frac{\setminus {\Delta}_{B}}{\setminus {\Delta}_{A}} = {\left(\frac{8}{8}\right)}^{2}$
$\setminus \frac{\setminus {\Delta}_{B}}{27} = 1$
$\setminus {\Delta}_{B} = 27$
Hence, the maximum possible area of triangle B $= 48$ &
the minimum possible area of triangle B $= 27$
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# 2021 JMPSC Accuracy Problems/Problem 6
## Problem
In quadrilateral $ABCD$, diagonal $\overline{AC}$ bisects both $\angle BAD$ and $\angle BCD$. If $AB=15$ and $BC=13$, find the perimeter of $ABCD$.
## Solution
Notice that since $\overline{AC}$ bisects a pair of opposite angles in quadrilateral $ABCD$, we can distinguish this quadrilateral as a kite.
$\linebreak$ With this information, we have that $\overline{AD}=\overline{AB}=15$ and $\overline{CD}=\overline{BC}=13$.
Therefore, the perimeter is $$15+15+13+13=\boxed{56}$$ $\square$
$\linebreak$ ~Apple321
## Solution 2
We note that triangle $ABC$ and $DAC$ are congruent due to $AA$ congruency. Therefore, $AD + DC = 28$ and the perimeter of the quadrilateral is $28+28 = \boxed{56}$
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# Which of the following is in ascending order?
Numbers are said to be in ascending order when they are arranged from the smallest to the largest number. E.g. 5, 9, 13, 17 and 21 are arranged in ascending order.
## Which of the following is in ascending order?
Numbers are said to be in ascending order when they are arranged from the smallest to the largest number. E.g. 5, 9, 13, 17 and 21 are arranged in ascending order.
## How do you structure this compare and contrast paper?
How to Write a Compare and Contrast Essay
1. Begin by Brainstorming With a Venn Diagram. The best compare and contrast essays demonstrate a high level of analysis.
2. Develop a Thesis Statement.
3. Create an Outline.
4. Write the Introduction.
5. Write the First Body Paragraph.
6. Repeat the Process for the Next Paragraphs.
7. Write the Conclusion.
## What are the two ways in ordering numbers?
When ordering numbers, we first start by comparing two numbers at a time. You can do this by either arranging the numbers in ascending or descending order. Each number has 4 in thousands digit, then move to hundreds digit; 4797 is the biggest and 4679 the smallest number.
## What do you mean by ascending order and descending order?
Arranging things, i.e., numbers, quantities, lengths, etc. from a larger value to lower value is known as descending order. The opposite of descending order is known as ascending order, in which the numbers are arranged from lower value to higher value.
## What do you mean by ascending order?
Ascending Order. Ascending Order. Arranging numbers (or other items) in ascending order means to arrange them from smallest to largest. Example 1 (with Numbers) The numbers 12, 5, 7, 10, 1, 160 arranged in ascending order are 1, 5, 7, 10, 12, 160.
## What is used to compare numbers?
Comparing Numbers
= When two values are equal, we use the “equals” sign example: 2+2 = 4
< When one value is smaller than another, we can use a “less than” sign. example: 3 < 5
> When one value is bigger than another, we can use a “greater than” sign example: 9 > 6
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Probability Concepts
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Probability is a measure $P(\bullet)$ which quantifies the dregree of (un)certainty associated with an event. We will discuss three common approaches to probability.
Classical Probability
Laplace’s classical definition of probability is based on equally likely outcomes. He postulates the following properties of events:
• the sample space is composed of a finite number of basic outcomes
• the random process generates exactly basic outcome and hence one elementary event
• the elementary events are equally likely, i.e. occur with the same probability
Accepting these assumptions, the probability of any event $A$ (subset of the sample space) can be computed as $P(A)=\frac{\#\left( \text{basic outcomes in }A\right) }{\#\left( \text{basic outcomes in }S\right) }=\frac{\#\left( \text{elementary events comprising }A\right) }{\#\left( \text{elementary events comprising }S\right) }$ Properties:
• $0\leq P(A) \leq1$
• $P(\emptyset)=0$
• $P(S)=1$
Example: Rolling a six-sided dieSample space: $S=\{1,2,3,4,5,6\}$ Define event $A=$ ‘even number’Elementary events in $A$: $\{2\}$,$\{4\}$,$\{6\}$$P(A)=\frac{3}{6}=0.5$
Statistical Probability
Richard von Mises originated the relative frequency approach to probability: The probability $P(A)$ for an event $A$ is defined as the limit of the relative frequency of $A$, i.e. the value the relative frequency will converge to if the experiment is repeated an infinite number of times. It is assumed that replications are independent of each other. Let $h_{n}(A)$ denote the absolute frequency of $A$ occurring in $n$ repetitions. The relative frequency of $A$ is then defined as $f_{n}(A)=\frac{h_{n}(A)}{n}$ According to the statistical concept of probability we have $P(A)=\lim_{n\rightarrow\infty}f_{n}(A)$ Since $0\leq f_{n}(A)\leq1$ it follows that $0\leq P(A)\leq1$. Example: Flipping a coin Denote by $T$ the event ‘a head appears’. Absolute and relative frequencies of $A$ after $n$ trials are listed in the table below. This particular sample displays a non monotonic convergence to $0.5$, the theoretical probability of a head occuring in repeated flips of a ’fair’ coin..
n $h_{n}(A)$ $f_{n}(A)$
10 7 0.700
20 11 0.550
40 17 0.425
60 24 0.400
80 34 0.425
100 47 0.470
200 92 0.460
400 204 0.510
600 348 0.580
800 404 0.505
1000 492 0.492
2000 1010 0.505
3000 1530 0.510
4000 2032 0.508
5000 2515 0.503
Visualizing the sequence of relative frequencies $f_{n}\left( A\right)$ as a function of sample size provides some intuition into the character of the convergence.
A central objective of statistics is to estimate or approximate probabilities of events using observed data. These estimates can then be used to make probabilistic statements about the process generating the data, (e.g., confidence intervals which we will study later), tto test propositions about the process and to predict the likelihood of future events
Axiomatic Foundation of Probability
$P$ is a probability measure. It is a function which assigns a number $P(A)$ to each event $A$ of the sample space $S$. Axiom 1$P(A)$ is real-valued with $P(A)\geq0$. Axiom 2$P(S)=1$. Axiom 3If two events $A$ and $B$ are mutually exclusive ($A\cap B=\emptyset$), then$P(A\cup B)=P(A)+P(B)$ Some basic properties of probabilityLet $A,B,A_{1},A_{2} ,\ldots\subset S$ be events and $P(\bullet)$ a probability measure. Then the following properties follow from the above three axioms
Properties
1. $P(\overline{A})=1-P(A)$
2. $\left( A \cap B = \emptyset\right) \Rightarrow P(A \cap B)=P(\emptyset) = 0$
3. If $A_{i}\cap A_{j}=\emptyset$ for $i\neq j$, then $P(A_{1}\cup A_{2}\cup\ldots)=P(A_{1})+P(A_{2})+\ldots$
Let $A$ and $B$ be any two events. Then$P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) -P\left( A\cap B\right)$
Extension to three events $A$, $B$, $C$:$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
Assume you have shuffled a standard deck of 52 playing cards. You are interested in the probability of a randomly drawn card being a queen or a ’heart’.We are thus interested in probability of the event $\left( \left\{ \text{Queen}\right\} \cup\left\{ \text{Heart}\right\} \right)$. Following Laplace’s notion of probability, we proceed as follows: There are 4 queens and 13 hearts in the deck. Hence,
• $P\left( \left\{ \text{Queen}\right\} \right) =\frac{4}{52}$
• $P\left( \left\{ \text{Heart}\right\} \right) =\frac{13}{52}$
But there is also one card which is both a queen and a heart. As this card is included in both counts, we would overstate the probability of either queen or heart appearing if we simply added both probabilities. In fact, the addition rule of probability requires one to deduct the probability of this joint event: $P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) -P\left( A\cap B\right)$ Here,
• $P\left( A\cap B\right) =P\left( \left\{ \text{Queen}\right\} \cap\left\{ \text{Heart}\right\} \right) =\frac{1}{52}$
Thus,$P\left( \left\{ \text{Queen}\right\} \cup\left\{ \text{Heart}\right\} \right) =P\left( \left\{ \text{Queen}\right\} \right) +P\left( \left\{ \text{Heart}\right\} \right) -P\left( \left\{ \text{Queen}\right\} \cap\left\{ \text{Heart}\right\} \right) =\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}$ The probability of drawing queen’s face and/or heart suit is $16/52$.
1. The event $B$ can be rewritten as a union of two disjoint sets $A \cap B$ and $\bar A \cap B$ as follows $B = (A \cap B) \cup(\bar A \cap B)$
as illustrated on the Venn diagram below:
The probability $P(B)$ is, according to axiom 3, $P(B)=P[(A\cap B)\cup(\bar{A}\cap B)]=P(A\cap B)+P(\bar{A}\cap B)$ which implies $P(\bar{A}\cap B)=P(B)-P(A\cap B)$
2. We rewrite the event $A \cup B$ as a union of two disjoint sets $A$ and $\bar A \cap B$ so that $A \cup B = A \cup(\bar A \cap B)$
The probability $P(A \cup B)$ follows from axiom 3 $P(A \cup B) = P[A \cup(\bar A \cap B)] = P(A) + P(\bar A \cap B)$ Now we obtain the desired result by calculating $P(\bar A \cap B)$ using the formula given in part one: $P(A \cup B) = P(A) + P(B) - P(A \cap B).$
Proof of Property 5:Let us show that for $A\subset B$ it follows that $P(A)\leq P(B)$. The event $B$ can be rewritten as $B = A \cup(B \setminus A)$, where $A$ and $B \setminus A$ are disjoint sets.According to axiom 3 we have the following: $P(B) = P(A) + P(B \setminus A)$. Nonnegativity of the probability $P(B \setminus A) \geq0$ implies that $P(B) \geq P(A)$. This rule can be illustrated using a Venn diagram:
Proof of Property 7:Let us prove that $P(A\setminus B)=P(A)-P(A\cap B)$. We have $A\setminus B=A\cap\bar{B}$ and $A=(A\cap B)\cup(A\cap\bar{B})$, where $(A\cap B)$ and $(A\cap\bar{B})$ are clearly disjoint.Using axiom 3 the probability of $A$ can be calculated as $P(A)=P[(A\cap B)\cup(A\cap\bar{B})]=P(A\cap B)+P(A\cap\bar{B})=P(A\cap B)+P(A\setminus B)$ This result is displayed on the following Venn diagram:
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# 2010 AMC 12A Problems/Problem 21
## Problem
The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$, where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
## Solutions
### Solution 1
The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$.
Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$.
Suppose we let $p$, $q$, and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$, $q$, and $r$.
\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\ (x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\ \sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}
In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$.
$$(x^3-ux^2+vx-w)^2$$ $$= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2$$
[Quick note: Since we don't know $a$, $b$, and $c$, we really don't even need the last 3 terms of the expansion.]
\begin{align*}&2u = 10\\ u^2+2v &= 29\\ 2uv+2w &= 4\\ u &= 5\\ v &= 2\\ w &= -8\\ &\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}
All that's left is to find the largest root of $x^3-5x^2+2x+8$.
\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\ &\boxed{\textbf{(A)}\ 4}\end{align*}
### Solution 2
The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$. We also know that this graph has 3 places tangent to the x-axis, which means that each root has to have a multiplicity of 2. Let the function be $(x-p)^2(x-q)^2(x-r)^2$.
Applying Vieta's formulas, we get $2p+2q+2r = 10$ or $p+q+r = 5$. Applying it again, we get, after simplification, $p^2+q^2+r^2+4pq+4pr+4qr = 29$.
Notice that squaring the first equation yields $p^2+q^2+r^2+2pq+2qr+2pr= 25$, which is similar to the second equation.
Subtracting this from the second equation, we get $2pq+2pr+2qr = 4$. Now that we have the $pq+pr+qr$ term, we can manpulate the equations to yield the sum of squares. $2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4$ or $2p^2+2q^2+2r^2+2pq+2qr+2pr = 46$. We finally reach $(p+q)^2+(q+r)^2+(p+r)^2 = 46$.
Since the answer choices are integers, we can guess and check squares to get $\{(p+q)^2, (q+r)^2, (p+r)^2\} = \{1, 9, 36\}$ in some order. We can check that this works by adding then and seeing $2p+2q+2r = 10$. We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get $\boxed{\textbf{(A)}\ 4}$.
Note: One could also multiply $2pq+2pr+2qr = 4$ by 2 and subtract from $p^2+q^2+r^2+4pq+4pr+4qr = 29$ to obtain $$p^2+q^2+r^2=21$$ The ordered triple {16,4,1} sums to 21, and the answer choices are all positive integers, therefore the answer is 4.
#### Alternative method:
After reaching $p+q+r = 5$ and $pq + qr + rp = 2$, we can algebraically derive $pqr$.
Applying Vieta's formulas on the $x^3$ term yields $2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4$.
Notice that $(p+q+r)(pq+qr+rp) = p^2q+pq^2+q^2r+qr^2+r^2p+rp^2+3pqr$, so $$2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+6pqr = 2(p+q+r)(pq+qr+rp) = 20.$$
Subtracting this from $2p^2q+2pq^2+2q^2r+2qr^2+2r^2p+2rp^2+8pqr = 4$ yields $2pqr = -16$, so $pqr = -8$, which means that $p$, $q$, and $r$ are the roots of the cubic $x^3 - 5x^2 + 2x + 8$, and it is not hard to find that these roots are $-1$, $2$, and $4$. The largest of these values is $\boxed{\textbf{(A)}\ 4}$.
## Solution 3
First, $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0$ has exactly $3$ roots. Therefore, $y = (kx^3+lx^2+mx+n)^2 = 0$.
So, $k^2x^6+2klx^5+(2km+l^2)x^4+2(kn+lm)x^3+ax^2-bx-c = 0$
By matching the coefficients of the first $4$ terms, we have $k^2 = 1, 2kl = -10, 2km+l^2 = 29, 2kn+2lm = -4$
Solving the equations above, we have $2$ sets of solutions; first set of which is $k = 1, l = -5, m = 2, n = 8$. Second set of which is $k = -1, l = 5, m = -2, n = -8$. After squaring both sets, they are the same i.e. $x^3-5x^2+2x+8 = 0$.
This is equal to $(x-4)(x-2)(x+1) = 0$. The largest root is $\boxed {\textbf{(A) 4}}$
~Arcticturn
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# There are two triangles. One has a base of 8 cm and a height of 10 cm. The other has a base of 5 cm and a height of 13 cm. Which triangle has a larger area?
May 6, 2016
The triangle with base 8 cm and height 10 cm.
#### Explanation:
$A r e a \triangle = \frac{1}{2} \cdot b a s e \cdot h e i g h t$
${\triangle}_{1} a r e a = \frac{1}{2} \cdot 8 \cdot 10$
$= 40$
${\triangle}_{2} a r e a = \frac{1}{2} \cdot 5 \cdot 13$
$= 32.5$
$40 > 32.5 \therefore$ Area of first triangle is larger.
May 6, 2016
base 8 , height 10
#### Explanation:
The area (A) of a triangle is found as follows.
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{A = \frac{1}{2} b h} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where b , is the base and h , the perpendicular height
For b = 8 and h = 10 :$A = \frac{1}{2} \times 8 \times 10 = 40 \text{ square cm}$
For b = 5 and h = 13 :$A = \frac{1}{2} \times 5 \times 13 = 32.5 \text{ square cm}$
Hence the triangle with b = 8 and h = 10 has the larger area.
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# If the polynomial x4 – 2x3 + 3x2 – ax + b is divided by (x – 1) and (x + 1) and the remainders are 5 and 19 respectively. But if that polynomial is divided by x + 2, then what will be the remainder —Let us calculate.
We have the polynomial x4 – 2x3 + 3x2 – ax + b.
Let it be P(x), such that
P(x) = x4 – 2x3 + 3x2 – ax + b …(i)
According to the question,
P(x) is divided by (x – 1) and (x + 1), and leaves the remainder 5 and 19 respectively.
We will use remainder theorem here.
By Remainder theorem that says, f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).
At first, let’s find the zero of the linear polynomial, (x – 1).
To find zero,
x – 1 = 0
x = 1
Now, let’s find the zero of the linear polynomial, (x + 1).
To find zero,
x + 1 = 0
x = -1
From Remainder theorem, we can say
When P(x) is divided by (x – 1), the remainder comes out to be 5.
We can also say that,
The remainder comes out to be P(1).
P(1) = 5
(1)4 – 2(1)3 + 3(1)2 – a(1) + b = 5
1 – 2 + 3 – a + b = 5
b – a + 2 = 5
b – a = 5 – 2
b – a = 3 …(ii)
And when P(x) is divided by (x + 1), the remainder comes out to be 19.
We can also say that,
The remainder comes out to be P(-1)
P(-1) = 19
(-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + b = 5
1 + 2 + 3 + a + b = 5
a + b + 6 = 5
a + b = 5 – 6
a + b = -1 …(iii)
Solving equations (ii) and (iii), we get
(b – a) + (a + b) = 3 + (-1)
2b = 3 – 1
2b = 2
b = 1
Putting b = 1 in equation (ii),
b – a = 3
1 – a = 3
a = 1 – 3
a = -2
The values of a = -2 and b = 1.
The polynomial = x4 – 2x3 + 3x2 – ax + b
The polynomial = x4 – 2x3 + 3x2 – (-2)x + 1
The polynomial = x4 – 2x3 + 3x2 + 2x + 1
So, when polynomial x4 – 2x3 + 3x2 + 2x + 1 is divided by (x + 2), the remainder can be calculated by using Remainder theorem.
First, we need to find zero of the linear polynomial, (x + 2).
To find zero,
Put (x + 2) = 0
x = -2
So, Required remainder = P(-2)
Required remainder = (-2)4 – 2(-2)3 + 3(-2)2 + 2(-2) + 1
Required remainder = 16 + 16 + 12 – 4 + 1
Required remainder = 32 + 12 – 3
Required remainder = 44 – 3
Required remainder = 41
Thus, remainder is 41.
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# A Deep Dive into Probability Distribution in Risk Management
Ever tossed a fair coin? I’d bet you have! At least in your childhood days while deciding which team would bat first in a baseball or a cricket match, or who would serve first in a badminton or tennis game. Every kid agrees to it because it’s unbiased. When you toss a fair coin, the chances of getting a head is 1/2 (0.5) or 50%. This is the division between the favorable outcome, which is a head and all possible outcomes (head and tail).
A coin toss is perhaps is the simplest introduction to probability, which informs chance or likelihood of occurrence of a random variable. The random variable here is getting a head. Let’s note the random variable as X and probability as P(X). Mathematically put:
Probability (Getting head) or P (X)
= Favorable number of outcomes of the event / Total number of possible outcomes
= 1 / 2 = 0.5 = 50%
Now, what’s the chance of getting a head when you toss a pair of coins together? In this case, the total number of possible outcomes is four: tail (first coin) and tail (second coin), tail and head, head and tail, and finally head and head. The values that the random variable can take are many, as shown in the below table:
As you can see, the possible values of X range from 0 to 2 (i.e., 0 head, 1 head, or 2 heads). This leads to the concept of distribution. Looking at the above table, we can see the frequencies of this random variable’s occurrences are 1, 2, and 1. This is distribution or frequency distribution. One could say: distribution is the possible values a random variable can take and how frequently these values occur.
Now, if I add probabilities to this random variable’s values, we get a probability distribution. This is depicted in below:
As shown in the above table:
• We have all favorable outcomes for 0, 1, or 2 heads. These are represented in the first and second columns.
• All possible outcomes are obviously four, and that’s shown in the third column.
• P(X) is shown in the final column, and the probabilities are 25%, 50%, and 25%, respectively for 0, 1, or 2 heads. This is probability distribution. Summed up, it equals one.
Hence, we can saythat probability distribution for a random variable describes how probabilities are distributed over the values of the random variable.
## Discrete and Continuous Distribution
Distribution can be discrete or continuous. Discrete means you are getting an integer number (1 head or 2 heads). You don’t say that you will get 0.33 head! Considering another example of counting the number of children in households of a locality, you will come-up with results such as 0 child, 1 child, 2 children, etc. You won’t get 0.57 child!
You may be laughing now – what’s 0.33 head or 0.57 child!? Good to see you smiling. Smiling lessens stress and helps in understanding.
All random variables; however, are not discrete. For example, let’s say you are determining the distribution of age, weight, or height of people in a locality. Considering height, it can be anything: 5 feet, 5.5 feet, 5.85 feet, 6.1 feet, and so on. In such a case, the distribution is continuous. So, this distinction is important: at a high-level, there are two types of random variables – discrete and continuous and respective probability distributions – discrete probability distribution and continuous probability distribution.
But, how does all of this fit into Risk Management? Risk Managers don’t toss coins or calculate heads/tails in an experiment. That’s kids’ games and not for grown-up men or women! Perhaps; although, child play teaches the basics neatly.
## Probability Distribution and Risk Management
With the above basics, let’s consider another example to understand probability distribution from the perspective of risk management. You are going to a friend’s house. It may take you one hour to reach your destination if you encounter no obstacles. If there is heavy traffic, it’s possible that you may not get there for three hours. With less traffic, it’s more likely to take 2 hours. Hence, you can say there are three possibilities:
1. Minimum (or Optimistic) travel duration = 1 hour
2. Most likely travel duration = 2 hours
3. Maximum (or Pessimistic) travel duration = 3 hours
In this case, the random variable (X) is the “travel duration.” Can you conclusively say which one of the estimates is correct? Unlikely, because other factors such as traffic conditions are involved. Now, if I add chances to these numbers, then we get probability distributions. I’ve prepared the below video to explain in more detail [Duration: 05m:33s]. For better audio-visual experience, you may want to go full HD and plug-in your earphones.
## Importance of Probability Distribution
In project risk management, the concept of probability distribution is applied to estimation. Continuing with our previous example, when we estimate, we take the most likely outcome of two hours, which is not correct because we’ve forgotten to consider other possibilities.
We can (and should!) consider possible scenarios, not just the most likely one. In other words, instead of saying an activity in a project is going to take “X” number of days, we also can consider other days using a distribution. For each duration in the distribution, there is a probability available.
This can be done for all the activities or tasks of the project, which in turn impacts the project schedule and cost. This enables us to build a more realistic plan.
Now that we have understood the basics of probability, distribution, and probability distribution, let’s look at the various types used in risk management.
## Triangular Distribution
Triangular distribution is the most common type of distribution used. Named triangular because of the shape of the curve, this refers to there being no pre-existing data, but only expert opinions or judgment.
### Symmetrical Triangular Distribution
The below distribution is triangular and symmetrical.
By looking at the graph above, we can say: There is approximately a 30% chance of the duration being 6 days, a full chance of the duration being 8 days’, and also a 30% chance of the duration being 10 days.”
### Asymmetrical Triangular Distribution
Do note that the triangle shown need not be symmetric. Asymmetrical diagrams are shown below:
From here, you can calculate the durations with respective chances or probabilities.
Let’s take another example of a project, once with a task of Product Requirement Documentation (PRD) Preparation with an estimated 5 days duration. This is the most likely estimate, but we do not have the minimum and maximum value.
By using the Primavera Risk Analysis (PRA) software tool, the triangular distribution is depicted as below:
The durations can be 4, 5, or 6 days (shown in the X-axis). The respective chance for minimum, likely, and maximum values are entered when you perform a duration risk analysis. This is demonstrated in a video in the later part of this article.
While building the schedule model, this triangular distribution can be noted as Triangle (4, 5, 6) or Triangle (4; 5; 6).
## Uniform Distribution
In rectangular distribution, you can use a maximum value and a minimum value, but not any most likely value. In the below example, we have a uniform (or rectangular) distribution.
Looking at it, we might say: The task has a minimum duration of 4 days, but a maximum duration of 12 days.
You can use Uniform Probability Distributions when you specify the extremes of uncertainty of the activity under consideration and when the intermediate values have equal chances of occurring. It is also possible when you cannot draw any inference on the possible distribution shape.
Taking our previous example of the PRD Preparation task, which is estimated to be 5 days, using PRA, we have the following values for Uniform distribution:
Like Triangular distribution, while building the schedule model, this distribution can be noted as Uniform (4, 6).
## Beta Distribution
Beta distribution, like triangular distribution has also three possible values – worst case, most likely, and best case. Like the triangular model, it also gives more weightage to the most likely case. We have seen one example of Beta distribution in the earlier video.
Unlike the triangular distribution, the shape for beta distribution is smoother and the tails in Beta distribution taper off less quickly. A sample beta distribution curve is shown below:
Beta distribution can also be symmetric or asymmetric in shape. The notations happen like Beta (6, 8, 10). As you can see above, there can be many values close to the most likely values, and it slowly tapers off towards the minimum or maximum ends.
Using the PRA software tool, for our task, PRD Preparation, a Beta (or BetaPert) distribution will come out as below:
Along with the triangular distribution, beta distribution is another frequently used probability distribution.
## Normal Distribution
Normal distribution is defined by the mean of a planned (or remaining duration) activity for an activity and standard deviation (SD) of the activity.
This distribution is used if there is historical information available. Normal distribution also has a bell-shaped curve like Beta, but considers SD to calculate the worst (and best) case scenarios.
For our example (task of PRD Preparation with a duration estimate of 5 days), we note the normal distribution as Normal (5,1), where 5 is the mean and 1 is the SD.
## Discrete Distribution
In a discrete distribution model, the duration of an activity under consideration can have a number of integer values, but without any intermediate values. In other words, the distribution is discrete, rather than continuous like in a triangle, beta, or uniform.
In the above sample, the activity has discrete distribution of values 6, 10, 18, and 20.
Considering our task of PRD Preparation, the discrete distribution will be seen as below with the PRA tool. The distributions are 2, 3, 4, and, 5 with respective weighting factors of 10, 20, 30 and, 50, respectively. This can be noted as Discrete ({2, 3, 4, 5}, {10, 20, 30, 50}).
## Practical Example and Demonstration
With this understanding, let’s take a practical look using MS Project and Primavera Risk Analysis. The video [Duration: 05m:42s] demonstrates a project plan with fixed activity estimates. It’s next imported to the PRA tool and analyzed with various probability distributions for the activities of the project.
## Conclusion
Probability distribution is very important when you use quantitative risk analysis, which involves a number of mathematical modeling and sampling. Managers or planners can also deploy advanced probability distributions such as lognormal distributions, cumulative distributions, general distributions, among others. The above video explains a few of these.
We have come a long way and seen a number of examples. I propose just one more exercise. I promise it won’t be difficult, provided you have read the content sincerely. Going back to our first examples of coin tosses, can you answer these:
• What’s the probability distribution of getting a head when you toss three coins?
• What are the values that the random variable can take?
If you are getting four values for the random variable of getting a head and when all your probability distributions are summed-up to equal one, then you have well understood the concept.
I welcome your thoughts, feedback, and suggestions in the comment section below.
*This article is dedicated to the memory of my father, the late Harendra Nath Dash, who passed away three years ago on June 11, 2019. He first introduced me to the concept of probability and statistics. It was mesmerizing then, and I still remember it. I wish this article to be a tribute to him and his teachings.
References:
[1] RMP Live Lessons – Guaranteed Pass or Your Money Back, by Satya Narayan Dash
[2] Practical RMP with Primavera Risk Analysis, by Satya Narayan Dash
[3] RMP 30 Contact Hours Online, by Satya Narayan Dash
[4] Book: I Want To Be A RMP, The Plain and Simple Way, Second Edition, by Satya Narayan Dash
#### Excel Data Visualization – Lesson 2 Transcription
Written by Satya Narayan Dash
Satya Narayan Dash is a management professional, coach, and author of multiple books. Under his guidance, over 2,000 professionals have successfully cracked PMP, ACP, RMP, and CAPM examinations – in fact, there are over 100 documented success stories written by these professionals. His course, PMP Live Lessons - Guaranteed Pass, has made many successful PMPs, and he’s recently launched RMP Live Lessons - Guaranteed Pass and ACP Live Lessons - Guaranteed Pass. His web presence is at https://managementyogi.com, and he can be contacted via email at managementyogi@gmail.com.
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Welcome To Basics In Maths
# These solutions designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.
### Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.
#### QUESTION 1
Find the unit vector in the direction of
Sol: Given vector is
The unit vector in the direction of a vector is given by
#### QUESTION 2
Find a vector in the direction of a where that has a magnitude of 7 units.
Sol: Given vector is
The unit vector in the direction of a vector is
The vector having the magnitude 7 and in the direction of is
#### QUESTION 3
Find the unit vector in the direction of the sum of the vectors, a = 2i + 2j – 5k and b = 2i + j + 3k
Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k
a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k
##### QUESTION 4
Write the direction cosines of the vector
Sol: Given vector is
∴ Direction cosines are
QUESTION 5
Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 ac are collinear when a, b, c are non-collinear vectors
Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 acA
B = OB – OA = a + 2b + 3c (– 2a + 3b + 5c)
AB = 3a b – 2c
AC = OC – OA = 7 ac(– 2a + 3b + 5c)
AC = 9a – 3b – 6c = 3(3a b – 2c)
AC = 3 AB
A, B and C are collinear
##### QUESTION 6
ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) AL and AM in terms of AB and AD (ii) 𝛌, if AM = 𝛌 AD – LM
Sol: Given, ABCD is a parallelogram and L and M are middle points of BC and CD
(i) Take A as the origin
M is the midpoint of CD
AM =
= AD + ½ AB (∵ AB = DC)
L is the midpoint of BC
AL =
(ii) AM = 𝛌 AD – LM
𝛌 = 3/2
#### QUESTION 7
If G is the centroid of the triangle ABC, then show that OG = when, are the position vectors of the vertices of triangle ABC.
Sol: OA = a, OB = b, OC = c and OD = d
D is the midpoint of BC
OD =
G divides median AD in the ratio 2: 1
OG =
OG =
##### QUESTION 8
If = , = are collinear vectors, then find m and n.
Sol: Given , are collinear vectors
= λ
Equating like vectors
2 = 4 λ; 5 = m λ; 1 = n λ
λ =
5 = m ⟹ m =10
1 = n ⟹ n = 2
∴ m = 10, n = 2
#### QUESTION 9
Let If , . Find the unit vector in the direction of a + b.
Sol: Given vectors are and
a + b =
The unit vector in the direction of a + b =
=
=
##### QUESTION 10
If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ.
Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k
⟹ a = tb
– 3i + 4j + λk = t (μi + 8j + 6k)
– 3i + 4j + λk = μt i + 8t j + 6t k
Equating like vectors
– 3 = μt; 4 = 8t, λ = 6t
4 = 8t
t =
– 3 = μ ⟹μ=– 6
λ = 6 ⟹ λ = 3
∴ μ=– 6, λ = 3
##### QUESTION 11
ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌
Sol: Given, ABCD is a pentagon
AB + AE + BC + + DC + ED + AC = 𝛌 AC
(AB + BC) + (AE + DC + ED) + AC = 𝛌 AC
AC + AC + AC = 𝛌 AC
3 AC = 𝛌 AC
𝛌 = 3
#### QUESTION 12
If the position vectors of the points A, B and C are – 2i + jk and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌
Sol: Given, OA = – 2i + jk , OB = –4i + 2j + 2k and OC = 6i – 3j – 13k
AB = OB – OA = –4i + 2j + 2k – (– 2i + jk)
= –4i + 2j + 2k +2ij + k
= –2i + j + 3k
AC = OC – OA = 6i – 3j – 13k – (– 2i + jk)
= 6i – 3j – 13k +2ij + k
= 8i –4 j –12k
= – 4 (2i + j + 3k)
AC = – 4 AB
Given AB = 𝛌 AC
𝛌 = – 1/4
#### QUESTION 13
If OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k then find the vector OD
Sol: Given OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k
OD = OA + AB + BC + CD
= i + j +k + 3i – 2j + k + i + 2j – 2k + 2i + j +3k
= 7i + 2j +4k
#### QUESTION 14
Let a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k, then find the unit vector in the opposite direction of a + b + c
Sol: Given, a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k
a + b + c = 2i +4 j –5 k + i + j+ k + j +2 k
= 3i +6j –2k
The unit vector in the opposite direction of a + b + c is
⟹
=
##### QUESTION 15
Is the triangle formed by the vectors 3i +5j +2k, 2i –3j –5k, 5i – 2j +3k
Sol: Let a =3i +5j +2k, b = 2i –3j –5k, c = 5i – 2j +3k
∴ Given vectors form an equilateral triangle.
##### QUESTION 16
Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are a, b, (3a – 2b) are collinear.
Sol: the vector equation of the straight line passing through two points a, b is
r = (1 – t) a+ t b
3a – 2b = (1 – t) a+ t b
Equating like vectors
1 – t = 3 and t = – 2
∴ Given points are collinear.
###### QUESTION 17
OABC is a parallelogram If OA = a and OC = c, then find the vector equation of the side BC
Sol: Given, OABC is a parallelogram and OA = a, OC = c
The vector equation of BC is a line which is passing through C(c) and parallel to OA
⟹ the vector equation of BC is r = c + t a
##### QUESTION 18
If a, b, c are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A
Sol: Given OA = a, OB = b, OC = c
D is mid of BC
OD = =
r = (1 – t) a + t ( )
#### QUESTION 19
Find the vector equation of the line passing through the point 2i +3j +k and parallel to the vector 4i – 2j + 3k
Sol: Let a =2i +3j +k, b = 4i – 2j + 3k
The vector equation of the line passing through a and parallel to the vector b is r = a + tb
r = 2i +3j +k + t (4i – 2j + 3k)
= (2 + 4t) i + (3 – 2t) j + (1 + 3t) k
##### QUESTION 20
Find the vector equation of the plane passing through the points i – 2j + 5k, 2j –k and – 3i + 5j
Sol: The vector equation of the line passing through a, b and cis r = (1 – t – s) a + tb + sc
r = (1 – t – s) (i – 2j + 5k) + t (2j –k) + s (– 3i + 5j)
= (1 – t – 4s) i + (– 2 – 3t + 7s) j + (5 – 6t – 5s) k
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## Go Math Grade 8 Answer Key Chapter 12 The Pythagorean Theorem
Chapter 12- Lesson 1:
Chapter 12- Lesson 2:
Chapter 12- Lesson 3:
### Guided Practice – The Pythagorean Theorem – Page No. 378
Question 1.
Find the length of the missing side of the triangle
a2 + b2 = c2 → 242 + ? = c2 → ? = c2
The length of the hypotenuse is _____ feet.
_____ feet
Answer: The length of the hypotenuse is 26 feet.
Explanation: According to Pythagorean Theorem, we shall consider values of a = 24ft, b = 10ft.
Therefore c = √(a2 +b2)
c = √(242 + 102)
= √(576 + 100)
= √676 = 26ft
Question 2.
Mr. Woo wants to ship a fishing rod that is 42 inches long to his son. He has a box with the dimensions shown.
a. Find the square of the length of the diagonal across the bottom of the box.
________ inches
Explanation: Here we consider the length of the diagonal across the bottom of the box as d.
Therefore, according to Pythagorean Theorem
W2 + l2 = d2
402 + 102 = d2
1600 + 100 = d2
1700 = d2
Question 2.
b. Find the length from a bottom corner to the opposite top corner to the nearest tenth. Will the fishing rod fit?
________ inches
Explanation: We denote by r, the length from the bottom corner to the opposite top corner. We use our Pythagorean formula to find r.
h2 + s2 = r2
102 + 1700 = r2
100 + 1700 = r2
1800 = r2, r = √1800 => 42.42 inches
ESSENTIAL QUESTION CHECK-IN
Question 3.
State the Pythagorean Theorem and tell how you can use it to solve problems.
Pythagorean Theorem: In a right triangle, the sum of squares of the legs a and b is equal to the square of the hypotenuse c.
a2 + b2 = c2
We can use it to find the length of a side of a right triangle when the lengths of the other two sides are known.
### 12.1 Independent Practice – The Pythagorean Theorem – Page No. 379
Find the length of the missing side of each triangle. Round your answers to the nearest tenth.
Question 4.
________ cm
Explanation: According to Pythagorean theorem we consider values of a = 4cm, b = 8cm.
c2 = a2 + b2
= 42 + 82
= 16 + 64
c2= 80, c= √80 => 8.944
After rounding to nearest tenth value c= 8.9cm
Question 5.
________ in.
Explanation: According to Pythagorean theorem we consider values of b = 8in, c= 14in
c2 = a2 + b2
142 = a2 + 82
196 = a2 + 64
a2 = 196 – 64
a = √132 => 11.4891
a = 11.5 in
Question 6.
The diagonal of a rectangular big-screen TV screen measures 152 cm. The length measures 132 cm. What is the height of the screen?
________ cm
Explanation: Let’s consider the diagonal of the TV screen as C = 152cm, length as A = 132 cm, and height of the screen as B.
As C2 = A2 + B2
1522 = 1322 + B2
23,104 = 17,424 + B2
B2 = 23,104 – 17,424
B = √5680 => 75.365
So the height of the screen B = 75.4cm
Question 7.
Dylan has a square piece of metal that measures 10 inches on each side. He cuts the metal along the diagonal, forming two right triangles. What is the length of the hypotenuse of each right triangle to the nearest tenth of an inch?
________ in.
Explanation:
Using the Pythagorean Theorem, we have:
a2 + b2 = c2
102 + 102 = c2
100 + 100 = c2
200 = c2
We are told to round the length of the hypotenuse of each right triangle to the nearest tenth of an inch, therefore: c = 14.1in
Question 8.
Represent Real-World Problems A painter has a 24-foot ladder that he is using to paint a house. For safety reasons, the ladder must be placed at least 8 feet from the base of the side of the house. To the nearest tenth of a foot, how high can the ladder safely reach?
________ ft
Explanation: Consider the below diagram. Length of the ladder C = 24ft, placed at a distance from the base B = 8ft, let the safest height be A.
By using Pythagorean Theorem:
C2 = A2 + B2
242 = A2 + 82
576 = A2 + 64
A2 = 576 – 64 => 512
A = √512 => 22.627
After rounding to nearest tenth, value of A = 22.6ft
Question 9.
What is the longest flagpole (in whole feet) that could be shipped in a box that measures 2 ft by 2 ft by 12 ft?
________ ft
Answer: The longest flagpole (in whole feet) that could be shipped in this box is 12 feet.
Explanation: From the above diagram we have to find the value of r, which gives us the length longest flagpole that could be shipped in the box. Where width w = 2ft, height h = 2ft and length l = 12ft.
First find s, the length of the diagonal across the bottom of the box.
w2 + l2 = s2
22 + 122 = s2
4 + 144 = s2
148 = s2
We use our expression for s to find r, since triangle with sides s, r, and h also form a right-angle triangle.
h2 + s2 = r2
22 + 148 = r2
4 + 148 = r2
152 = r2
r = 12.33ft.
Question 10.
Sports American football fields measure 100 yards long between the end zones, and are 53 $$\frac{1}{3}$$ yards wide. Is the length of the diagonal across this field more or less than 120 yards? Explain.
____________
Answer: The diagonal across this field is less than 120 yards.
Explanation: From the above details we will get a diagram as shown below.
We are given l = 100 and w = 53 = . If we denote with d the diagonal of the field, using the Pythagorean Theorem, we have:
l2 + w2 = d2
1002 + (160/3)2 = d2
10000 + (25600/9) = d2
9*10000 + 9*(25600/9) = 9* d2
90000 + 25600 = 9 d2
(115600/9) = d2
(340/9) = d2
d = 113.3
Hence the diagonal across this field is less than 120 yards.
Question 11.
Justify Reasoning A tree struck by lightning broke at a point 12 ft above the ground as shown. What was the height of the tree to the nearest tenth of a foot? Explain your reasoning.
________ ft
Answer: The total height of the tree was 52.8ft
Explanation:
By using the Pythagorean Theorem
a2 + b= c2
122 + 392 = c2
144 + 1521 = c2
1665 = c2
We are told to round the length of the hypotenuse to the nearest tenth of a foot, therefore: c = 40.8ft.
Therefore, the total height of the tree was:
height = a+c
height = 12 +40.8
height = 52.8 feet
### FOCUS ON HIGHER ORDER THINKING – The Pythagorean Theorem – Page No. 380
Question 12.
Multistep Main Street and Washington Avenue meet at a right angle. A large park begins at this corner. Joe’s school lies at the opposite corner of the park. Usually Joe walks 1.2 miles along Main Street and then 0.9 miles up Washington Avenue to get to school. Today he walked in a straight path across the park and returned home along the same path. What is the difference in distance between the two round trips? Explain.
________ mi
Answer: Joe walks 1.2 miles less if he follows the straight path across the park.
Explanation: Using the Pythagorean Theorem, we find the distance from his home to school following the straight path across the park:
a2 + b= c2
1.22 + 0.92 = c2
1.44 + 0.81 = c2
2.25 = c2
1.5 = c
Therefore, the distance of Joe’s round trip following the path across the park is 3 miles (dhome-school + dschool-home = 1.5 + 1.5). Usually, when he walks along Main Street and Washington Avenue, the distance of his round trip is 4.2 miles (dhome-school + dschool-home = (1.2 + 0.9) + (0.9+1.2)). As we can see, Joe walks 1.2 miles less if he follows the straight path across the park.
Question 13.
Analyze Relationships An isosceles right triangle is a right triangle with congruent legs. If the length of each leg is represented by x, what algebraic expression can be used to represent the length of the hypotenuse? Explain your reasoning.
Explanation: From the Pythagorean Theorem, we know that if a and b are legs and c is the hypotenuse, then a2 + b= c2. In our case, the length of each leg is represented by x, therefore we have:
a2 + b= c2
x2 + x2 = c2
2x2 = c2
c = x√ 2
Question 14.
Persevere in Problem Solving A square hamburger is centered on a circular bun. Both the bun and the burger have an area of 16 square inches.
a. How far, to the nearest hundredth of an inch, does each corner of the burger stick out from the bun? Explain.
________ in
Answer: Each corner of the burger sticks out 0.57 inches from the bun.
Explanation: Frist, we need to find the radius r of the circular bun. We know that its area A is 16 square inches, therefore:
A = πr2
16 = 3.14*r2
r2 = (16/3.14)
r = 2.26
Then, we need to find the side s of the square hamburger. We know that its area A is 16 square inches, therefore:
A = s2
16 = s2
s = 4
Using the Pythagorean Theorem, we have to find diagonal d of the square hamburger:
s2 + s2 = d2
42 + 42 = d2
16 + 16 = d2
32 = d2
d = 5.66
To find how far does each corner of the burger stick out from the bun, we denote this length by a and we get:
a = (d/2) – r => (5.66/2) – 2.26
a = 0.57.
Therefore, Each corner of the burger sticks out 0.57 inches from the bun.
Question 14.
b. How far does each bun stick out from the center of each side of the burger?
________ in
Answer: Each bun sticks out 0.26 inches from the center of each side of the burger.
Explanation:
We found that r = 2.26 and s = 4. To find how far does each bun stick out from the center of each side of the burger, we denote this length by b and we get:
b = r – (s/2) = 2.26 – (4/2)
b = 0.26 inches.
Question 14.
c. Are the distances in part a and part b equal? If not, which sticks out more, the burger or the bun? Explain.
Answer: The distances a and b are not equal. From the calculations, we found that the burger sticks out more than the bun.
### Guided Practice – Converse of the Pythagorean Theorem – Page No. 384
Question 1.
Lashandra used grid paper to construct the triangle shown.
a. What are the lengths of the sides of Lashandra’s triangle?
_______ units, _______ units, _______ units,
Answer: The length of Lashandra’s triangle is 8 units, 6 units, 10 units.
Question 1.
b. Use the converse of the Pythagorean Theorem to determine whether the triangle is a right triangle.
The triangle that Lashandra constructed is / is not a right triangle.
_______ a right triangle
Answer: Lashandra’s triangle is right angled triangle as it satisfied Pythagorean theorem
Explanation:
Verifying with Pythagorean formula a2 + b= c2
82 + 62 = 102
64 + 36 =100
100 = 100.
Question 2.
A triangle has side lengths 9 cm, 12 cm, and 16 cm. Tell whether the triangle is a right triangle.
Let a = _____, b = _____, and c = ______.
By the converse of the Pythagorean Theorem, the triangle is / is not a right triangle.
_______ a right triangle
Answer: The given triangle is not a right-angled triangle
Explanation: Verifying with Pythagorean formula a2 + b= c2
92 + 122 = 162
81 + 144 = 256
225 ≠ 256.
Hence given dimensions are not from the right angled triangle.
Question 3.
The marketing team at a new electronics company is designing a logo that contains a circle and a triangle. On one design, the triangle’s side lengths are 2.5 in., 6 in., and 6.5 in. Is the triangle a right triangle? Explain.
_______
Answer: It is a right-angled triangle.
Explanation: Let a = 2.5, b = 6 and c= 6.5
Verifying with Pythagorean formula a2 + b= c2
2.52 + 62 = 6.52
6.25 + 36 = 42.25
42.25 = 42.25.
Hence it is a right-angled triangle.
ESSENTIAL QUESTION CHECK-IN
Question 4.
How can you use the converse of the Pythagorean Theorem to tell if a triangle is a right triangle?
Answer: Knowing the side lengths, we substitute them in the formula a2 + b= c2, where c contains the biggest value. If the equation holds true, then the given triangle is a right triangle. Otherwise, it is not a right triangle.
### 12.2 Independent Practice – Converse of the Pythagorean Theorem – Page No. 385
Tell whether each triangle with the given side lengths is a right triangle.
Question 5.
11 cm, 60 cm, 61 cm
______________
Answer: Since 112 + 602 = 612, the triangle is a right-angled triangle.
Explanation: Let a = 11, b = 60 and c= 61
Using the converse of the Pythagorean Theorem a2 + b= c2
112 + 602 = 612
121 + 3600 = 3721
3721 = 3721.
Since 112 + 602 = 612, the triangle is a right-angled triangle.
Question 6.
5 ft, 12 ft, 15 ft
______________
Answer: Since 52 + 122 ≠ 152, the triangle is not a right-angled triangle.
Explanation: Let a = 5, b = 12 and c= 15
Using the converse of the Pythagorean Theorem a2 + b= c2
52 + 122 = 152
25 + 144 = 225
169 ≠ 225.
Since 52 + 122 ≠ 152, the triangle is not a right-angled triangle.
Question 7.
9 in., 15 in., 17 in.
______________
Answer: Since 92 + 152 ≠ 172, the triangle is not a right-angled triangle.
Explanation: Let a = 9, b = 15 and c= 17
Using the converse of the Pythagorean Theorem a2 + b= c2
92 + 152 = 172
81 + 225 = 225
306 ≠ 225.
Since 92 + 152 ≠ 172, the triangle is not a right-angled triangle.
Question 8.
15 m, 36 m, 39 m
______________
Answer: Since 152 + 362 = 392, the triangle is a right-angled triangle.
Explanation: Let a = 15, b = 36 and c= 39
Using the converse of the Pythagorean Theorem a2 + b= c2
152 + 362 = 392
225 + 1296 = 1521
1521 = 1521.
Since 152 + 362 = 392, the triangle is a right-angled triangle.
Question 9.
20 mm, 30 mm, 40 mm
______________
Answer: Since 202 + 302 ≠ 402, the triangle is not a right-angled triangle.
Explanation: Let a = 20, b = 30 and c= 40
Using the converse of the Pythagorean Theorem a2 + b= c2
202 + 302 = 402
400 + 900 = 1600
1300 ≠ 1600.
Since 202 + 302 ≠ 402, the triangle is not a right-angled triangle.
Question 10.
20 cm, 48 cm, 52 cm
______________
Answer: Since 202 + 482 = 522, the triangle is a right-angled triangle.
Explanation: Let a = 20, b = 48 and c= 52
Using the converse of the Pythagorean Theorem a2 + b= c2
202 + 482 = 522
400 + 2304 = 2704
2704 = 2704.
Since 202 + 482 = 522, the triangle is a right-angled triangle.
Question 11.
18.5 ft, 6 ft, 17.5 ft
______________
Answer: Since 62 + 17.52 = 18.52, the triangle is a right-angled triangle.
Explanation: Let a = 6, b = 17.5 and c= 18.5
Using the converse of the Pythagorean Theorem a2 + b= c2
62 + 17.52 = 18.52
36 + 306.25 = 342.25
342.5 = 342.25.
Since 62 + 17.52 = 18.52, the triangle is a right-angled triangle.
Question 12.
2 mi, 1.5 mi, 2.5 mi
______________
Answer: Since 22 + 1.52 = 2.52, the triangle is a right-angled triangle.
Explanation: Let a = 2, b = 1.5 and c= 2.5
Using the converse of the Pythagorean Theorem a2 + b= c2
22 + 1.52 = 2.52
4 + 2.25 = 6.25
6.25 = 6.25.
Since 22 + 1.52 = 2.52, the triangle is a right-angled triangle.
Question 13.
35 in., 45 in., 55 in.
______________
Answer: Since 352 + 452 ≠ 552, the triangle is not a right-angled triangle.
Explanation: Let a = 35, b = 45 and c= 55
Using the converse of the Pythagorean Theorem a2 + b= c2
352 + 452 = 552
1225 + 2025 = 3025
3250 ≠ 3025.
Since 352 + 452 ≠ 552, the triangle is not a right-angled triangle.
Question 14.
25 cm, 14 cm, 23 cm
______________
Answer: Since 142 + 232 ≠ 252, the triangle is not a right-angled triangle.
Explanation: Let a = 14, b = 23 and c= 25 (longest side)
Using the converse of the Pythagorean Theorem a2 + b= c2
142 + 232 = 252
196 + 529 = 625
725 ≠ 625.
Since 142 + 232 ≠252, the triangle is not a right-angled triangle.
Question 15.
The emblem on a college banner consists of the face of a tiger inside a triangle. The lengths of the sides of the triangle are 13 cm, 14 cm, and 15 cm. Is the triangle a right triangle? Explain.
________
Answer: Since 132 + 142 ≠ 152, the triangle is not a right-angled triangle.
Explanation: Let a = 13, b = 14 and c= 15
Using the converse of the Pythagorean Theorem a2 + b= c2
132 + 142 = 152
169 + 196 = 225
365 ≠ 225.
Since 132 + 142 ≠ 152, the triangle is not a right-angled triangle.
Question 16.
Kerry has a large triangular piece of fabric that she wants to attach to the ceiling in her bedroom. The sides of the piece of fabric measure 4.8 ft, 6.4 ft, and 8 ft. Is the fabric in the shape of a right triangle? Explain.
________
Answer: The triangular piece of fabric that Kerry has is in the shape of a right angle since it follows the Pythagorean theorem.
Explanation: Let a = 4.8, b = 6.4 and c= 8
Using the converse of the Pythagorean Theorem a2 + b= c2
4.82 + 6.42 = 82
23.04 + 40.96 = 64
64 = 64.
Since 4.82 + 6.42 = 82, the triangle is a right-angled triangle.
Question 17.
A mosaic consists of triangular tiles. The smallest tiles have side lengths 6 cm, 10 cm, and 12 cm. Are these tiles in the shape of right triangles? Explain.
________
Answer: Since 62 + 102 ≠ 122, by the converse of the Pythagorean Theorem, we say that the tiles are not in the shape of right-angled triangle.
Explanation: Let a = 6, b = 10 and c= 12
Using the converse of the Pythagorean Theorem a2 + b= c2
62 + 102 = 122
36 + 100 = 144
136 ≠ 144.
Since 62 + 102 ≠ 122, by the converse of the Pythagorean Theorem, we say that the tiles are not in the shape of right-angled triangle.
Question 18.
History In ancient Egypt, surveyors made right angles by stretching a rope with evenly spaced knots as shown. Explain why the rope forms a right angle.
Answer: The rope has formed a right-angled triangle because the length of its sides follows Pythagorean Theorem.
Explanation: The knots are evenly placed at equal distances
The lengths in terms of knots are a=4 knots, b = 3knots, c = 5 knots
Therefore a2 + b= c2
42 + 3= 52
16+9 = 25
25 = 25.
Hence rope has formed a right-angled triangle because the length of its sides follows Pythagorean Theorem.
### Converse of the Pythagorean Theorem – Page No. 386
Question 19.
Justify Reasoning Yoshi has two identical triangular boards as shown. Can he use these two boards to form a rectangle? Explain.
Answer: Since it was proved that both can form a right-angled triangle, we can form a rectangle by joining them.
Explanation: Given both triangles are identical, if both are right-angled triangles then we can surely join to form a rectangle.
Let’s consider a = 0.75, b= 1 and c=1.25.
By using converse Pythagorean Theorem a2 + b= c2
0.752 + 12 = 1.252
0.5625 + 1 = 1.5625
1.5625 = 1.5625.
Since it was proved that both can form right angled triangle, we can form a rectangle by joining them.
Question 20.
Critique Reasoning Shoshanna says that a triangle with side lengths 17 m, 8 m, and 15 m is not a right triangle because 172 + 82 = 353, 152 = 225, and 353 ≠ 225. Is she correct? Explain
_______
Answer: She is not right, A triangle with sides 15, 8, and 17 is a right-angled triangle.
Explanation: Lets consider a =15, b= 8 and c = 17 (which is long side)
We will verify by using converse Pythagorean Theorem a2 + b= c2
152 + 82 = 172
225 + 64 = 289
289 = 289.
Since the given dimensions satisfied Pythagorean Theorem, we can say it is a right-angled triangle. In the given above statement what Shoshanna did was c2 + b2 = a2, which is not the correct definition of the Pythagorean Theorem.
FOCUS ON HIGHER ORDER THINKING
Question 21.
Make a Conjecture Diondre says that he can take any right triangle and make a new right triangle just by doubling the side lengths. Is Diondre’s conjecture true? Test his conjecture using three different right triangles.
_______
Answer: Yes, Diondre’s conjecture is true. By doubling the sides of a right triangle would create a new right triangle.
Explanation: Given a right triangle, the Pythagorean Theorem holds. Therefore, a2 + b= c2
If we double the side lengths of that triangle, we get:
(2a)2 + (2b)= (2c)2
4a2 + 4b2 = 4c2
4(a2 + b2) = 4c2
a2 + b= c2
As we can see doubling the sides of a right triangle would create a new right triangle.We can test that by using three different right triangles.
The triangle with sides a = 6, b = 8 and c = 10 is a right triangle. We double its sides and check if the new triangle is a right triangle. After doubling value of a = 12, b = 16 and c = 20.
122 + 162 = 202
144 + 256 = 400
400 = 400
Hence proved!
Since 122 + 162 = 202, the new triangle is a right triangle by the converse of the Pythagorean Theorem.
The triangle with sides a = 3, b = 4 and c = 5 is a right triangle. We double its sides and check if the new triangle is a right triangle. After doubling value of a = 6, b = 8 and c = 10.
62 + 82 = 102
36 + 64 = 100
100 = 100
Hence proved!
Since 62 + 82 = 102, the new triangle is a right triangle by the converse of the Pythagorean Theorem.
The triangle with sides a = 12, b = 16 and c = 20 is a right triangle. We double its sides and check if the new triangle is a right triangle. After doubling value of a = 24, b = 32 and c = 40.
242 + 322 = 402
576 + 1024 = 1600
1600 = 1600
Hence proved!
Since 242 + 322 = 402, the new triangle is a right triangle by the converse of the Pythagorean Theorem.
Question 22.
Draw Conclusions A diagonal of a parallelogram measures 37 inches. The sides measure 35 inches and 1 foot. Is the parallelogram a rectangle? Explain your reasoning.
_______
Answer: Since 122 + 352 = 372, the triangle is right triangle. Therefore, the given parallelogram is a rectangle.
Explanation: A rectangle is a parallelogram where the interior angles are right angles. To prove if the given parallelogram is a rectangle, we need to prove that the triangle formed by the diagonal of the parallelogram and two sides of it, is a right triangle. Converting all the values into inches, we have a = 12, b = 35 and c = 37. Using the converse of the Pythagorean Theorem, we have:
a2 + b= c2
122 + 352 = 372
144 + 1225 = 1369
1369 = 1369.
Since 122 + 352 = 372, the triangle is right triangle. Therefore, the given parallelogram is a rectangle.
Question 23.
Represent Real-World Problems A soccer coach is marking the lines for a soccer field on a large recreation field. The dimensions of the field are to be 90 yards by 48 yards. Describe a procedure she could use to confirm that the sides of the field meet at right angles.
Answer: To confirm that the sides of the field meet at right angles, she could measure the diagonal of the field and use the converse of the Pythagorean Theorem. If a2 + b= c2 (where a = 90, b = 48 and c is the length of the diagonal), then the triangle is right triangle. This method can be used for every corner to decide if they form right angles or not.
### Guided Practice – Distance Between Two Points – Page No. 390
Question 1.
Approximate the length of the hypotenuse of the right triangle to the nearest tenth using a calculator.
_______ units
Answer: The length of the hypotenuse of the right triangle to the nearest tenth is 5.8 units.
Explanation: From the above figure let’s take
Length of the vertical leg = 3 units
Length of the horizontal leg = 5 units
let length of the hypotenuse = c
By using Pythagorean Theorem a2 + b= c2
c2 = 32 + 52
c2 = 9 +25
c = √34 => 5.830.
Therefore Length of the hypotenuse of the right triangle to the nearest tenth is 5.8 units.
Question 2.
Find the distance between the points (3, 7) and (15, 12) on the coordinate plane.
_______ units
Answer: Distance between points on the coordinate plane is 13
Explanation: So (x1, y1) = (3,7) and (x2, y2) = (15, 12)
distance formula d = √( x2 – x1)2 + √( y2 – y1)2
d = √(15 -3)2 + √(12 – 7)2
d = √122 + 52
d = √144 + 25
d = √169 => 13
Therefore distance between points on the coordinate plane is 13.
Question 3.
A plane leaves an airport and flies due north. Two minutes later, a second plane leaves the same airport flying due east. The flight plan shows the coordinates of the two planes 10 minutes later. The distances in the graph are measured in miles. Use the Pythagorean Theorem to find the distance shown between the two planes.
_______ miles
Answer: The distance between the two planes is 103.6 miles.
Explanation:
Length of the vertical dv = √(80 -1)2 + √(1-1)2
= √792 => 79.
Length of the horizontal dh = √(68 -1)2 + √(1-1)2
= √672 => 67.
Distance between the two planes D = √(792 + 672)
= √(6241+4489) => √10730
= 103.5857 => 103.6 miles.
ESSENTIAL QUESTION CHECK-IN
Question 4.
Describe two ways to find the distance between two points on a coordinate plane.
Explanation: We can draw a right triangle whose hypotenuse is the segment connecting the two points and then use the Pythagorean Theorem to find the length of that segment. We can also the Distance formula to find the length of that segment.
For example, plot three points; (1,2), (20,2) and (20,12)
Using the Pythagorean Theorem:
The length of the horizontal leg is the absolute value of the difference between the x-coordinates of the points (1,2) and (20,2).
|1 – 20| = 19
The length of the horizontal leg is 19.
The length of the vertical leg is the absolute value of the difference between the y-coordinates of the points (20,2) and (20,12).
|2 – 12| = 10
The length of the vertical leg is 10.
Let a = 19, b = 10 and let c represent the hypotenuse. Find c.
a2 + b= c2
192 + 10= c2
361 + 100 = c2
461 = c2
distance is 21.5 = c
Using the Distance formula:
d= √( x2 – x1)2 + √( y2 – y1)2
The length of the horizontal leg is between (1,2) and (20,2).
d= √( x2 – x1)2 + √( y2 – y1)2
= √(20 -1)2 + √(2-2)2
= √(19)2 + √(0)2
= √361 => 19
The length of the vertical leg is between (20,2) and (20,12).
d= √( x2 – x1)2 + √( y2 – y1)2
= √(20 -20)2 + √(12-2)2
= √(0)2 +√(10)2
= √100 => 10
The length of the diagonal leg is between (1,2) and (20,12).
d= √( x2 – x1)2 + √( y2 – y1)2
= √(20 -1)2 + √(12-2)2
= √(19)2 + √(10)2
= √(361+100) => √461 = 21.5
### 12.3 Independent Practice – Distance Between Two Points – Page No. 391
Question 5.
A metal worker traced a triangular piece of sheet metal on a coordinate plane, as shown. The units represent inches. What is the length of the longest side of the metal triangle? Approximate the length to the nearest tenth of an inch using a calculator. Check that your answer is reasonable.
_______ in.
Answer: The length of the longest side of the metal triangle to the nearest tenth is 7.8 units.
Explanation: From the above figure let’s take
Length of the vertical leg = 6 units
Length of the horizontal leg = 5 units
let length of the hypotenuse = c
By using Pythagorean Theorem a2 + b= c2
c2 = 62 + 52
c2 = 36 +25
c = √61 => 7.8
Therefore Length of the longest side of the metal triangle to the nearest tenth is 7.8 units.
Question 6.
When a coordinate grid is superimposed on a map of Harrisburg, the high school is located at (17, 21) and the town park is located at (28, 13). If each unit represents 1 mile, how many miles apart are the high school and the town park? Round your answer to the nearest tenth.
_______ miles
Answer: The high school and the town park are 13.6 miles apart.
Explanation: The coordinates of the high school are said to be (17,21), where as the coordinates of the park are (28,13). In a coordinate plane, the distance d between the points (17,21) and (28,13) is:
d= √( x2 – x1)2 + √( y2 – y1)2
= √(28 -17)2 + √(13-21)2
= √(11)2 + √(-8)2
= √(121+64) => √185 = 13.6014
Rounding the answer to the nearest tenth:
d = 13.6.
Taking into consideration that each unit represents 1 mile, the high school and town park are 13.6 miles apart.
Question 7.
The coordinates of the vertices of a rectangle are given by R(- 3, – 4), E(- 3, 4), C (4, 4), and T (4, – 4). Plot these points on the coordinate plane at the right and connect them to draw the rectangle. Then connect points E and T to form diagonal $$\overline { ET }$$.
a. Use the Pythagorean Theorem to find the exact length of $$\overline { ET }$$.
Explanation:
Taking into consideration the triangle TRE, the length of the vertical leg (ER) is 8 units. The length of the horizontal leg (RT) is 7 units. Let a = 8 and b =7. Let c represent the length of the hypotenuse, the diagonal ET. We use the Pythagorean Theorem to find c.
a2 + b= c2
c2 = 82 + 72
c2 = 64 +49
c = √113 => 10,63.
The diagonal ET is about 10.63 units long.
Question 7.
b. How can you use the Distance Formula to find the length of $$\overline { ET }$$ ? Show that the Distance Formula gives the same answer.
Answer: The diagonal ET is about 10.63 units long. As we can see the answer is the same as the one we found using the Pythagorean Theorem.
Explanation: Using the distance formula, in a coordinate plane, the distance d between the points E(-3,4) and T(4, -4) is:
d= √( x2 – x1)2 + √( y2 – y1)2
= √(4 – (-3))2 + √(- 4 – 4)2
= √(7)2 + √(-8)2
= √(49+64) => √113 = 10.63.
The diagonal ET is about 10.63 units long. As we can see the answer is the same as the one we found using the Pythagorean Theorem.
Question 8.
Multistep The locations of three ships are represented on a coordinate grid by the following points: P(- 2, 5), Q(- 7, – 5), and R(2, – 3). Which ships are farthest apart?
Answer: Ships P and Q are farthest apart
Explanation: Distance Formula: In a coordinate plane, the distance d between two points (x1,y1) and (x2,y2) is:
d= √( x2 – x1)2 + √( y2 – y1)2
The distance d1 between the two points P(-2,5) and Q(-7,-5) is:
d1 = √( xQ – xP)2 + √( yQ – yP)2
= √(-7 – (-2))2 + √(- 5 – 5)2
= √(-5)2 + √(-10)2
= √(25+100) => √125 = 11.18
The distance d2 between the two points Q(-7,-5) and R(2,-3) is:
d3 = √( xR – xQ)2 + √( yR – yQ)2
= √(2 – (-7))2 + √(- 3 – 5)2
= √(9)2 + √(2)2
= √(81+4) => √85 = 9.22
The distance d3 between the two points P(-2,5) and R(2,-3) is:
d3 = √( xR – xP)2 + √( yR – yP)2
= √(2 – (-2))2 + √(- 3 – 5)2
= √(4)2 + √(-8)2
= √(16+64) => √80 = 8.94.
As we can see, the greatest distance is d1 11.8, which means that ships P and Q are farthest apart.
### Distance Between Two Points – Page No. 392
Question 9.
Make a Conjecture Find as many points as you can that are 5 units from the origin. Make a conjecture about the shape formed if all the points 5 units from the origin were connected.
Explanation: Some of the points that are 5 units away from the origin are: (0,5), (3,4), (4,3),(5,0),(4,-3),(3,-4),(0,-5),(-3,-4),(-4,-3),(-5,0),(-4,3),(-3,4) etc, If all the points 5 units away from the origin are connected, a circle would be formed.
Question 10.
Justify Reasoning The graph shows the location of a motion detector that has a maximum range of 34 feet. A peacock at point P displays its tail feathers. Will the motion detector sense this motion? Explain.
Answer: Considering each unit represents 1 foot, the motion detector, and peacock are 33.5 feet apart. Since the motion detector has a maximum range of 34 feet, it means that it will sense the motion of the peacock’s feathers.
Explanation: The coordinates of the motion detector are said to be (0,25), whereas the coordinates of the peacock are (30,10). In a coordinate plane, the distance d between the points (0,25) and (30,10) is:
d = √( x2 – x1)2 + √( y2 – y1)2
= √(30 – 0)2 + √(10 – 25)2
= √(30)2 + √(-15)2
= √(900+225) => √1125.
Rounding answer to the nearest tenth:
d = 33.5 feet.
Considering each unit represents 1 foot, the motion detector and peacock are 33.5 feet apart. Since the motion detector has a maximum range of 34 feet, it means that it will sense the motion of the peacock’s feathers.
FOCUS ON HIGHER ORDER THINKING
Question 11.
Persevere in Problem Solving One leg of an isosceles right triangle has endpoints (1, 1) and (6, 1). The other leg passes through the point (6, 2). Draw the triangle on the coordinate plane. Then show how you can use the Distance Formula to find the length of the hypotenuse. Round your answer to the nearest tenth.
Explanation:
One leg of an isosceles right triangle has endpoints (1,1) and (6,1), which means that the leg is 5 units long. Since the triangle is isosceles, the other leg should be 5 units long too, therefore the endpoints of the second leg that passes through the point (6,2) are (6,1) and (6,6).
In the coordinate plane, the length of the hypotenuse is the distance d between the points (1,1) and (6,6).
d = √( x2 – x1)2 + √( y2 – y1)2
= √(6 – 1)2 + √(6 – 1)2
= √(5)2 + √(5)2
= √(25+25) => √50.
d = 7.1.
The hypotenuse is around 7.1 units long.
Question 12.
Represent Real-World Problems The figure shows a representation of a football field. The units represent yards. A sports analyst marks the locations of the football from where it was thrown (point A) and where it was caught (point B). Explain how you can use the Pythagorean Theorem to find the distance the ball was thrown. Then find the distance.
_______ yards
Answer: The distance between point A and B is 37 yards
Explanation:
To find the distance between points A and B, we draw segment AB and label its length d. Then we draw vertical segment AC and Horizontal segment CB. We label the lengths of these segments a and b. triangle ACB is a right triangle with hypotenuse AB.
Since AC is vertical segment, its length, a, is the difference between its y-coordinates. Therefore, a = 26 – 14 = 12 units.
Since CB is horizontal segment, its length b is the difference between its x-coordinates. Therefore, b = 75 – 40 = 35units.
We use the Pythagorean Theorem to find d, the length of segment AB.
d2 = a2 + b2
d2 = 122 + 352
d2 = 144 + 1225
d2 = 1369 => d = √1369 => 37
The distance between point A and B is 37 yards
### Ready to Go On? – Model Quiz – Page No. 393
12.1 The Pythagorean Theorem
Find the length of the missing side.
Question 1.
________ meters
Answer: Length of missing side is 28m
Explanation: Lets consider value of a = 21 and c = 35.
Using Pythagorean Theorem a2 + b= c2
212 + b2 = 352
441 + b2 = 1225
b2= 784 => b = √784 = 28.
Therefore length of missing side is 28m.
Question 2.
________ ft
Answer: Length of missing side is 34ft
Explanation: Let’s consider value of a = 16 and b = 30.
Using Pythagorean Theorem a2 + b= c2
162 + 302 = c2
256 + 900 = c2
c2= 1156 => c = √1156 = 34.
Therefore length of missing side is 34ft.
12.2 Converse of the Pythagorean Theorem
Tell whether each triangle with the given side lengths is a right triangle.
Question 3.
11, 60, 61
____________
Answer: Since 112 + 602 = 612, by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.
Explanation: Let a = 11, b = 60 and c= 61
Using the converse of the Pythagorean Theorem a2 + b= c2
112 + 602 = 612
121 + 3600 = 3721
3721 = 3721
Since 112 + 602 = 612, by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.
Question 4.
9, 37, 40
____________
Answer: Since 92 + 372 ≠ 402, by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.
Explanation: Let a = 9, b = 37 and c= 40
Using the converse of the Pythagorean Theorem a2 + b= c2
92 + 372 = 402
81 + 1369 = 1600
1450 ≠ 3721.
Since 92 + 372 ≠ 402, by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.
Question 5.
15, 35, 38
____________
Answer: Since 152 + 352 ≠ 382, by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.
Explanation: Let a = 15, b = 35 and c= 38
Using the converse of the Pythagorean Theorem a2 + b= c2
152 + 352 = 382
225 + 1225 = 1444
1450 ≠ 1444
Since 152 + 352 ≠ 382, by the converse of the Pythagorean Theorem, we say that the given sides are not in the shape of right-angled triangle.
Question 6.
28, 45, 53
____________
Answer: Since 282 + 452 = 532, by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.
Explanation: Let a = 28, b = 45 and c= 53
Using the converse of the Pythagorean Theorem a2 + b= c2
282 + 452 = 532
784 + 2025 = 2809
2809 = 2809
Since 282 + 452 = 532, by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.
Question 7.
Keelie has a triangular-shaped card. The lengths of its sides are 4.5 cm, 6 cm, and 7.5 cm. Is the card a right triangle?
____________
Answer: Since 4.52 + 62 = 7.52, by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.
Explanation: Let a = 4.5, b = 6 and c= 7.5
Using the converse of the Pythagorean Theorem a2 + b= c2
4.52 + 62 = 7.52
20.25 + 36 = 56.25
56.25= 56.25
Since 4.52 + 62 = 7.52, by the converse of the Pythagorean Theorem, we say that the given sides are in the shape of right-angled triangle.
12.3 Distance Between Two Points
Find the distance between the given points. Round to the nearest tenth.
Question 8.
A and B
________ units
Answer: Distance between A and B is 6.7 units
Explanation: A= (-2,3) and B= (4,6)
Distance between A and B is d = √( x2 – x1)2 + √( y2 – y1)2
= √(4 – (-2)2 + √(6 – 3)2
= √(6)2 + √(3)2
= √(36+9) => √45 = 6.7 units
Question 9.
B and C
________ units
Answer: Distance between B and C is 7.07 units
Explanation: B= (4,6) and C= (3,1)
Distance between B and C is d = √( x2 – x1)2 + √( y2 – y1)2
= √(4 – 3)2 + √(6 – (-1))2
= √(1)2 + √(7)2
= √(1+49) => √50 = 7.07 units
Question 10.
A and C
________ units
Answer: Distance between A and C is 6.403 units
Explanation: A= (-2,3) and C= (3, -1)
Distance between A and C is d = √( x2 – x1)2 + √( y2 – y1)2
= √(3 – (-2)2 + √(-1 – 3)2
= √(5)2 + √(-4)2
= √(25+16) => √41 = 6.403 units
ESSENTIAL QUESTION
Question 11.
How can you use the Pythagorean Theorem to solve real-world problems?
Answer: We can use the Pythagorean Theorem to find the length of a side of a right triangle when we know the lengths of the other two sides. This application is usually used in architecture or other physical construction projects. For example, it can be used to find the length of a ladder, if we know the height of the wall and distance on the ground from the wall of the ladder.
### Selected Response – Mixed Review – Page No. 394
Question 1.
What is the missing length of the side?
A. 9 ft
B. 30 ft
C. 39 ft
D. 120 ft
Explanation:
Given a= 80 ft
b= ?
c= 89 ft
As a2+b2=c 2
802+b2= 892
6,400+b2= 7,921
b2= 7,921-6,400
b= √1,521
b= 39 ft.
Question 2.
Which relation does not represent a function?
Options:
A. (0, 8), (3, 8), (1, 6)
B. (4, 2), (6, 1), (8, 9)
C. (1, 20), (2, 23), (9, 26)
D. (0, 3), (2, 3), (2, 0)
Explanation: The value of X is the same for 2 points and 2 values of Y [(2, 3), (2, 0)]. The value of X is repeated for a function to exist, no two points can have the same X coordinates.
Question 3.
Two sides of a right triangle have lengths of 72 cm and 97 cm. The third side is not the hypotenuse. How long is the third side?
Options:
A. 25 cm
B. 45 cm
C. 65 cm
D. 121 cm
Explanation:
Given a= 72 cm
b= ?
c= 97 cm
As a2+b2=c 2
722+b2= 972
5,184+b2= 9,409
b2= 9,409-5,184
b= √4,225
b= 65 cm.
Question 4.
To the nearest tenth, what is the distance between point F and point G?
Options:
A. 4.5 units
B. 5.0 units
C. 7.3 units
D. 20 units
Explanation:
Given F= (-1,6) =(x1,y1).
G= (3,4) = (x2,y2).
The difference between F&G points is
d= √(x2-x1)2 + (y2-y1)2
= √(3 – (-1))2 + (4 – 6)2
= √(4)2 + (-2)2
= √16+4
= √20
= 4.471
= 4.5 units.
Question 5.
A flagpole is 53 feet tall. A rope is tied to the top of the flagpole and secured to the ground 28 feet from the base of the flagpole. What is the length of the rope?
Options:
A. 25 feet
B. 45 feet
C. 53 feet
D. 60 feet
Explanation:
By Pythagorean theorem
a2+b2=c 2
532+282= C2
2,809+784= C2
C2 = 9,409-5,184
C2 = 3,593
C= √3,593
C= 59.94 feet
=60 feet.
Question 6.
Which set of lengths are not the side lengths of a right triangle?
Options:
A. 36, 77, 85
B. 20, 99, 101
C. 27, 120, 123
D. 24, 33, 42
Explanation:
Check if side lengths in option A form a right triangle.
Let a= 36, b= 77, c= 85
By Pythagorean theorem
a2+b2=c 2
362+772= 852
1,296+ 5,929= 7,225
7,225= 7,225
As 362+772= 852 the triangle is a right triangle.
Check if side lengths in option B form a right triangle.
Let a= 20, b= 99, c= 101
By Pythagorean theorem
a2+b2=c 2
202+992= 1012
400+ 9,801= 10,201
10,201= 10,201
As 202+992= 1012 the triangle is a right triangle.
Check if side lengths in option B form a right triangle.
Let a= 27, b= 120, c= 123
By Pythagorean theorem
a2+b2=c 2
272+1202= 1232
729+ 14,400= 15,129
15,129= 15,129
As 272+1202= 1232 the triangle is a right triangle.
Check if side lengths in option B form a right triangle.
Let a= 27, b= 120, c= 123
By Pythagorean theorem
a2+b2=c 2
242+332= 422
576+ 1,089= 1,764.
1,665= 1,764
As 242+332 is not equal to 422 the triangle is a right triangle.
Question 7.
A triangle has one right angle. What could the measures of the other two angles be?
Options:
A. 25° and 65°
B. 30° and 15°
C 55° and 125°
D 90° and 100°
Explanation:
The sum of all the angles of a triangle is 180
<A+<B+<C= 180°
<A+<B+ 90°= 180°
<A+<B= 180°-90°
<A+<B= 90, here we will verify with the given options.
25°+65°= 90°
So, the measure of the other two angles are 25° and 65°
Question 8.
A fallen tree is shown on the coordinate grid below. Each unit represents 1 meter.
a. What is the distance from A to B?
_______ meters
Explanation:
A= (-5,3)
B= (8,0)
Distance between A & B is
D= √{8-(-5)2 + (0-3)2
= √(13)2 + (-3)2
= √169+9
= √178
= 13.34 m.
Question 8.
b. What was the height of the tree before it fell?
_______ meters
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Section - Collinear
Chapter 12 Class 11 - Intro to Three Dimensional Geometry
Concept wise
Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
### Transcript
Example 8 Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear. Let points be A (– 4, 6, 10) , B (2, 4, 6) , C (14, 0, – 2) Point A, B, & C are collinear if point C divides AB in some ratio externally & internally We know that Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is (x, y ,z) = ((mx2 + nx1)/(m + n),(my2 + ny1)/(m + n), (〖𝑚𝑧〗_2 + 〖𝑛𝑧〗_1)/(𝑚 + 𝑛)) Here, let point C(14, 0, – 2) divide A(– 4, 6, 10) , B(2, 4, 6) in the ratio k : 1 Here, x1 = – 4, y1 = 6, z1 = 10 x2 = 2, y2 = 4, z2 = 6 & m = k , n = 1 Putting values (14, 0, – 2) = ((𝑘(2) + 1(−4))/(𝑘 + 1),(𝑘(4) + 1 (6))/(𝑘 + 1),(𝑘 (6) + 1 (10))/(𝑘 +1)) (14, 0, – 2) = ((2𝑘 − 4 )/( 𝑘 + 1),(4𝑘 + 6)/(k +1),(6𝑘 + 10)/( k + 1)) Comparing x – coordinate Comparing x – coordinate 14 = (2𝑘 − 4)/(𝑘 + 1) (k + 1) (14) = 2k – 4 14k + 14 = 2k – 4 14k – 2k = – 4 – 14 12k = – 18 k = (−18)/12 k = (−3)/2 Since k is negative Point C divides line segment AB externally in the ratio 3 : 2 Thus, A, B & C are collinear
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# Ex.2.4 Q3 fractions-and-decimals Solutions-Ncert Maths Class 7
Go back to 'Ex.2.4'
## Question
Find:
(i) \begin{align} \frac{7}{3} \div 2\end{align}
(ii) \begin{align} \frac{4}{9} \div 5 \end{align}
(iii) \begin{align} \frac{6}{{13}} \div 7\end{align}
(iv) \begin{align} 4\frac{1}{3} \div 3\end{align}
(v) \begin{align} 3\frac{1}{2} \div 4\end{align}
(vi) \begin{align} 4\frac{3}{7} \div 7\end{align}
Video Solution
Fractions And Decimals
Ex 2.4 | Question 3
## Text Solution
What is known?
Expression.
What is unknown?
Value of the expression.
Reasoning:
To divide fractions take the reciprocal of the divisor and multiply it with dividend.
Steps:
(i) \begin{align} \frac{7}{3} \div 2\end{align}
\begin{align}&= \frac{7}{3} \times \frac{1}{2}\\&= \frac{7}{6}\;({\text{improper fraction}})\end{align}
Converting into mixed fraction,
we get \begin{align} = 1\frac{1}{6} \end{align}
(ii) \begin{align} \frac{4}{9} \div 5 \end{align}
\begin{align}&= \frac{4}{9} \times \frac{1}{5}\\&= \frac{4}{{45}}\end{align}
(iii) \begin{align} \frac{6}{{13}} \div 7\end{align}
\begin{align}& = \frac{6}{{13}} \times \frac{1}{7}\\ &= \frac{6}{{91}}\end{align}
(iv) \begin{align} 4\frac{1}{3} \div 3\end{align}
\begin{align}&= \frac{{13}}{3} \times \frac{1}{3}\\&= \frac{{13}}{9}\;({\text{improper fraction}}) \end{align}
Converting into mixed fraction,
we get \begin{align} = 1\frac{4}{9} \end{align}
(v) \begin{align} 3\frac{1}{2} \div 4\end{align}
\begin{align}&= \frac{7}{2} \times \frac{1}{4}\\&= \frac{7}{8}\end{align}
(vi) \begin{align} 4\frac{3}{7} \div 7\end{align}
\begin{align}&= 4\frac{3}{7} \times \frac{1}{7}\\&= \frac{{31}}{7} \times \frac{1}{7}\\&= \frac{{31}}{{49}}\end{align}
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## 4x4 Matrix
Find the determinant of the following 4 x 4 matrix:
Hint
$$\begin{bmatrix}a & b & c & d\\ e & f & g & h\\ i & j & k & l\\ m & n & o & p\end{bmatrix}=a\cdot \begin{bmatrix}f & g & h\\ j & k & l\\ n & o &p \end{bmatrix}-b\cdot \begin{bmatrix}e & g & h\\ i & k & l\\ m & o & p\end{bmatrix}+c\cdot \begin{bmatrix}e & f & h\\ i & j & l\\ m & n & p\end{bmatrix}-d\cdot \begin{bmatrix}e & f & g\\ i & j & k\\ m & n & o\end{bmatrix}$$$Hint 2 Remember, to solve a 3 x 3 matrix: $$\begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$$$
The formula to solve a 4 x 4 matrix, which breaks down the problem to multiple 3 x 3 matrices:
$$\begin{bmatrix}a & b & c & d\\ e & f & g & h\\ i & j & k & l\\ m & n & o & p\end{bmatrix}=a\cdot \begin{bmatrix}f & g & h\\ j & k & l\\ n & o &p \end{bmatrix}-b\cdot \begin{bmatrix}e & g & h\\ i & k & l\\ m & o & p\end{bmatrix}+c\cdot \begin{bmatrix}e & f & h\\ i & j & l\\ m & n & p\end{bmatrix}-d\cdot \begin{bmatrix}e & f & g\\ i & j & k\\ m & n & o\end{bmatrix}$$$Remember, to solve a 3 x 3 matrix: $$\begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$$$
For our problem statement:
$$\begin{bmatrix}0 & 0 & 0 & 0\\ 12 & 11 & 10 & 9\\ 8 & 7 & 6 & 5\\ 4 & 3 & 2 & 1\end{bmatrix}=0\cdot \begin{bmatrix}11 & 10 & 9\\ 7 & 6 & 5\\ 3 & 2 & 1\end{bmatrix}-0\cdot \begin{bmatrix}12 & 10 & 9\\ 8 & 6 & 5\\ 4 & 2 & 1\end{bmatrix}+0\cdot \begin{bmatrix}12 & 11 & 9\\ 8 & 7 & 5\\ 4 & 3 & 1\end{bmatrix}-0\cdot \begin{bmatrix}12 & 11 & 10\\ 8 & 7 & 6\\ 4 & 3 & 2\end{bmatrix}$$$Thus, $$\begin{bmatrix}0 & 0 & 0 & 0\\ 12 & 11 & 10 & 9\\ 8 & 7 & 6 & 5\\ 4 & 3 & 2 & 1\end{bmatrix}=0-0+0-0=0$$$
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# Arithmetic sequence
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## Arithmetic sequencePresentation Transcript
• ARITHMETIC SEQUENCE
BY: LEAH V. MELENDRES
By PresenterMedia.com
• This lesson will work with arithmetic sequences, their recursive and explicit formulas and finding terms in a sequence. In this lesson, it is assumed that you know what a sequence is.
• Let’s look at the arithmetic sequence 20, 24, 28, 32, 36, . . .
This arithmetic sequence has a common difference of 4, meaning that we add 4 to a term in order to get the next term in the sequence.
• The recursive formula for an arithmetic sequence is written in the form
an = an-1 + d
in our example, since the common difference (d)
is 4, we would write:
an = an-1 + 4
• So once you know the common difference in an arithmetic sequence you can write the recursive form for that sequence. However, the recursive formula can become difficult to work with if we want to find the 50th term. Using the recursive formula, we would have to know the first 49 terms in order to find the 50th. This sounds like a lot of work. There must be an easier way. And there is!
• Rather than write a recursive formula, we can write an explicit formula.
Explicit formula is also sometimes called the closed form. To write the explicit or closed form of an arithmetic sequence, we use
an = a1 + (n-1)d
an – is the nth term of the sequence. When writing the general expression for an arithmetic sequence, you will not actually find a value for this. (It will be part of your much in the same way x’s and y’s are part of algebraic equations). a1 – is the first term in the sequence. n – is treated like the variable in a sequence. d – is the common difference for the arithmetic sequence. You will either be given this value or be given enough information to it. You must substitute a value for d into the formula
• Let's Practice
• Given an arithmetic sequence:-6, -3, 0, 3, ..., 147.What is the term number for the last number in the sequence? [a? = 147]
Given: 8, 6, 4, 2, ...For the formula that represents the general term, an = dn + c, what are the correct values for d and c ?
• THANK YOU!!!
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# sin2x – cos2x = 1 for all values of x
Prove the identity, ?
Unit Circle’s equation is x² + y² = 1
All the points on the circle contains coordinates which make the equation x² + y² = 1, true!
On the unit circle, from any arbitrary point (x, y), the representation of coordinates can be given by (sin + cos ), where, is rotation’s degree from the x-axis which is positive.
On substitution, sin = x & cos = y into the unit circle’s equation, thus, we can observe that sin ² + cos ² = 1.
Generally, the identity given is just proven with the help of the Pythagoras theorem and the unit circle, either range of different methods or calculus.
Although if one proves it with the help of different methods, it’s important to remember that we can’t prove all the trigonometric identities with the use of alternative trigonometric identities, as that would be depending on a circular reasoning, in addition to that even if we prove it using different trigonometric identities, we should be using different methods in some or the other way.
It so appears that sin²(x)+cos²(x)=1 is known to be one of the simpler identities to verify with the use of alternative methods, and therefore, it’s usually done in this way.
Nevertheless, let’s now switch on to the proof with the formula of angle addition use for cosine:
cos(α + β)= cos(α)cos(β)−sin(α)sin(β)
Further, it’s important to note that odd function is sine and the even function is cosine, which means,
sin(−x) = −sin(x);
cos(−x) = cos(x)
Now, proceeding for the proof:
Let α = x; β = −x
Cos ( x + (-x)) = Cos (x) Cos (-x) – Sin (x) Sin (-x) ⇒ Cos (0) = cos (x) cos ( x – (-sin (x) sin (x)) …………………………eq.1
Considering the above equation
sin(−θ)= −sin(θ)
cos(−θ) = cos(θ)
Therefore, 1 = cos²(x) + sin²(x)
Another simplest method to prove this identity is:
Proof:
Let’s consider the right angled triangle, having θ as the internal angle:
Then: sinθ= a / c,
cosθ =b / c
Therefore,
sin²θ + cos²θ = a² / c² + b² / c² = a² + b² / c²
By the Pythagoras theorem:
As a² + b² = c²,
Accordingly, a² + b² / c² = 1 ; (a² + b² – c² = 0)
So, by the given Pythagoras, which proves this identity for
θ∈ (0, π / 2) or (0, 90º)
For angles that are present outside this range then, we can make use of:
1. sin(−θ)= −sin(θ)
2. Sin (θ+π) = −sin(θ)
3. cos(−θ) = cos(θ)
4. cos(θ+π) = −cos(θ)
Hence, for eg:
sin²(θ+π) + cos²(θ+π) = (−sinθ)² + (−cosθ)² = sin²θ + cos²θ = 1
Or,
sin²(θ+180º) + cos²(θ+180º) = (−sinθ)² + (−cosθ)² = sin²θ + cos²θ = 1
Considering Pythagoras theorem
In a right angle triangle given having a, b, and c as their sides
So, consider this below diagram:
Area of larger square = (a+b)²
Area of smaller square is = c²
Area of each triangle = 1 / 2 a.b
Therefore, we have:
(a+b)² = c² + 4 ⋅ 1 / 2 a*b ; i.e.
a² + 2ab + b² = c² + 2ab
Now,
From both sides, subtract “2a” to get:
a²+ b² = c2
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# What is the sum of first 50 even numbers?
Arithmetic is a part of mathematics that works with different types of numbers, fractions, applied different operations on numbers like addition, multiplication, etc. The word Arithmetic comes from the Greek word arithmos, which means number. Also Arithmetic involves exponentiation, the calculation of percentages, finding the value of number series, logarithmic functions, and square roots, etc.
There is series in arithmetic called Arithmetic Progression (AP), this is a sequence of numbers, where the difference between any two consecutive terms is always the same. Let’s say, a series is 2,4,6,8,10,12,….., in this series, the difference between any two consecutive numbers is 2. If we add this 2 with the previous number then we get the next number in the series, similarly, if we subtract 2 from the next number, we get the previous number.
In this article, we are finding the sum of the first 50 even numbers. To work with this series there are some formulas available, that are,
Let’s say a series A consist of some element a1, a2, a3, a4, a5, a6,…an
A = {a1, a2, a3, a4, a5, a6,…an}
• Common difference between two terms (d) = (a1-a2)
• Sum of the series(S) = (n/2)[2a + (n – 1)d]
• First term = a
• 2nd term = a+d
• 3rd term = a+2d
• Similarly, Nth term = a+(n-1)d
Now let’s take a series that has 5 terms, then the sum of the series is as below,
1st term, a1 = a
2nd term, a2 = a+d
3rd term, a3 = a+2d
4th term, a4 = a+3d
5th term, a5 = a+4d
so, sum of this series, S5= (a1 +a2 +a3 +a4+a5) = (a + a+d + a+2d + a+3d + a+4d)
It can also be written as,
=> S5 = a+a+a+a+a + d+2d+3d+4d
=> S5 = 5a+10d
=> S5 = 5×(a + 2d) [take 5 as common] ——————-(1)
we can this equation as,
2×S5 = 5 × 2 × [a + 2d] [multiply 2 in both side]
=> 2×S5 = 5×[2a+4d]
=> S5 = (5/2)×[2a + (5-1)d] Here n = 5,
Now let’s take a series that has 4 terms, then the sum of the series is as below,
1st term, a1 = a
2nd term, a2 = a+d
3rd term, a3 = a+2d
4th term, a4 = a+3d
so, sum of this series, S4 = (a + a+d + a+2d + a+3d)
It can also be written as,
=> S4 = a+a+a+a + d+2d+3d
=> S4 = 4a+6d
=> S4 = 2×(2a + 3d) [take 2 as common]
we can this equation as,
S4 = (4/2) × [2a + (4-1)*d] Here n = 4,
So, for finding the sum of series which has N term, the formula looks like
Sn = (n/2)×[2a + (n-1)×d]
### What is the sum of the first 50 even numbers?
First of all, we have to find the series of even numbers, let’s find that
The first even number is 2,
2nd even number is 4
3rd even number is 6 and so on….. so the last term will be 50 × 2 = 100
Secondly, find the common difference between them, d = 4 – 2 = 2 or 6 – 4 = 2
So, d = 2,
Thirdly, find the sum of the first 50 even terms,
Use the formula for finding the sum of series, that is Sn = (n/2)×[2a + (n-1)×d]
here, n=50, a=2, d=2
therefore, S50 = (50/2)×[2×2 + (50-1)×2]
=> S50 = 25×[4+ 49×2]
=> S50 = 25×102
=> S50 = 2550
The sum of first 50 even number is 2550
Another Method of Solving:
To find the sum of even numbers up to N we can use another formula, that is, N×(N+1)
This formula only works with consecutive even numbers, that is, 2,4,6,8,10,12,14,16…… like this.
Let’s say a series has the first 5 even numbers, 2,4,6,8,10
so the sum of this series,
S5 = 5×(5+1) [N =5]
S5 = 5×6 = 30
To check if the sum is correct or not we can sum up the numbers, 2+4+6+8+10 = 30, so the formula works fine
Here we are finding the sum of first 50 even numbers,
The sum, S50 = N*(N+1)
S50 = 50*(50+1) = 50*51
S50 = 2550
The sum of first 50 even number is 2550
Another Method of Solving:
For this method, we need to find the last term of the series.
To find the last term we a formula,
Tn = a+(n-1)d
=> T50 = 2+(50-1)×2 [Here, a=2, d=2. n=50]
=> T50 = 2+49×2 = 2+98
=> T50 = 100 ———–(2)
Now to find the sum, the formula is, Sn=(N/2) × (a + Tn), This formula is correct for all series,
let’s say a series has 4 terms 1,5,9,13. the sum of this series according to this formula will be,
S4 = (4/2)×(1+13) = 2 × 14 = 28
Check if it is correct or not, 1+5+9+13 = 28, so the formula is correct.
Now find the sum of the first 50 even numbers,
S50 = (50/2) × (2 + 100) [Here, a=2, Tn=100 — from (2)]
S50 = 25 × 102
S50 = 2550
The sum of first 50 even number is 2550
### Similar Questions
Question 1. Find the sum of the first 30 even numbers.
Solution:
To find this sum we can use any previously defined method,
Let’s use the 2nd method, which is N×(N+1)
S30 = 30×(30+1)
=> S30 = 30×31
=>S30 = 930
The sum of the first 30 even numbers is 930
Question 2. Find the sum of a series whose first term is 6 and the common difference is 4 and the number of terms in the series is also 6.
Solution:
To solve this problem we can use the 3rd and 1st methods.
Using 1st method:
Given, d=4,a=6,N=6
put all the values in this formula Sn = (n/2)×[2a + (n-1)×d]
S6 = (6/2)×[2×6+(6-1)×4] = 3×(12+20)
S6 = 3×32
S6 = 96
Using 3rd method:
Given, d=4,a=6,N=6
Find the nth term,
Tn=a+(n-1)d
T6 = 6 + (6-1)×4
T6 = 6+ 20
T6 = 26——–nth term
Now put all the values in this formula Sn=(N/2) × (a + Tn)
S6 = (6/2) × (6+26)
S6 = 3×32
S6 = 96
Now find the series, first term is 6 and common difference is 4
so the series will be,
a1 = 6, a2 = 6+4 = 10, a3 = 10+4 =14,
a4 = 14+4 = 18, a5= 18+4= 22, a6= 22+4 = 26
the series is 6,10,14,18, 22, 26
the sum is 6+10+14+18+22+26 = 96
So the solution is correct. and the sum of this series is 96
Question 3. Find the series whose sum is 147, the last term is 33 and the number of terms in the series is 7.
Solution:
Given, Sn =147, Tn = 33, N=7
we can use this Sn=(N/2) × (a + Tn) formula here to find the first term in that series
Putting all the given values, we get
147 = (7/2)×(a + 33)
=> 147×2 = 7×a + 7×33
=> 294 = 7a + 231
=> 7a = 294-231
=>7a = 63
=> a = 63/7
=>a = 9
So the first term is 9.
Now find the common difference d, to find d we can use any formula which contains d
Here we are using the formula for finding the Nth term, that is Tn=a+(n-1)d
Putting all the values, we get,
33 = 9 + (7-1)×d [Given Tn = 33, a=9, n=7]
=> 33-9 = 6×d
=>6d = 24
=>d=24/6
=> d=4
the common difference is 4
Now find the series,
a1 = 9, a2 = 9+4 = 13, a3 = 13+4 = 17, a4 = 17+4= 21, a5 = 21+4=25, a6 = 25+4 = 29, a7 = 29+4 = 33
So the final series is 9, 13, 17, 21, 25, 29, 33
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Related Tutorials
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# The sum of three terms of an A.P. is 21 and the product
Question:
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Solution:
In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.
We need to find the three terms.
Here,
Let the three terms be $(a-d), a,(a+d)$ where, $a$ is the first term and $d$ is the common difference of the A.P
So.
$(a-d)+a+(a+d)=21$
$3 a=21$
$a=7$ ................(1)
Also,
$(a-d)(a+d)=a+6$
$a^{2}-d^{2}=a+6$ (Using $a^{2}-b^{2}=(a+b)(a-b)$ )
$(7)^{2}-d^{2}=7+6$ (Using 1)
$49-13=d^{2}$
Further solving for d,
$d^{2}=36$
$d=\sqrt{36}$
$d=6 \quad \ldots \ldots(2)$
Now, using the values of and d in the expressions of the three terms, we get,
First term $=a-d$
So,
$a-d=7-6$
$=1$
Second term = a
So,
$a=7$
Also,
Third term $=a+d$
So,
$a+d=7+6$
$=13$
Therefore, the three terms are 1,7 and 13 .
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\left\{2, 8\right\}$
Subtract $10x$ to both sides: $x^2+16-10x=10x-10x \\x^2-10x+16=0$ RECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The trinomial above has $b=-10$ and $c=16$. Note that $16=-8(-2)$ and $-10= -8+(-2)$. This means that $d=-8$ and $e=-2$ Thus, the factored form of the trinomial is: $[x+(-8)][x+(-2)] = (x-8)(x-2)$ The given equation may be written as: $(x-8)(x-2)=0$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: $\begin{array}{ccc} &x-8= 0 &\text{ or } &x-2=0 \\&x=8 &\text{ or } &x=2 \end{array}$ Thus, the solution set is $\left\{2, 8\right\}$.
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2nd Gear Solution
Second Gear refers to Figure 9-47 in Norton.
In this configuration, clutch C2 is engaged and Brake B2 is also engaged.
The gear train will be solved with 5 as the input and 4 as the output.
There will be two equations for this case. The first, last and arm for each will be defined as
f = 5
l = 7
a = 3
and
f = 5
l = 3
a = 4
The first step is to solve each equation for omega 3.
(1)
\begin{align} \frac {\omega_{7/3}}{\omega_{5/3}}& =\frac {\omega_7-\omega_3}{\omega_5-\omega_3}& = \left(\frac{-N_5}{N_7}\right)\\ \end{align}
from (1)
(2)
\begin{align} (\omega_{7}-\omega_{3})=(\omega_{5}-\omega_{3})&\left(\frac{-N_5}{N_7}\right) \end{align}
simplifying (2), the following expression for omega 3 is obtained
(3)
\begin{align} \omega_{3}=\frac{\omega_{7}+\left(\frac {N_5}{N_7}\right)\omega_{5}}{\left(1-\frac{N_5}{N_7}\right)} \end{align}
but with B2 engaged,
(4)
\begin{align} \omega_{7} = 0 \end{align}
so (3) becomes
(5)
\begin{align} \omega_{3}=\frac{N_5}{N_7+N_5}*\omega_5 \end{align}
Now, solving the second equation for omega 3
(6)
\begin{align} \frac {\omega_{3/4}}{\omega_{5/4}}& =\frac {\omega_3-\omega_4}{\omega_5-\omega_4}& = \left(\frac{-N_5}{N_3}\right)\\ \end{align}
from (4)
(7)
\begin{align} (\omega_{3}-\omega_{4})=(\omega_{5}-\omega_{4})&\left(\frac {-N_5}{N_3}\right) \end{align}
Solving for omega 3,
(8)
\begin{align} \omega_{3}=\omega_{5}&\left(\frac{-N_5}{N_3}\right)+\left(1+\frac{N_5}{N_3}\right)&\omega_{4}\\ \end{align}
Substituting (5) into (8)
(9)
\begin{align} \omega_{3}=\frac{N_5}{N_7+N_5}*\omega_5=-\omega_{5}&\left(\frac{N_5}{N_3}\right)+\left(1+\frac{N_5}{N_3}\right)&\omega_{4}\\ \end{align}
solving for the output…
(10)
\begin{align} \omega_{4}=\frac{\left(\frac{N_5}{N_7+N_6}\right)+\frac{N_5}{N_3}}{\left(1+\frac{N_5}{N_3}\right)}*\omega_5 \end{align}
page revision: 40, last edited: 24 Sep 2010 19:03
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# Recursive functions in discrete mathematics
A recursive function is a function that its value at any point can be calculated from the values of the function at some previous points. For example, suppose a function f(k) = f(k-2) + f(k-3) which is defined over non negative integer. If we have the value of the function at k = 0 and k = 2, we can also find its value at any other non-negative integer. In other words, we can say that a recursive function refers to a function that uses its own previous points to determine subsequent terms and thus forms a terms sequence. In this article, we will learn about recursive functions along with certain examples.
## What is Recursion?
Recursion refers to a process in which a recursive process repeats itself. Recursive is a kind of function of one and more variables, usually specified by a certain process that produces values of that function by continuously implementing a particular relation to known values of the function.
Here, we will understand the recursion with the help of an example.
Suppose you are going to take a stair to reach the first floor from the ground floor. So, to do this, you have to take one by one steps. There is only a way to go to the second step that is to the steeped first step. Suppose you want to go to the third step; you need to take the second step first. Here, you can clearly see the repetition process. Here, you can see that with each next step, you are adding the previous step like a repeated sequence with the same difference between each step. This is the actual concept behind the recursive function.
Step 2: Step 1 + lowest step.
Step 3: Step 2 + Step 1 + lowest step.
Step 4: Step 3 + step 2 + step 1+ lowest step, and so on.
A set of natural numbers is the basic example of the recursive functions that start from one goes till infinity, 1,2,3,4,5,6,7,8, 9,…….infinitive. Therefore, the set of natural numbers shows a recursive function because you can see a common difference between each term as 1; it shows each time the next term repeated itself by the previous term.
### What is a recursively defined function?
The recursively defined functions comprise of two parts. The first part deals with the smallest argument definition, and on the other hand, the second part deals with the nth term definition. The smallest argument is denoted by f (0) or f (1), whereas the nth argument is denoted by f (n).
Follow the given an example.
Suppose a sequence be 4,6,8,10
The explicit formula for the above sequence is f (n)= 2n + 2
The explicit formula for the above sequence is given by
f (0) = 2
f(n) = f (n-1) + 2
Now, we can get the sequence terms applying the recursive formula as follows f(2 ) f (1) + 2
f(2) = 6
f (0) = 2
f(1) = f(0) + 2
f (1) = 2 + 2 = 4
f(2 ) = f (1) + 2
f(2) = 4 + 2 = 6
f(3 ) = f (2) + 2
f(3 ) = 6 + 2 = 8
With the help of the above recursive function formula, we can determine the next term.
### What makes the function recursive?
Making any function recursive needs its own term to calculate the next term in the sequence. For example, if you want to calculate the nth term of the given sequence, you first need to know the previous term and the term before the previous term. Hence, you need to know the previous term to find whether the sequence is recursive or not recursive. So we can conclude that if the function needs the previous term to determine the next term in the sequence, the function is considered a recursive function.
### The formula of the Recursive function
If a1, a2, a3, a4, a5, a6, ……..an,……is many sets or a sequence, then a recursive formula will need to compute all the terms which are existed previously to calculate the value of an
an = an-1 + a1
The above formula can also be defined as Arithmetic Sequence Recursive Formula. You can see clearly in the sequence mentioned above that it is an arithmetic sequence, which comprises the first term followed by the other terms and a common difference between each term. The common difference refers to a number that you add or subtract to them.
A recursive function can also be defined as the geometric sequence, where the number sets or a sequence have a common factor or common ratio between them. The formula for the geometric sequence is given as
an = an-1 * r
Usually, the recursive function is defined in two parts. The first one is the statement of the first term along with the formula, and another is the statement of the first term along with the rule related to the successive terms.
### How to write a Recursive formula for arithmetic sequence
To write the Recursive formula for arithmetic sequence formula, follow the given steps
Step 1:
In the first step, you need to ensure whether the given sequence is arithmetic or not (for this, you need to add or subtract two successive terms). If you get the same output, then the sequence is taken as an arithmetic sequence.
Step 2:
Now, you need to find the common difference for the given sequence.
Step 3:
Formulate the recursive formula using the first term and then create the formula by using the previous term and common difference; thus you will get the given result
an = an-1 + d
now, we understand the given formula with the help of an example
suppose 3,5,7,9,11 is a given sequence
In the above example, you can easily find it is the arithmetic sequence because each term in the sequence is increases by 2. So, the common difference between two terms is 2. We know the formula of recursive sequence
an = an-1 + d
Given,
d = 2
a1 = 3
so,
a2 = a(2-1) + 2 = a1 + 2 = 3+2 = 5
a3 = a(3-1) + 2 = a2+ 2 = 5+2 = 7
a4 = a(4-1) + 2 = a3 + 2 = 7+2 = 9
a5 = a(5-1) + 2 = a + 2 = 9+2 = 11, and the process continues.
### How to write a Recursive formula for Geometric sequence?
To write the Recursive formula for Geometric sequence formula, follow the given steps:
Step 1
In the first step, you need to ensure whether the given sequence is geometric or not (for this, you need to multiply or divide each term by a number). If you get the same output from one term to the next term, the sequence is taken as a geometric sequence.
Step 2
Now, you need to find the common ratio for the given sequence.
Step 3
Formulate the recursive formula using the first term and then create the formula by using the previous term and common ratio; thus you will get the given result
an = r * an-1\
Now, we understand the given formula with the help of an example
suppose 2,8,32, 128,.is a given sequence
In the above example, you can easily find it is the geometric sequence because the successive term in the sequence is obtained by multiplying 4 to the previous term. So, the common ratio between two terms is 4. We know the formula of recursive sequence
an = r * an-1\
an = 4
an-1 = ?
Given,
r = 4
a1 = 2
so,
a2 = a (2-1) * 4 = a1 + * 4 = 2* 4 = 8
a3 = a (3-1) * 4 = a2* 4 = 8 * 4 = 32
a4 = a (4-1) * 4 = a3 * 4 = 32* 4 = 128, and the process continues.
### Example of recursive function
Example 1:
Determine the recursive formula for the sequence 4,8,16,32,64, 128,….?
Solution:
Given sequence 4,8,16,32,64,128,…..
The given sequence is geometric because if we multiply the preceding term, we get the successive terms.
To determine the recursive formula for the given sequence, we need to write it in the tabular form
Term Numbers Sequence Term Function Notation Subscript Notation
1 4 f(1) a1
2 8 f(2) a2
3 16 f(3) a3
4 32 f(4) a4
5 64 f(5) a5
6 128 f(6) a6
n . f(n) an
Hence, the recursive formula in function notion is given by
f(1) = 4, f(n) . f(n- 1)
In subscript notation, the recursive formula is given by
a1 = 4, an = 2. an-1
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# Texas Go Math Grade 6 Benchmark Test Answer Key Part 1
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Benchmark Test Answer Key Part 1.
## Texas Go Math Grade 6 Benchmark Test Answer Key Part 1
Selected Response
Question 1.
Which temperature is coldest?
(A) – 13° F
(B) 20° F
(C) -20° F
(D) 13° F
(C) -20°F
Explanation:
The smallest number of all, of those is -20, so, we can conclude that the coldest temperature is -20° F.
Question 2.
Which group of numbers is in order from least to greatest?
(A) 2.58, 2$$\frac{5}{8}$$, 2.6, 2$$\frac{2}{3}$$
(B) 2$$\frac{2}{3}$$, 2$$\frac{5}{8}$$, 2.6, 2.58
(C) 2$$\frac{5}{8}$$, 2$$\frac{2}{3}$$, 2.6, 2.58
(D) 2.58, 2.6, 2$$\frac{5}{8}$$, 2$$\frac{2}{3}$$
(D) 2.58, 2.6, 2$$\frac{5}{8}$$, 2$$\frac{2}{3}$$
Explanation:
The group D is in required order because:
2$$\frac{5}{8}$$ = 2.625
2$$\frac{5}{8}$$ = 2.667
2.58 < 2.6 < 2.625 < 2.667
Question 3.
Evaluate a + b for a = -46 and b = 34.
(A) -12
(B) 80
(C) -80
(D) 12
(A) -12
Explanation:
We will, substitute 46 for a and 34 for b in expression a + b in order to evaluate it, so, we have the following:
46 + 34= 12
Question 4.
One winter day, the temperature ranged from a high of 40°F to a low of -5°F. By how many degrees did the temperature change?
(A) 45° F
(B) 35° F
(C) 25° F
(D) 55° F
(A) 45° F
Explanation:
In order to find how many degrees the temperature changed, we need to subtract those two values, and get:
40 – (-5) = 40 + 5
= 45
So, the temperature changed for 45° F
Question 5.
Find the quotient. 9$$\frac{3}{5}$$ ÷ $$\frac{8}{15}$$
(A) 2
(B) 18
(C) 16$$\frac{7}{8}$$
(D) 18$$\frac{3}{4}$$
(B) 18
Explanation:
In order to find the final quotient we need to simplify given expression:
$$\frac{48}{5}$$ ∙ $$\frac{15}{8}$$ = $$\frac{6}{1}$$ ∙ $$\frac{3}{1}$$
= 18
So, the answer is that result is (B) 18.
Question 6.
Juan purchased 3.4 pounds of nails at a cost of $4.51 per pound. What was the cost of the nails? (A)$153.34
(B) $1.53 (C)$7.91
(D) $15.33 Answer: (D)$15.33
Explanation:
To find what was the cost of the nails, we need to multiply those values. So, we have the following:
3.4 ∙ 4.51 = 15.33
So, the cost of the nails was $15.33. Question 7. You are working as an assistant to a chef. The chef has 6 cups of berries and will use $$\frac{2}{3}$$ cup of berries for each dessert he makes. How many desserts can he make? (A) 4 desserts (B) 6$$\frac{2}{3}$$ desserts (C) 9 desserts (D) 12 desserts Answer: (C) 9 desserts Explanation: If we want to find how many deserts the chef can make, we need to divide 6 by $$\frac{2}{3}$$. So, we have the following: 6 : $$\frac{2}{3}$$ = 6 ∙ $$\frac{3}{2}$$ = $$\frac{18}{2}$$ = 9 So, correct answer is C, he can make 9 desserts. Question 8. Jorge is building a table out of boards that are 3.75 inches wide. He wants the table to be at least 36 inches wide. How many boards does he need? (A) 9 (B) 9.6 (C) 10 (D) 135 Answer: (B) 9.6 Explanation: In order to find how many boards Jorge needs for table to be at least 36 inches, we need to divide 36 by 3.75 and get: 36 ÷ 3.75 = 9.6 So, he needs 9.6 boards. Question 9. The fuel for a chainsaw is a mix of oil and gasoline. The label says to mix 6 ounces of oil with 16 gallons of gasoline. How much oil would you use if you had 24 gallons of gasoline? (A) 3 ounces (B) 9 ounces (C) 12 ounces (D) 85.3 ounces Answer: (B) 9 ounces Explanation: Because, according to label, it is mixed 6 ounces of oil with 16 gallons of gasoline, we need to calculate x, which is oil in ounces needed for mix with 24 gallons of gasoline So, we have the following: 6 → 16 x → 24 x = $$\frac{6 \cdot 24}{16}$$ = 9 So, conclusion is that it is 9 ounces of oil needed for 24 gallons of gasoline. Question 10. A grocery store sells the brands of yogurt shown in the table. Which brand of yogurt has the lowest unit price? (A) Sunny (B) Fruity (C) Smooth (D) Yummy Answer: (B) Fruity Explanation: In order to find the lowest unit price, we need to find quotient between size and price for each brand of yogurt Sunny: $$\frac{12}{2}$$ = 6 Fruity: $$\frac{14}{2.34}$$ = 5.98 Smooth: $$\frac{18}{2.6}$$ = 6.92 Yummy: $$\frac{16}{2.24}$$ = 7.14 We can conclude that the lowest unit price has fruity yogurt. So, correct answer is B. Question 11. Find the unit rate. Patricia paid$385 for 5 nights at a hotel.
(A) $$\frac{\ 77}{1 \text { night }}$$
(B) $$\frac{\ 154}{1 \text { night }}$$
(C) $$\frac{\ 385}{1 \text { night }}$$
(D) $$\frac{\ 39}{1 \text { night }}$$
(A) $$\frac{\ 77}{1 \text { night }}$$
Explanation:
In order to find the unit rate we have to divide $35 which is the sum of money Patricia paid for 5 nights by 5. On that way we will find how much costs one day: $$\frac{385}{5}$$ = 77 So, Patricia paid$ 77 for one night and required unit rate is $$\frac{\ 77}{1 \text { night }}$$.
Question 12.
Which of the following does NOT show a pair of equivalent ratios?
(A) $$\frac{3}{7}$$, $$\frac{9}{21}$$
(B) $$\frac{24}{56}$$, $$\frac{3}{7}$$
(C) $$\frac{3}{7}$$, $$\frac{9}{28}$$
(D) $$\frac{3}{7}$$, $$\frac{12}{28}$$
(C) $$\frac{3}{7}$$, $$\frac{9}{28}$$
Explanation:
We can see that only at C we do not have pair of equivalent ratios.
Question 13.
Gina paid $129 for a bicycle that was on sale for 75% of its original price. What was the original price of the bicycle? (A)$54.00
(B) $96.75 (C)$172
(D) $204 Answer: (C)$172
Explanation:
First we need to form next proportion in order to find the originaL price of the bicycle:
x = $$\frac{129 \cdot 100}{75}$$ = $172 So, conclusion is$172 was the original price of the bicycle.
Question 14.
Write the fraction $$\frac{9}{50}$$ as a percent. If necessary, round your answer to the nearest hundredth.
(A) 0.18%
(B) 0.45%
(C) 18%
(D) 45%
(C) 18%
Explanation:
If we want to write given fraction as a percent, we need multiply it by 100 and get:
$$\frac{9}{50}$$ ∙ 100 = $$\frac{900}{50}$$ = 18
So, $$\frac{9}{50}$$ is 18%. Correct answer is C.
Question 15.
Write an equation you can use to find the missing value in the table.
(A) t = a+1
(B) t = a + 15
(C) t = a + 3
(D) t = a + 10
(C) t = a + 3
Explanation:
From the table we can conclude that we can use equation C, that is t = a + 3, in order to find missing value it the table.
Question 16.
Which expression is NOT equivalent to the expression 38 – 14?
(A) 2(19 – 7)
(B) 2(19) – 2(7)
(C) (19 – 7)2
(D) 2(36 – 12)
(D) 2(36 – 12)
Explanation:
We can notice the following:
2(19 – 7) = (19 – 7)2
= 2(19) – 2(7)
= 38 – 14
Also, we can see that only D is not equivalent to the expression 38 – 14.
Question 17.
Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 9 times brighter than the next category. For example, a day that is dazzling is 9 times brighter than a day that is radiant. How many times brighter is a dazzling day than an illuminated day?
(A) 2 times brighter
(B) 9 times brighter
(C) 81 times brighter
(D) 729 times brighter
(C) 81 times brighter
Explanation:
Because each category in the table is 9 times brighter than the next category and we already have that dazzling day is 9 times brighter than radiant day, and we can conclude that radiant day is 9 times brighter than a day that is illuminated, the conclusion is that dazzling day is 9 ∙ 9 = 81 times brighter than an illuminated day.
Question 18.
Evaluate the expression. (28 ÷ 4) ∙ 5 – 6 + 42
(A) 9
(B) 13
(C) 37
(D) 45
(D) 45
Explanation:
In this expression first we caLcuLate part of it which is in brackets and squating, then muLtipLy, add or subtract at the end.
So, we have the following:
7.5 – 6 + 42 = 35 – 6 + 16
= 29 + 16
= 45
Conclusion is that result is 45, so, correct answer is D.
Gridded Response
Question 19.
Shari bought 2$$\frac{1}{2}$$ yards of fabric for $7.99 per yard. She used $$\frac{3}{4}$$ of the fabric to make decorative pillows. How much money could Shari have saved by only buying the amount of fabric that she used? Write your answer in dollars and round to the nearest tenth. Answer: For 2 ∙ $$\frac{1}{2}$$ yards of fabric, Shari paried: 2$$\frac{1}{2}$$ ∙ 7.99 = $$\frac{5}{2}$$ ∙ 7.99 = 19.975 But , she used only $$\frac{3}{4}$$, and for $$\frac{3}{4}$$ yards of fabric, she would pay: $$\frac{3}{4}$$ ∙ 7.99 = 5.9923 So. she could save 19.975 – 5.9923 ≈$14.
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# Texas Go Math Grade 6 Module 10 Answer Key Generating Equivalent Numerical Expressions
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 10 Answer Key Generating Equivalent Numerical Expressions.
## Texas Go Math Grade 6 Module 10 Answer Key Generating Equivalent Numerical Expressions
Texas Go Math Grade 6 Module 10 Are You Ready? Answer Key
Find the product.
Question 1.
992 × 16
Answer:
Given expression = 992 × 16
Rewrite the given expression as a product of a number and a difference, therefore = 16(1000 – 8)
Apply distributive property to expand the parenthesis = 16000 – 128
Evaluate = 15872
9920 × 16 = 15872
Question 2.
578 × 27
Answer:
Solution to this example is given below
Final Solution = 15,606
Question 3.
839 × 65
Answer:
Given expression = 839 × 65
Rewrite the given expression as a product of a number and a difference, therefore = 65(800 + 39)
Apply distributive property to expand the parenthesis = 52000 + 2535
Evaluate = 54535
839 × 65 = 54535
Question 4.
367 × 23
Answer:
Solution to this example is given below
Final Solution = 8,441
Find the product.
Question 5.
7 × 7 × 7
Answer:
Solution to this example is given below
Question 6.
3 × 3 × 3 × 3
Answer:
Solution to this example is given below
Question 7.
6 × 6 × 6 × 6 × 6
Answer:
Solution to this example is given below
Question 8.
2 × 2 × 2 × 2 × 2 × 2 × 2
Answer:
Solution to this example is given below
Divide.
Question 9.
20 ÷ 4
Answer:
Solution to this example is given below
20 ÷ 4 = ?
Think: 4 times what number equals 20?
⇒ 4 × 5 = 20
⇒ 20 ÷ 4 = 5
So, 20 ÷ 4 = 5.
Question 10.
21 ÷ 7
Answer:
Solution to this example is given below
21 ÷ 7 = ?
Think: 7 times what number equals 20?
⇒ 7 × 3 = 21
⇒ 21 ÷ 7 = 3
So, 21 ÷ 7 = 3.
Question 11.
42 ÷ 7
Answer:
Solution to this example is given below
42 ÷ 7 = ?
Think: 7 times what number equals 20?
⇒ 7 × 6 = 42
⇒ 42 ÷ 7 = 6
So, 42 ÷ 7 = 6.
Question 12.
56 ÷ 8
Answer:
Solution to this example is given below
56 ÷ 8 = ?
Think: 8 times what number equals 20?
⇒ 8 × 7 = 56
⇒ 56 ÷ 8 = 7
So, 56 ÷ 8 = 7.
Texas Go Math Grade 6 Module 9 Reading Start-Up Answer Key
Visualize Vocabulary
Use the words to complete the graphic. You may put more than one word in each box.
Understand Vocabulary
Complete the sentences using the preview words.
Question 1.
A number that is formed by repeated multiplication by the same factor is a _________________ .
Answer:
Power
Question 2.
A rule for simplifying expressions is _____________________
Answer:
Order of Operations.
It can consists of multiple operations but follow a certain order.
Question 3.
The _________________ is a number that is multiplied. The number that indicates how many times this number is used as a factor is the ____________ .
Answer:
base and exponent.
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Home | | Maths 7th Std | Exercise 1.4 (Division of Decimal Numbers)
# Exercise 1.4 (Division of Decimal Numbers)
7th Maths : Term 3 Unit 1 : Number System : Division of Decimal Numbers : Exercise 1.4
Operations on Decimal Numbers
Division of Decimal Numbers
Exercise 1.4
1. Simplify the following.
(i) 0.6 ÷ 3
(ii) 0.90 ÷ 5
(iii) 4.08 ÷ 4
(iv) 21.56 ÷ 7
(v) 0.564 ÷ 6
(vi) 41.36 ÷ 4
(vii) 298.2 ÷ 3
Solution:
2. Simplify the following.
(i) 5.7 ÷ 10
(ii) 93.7 ÷ 10
(iii) 0.9 ÷ 10
(iv) 301.301 ÷ 10
(v) 0.83 ÷ 10
(vi) 0.062 ÷ 10
Solution:
3. Simplify the following.
(i) 0.7 ÷ 100
(ii) 3.8 ÷ 100
(iii) 49.3 ÷ 100
(iv) 463.85 ÷ 100
(v) 0.3 ÷ 100
(vi) 27.4 ÷ 100
Solution:
4. Simplify the following.
(i) 18.9 ÷ 1000
(ii) 0.87 ÷ 1000
(iii) 49.3 ÷ 1000
(iv) 0.3 ÷ 1000
(v) 382.4 ÷ 1000
(vi) 93.8 ÷ 1000
Solution:
5. Simplify the following.
(i) 19.2 ÷ 2.4
(ii) 4.95 ÷ 0.5
(iii) 19.11÷ 1.3
(iv) 0.399 ÷ 2.1
(v) 5.4 ÷ 0.6
(vi) 2.197 ÷ 1.3
Solution:
6. Divide 9.55 kg of sweet among 5 children. How much will each child get?
Solution:
7. A vehicle covers a distance of 76.8 km for 1.2 litre of petrol. How much distance will it cover for one litre of petrol?
Solution:
8. Cost of levelling a land at the rate of ₹ 15.50 sq. ft is ₹ 10,075. Find the area of the land.
Solution:
9. The cost of 28 books are ₹ 1506.4. Find the cost of one book.
Solution:
10. The product of two numbers is 40.376. One number is 14.42. Find the other number.
Solution:
Objective type questions
11. 5.6÷ 0.5= ?
(i) 11.4
(ii) 10.4
(iii) 0.14
(iv) 11.2
Hint:
5.6 / 0.5 = 56 / 5
= 11.2
12. 2.01÷ 0.03 = ?
(i) 6.7
(ii) 67.0
(iii) 0.67
(iv) 0.067
Hint:
2.01 / 0.03 = 201 / 3
= 67
13. 0.05 ÷ 0.5 = ?
(i) 0.01
(ii) 0.1
(iii) 0.10
(iv) 1.0
Solution: 0.05 / 0.5 = (5/100) / (5 / 10) = 5 /100 × 10/5 = 1 /10 = 0.1
Exercise 1.4
1. (i) 0.2 (ii) 0.18 (iii) 1.02 (iv) 3.08 (v) 0.094 (vi) 10.34 (vii) 99.4
2. (i) 0.57 (ii) 9.37 (iii) 0.09 (iv) 30.1301 (v) 0.083 (vi) 0.0062
3. (i) 0.007 (ii) 0.038 (iii) 0.493 (iv) 4.6385 (v) 0.003 (vi) 0.274
4. (i) 0.0189 (ii) 0.00087 (iii) 0.0493 (iv) 0.0003 (v) 0.3824 (vi) 0.0938
5. (i) 8 (ii) 9.9 (iii) 14.7 (iv) 0.19 (v) 9 (vi) 1.69
6. 1.91 kg
7. 64 km
8. 650 sq.ft
9. ₹ 53.80
10. 2.8
Objective type questions
11. (iv) 11.2
12. (ii) 67.0
13. (ii) 0.1
Operations on Decimal Numbers
Division of Decimal Numbers
Exercise 1.4
1. Simplify the following.
(i) 0.6 ÷ 3
(ii) 0.90 ÷ 5
(iii) 4.08 ÷ 4
(iv) 21.56 ÷ 7
(v) 0.564 ÷ 6
(vi) 41.36 ÷ 4
(vii) 298.2 ÷ 3
Solution:
i) 0.6 ÷ 3 = 6/10 × 1/3 = 6/3 × 1/10 = 2 × 1/10 = 2/10 = 0.2
(ii) 0.90 ÷ 5 = 90/100 × 1/5 = 90/5 × 1/100 = 18 × ( 1/100) = 18/100 = 0.18
(iii) 4.08 ÷ 4 = 408/100 × 1/4 = 408/4 × 1/100 = 102 × ( 1/100) = 102/100 = 1.02
(iv) 21.56 ÷ 7 = 2156/100 × 1/7 = 2156/7 × 1/100 = 102 × (1/100) = 308/100 = 3.08
(v) 0.564 ÷ 6 = 564/1000 × 1/6 = ( 564/6) × ( 1/1000) = 94/1000 = 0.094.
(vi) 41.36 ÷ 4 = 4136 /100 × 1/4 = (4136/4) × 1/100 = 1034/100 = 10.34
(vii) 298.2 ÷ 3 = 2982/10 × 1/3 =( 2982 /3) × 1/10 = 994/10 = 99.4
2. Simplify the following.
(i) 5.7 ÷ 10
(ii) 93.7 ÷ 10
(iii) 0.9 ÷ 10
(iv) 301.301 ÷ 10
(v) 0.83 ÷ 10
(vi) 0.062 ÷ 10
Solution:
(i) 5.7 ÷ 10 = 57/10 × 1/10 = 57/100 = 0.57
(ii) 93.7 ÷ 10 = 937/10 × 1/10 = 937/100 = 9.37
(iii) 0.9 ÷ 10 = 9/10 × 1/10 = 9/100 = 0.09
(iv) 301.301 ÷ 10 = (301301/1000 )x 1/10 = 301301/10000 = 30.1301
(v) 0.83 ÷ 10 = (83/100 )x 1/10 = 83/1000 = 0.083
(vi) 0.062 ÷10 =( 62/1000) × 1/10 = 62/10000 = 0.0062
3. Simplify the following.
(i) 0.7 ÷ 100
(ii) 3.8 ÷ 100
(iii) 49.3 ÷ 100
(iv) 463.85 ÷ 100
(v) 0.3 ÷ 100
(vi) 27.4 ÷ 100
Solution:
(i) (i) 0.7 ÷ 100 = 7/10 × 1/100 = 7/1000 = 0.007
(ii) 3.8 ÷ 100 = 38/10 × 1/100 = 38/1000 = 0.038
(iii) 49.3 ÷ 100 = 493/10 × 1/100 = 493/1000 = 0.493
(iv) 463.85 ÷ 100 = (46385/100) × 1/100 = 46385/10000 = 4.6385
(v) 0.3 ÷ 100 = 3/10 × 1/100 = 3/ 1000 = 0.003
(vi) 27.4 ÷ 100 = 274/10 × 1/100 = 274/1000 = 0.274
4. Simplify the following.
(i) 18.9 ÷ 1000
(ii) 0.87 ÷ 1000
(iii) 49.3 ÷ 1000
(iv) 0.3 ÷ 1000
(v) 382.4 ÷ 1000
(vi) 93.8 ÷ 1000
Solution:
(i) 18.9 ÷ 1000 = 189/10 × 1/1000 = 189/10000 = 0.0189
(ii) 0.87÷ 1000 = 87/100 × 1/1000 = 87/ 100000 = 0.00087
(iii) 49.3 ÷1000 = 493/10 × 1/100 = 493/ 1000 = 0.493
(iv) 0.3÷ 1000 = 3/10 × 1/1000 = 3/10000 = 0.0003
(v) 382.4 ÷1000 = 3824/10 × 1/1000 = 3824/10000 = 0.3824
(vi) 93.8 ÷1000 = 938/10 × 1/1000 = 938/10000 = 0.0938
5. Simplify the following.
(i) 19.2 ÷ 2.4
(ii) 4.95 ÷ 0.5
(iii) 19.11÷ 1.3
(iv) 0.399 ÷ 2.1
(v) 5.4 ÷ 0.6
(vi) 2.197 ÷ 1.3
Solution:
(i) 19.2 ÷ 2.4 = ( 192/10) / ( 24/10 ) = 192/10 × 10/24 = 192/24 = 8
(ii) 4.95 ÷ 0.5 = ( 495/100) / ( 5/10 ) = 495/100 × 10/5 = 495/5 × 10/100 = 99 × 1/10 = 99/10 = 9.9
(iii) 19.11 ÷ 1.3 = ( [19 × 11]/100)/( 13/10 ) = [1911/100] × [10/13]
= 1911/13 × 10/100 = 147 × 1/10 = 147/10 = 14.7
(iv) 0.399 ÷ 2.1 = (399/1000) / (21/10) = 399/1000 × 10/21
= 399/21 × 10/1000
= 19 × 1/100 = 19/100 = 0.19
(v) 5.4 ÷ 0.6
= (54/10) / (6/10) = 54/10 × 10/6
= 54 / 6 = 9
(vi) 2.197 ÷ 1.3
= (2197/1000) / (13/10) = 2197/1000 × 10/13
= 169 × 1/100
= 169/100 = 1.69
6. Divide 9.55 kg of sweet among 5 children. How much will each child get?
Solution:
Weight of the sweet = 9.55 kg
Weight of sweet for 5 children = 955/100 kg
Weight of sweet for 1 child = (955/100) / 5 = ( 955/100) × 1/5 =( 955 / 5) × 1/100
= 191/100 = 1.91
Each child will get 1.91 kg sweet.
7. A vehicle covers a distance of 76.8 km for 1.2 litre of petrol. How much distance will it cover for one litre of petrol?
Solution:
For 1.2 litre of petrol the distance covered = 76.8 km = 768 / 10 km
For 1 litre of petrol distance covered = [ (768 / 10) / 1.2 ] km
= (768 / 10) / (12 / 10) = ( 768 / 10) × (10 / 12)
= (768 / 12) × (10 / 10) = 64 km
For 1 litre of petrol distance covered = 64 km.
8. Cost of levelling a land at the rate of ₹ 15.50 sq. ft is ₹ 10,075. Find the area of the land.
Solution:
Cost of levelling the entire land = ₹ 10, 075
Cost of levelling 1 sq. ft = ₹ 15.50
Area of the land = Cost of levelling the entire land / Cost of levelling 1 sq. ft.
=> (10075 / 15.50) × (100 / 100) = ( 10075 × 100) / (15.50 × 100) = 1007500 / 1550
= 100750 / 155 = 650
Area of the land = 650 sq.ft.
9. The cost of 28 books are ₹ 1506.4. Find the cost of one book.
Solution:
Cost of 28 books = ₹ 1506.4
Cost of 1 book = 1506.4 / 28 = (15064 / 10) / ( 28 / 1) = (15064 / 10) × (1 / 28)
= (1506.4 / 28) × (1 / 10) = 538 × ( 1 / 10 ) = 538 / 10 = ₹ 53.80
Cost of 1 book = ₹ 53.80
10. The product of two numbers is 40.376. One number is 14.42. Find the other number.
Solution:
Product of two numbers = 40.376
One number = 14.42
Another number = 40376 / 14.42 = (40376 / 1000) / ( 1442 / 100)
= (40376 / 1000) × (100 / 1442) = ( 40376 / 1442) × (100 / 1000)
= 28 × (1 /10) = 28 / 10 = 2.8
Other number = 2.8
Objective type questions
11. 5.6÷ 0.5= ?
(i) 11.4
(ii) 10.4
(iii) 0.14
(iv) 11.2
Hint:
5.6 / 0.5 = 56 / 5
= 11.2
12. 2.01÷ 0.03 = ?
(i) 6.7
(ii) 67.0
(iii) 0.67
(iv) 0.067
Hint:
2.01 / 0.03 = 201 / 3
= 67
13. 0.05 ÷ 0.5 = ?
(i) 0.01
(ii) 0.1
(iii) 0.10
(iv) 1.0
Solution: 0.05 / 0.5 = (5/100) / (5 / 10) = 5 /100 × 10/5 = 1 /10 = 0.1
Exercise 1.4
1. (i) 0.2 (ii) 0.18 (iii) 1.02 (iv) 3.08 (v) 0.094 (vi) 10.34 (vii) 99.4
2. (i) 0.57 (ii) 9.37 (iii) 0.09 (iv) 30.1301 (v) 0.083 (vi) 0.0062
3. (i) 0.007 (ii) 0.038 (iii) 0.493 (iv) 4.6385 (v) 0.003 (vi) 0.274
4. (i) 0.0189 (ii) 0.00087 (iii) 0.0493 (iv) 0.0003 (v) 0.3824 (vi) 0.0938
5. (i) 8 (ii) 9.9 (iii) 14.7 (iv) 0.19 (v) 9 (vi) 1.69
6. 1.91 kg
7. 64 km
8. 650 sq.ft
9. ₹ 53.80
10. 2.8
Objective type questions
11. (iv) 11.2
12. (ii) 67.0
13. (ii) 0.1
Tags : Questions with Answers, Solution | Number System | Term 3 Chapter 1 | 7th Maths , 7th Maths : Term 3 Unit 1 : Number System
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 3 Unit 1 : Number System : Exercise 1.4 (Division of Decimal Numbers) | Questions with Answers, Solution | Number System | Term 3 Chapter 1 | 7th Maths
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# What Are Consecutive Numbers? - Definition & Examples
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Instructor: T.J. Hoogsteen
T.J. is currently a grade 5 teacher and Vice-Principal. He has a master's degree in Educational Administration and is working toward an Ed.D. in Educational Leadership.
When numbers are in counting order, they are also in consecutive order. This lesson will review the three different types of consecutive numbers (counting order, even, and odd),
## What Are Consecutive Numbers?
Jeff and Omar were sitting around and talking about great sports teams. Jeff's friend brought up the New York Yankees dynasty from the late 1990s. He said that the Yankees won the World Series for three consecutive years starting in 1998. Omar, knowing the first World Series was won in 1998, tried to think of the years that the other two titles were won.
How could he figure that out? First, he would have to figure out what the consecutive numbers are that come after 1998. To do that, knowing what consecutive means would help. Consecutive means to follow continuously in a series or sequence. So that means consecutive numbers are numbers that follow each other in order.
Another thing to think about when solving the above problem is that there are three different kinds of consecutive numbers: consecutive numbers, even consecutive numbers, and odd consecutive numbers.
## Consecutive Numbers
Consecutive numbers are numbers that follow each other in order from smallest to largest, what we call regular counting order. Let's take a look at some examples:
You can see that in each of these three series, the numbers go up by one each time, in sequence and without skipping any numbers in between. They are all series of consecutive numbers.
Even consecutive numbers are even numbers that follow in order from smallest to largest when you are counting by twos. Some examples of even consecutive numbers include:
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Unexpectedly Intriguing!
20 December 2019
Sometimes, great insights come from tinkering. For many inventors, tinkering with older inventions is a time-tested method for gaining a better understanding of how they work and how they could work better. The insights they gain can then be applied in new ways, or perhaps combined with other insights in new ways, leading to new innovations.
Po-Shen Loh, a maths professor at Carnegie Mellon, has been tinkering with quadratic equations. In doing that, he has come up with a simpler proof of the quadratic formula. He explains the more intuitive approach he developed for finding the roots of quadratic equations in the following video:
As the video makes clear, the insights that Loh has put together in a new way have all been around for hundreds, if not thousands, of years. What's new and innovative is how Loh has combined them, which offers the promise of making it easier for students to understand how to solve quadratic equations, which ranks as one of the hardest things students learn in algebra classes.
If you didn't catch the 15 seconds near the end of the video with Loh's instructions for using his method for solving the quadratic equation, here they are:
### Alternative Method of Solving Quadratics
1. If you find r and s with sum -B and product C, then + Bx + C = (x - r)(x - s), and they are all the roots
2. Two numbers sum to -B when they are -(B/2) ± u
3. Their product is C when (/4) - = C
4. Square root always gives value u
5. Thus, -(B/2) ± u work as r and s, and are all the roots
If you're used to seeing quadratic equations of the form: ax² + bx + c = 0, we should note that Loh's method divides each term of this equation by a, where in his formulation, B = b/a and C = c/a. That leaves the term by itself, and of course, 0/a = 0, which all but eliminates the need to track the a in the traditional formulation any further.
On the other hand, if you have the equation in that traditional form, and you already know the quadratic formula, you can still use it - it hasn't changed and it still produces valid results. It may even be easier for given values.
If you want to go on a deeper dive into Loh's methods for solving quadratic equations, here's a 40-minute video with lots of examples and discussion of the insights involved.
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# Area of Irregular Polygon
3 teachers like this lesson
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## Objective
SWBAT apply their knowledge of arrays to find the area of an irregular polygon.
#### Big Idea
Students look for break apart irregular polygons to make arrays to find the area of these common shapes.
## Warm Up
5 minutes
Students are seated on the carpet using whiteboards to practice area problems. I purposely choose problems including multiplication facts that will be easy to computer for area. The reason I choose these familiar facts is because I want the students to feel confident in determining area.
I begin by drawing a rectangle with side dimensions of 3 and 2. I ask the students to draw a rectangle, write the matching multiplication sentence, and determine the product of square units. This attention to detail with labeling is something that is necessary as students move between understanding measurements in the Common Core Standards.
This process of drawing rectangles is repeated with three similar problems. The challenge for the students during the warm up is to have them explain the process and their solution to a partner seated next to them on the carpet. This could be differentiated for students by challenging them with more complex problems, or it could be modified for students needing a visual to create an array next to the rectangle for visual support. This would also be a language support for ELL students, who can refer to/create arrays to demonstrate understanding.
## Mini Lesson
10 minutes
Once students solve the basic rectangle areas, and they can confidently explain the concept of their work, I draw an L-shaped polygon on a whiteboard. I ask the students to think about what is needed to find the area of this shape since it is not just a rectangle. The students share with their partners, and the students are able to identify the shape is created from two rectangles with one shared line.
Following this realization, I draw two rectangles separated from each other to provide the visual model to the students. I then mark the separated rectangles with numbers such as 7 and 5, and I ask the students to determine the area. I again use factors that are easy for the students to multiply.
I then mark the L-shaped polygon with the same numbers that were used for the separated rectangles in the example above (7 and 5), and I ask the students if the area has stayed the same. It is important to ask the students to make this determination so that they are using their analytical thinking skills required within the Common Core standards as this new complex shape is considered. The opinions of my students are split and it was an important moment for my students to discuss with each other why or why not the area was the same. Based on their discussions, one student requests centimeter grid paper to prove, perhaps understand, the area.
The entire class uses grid paper to draw an L-shape, and create the two identical rectangles separately to see if the area remains the same. Once it is determined the area remains the same, the students enthusiastically want to try more of these type of shapes.
I chose at this time to mark all sides of the polygon, so they focus on the area. I didn't want to lose the momentum by creating multiple step problems. I provide more problems, using different L-shapes, and then I introduce a U-shaped polygon. Students work with their whiteboards and partners to solve.
I think they're ready, so I challenge the students with a random shaped polygon similar to the teeth on a jack-o-lantern with an up - down pattern of rectangles.
## Try It On Your Own
20 minutes
Before the students begin working on the challenge of the irregular shape, I have students use their math journals with the following steps to plan out a strategy to solve. I ask the students to include in their plan:
1. Figure out the lengths of each side
2. Draw lines to create rectangles
3. Create arrays to model multiplication, if needed
4. Multiply sides to find the area
5. Add area measurements together
I display four complex shapes on the projector, with the length of the sides labeled. Students are given the opportunity to work with a partner and solve the area, and also determine if any of the shapes had missing numerical values. Students use their whiteboards, and centimeter grid paper as needed for support and success. Below is demonstration of how one student solved the area of one of the polygons.
## Closure
5 minutes
To close the lesson, I ask the students to explain to me what I would need to do to find the area of a similar shape. I randomly draw another shape, asking the students to apply the same steps they had used when they had solved the prior problems. Verbalizing this process requires students to dig in to and explain their thinking, and also to consider the thinking of others.
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# How to easily determine the results distribution for multiple dice?
I want to calculate the probability distribution for the total of a combination of dice.
I remember that the probability of is the number of combinations that total that number over the total number of combinations (assuming the dice have a uniform distribution).
What are the formulas for
• The number of combinations total
• The number of combinations that total a certain number
• I think you should treat $(X_1=1 , X_2=2)$ and $(X_1=2 , X_2=1)$ as different events. Sep 21, 2015 at 6:23
Exact solutions
The number of combinations in $$n$$ throws is of course $$6^n$$.
These calculations are most readily done using the probability generating function for one die,
$$p(x) = x + x^2 + x^3 + x^4 + x^5 + x^6 = x \frac{1-x^6}{1-x}.$$
(Actually this is $$6$$ times the pgf--I'll take care of the factor of $$6$$ at the end.)
The pgf for $$n$$ rolls is $$p(x)^n$$. We can calculate this fairly directly--it's not a closed form but it's a useful one--using the Binomial Theorem:
$$p(x)^n = x^n (1 - x^6)^n (1 - x)^{-n}$$
$$= x^n \left( \sum_{k=0}^{n} {n \choose k} (-1)^k x^{6k} \right) \left( \sum_{j=0}^{\infty} {-n \choose j} (-1)^j x^j\right).$$
The number of ways to obtain a sum equal to $$m$$ on the dice is the coefficient of $$x^m$$ in this product, which we can isolate as
$$\sum_{6k + j = m - n} {n \choose k}{-n \choose j}(-1)^{k+j}.$$
The sum is over all nonnegative $$k$$ and $$j$$ for which $$6k + j = m - n$$; it therefore is finite and has only about $$(m-n)/6$$ terms. For example, the number of ways to total $$m = 14$$ in $$n = 3$$ throws is a sum of just two terms, because $$11 = 14-3$$ can be written only as $$6 \cdot 0 + 11$$ and $$6 \cdot 1 + 5$$:
$$-{3 \choose 0} {-3 \choose 11} + {3 \choose 1}{-3 \choose 5}$$
$$= 1 \frac{(-3)(-4)\cdots(-13)}{11!} + 3 \frac{(-3)(-4)\cdots(-7)}{5!}$$
$$= \frac{1}{2} 12 \cdot 13 - \frac{3}{2} 6 \cdot 7 = 15.$$
(You can also be clever and note that the answer will be the same for $$m = 7$$ by the symmetry 1 <--> 6, 2 <--> 5, and 3 <--> 4 and there's only one way to expand $$7 - 3$$ as $$6 k + j$$; namely, with $$k = 0$$ and $$j = 4$$, giving
$${3 \choose 0}{-3 \choose 4} = 15 \text{.}$$
The probability therefore equals $$15/6^3$$ = $$5/36$$, about 14%.
By the time this gets painful, the Central Limit Theorem provides good approximations (at least to the central terms where $$m$$ is between $$\frac{7 n}{2} - 3 \sqrt{n}$$ and $$\frac{7 n}{2} + 3 \sqrt{n}$$: on a relative basis, the approximations it affords for the tail values get worse and worse as $$n$$ grows large).
I see that this formula is given in the Wikipedia article Srikant references but no justification is supplied nor are examples given. If perchance this approach looks too abstract, fire up your favorite computer algebra system and ask it to expand the $$n^{\text{th}}$$ power of $$x + x^2 + \cdots + x^6$$: you can read the whole set of values right off. E.g., a Mathematica one-liner is
With[{n=3}, CoefficientList[Expand[(x + x^2 + x^3 + x^4 + x^5 + x^6)^n], x]]
• Will that mathematica code work with wolfram alpha?
– user28
Oct 14, 2010 at 23:56
• That works. I tried your earlier version but could not make any sense of the output.
– user28
Oct 15, 2010 at 14:24
• @Srikant: Expand[Sum[x^i,{i,1,6}]^3] also works in WolframAlpha
– A. N. Other
Oct 18, 2010 at 7:04
• @A.Wilson I believe many of those references provide a clear path to the generalization, which in this example is $(x+x^2+\cdots+x^6)(x+x^2+x^3+x^4)^3$. If you would like R code to compute these things, see stats.stackexchange.com/a/116913 for a fully implemented system. As another example, the Mathematica code is Clear[x, d]; d[n_, x_] := Sum[x^i, {i, 1, n}]; d[6, x] d[4, x]^3 // Expand
– whuber
Dec 1, 2015 at 14:41
• Note that @whuber's clarification is for 1d6 + 3d4, and that should get you there. For an arbitrary wdn + vdm, (x + x^2 + ... + x^w)^n(x + x^2 + ... + x^v)^m. Additional terms are polynomials constructed and multiplied with the product in the same way. Jan 5, 2016 at 0:58
Yet another way to quickly compute the probability distribution of a dice roll would be to use a specialized calculator designed just for that purpose.
Torben Mogensen, a CS professor at DIKU has an excellent dice roller called Troll.
The Troll dice roller and probability calculator prints out the probability distribution (pmf, histogram, and optionally cdf or ccdf), mean, spread, and mean deviation for a variety of complicated dice roll mechanisms. Here are a few examples that show off Troll's dice roll language:
Roll 3 6-sided dice and sum them: sum 3d6.
Roll 4 6-sided dice, keep the highest 3 and sum them: sum largest 3 4d6.
Roll an "exploding" 6-sided die (i.e., any time a "6" comes up, add 6 to your total and roll again): sum (accumulate y:=d6 while y=6).
Troll's SML source code is available, if you want to see how its implemented.
Professor Morgensen also has a 29-page paper, "Dice Rolling Mechanisms in RPGs," in which he discusses many of the dice rolling mechanisms implemented by Troll and some of the mathematics behind them.
A similar piece of free, open-source software is Dicelab, which works on both Linux and Windows.
$\newcommand{red}{\color{red}}$ $\newcommand{blue}{\color{blue}}$
Let the first die be red and the second be black. Then there are 36 possible results:
\begin{array}{c|c|c|c|c|c|c} &1&2&3&4&5&6\\\hline \red{1}&\red{1},1&\red{1},2&\red{1},3&\red{1},4&\red{1},5&\red{1},6\\ &\blue{^2}&\blue{^3}&\blue{^4}&\blue{^5}&\blue{^6}&\blue{^7}\\\hline \red{2}&\red{2},1&\red{2},2&\red{2},3&\red{2},4&\red{2},5&\red{2},6\\ &\blue{^3}&\blue{^4}&\blue{^5}&\blue{^6}&\blue{^7}&\blue{^8}\\\hline \red{3}&\red{3},1&\red{3},2&\red{3},3&\red{3},4&\red{3},5&\red{3},6\\ &\blue{^4}&\blue{^5}&\blue{^6}&\blue{^7}&\blue{^8}&\blue{^9}\\\hline \red{4}&\red{4},1&\red{4},2&\red{4},3&\red{4},4&\red{4},5&\red{4},6\\ &\blue{^5}&\blue{^6}&\blue{^7}&\blue{^8}&\blue{^9}&\blue{^{10}}\\\hline \red{5}&\red{5},1&\red{5},2&\red{5},3&\red{5},4&\red{5},5&\red{5},6\\ &\blue{^6}&\blue{^7}&\blue{^8}&\blue{^9}&\blue{^{10}}&\blue{^{11}}\\\hline \red{6}&\red{6},1&\red{6},2&\red{6},3&\red{6},4&\red{6},5&\red{6},6\\ &\blue{^7}&\blue{^8}&\blue{^9}&\blue{^{10}}&\blue{^{11}}&\blue{^{12}}\\\hline \end{array}
Each of these 36 ($\red{\text{red}},\text{black}$) results are equally likely.
When you sum the numbers on the faces (total in $\blue{\text{blue}}$), several of the (red,black) results end up with the same total -- you can see this with the table in your question.
So for example there's only one way to get a total of $2$ (i.e. only the event ($\red{1},1$)), but there's two ways to get $3$ (i.e. the elementary events ($\red{2},1$) and ($\red{1},2$)). So a total of $3$ is twice as likely to come up as $2$. Similarly there's three ways of getting $4$, four ways of getting $5$ and so on.
Now since you have 36 possible (red,black) results, the total number of ways of getting all the different totals is also 36, so you should divide by 36 at the end. Your total probability will be 1, as it should be.
• Wow, the table is beautiful! Sep 21, 2015 at 10:46
• Very pretty indeed Sep 21, 2015 at 12:29
There's a very neat way of computing the combinations or probabilities in a spreadsheet (such as excel) that computes the convolutions directly.
I'll do it in terms of probabilities and illustrate it for six sided dice but you can do it for dice with any number of sides (including adding different ones).
(btw it's also easy in something like R or matlab that will do convolutions)
Start with a clean sheet, in a few columns, and move down a bunch of rows from the top (more than 6).
1. put the value 1 in a cell. That's the probabilities associated with 0 dice. put a 0 to its left; that's the value column - continue down from there with 1,2,3 down as far as you need.
2. move one column to the right and down a row from the '1'. enter the formula "=sum(" then left-arrow up-arrow (to highlight the cell with 1 in it), hit ":" (to start entering a range) and then up-arrow 5 times, followed by ")/6" and press Enter - so you end up with a formula like =sum(c4:c9)/6 (where here C9 is the cell with the 1 in it).
Then copy the formula and paste it to the 5 cells below it. They should each contain 0.16667 (ish).
Don't type anything into the empty cells these formulas refer to!
3. move down 1 and to the right 1 from the top of that column of values and paste ...
... a total of another 11 values. These will be the probabilities for two dice.
It doesn't matter if you paste a few too many, you'll just get zeroes.
4. repeat step 3 for the next column for three dice, and again for four, five, etc dice.
We see here that the probability of rolling $12$ on 4d6 is 0.096451 (if you multiply by $4^6$ you'll be able to write it as an exact fraction).
If you're adept with Excel - things like copying a formula from a cell and pasting into many cells in a column, you can generate all tables up to say 10d6 in about a minute or so (possibly faster if you've done it a few times).
If you want combination counts instead of probabilities, don't divide by 6.
If you want dice with different numbers of faces, you can sum $k$ (rather than 6) cells and then divide by $k$. You can mix dice across columns (e.g. do a column for d6 and one for d8 to get the probability function for d6+d8):
• This is very useful for someone like me who just wants a way to do it, without having to understand! If you don't mind the volatility of the OFFSET() function, you can make this dynamic using a named range. For eg I made a range called DiceSize to hold the number of sides and put the first "1" probability in B23. I used a dynamic named range called KingSum which refers to =OFFSET('Dice Rolls'!$A$22,-1*DiceSize,,DiceSize,1). I could then used the formula =SUM(OFFSET(KingSum,ROW(A1),COLUMN(A1)))/DiceSize in Cell C23, dragged all over a large area to give a table reliant on DiceSize. Jan 23, 2021 at 11:38
This is actually a suprisingly complicated question. Luckily for you, there exist an exact solution which is very well explained here:
http://mathworld.wolfram.com/Dice.html
The probability you are looking for is given by equation (10): "The probability of obtaining p points (a roll of p) on n s-sided dice".
In your case: p = the observed score (sum of all dice), n = the number of dice, s = 6 (6-sided dice). This gives you the following probability mass function:
$$P(X_n = p) = \frac{1}{s^n} \sum_{k=0}^{\lfloor(p-n)/6\rfloor} (-1)^k {n \choose k} {p-6k-1 \choose n-1}$$
• Welcome to our site, Felix!
– whuber
Feb 26, 2018 at 20:30
Characteristic functions can make computations involving the sums and differences of random variables really easy. Mathematica has lots of functions to work with statistical distributions, including a builtin to transform a distribution into its characteristic function.
I'd like to illustrate this with two concrete examples: (1) Suppose you wanted to determine the results of rolling a collection of dice with differing numbers of sides, e.g., roll two six-sided dice plus one eight-sided die (i.e., 2d6+d8)? Or (2) suppose you wanted to find the difference of two dice rolls (e.g., d6-d6)?
An easy way to do this would be to use the characteristic functions of the underlying discrete uniform distributions. If a random variable $X$ has a probability mass function $f$, then its characteristic function $\varphi_X(t)$ is just the discrete Fourier Transform of $f$, i.e., $\varphi_X(t) = \mathcal{F}\{f\}(t) = E[e^{i t X}]$. A theorem tells us:
If the independent random variables $X$ and $Y$ have corresponding probability mass functions $f$ and $g$, then the pmf $h$ of the sum $X + Y$ of these RVs is the convolution of their pmfs $h(n) = (f \ast g)(n) = \sum_{m=-\infty}^\infty f(m) g(n-m)$.
We can use the convolution property of Fourier Transforms to restate this more simply in terms of characteristic functions:
The characteristic function $\varphi_{X+Y}(t)$ of the sum of independent random variables $X$ and $Y$ equals the product of their characteristic functions $\varphi_{X}(t) \varphi_{Y}(t)$.
This Mathematica function will make the characteristic function for an s-sided die:
MakeCf[s_] :=
Module[{Cf},
Cf := CharacteristicFunction[DiscreteUniformDistribution[{1, s}],
t];
Cf]
The pmf of a distribution can be recovered from its characteristic function, because Fourier Transforms are invertible. Here is the Mathematica code to do it:
RecoverPmf[Cf_] :=
Module[{F},
F[y_] := SeriesCoefficient[Cf /. t -> -I*Log[x], {x, 0, y}];
F]
Continuing our example, let F be the pmf that results from 2d6+d8.
F := RecoverPmf[MakeCf[6]^2 MakeCf[8]]
There are $6^2 \cdot 8 = 288$ outcomes. The domain of support of F is $S=\{3,\ldots,20\}$. Three is the min because you're rolling three dice. And twenty is the max because $20 = 2 \cdot 6 + 8$. If you want to see the image of F, compute
In:= F /@ Range[3, 20]
Out= {1/288, 1/96, 1/48, 5/144, 5/96, 7/96, 13/144, 5/48, 1/9, 1/9, \
5/48, 13/144, 7/96, 5/96, 5/144, 1/48, 1/96, 1/288}
If you want to know the number of outcomes that sum to 10, compute
In:= 6^2 8 F[10]
Out= 30
If the independent random variables $X$ and $Y$ have corresponding probability mass functions $f$ and $g$, then the pmf $h$ of the difference $X - Y$ of these RVs is the cross-correlation of their pmfs $h(n) = (f \star g)(n) = \sum_{m=-\infty}^\infty f(m) g(n+m)$.
We can use the cross-correlation property of Fourier Transforms to restate this more simply in terms of characteristic functions:
The characteristic function $\varphi_{X-Y}(t)$ of the difference of two independent random variables ${X,Y}$ equals the product of the characteristic function $\varphi_{X}(t)$ and $\varphi_{Y}(-t)$ (N.B. the negative sign in front of the variable t in the second characteristic function).
So, using Mathematica to find the pmf G of d6-d6:
G := RecoverPmf[MakeCf[6] (MakeCf[6] /. t -> -t)]
There are $6^2 = 36$ outcomes. The domain of support of G is $S=\{-5,\ldots,5\}$. -5 is the min because $-5=1-6$. And 5 is the max because $6-1=5$. If you want to see the image of G, compute
In:= G /@ Range[-5, 5]
Out= {1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36}
• Of course, for discrete distributions, including distributions of finite support (like those in question here), the cf is just the probability generating function evaluated at x = exp(i t), making it a more complicated way of encoding the same information.
– whuber
Oct 17, 2010 at 21:24
• @whuber: As you say, the cf, mgf, and pgf are more-or-less the same and easily transformable into one another, however Mathematica has a cf builtin that works with all the probability distributions it knows about, whereas it doesn't have a pgf builtin. This makes the Mathematica code for working with sums (and differences) of dice using cfs particularly elegant to construct, regardless of the complexity of dice expression as I hope I demonstrated above. Plus, it doesn't hurt to know how cfs, FTs, convolutions, and cross-correlations can help solve problems like this.
– A. N. Other
Oct 18, 2010 at 2:38
• @Elisha: Good points, all of them. I guess what I wonder about the most is whether your ten or so lines of Mathematica code are really more "elegant" or efficient than the single line I proposed earlier (or the even shorter line Srikant fed to Wolfram Alpha). I suspect the internal manipulations with characteristic functions are more arduous than the simple convolutions needed to multiply polynomials. Certainly the latter are easier to implement in most other software environments, as Glen_b's answer indicates. The advantage of your approach is its greater generality.
– whuber
Oct 18, 2010 at 3:35
Approximate Solution
I explained the exact solution earlier (see below). I will now offer an approximate solution which may suit your needs better.
Let:
$X_i$ be the outcome of a roll of a $s$ faced dice where $i=1, ... n$.
$S$ be the total of all $n$ dice.
$\bar{X}$ be the sample average.
By definition, we have:
$\bar{X} = \frac{\sum_iX_i}{n}$
In other words,
$\bar{X} = \frac{S}{n}$
The idea now is to visualize the process of observing ${X_i}$ as the outcome of throwing the same dice $n$ times instead of as outcome of throwing $n$ dice. Thus, we can invoke the central limit theorem (ignoring technicalities associated with going from discrete distribution to continuous), we have as $n \rightarrow \infty$:
$\bar{X} \sim N(\mu, \sigma^2/n)$
where,
$\mu = (s+1)/2$ is the mean of the roll of a single dice and
$\sigma^2 = (s^2-1)/12$ is the associated variance.
The above is obviously an approximation as the underlying distribution $X_i$ has discrete support.
But,
$S = n \bar{X}$.
Thus, we have:
$S \sim N(n \mu, n \sigma^2)$.
Exact Solution
Wikipedia has a brief explanation as how to calculate the required probabilities. I will elaborate a bit more as to why the explanation there makes sense. To the extent possible I have used similar notation to the Wikipedia article.
Suppose that you have $n$ dice each with $s$ faces and you want to compute the probability that a single roll of all $n$ dice the total adds up to $k$. The approach is as follows:
Define:
$F_{s,n}(k)$: Probability that you get a total of $k$ on a single roll of $n$ dices with $s$ faces.
By definition, we have:
$F_{s,1}(k) = \frac{1}{s}$
The above states that if you just have one dice with $s$ faces the probability of obtaining a total $k$ between 1 and s is the familiar $\frac{1}{s}$.
Consider the situation when you roll two dice: You can obtain a sum of $k$ as follows: The first roll is between 1 to $k-1$ and the corresponding roll for the second one is between $k-1$ to $1$. Thus, we have:
$F_{s,2}(k) = \sum_{i=1}^{i=k-1}{F_{s,1}(i) F_{s,1}(k-i)}$
Now consider a roll of three dice: You can get a sum of $k$ if you roll a 1 to $k-2$ on the first dice and the sum on the remaining two dice is between $k-1$ to $2$. Thus,
$F_{s,3}(k) = \sum_{i=1}^{i=k-2}{F_{s,1}(i) F_{s,2}(k-i)}$
Continuing the above logic, we get the recursion equation:
$F_{s,n}(k) = \sum_{i=1}^{i=k-n+1}{F_{s,1}(i) F_{s,n-1}(k-i)}$
See the Wikipedia link for more details.
• @Srikant Excellent answer, but does that function resolve to something arithmetic (ie: not recursive)? Oct 14, 2010 at 20:06
• @C. Ross Unfortunately I do not think so. But, I suspect that the recursion should not be that hard as long as are dealing with reasonably small n and small s. You could just build-up a lookup table and use that repeatedly as needed.
– user28
Oct 14, 2010 at 20:10
• The wikipedia page you linked has a simple nonrecursive formula which is a single sum. One derivation is in whuber's answer. Jul 17, 2012 at 23:44
• The wiki link anchor is dead, do you know of a replacement? Apr 13, 2014 at 20:21
Here's another way to calculate the probability distribution of the sum of two dice by hand using convolutions.
To keep the example really simple, we're going to calculate the probability distribution of the sum of a three-sided die (d3) whose random variable we will call X and a two-sided die (d2) whose random variable we'll call Y.
You're going to make a table. Across the top row, write the probability distribution of X (outcomes of rolling a fair d3). Down the left column, write the probability distribution of Y (outcomes of rolling a fair d2).
You're going to construct the outer product of the top row of probabilities with the left column of probabilities. For example, the lower-right cell will be the product of Pr[X=3]=1/3 times Pr[Y=2]=1/2 as shown in the accompanying figure. In our simplistic example, all the cells equal 1/6.
Next, you're going to sum along the oblique lines of the outer-product matrix as shown in the accompanying diagram. Each oblique line passes through one-or-more cells which I've colored the same: The top line passes through one blue cell, the next line passes through two red cells, and so on.
Each of the sums along the obliques represents a probability in the resulting distribution. For example, the sum of the red cells equals the probability of the two dice summing to 3. These probabilities are shown down the right side of the accompanying diagram.
This technique can be used with any two discrete distributions with finite support. And you can apply it iteratively. For example, if you want to know the distribution of three six-sided dice (3d6), you can first calculate 2d6=d6+d6; then 3d6=d6+2d6.
There is a free (but closed license) programming language called J. Its an array-based language with its roots in APL. It has builtin operators to perform outer products and sums along the obliques in matrices, making the technique I illustrated quite simple to implement.
In the following J code, I define two verbs. First the verb d constructs an array representing the pmf of an s-sided die. For example, d 6 is the pmf of a 6-sided die. Second, the verb conv finds the outer product of two arrays and sums along the oblique lines. So conv~ d 6 prints out the pmf of 2d6:
d=:$% conv=:+//.@(*/) |:(2+i.11),:conv~d 6 2 0.0277778 3 0.0555556 4 0.0833333 5 0.111111 6 0.138889 7 0.166667 8 0.138889 9 0.111111 10 0.0833333 11 0.0555556 12 0.0277778 As you can see, J is cryptic, but terse. Love the username! Well done :) The outcomes you should count are the dice rolls, all$6\times 6=36$of them as shown in your table. For example,$\frac{1}{36}$of the time the sum is$2$, and$\frac{2}{36}$of the time the sum is$3$, and$\frac{4}{36}$of the time the sum is$4$, and so on. • I'm really confused by this. I answered a very recent newbie question from someone called die_hard, who apparently no longer exists, then found my answer attached to this ancient thread! Sep 21, 2015 at 19:10 • Your answer to the question at stats.stackexchange.com/questions/173434/… was merged with the answers to this duplicate. – whuber Feb 16, 2017 at 18:47 You can solve this with a recursive formula. In that case the probabilities of the rolls with$n$dice are calculated by the rolls with$n-1$dice. $$a_n(l) = \sum_{l-6 \leq k \leq l-1 \\ \text{ and } n-1 \leq k \leq 6(n-1)} a_{n-1}(k)$$ The first limit for k in the summation are the six preceding numbers. E.g if You want to roll 13 with 3 dice then you can do this if your first two dice roll between 7 and 12. The second limit for k in the summation is the limits of what you can roll with n-1 dice The outcome: 1 1 1 1 1 1 1 2 3 4 5 6 5 4 3 2 1 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1 1 4 10 20 35 56 80 104 125 140 146 140 125 104 80 56 35 20 10 4 1 1 5 15 35 70 126 205 305 420 540 651 735 780 780 735 651 540 420 305 205 126 70 35 15 5 1 edit: The above answer was an answer from another question that was merged into the question by C.Ross The code below shows how the calculations for that answer (to the question asking for 5 dice) were performed in R. They are similar to the summations performed in Excel in the answer by Glen B. # recursive formula nextdice <- function(n,a,l) { x = 0 for (i in 1:6) { if ((l-i >= n-1) & (l-i<=6*(n-1))) { x = x+a[l-i-(n-2)] } } return(x) } # generating combinations for rolling with up to 5 dices a_1 <- rep(1,6) a_2 <- sapply(2:12,FUN = function(x) {nextdice(2,a_1,x)}) a_3 <- sapply(3:18,FUN = function(x) {nextdice(3,a_2,x)}) a_4 <- sapply(4:24,FUN = function(x) {nextdice(4,a_3,x)}) a_5 <- sapply(5:30,FUN = function(x) {nextdice(5,a_4,x)}) • @user67275 your question got merged into this question. But I wonder what your idea was behind your formula: "I used the formula: no of ways to get 8: 5_H_2 = 6_C_2 = 15" ? Dec 12, 2017 at 15:43 One approach is to say that the probability$X_n=k$is the coefficient of$x^{k}$in the expansion of the generating function $$\left(\frac{x^6+x^5+x^4+x^3+x^2+x^1}{6}\right)^n=\left(\frac{x(1-x^6)}{6(1-x)}\right)^n$$ So for example with six dice and a target of$k=22$, you will find$P(X_6=22)= \frac{10}{6^6}\$. That link (to a math.stackexchange question) gives other approaches too
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# Addition And Subtraction Math Facts Worksheets latest 2023
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## Basic Math Facts – Combining Like Terms
In mathematics, you have four main operations: addition, subtraction, multiplication and division. Since subtraction is the inverse of addition, multiplication is repeated addition, and division is the inverse of multiplication, you see that the other three operations are indirectly derived from addition. In this sense, there is really only one binary operation in mathematics: addition. Binary operation refers to the use of a mathematical operator, such as addition, on two numbers or variables, as in x + y. Since we see how important addition is now, we should thoroughly understand one of the most important tasks in all of mathematics—that of combining like terms.
Similar terms are expressions that involve the same combination of variables and their respective exponents but different numerical coefficients. The coefficients, if you remember, are the numbers in front of the variable. To put it in simple terms, like terms are like apples and apples, oranges and oranges. Examples of similar terms are 4x and 2xWhere 3 years and 9 years. To eliminate the abstraction of this whole business, the student must keep in mind that as long as the expressions are similar without regard to the coefficients, the terms can be added or subtracted. Thus 3xy and 4xy are like terms and can be combined to give 7xy. Take away the coefficients 3 and 4, and what’s left? xy.
Often a student will not be able to get the final answer to an algebra problem because at some point similar terms were combined incorrectly. In more complicated math problems, expressions can get a bit more complicated. However, if you keep in mind that like terms are like “animals”, so to speak, then, like animals, they can safely mate. If the terms don’t look alike, you can never combine them. The results are always disastrous. What generally helps students is to take them out of abstraction and put them face to face with concrete facts: if two algebraic expressions, after removing the numbers in front, resemble each other, then they are like terms and can be add and subtract. Note that we are only talking about the two operations of addition and subtraction as these are the two operations that require the terms to be the same before they combine. Multiplication and division do not have this requirement.
Let’s take a look at some examples to make this crystal clear and to see where some possible issues might arise. Let’s do the examples below.
1) 3x + 18x
2) 8xyw – 3xyw + xyw
3) 3x^2 – x^2 + 6x
The first example can be thought of as 3 x and 18 x. Think of the actual letter in plastic form in a child’s play. Obviously you have 21x or 21x as an answer.
The second example gives an indication of when students might start having problems. The minute more than one letter or variable is introduced, students quickly become intimidated. Do not be. If you remove the coefficients of each of the terms, you see that they are all xyw terms. The last term has a coefficient of 1, which is understood. Combining, we have 6xyw.
The third example introduces an expression with exponents. Remember that the exponent, or power, simply tells us how many times to use the number as a factor when multiplying by itself. So x^2 tells us to multiply x by itself, i.e. x^2 = x*x. If you take away the coefficients in this example, you see you have 2 x^2 terms and one x term. So you can only combine x^2 terms. The answer becomes 2x^2 + 6x. Note that terms that cannot be combined remain as they are.
The information here should make you a master of combining similar terms because, in reality, it is a very easy—but extremely important—task. If you follow the precepts laid out here, you should have no more trouble simplifying basic algebraic expressions.
If you have any questions about Addition And Subtraction Math Facts Worksheets, please let us know, all your questions or suggestions will help us improve in the following articles!
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# Addition and Subtraction of Integers
Addition and subtraction are two very common operations performed with integers. There are some rules followed while performing them.
There are some rules followed while adding two or more integers. A number line is often used to show the sum of two integers.
### Adding Integers with Same Sign
Adding 2 positive integers gives an integer with a positive sign
For example, (+5) + (+4) = +9
The addition of 2 negative integers gives an integer with a negative sign
For example, (-6) + (-4) = -10
### Adding Integers with Different Signs
While adding a positive and a negative integer, first, we need to find the absolute values of the number and then subtract the smaller absolute value from the larger one, and finally, we add the sign of the addend with the largest value.
For example,
(-9) + (+3) = -6
(+9) + (-3) = +6
### Properties
The properties of addition of whole numbers also hold true for integers.
1. Closure Property: It states that the sum of any 2 integers also gives an integer.
For example, (+15) + (+7) = 22 is also an integer
2. Commutative Property: It states that the addition result remains even if the position of the numbers is interchanged.
For example, (-15) + (+7) = (+7) + (- 15) = – 8
3. Associative Property: It states that adding 3 or more integers gives the same result regardless of how it is grouped.
For example, (-11) + ((- 9) + 7) = (-11 + (- 9)) + 7 = – 13
For example, 0 + 15 = 15
5. Additive Inverse: When an integer is added to its negative inverse, the result is always 0. The two converse integers are termed additive inverses of one another.
For example, 15 + (-15) = 0
Add:Â (-2) + (-3) + (-1) + (-2) + (-1)
Solution:
(-2) + (-3) + (-1) + (-2) + (-1)
= (-5) + (-1) +(-2) + (-1)
= (-6) +(-2) + (-1)
= (-8) + (-1)
= -9
## Subtraction of Integers
There are 2 general steps to be followed while performing subtraction. They are:
2. Taking the inverse of the number which comes after the sign and then performing the addition
The above steps are how subtracting integers is related to adding integers.
For example, let us subtract (-8) – (3)
Changing the sign, we get
=> (-8) + (3)
Taking the inverse of the number which comes after the sign
⇒ (-8) + (-3) (inverse of 3 is -3)
Adding and putting the sign of the greater number
⇒ (-8) + (-3) = -11
Like addition, the subtraction of integers also has 3 possibilities.
### Subtracting Integers with Same Sign
Subtraction between 2 positive integers is a normal subtraction.
For example, (+8) – (+1) => 8 – 1 = 7
When subtracting between 2 negative integers, the subtraction sign and the negative sign of the second digit become a positive sign.
For example, (-2) – (-8) => (-2) + 8 = 6
### Subtracting Integers with Different Signs
When subtracting a positive from a negative number, we add the two numbers and put a negative sign before it.
For example, (-2) – (+8) => -2 – 8 = -10
Subtracting a negative from a positive number is just a simple addition.
For example, (+2) – (-3) => 2 + 3 = 5
### Properties
1. Closure Property: It states that the difference between any 2 given integers also results in an integer.
For example, (+12) – (+7) = 5 is an integer
2. Commutative Property: It states that the difference between any 2 given integers changes if the order of the integers is interchanged. So, it is not true for subtraction.
For example, (+12) – (+7) = 5 but (+7) – (+12) = -5
So, (+12) – (+7) ≠(+7) – (+12)
3. Associative Property: It states that subtraction between 3 or more integers changes the result if the grouping of the integers is changed. So, it is not true for subtraction.
For example, ((+60) – (+20)) – (+30) = – 10 but [(+60) – ((+20) – (+30)] = 70
So, ((+60) – (+20)) – (+30) ≠[(+60) – ((+20) – (+30)]
Subtract: (-2) – (5)
Solution:
Given (-2) – (5)
=> (-2) + 5
Taking the inverse of the number which comes after the sign and then performing the addition
=> (-2) + (-5)
Adding and putting the sign of the greater number
=> -7
Evaluate: 8 – 12 + (-8) + 5
Solution:
8 – 12 + (-8) + 5
Opening the bracket
=> 8 – 12 – 8 + 5
Rearranging
=> 8 + 5 – 12 – 8
=> 13 – 20
=> -7
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Eureka Math Algebra 1 Module 3 Lesson 8 Answer Key
Engage NY Eureka Math Algebra 1 Module 3 Lesson 8 Answer Key
Eureka Math Algebra 1 Module 3 Lesson 8 Exercise Answer Key
Opening Exercise
The sequence of perfect squares {1, 4, 9, 16, 25,…} earned its name because the ancient Greeks realized these quantities could be arranged to form square shapes.
If S(n) denotes the nth square number, what is a formula for S(n)?
S(n) = n2
Exercises 1–11
Exercise 1.
Prove whether or not 169 is a perfect square.
169 is a perfect square because it can be arranged into a 13 row by 13 column array of dots.
Exercise 2.
Prove whether or not 200 is a perfect square.
If 200 is a perfect square, then there is a positive integer a such that a2 = 200. But since 142 = 196, and
152 = 225, we have 14 < a < 15, which means a cannot be an integer.
Hence, 200 is not a perfect square because it cannot be arranged into an n row by n column array of dots.
Exercise 3.
If S(n) = 225, then what is n?
n = 15 because 15 × 15 = 225.
Exercise 4.
Which term is the number 400 in the sequence of perfect squares? How do you know?
Since 400 = 20 × 20, this number would be the 20th term in the sequence.
Instead of arranging dots into squares, suppose we extend our thinking to consider squares of side length x cm.
Exercise 5.
Create a formula for the area A(x) cm2 of a square of side length x cm: A(x) = ___________.
A(x) = x2
Exercise 6.
Use the formula to determine the area of squares with side lengths of 3 cm, 10.5 cm, and π cm.
A(3) = 9 The area of a square with side lengths of 3 cm is 9 cm2.
A(10.5) = 110.25 The area of a square with side lengths of 10.5 cm is 0.25 cm2.
A(π) = π2 The area of a square with side lengths of π cm is π2 cm2.
Exercise 7.
What does A(0) mean?
In this situation, A(0) has no physical meaning since you cannot have a square whose sides measure cm.
Exercise 8.
What does A( – 10) and A($$\sqrt{2}$$) mean?
In this situation, A( – 10) has no physical meaning since a square cannot have sides whose measure is negative. A($$\sqrt{2}$$) does have meaning although it would be impossible to measure that side length physically with a ruler.
The only way for the input to be negative is if we redefine what the input value represents—perhaps we just want to find the square of a number and not relate it to a geometric figure at all.
The triangular numbers are the numbers that arise from arranging dots into triangular figures as shown:
Exercise 9.
What is the 100th triangular number?
It would be 100(101) divided by 2 or 5050 using the idea displayed in the picture in Exercise 12 below.
Exercise 10.
Find a formula for T(n), the nth triangular number (starting with n = 1).
T(n) = $$\frac{n(n + 1)}{2}$$
Exercise 11.
How can you be sure your formula works?
By substituting a term number into the formula, you get the correct number of dots in that figure. The formula also works because each triangular number is exactly half of a rectangular arrangement of dots whose dimensions are n by n + 1. For example:
Exercises 12–14
Exercise 12.
Create a graph of the sequence of triangular numbers T(n) = $$\frac{n(n + 1)}{2}$$, where n is a positive integer.
Exercise 13.
Create a graph of the triangle area formula T(x) = $$\frac{x(x + 1)}{2}$$, where x is any positive real number.
Exercise 14.
How are your two graphs alike? How are they different?
The graph of Exercise 12 is not connected because the input values must be positive integers.
The graph of Exercise 13 is connected because the area of a triangle can be any positive real number. Both graphs have points in common at the positive integer input values. Neither graph is a linear or an exponential function.
Eureka Math Algebra 1 Module 3 Lesson 8 Problem Set Answer Key
Question 1.
The first four terms of two different sequences are shown below. Sequence A is given in the table, and sequence B is graphed as a set of ordered pairs.
a. Create an explicit formula for each sequence.
A(n) = 15 + 16(n – 1), where n is an integer greater than 0. B(n) = 3(2)(n – 1), where n is an integer greater than 0.
b. Which sequence will be the first to exceed 500? How do you know?
Students may use observation of the rate of growth in the table and graph or use the formulas they wrote in part (a) to describe that the growth rate of sequence B will cause it to eventually exceed the values of sequence A and to reach 500 before sequence A does. They may also choose a more definitive proof by extending tables of values or graphs of the terms to see which sequence exceeds 500 first. Alternatively, they could choose to use trial and error or algebra skills to find the smallest value of n for which each formula yields a value of 500 or greater. For sequence A, n = 32 is the first term to yield a value greater than 500. For sequence B, n = 9 is the first term to yield a value greater than 500.
Question 2.
A tile pattern is shown below.
a. How is this pattern growing?
Each figure contains three more tiles than the previous figure.
b. Create an explicit formula that could be used to determine the number of squares in the nth figure.
F(n) = 3n, where n is a positive integer.
c. Evaluate your formula for n = 0 and n = 2.5. Draw Figure 0 and Figure 2.5, and explain how you decided to create your drawings.
F(0) = 0, and F(2.5) = 7.5. You could draw no squares for Figure 0.
You could draw 7 squares and $$\frac{1}{2}$$ of a square for Figure 2.5.
Question 3.
The first four terms of a geometric sequence are graphed as a set of ordered pairs.
a. What is an explicit formula for this sequence?
A(n) = 4.5(2)(n – 1) for n > 0.
b. Explain the meaning of the ordered pair (3,18).
It means that the 3rd term in the sequence is 18, or A(3) = 18.
c. As of July 2013, Justin Bieber had over 42,000,000 Twitter followers. Suppose the sequence represents the number of people that follow your new Twitter account each week since you started tweeting. If your followers keep growing in the same manner, when will you exceed 1,000,000 followers?
This sequence will exceed 1,000,000 followers when n > 19. At some time in the 19th week, I would have 1,000,000 followers.
Eureka Math Algebra 1 Module 3 Lesson 8 Exit Ticket Answer Key
Recall that an odd number is a number that is one more than or one less than twice an integer. Consider the sequence formed by the odd numbers {1, 3, 5, 7,…}.
Question 1.
Find a formula for O(n), the nth odd number starting with n = 1.
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Sampling Without Replacement
The draws in a simple random sample aren't independent of each other. This makes calculating variances a little less straightforward than in the case of draws with replacement.
In this section we will find the variance of a random variable that has a hypergeometric distribution. Then we will use the variance to examine the accuracy of polls.
As a preliminary, let's do some calculations involving indicator random variables.
Squares and Products of Indicators
Let $I_A$ be the indicator of the event $A$. Then the distribution of $I_A$ is given by
value $0$ $1$
probability $1-P(A)$ $P(A)$
We know that $E(I_A) = P(A)$.
Now let's look at the random variable $I_A^2$. As we have seen before, this is a function of $I_A$ that is in fact equal to $I_A$: when $I_A = 0$ then $I_A^2 = 0$ and when $I_A = 1$ then $I_A^2 = 1$.
So $I_A^2 = I_A$ and hence $E(I_A^2) = P(A)$.
Now let $I_B$ be the indicator of the event $B$, and consider the product $I_AI_B$.
This product is itself an indicator. It has the value 1 when both A and B occur, and it is 0 otherwise. In other words, it is the indcator of the event $AB$.
Therefore $E(I_AI_B) ~ = ~ P(AB)$.
The expectation of the product of two indicators is the probability that both the events being indicated occur.
SD of the Hypergeometric
Let $X$ have the hypergeometric $(N, G, n)$ distribution. That is, let $X$ be the number of good elements in a simple random sample of size $n$ drawn from a population of $N$ elements of which $G$ are good.
As we have seen before,
$$X ~ = ~ I_1 + I_2 + \cdots + I_n$$
where $I_j$ is the indicator of the event that the $j$th draw is good.
We know that $I_1, I_2, \ldots, I_n$ are identically distributed, and that each has chance $G/N$ of being 1. That was how we showed that
$$E(X) ~ = ~ n\frac{G}{N}$$
But we can't just add the variances of the indicators to get the variance of $X$. Unlike the binomial case, these indicators aren't independent.
So let's get back to basics and try to use
$$Var(X) ~ = ~ E(X^2) - (E(X)^2) ~ = ~ E(X^2) - \big{(} n\frac{G}{N} \big{)}^2$$
To find $E(X^2)$ note that
$$X^2 ~ = ~ (I_1 + I_2 + \cdots + I_n)^2 ~ = ~ \sum_{j=1}^n I_j^2 + \mathop{\sum_{j=1}^n \sum_{k=1}^n}_{j \ne k} I_jI_k$$
So
\begin{align*} E(X^2) ~ &= ~ E(\sum_{j=1}^n I_j^2) + E(\mathop{\sum_{j=1}^n \sum_{k=1}^n}_{j \ne k} I_jI_k)\\ &=~ \sum_{j=1}^n E(I_j^2) + \mathop{\sum_{j=1}^n \sum_{k=1}^n}_{j \ne k} E(I_jI_k)\\ &=~ nE(I_1^2) + n(n-1)E(I_1I_2) \end{align*}
by the symmetry of simple random sampling.
Apply our results about indicators to see that
$$E(X^2) ~ = ~ n \frac{G}{N} + n(n-1)\frac{G}{N}\cdot\frac{G-1}{N-1}$$
and therefore
$$Var(X) ~ = ~ n \frac{G}{N} + n(n-1)\frac{G}{N}\cdot\frac{G-1}{N-1} - \big{(} n\frac{G}{N} \big{)}^2$$
That looks awful but it's actually not so bad. Pull out a common factor:
$$Var(X) ~ = ~ n\frac{G}{N}\big{(} 1 + (n-1)\frac{G-1}{N-1} - n\frac{G}{N}\big{)}$$
After a little manipulation this becomes
$$Var(X) ~ = ~ n\frac{G}{N} \cdot \frac{N-G}{N}\cdot\frac{N-n}{N-1}$$
The initial part of this formula is the binomial variance $npq$. To see this more clearly, write $B = N-G$ for the number of bad elements. Then
$$Var(X) ~ = ~ \big{(} n\frac{G}{N} \cdot \frac{B}{N}\big{)}\frac{N-n}{N-1}$$
and
$$SD(X) ~ = ~ \sqrt{n\frac{G}{N} \cdot \frac{B}{N}}\sqrt{\frac{N-n}{N-1}} ~ = ~ \sqrt{npq} \sqrt{\frac{N-n}{N-1}}$$
for $p = \frac{G}{N}$.
For example, the number of hearts in a 5-card poker hand is expected to be
$$5 \times \frac{13}{52} ~ \approx ~ 1.25$$
with an SD of
$$\sqrt{5 \times \frac{13}{52} \times \frac{39}{52}}\sqrt{\frac{52 - 5}{52 - 1}} ~ \approx ~ 0.93$$
The Size of the FPC
We have shown that the SD of the number of good elements when drawing without replacement is the same as though we had been drawing with replacement, times the finite population correction or fpc given by
$$\text{fpc} ~ = ~ \sqrt{\frac{N-n}{N-1}}$$
Since the sample size is typically greater than 1, the fpc is typically less than 1.
This implies that the hypergeoemtric SD is less than the corresponding binomial SD if the draws had been made with replacement.
The figure below shows the histogram of the hypergeometric $(52, 26, 5)$ distribution (the distribution of the number of red cards in a poker hand) and the binomial $(5, 26/52)$ distribution.
Both histograms are centered at 2.5. You can see that the hypergeometric histogram is a bit taller and hence a bit less spread out than the binomial.
But not by much! Let's see why.
As we have observed before, sampling with and without replacement are essentially the same when the sample size is small relative to the population size. We now have another confirmation of this.
When the sample size is small relative to the population, the finite population correction is close to 1. That is because
$$\frac{N-n}{N-1} ~ = ~ 1 - \frac{n-1}{N-1} ~ \approx ~ 1 - \frac{n}{N} ~ \approx ~ 1$$
when $\frac{n}{N}$ is small.
The SD of the number of good elements in the sample is the same for sampling with and without replacement, apart from the fpc. When the fpc is close to 1, the two SDs are essentially equal.
Data scientists often have to work with relatively small samples drawn from large populations. We see once more that in such situations they can treat the draws as if they had been made with replacement.
The Accuracy of Simple Random Samples
When politicans don't like the result of a poll, they sometimes say they don't trust polls that are based on tiny percents of the population. Let's see what we think of that.
Suppose a poll is based on a simple random sample drawn from a huge population of voters of whom a proportion $p$ favor a politician. Then the SD of the number of voters who favor the politician is
$$\sqrt{npq}\sqrt{\frac{N-n}{N-1}} ~ \approx ~ \sqrt{npq}$$
because the fpc is close to 1.
Essentially, the SD depends on $p$ and on the sample size $n$, not on the population size $N$. The population size appears only in the fpc which is close to 1. In effect, the population size is so large in comparison to the sample size that it might as well be infinite.
Thus if all other things (such as $p$) are equal, a simple random sample of size 100 taken from Berkeley (population about 120,000) is just about as accurate as a simple random sample of size 100 taken from San Francisco (population about 880,000).
As for the doubting politicians, we have to remind them that when the population is large, it's the sample size that matters, not the population size. A a 0.1% sample of one million voters is a small percent of the population but it consists of $n = 1000$ voters. Simple random samples that big are pretty accurate, as we will see in the next section.
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# 2005 AIME II Problems/Problem 4
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$
## Solution 1
$10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors.
$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.
$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.
Now, we use the Principle of Inclusion-Exclusion. We have $121 + 64 + 276$ total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers. Thus, we must subtract off the divisors of their pair-wise greatest common divisors.
$\gcd(10^{10},15^7) = 5^7$ which has 8 divisors.
$\gcd(15^7, 18^{11}) = 3^7$ which has 8 divisors.
$\gcd(18^{11}, 10^{10}) = 2^{10}$ which has 11 divisors.
So now we have $121 + 64 + 276 - 8 -8 -11$ potential divisors. However, we've now undercounted those factors which divide all three of our numbers. Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of $121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}$.
## Solution 2
We can rewrite the three numbers as $10^{10} = 2^{10}\cdot 5^{10}$, $15^7 = 3^7\cdot5^7$, and $18^{11} = 2^{11}\cdot3^{22}$. Assume that $n$ (a positive integer) is a divisor of one of the numbers. Therefore, $n$ can be expressed as ${p_1}^{e_1}$ or as ${p_2}^{e_2}{p_3}^{e_3}$ where $p_1$, $p_2$ are in $\{2,3,5\}$ and $e_1$, $e_2$ are positive integers.
If $n$ is the power of a single prime, then there are 11 possibilities ($2^1$ to $2^{11}$) for $p_1=2$, 22 possibilities ($3^1$ to $3^{22}$) for $p_1=3$, 10 possibilities ($5^1$ to $5^{10}$) for $p_1=5$, and 1 possibility if $n=1$. From this case, there are $11+22+10+1=44$ possibilities.
If $n$ is the product of the powers of two primes, then we can just multiply the exponents of each rewritten product to get the number of possibilities, since each exponent of the product must be greater than 0. From this case, there are $10*10+11*22+7*7=100+242+49=391$ possibilities.
Adding up the two cases, there are $44+391=\boxed{435}$ positive integers.
-alpha_2
2005 AIME II (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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The Way To Prime
This animation image illustrates how to find prime numbers between 2 to 120. Not too mathematical this time, but sticky one. Hope you will enjoy it.
Finding Primes
Source: Unknown
Two Interesting Math Problems
Problem1: Smallest Autobiographical Number:
A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on.
For example: 42101000 is autobiographical.
Find, with explanation, the smallest autobiographical number.
Solution of Problem 1
Problem 2: Fit Rectangle:
A rectangle has dimensions $39.375$ cm $\times 136.5$ cm.
• Find the least number of squares that will fill the rectangle.
• Find the least number of squares that will fill the rectangle, if every square must be the same size and Find the largest square that can be tiled to completely fill the rectangle.
Solution of Problem 2
Solutions of Problem 1:
The restrictions which define an autobiographical number make it straightforward to find the lowest one. It cannot be 0, since by
definition the first digit must indicate the number of zeros in the number. Presumably then, the smallest possible autobiographical number will contain only one 0.If this is the case, then the first digit must be 1. 10 is not a candidate because the second digit must indicate the number of 1s in the number–in this case, 1. So If the
number contains only one zero, it must contain more than one 1.
(If it contained one 1 and one 0, then the first two digits
would be 11, which would be contradictory since it actually contains two 1s).
Again, presumably the lowest possible such number will contain the lowest
possible number of 1s, so we try a number with one 0 and two 1s. It will be of the form: 12-0–..
Now, there is one 2 in this number, so the first three digits must be 121. To meet all the conditions discussed above, we can simply take a 0 onto the end of this to obtain 1210, which is
the smallest auto-biographical number.
Solution of Problem 2:
We solve the second and third parts of the question
first:
We convert each number to a fraction and get a common denominator, then find the gcd (greatest common divisor) of the numerators.
That is, with side lengths $39.375$ cm and $136.5$ cm , we convert those numbers to fractions (with a common
denominator):
$39.375 = \dfrac{315}{8}$.
$136.5 = \dfrac{273}{2} = \dfrac{1092}{8}$.
Now we need to find the largest common factor of 315 and 1092.
Which is 21. So $\dfrac{21}{8}=2.625$ is the largest number that divides evenly into the two numbers $39.375$ and $136.5$.
There will be $\dfrac{1092}{21} \times \dfrac{315}{21} = 52 \times 15 = 780$ squares, each one a $2.625$ cm $\times 2.625$ cm square needed to fill the rectangle (52 in each row,with 15 rows).
Now we shall solve the first part.
Number of squares lengthwise is 52 and breadthwise is 15. Now we will combine these squares in order to find least number of squares to fill the rectangle. First three squares would be of
dimension 15 by 15. In this way length of 45 units is utilized. Now the rectangle which is left with us excluding three squares is 7 by 15. Again in the same way we can make two squares of dimension 7 by 7. In this way breadth of 14 units is utilized.
Now we are left with the rectangle of dimension 7 by 1.
These can further be subdivided into seven squares each of
dimension 1 by 1. In this way the least number of squares to fill the
rectangle is 3 + 2+ 7 = 12. The required answer is 12.
Note that the three numbers 3, 2, and 7 are involved in the Euclidean Algorithm for finding the g.c.d.!
Source: Internet
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# Triangles, Rectangles, Squares, and Circles
Size: px
Start display at page:
Transcription
1 LESSON Name 2 Teacher Notes: page 27 Triangles, Rectangles, Squares, and Circles Refer students to Circle on page 4 in the Student Reference Guide. Post Reference Chart Circle. Use the compasses from the manipulative kit to illustrate concepts in this lesson. New Concept Triangle 3 sides Rectangle 4 sides width length Square 4 equal sides (special kind of rectangle) Circle Radius = of diameter 2 A compass is a tool used to draw circles. radius cm in Diameter = 2 radius diameter Saxon Math Intermediate 4 33 Adaptations Lesson 2
2 New Concept, continued Activity page 29 Drawing a Circle Use your textbook to complete this activity. Lesson Practice a. Draw a TRIANGLE with two sides that are the same length. b. Draw a RECTANGLE that is about twice as long as it is wide. c. Use a compass to draw a CIRCLE with a radius of inch. d. What is the diameter of a circle that has a 3-cm radius? 3 cm e. What is another name for a rectangle whose length is equal to its width? s Saxon Math Intermediate 4 34 Adaptations Lesson 2
3 Written Practice page Hiroshi Harry in all Tisha n Jane total n = 3. The ones digit is 5. The number is greater than 640. It is less than seven hundred fifty-three expanded form: 5. x 0 y 6. average daily high and low temperatures x + y Use a centimeter ruler. a. length b. width c. perimeter Perimeter Add all sides. a. b. c. Saxon Math Intermediate 4 35 Adaptations Lesson 2
4 Written Practice, continued page \$486 + \$ \$524 + \$ Perimeter Add all sides. 3. bottom missing 7 a bottom missing 5 b 6 a = b = d bottom missing 4 b 2 d = b = Saxon Math Intermediate 4 36 Adaptations Lesson 2
5 Written Practice, continued page 3 9. top missing y bottom missing 75 p y = p = a., 28, 35, 42,,,, b., 40, 30, 20,,,, 26. Add the two l 27. d = 2r plus the two w 4 cm to fine the p. A 8 in. B 2 in. C 8 cm D 2 cm Saxon Math Intermediate 4 37 Adaptations Lesson 2
6 Written Practice, continued page Use your student clock. Count forward minutes 7:35 a.m. = minutes start time end time 29. Round to the nearest ten: a. 7 6 b. 7 3 c. 7 5 a. b c. 30. Round to the nearest 25 cents: a. \$6.77 \$6.77 \$6.75 \$7.00 b. \$7.97 \$7.97 \$7.75 \$8.00 Saxon Math Intermediate 4 38 Adaptations Lesson 2
8 Lesson Practice What fraction of each shape is shaded? (Name the number shaded out of the total number of parts.) a. b. Order from least to greatest. c. 3, 2 0, 4,, d. 2 5, 3 8, 2,, e. Order from greatest to least. f. Order from greatest to least. Act it out using fraction manipulatives. Act it out using fraction manipulatives. 3 4, 5, 2 3,, 6 0, 4 5, 2,, g. Three dimes are what fraction of a h. \$ \$2.75 dollar? \$ i. \$ \$4.28 j. Estimate. Use compatible numbers. \$.49 + \$.49 \$ + \$ \$ Saxon Math Intermediate 4 40 Adaptations Lesson 22
9 Written Practice page l in. beginning grew end l = 3. Less than , 80, 72, 64,,,, odd 5., 60, 54, 48,, 6.,, Perimeter add all sides 7. See page in the Student Reference Guide. 8. What place? 89 Saxon Math Intermediate 4 4 Adaptations Lesson 22
10 Written Practice, continued page expanded form: words: + +. Eighteen is greater than negative twenty. 2. a. 2 8 b. \$5.95 a. b. 3. Estimate. Desk = meters high cm d = 2r 6. Saxon Math Intermediate 4 42 Adaptations Lesson 22
11 Written Practice, continued page meter = cm 2 meters = \$ \$ \$ \$ n x p 4 n = x = p = If the equation 20 + n = 60 is true, then which of the following equations is not true? a = n b. 60 n = n c. n 20 = 60 d. n + 20 = 60 Saxon Math Intermediate 4 43 Adaptations Lesson 22
12 Written Practice, continued page Estimate. compatible numbers \$.26 \$ + \$ \$ \$,26 can be rounded to, and \$3.73 can be rounded to. It s easier to estimate the sum using c n. 29. Waterfalls Name Location Height (ft) Multnomah Oregon 620 Maletsunyane Lesotho, Africa 630 Wentworth Australia 64 Reichenbach Switzerland 656 Least to greatest. Look at the hundreds. Look at the tens. Wentworth,,, 30. tenth number 4, 8, 2, 6, 20,,,,, Saxon Math Intermediate 4 44 Adaptations Lesson 22
13 LESSON Name 23 Teacher Notes: page 40 Lines, Segments, Rays, and Angles Introduce Hint #5 Geometry Vocabulary. Refer students to Types of Lines and Types of Angles on pages 5 6 in the Student Reference Guide. New Concept Lines and Segments Math Language Parallel lines never cross. When lines cross, they intersect. Intersecting lines that form square corners are perpendicular. line A B ray F G segment D E parallel lines intersecting lines intersecting and perpendicular Math Language A vertex is where 2 sides of an angle meet. Angles are formed where lines or segments intersect or where two or more rays begin. Angles side vertex side Straight Obtuse Acute Right Activity page 43 Real-World Segments and Angles This activity is optional. Saxon Math Intermediate 4 45 Adaptations Lesson 23
14 Lesson Practice a. Draw two SEGMENTS that intersect b. Draw two LINES that are but are not perpendicular. perpendicular. c. Draw a RAY. d. Are the rails of a train track parallel or perpendicular? Train track rails are p because they never i. Locate and describe the angles of items in or around your classroom. item type of angle item type of angle item type of angle e. A triangle has how many angles? f. Which of these angles does not look like a right angle? A B C D Saxon Math Intermediate 4 46 Adaptations Lesson 23
15 Written Practice page first second both 2. 2 first n second both 3. Greater than a., 40, 36, 32,,,, odd b., 30, 27, 24,,,, a. b Six hundred thirty-eight is less than six hundred eighty-three a. 9 2 b. \$9.67 a. b. Saxon Math Intermediate 4 47 Adaptations Lesson 23
16 Written Practice, continued page If a nickel s radius is cm, its diameter is cm. Ten nickels in a row is cm, or cm Which of these shapes has four right angles? a. Length A B b. Width c. Perimeter Perimeter Add all sides. C D. 2. Use your student clock. Count forward minutes. = start time end time minutes 3. \$83 \$27 Check: \$27 + \$ Check: Saxon Math Intermediate 4 48 Adaptations Lesson 23
17 Written Practice, continued page Check: \$ \$ \$ \$ k b 6 k = b = 20. k k = = 25. a. \$ = nickels b. nickel = of a dollar c. 7 nickels = of a dollar Saxon Math Intermediate 4 49 Adaptations Lesson 23
18 Written Practice, continued page If 26 + m = 63, then which of these equations is not true? m A m + 26 = 63 B m 63 = 26 C 63 m = 26 D = m 27. Which of these figures illustrates a ray? A B C D 28. \$.99 \$ A reasonable estimate would be \$ = \$. 29. Look at the hundreds. Look at the tens. Look at the ones v a b c v = Saxon Math Intermediate 4 50 Adaptations Lesson 23
19 LESSON Name 24 Teacher Note: page 47 Review Missing Numbers on page 7 in the Student Reference Guide. Inverse Operations New Concept ADDITION: to find missing addend 5 + a a = 3 x SUBTRACTION: to find missing top t t = 8 SUBTRACTION: to find missing bottom 8 b b = 6 subtract 7 3 x = 4 add subtract Lesson Practice Find each missing number: a m 42 b. q c. 53 w 28 m = q = w = d. n e y 63 f. 62 a 26 n = y = a = Saxon Math Intermediate 4 5 Adaptations Lesson 24
20 Written Practice page 49. in in ft = in. 2 ft = 2. 2 commercials p not commercials 3. Does equal mean odd or even? So which is not? total A 36 B 45 C 60 D 24 p = 4., 9, 8,,, 45,, 5. 7, 4, 2,,,, Saxon Math Intermediate 4 52 Adaptations Lesson 24
21 Written Practice, continued page a. 7 7 b. \$29.39 c. \$9.9 \$9.9 a. \$9.00 \$9.25 b. c. 8. meter is a little more than 3 feet Use your student clock. Count forward hours: 8:0 p.m. = hours start time end time Count forward minutes: 8:0 p.m. 9:05 p.m. = minutes Add. start time end time hours + minutes = hours and minutes Saxon Math Intermediate 4 53 Adaptations Lesson 24
22 Written Practice, continued page Which street is parallel to Elm? Parallel looks like railroad tracks. Elm Oak Broadway W N S E. a. \$ = dimes 2. 5 cm long 2 cm wide b. dime = of a dollar c. 9 dimes = of a dollar a. b. c. 3. Name each type of angle shown below. 4. \$3 \$4 a. b. c. a. b. c. Saxon Math Intermediate 4 54 Adaptations Lesson 24
23 Written Practice, continued page 5 5. \$468 + \$ \$ \$ c 9 9. b d 93 c = b = d = 2. n n = See page 5. Saxon Math Intermediate 4 55 Adaptations Lesson 24
24 Written Practice, continued page A segment has two e and is part of a l. A has no e. 28. Round to nearest dollar. \$4.25 \$ + \$ \$ + \$ one ticket two tickets One ticket costs about \$ A reasonable estimate of the cost of two tickets is \$ + \$, or \$ = Jasmine found found seashells. Then she more seashells. How many F did she find in all? Saxon Math Intermediate 4 56 Adaptations Lesson 24
25 LESSON Name 25 Teacher Note: page 52 Subtraction Word Problems Review Missing Numbers on page 7 in the Student Reference Guide. New Concept Lesson Practice Subtraction word problems have a missing number. If the top number is missing Add. Some Some went away What is left p pencils 5 pencils 22 pencils p = 37 If the bottom number is missing Subtract. Some 42 seashells Some went away s seashells s = 3 What is left 29 seashells If the difference is missing Subtract as usual. Some 65 beads Some went away What is left 3 beads b beads b = 52 a. 42 had n mailed now b. n beginning donated now n = n = c. 75 \ c had spent n \ c now n = Saxon Math Intermediate 4 57 Adaptations Lesson 25
26 Written Practice page had had gave gave now now 3. n had 4. Half Divide by 2. used now 5. a., 5, 0,,, 25,, b., 5, 0,,, 5,, a. b. 6. Seven hundred sixty-two is less than eight hundred twenty-six. Saxon Math Intermediate 4 58 Adaptations Lesson 25
27 Written Practice, continued page a. 7 8 b. \$7.80 c. \$7.80 \$7.80 \$7.75 \$8.00 a. b. c in. d = 2r Use your student clock. Count forward minutes: :35 p.m. = minutes start time end time Saxon Math Intermediate 4 59 Adaptations Lesson 25
28 Written Practice, continued page Which street is perpendicular to Elm? Perpendicular looks like this:. Elm Oak Broadway W N S E 2. 4 cm Find the pattern. Try ones. Try twos. 4. \$52 \$ \$ \$2.03 Saxon Math Intermediate 4 60 Adaptations Lesson 25
29 Written Practice, continued page g b d g = b = d = 2. y y = Which of these is not equivalent to one meter? A 000 mm B 00 cm C 000 km D m meter = mm meter = cm km = m Saxon Math Intermediate 4 6 Adaptations Lesson 25
30 Written Practice, continued page A ray has one e. A segment has t endpoints Illinois Potomac total 29. Round to nearest 0 cents. \$ \$0.59 \$ + \$ To get a reasonable estimate, I used com numbers. \$3.39 is close to \$, and \$0.59 is close to \$. Adding these together gives a sum of \$. 30. from F 0 F = from 0 F 2 F = Saxon Math Intermediate 4 62 Adaptations Lesson 25
31 LESSON Name 26 Teacher Notes: page 58 Drawing Pictures of Fractions Introduce Hint #6 Drawing Fractional Parts. For additional practice, students may complete Fraction Activity 26. New Concept Example To draw a picture of a fraction:. Draw the figure. 2. Divide into equal parts (bottom number). 3. Shade the correct number of parts (top number). Draw a rectangle, and shade 2 of it. 3 Rectangle 3 equal parts 2 parts shaded Here are some pictures of other fractions: Example To divide a circle into thirds:. Draw a dot in the center. 2. Make a Y from the dot. 3. There are 3 equal parts. These are NOT equal parts: Saxon Math Intermediate 4 63 Adaptations Lesson 26
33 Written Practice, continued page Less than cm 6 cm 8 cm even Perimeter Add all sides. 7. Negative twenty is less than negative twelve. 8. a. 9 b. \$0.90 a. b. 9. meter = cm See page in the Student Reference Guide. 0. Before noon Which street makes a right angle with Oak? A right angle looks like this: Elm Oak Broadway W N S E Saxon Math Intermediate 4 65 Adaptations Lesson 26
34 Written Practice, continued page F 4. y \$486 + \$277 y = 6. \$68 \$39 7. \$ \$ n n 65 n = n = Saxon Math Intermediate 4 66 Adaptations Lesson 26
35 Written Practice, continued page c e c = e = In which figure is 2 not shaded? A B C D 27. Is the largest angle of this triangle acute, right, or obtuse? Saxon Math Intermediate 4 67 Adaptations Lesson 26
36 Written Practice, continued page How many three-digit numbers can you write with 0, 7, and 3? Write even or odd. Act it out. 3 7 odd can t be 0 7 different numbers. 29. Round to nearest fifty cents. \$ \$8.59 \$ + \$ Yes, \$4 is reasonable because \$5.45 is about \$ and \$8.59 is about \$, and the sum of these is \$ Saxon Math Intermediate 4 68 Adaptations Lesson 26
37 LESSON Name 27 Teacher Notes: page 62 Review Hint # Elapsed Time. Review Hint #0 Reading Clocks. Multiplication as Repeated Addition Elapsed Time New Concept Multiplication as Repeated Addition Change + problems to problems : = = 20 Use multiplication to show repeated addition. 4 5 = 20 or 5 4 Four groups of five equals twenty. 20 Elapsed Time Example Elapsed time: will it be ago forward backward It is afternoon. What time will it be in hour and 50 minutes?. The time now is :45 p.m. 2. Count forward hour to 2:45 p.m. 3. Count forward 50 minutes to 3:35 p.m Saxon Math Intermediate 4 69 Adaptations Lesson 27
38 New Concept, continued Activity page 64 Finding Time Use your textbook to complete this activity. Lesson Practice a. 3 b. 9 c. 7 d. 5 How many times? Remember to write a.m. or p.m. e. If it is morning, what time will it be in 2 hours and 25 minutes? Time now: Count forward 2 hours. Count forward 25 minutes. f. If it is morning, what time was it 6 hours and 30 minutes ago? Time now: Count back 6 hours. Count back 30 minutes. Saxon Math Intermediate 4 70 Adaptations Lesson 27
39 Written Practice page before noon 2. ft a. ft = left at noon b. Perimeter Perimeter Add all sides. a. b. 3. Even numbers between 3 and 39 4., 2, 5, 8,,,,,,, 5., 2, 24, 36,,, , expanded form: a. 6 3 words: b. \$6.30 a. b. Saxon Math Intermediate 4 7 Adaptations Lesson 27
40 Written Practice, continued page Look at hundreds. Look at tens. 0. a b Find the pattern. Try fifties a. b.. Shade one fourth of the square. 2. out of parts are shaded, so is the num and is the den. 3. afternoon What time will it be 3 hours from now? Time now: Count forward 3 hours Saxon Math Intermediate 4 72 Adaptations Lesson 27
41 Written Practice, continued page \$67 \$ \$5.88 \$ d d = f bottom missing 87 r top missing b 4 27 f = r = b = Change to multiplication: a. \$ = pennies b. penny = of a dollar c. pennies = of a dollar a. b. c. Saxon Math Intermediate 4 73 Adaptations Lesson 27
42 Written Practice, continued page If 3 and 4, then what does equal? A 343 B 7 C 0 D 27. Draw 2 perpendicular rays from the dot. 28. v total 29. Round to nearest dollar. \$ \$5.95 \$ + \$ No, \$40 is reasonable because \$5.95 is close to \$, and the sum of \$ plus \$ is \$. 30. Show six different ways to add 2, 4, and Saxon Math Intermediate 4 74 Adaptations Lesson 27
43 LESSON Name 28 Teacher Notes: page 68 Introduce Hint #7 Multiplication/ Division Fact Families. Refer students to Multiplication Multiplication Table Table on page 5 in the Student Reference Guide. Review Properties of Operations on page 22 in the Student Reference Guide. New Concept One fact family four facts ) ) 2 Properties of Multiplication Commutative Property m n = n m Identity Property n = n Zero Property 0 n = 0 2s facts 5s facts double the other number count by 5s Lesson Practice a. 9 3 b. 3 9 c. 6 4 d. 4 6 e. 7 8 f. 8 7 g. 5 8 h. 8 5 i. 0 0 j. 0 8 k. 9 l. 2 2 Saxon Math Intermediate 4 75 Adaptations Lesson 28
44 Lesson Practice, continued m. Which property of multiplication is shown below? 2 = 2 C Property of Multiplication n. Use the Zero Property of Multiplication to find the product: 0 25 = o. Use the Identity Property of Multiplication to find the product: 25 = Written Practice page morning afternoon in all 2. \$8 saved + n more needs 3. Perimeter Add all sides. 4 cm 4., 2,,, 30, 36,, 5., 36,,, 24, 20,, 6. Change to multiplication: = Saxon Math Intermediate 4 76 Adaptations Lesson 28
45 Written Practice, continued page a. 2 8 b. \$2.29 c. \$2.29 \$2.29 \$2.25 \$2.50 a. b. c. 8. Finish drawing the right triangle (one right angle). The two perpendicular sides should be 3 cm long and 4 cm long. 4 cm 9. Morning. What time will it be 90 minutes from now? Time now: Count forward 90 minutes Saxon Math Intermediate 4 77 Adaptations Lesson 28
46 Written Practice, continued page expanded form: words: a. temperature b. increase by 0 degrees Count by twos. 0 C a. b \$286 + \$ \$73 \$39 6. \$ \$ c 9 8. n c = n = Saxon Math Intermediate 4 78 Adaptations Lesson 28
47 Written Practice, continued page y d y = d = a. 8 b. 0 7 c yard meter a. b. c. 26. Which of the following shows 3 ones and 4 hundreds? A 304 B 403 C 4003 D 3400 Saxon Math Intermediate 4 79 Adaptations Lesson 28
48 Written Practice, continued page See the multiplication table on page 70. How many times? Property: C 28. cheetah elk s difference Property of Multiplication 29. Round to the nearest five. \$37 \$ + \$54 + \$ \$ is a reasonable e because \$37 is close to, \$54 is close to \$, and the sum of these is \$ , 80, 70, 60, 50,,,,,, Saxon Math Intermediate 4 80 Adaptations Lesson 28
49 LESSON 29 page 75 Name Multiplication Facts: 0s, s, 2s, 5s New Concept Lesson Practice 0 any number = 0. any number = the same number. 2 any number = double the number. 5 any number = a number that ends in 0 or in 5. Instead of completing Power Up C, complete the multiplication facts below. a. 0 3 = b. 4 0 = c. 7 0 = d. 5 = e. 6 = f. 4 = g. 5 9 = h. 5 4 = i. 5 6 = j. 2 5 = k. 7 5 = l. 3 5 = m. 2 3 = n. 5 5 = o. 7 2 = p. 6 2 = q. 2 8 = r. 2 2 = Saxon Math Intermediate 4 8 Adaptations Lesson 29
50 Written Practice page made Rochelle + n Zuri sold remained in all 3. a. radius in feet b. diameter in feet d = 2r 4. Subtract to find the pattern., 8,,, 32, 40,, 2 yd a. b. 5., 4,,, 35, 42, 6. Less than Two hundred nine is greater than one hundred ninety. odd Saxon Math Intermediate 4 82 Adaptations Lesson 29
51 Written Practice, continued page Afternoon. Time now: 9. Shade two thirds of the 3-cm-by-- cm rectangle. 3 cm Count forward 5 minutes. Count forward 35 minutes a. 2 8 =. A B b. 5 7 = c. 2 7 = D C d. 5 8 = It is it is b r angle. because than a a. b. c. d. Saxon Math Intermediate 4 83 Adaptations Lesson 29
52 Written Practice, continued page Find the pattern See page 2 in the Student Reference Guide. At what temperature does water freeze a. on the Fahrenheit scale? b. on the Celsius scale? a. b. 4. \$83 \$9 5. \$286 + \$ \$ \$ q n 37 q = n = Saxon Math Intermediate 4 84 Adaptations Lesson 29
53 Written Practice, continued page m g m = g = Change to multiplication: = 25. sets of ten = 26. Which of these does not equal 24? A 3 8 B 4 6 C 2 2 D Use Properties of Operations table on page 22 in the Student Reference Guide. a. Z Property of Multiplication b. C Property of Multiplication c. I Property of Multiplication Saxon Math Intermediate 4 85 Adaptations Lesson 29
54 Written Practice, continued page a. \$3.49 b. \$3.49 \$3.49 \$3.25 \$3.50 a. b. 29. equation with product of 8 = z x y x + y z + Saxon Math Intermediate 4 86 Adaptations Lesson 29
55 LESSON 30 page 79 Name Subtracting Three-Digit Numbers with Regrouping New Concept To subtract three-digit numbers: \$365 \$87? 5 \$36 5 \$ \$3 6 5 \$ \$3 6 5 \$ 8 7 \$ 7 8 Line up last digits. Exchange ten for 0 ones; subtract ones. Exchange hundred for 0 tens; subtract tens. Subtract hundreds. To subtract three-digit decimal numbers: \$4.0 \$.2 0 \$4. 0 \$ \$4. 0 \$ \$4. 0 \$. 2 \$2.98 Line up the decimal points. Subtract pennies. Subtract dimes. Subtract dollars. Lesson Practice Subtract. a. \$365 \$287 b. \$4.30 \$.8 c d. 240 e. 459 f. 57 Saxon Math Intermediate 4 87 Adaptations Lesson 30
56 Lesson Practice, continued g. Round to nearest 25 cents before adding. \$ \$2.27 \$ + \$ \$ is a reasonable e because \$8.24 is about \$ and 2.27 is about \$. The dif between the rounded numbers is \$. page 82 Written Practice. n bell rang left black shoes n no black shoes 73 altogether 73 remained 3. A nickel is worth morning Gilbert has an even number of nickels. Which could not be the value of his nickels? Any even number 5 will end in. Time now: Count forward 5 minutes. A 45 B 70 C 20 D Saxon Math Intermediate 4 88 Adaptations Lesson 30
57 Written Practice, continued page , 2, 8,,,, 6. Find the pattern expanded form: words: 9. a. 6 8 = b. 4 2 = c. 4 5 = d. 6 0 = a. b. c. d. Saxon Math Intermediate 4 89 Adaptations Lesson 30
58 Written Practice, continued page Use a centimeter ruler to find the: 2. a. length. b. width. c. perimeter. Perimeter Add all sides. a. b. c \$3.86 +\$ \$4.86 \$ m a 43 m = a = Saxon Math Intermediate 4 90 Adaptations Lesson 30
59 Written Practice, continued page y q y = q = 2. Since 89 is g than 85, = it is closer to than to. 23. Find half of Change to multiplication: = = 26. Which of these sets of numbers is not an addition/subtraction fact family? A, 2, 3 B 2, 3, 5 C 2, 4, 6 D 3, 4, Saxon Math Intermediate 4 9 Adaptations Lesson 30
60 Written Practice, continued page See Multiplication Table on page 5 in the Student Reference Guide. a. 0 0 = b. = c. 2 2 = 28. Tabitha bought a package of pencils. She sharpened them. of How many were not s? a. = b. c. 29. Round to the nearest 50. \$500 is r because \$749 is about \$, \$259 is about \$, and \$ \$ = \$. 30. c a b a + b = c c = + = = Saxon Math Intermediate 4 92 Adaptations Lesson 30
61 INVESTIGATION 3 Name Teacher Notes: page 85 Focus on Multiplication Patterns Area Squares and Square Roots Introduce Hint #8 Area and Perimeter Vocabulary and Hint #9, Square Roots. Review Perimeter, Area, Volume on page 7 in the Student Reference Guide. Students will need copies of Lesson Activity 20. Multiplication Patterns An array is a rectangular model of multiplication. 3 rows and 5 columns 3 5 = 5 3 and 5 are factors of 5 5 columns 3 rows Refer to this array of Xs to answer problems 4 below.. How many rows are in the array? 2. How many columns are in the array? 3. How many Xs are in the array? 4. What multiplication fact is illustrated by the array? = Saxon Math Intermediate 4 93 Adaptations Investigation 3
62 INVESTIGATION 3 continued 5. Represent Draw an array of 2 Xs in 2 rows. 2 2 =? 6. How many columns of Xs are in the array you drew? 7. What multiplication fact is illustrated by the array you drew? = Here is an array of 0 Xs. 8. Which two factors of 0 are shown by this array? and 9. Can you draw a rectangular array of ten Xs with three rows? Act it out. 0. Can you draw a rectangular array of ten Xs with four rows? Act it out.. Can you draw a rectangular array of ten Xs with five rows? Act it out. 2. Represent Draw an array of Xs in 3 rows and 6 columns. Write the multiplication fact. = Saxon Math Intermediate 4 94 Adaptations Investigation 3
63 INVESTIGATION 3 continued 3. Represent The chairs in a room were arranged in 6 rows, with 4 chairs in each row. Draw an array. Write the multiplication fact. = Area An area model is an array of connected squares. 4 6 = 24 6 squares on this side 4 squares on this side 4. How many small squares are in the rectangle above? Represent Use Lesson Activity 20 for problems 5 and Outline a 3-cm-by-8-cm rectangle. How many small squares are in the rectangle? What multiplication fact is illustrated by the rectangle? = 6. Outline a rectangle that is made up of 24 squares. Make the rectangle 2 cm wide. How long is the rectangle? What multiplication fact is illustrated by the rectangle? = Saxon Math Intermediate 4 95 Adaptations Investigation 3
64 INVESTIGATION 3 continued Math Language Area is the space inside the perimeter of a flat figure. Trace the edge of this paper with your finger. You are tracing the paper s perimeter. Rub the surface of this paper with your palm. You are rubbing the paper s area. 7. Use your finger to trace the perimeter of your desktop. 8. Use the palm of your hand to sweep over the area of your desktop. To measure the area of a shape cover its surface. count how many squares cm in. cm one square centimeter ( sq. cm) in. one square inch ( sq. in.) 9. How many square centimeters cover the area of this rectangle? 3 cm 2 cm square centimeters 20. Represent On Lesson Activity 20, outline a 4 cm by 3 cm rectangle. Area Perimeter Problems 2 23 are optional. Saxon Math Intermediate 4 96 Adaptations Investigation 3
65 INVESTIGATION 3 continued Activity page 88 Finding Perimeter and Area This activity is optional. Activity page 89 Estimating Perimeter and Area Use your textbook to complete this activity. Squares and Square Roots Use Lesson Activity 20 for problems Represent Outline these four squares and write the multiplication fact for each square. by Multiplication fact: 2 by 2 Multiplication fact: 3 by 3 Multiplication fact: 4 by 4 Multiplication fact: When we square a number, we multiply a number by itself: 3 squared = = Outline a 6 by 6 square. 6 squared = 26. Outline a 7 by 7 square. 7 squared = Problems 27 and 28 are optional. Saxon Math Intermediate 4 97 Adaptations Investigation 3
66 INVESTIGATION 3 continued To find the square root of a number, ask: What factor multiplied by itself equals the original number? The square root of 25 equals 5 because 5 5 = = 5 25 squares in all 5 squares on each side 29. a. 9 squared = b. What is the square root of 9? 30. Find each square root: a. 4 = b. 6 = c. 64 = 3. If the area of a square is 49 square centimeters, how long is each side of the square? Saxon Math Intermediate 4 98 Adaptations Investigation 3
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# 🧚🏽♀️Abstract Linear Algebra I Unit 5 – Determinants: Properties and Applications
Determinants are powerful tools in linear algebra, encoding crucial matrix properties. They're scalar values associated with square matrices, calculated using specific formulas. Determinants play a key role in solving systems of equations, finding matrix inverses, and determining geometric transformations. Understanding determinant properties is essential for mastering linear algebra. These properties include the effects of row operations, relationships between matrix operations and determinants, and geometric interpretations. Determinants also have important applications in advanced mathematics and various scientific fields.
## What Are Determinants?
• Determinants are scalar values associated with square matrices that encode important properties of the matrix
• Denoted as $det(A)$ or $|A|$ where $A$ is a square matrix
• Calculated using a specific formula involving the entries of the matrix
• Determinant of a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $ad-bc$
• Determinant of a 3x3 matrix can be calculated using the Laplace expansion or Sarrus' rule
• Laplace expansion involves cofactors and minors of the matrix
• Sarrus' rule uses a mnemonic device to calculate the determinant
• Determinant of an $n \times n$ matrix can be calculated recursively using cofactor expansion along any row or column
• The value of the determinant is independent of the choice of row or column for cofactor expansion
• Determinants have important applications in linear algebra, such as determining the invertibility of a matrix and solving systems of linear equations
## Key Properties of Determinants
• Determinant of the identity matrix is always 1, i.e., $det(I_n) = 1$
• Determinant of a matrix is equal to the determinant of its transpose, i.e., $det(A) = det(A^T)$
• If two rows (or columns) of a matrix are interchanged, the determinant changes sign
• Swapping any two rows (or columns) of a matrix multiplies the determinant by -1
• If a matrix has a row (or column) of zeros, its determinant is 0
• Multiplying a row (or column) of a matrix by a scalar $k$ multiplies the determinant by $k$
• $det(kA) = k^n det(A)$ where $A$ is an $n \times n$ matrix
• If two rows (or columns) of a matrix are proportional, the determinant is 0
• The determinant of a triangular matrix (upper or lower) is the product of its diagonal entries
• The determinant of a block diagonal matrix is the product of the determinants of its diagonal blocks
## Calculating Determinants
• For a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is calculated as $det(A) = ad - bc$
• For a 3x3 matrix, the determinant can be calculated using the Laplace expansion or Sarrus' rule
• Laplace expansion: $det(A) = a_{11}C_{11} - a_{12}C_{12} + a_{13}C_{13}$ where $C_{ij}$ are the cofactors
• Sarrus' rule: Multiply the entries along the main diagonal and the two parallel diagonals, then subtract the products of entries along the other three diagonals
• For larger matrices, cofactor expansion can be used recursively along any row or column
• Cofactor expansion along the $i$-th row: $det(A) = \sum_{j=1}^n (-1)^{i+j} a_{ij} det(M_{ij})$ where $M_{ij}$ is the minor of entry $a_{ij}$
• Cofactor expansion along the $j$-th column: $det(A) = \sum_{i=1}^n (-1)^{i+j} a_{ij} det(M_{ij})$
• Determinants can also be calculated using row reduction to convert the matrix into an upper triangular form
• Elementary row operations (except for row swapping) do not change the determinant
• Once in upper triangular form, the determinant is the product of the diagonal entries
## Determinants and Matrix Operations
• The determinant of a product of matrices is the product of their determinants, i.e., $det(AB) = det(A) \cdot det(B)$
• This property holds for any number of matrices: $det(ABC \cdots) = det(A) \cdot det(B) \cdot det(C) \cdots$
• The determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix, i.e., $det(A^{-1}) = \frac{1}{det(A)}$
• This property is only valid for invertible matrices (matrices with non-zero determinants)
• The determinant of a matrix raised to a power is the determinant of the original matrix raised to that power, i.e., $det(A^n) = (det(A))^n$
• If $A$ and $B$ are similar matrices, i.e., $B = P^{-1}AP$ for some invertible matrix $P$, then $det(A) = det(B)$
• The determinant of a matrix is invariant under similarity transformations
• The determinant of a Kronecker product of two matrices is the product of the determinants raised to the power of the size of the other matrix, i.e., $det(A \otimes B) = (det(A))^m \cdot (det(B))^n$ where $A$ is $n \times n$ and $B$ is $m \times m$
## Applications in Linear Systems
• A square matrix $A$ is invertible if and only if its determinant is non-zero
• If $det(A) \neq 0$, then $A$ has a unique inverse $A^{-1}$
• If $det(A) = 0$, then $A$ is singular (non-invertible) and the linear system $Ax = b$ may have no solution or infinitely many solutions
• Cramer's rule can be used to solve a system of linear equations $Ax = b$ using determinants
• The solution for the $i$-th variable is given by $x_i = \frac{det(A_i)}{det(A)}$ where $A_i$ is the matrix formed by replacing the $i$-th column of $A$ with the vector $b$
• Cramer's rule is practical for small systems but becomes computationally expensive for larger ones
• The determinant can be used to find the volume of a parallelepiped spanned by the columns (or rows) of a matrix
• For a 3x3 matrix $A$, the volume of the parallelepiped spanned by its columns is $|det(A)|$
• Determinants appear in the formulas for the cross product and triple product of vectors in 3D space
• Cross product: $\vec{a} \times \vec{b} = det(\begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix})$
• Triple product: $\vec{a} \cdot (\vec{b} \times \vec{c}) = det(\begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix})$
## Geometric Interpretation
• The determinant of a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ represents the signed area of the parallelogram spanned by the column vectors of $A$
• The absolute value of the determinant gives the area, while the sign indicates the orientation (clockwise or counterclockwise)
• For a 3x3 matrix, the determinant represents the signed volume of the parallelepiped spanned by the column vectors of the matrix
• The absolute value of the determinant gives the volume, while the sign indicates the orientation (right-handed or left-handed)
• In higher dimensions, the determinant of an $n \times n$ matrix represents the signed $n$-dimensional volume of the $n$-dimensional parallelepiped spanned by its column vectors
• A matrix with a determinant of 0 corresponds to a transformation that collapses the space onto a lower-dimensional subspace
• For example, a 3x3 matrix with a determinant of 0 may map a 3D space onto a 2D plane or a 1D line
• The sign of the determinant indicates whether the linear transformation represented by the matrix preserves or reverses the orientation of the space
• A positive determinant indicates an orientation-preserving transformation (e.g., rotation)
• A negative determinant indicates an orientation-reversing transformation (e.g., reflection)
• Cauchy-Binet formula: For matrices $A$ and $B$ of compatible sizes, $det(AB) = \sum_{S} det(A_S) \cdot det(B_S)$ where $S$ ranges over all subsets of size $n$ chosen from $\{1, 2, \ldots, m\}$ and $A_S, B_S$ are the $n \times n$ submatrices of $A$ and $B$ with columns indexed by $S$
• Jacobi's formula: For an invertible matrix $A(t)$ that depends on a parameter $t$, $\frac{d}{dt} det(A(t)) = tr(adj(A(t)) \cdot \frac{dA(t)}{dt})$ where $adj(A)$ is the adjugate matrix of $A$ and $tr$ denotes the trace
• Sylvester's determinant theorem: For matrices $A$ and $B$ of sizes $m \times n$ and $n \times m$ respectively, $det(I_m + AB) = det(I_n + BA)$
• Vandermonde determinant: The determinant of a Vandermonde matrix $V = (a_i^{j-1})_{1 \leq i,j \leq n}$ is given by $det(V) = \prod_{1 \leq i < j \leq n} (a_j - a_i)$
• Determinantal identities: There are various identities involving determinants, such as the Dodgson condensation formula, Plücker relations, and Laplace's expansion for bordered determinants
• Determinants in multilinear algebra: Determinants can be generalized to the context of multilinear algebra, where they are related to the exterior algebra and the notion of orientation of vector spaces
• Determinants in differential geometry: Determinants play a role in the definition of volume forms and the Jacobian determinant, which relates the volume elements of different coordinate systems
## Practice Problems and Examples
1. Calculate the determinant of the matrix $A = \begin{bmatrix} 2 & -1 & 0 \\ 3 & 4 & -2 \\ 1 & 0 & 5 \end{bmatrix}$
• Solution: $det(A) = 2 \cdot (20 - 0) - (-1) \cdot (15 - (-2)) + 0 \cdot (3 - 0) = 40 + 17 + 0 = 57$
2. Determine if the matrix $B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ is invertible.
• Solution: $det(B) = 1 \cdot (45 - 48) - 2 \cdot (28 - 42) + 3 \cdot (32 - 35) = -3 + 28 - 9 = 0$. Since $det(B) = 0$, $B$ is not invertible.
3. Find the area of the parallelogram spanned by the vectors $\vec{u} = (2, 3)$ and $\vec{v} = (1, -1)$.
• Solution: The area is given by the absolute value of the determinant $\begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix} = -2 - 3 = -5$. Thus, the area is $|-5| = 5$.
4. Use Cramer's rule to solve the system of linear equations: $2x + 3y = 5$ $x - y = 1$
• Solution: Let $A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$, $A_x = \begin{bmatrix} 5 & 3 \\ 1 & -1 \end{bmatrix}$, and $A_y = \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix}$. Then, $det(A) = -5$, $det(A_x) = -10$, and $det(A_y) = 7$. By Cramer's rule, $x = \frac{det(A_x)}{det(A)} = \frac{-10}{-5} = 2$ and $y = \frac{det(A_y)}{det(A)} = \frac{7}{-5} = -\frac{7}{5}$.
5. Prove that if $A$ is an $n \times n$ matrix and $B$ is obtained from $A$ by adding a multiple of one row to another row, then $det(B) = det(A)$.
• Proof: Let $A = (a_{ij})$ and suppose $B$ is obtained from $A$ by adding $k$ times the $i$-th row to the $j$-th row. Then, $B = (b_{ij})$ where $b_{ij} = a_{ij}$ for all $i \neq j$ and $b_{jk} = a_{jk} + ka_{ik}$ for all $k$. Using the linearity of the determinant along the $j$-th row, we have: $det(B) = det(a_{j1}, \ldots, a_{jn}) + k \cdot det(a_{i1}, \ldots, a_{in}) = det(A) + k \cdot 0 = det(A)$ since the determinant of a matrix with two identical rows (the $i$-th and $j$-th rows in this case) is 0.
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# Solving 1-step Equations
## What is an equation? What is a variable?
An EQUATION says that two things are equal. It will have an equals sign: 7 + 2 = 10 − 1
That equation says: what is on the left (7 + 2) is equal to what is on the right (10 − 1)
So an equation is like a statement "this equals that".
A VARIABLE is a symbol for a number we don't know yet. It is usually a letter like x or y, but it can be any letter.
## The GOAL
The goal in solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. The second goal is to have the number in front of the variable equal to one.
The most important thing to remember in solving a linear equation is that whatever operation you do to one side of the equation, you MUST do to the other side. So if you subtract a number from one side, you MUST subtract the same value from the other side. You will see how this works in the examples.
The best thing about working equations is that you can check your answer each time to see if it is correct. Substitute your answer in for the variable, and solve. Do both sides of your equation match? Then you have correctly solved your equation!
## Video Lesson
Watch the videos, Adding and Subtracting, and complete video notes in your packet.
## ADDITION
Read the math equation from above, "y plus 14 equals 20". To get the variable alone, we must do the opposite, or subtract 14. If we subtract 14 from the left side of the equation, we MUST also subtract 14 from the right side. See the picture to see EXACTLY how work must be shown.
Don't forget to CHECK your work. To check this answer, substitute 6 in for the y.
6 + 14 = 20
20 = 20 CORRECT!
## Subtraction Example
Read the math equation from above, "x minus one hundred twenty equals eighty". To get the variable alone, we will have to ADD 120. If we add 120 to one side of the equation, we MUST add 120 to the other side of the equation. See the picture above to show EXACTLY how work must be shown.
Don't forget to CHECK your work. To check this answer, substitute 200 in for the x.
200 - 120 = 80
80 = 80 CORRECT!
## Video Lesson
Watch the videos, Multiplying and Dividing, and complete video notes in your packet.
## MULTIPLICATION
Read the equation from above, "3 times n equals 12". To get the variable alone, we must do the opposite of times, which is divide. If we divide 3n by 3, we must do the same to the other side, which is divide 12 by 3. See above example to show EXACT steps which must be shown.
Don't forget to check your work. To check, substitute 4 in for n:
3(4) = 12
12 = 12 CORRECT!
## DIVISION
Read the equation from above, "k divided by 2 equals 16". To get the variable alone, we must do the opposite of divide, which is multiply. We multiply by the reciprocal (make it a fraction and flip it upside down). If we multiply by 2/1 on the left side, we must do the same on the right side of the equals sign. See above example to show EXACT steps which must be shown.
Don't forget to check your work. To check, substitute 32 in for k:
32 = 16
2
16 = 16 CORRECT!
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# How do you find the product of (7x^2 + 5y + 7) (y + 5) ?
Oct 24, 2017
See a solution process below:
#### Explanation:
To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.
$\left(\textcolor{red}{7 {x}^{2}} + \textcolor{red}{5 y} + \textcolor{red}{7}\right) \left(\textcolor{b l u e}{y} + \textcolor{b l u e}{5}\right)$ becomes:
$\left(\textcolor{red}{7 {x}^{2}} \times \textcolor{b l u e}{y}\right) + \left(\textcolor{red}{7 {x}^{2}} \times \textcolor{b l u e}{5}\right) + \left(\textcolor{red}{5 y} \times \textcolor{b l u e}{y}\right) + \left(\textcolor{red}{5 y} \times \textcolor{b l u e}{5}\right) + \left(\textcolor{red}{7} \times \textcolor{b l u e}{y}\right) + \left(\textcolor{red}{7} \times \textcolor{b l u e}{5}\right)$
$7 {x}^{2} y + 35 {x}^{2} + 5 {y}^{2} + 25 y + 7 y + 35$
We can now combine like terms:
$7 {x}^{2} y + 35 {x}^{2} + 5 {y}^{2} + \left(25 + 7\right) y + 35$
$7 {x}^{2} y + 35 {x}^{2} + 5 {y}^{2} + 32 y + 35$
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## Base of Number System-
• Base of a number system is the total number of digits used in that number system.
• Number system with base ‘b’ has its digits in the range [0 , b-1].
• It is also called as radix of a number system.
## Examples-
Consider the following examples-
### Base 10 Number System-
Consider base 10 number system popularly called as decimal number system-
• Total number of digits used in this number system are 10 since it has base 10.
• These digits are 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9.
• These digits clearly lie in the range [0 , base-1] = [0 , 9].
### Base 2 Number System-
Consider base 2 number system popularly called as binary number system-
• Total number of digits used in this number system are 2 since it has base 2.
• These digits are 0 and 1.
• These digits clearly lie in the range [0 , base-1] = [0 , 1].
## Important Note-
It is important to note that-
• All the digits of any number system with base ‘b’ are always less than ‘b’.
• This is clear from the range [0 , base-1] in which the digits of any number system lie.
## Types of Number System-
Four most commonly used number systems are-
1. Decimal Number System
2. Binary Number System
3. Octal Number System
The following table shows the base and digits used in these number systems-
Number System Base Digits Used Decimal Number System 10 0, 1, 2, 3, 4, 5, 6, 7, 8, 9(Total 10 digits) Binary Number System 2 0, 1(Total 2 digits) Octal Number System 8 0, 1, 2, 3, 4, 5, 6, 7(Total 8 digits) Hexadecimal Number System 16 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F(Total 16 digits)
To gain better understanding about Basics of Number System,
Watch this Video Lecture
Next Article- Converting Any Base to Base 10
Get more notes and other study material of Number System.
Watch video lectures by visiting our YouTube channel LearnVidFun.
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# Complex Numbers Loci Questions Worksheet
A Logical Amounts Worksheet may help your child be a little more acquainted with the principles powering this rate of integers. In this particular worksheet, students should be able to remedy 12 distinct difficulties linked to logical expressions. They are going to learn to increase several amounts, team them in pairs, and figure out their products and services. They will likely also practice simplifying realistic expressions. Once they have enhanced these concepts, this worksheet is a useful instrument for furthering their reports. Complex Numbers Loci Questions Worksheet.
## Realistic Phone numbers certainly are a ratio of integers
There are two varieties of figures: irrational and rational. Realistic phone numbers are considered complete numbers, whilst irrational amounts usually do not recurring, and possess an endless variety of numbers. Irrational phone numbers are non-absolutely nothing, low-terminating decimals, and rectangular origins that are not ideal squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To outline a realistic number, you need to understand what a reasonable amount is. An integer is actually a total variety, as well as a realistic amount is really a rate of two integers. The rate of two integers is definitely the amount on top split by the number on the bottom. For example, if two integers are two and five, this would be an integer. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They are often created in a portion
A realistic quantity includes a denominator and numerator which are not absolutely no. This means that they can be indicated being a small fraction. Together with their integer numerators and denominators, reasonable figures can also have a bad worth. The adverse importance should be placed left of and its particular absolute importance is its distance from absolutely nothing. To make simpler this illustration, we are going to state that .0333333 is actually a fraction which can be composed as a 1/3.
As well as bad integers, a logical quantity can be produced in a portion. For example, /18,572 is a logical number, while -1/ is not. Any portion made up of integers is logical, provided that the denominator is not going to consist of a and might be created for an integer. Furthermore, a decimal that leads to a level is yet another logical number.
## They create feeling
Even with their name, realistic phone numbers don’t make very much feeling. In math, they are one entities by using a exclusive size around the quantity collection. Because of this when we matter one thing, we can order the dimensions by its proportion to its unique amount. This keeps accurate even if there are actually endless rational amounts involving two certain amounts. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To obtain the duration of a pearl, by way of example, we could matter its thickness. An individual pearl is 10 kilograms, and that is a realistic quantity. Furthermore, a pound’s excess weight is equal to ten kilos. Thus, we must be able to break down a lb by twenty, without the need of be concerned about the size of one particular pearl.
## They can be indicated like a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal number may be written as a several of two integers, so 4 times five is the same as 8. A similar issue involves the frequent small fraction 2/1, and both sides should be divided by 99 to have the correct response. But how do you create the conversion? Here are several good examples.
A realistic variety can also be written in various forms, including fractions as well as a decimal. A great way to signify a reasonable amount in the decimal would be to split it into its fractional comparable. You can find three ways to divide a reasonable quantity, and every one of these approaches yields its decimal comparable. One of these simple approaches is always to split it into its fractional counterpart, and that’s what’s known as a terminating decimal.
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# POLYAS FOUR STAGES OF PROBLEM SOLVING
How many students passed the last math test? Once the problem is read, you need to list all the components and data that are involved. Find the dimensions if the perimeter is to be 26 inches. Keep in mind that x is representing an ODD number and that the next odd number is 2 away, just like 7 is 2 away form 5, so we need to add 2 to the first odd number to get to the second consecutive odd number. The two angles are 30 degrees and degrees. The sum of 85, 86 and 87 does check to be This is where you solve the equation you came up with in your ‘devise a plan’ step.
Sometimes the problem lies in understanding the problem. These are practice problems to help bring you to the next level. Consecutive EVEN integers are even integers that follow one another in order. Look back check and interpret. George Polya , known as the father of modern problem solving, did extensive studies and wrote numerous mathematical papers and three books about problem solving.
In fact there is no such thing as too much practice.
Practice Problems 1a – 1g: This is where you will be assigning your variable. Even the best athletes and musicians had some coaching along the way and lots of practice.
OCDSB HOMEWORK GUIDELINES
# Implementing Polya’s four steps.
You may be familiar with the expression ‘don’t look back’. Since length can be written in terms of width, we will let.
The perimeter of a rectangle with width of 3 inches and length of 10 inches does come out to be Sounds simple enough, but some people jump the gun and try to start solving the problem before they have read the whole problem. In fact there is no such thing as too much practice. In order to fpur an understanding of the problem, you, of course, need to read the problem carefully. The sum of 3 consecutive integers is Just read and translate it left to right to set up your equation.
If you add on 8. Carry out the plan solve. Since length can be written in terms of width, we will let.
In this tutorial, we will be setting up equations for each problem. Even the best athletes and musicians had some coaching along the way and lots of practice. The two angles are 30 degrees and degrees.
The sum of a number and 2 is 6 less than twice that number. If you need help solving them, by all means, go back to Tutorial 7: Intermediate Algebra Tutorial 8: If you add on 8. How many students passed the last math test? The perimeter of a rectangle with width of 3 inches or length of 10 inches does come out to be Length is 10 inches.
LANCIA THESIS 2.4 JTD EMBLEMA CONSUMI
One number is 3 less than another number. If you need a review on these translations, you can go back to Tutorial 2: When x is 5 the cost and the revenue both equal How much would you save if you bought it at this sale? If you need help solving them, by all means, go back to Tutorial 7: Note that 6 is two more than 4, prohlem first even integer.
Setting up an equation, drawing a diagram, and making a chart are all ways that you can go about solving your problem. Find the dimensions if the perimeter is to be 26 inches.
If your answer does check out, make sure that you write your final answer with the correct labeling. Math works just like anything else, if you want to get good at it, then you need to practice it.
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# Monthly Archives: August 2012
## Konig’s lemma
This post is about trees. Not real-life ones but mathematical trees, or more formally about acyclic connected graphs. Just as actual big skinny trees are tall, graph theoretic big skinny trees are also tall. The goal here is to prove this formally.
But first some definitions. A tree is called labeled if every vertex has been assigned a symbol, called its label. A fixed vertex in a tree may be labelled as $r$ and referred to as its root. The level $i$ in a tree is the collection of all vertices at a fixed distance $i$ from the root. Admittedly these definitions are informal and may be formalized in terms of directed graphs.
An example of a labelled tree is given below:
Here $r$ is the root and the vertex sets $\{a,b\}$ and $\{1,2,3\}$ are at level $1$ and $2$ respectively. Clearly this labeling is arbitrary and any other labeling is also possible.
We now prove our result. It is due to Konig, a Hungarian professor of mathematics, who had Erdos amongst his students. It is called a lemma because it was initially used by Konig to investigate another theorem (see this) . As often happens it has turned out to be a powerful tool to use on many other problems as well.
Lemma: If $T$ is an infinite tree and each level of $T$ is finite then $T$ contains an infinite path.
Proof: Suppose $T$ is a labeled tree with root $r$. Let $L_0=\{r\}, L_1,L_2,\cdots$ be the levels of $T$. We construct our path by starting with $r$. Next we partition all vertices in levels higher then $L_0$ (i.e. all vertices in levels $L_i$ where $i>0$) into as many finite parts as is the cardinality of $L_1$. In other words if $L_1=\{v_1,\cdots v_n\}$ then we partition the vertices (there are infinitely many of them) in $n$ parts: $P_1,P_2\cdots P_n$. This partitioning is done as follows: Any vertex $w$ will be connected to $r$ via some $v_i$ owing to connectedness. We put $w$ in $P_i$.
Now since we have partitioned infinitely many vertices in finitely many parts, one of the parts will contain infinitely many vertices.(This is also called the infinite pigeonhole principle). Let that part be $P_j$. Choose $w_1=v_j$ as the next vertex in our under-construction path. Next we delete the vertex $r$ and consider the component containing $w_1$. It is also an infinite tree with each level finite and having $w_1$ as its root. By a similar piece of reasoning, we get a vertex $w_2$ which is adjacent to $w_1$ and has infinitely many vertices corresponding to it. We continue building our path by choosing $w_2$ as a part of it. By induction we can continue and get an infinite path $rw_1w_2\cdots$. This proves the theorem.$\Box$
It should be noted that at each of the infinitely many stages of selecting the $w_i$ we choose it out of a finite set: the finite set of vertices at that level each of which correspond to infinite parts in the current partitioning. There is no canonical way to make this choice. To ensure that this can be done we need to use a weak form of the axiom of choice: Let $X\ne\emptyset$ and $\mathcal{F}(X)$ be the collection of all finite subsets of it. Then there is a function $f:\mathcal{F}(X)-\emptyset\to X$ such that $f(x)\in x\forall x\in \mathcal{F}(X)-\emptyset$. A curious fact is that Konig’s lemma in turn applies this weak form of the axiom of choice and so both of them are equivalent. While their equivalence can be established using the ZF axioms, obviously neither can be proved within ZF.
Filed under Combinatorics, Graph theory
## 1+2+3+…(n-1)=n(n-1)/2
Each unordered pair of green balls corresponds uniquely with a blue one. So the set of pairs of green balls has the same order as the set of blue balls. Since $\tbinom{5}{2}$ pairs are possible and the blue balls are $1+2+3+4$ in number so $1+2+3+4 = \tbinom{5}{2}=5(5-1)/2$
Filed under Combinatorics, Miscellaneous
## The van der Waerden theorem
In this post I aim to prove van der Waerden’s theorem on arithmetic progressions.
The theorem states that $\forall k,r\in\mathbb{N}\exists W(k,r)\in\mathbb{N}$ so that if the set $\{1,2\cdots W(k,r)\}$ is partitioned into $r$ classes, one of the classes is guaranteed to contain an arithmetic progression of length $k$.
Before proceeding further, lets look at a bit of history regarding this result. As per a recent book the theorem was conjectured independently by the mathematicians Schur and Baudet sometime around the 1920’s. (The exact dates are unknown). In 1926 van der Waerden, then a young scholar in Hamburg, heard of this conjecture (he believed it to be solely by Baudet) and proved it in the course of a remarkable discussion with his teacher Emil Artin and the mathematician Otto Schreier. The beauty of the proof was that it did not use any tools of advanced mathematics and yet was not simple by any means. Khinchin has referred to it as one of the three pearls of number theory.
Our proof here is not the same proof as originally given by van der Waerden, but a slightly slicker version based on the same idea. We start by making some preparations for the proof. For convenience we will take $[n]=\{1,2\cdots n\}$. Further we can very well uniquely identify a partitioning $\{W_1,W_2,\cdots W_r\}$ of a set $[W(k,r)]=\cup_{i=1}^rW_i$ by a function $f$ with domain $[W(k,r)]$ and range $[r]$ where $f^{-1}(i)$ equals $W_i$. Traditionally the numbers $\{1,2\cdots r\}$ are called colors, every such function is called a $r$-coloring and the sets $f^{-1}(i)$ are referred to as being colored $i$. Given a $r$-coloring a subset of $[W(k,r)]$ is called monochromatic if every element in it corresponds with the same color. With these new definitions the van der Waerden theorem takes the form:
Theorem: $\forall k,r\in\mathbb{N}\exists W(k,r)\in\mathbb{N}$ so that if $[W(k,r)]$ is $r$-colored then there is a monochromatic arithmetic progression of length $k$.
We shall also use the notation $a + [0,k)d$ for the arithmetic progression of length $k$ (also abbreviated as a $k$-AP) given by the ordered set $\{a
And now, on to the proof. As a first step towards proving the result we shall first introduce the concept of wheels. The next step will establish some technical lemmas concerning them and after that we will prove the van der Waerden theorem.
A wheel of radius $k$, degree $n$ and base point or origin $a$ is a $n$ tuple: $W= (a + [0,k) d_1,\cdots, a + [0,k)d_n)$ of $k$-AP's each of which has base point $a$. The AP's $a + [1,k)d_i, 1 \le i \le n$ are called the spokes of the wheel. Assuming that all numbers under consideration have been colored a wheel is weakly polychromatic if all of it’s $n$ spokes are monochromatic with distinct colors, and strongly polychromatic if it’s $n$ spokes and origin are all monochromatic with distinct colors.
An example of a strongly polychromatic wheel of degree $8$ and radius $5$ is given below. In case of a weakly polychromatic wheel the origin, i.e. $1$, can be of any of the colors of the spokes.
The $8$ AP’s are:
$1,2,3,4,5$
$1,10,19,28,37$
$1,4,7,10,13$
$1,7,13,19,25$
$1,6,11,16,21$
$1,8,15,22,29$
$1,5,9,13,17$
$1,3,5,7,9$
A translation of a wheel $W$ by a number $r$ defined as the wheel obtained by adding $a$ to each number in $W$.
We now move on to the next stage in our proof, where we shall establish some technical lemmas concerning wheels.
Lemma 1: If $W= (a + [0,k) r_1,\cdots, a + [0,k)r_n)$ is a wheel of radius $k$ and degree $n$ and the $k-1$ translations $W + r ,\cdots W+(k-1)r$ of $W$ are all strongly polychromatic with the same colors, then the wheel $\hat{W} = (a + [0,k)r , a + [0,k)(r_1 + r),\cdots a + [0,k)(r_n + r))$ of radius $k$ and degree $n+1$ is weakly polychromatic.
Proof: The integers $a + (r_i + r), a + 2(r_i + r), \cdots a + (k-1)(r_i + r)$ respectively picked from $W+r, W + 2r ,\cdots W+(k-1)r$ are identical in color because the ith component of each translation is colored identically. So $\forall i, 1 \le i \le n, a + [1,k)(r_i + r)$ is monochromatically colored and distinct from others. Similarly the AP of the base points $a+r,a+2r,\cdots a+(k-1)r$ also has a distinct color. Thus we have a $n+1$ tuple of AP’s each of which is distinctly monochromatic if we ignore the base point $a$, or that $\hat{W}$ is weakly polychromatic.
Two more lemmas whose proofs are obvious are:
Lemma 2: If $W$ is a weakly polychromatic wheel of radius $k$, then either $W$ is strongly polychromatic, or contains a monochromatic AP of length $k$.
Lemma 3: A strongly polychromatic wheel cannot have degree $r$ where $r$ is the total number of colors.
With the necessary machinery in our arsenal we now attack the van der Waerden theorem proper.
Proof of the van der Waerden theorem: We shall use double induction. The first (outer) induction is on $k$. Clearly for $k = 1,2$ and any $r$ the result holds. Let us suppose that $W(k-1,r)$ exists for all r. Now we
Claim: For any $n\in\mathbb{N},\exists N=N(k-1,r,n)\in\mathbb{N}$ such that given any $r$-coloring of $[N]$, there exists either a monochromatic AP of length $k$, or there exists a strongly polychromatic wheel of radius $k$ and degree $n$.
To prove the claim we use (an inner) induction on $n$. The case $n = 1$ is trivial if we use the outer induction hypothesis. Suppose the result holds for $n-1$, that is, $N_1 = N(k-1,r,n-1)$ exists. Now by the inner hypothesis any block of integers of length $N_1$ contains either a monochromatic AP of length $k$ (in which case we are done) or a strongly polychromatic wheel of radius $k$ and degree $n-1$. In the second case let $N_2 = 2W(k-1, r^{N_1})$ which exists by the outer hypothesis, color the first $N_1N_2$ integers and consider $N_2$ consecutive blocks of length $N_1$. Since there are $r^{N_1}$ colorings of each block, by the outer hypothesis there must be an AP of $k-1$ identically colored blocks: $B+d, B+2d,\cdots B+(k-1)d$.
Now as $|B| = N_1 = N(k-1,r,n-1)$ so inside each block, there is either a monochromatic AP of length $k$ in it, (which proves the theorem) or a strongly polychromatic wheel of radius $k$ and degree $n-1$. Ruling out the former case means that each block $B + d, \cdots B +(k-1)d$ contains a strongly polychromatic wheel of degree $n-1$. Since these blocks are identical in color so we have a AP of strongly polychromatic identical wheels. In this case, by lemma 1, the AP of blocks contains a weakly polychromatic wheel $W$ of radius $k$ and degree $n$. By lemma 2, $W$ is either strongly polychromatic or contains a monochromatic AP of length k. In either case we have proved our claim.
The proof of van der Waerden’s theorem is trivial now. Choose $W(k,r) = N(k-1,r,r)$. Now an $r$-coloring of $[W(k,r)]$ contains either a monochromatic AP of length $k$ or a strongly polychromatic wheel of degree $r$. The latter case is not possible due to lemma 3.$\Box$
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Filed under Combinatorics, Ramsey theory
## Mathematics and brainvita
In this post, I will analyse the single player game brainvita (also known as peg solitaire). The game is described by having a board with holes having the pattern:
Initially all holes except the central one are filled with pegs. The player may jump a peg horizontally or vertically over another peg into an empty hole. The jumped peg is then removed by the player. The objective is to finish the game with just one peg left.
The aim of this post is to show that there are (at most) $5$ ending board positions in brainvita. In other words, assuming the player wins, his last peg left on the board can be only among $5$ fixed board positions.
The proof uses abstract algebra. Suppose that the pegs correspond to coordinates in the plane in the natural fashion described below: The central peg is at $(0,0)$, and all others have integer coordinates as well. For convenience adjacent pegs may be taken a unit distance apart. Some coordinates are described below:
Now consider all possible ways of placing pegs on the board. There are $2^{33}$ of them. Many of these ways are impossible to reach in a valid game. Regardless of this fact, let $W$ be the set of all possible ways and for a given $w\in W$, let $C_w$ denote the set of all the coordinates where there is a peg in $w$. Now define a function $f:W\to \frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ by $f(w)=\sum_{(m,n)\in C_w}x^{m+n}$ where $\frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ is the finite field on $4$ elements. (If there are no pegs on the board we define $f$ to be $0$.)
The function $f$ has the property that for any given way $w$ of placing pegs on the board, if a legal move is performed to yield another way $w'$ then $f(w)=f(w')$. To see this, suppose that in the way $w$ there were pegs at $(m,n)$ and $(m+1,n)$ positions and the hole at the $(m+2,n)$ position was empty. A legal move may be performed by jumping the $(m,n)$ peg over the $(m+1,n)$ peg to yield a way $w'$ where the holes at $(m,n)$ and $(m+1,n)$ are empty and there is a peg at $(m+2,n)$. Clearly the only change between $f(w)$ and $f(w')$ occurs on account of these three coordinates and $f(w)-f(w')=x^{m+n}+x^{m+1+n}-x^{m+2+n}$. But as $x^{m+n}+x^{m+1+n}-x^{m+2+n}=x^{m+n}(1+x-x^2)=x^{m+n}.0=0$ in $\frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ so $f(w)-f(w')=0$ following which $f(w)=f(w')$. The same holds if instead of moving a peg to the right we had moved it above, below or to the left. Moreover all this is regardless of the fact whether the way $w$ can actually be reached by a valid sequence of moves in the game.
Now we define another function $g:W\to \frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ by $g(w)=\sum_{(m,n)\in C_w}x^{m-n}$ (the no pegs way means $g$ is $0$) and by a similar piece of reasoning conclude that $g$ also has the same property enjoyed by $f$. It is also easy to observe that both $f$ and $g$ are onto functions and so $f(W)\times g(W)$ has $16$ elements.
Next we partition the set $W$ as follows. Any $w\in W$ corresponds with a pair $(f(w),g(w))$ and so with exactly one among the $16$ elements of $f(W)\times g(W)$. In this way all the $2^{33}$ elements of $W$ are partitioned in $16$ parts. Further each part has the curious property that for any $w$ in that part, any sequence of legal moves will never change the value of $(f(w),g(w))$ and so the resultant $w'$ will also be in the same part. It is now obvious that among these $16$ parts one and only part consists of all the $w's$ which can actually occur in a game of brainvita. All other $w's$ correspond to ways which can never occur in a legal game.
What is the value of $(f(w),g(w))$ that this one special part corresponds to? The initial board (assuming it corresponds with $i\in W$) is set up so that $f(i)=g(i)=1$. So any way $w$ that arises during the game also has $f(w)=g(w)=1$. Assume now that the game has been successfully finished with a single peg at $(m,n)$, and that this corresponds to the way $w^*$. We therefore have $f(w^*)=g(w^*)=1$. But by definition $f(w^*)=x^{m+n}$ and $g(w^*)=x^{m-n}$. So we must have $x^{m+n}=1$. Since the multiplicative group $\{1,x,x^2\}$ is cyclic of order $3$ so as a consequence of Lagrange’s theorem we must have $3|(m+n)$. Similarly we have $3|(m-n)$. Combining these facts together we have $3$ dividing both $m$ and $n$. The only five coordinates which satisfy this condition are $(-3,0),(0,-3),(3,0),(0,3)$ and $(0,0)$. So the player has to end up at one among these $5$ positions. This completes the proof.
It should be noted that we have only found an upper bound on the number of positions the final peg can end up at. However by explicitly giving a sequence of moves for each of these $5$ coordinates it is possible to show that all these endings are indeed possible.
1 Comment
Filed under Algebra, Game Theory
## Two one sentence proofs
Continuing in the same vein here are two one sentence proofs.
Theorem: For $x,y\in\mathbb{C}, n\in\mathbb{N}$ we have $(x+y)^n=\sum_{k=0}^{n}\tbinom{n}{k}x^ky^{n-k}$.
Proof: The coefficient of $x^ky^{n-k}$ in the left hand side of the given expression, i.e. in $\underbrace{(x+y)(x+y)\cdots (x+y)}_{\text{n terms}}$, is obtained by selecting exactly $k$ parenthesis out of the total $n$ available to yield the $x's$, (the remaining parenthesis yield the $y's$), which can be done in $\tbinom{n}{k}$ ways following which the coefficient is exactly $\tbinom{n}{k}$.$\Box$
Theorem: $\sqrt{2}$ is irrational.
Proof: If $\sqrt{2}=m/n$ is in lowest terms then $\sqrt{2}=(2n-m)/(m-n)$ is in lower terms.$\Box$
Remark: The above proof generalizes to the case when $k$ is a non-square positive integer. Indeed, if $j<\sqrt{k} then assuming that $\sqrt{k}=m/n$ is in lowest terms also implies that $\sqrt{k}=(kn-jm)/(m-jn)$ is in lower terms.
Filed under Combinatorics, Miscellaneous
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# Area
## Teaching Area and Perimeter
Since they are introduced together, one frequently finds children mixing up the two concepts. Also formulae for arriving at these measures are brought in too quickly - well before the concepts are fully understood. One can avoid this problem by spacing out these two concepts. Area could be explored in the first stage as it occurs frequently in a child's everyday experiences.
## Pythagorean Triples part 4
Shailesh Shirali takes you to a wonderful tale of Pythagoren triples one more time.
## Octagon in a Square
An octagon is constructed within a square by joining each vertex of the square to the midpoints of the two sides remote from that vertex. Eight line segments are thus drawn within the square, creating an octagon (shown shaded). The following two questions had been posed: (i) Is the octagon regular? (ii) What is the ratio of the area of the octagon to that of the square?
## Practice Questions on Squares and Cubes
Let your learners figure out the questions and solutions to the worksheet that covers squares, cubes and their roots.
1. Make a question on square root by using following information.
a. 196 sq ft
## Weekend Activity Balloon Bench
You have blown and burst a balloon. Have you ever made a bench out of them? Check this activity from Arvind Gupta and have fun.
## How to Prove it?
In this article, we offer a second proof of the triangle-in-a-triangle theorem, using the principles of similarity geometry. Then, using vectors, we prove a result which is a generalisation of that theorem.
## When does it have Maximum Area? ~ A GeoGebra Exploration
Use GeoGebra to check when will a rectangle have maximum area when it is inscribed in a triangle.
## How to Prove It
In this article we examine how to prove a result obtained after careful GeoGebra experimentation. It was featured in the March 2015 issue of At Right Angles, in the ‘Tech Space’ section.
## A Flower with Four Petals
Find the area marked x.
## How To Prove It - IV
This continues the ‘Proof’ column begun earlier. In this ‘episode’ we study some results from geometry related to the theme of concurrence.
## Pages
19019 registered users
7425 resources
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# Evaluate the following integrals. \int e^{3-4x}dx
Evaluate the following integrals.
$\int {e}^{3-4x}dx$
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Bubich13
Step 1
Given integral:
$\int {e}^{3-4x}dx$
Substitute t=3-4x and differentiate it w.r.t "x"
$\frac{dt}{dx}=\frac{d}{dx}\left(3-4x\right)$
$\frac{dt}{dx}=-4$
$\frac{-dt}{4}=dx$
Therefore, the given integral becomes,
$\frac{-1}{4}\int {e}^{t}dt$
Step 2
We know that
$\int {e}^{x}dx={e}^{x}+c$
Where, "c" is integration constant.
$⇒\frac{-1}{4}{e}^{t}+c$
Substitute t=3-4x in above equation, we get
$⇒\frac{-1}{4}{e}^{\left(3-4x\right)}+c$
$⇒\frac{-{e}^{\left(3-4x\right)}}{4}+c$
###### Not exactly what you’re looking for?
Bubich13
Given
$\int {e}^{3-4x}dx$
$\int -\frac{{e}^{t}}{4}dt$
$-\frac{1}{4}\cdot \int {e}^{t}dt$
Solve
$-\frac{1}{4}{e}^{t}$
$-\frac{1}{4}{e}^{3-4x}$
$-\frac{{e}^{3-4x}}{4}$
$-\frac{{e}^{3-4x}}{4}+C$
###### Not exactly what you’re looking for?
Vasquez
$\begin{array}{}\int {e}^{3-4x}dx\\ =-\frac{1}{4}\int {e}^{u}du\\ \int {e}^{u}du\\ ={e}^{u}\\ -\frac{1}{4}\int {e}^{u}du\\ =-\frac{{e}^{u}}{4}\\ u=3-4x:\\ =-\frac{{e}^{3-4x}}{4}\\ Solution:\\ =-\frac{{e}^{3-4x}}{4}+C\end{array}$
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Ace the Trig Exam With This Study Guide
By Kayla Griffin
Most of trigonometry involves lots of memorization, or lots of reference checking. Practice some basic formulas and rules that will get you started on nearly everything.
Before you start reading, you should know that this is only a review. If you don't have any experience with the subject, there is not enough information here to teach it to you. For those of you that have studied the material before, this is a condensed formula sheet to help you prepare for exams. It is also helpful as a handy homework reference.
Angles
Until now, you probably measured angles in degrees. Most of trig uses radians.
Unit Circle
The unit circle is segmented into 360 degrees or 2*pi radians. (For this article, p will denote pi=3.14159...)
0 degrees = 0 radians = 2p radians = 360 degrees (Note that 0 and 2p are often used interchangeably)
The unit circle below gives the commonly used angles and their sines/cosines.
If Θ is an angle made by connecting (0,0), (1,0) and any other point on the unit circle M = (a,b), then cosΘ = a, sinΘ = b, or equivalently: (a,b) = (cosΘ, sinΘ)
For example: the line for Θ=p/4 radians crosses the circle at (sqrt(2)/2, sqrt(2)/2). Therefore x = cos Θ = sqrt(2)/2 = y =sinΘ, or equivalently: (sqrt(2)/2, sqrt(2)/2) = (cos(p/4), sin(p/4)).
Negative Angles
Counterclockwise movement on the unit circle is measured in positive units and clockwise movement is measured in negative units. Therefore, the distance from p to 2p can also be notated as 0 to -p, depending on which way you travel the circle.
Relationships between angles are important:
Supplementary Angles: x+y = p = 90 degrees
Complementary Angles: x+y = 2p = 180 degrees
Triangles
For the purpose of the trig cheat sheet, and conventionally, if the sides of a triangle are labeled a, b, c then the angles are ABC, where A is the angle made by b and c, B is the angle made by a and c, and C is the angle made by a and b. A is called the opposite angle of a, and a is the opposite side of A.
Much of trigonometry is based on triangles, usually right or acute.
Triangle area formulas:
½*base*height=Area
½*a*b*sinC=Area
½*b*c*sinA=Area
½*c*a*sinB=Area
s*(s-a)*(s-b)*(s-c)=(a+b+c)*(b+c)*(a+c)*(a+b)=Area^2 [s=a+b+c]
More With Triangles
This trig cheat sheet uses Θ, A and B to represent angles. Your notes or textbook may use x and y or other variables, but the meanings are the same.
Right Triangle and Sine, Cosine, Tangent
An easy pneumonic for the sine, cosine, tangent right triangle definitions is SOHCAHTOA. This stands for Sine=Opposite/Hypotenuse (SOH); Cosine=Adjacent/Hypotenuse (COH); Tangent=Opposite/Adjacent (TOA).
The chart below gives a breakdown of the trig functions and their relationship to right triangles and the unit circle.
Here is a simple way to remember the right triangle relationships: Hold up your left hand, palm facing out, fingers together and thumb sticking out the right like an L. Now, imagine a line from the tip of your forefinger to the tip of your thumb. If you pretend the curve where your thumb meets your hand is a right angle, then you have a right triangle. Your thumb is a, your forefinger is b, and the imaginary line (the hypotenuse) is c. Then the angle at the tip of your forefinger is A, the tip of your thumb is B and the right angle is C. Now, use that to memorize the following definitions:
sinA=opp/hyp = a/c
cscA = hyp/opp = c/a = 1/sinA
secA = hyp/adj = c/b = 1/cosA
cotA = adj/opp = c/b = 1/tanA
Laws and Theorems
The trig cheat sheet only gives the formulas for these theorems. If you are required to use the full definition, or are unsure when to use each formula, you will need additional study material.
Pythagorean Theorem, Law of Sines, Law of Cosines, Law of Tangents are all about the relationships between the sides and angles of triangles.
~Pythagorean theorem (only for right triangles with sides a, b, c, where c is the hypotenuse):
Using the definition of a right triangle on your hand (as explained above) the lengths of the sides are related by the following formula:
a2 +b2 = c2
~Law of sines (for all triangles with sides lengths a, b, c and with angles A, B, C)
sin(A)/a = sin(B)/b = sin(C)/c
~Law of cosines (for all triangles with sides lengths a, b, c and with angles A, B, C)
c²=a²+b²-2*a*b*cos(C)
b²=a²+c²-2*a*c*cos(B)
a²=b²+c²-2*b*c*cos(A)
When the angle C is p, 0, or 2p, the law of cosines reduces to a simpler formula. If it is easier for you to memorize these too, go ahead, but usually I just solve for them.
When C = p: c²=a²+b² (the Pythagorean theorem)
When C=0, cos(0)=1: c2 = (a-b)2
When C=2p, cos(2)=-1: c 2 = (a+b)2
~Law of Tangents (for all triangles with sides a, b, c and with angles A, B, C)
(a+b)/(a-b) = [tan(1/2(A+B))]/[tan(1/2(A-B))]
Important Trig Formulas (More Memorization)
Here is a list of the most important trig formulas. It is best to memorize as many as you can, but remember that some can be derived from others (if that’s easier for you) and many are related. The trig cheat sheet uses Greek letters such as Θ to represent angles. Please note that where marked below, a and b are angles, and p represents pi=3.14159....
Squared formulas
sin2(Θ)+cos2(Θ)=1
1+tan2(Θ) = sec2(Θ)
1+cot2(Θ) = csc2(Θ)
Symmetry
sin(-Θ) = -sinΘ
cos(-Θ) = cosΘ
tan(-Θ) = -tanΘ
Periodic Functions (p = pi = 3.14159...)
sin(Θ) = sin(Θ+2kp)
cos(Θ) = cos(Θ+2kp)
where k is any integer, positive or negative.
Sum and Difference formulas (a and b are angles)
sin(a+b)=sin(a)cos(b)+sin(b)cos(a)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
tan(a+b)=[tan(a)+tan(b)]/[1-tan(a)tan(b)]
sin(a-b)=sin(a)cos(b)-sin(b)cos(a)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
tan(a-b)=[tan(a)-tan(b)]/[1+tan(a)tan(b)]
tan([a+b]/2) = [sin(a)+sin(b)]/[cos(a)+cos(b)] = -[cos(a)-cos(b)]/[sin(a)-sin(b)]
Power Reduction Formulas
sin2(Θ) = [1-cos(2Θ)]/2
cos2(Θ) = [1+cos(2Θ)]/2
Product to Sum Formulas (a and b are angles)
cos(a)cos(b) = ½[cos(a-b) + cos(a+b)]
sin(a)sinb() = ½ [cos(a-b) – cos(a+b)]
sin(a)cos(b) = ½[sin(a+b) + sin(a-b)]
cos(a)sin(b) = ½[sin(a+b) – sin(a-b)]
Sum to Product Formulas (a and b are angles)
sin(a)+sin(b) = 2sin([a+b]/2)cos([a+b]/2)
cos(a)+cos(b) = 2cos([a+b]/2)cos([a-b]/2)
cos(a)-cos(b) = -2sin([a+b]/2)sin([a-b]/2)
Half, Double, and Triple Angle formulas (SQRT(arg) means take the square root of arg)
~Half Angle
sin(Θ/2) = SQRT([1-cos(Θ)]/2)
cos(Θ/2) = SQRT([1+cos(Θ)]/2)
tan(Θ/2) = sin(Θ)/[1+cos(Θ)] = [1-cos(Θ)]/sin(Θ)
~Double Angle
sin(2Θ)=2sin(Θ)cos(Θ) = 2tan(Θ)/[1+tan2(Θ)]
cos(2Θ)=cos2(Θ)-sin2(Θ) = 2cos2(Θ) – 1 = 1-2sin2(Θ) = [1-tan2(Θ)]/[1+tan2(Θ)]
tan(2Θ)=2tan(Θ)/[1-tan2(Θ)]
~Triple Angle
sin(3Θ) = 3sin(Θ)-4sin3(Θ)
cos(3Θ) = 4cos3(Θ) – 3cos(Θ)
tan(3Θ) = [3tan(Θ) – tan3(Θ)]/[1-3tan2(Θ)]
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# Lesson 7
Practice with Rational Bases
Let's practice with exponents.
### 7.1: Which One Doesn’t Belong: Exponents
Which expression doesn’t belong?
$$\frac{2^{8}}{2^5}$$
$$\left(4^{\text-5}\right)^{8}$$
$$\left( \frac34 \right)^{\text-5} \boldcdot \left( \frac34 \right)^{8}$$
$$\frac{10^{8}}{5^5}$$
### 7.2: Exponent Rule Practice
1. Choose 6 of the equations to write using a single exponent:
• $$7^5 \boldcdot 7^6$$
• $$3^{\text-3} \boldcdot 3^8$$
• $$2^{\text-4} \boldcdot 2^{\text-3}$$
• $$\left(\frac{5}{6}\right)^4 \left(\frac{5}{6}\right)^5$$
• $$\frac{3^5}{3^{28}}$$
• $$\frac{2^{\text-5}}{2^4}$$
• $$\frac{6^5}{6^{\text-8}}$$
• $$\frac{10^{\text-12}}{10^{\text-20}}$$
• $$\left(7^2\right)^3$$
• $$\left(4^3\right)^{\text-3}$$
• $$\left(2^{\text-8}\right)^{\text-4}$$
• $$\left(6^{\text-3}\right)^5$$
2. Which problems did you want to skip in the previous question? Explain your thinking.
3. Choose 3 of the following to write using a single, positive exponent:
• $$2^{\text-7}$$
• $$3^{\text-23}$$
• $$11^{\text-8}$$
• $$4^{\text-9}$$
• $$2^{\text-32}$$
• $$8^{\text-3}$$
4. Choose 3 of the following to evaluate:
• $$\frac{10^5}{10^5}$$
• $$\left(\frac{2}{3}\right)^3$$
• $$2^8 \boldcdot 2^{\text-8}$$
• $$\left(\frac{5}{4}\right)^2$$
• $$\left(3^4\right)^0$$
• $$\left(\frac{7}{2}\right)^2$$
### 7.3: Inconsistent Bases
Mark each equation as true or false. What could you change about the false equations to make them true?
1. $$\left(\frac{1}{3}\right)^2 \boldcdot \left(\frac{1}{3}\right)^4 = \left(\frac{1}{3}\right)^6$$
2. $$3^2 \boldcdot 5^3 = 15^5$$
3. $$5^4 + 5^5 = 5^9$$
4. $$\left(\frac{1}{2}\right)^4 \boldcdot 10^3 = 5^7$$
5. $$3^2 \boldcdot 5^2 = 15^2$$
Solve this equation: $$3^{x-5} = 9^{x+4}$$. Explain or show your reasoning.
### Summary
In the past few lessons, we found rules to more easily keep track of repeated factors when using exponents. We also extended these rules to make sense of negative exponents as repeated factors of the reciprocal of the base, as well as defining a number to the power of 0 to have a value of 1. These rules can be written symbolically as:
$$\displaystyle x^n \boldcdot x^m = x^{n+m},$$ $$\displaystyle \left(x^n\right)^m = x^{n \boldcdot m},$$ $$\displaystyle \frac{x^n}{x^m} = x^{n-m},$$ $$\displaystyle x^{\text-n} = \frac{1}{x^n},$$ and $$\displaystyle x^0 = 1,$$
where the base $$x$$ can be any positive number. In this lesson, we practiced using these exponent rules for different bases and exponents.
### Glossary Entries
• base (of an exponent)
In expressions like $$5^3$$ and $$8^2$$, the 5 and the 8 are called bases. They tell you what factor to multiply repeatedly. For example, $$5^3$$ = $$5 \boldcdot 5 \boldcdot 5$$, and $$8^2 = 8 \boldcdot 8$$.
• reciprocal
Dividing 1 by a number gives the reciprocal of that number. For example, the reciprocal of 12 is $$\frac{1}{12}$$, and the reciprocal of $$\frac25$$ is $$\frac52$$.
|
# A First Course in Statistics (12th Edition) View more editions 79% (1296 ratings) for this book
• 1379 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 7
a.
Use the Random Numbers applet to generate random numbers from 1 to 10 and to observe the repeated numbers and the number of times each of these numbers occurs.
Applet procedure:
Step 1: Enter Minimum value as 1 and Maximum value as 10.
Step 2: Enter Sample size as 10.
Step 3: Click Sample.
• Step 2 of 7
Applet output:
From the Applet output, the repeated numbers are 9, 1, and 8. Also, the number 9 is repeated for 2 times, the number 1 is repeated for 2 times and the number 8 is repeated for 2 times.
• Step 3 of 7
b.
Use the Random Numbers applet to generate random numbers from 1 to 20 and to observe whether there are repeated numbers.
Applet procedure:
Step 1: Enter Minimum value as 1 and Maximum value as 20.
Step 2: Enter Sample size as 10.
Step 3: Click Sample.
• Step 4 of 7
Applet output:
From the Applet output, it is clear that the repeated numbers are 19, 12, 1, and 9. Hence, repeat the same procedure by increasing the maximum value by 30.
• Step 5 of 7
Use the Random Numbers applet to generate random numbers from 1 to 30 and to observe whether there are repeated numbers.
Applet procedure:
Step 1: Enter Minimum value as 1 and Maximum value as 30.
Step 2: Enter Sample size as 10.
Step 3: Click Sample.
• Step 6 of 7
Applet output:
From the Applet output, it is clear that none of the numbers are repeated. Thus, the smallest maximum value for which the repeated numbers is not obtained is 30.
• Step 7 of 7
c.
Description:
From part a, when the population size is 10, sample size is 10 and the 3 numbers are repeated.
From part b, when the population size is 20, sample size is 10 and the 3 numbers are repeated. Also, when the population size is 30, sample size is 10; there are no repetitions in numbers.
Hence, it can be concluded that there will be no repetitions when a relatively small sample is chosen from a large population.
Corresponding Textbook
A First Course in Statistics | 12th Edition
9780134080628ISBN-13: 0134080629ISBN: Authors:
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MT2002 Analysis
Previous page (Cauchy sequences) Contents Next page (Limits of functions)
## Continuity for Real functions
We now introduce the second important idea in Real analysis.
It took mathematicians some time to settle on an appropriate definition. See Some definitions of the concept of continuity.
Continuity can be defined in several different ways which make rigorous the idea that a continuous function has a graph with no breaks in it or equivalently that "close points" are mapped to "close points".
For example, is the graph of a continuous function on the interval (a, b)
while is the graph of a function with a discontinuity at c.
To understand this, observe that some points close to c (arbitrarily close to the left) are mapped to points which are not close to f(c).
We will give a definition in terms of convergence of sequences and show later how it can be reformulated in terms of the above description.
Definition
A function f : R R is said to be continuous at a point p R if whenever (an) is a real sequence converging to p, the sequence (f(an)) converges to f(p).
Definition
A function f defined on a subset D of R is said to be continuous if it is continuous at every point p D.
Example
In the discontinuous function above take a sequence of reals converging to c from below. (That is, all the terms are < c.) Then the image of these gives a sequence which does not converge to f(c).
We also have the following.
Definition
A real valued function f defined on a subset S of R is said to be continuous if it is continuous at all points of S.
It will be easier to give (a lot of) examples of continuous functions after we have proved the following two results.
Definition
If f and g are functions from R to R, we define the function f + g by (f + g)(x) = f(x) + g(x) for all x in R.
Similarly we may define the difference, product and quotient of functions.
Theorem
If f and g are continuous a point p of R, then so are f + g, f - g, f.g and (provided g(p) 0) f /g .
Proof
This follows directly from the corresponding arithmetic properties of sequences.
For example: to prove that f + g is continuous at p R
Suppose (xn) p. We are told that (f(xn)) f(p) and (g(xn)) g(p) and we must prove that (f + g)(xn)) (f + g)(p).
But the LHS of this expression is f(xn) + g(xn) and the RHS is f(p) + g(p) and so the result follows from the arithmetic properties of sequences.
Theorem
The composite of continuous functions is continuous.
Proof
Suppose f: R R and g: R R. Then the composition g f is defined by g f(x) = g(f(x)).
We assume that f is continuous at p and that g is continuous at f(p). So suppose that (xi) p. Then (f(xi)) f(p) and then (g(f(xi))) g(f(p)) which is what we need.
Examples
1. Clearly the identity function which x x is continuous.
Hence, using the above, any polynomial function is continuous and hence any rational function (a ratio of polynomial functions) is continuous at any point where the denominator is non-zero.
2. We will see later that functions like , sin, cos, exp, log, ... are continuous. It follows that , for example sin2(x + 5), exp(-x2), (1 + x4), ... are continuous since they are made by composing continuous functions.
Previous page (Cauchy sequences) Contents Next page (Limits of functions)
JOC September 2002
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# The length of a rectangle is 8cm greater than its width. How do you find the dimensions of the rectangle if its area is 105cm²?
Apr 1, 2018
Dimensions: $15$cm $\times$ $7$ cm
#### Explanation:
Let the length of the rectangle be $l$ and the width of the rectangle be $w$,
$l \cdot w = 105$
$l = w + 8$
Substitute $l = w + 8$ into $l \cdot w = 105$,
$\left(w + 8\right) \cdot w = 105$
Expand,
${w}^{2} + 8 w - 105 = 0$
Factor,
$\left(w - 7\right) \left(w + 15\right) = 0$
Solve,
w=7 or cancel(-15 ( reject $- 15$ as $w > 0$ )
When $w = 7$,
$l = 7 + 8$
$l = 15$
Hence, the length is $15$cm and the width is $7$cm.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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Trapezoids
Quadrilaterals with exactly one pair of parallel sides.
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Trapezoids
What if you were told that the polygon ABCD is an isoceles trapezoid and that one of its base angles measures 38\begin{align*}38^\circ\end{align*} ? What can you conclude about its other angles? After completing this Concept, you'll be able to find the value of a trapezoid's unknown angles and sides given your knowledge of the properties of trapezoids.
Guidance
A trapezoid is a quadrilateral with exactly one pair of parallel sides. Examples look like:
An isosceles trapezoid is a trapezoid where the non-parallel sides are congruent. The third trapezoid above is an example of an isosceles trapezoid. Think of it as an isosceles triangle with the top cut off. Isosceles trapezoids also have parts that are labeled much like an isosceles triangle. Both parallel sides are called bases.
Recall that in an isosceles triangle, the two base angles are congruent. This property holds true for isosceles trapezoids.
Theorem: The base angles of an isosceles trapezoid are congruent.
The converse is also true: If a trapezoid has congruent base angles, then it is an isosceles trapezoid. Next, we will investigate the diagonals of an isosceles trapezoid. Recall, that the diagonals of a rectangle are congruent AND they bisect each other. The diagonals of an isosceles trapezoid are also congruent, but they do NOT bisect each other.
Isosceles Trapezoid Diagonals Theorem: The diagonals of an isosceles trapezoid are congruent.
The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides. There is only one midsegment in a trapezoid. It will be parallel to the bases because it is located halfway between them. Similar to the midsegment in a triangle, where it is half the length of the side it is parallel to, the midsegment of a trapezoid also has a link to the bases.
Investigation: Midsegment Property
Tools Needed: graph paper, pencil, ruler
1. Draw a trapezoid on your graph paper with vertices A(1,5), B(2,5), C(6,1)\begin{align*}A(-1, 5), \ B( 2, 5), \ C(6, 1)\end{align*} and D(3,1)\begin{align*}D(-3, 1)\end{align*}. Notice this is NOT an isosceles trapezoid.
2. Find the midpoint of the non-parallel sides either by using slopes or the midpoint formula. Label them E\begin{align*}E\end{align*} and F\begin{align*}F\end{align*}. Connect the midpoints to create the midsegment.
3. Find the lengths of AB, EF\begin{align*}AB, \ EF\end{align*}, and CD\begin{align*}CD\end{align*}. Can you write a formula to find the midsegment?
Midsegment Theorem: The length of the midsegment of a trapezoid is the average of the lengths of the bases, or EF=AB+CD2\begin{align*}EF=\frac{AB+CD}{2}\end{align*}.
Example A
Look at trapezoid TRAP\begin{align*}TRAP\end{align*} below. What is mA\begin{align*}m \angle A\end{align*}?
TRAP\begin{align*}TRAP\end{align*} is an isosceles trapezoid. So, mR=115\begin{align*}m \angle R = 115^\circ\end{align*}. To find mA\begin{align*}m \angle A\end{align*}, set up an equation.
115+115+mA+mP230+2mA2mAmA=360=360mA=mP=130=65\begin{align*}115^\circ + 115^\circ + m \angle A + m \angle P & = 360^\circ\\ 230^\circ + 2m \angle A & = 360^\circ \rightarrow m \angle A = m \angle P\\ 2m \angle A & = 130^\circ\\ m \angle A & = 65^\circ\end{align*}
Notice that \begin{align*}m \angle R + m \angle A = 115^\circ + 65^\circ = 180^\circ\end{align*}. These angles will always be supplementary because of the Consecutive Interior Angles Theorem. Therefore, the two angles along the same leg (or non-parallel side) are always going to be supplementary. Only in isosceles trapezoids will opposite angles also be supplementary.
Example B
Write a two-column proof.
Given: Trapezoid \begin{align*}ZOID\end{align*} and parallelogram \begin{align*}ZOIM\end{align*}
\begin{align*}\angle D \cong \angle I\end{align*}
Prove: \begin{align*}\overline{ZD} \cong \overline{OI}\end{align*}
Statement Reason
1. Trapezoid \begin{align*}ZOID\end{align*} and parallelogram \begin{align*}ZOIM, \ \angle D \cong \angle I\end{align*} Given
2. \begin{align*}\overline{ZM} \cong \overline{OI}\end{align*} Opposite Sides Theorem
3. \begin{align*}\angle I \cong \angle ZMD\end{align*} Corresponding Angles Postulate
4. \begin{align*}\angle D \cong \angle ZMD\end{align*} Transitive PoC
5. \begin{align*}\overline{ZM} \cong \overline{ZD}\end{align*} Base Angles Converse
6. \begin{align*}\overline{ZD} \cong \overline{OI}\end{align*} Transitive PoC
Example C
Find \begin{align*}x\end{align*}. All figures are trapezoids with the midsegment.
1. a) \begin{align*}x\end{align*} is the average of 12 and 26. \begin{align*}\frac{12+26}{2}=\frac{38}{2}=19\end{align*}
b) 24 is the average of \begin{align*}x\end{align*} and 35.
\begin{align*}\frac{x+35}{2}&=24\\ x+35&=48\\ x&=13\end{align*}
c) 20 is the average of \begin{align*}5x-15\end{align*} and \begin{align*}2x-8\end{align*}.
\begin{align*}\frac{5x-15+2x-8}{2}&=20\\ 7x-23& =40\\ 7x&=63\\ x&=9\end{align*}
Watch this video for help with the Examples above.
Concept Problem Revisited
Given an isosceles trapezoid with \begin{align*} m \angle B = 38^\circ \end{align*}, find the missing angles.
In an isosceles trapezoid, base angles are congruent.
\begin{align*} \angle B \cong \angle C\end{align*} and \begin{align*}\angle A \cong \angle D\end{align*}
\begin{align*}m \angle B = m \angle C = 38^\circ\\ 38^\circ + 38^\circ + m \angle A + m \angle D = 360^\circ\\ m \angle A = m \angle D = 142^\circ\end{align*}
Vocabulary
A trapezoid is a quadrilateral with exactly one pair of parallel sides. An isosceles trapezoid is a trapezoid where the non-parallel sides and base angles are congruent. The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides.
Guided Practice
\begin{align*}TRAP\end{align*} an isosceles trapezoid.
Find:
1. \begin{align*}m \angle TPA\end{align*}
2. \begin{align*}m \angle PTR\end{align*}
3. \begin{align*}m \angle PZA\end{align*}
4. \begin{align*}m \angle ZRA\end{align*}
1. \begin{align*}\angle TPZ \cong \angle RAZ\end{align*} so \begin{align*}m\angle TPA=20^\circ + 35^\circ = 55^\circ\end{align*}.
2. \begin{align*}\angle TPA\end{align*} is supplementary with \begin{align*}\angle PTR\end{align*}, so \begin{align*}m\angle PTR=125^\circ\end{align*}.
3. By the Triangle Sum Theorem, \begin{align*}35^\circ + 35^\circ + m\angle PZA=180^\circ\end{align*}, so \begin{align*}m\angle PZA=110^\circ\end{align*}.
4. Since \begin{align*}m\angle PZA = 110^\circ\end{align*}, \begin{align*}m\angle RZA=70^\circ\end{align*} because they form a linear pair. By the Triangle Sum Theorem, \begin{align*}m\angle ZRA=90^\circ\end{align*}.
Practice
1. Can the parallel sides of a trapezoid be congruent? Why or why not?
For questions 2-7, find the length of the midsegment or missing side.
Find the value of the missing variable(s).
Find the lengths of the diagonals of the trapezoids below to determine if it is isosceles.
1. \begin{align*}A(-3, 2), B(1, 3), C(3, -1), D(-4, -2)\end{align*}
2. \begin{align*}A(-3, 3), B(2, -2), C(-6, -6), D(-7, 1)\end{align*}
3. \begin{align*}A(1, 3), B(4, 0), C(2, -4), D(-3, 1)\end{align*}
4. \begin{align*}A(1, 3), B(3, 1), C(2, -4), D(-3, 1)\end{align*}
1. Write a two-column proof of the Isosceles Trapezoid Diagonals Theorem using congruent triangles.
Given: \begin{align*}TRAP\end{align*} is an isosceles trapezoid with \begin{align*}\overline{TR} \ || \ \overline{AP}\end{align*}.
Prove: \begin{align*}\overline{TA} \cong \overline{RP}\end{align*}
1. How are the opposite angles in an isosceles trapezoid related?.
2. List all the properties of a trapezoid.
Vocabulary Language: English
Diagonal
Diagonal
A diagonal is a line segment in a polygon that connects nonconsecutive vertices
midsegment
midsegment
A midsegment connects the midpoints of two sides of a triangle or the non-parallel sides of a trapezoid.
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# What Is Differentiation In Maths Assignment Help
## Find Someone to do Project
What is a new object? First of all, let‘s say that you want to make a new car. Then, let“s say that your car is new. Since you want to change the shape, you can do it by going into a new car, or by making a new car by making the shape of the car. I said that I want to make something new. And I“m sure you want to do it by making something new. That’s the reason why I“ve given you the name differentiation. (The name that I gave you is called “the differentiation of the new object in the new car”). Even though it is a mathematical process, differentiation is not a physical concept. Rather, it is a logical concept. It is created by the process. In this process, you create the new object. Your new object is the new car. The car. The cell, the shape, the shape is the new object, which is the new cell. You are going to make a car by making a car by doing the same thing that was made by your original car. (AllWhat Is Differentiation In Maths? In this short paper, we will show that differentiation of $2$-dimensional $n$-dimensional vector space can be represented by sum of terms. In the following, we will first define the representation of differentiation of $n$ dimensions in terms of $2n$-dimensions. Then, we will introduce the main tools for differentiation from $2$ to $n$ dimension. Definition of differentiation in $2$ ==================================== Let $V$ be a 2-dimensional vector over $\mathbb{R}$ with respect to the basis $a_1,\ldots,a_n$ and $b_1, \ldots,b_n$. We will denote by $D=K\oplus \mathbb{C}[\sqrt{2}]$.
## Assignment Help Websites
$def:D2$ A $2$ dimensional vector $V$ is called of $D$-type if $K$ is the dual vector space of $D$, and $D^*=K\cap D$ and $D\cap D^*=\{0\}$. We say that a 2-dimension $D$ of $D_2$ is a $2$-$D$-dimensional (i.e. $D\in D_2$) if there exists a unique $1$-$D_2$, which is the $D$-$D^*$-dimension, such that $D\subset D_2$. $D_2\in D$ if and only if $K=\mathbb{Z}$. We will show that $D_1\in D$, $D_3\in D-D$, $D^2\in D$ and $\mathbb C[\sqr{4}]\subset \mathbb R$. Let us first define a set of vectors in $D$ that satisfy the following property : \(i) if $K\cap \mathbb C [\sqrt p]\neq \emptyset$, then $K\subset K^*$; \]$defi:D2-D$ Let $D$ be a $2*$-dimensional subspace of $D$. We will call the set $D$ the *zero dimensional vector space*. \ **1.** The set of vectors $D\times K\in D\times D$ that satisfy **(i)**$D$-dimension condition. $[@D2-2]$ \($theo:D21$)$def1$ Let $(K, D)$ be a vector space with visit their website $n\ge 2$. Assume that the set $S(\mathbb{Q},\mathbb C)$ is finite. Then $S(\sqrt{n},\mathcal{V})$ admits a finite-dimensional vector representation $\widetilde{V}$ of $S(\sqrt n, \mathcal{R})$. In particular, $S(\widetilde{\mathbb{F}},\mathfrak{S})$ admits an equivalent representation $\widit{V}$, i.e., $\widit{\widetilde V}=\widetilde V$ for any $V$. The following result is an extension of the previous result. Let $\mathbb R$ be a real vector space. Let $V$ and $\widet{V}_1$ be two $2$dimensional vector spaces satisfying $\widet{\widet{D}^*}=\mathcal N(\mathbb R,\mathbb Q)\cap\mathcal V$, $\widet\widet{S}(\widet{R},\mathbf{\mathrm{d}})$ and $\widehat{\widehat V}= \widehat V$ be the corresponding representations of $\mathbb Q\times \mathbb Q$. Then $\widet{{\mathbf{R}}}$ and $\overline{{\mathbb R}}$ are equivalent representations in the same $2*\mathbb Z$.
## Assignment Help Services
What Is Differentiation In Maths? Differentiation is a very important topic in mathematics and is studied by many mathematicians, including mathematicians. It is one of the most important topics in math. Although it is not a part of the scientific community, it is an important topic in the scientific community too. Differentiation results from different mathematical models. Before you start, I want to make a couple of comments. First, it is important to understand the differentiations. Differentiation is an important subject in mathematics. When you are studying different techniques, you are studying a mathematical model. When check that study the mathematical model, you are making a distinction between different types of models. Differentiating is very important to study different models. Differentiation can be done in two ways. One is by applying different principles. The other way is to apply the same principle. This is called differentiation by Riemannian geometry. In the first way, the more background you know about the mathematical model and the part of the model in which you are studying, the more you know about differentiation. In the second way, the most important part of differentiation is the proof, and this is called proof of differentiation by Rappelsti. In the third way, the proof of differentiation is done by applying differentiation by Remker. This is very important for your study of different types of mathematical models. In the last line, you are working on the proof of the differentiation of the model by Rappelman. This is another way to study differential forms.
## Pay Someone to do Assignment
In the fourth way, you are also working on the different forms of the model. In the fifth way, you can apply the differentiation on the model by applying differentiation at every level of the model, so the proof of something is much more difficult. Let us start with the basics. Differentiate by Riemmanian geometry. Riemmanian Geometry. You can think of Riemmanians as being the points of a three dimensional space. For example, a plane is a 3 dimensional space. Thus, you can think of the plane as being a horizontal plane. For the next example, let’s think of a plane as a horizontal plane, and let’s think about a plane as being between two horizontal planes. A plane can be seen as a set of horizontal lines in a space. Now, we can think of each point of a plane, and its vertical and horizontal lines, as horizontal lines in the plane. Thus, the point of the line between two horizontal lines is a horizontal line. But what about the point of a line between two points? Now we can think about each point of the horizontal line, and its horizontal and vertical line, as horizontal and browse around this site lines in the horizontal plane. Thus the horizontal and vertical points of the line are horizontal and vertical. We are going to use Riemman’s theorem to show that a point is an edge of a three-dimensional space. 1. For a line through the origin, its length is proportional to the area of the line, and therefore the area of a line is proportional to its length. 2. To find a point in a three-dimensions space, we must find a point at the origin, and its distance from the origin. We can do this by taking the distance to the origin.
## College Homework Assignments
Then we know that a point in the plane is the point at the same
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+0
# geometry problem
0
53
1
A certain square and a certain equilateral triangle have the same perimeter. The square and triangle are inscribed in circles, as shown below. If $A$ is the area of the circle containing the square, and $B$ is the area of the circle containing the triangle, then find $A/B.$
Apr 14, 2020
#1
+7826
+1
For the square one, construct two radii to two consectuive vertices.
Let O be the center, and A, B be the vertices.
Let r be the radii of the circle on the left.
$$\angle AOB = 90^\circ$$
By Pythagoras theorem,
$$r^2 + r^2 = AB^2\\ AB = \sqrt2 r$$
Perimeter of square = $$4 \sqrt2 r$$
Do the same thing for the circle on the right, but notice that this time $$\angle AOB = 120^\circ$$.
Let R be the radii of the circle on the right.
$$AB^2 = R^2 + R^2 - 2(R)(R)\cos 120^\circ\\ AB = \sqrt3R$$
Perimeter of triangle = $$3\sqrt 3 R$$
If the perimeters are the same, then
$$3\sqrt 3 R = 4\sqrt2 r\\ \dfrac{R}{r} = \dfrac{4\sqrt6}{9} --- (1)$$
We proceed to calculate the areas.
Using the notations in the problem,
$$A = \left(\sqrt2 r\right)^2 = 2r^2$$
$$B = \dfrac{\sqrt3}{4} \left(\sqrt3R\right)^2 = \dfrac{3\sqrt3}{4}R^2$$
By (1), we get $$\dfrac AB = \dfrac{2r^2}{\dfrac{3\sqrt 3}{4}R^2} = \dfrac{8}{3\sqrt3} \left(\dfrac{R}{r}\right)^{-2} = \dfrac{8\sqrt3}{9}\left(\dfrac{27}{32}\right) = \boxed{\dfrac{3\sqrt 3}{4}}$$
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Apr 14, 2020
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# Question 6d652
Sep 30, 2017
Here's what I got.
#### Explanation:
As you know, the molarity of a solution tells you the number of moles of solute present for every $\text{1 L}$ of the solution.
This implies that you can increase the molarity of a solution by decreasing the amount of solvent it contains.
Now, let's say that the molarity of the initial solution is
${c}_{1} = \frac{n}{V} _ 1$
Here
• $n$ is the number of moles of solute present in the solution
• ${V}_{1}$ is the volume of the initial solution
After you remove some of the solvent, the molarity of the solution increases by 40%. If you take ${c}_{2}$ to be the new molarity of the solution, you can say that
${c}_{2} = \frac{140}{100} \cdot {c}_{1}$
Since the number of moles of solute remained constant, you can say that
${c}_{2} = \frac{n}{V} _ 2$
Here ${V}_{2}$ is the volume of the second solution.
This means that you have
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}}}{V} _ 2 = \frac{140}{100} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}}}{V} _ 1$
This is equivalent to
$\frac{1}{V} _ 2 = \frac{7}{5} \cdot \frac{1}{V} _ 1$
Rearrange to solve for ${V}_{2}$
${V}_{2} = \frac{5}{7} \cdot {V}_{1}$
You can thus say that the volume of the solution decreased by
${V}_{1} - \frac{5}{7} \cdot {V}_{1} = \frac{2}{7} \cdot {V}_{1}$
which is equivalent to a percent decrease of
"% decrease" = (2/7 * color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * 100% = color(darkgreen)(ul(color(black)(29%)))
I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the increase in molarity.
Notice that this is equivalent to saying that
"% change" = (5/7 * color(red)(cancel(color(black)(V_1))) - color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * 100% = - 29%#
Here the minus sign symbolizes percent decrease.
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# Slope|Definition & Meaning
## Definition
The slope (or gradient) is defined as the ratio of the difference between two points’ vertical and horizontal coordinates. Physically, it represents the steepness of the line joining the two points (how much movement occurs along the y-axis for a given movement in the x-axis and vice versa). Mathematically:
$\text{slope} = m = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}$
Figure 1 shows two lines with different slopes. Line B has a greater slope as it is steeper than line A.
Figure 1 – Two Lines A and B with Different Slopes
## Line
A line is a one-dimensional object which is infinitely long and has no width, depth, or curvature. It is made with an infinite number of points extending in opposite directions. The points lying on the same line are known as collinear points.
## Calculating the Slope of a Line
Two points (x1, y1) and (x2, y2) are required to find the slope of a line. The difference between the y-coordinates gives the rise(vertical change), and the difference between the x-coordinates gives the run(horizontal change).
Dividing the rise by run gives the slope “m” of the line as:
m = rise/run = (y2 – y1)/(x2 – x1)
For example, given two points (1, 2) and (3, 5) of the line, here:
x1 = 1, y1 = 2, x2 = 3, y2 = 5
Calculating the slope gives:
m = (5 – 2) / (3 – 1) = 3/2
### Slope in Trigonometry
The slope m of a line can also be defined as the tangent of the angle θ that the line makes with the x-axis. Hence,
m = tan θ
Where θ lies between -90° and 90°. If the slope is known, the value of θ can be calculated by using the equation:
θ = tan-1 (m)
For example, if a line’s slope is 10, the angle θ will be:
θ = tan-1 (10) = 84.3°
## Infinite and Zero Slope
A vertical line has an infinite slope, and a horizontal line has a zero slope. An undefined slope means that with no change in x-values, there is a change in y-values. A zero slope means that no change occurs in y-values for changing x-values.
## Positive and Negative Slope
A positive slope means that a line moves in the upward direction from left to right. The y-values increase with increasing x-values.
A negative slope of a line means that y-values decrease with increasing x-values. The line goes in a downward direction from left to right.
Figure 2 shows lines with zero, infinite, positive, and negative slopes.
Figure 2 – Demonstration of Zero, Infinite, Positive, and Negative Slopes
## Different Forms of Equations of a Line
The equation of a line is known as a linear equation having a degree of one. The following are three different ways to represent a linear equation.
### Standard Form
A linear equation’s standard form of two variables, x and y, is:
Px + Qy = R
Where “P” and “Q” are the coefficients of the variables x and y, respectively, and R is a constant. For example, the equation 2x – 3y = 9 is in standard form.
### Slope-intercept Form
A line’s slope-intercept form is given as
y = mx + a
Where “m” is the line’s slope, and “a” is the y-intercept.
For example, if a line’s slope is 2 and the y-intercept is -1, its slope-intercept form will be:
y = 2x – 1
The y-intercept is the value of y at the point (0, a) where the line crosses the y-axis. It can be found by placing x = 0 in the linear equation and solving for y.
The standard form can be converted into the slope-intercept form by rearranging the equation. For example, to convert the equation:
x + 2y = 8
into slope-intercept form, subtracting x on both sides gives:
2y = 8 – x
Dividing by 2 into both sides gives:
y = 4 – (1/2)x
Hence, the slope-intercept form is:
y = – (1/2)x + 4
#### Graphing a Line Using Slope-Intercept Form
The slope-intercept form provides the two essential parameters to graph a line, i.e., the slope and the y-intercept. Consider the equation:
y = (1/2)x + 4
The slope of the line is 1/2, and its y-intercept is 4. A slope of 1/2 means a rise of 1 and a run of 2. A positive slope means that the rise will be in the upward direction from left to right.
Figure 3 shows the line obtained by using the slope-intercept form.
Figure 3 – Graphing a Line Using the Slope-intercept Form of a Line
Starting from the y-intercept 4, i.e., the point (0,4) and moving 1 unit up(rise) and 2 units right(run), the point reached is (2, 5). A line can be drawn by joining the two points (0, 4) and (2, 5) and extending the ends infinitely.
### Point-slope Form
A line’s point-slope form is given as:
y – y1 = m(x – x1)
Where x1 and y1 are the x and y-coordinates of a point on the line. The point-slope is used to find a line’s equation when a point (x1, y1) on the line and its slope are given. It is also used when two points are given with which slope can be calculated.
#### Derivation of Point-slope Form
The point-slope form is derived from the slope-intercept form. Figure 4 shows a line having any points (x1, y1) and (x, y) on it.
Figure 5 – Demonstration of Two Points on a Line and their Horizontal and Vertical Difference
For the point (x, y), the slope-intercept form is:
y = mx + a
For the point (x1, y1), the slope-intercept form is:
y1 = mx1 + a
Subtracting y1 from y gives the vertical difference as:
y – y1 = (mx + a) – (mx1 + a)
y – y1 = mx + a – mx1 – a
y – y1 = mx – mx1
Taking m as common gives the point-slope form:
y – y1 = m(x – x1)
## Slope of Two Parallel Lines
Two lines are parallel if and only if their slopes are equal. For example, the two lines:
y = 3x + 2 and y = 3x – 5
are parallel as both have the same slope, i.e., 3 as shown in figure 5.
Figure 5 – Demonstration of Slopes of Two Parallel Lines
## Slope of Two Perpendicular Lines
Two lines are perpendicular to each other if the product of their slopes is -1. For example, the two lines:
y = (-1/2)x – 3 and y = 2x + 1
have the slopes m1 = -1/2 and m2 = 2 respectively. The product of their slopes gives:
m1.m2 = (-1/2)(2) = -1
Hence, the two lines are perpendicular, as shown in figure 6.
Figure 6 – Demonstration of Slopes of Two Perpendicular Lines
A line with a zero slope is also perpendicular to a line with an infinite slope.
## Slope of a Function
The slope at a point (l, m) on the function’s curve can be found by taking its first derivative and finding its value by substituting the point (l, m). For example, the function:
y = x3
has the first derivative as:
dy/dx = d(x3)/dx = 3x2
To find the slope at the point (2, 3) on the function, substituting (2, 3) in the first derivative gives:
dy/dx at P(2, 3) = 3(2)2 = 3(4) = 12
Hence, the slope at the point (2, 3) is 12.
### Exponential Function
The exponential function ex has the first derivative as:
dy/dx = d(ex)/dx = ex
The function f(x) = ex is a special function as it is the only function whose slope at each point on the curve is equal to the value of the function at that point. For example, at x = 1, the function’s value is:
f(1) = e1 = e ≈ 2.178
which is also the slope of the curve at the point (1, e), as shown in figure 7.
Figure 7 – The Exponential Function and its Slope at the Point (1, e)
A function’s curve has a varying slope at each point. A tangent line is drawn at the point to find the slope at that point on the curve. This tangent line’s slope gives the curve’s slope at that point.
## Examples Related to Slope
### Example 1 – Finding the Slope of a Line
Given two points (0, 0) and (5, -2) on the line, find the slope.
### Solution
Slope = (y2 – y1) / (x2 – x1) = (-2 – 0) / (5 – 0) = -2/5
### Example 2 – Using the Slope-intercept Form
A line has a slope of -1 and a y-intercept of 2. Find the equation of the line and graph the line.
### Solution
As the slope and y-intercept are given, the slope-intercept form of a line is:
y = mx + a
Here,
m = -1 and a = 2
Putting the values gives the equation as:
y = (-1)x + 2
y = – x + 2
The slope is -1 with a rise of -1 and a run of 1. To graph the line, starting from point (0, 2) and moving 1(rise) unit down(as the slope is negative) and 1(run) unit left. The point reached is (1, 1).
Joining the two points (0, 2) and (1, 1) and extending them in the opposite direction is the required line, as shown in figure 8.
Figure 8 – Graphing a Line Using Slope and y-intercept
### Example 3 – Using the Point-slope Form
Find the equation of a line given the slope as -2 and a point (5, 0) that goes through the line. Also, convert the equation into standard form.
### Solution
As the slope and a point are given, the point-slope form is:
y – y1 = m(x – x1)
Here,
(x1, y1) = (5, 0) and m = -2
Putting the values in the above equation gives:
y – 0 = -2(x – 5)
y = -2x + 10
Adding 2x on both sides to convert the above equation in standard form as:
2x + y = 10
All the images are created using GeoGebra.
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# Bridges in Mathematics Grade 5 Home Connections Unit 6 Module 3 Answer Key
Students looking for the Bridges in Mathematics Grade 5 Home Connections Answer Key Unit 6 Module 3 can find a better approach to solve the problems.
## Bridges in Mathematics Grade 5 Home Connections Answer Key Unit 6 Module 3
Bridges in Mathematics Grade 5 Home Connections Unit 6 Module 3 Session 1 Answer Key
Measurement & Multiplication Review
Question 1.
a. How many meters are in 1 kilometer?
1,000 meters,
Explanation:
In 1 kilometer there are 1,000 meters.
b. How many meters are in 3 kilometers?
3,000 meters,
Explanation:
In 1 kilometer there are 1,000 meters so in 3 kilometers we have 3 X 1,000 = 3,000 kilometers.
Question 2.
Our school’s swimming pool is 25 meters long. If our coach wants us to swim 3 kilometers, how many lengths of the pool will we need to swim? Show all your work.
120 meters lengths of the pool will be needed to swim,
Explanation:
Given school’s swimming pool is 25 meters long. If our coach wants us to swim 3 kilometers, the lengths of the pool will we need to swim is 3 X 1,000 meter/25 = 3,000/25 meters = 120 meters.
Question 3.
The distance around our school’s playing field is 300 meters. If our coach wants us to run 3 kilometers, how many times will we need to run around the field?
10 times he will need to run around the field,
Explanation:
Given the distance around our school’s playing field is 300 meters. If our coach wants us to run 3 kilometers, number of times will we need to run around the field is 3 X 1,000 / 300 = 3,000/300 = 10.
Question 4.
CHALLENGE How many centimeters are there in 1 meter?
a. How many square centimeters are in 1 square meter?
10,000 square centimeters,
Explanation:
As we know 1 square meter is equal to 10,000 square centimeters, so 10,000 square centimeters are in 1 square meter.
b. How many cubic centimeters are in 1 cubic meter?
1,000,000 cubic centimeters,
Explanation:
As 1 meter = 100 cm, 1 cubic meter = 100 cm X 100 cm X 100 cm = 1,000,000 cubic centimeters, therefore are 1,000,000 cubic centimeters are in 1 cubic meter.
Question 5.
Think about rounding to estimate the answers to the problems below. Then rewrite each problem vertically and solve it using the standard algorithm. Check your answer against your estimate to make sure that it is reasonable.
Explanation:
Thought about rounding to estimate the answers to the problems above. Then rewritten each problem vertically and solved it using the standard algorithm. Checked my answer against my estimate to make sure that it is reasonable.
Question 6.
Circle the two numbers whose product is 627.
13
19
33
49
Explanation:
Asked to circle the two numbers whose product is 627, We will check with all the numbers
1. 13 X 19 = 247, 2. 13 X 33 = 429, 3. 13 X 49 = 637, 4. 19 X 33 = 627, 5. 19 X 49 = 931, 6. 33 X 49 = 1,617. Therefore as 19 X 33 = 627 the numbers are 19 and 33 so circle them.
Bridges in Mathematics Grade 5 Home Connections Unit 6 Module 3 Session 3 Answer Key
Camping Trip
The Zamora family is going on a camping trip next week. There are four people in the family: Mr. and Mrs. Zamora and the 11-year-old twins, Ramon and Daria. Help them do some planning for their trip. Fill in the bubble beside the correct answer to each question below.
Question 1.
Mrs. Zamora wants to cut a piece of rope that’s long enough to dry the family’s laundry on every day. Which of these units should she use to measure the rope?
inches
feet
yards
miles
Explanation:
Given Zamora family is going on a camping trip next week. There are four people in the family: Mr. and Mrs. Zamora and the 11-year-old twins, Ramon and Daria. Mrs. Zamora wants to cut a piece of rope that’s long enough to dry the family’s laundry on every day. So units should she use to measure the rope is 1 inches = 2.54 centimeters, 1 feet = 30.48 centimeters which is small for family laundry, 1 yards = 91.44 centimeters, 1 miles = 1,60,934 centimeters,
therefore best fit will be in yards. Filled in the bubble beside the correct answer.
Question 2.
Mr. Zamora wants to figure out how far they’ll have to drive to get to the campsite. He already knows it will take about a day to get there. Which of these units should he use?
inches
feet
yards
miles
Explanation:
As Mr. Zamora wants to figure out how far they’ll have to drive to get to the campsite. He already knows it will take about a day to get there. Out of these units should he use are as 1 inches = 2.54 centimeters, 1 feet = 30.48 centimeters which is small for family laundry, 1 yards = 91.44 centimeters, 1 miles = 1,60,934 centimeters therefore best fit will be in miles. Filled in the bubble beside the correct answer.
Question 3.
Ramon wants to find the area of his sleeping bag to see how much room he’ll have in the family’s tent. Which of these units should he use?
square inches
square feet
square yards
square miles
Explanation:
As Ramon wants to find the area of his sleeping bag to see how much room he’ll have in the family’s tent. Out of these units should he use are as 1 square inches = 6.4516 square centimeters, 1 square feet = 929.03 square centimeters which is small for family laundry, 1 square yards = 8361.27 square centimeters, 1 square miles = 2.59e+10 square centimeters therefore best fit will be in square feet filled in the bubble beside the correct answer.
Question 4.
Daria says that when they arrive she’s going to measure the area of their campsite. Mrs. Zamora says the campsite is big enough for their car, their tent, their picnic table and chairs, and their campfire, with a little room left over. Which of these units should Daria use?
square inches
square feet
square yards
square miles
Explanation:
As Daria says that when they arrive she’s going to measure the area of their campsite. Mrs. Zamora says the campsite is big enough for their car, their tent, their picnic table and chairs, and their campfire, with a little room left over. Out of these units should Daria use are 1 square inches = 6.4516 square centimeters, 1 square feet = 929.03 square centimeters which is small for family laundry, 1 square yards = 8,361.27 square centimeters, 1 square miles = 2.59e+10 square centimeters therefore best fit will be in square miles filled in the bubble beside the correct answer.
Question 5.
Mr. Zamora wants to find the volume of the car’s trunk so he’ll know how much luggage will fit. Which of these units should he use?
cubic inches
cubic feet
cubic yards
Explanation:
Given Mr. Zamora wants to find the volume of the car’s trunk so he’ll know how much luggage will fit. Out of these units should he use are as 1 cubic inches = 16.3871 cubic centimeters, 1 cubic feet = 28,316.8 cubic centimeters, 1 cubic yards = 7,64,555 cubic centimeters so the best fit will be cubic yards filled in the bubble beside the correct answer.
Question 6.
Daria is going to collect pebbles at the lake. She wants to measure the volume of a metal lunch box to keep them in. Which of these units should she use?
cubic inches
cubic feet
cubic yards
Explanation:
As Daria is going to collect pebbles at the lake. She wants to measure the volume of a metal lunch box to keep them in. Out of these units should she use are as 1 cubic inches = 16.3871 cubic centimeters, 1 cubic feet = 28,316.8 cubic centimeters, 1 cubic yards = 7,64,555 cubic centimeters so the best fit will be cubic feet filled in the bubble beside the correct answer.
Review
Question 7.
Jasmine is planning a large family gathering. She needs to rent at least 200 chairs. Company A charges $0.50 per chair for the first 100 and then$0.25 for every chair after that. Company B charges $0.40 per chair. a. Which company should Jasmine rent from? Explain your answer. Answer: Company B, Explanation: As Jasmine is planning a large family gathering. She needs to rent at least 200 chairs. Company A charges$0.50 per chair for the first 100 and then $0.25 for every chair after that. Company B charges$0.40 per chair. If Jasmine is planning for company A then the cost is
(100 X $0.50) + (100 X$0.75) = $50 +$75 = $125 and if planning for company B then the cost is (200 X$0.40) = $80 so$125 > $80 the company should Jasmine rent from is Company B. b. How much money will she save by using that company? Show your work. Answer:$45,
Explanation:
As Jasmine is planning a large family gathering. She needs to rent at least 200 chairs. Company A charges $0.50 per chair for the first 100 and then$0.25 for every chair after that. Company B charges $0.40 per chair. If Jasmine is planning for company A then the cost is (100 X$0.50) + (100 X $0.75) =$50 + $75 =$125 and if planning for company B then the cost is (200 X $0.40) =$80 so $125 >$80 much money will she save by using that company B is $125 –$80 = \$45.
Question 8.
Frank bought several items in the produce department of his grocery store for a family gathering. He purchased 13.25 pounds of apples and twice that amount of oranges. What was the total weight of the fruit that Frank bought? Show your work.
The total weight of the fruit that Frank bought are 39.75 pounds,
Explanation:
As Frank bought several items in the produce department of his grocery store for a family gathering. He purchased 13.25 pounds of apples and twice that amount of oranges. The total weight of the fruit that Frank bought are 13.25 pounds + 2 X 13.25 pounds = 13.25 pounds + 26.50 pounds = 39.75 pounds.
Question 9.
CHALLENGE Frank also made punch to take to the family gathering. He filled a jar with 3$$\frac{3}{4}$$ liters of punch and another jar with 5$$\frac{3}{4}$$ liters of punch. On his way to the family gathering, some of the punch spilled out of the jars in his car. Only 1$$\frac{1}{2}$$ liters were left in the first jar, and 3$$\frac{1}{4}$$ liters were left in the second jar. How much of the punch was spilled?
Much of the punch was spilled is 4$$\frac{3}{4}$$ liters,
Explanation:
Given Frank also made punch to take to the family gathering. He filled a jar with 3$$\frac{3}{4}$$ liters of punch and another jar with 5$$\frac{3}{4}$$ liters of punch. On his way to the family gathering, some of the punch spilled out of the jars in his car. Only 1$$\frac{1}{2}$$ liters were left in the first jar and 3$$\frac{1}{4}$$ liters were left in the second jar. Total liters of punch are 3$$\frac{3}{4}$$ liters + 5$$\frac{3}{4}$$ liters first we make mixed fraction into fractions as $$\frac{3 X 4 + 3}{4}$$ liters + $$\frac{5 X 4 + 3}{4}$$ liters = $$\frac{15}{4}$$ liters +$$\frac{23}{4}$$ liters = $$\frac{15 + 23}{4}$$ liters = $$\frac{38}{4}$$ liters both goes in 2 we get $$\frac{19 X 2}{2 X 2}$$ liters = $$\frac{19}{2}$$ liters, Now left in two jars are 1$$\frac{1}{2}$$ liters + 3$$\frac{1}{4}$$ liters first we make mixed fraction into fractions as $$\frac{1 X 2 + 1}{2}$$ liters + $$\frac{3 X 4 + 1}{4}$$ liters = $$\frac{3}{2}$$ liters + $$\frac{13}{4}$$ liters as denominators are not the same we make first common denominator as $$\frac{3 X 2}{4}$$ liters + $$\frac{13}{4}$$ liters we get $$\frac{6 + 13}{4}$$ liters = $$\frac{19}{4}$$ liters therefore much of the punch was spilled is
$$\frac{19}{2}$$ liters – $$\frac{19}{4}$$ liters as denominators are not the same we make first common denominator as $$\frac{19 x 2}{4}$$ liters – $$\frac{19}{4}$$ = $$\frac{38}{4}$$ liters – $$\frac{19}{4}$$ we get
$$\frac{38 – 19}{4}$$ liters = $$\frac{19}{4}$$ liters as numerator is greater we write in mixed fraction as $$\frac{4 X 4 + 3}{4}$$ liters = 4$$\frac{3}{4}$$ liters.
Bridges in Mathematics Grade 5 Home Connections Unit 6 Module 3 Session 5 Answer Key
Another Camping Trip
The Eng family is going on a camping trip next week. There are 4 people in the family: Mr. and Mrs. Eng and their children, Jason and Kristen. Help them do some planning for their trip. Fill in the bubble beside the correct answer to each question below.
Question 1.
The shoelaces on Jason’s tennis shoes are almost worn out. He has to measure them so he gets the right length at the store. Which of these units should he use?
millimeters
centimeters
meters
kilometers
Explanation:
As the Eng family is going on a camping trip next week. There are 4 people in the family: Mr. and Mrs. Eng and their children, Jason and Kristen. The shoelaces on Jason’s tennis shoes are almost worn out. He has to measure them so he gets the right length at the store. Out of these units should he use are 1 millimeters = 0.1 centimeters, 1 meters = 100 centimeters and 1 kilometers = 1,00,000 centimeters so it will be centimeters. Filled in the bubble beside the correct answer.
Question 2.
Mrs. Eng says it will be a 3-minute walk from their tent to the lake. Kristen wants to measure the distance. Which of these units should she use?
millimeters
centimeters
meters
kilometers
Explanation:
As Mrs. Eng says it will be a 3-minute walk from their tent to the lake. Kristen wants to measure the distance. Out of these units should he use are 1 millimeters = 0.1 centimeters, 1 meters = 100 centimeters and 1 kilometers = 1,00,000 centimeters so it will be meters. Filled in the bubble beside the correct answer.
Question 3.
Kristen wants to find the area of the floor of her family’s tent to make sure everything will fit. Which of these units should she use?
square centimeters
square meters
square kilometers
Explanation:
As Kristen wants to find the area of the floor of her family’s tent to make sure everything will fit. Out of these units should she use are as 1 square meters = 10,000 centimeters and 1 square kilometers = 1e+10 centimeters therefore it is square meters. Filled in the bubble beside the square meters of the correct answer.
Question 4.
Jason says that when they arrive, he’s going to measure the area of the parking space to make sure their truck and bikes will fit. Which of these units should he use?
square centimeters
square meters
square kilometers
Explanation:
Given Jason says that when they arrive, he’s going to measure the area of the parking space to make sure their truck and bikes will fit. Out of these units should he use are as 1 square meters = 10,000 centimeters and 1 square kilometers = 1e+10 centimeters therefore it is square kilometers. Filled in the bubble beside the square kilometers of correct answer.
Question 5.
Which formula should Kristen use to find the area of the campsite?
A = 2l × 2w
A = l × w × h
A = l × w
A = B × h
Explanation:
To apply which formula should Kristen use to find the area of the campsite is we got square meters so Area = length X width as length is measured in meter and width in meter it is
A = l X W filled in the bubble beside the correct answer.
Question 6.
Mr. Eng wants to find the volume of the family car trunk so he’ll know how much luggage will fit back there. Which of these units should he use?
cubic millimeters
cubic centimeters
cubic meters
Explanation:
Given Mr. Eng wants to find the volume of the family car trunk so he’ll know how much luggage will fit back there. Out of these units should he use 1 cubic millimeters = 1 cubic centimeters, 1 cubic meters = 10,00,000 cubic centimeters filled in the bubble beside the cubic meters of the correct answer.
Question 7.
Jason wants to measure the volume of a shoe box to find out how many comic books he can fit into it for the trip. Which of these units should he use?
cubic millimeters
cubic centimeters
cubic meters
Explanation:
As Jason wants to measure the volume of a shoe box to find out how many comic books he can fit into it for the trip. Out of these units should he use 1 cubic millimeters = 1 cubic centimeters, 1 cubic meters = 10,00,000 cubic centimeters filled in the bubble beside the cubic centimeters of the correct answer.
Question 8.
Which formula should Jason use to find the volume of the shoe box?
V = l × w
V = 2l + 2w
V = l × w × h
V = l + w + h
Explanation:
We know 1 cubic centimeter is equal to the volume of a cube with edges that measure 1 cm each. Since the volume of a cube is calculated as length X width X height, the volume of the cube is 1 cm3. Therefore formula should Jason use to find the volume of the shoe box is
V = l X w X h filled in the bubble beside the volume of the correct answer.
Question 9.
Solve the two-story problems below. Show your thinking using words, numbers, or labeled sketches.
a. The regular-sized box of cereal measures 6 cm by 30 cm by 20 cm. What is the volume of the cereal box?
The volume of the cereal box is 3,600 cubic centimeters,
Explanation:
Given the regular-sized box of cereal measures 6 cm by 30 cm by 20 cm. So the volume of the cereal box is 6 cm X 30 cm X 20 cm = 3,600 cubic centimeters.
b. The large cereal box is twice as wide as the regular.
• Do you have enough information to find the volume?
• If not, what else do you need? If so, what is the volume of the large cereal box?
Yes, Volume of the large cereal box is 7,200 cubic centimeters,
Explanation:
Given the large cereal box is twice as wide as the regular means its length and height are same but width is twice, Yes we can find the volume it is V = 6 cm X (2 X 30) cm X 20 cm =
7,200 cubic centimeters.
Question 10.
CHALLENGE Graph and label the points that represent cubes with side lengths of 1, 2, 3, and 4 centimeters and their volumes. The first one is done for you. What does the point (1, 1) represent?
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# How do you solve (x+2)/(x+1)>0?
Nov 17, 2016
The "critical" values are $x = - 2$ and $x = - 1$. When $x < - 2$, we get $\frac{-}{-}$$= +$. When $- 2 < x < - 1$, we get $\frac{+}{-}$$= -$. When $x > - 1$, we get $\frac{+}{+}$$= +$. Thus, our solution is $x \in \left(- \infty , - 2\right) \cup \left(- 1 , \infty\right) .$
#### Explanation:
In order for the fraction as a whole to be greater than 0, both the numerator and the denominator will need to be greater than 0 (or less than 0), so that the division will yield a positive value.
We can think of the LHS as the product of two factors: $x + 2$ and $\frac{1}{x + 1}$. That is,
$\left(x + 2\right) \left(\frac{1}{x + 1}\right) > 0$
The first factor $x + 2$ will be greater than 0 when x is greater than -2:
$x + 2 > 0$
$\iff x > - 2$
The second factor $\frac{1}{x + 1}$ is a little more tricky, but it's easy when you realize that a fraction like this will be positive only when its denominator is the same sign as the numerator (in this case, positive). The numerator is a constant; it cannot affect the fraction's sign. In other words:
$\frac{1}{x + 1} > 0$
$\iff x + 1 > 0$
$\iff x > - 1$
Next, we create a sign table for the factors and fill it in with $+$ or$-$ signs depending on where each factor is positive or negative. Our "points of interest" are -1 and -2.
Factor - - - - - - - - - - - Sign - - - - - - -
. . . . . . . . .------- -2 ---------- -1 -----------
$x + 2$ . . . $- - 0 + + + + +$
$x + 1$ . . . $- - - - - \emptyset + +$
. . . . . . . . .-------------------------------------
$\frac{x + 2}{x + 1}$ . . $+ + 0 - - - \emptyset + +$
To fill in the bottom row, simply multiply the signs of all the factor rows together.
Notice the $\emptyset$ for the factor $x + 1$. That's there because $x + 1$ is a denominator factor, and any $x$-value that makes such a factor equal to 0 will make the whole expression undefined.
Finally, we pluck out the interval(s) over which the whole expression is greater than 0. After multiplying the signs in the sign table, we see that we get:
$\frac{x + 2}{x + 1} > 0$ when $x < - 2$ and $x > - 1$.
Our solution is: $\left\{x | x < - 2 \cup x > - 1\right\}$, or
$x \in \left(- \infty , - 2\right) \cup \left(- 1 , \infty\right)$.
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The Trick
After discussing multiplication of numbers by 11, let us discuss another math shortcut method which is very popular among kids and teenagers. This is squaring numbers ending in 5.
For kids, you can read the first portion of this post if you want to learn the trick, and read further if you want to learn the mathematics behind the trick.
Remember: it is cool if you know the trick, but it is even cooler if you know why it works. Mathematics is not just knowing the method, but also knowing the reasons.
Math Shortcut on Squaring Numbers Ending in 5
The algorithm is squaring numbers ending in 5 is the following.
1.) Multiply the ones digits. That is 5 x 5 since they both of them in 5. Write 25 as product as shown below. The last two digits of this trick is always 25.
2.) Add 1 to one of the tens digit.
3.) Multiply the sum into by the tens digit of the other number. Here, we multiply 3 in (2) by 2.
That means that the 45× 45 = 2025.
Example 2: What is 35 × 35?
First, our last two digits is 5 × 5 = 25. Then, instead of 3× 3, we increase one of the factors by 1. That is 4 × 3 = 12. So, the correct answer 35 × 35 = 1225.
Now, why does this trick work? Will it work in all possible cases? Try other 2-digit ending in 5 and check the product manually or use a calculator.
The Secret
The secret of this trick needs a little Algebra. If you have taken Algebra 1, then this explanation is fine for you.
Two digit numbers can be expressed as 10t + u, where t and u are the tens and units respectively. Here are a few examples.
28 = 20 + 8 = 10(2) + 8
74 = 70 + 4 = 10(7) + 4
96 = 90 + 6 = 10(9) + 6
This means that if we have a number which has tens digit as n, then squaring a number ending in 5 will look like
(10n + 5)(10n + 5).
Multiplying, we have
100n2 + 100n + 25.
Since our last two digits in the multiplication above is 25, if we get 25, we are only left 100n2 + 100n which is hundreds digit.
We can factor 100n2 + 100n as 100 n(n + 1).
Notice: n is equivalent to 4 in our first example and the n + 1 is the equivalent to 5. This means that we always have to add 1 to n in one of the tens digit and then multiply to get the product.
Next post >>
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# How To Multiply Exponents – Multiplying Exponents Rules
Learners can learn how to multiply exponents here, as you need to just have the same base and then add the exponents. For example: a^m * a^n = a^(m+n).
Multiplying exponents is a fundamental operation in mathematics, particularly in algebra. When you multiply two exponential expressions with the same base, you can simplify the expression by adding the exponents. This means that if you have two expressions like a^m and a^n, where a is the base and m and n are the exponents, you can simplify the expression to a^(m+n).
This concept is also known as the “power of a power” rule and is a key building block for understanding more advanced mathematical concepts like logarithms and exponential functions. Understanding how to multiply exponents can help you simplify complex expressions and solve problems more efficiently, making it a valuable skill to have in mathematics and other related fields.
## How To Multiply Exponents With Example
Here’s an example of how to multiply exponents step by step:
Example: (2^3) * (2^4)
• Identify the base, which is 2 in this case.
• Add the exponents, which are 3 and 4.
• Write the product with the base and the sum of the exponents as the new exponent.
Result: (2^3) * (2^4) = 2^(3 + 4) = 2^7 = 128.
So the product of (2^3) * (2^4) is 128.
Multiplying exponents is a useful mathematical operation because it allows you to simplify expressions and solve problems more easily. It is a basic concept in algebra and is used in many different fields, such as science, engineering, finance, and computer science.
Understanding how to multiply exponents can also help you understand more advanced mathematical concepts, such as logarithms and exponential functions. Overall, having a good understanding of multiplying exponents can help you in a variety of applications and lead to a better understanding of mathematics as a whole.
#### Multiplying Exponents Rules
Here are a few examples of multiplying exponents:
1. (2^3) * (2^4) = 2^(3+4) = 2^7 = 128
2. (5^2) * (5^3) = 5^(2+3) = 5^5 = 3125
3. (3^3) * (3^2) = 3^(3+2) = 3^5 = 243
4. (x^2) * (x^3) = x^(2+3) = x^5
5. (a^4) * (a^5) = a^(4+5) = a^9
In conclusion, exponents play a crucial role in mathematics and are used to represent repeated multiplication of a number by itself. Understanding the rules of multiplying exponents is essential for success in many areas of mathematics, including algebra and calculus.
When multiplying exponents with the same base, one can add the exponents to find the result. The multiply exponents with different bases, one must multiply the bases and keep the exponents separate. When multiplying exponential expressions with different bases and exponents, one must multiply each term separately, and then simplify the expression by combining like terms. Additionally, it is important to understand the laws of exponents, such as the product of powers law and the power of a power law, which allow for the simplification of exponential expressions.
In summary, multiplying exponents is a fundamental mathematical operation that plays a critical role in many areas of mathematics, and a thorough understanding of the process, rules, and applications is essential for success in related fields.
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# How to Find Values of Functions from Graphs?
In a few simple steps, we can find the value of the function from the graph. In this step-by-step guide, you will learn more information about finding values of functions from graphs.
A function in mathematics is represented as a rule, which gives a unique output for each input $$x$$.
## Step by step guide tofinding values of functions from graphs
We can find the value of the function from the graph in a few simple steps. Note this example to learn how to find a function from a graph. For example, find the value of a function $$f(x)$$ when $$x = a$$.
1. Draw a vertical line through the value $$a$$ on the $$x$$-axis.
2. Mark the point of intersection of the line $$x = a$$ and the graph of $$f(x)$$.
3. Draw a horizontal line from the point of intersection to the $$y$$-axis.
4. Let the horizontal line meet the $$y$$-axis at $$b$$.
The value of the function $$f(x)$$ at $$x = a$$ is $$b$$. That is, $$f(a) = b$$.
### Finding Values of Functions from Graphs– Example 1:
Use the graph of $$f(x)$$ shown below to find $$f(4)$$.
Solution:
First, draw a vertical line through $$4$$ on the $$x$$-axis. Then mark the intersection of the vertical line $$x = 4$$ and the graph $$f (x)$$. Now draw a horizontal line from the point of intersection to the $$y$$-axis.
So, the value of the function $$f(x)$$ at $$x = 4$$ is $$7$$.
## Exercises forFinding Values of Functions from Graphs
• Use the graph of $$f(x)$$ shown to find $$f(8)$$.
• Use the graph of $$f(x)$$ shown to find $$f(2)$$.
1. $$\color{blue}{7}$$
2. $$\color{blue}{10}$$
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# 2011 AMC 12B Problems/Problem 13
## Problem
Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$. What is the sum of the possible values of $w$?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$
## Solution
Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. This means $a+b+c=9$. $a,b,c$ must be in the set ${1,3,4,5,6}$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$.
\begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ if \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*}
\begin{align*} if \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*}
The sum of the two w's is $15+16=31$ $\boxed{B}$
2011 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Involute Function Calculator
Roulette and involuteInvolute gearsInvolute gear design and involute functionHow does our involute function calculator work?FAQs
Not happy with simple curves, mathematicians decided to see what happens when two curves slide on one another, and the result made many engineers extremely happy: discover why with our involute function calculator!
Here you will learn:
• What the involute function is;
• What it has to do with involute curves; and
• How it is related to the gear pressure angle, which is a very important concept in engineering. ⚙️
## Roulette and involute
When mathematicians talk about roulette, they are not talking of the casino game but rather of a particular family of curves obtained by rolling a curve on another fixed curve and following the trajectory of a given, fixed point integral with the rolling curve. The other conditions are that:
• Both curves must be differentiable — this means that it is "smooth" in its entirety, without points like cusps or interruptions where it would be impossible to define its slope.
• The curves must be tangent at every moment.
🔎 There is a drawing device that has fascinated kids for decades, which bases its attractiveness on roulette curves: the spirograph. By rolling geared shapes (like circles and ellipses) with pencil holes around each other, it is possible to draw all sorts of mesmerizing curves.
A particular class of roulette curves is obtained by following a line wrapped around a curve (taut string) as it unwraps, or vice-versa: this type of roulette is called the involute.
When we talk of a string wrapped around a circle, the resulting curve, resembling a spiral, is called the involute of a circle. Here it is, with its construction.
The red straight line represents the tensioned string as it wraps/unwraps. The tip of the string draws the involute.
The involute is described by this parametric equation:
$\begin{split} x &=a(\cos{\theta}+\theta\cdot\sin{\theta}) \\ y &=a(\sin{\theta}−\theta\cdot\cos{\theta}) \end{split}$
where $a$ is the radius of the circle and $\theta$ is the angle at which the string separates from the circle (where circle and line are tangent).
## Involute gears
Involute functions are interesting for mathematicians but fundamental for engineers: the main application of the involute function is the construction of involute gears.
It is necessary to introduce a few elements of involute gears:
• The line of action and the line of centers;
• The pitch point;
• The pressure angle.
The line of centers connects the centers of the two gears. While a couple of teeth are in contact as the gears roll, the point of contact moves along an imaginary line called line of action.
This is the core point of the peculiarity of involute gears: the contact between teeth happens only on a straight line — obviously, if the teeth' profiles match.
The pitch point is the point on the line of centers where the circumferential speeds of the gears (you can learn how to calculate it with the circular motion calculator) match: this means that points on gears' edges travel at the same speed. In involute gears, this is also the point of intersection between the action line and the line of centers.
Let's draw the perpendicular to the line of centers passing through the pitch point.
At this point, the angle between this line and the action line is called the pressure angle. Its value remains constant during the operation of the gears; hence, it is characteristic of a given design.
## Involute gear design and involute function
The involute gear pressure angle is the fundamental parameter when designing the gear itself, since its value affects the tilting of the line of action and the behavior of the gear during its operations.
Here is how to calculate the involute as a function of the pressure angle.
We start by building the circle with center $O$ and radius $OS$. Then, we draw a part of the involute of the circle. We call the origin point $S$, then choose a point $P$ on this curve. The taut string touches the circumference in the point $T$.
By construction, the segment $TP$ and the arc $ST$ have the same length. It is then possible to write: $\overline{TP} =\stackrel{\frown}{ST}$.
In this construction, the angle of pressure is the angle $\hat{TOP} = \alpha$. The other angle, $\hat{SOP} = \varphi$, can be found following the next steps:
$\overline{TP} =\ \stackrel{\frown}{ST} \\ r\cdot\tan{\alpha}= r \left( \varphi + \alpha\right) \\ \tan \alpha = \varphi +\alpha\\ \varphi = \operatorname{Inv}(\alpha) =\tan{\alpha} - \alpha$
The last expression is the involute function $\tan$ is the tangent function. The resulting angle is extremely important in the design of involute gears, for example, in the calculation of the tooth thickness.
From the geometrical construction, you can see that $\varphi$ (the value of the involute function for a given pressure angle $\alpha$) is somehow connected to the "thickness" of the tooth of an involute gear. This explains why the involute function is so important.
An increase in the pressure angle equals an increase in the width of the tooth. A wider tooth is a stronger tooth that can withstand bigger loads. On the other hand, smaller teeth obtained by reducing the pressure angle $\alpha$ give advantages to the smoothness of the operation.
## How does our involute function calculator work?
Using our involute function calculator is really easy. You can input the value of the gear pressure angle $\alpha$ to obtain right away the value of the relative involute function.
🙋 You can't use this calculator in "reverse" due to the presence of the tangent function in the involute!
Let's try with an example.
The most common gear pressure angle currently used is $20\degree$. In the involute function calculator, change the units of $\alpha$ to degrees and input $20$. The result is dimensionless and equal to $\thicksim 0.0149$.
If you are interested in other calculators, check our gear ratio calculator, the gear ratio speed calculator, or discover the sister curve of the involute with the catenary curve calculator!
FAQs
### What is an involute?
An involute is a parametric curve that describes the wrapping/unwrapping of a taut string around a generating curve. The most common involute is the involute of the circle, which returns a curve similar to a spiral.
### How do I calculate the involute function?
The involute function is a function that takes as argument an angle — the pressure angle — and returns a value that corresponds to the width of the involute curve at that point. The formula for the involute function is φ = tan(α) − α.
### What is the relation between involute function and angle of action?
The angle of action of an involute gear is associated with the width of the teeth of the gear itself through the involute function.
### Why is the involute important in engineering?
Engineers use the shape of the involute of a circle to design the teeth of gears that touch only at one point during the entire contact time. Involute gears are widespread and compose the vast majority of gears produced worldwide.
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# Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.
$\boldsymbol{ \left[ \begin{array}{ c c | c } 1 & 3 & -8 \\ -4 & h & 1 \end{array} \right] }$
The aim of this question is to understand the solution of the system of linear equations using the row operations and row echelon form.
Any matrix is said to be in the row echelon form if it fulfills three requirements. First, the first non-zero number in every row must be 1 (called the leading 1). Second, each leading 1 must be on the right of the leading 1 in the previous row. Third, all non-zero rows must precede the zero rows. For example:
$\left[ \begin{array}{ c c c | c } 1 & x & x & x \\ 0 & 0 & 1 & x \\ 0 & 0 & 0 & 0 \end{array} \right]$
Where x can have any value.
The row echelon form can be used to solve a system of linear equations. We simply write the augmented matrix and then convert it to the row echelon form. Then we convert it back to the equation form and find the solutions by back substitution.
The linear system of equations represented by an augmented matrix will have a unique solution (consistency) if the following condition is satisfied:
$\text{ no. of non-zero rows } \ = \ \text{ no. of unknown variables }$
Given:
$\left[ \begin{array}{ c c | c } 1 & 3 & -8 \\ -4 & h & 1 \end{array} \right]$
Reducing to row echelon form:
$R_2 \ + \ 4R_1 \rightarrow \left[ \begin{array}{ c c | c } 1 & 3 & -8 \\ 0 & h-12 & -31 \end{array} \right]$
It can be deduced from the above matrix that the system of linear equations formed by these coefficients will have a unique solution on all possible values of $R^n$ except when h = 12 (because this nullifies the 2nd equation and the system reduces to a single equation describing two variables).
## Numerical Result
$h$ can have all possible values of $R^n$ excluding $h = 12$.
## Example
Find all the possible values of $y$ such that the following augmented matrix represents a consistent system of linear equations:
$\boldsymbol{ \left[ \begin{array}{ c c | c } 9 & 18 & 0 \\ 5 & y & 1 \end{array} \right] }$
Reducing the given matrix to row echelon form via row operations:
$\dfrac{ 1 }{ 9 } R_1 \rightarrow \left[ \begin{array}{ c c | c } 1 & 2 & 0 \\ 5 & y & 1 \end{array} \right]$
$R_2 – 5 R_1 \rightarrow \left[ \begin{array}{ c c | c } 1 & 2 & 0 \\ 0 & y-10 & 1 \end{array} \right]$
It can be deduced from the above matrix that the system of linear equations formed by these coefficients will have a unique solution on all possible values of $R^n$ except when y = 10.
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Comparison Theorems for Sequences
# Comparison Theorems for Sequences
We will now look at a few very important comparison theorems for both divergent and convergent sequences.
Theorem 1: If $a_n ≤ b_n$ for all $n ≥ N$ and $\lim_{n \to \infty} a_n = \infty$ then $\lim_{n \to \infty} b_n = \infty$.
What theorem 1 essentially says is that if a sequence $\{ a_n \}$ races off to infinity, and the sequence $\{ b_n \}$ is above the sequence $\{ a_n \}$ when $n ≥ N$ (that is $\{ b_n \}$ is ultimately above $\{ a_n \}$), then the sequence $\{ b_n \}$ must also race off to infinity.
• Proof of Theorem 1: We know that $\lim_{n \to \infty} a_n = \infty$ implies that $\forall k \in \mathbb{R} \: \exists N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n > k$.
• But if $a_n ≤ b_n$ for all $n ≥ N$, then we get that $k < a_n ≤ b_n$ if $n ≥ N$ or rather $k < b_n$. So by the definition, $\lim_{n \to \infty} b_n = \infty$. $\blacksquare$
Theorem 2: If $a_n ≥ b_n$ for all $n ≥ N$ and $\lim_{n \to \infty} a_n = -\infty$ then $\lim_{n \to \infty} b_n = -\infty$.
This theorem is very similar to theorem 1. It says that if the sequence $\{ a_n \}$ races off to negative infinity, and the seqeunce $\{ b_n \}$ is below the sequence $\{ a_n \}$ when $n ≥ N$, then the sequence $\{ b_n \}$ must also race off to negative infinity.
• Proof of Theorem 2: We know that $\lim_{n \to \infty} a_n = -\infty$ implies that if $\forall k \in \mathbb{R} \: \exists N \in \mathbb{N}$}] such that if [[$n ≥ N$ then $a_n < k$.
• But if $a_n ≥ b_n$ or rather $b_n ≤ a_n < k$ for all $n ≥ N$ , then we get that $b_n < k$ if $n ≥ N$. So by the definition, $\lim_{n \to \infty} b_n = -\infty$. $\blacksquare$
Theorem 3: If $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ and $a_n ≤ b_n$ for all $n ≥ N$ then $A ≤ B$.
What this theorem says is that if the sequence $\{ b_n \}$ is ultimately above the sequence $\{ a_n \}$ when $n ≥ N$, and both sequences are convergent, then the point to which $\{ b_n \}$ converges must be above or equal to the point to which $\{ a_n \}$ converges.
• Proof of Theorem 3: Suppose that in fact $A > B$. We want to establish a contradiction here so we will let $\epsilon = \frac{A - B}{3}$.
• So $\lim_{n \to \infty} a_n$ implies that $\forall \epsilon > 0 \: \exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \frac{A - B}{3} = \epsilon$.
• Similarly $\lim_{n \to \infty} b_n$ implies that $\forall \epsilon > 0 \: \exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \frac{A - B}{3} = \epsilon$.
• Now we want to pick an $N \in \mathbb{N}$ such that $N ≥ N_1$ and $N ≥ N_2$ and thus we choose $N = \mathrm{max} \{ N_1, N_2 \}$.
• So we know that $\mid a_n - A \mid < \frac{A - B}{3}$ implies that $- \frac{A-B}{3} < a_n - A < \frac{A-B}{3}$. So then $A - \frac{A - B}{3} < a_n$.
• Similarly we know that $\mid b_n - B \mid < \frac{A - B}{3}$ implies that $- \frac{A-B}{3} < b_n - B < \frac{A-B}{3}$. So then $b_n < B + \frac{A-B}{3}$.
• So we know that if $n ≥ N$, then $a_n ≤ b_n$ and so:
(1)
\begin{align} \quad A - \frac{A - B}{3} < a_n ≤ b_n < B + \frac{A - B}{3} \\ A - \frac{A - B}{3} ≤ B + \frac{A - B}{3} \\ 3A - (A - B) ≤ 3B + (A - B) \\ 2A + B ≤ A + 2B \\ A + B ≤ 2B \\ A ≤ B \end{align}
• But that's a contradiction since we assumed that $A > B$. Therefore $A ≤ B$. $\blacksquare$
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# What is the answer?
## Imaginary numbers:
Nov 14, 2017
${i}^{0} = 1$
${i}^{1} = i$
${i}^{2} = - 1$
${i}^{3} = - i$
${i}^{4} = 1$
#### Explanation:
Generally if you want to get the solution to ${i}^{x}$ where $x$ is any non-negative integer, you need to remember the following:
${i}^{0} = 1$
${i}^{1} = i$
${i}^{2} = - 1$
${i}^{3} = - i$
${i}^{4} = 1$
For all non-negative integers, if it is an exponent of $i$ then you simply have to divide $x$ by 4, 3, or 2.
If $x$ is divisible by 4 then ${i}^{x} = 1$
If $x$ is divisible by 3 then ${i}^{x} = - i$
If $x$ is divisible by 2 then ${i}^{x} = - 1$
If $x$ is not divisible by any of the above numbers then ${i}^{x} = i$
If $x$ is 0 then ${i}^{x} = 1$ since anything raised to 0 is equal to 1
Note that for this you must choose the highest number among 4, 3, and 2 that $x$ is divisible with.
Similarly,
${i}^{2} = i \cdot i = - 1$, so it also goes that:
${i}^{3} = i \cdot i \cdot i = {i}^{2} \cdot i = - 1 \cdot i = - i$
${i}^{4} = i \cdot i \cdot i \cdot i = {i}^{2} \cdot {i}^{2} = - 1 \cdot - 1 = 1$
Nov 14, 2017
${i}^{0} = 1$
${i}^{1} = i = \sqrt{- 1}$
${i}^{2} = - 1$
${i}^{3} = - i = - \sqrt{- 1}$
${i}^{4} = 1$
#### Explanation:
$i = \sqrt{- 1}$
${i}^{0} = \left({x}^{0} = 1 | x \in C\right)$ ... any number (x) power 0 is 1
${i}^{1} = i$ ... any number (x) power 1 is the number (x)
${i}^{2} = i \cdot i = \left(\sqrt{- 1}\right) \left(\sqrt{- 1}\right) = - 1$
${i}^{3} = i \cdot {i}^{2} = \sqrt{- 1} \cdot 1 = \sqrt{- 1} = i$
${i}^{4} = {i}^{2 + 2} = {i}^{2} \cdot {i}^{2} = \left(- 1\right) \cdot \left(- 1\right) = 1$
note: for any ${i}^{a} = b$, one can write
$\left({i}^{2 n}\right) \left({i}^{a - 2 n}\right) = b$
when
$2 n$ is Even
AND
$2 n - a = 1 \mathmr{and} 2 n - a = 3$
for exmple:
i^35=?
We will take $\left(2 n = 32 , 2 n - a = 3\right)$
${i}^{35} = {i}^{32} \cdot {i}^{3} = \left({\left({i}^{4}\right)}^{8}\right) \cdot {i}^{3} = {\left(1\right)}^{8} \cdot {i}^{3} = 1 \cdot {i}^{3} = {i}^{3} = i$
end note
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# Deriving the Poisson Distribution
The Poisson distribution can be derived from the binomial distribution by doing two steps:
1. substitute $\frac{\mu}{n}$ for p
2. Let n increase without bound
Step one is possible because the mean of a binomial distribution is $\mu = np$. So another way of expressing p, the probability of success on a single trial, is $\frac{\mu}{n}$. This has some intuition. Recall that a binomial distribution gives the probability of a number of successes (x) in a fixed number of trials (n) for some probability (p). So if an event has a probability of 0.2, and we observe 10 trials (where the trials are independent), the expected value of seeing the event occur is 10(0.2) = 2. On average we would see the event happen twice in 10 trials. Which means we can state the probability as a rate, $\frac{2}{10}$. Success occurs, on average, 2 times per 10 trials.
The next step leads to the Poisson. We let the number of trials increase without bound, which means the probability shrinks. Since n is in the denominator of $\frac{\mu}{n}$, this gives a small chance of success in a huge number of trials. So whereas the binomial deals with a fixed number of trials, the Poisson accounts for an unlimited number of trials in time or space. Taking the limit as n tends to infinity gives us the Poisson. Here’s how.
Recall the binomial probability mass function:
$f(x) = \frac{n!}{x!(n-x)!}p^{x}(1-p)^{n-x}$
Substitute $\frac{\mu}{n}$ for p:
$f(x) = \frac{n!}{x!(n-x)!}(\frac{\mu}{n})^{x}(1-\frac{\mu}{n})^{n-x}$
Now the fun begins. Let n grow very large:
$\lim_{n \to \infty}\frac{n!}{x!(n-x)!}(\frac{\mu}{n})^{x}(1-\frac{\mu}{n})^{n-x}$
But wait! Don’t let n blow up just yet. Let’s do some rearranging first to help us take the limit:
$\lim_{n \to \infty}\frac{n!}{(n-x)!n^{x}}\frac{\mu^{x}}{x!}(1-\frac{\mu}{n})^{n}(1-\frac{\mu}{n})^{-x}$
And let’s also simplify $\frac{n!}{(n-x)!n^{x}}$:
$\frac{n!}{(n-x)!n^{x}} = \frac{n(n-1)(n-2)\dots(n-x+1)}{n^{x}}$ $=1(1-\frac{1}{n})\dots(1-\frac{x-2}{n})(1-\frac{x-1}{n})$
Before we flip the switch on the limit-to-infinity machine, let’s assess what we got:
$\lim_{n \to \infty}1(1-\frac{1}{n})\dots(1-\frac{x-2}{n})(1-\frac{x-1}{n})\frac{\mu^{x}}{x!}(1-\frac{\mu}{n})^{n}(1-\frac{\mu}{n})^{-x}$
Notice this is basically four things multiplied together:
1. $1(1-\frac{1}{n})\dots(1-\frac{x-2}{n})(1-\frac{x-1}{n})$
2. $\frac{\mu^{x}}{x!}$
3. $(1-\frac{\mu}{n})^{n}$
4. $(1-\frac{\mu}{n})^{-x}$
We can make life easier by taking the limit of each of those four pieces individually:
1. $\lim_{n \to \infty}1(1-\frac{1}{n})\dots(1-\frac{x-2}{n})(1-\frac{x-1}{n})=1$
2. $\lim_{n \to \infty}\frac{\mu^{x}}{x!}=\frac{\mu^{x}}{x!}$ (no n to blow up!)
3. $\lim_{n \to \infty}(1-\frac{\mu}{n})^{n}=e^{-\mu}$
4. $\lim_{n \to \infty}(1-\frac{\mu}{n})^{-x}=1$
1, 2, and 4 are easy. Number 3, though, requires a long forgotten formula you probably learned in calculus just long enough to take an exam:
$\lim_{n \to \infty}(1+\frac{b}{n})^{n}=e^{b}$
Set b = $-\mu$ and you get the result.
Now put it all back together and you have the probability mass function of the Poisson distribution:
$f(x) = \frac{\mu^{x}e^{-\mu}}{x!}$
Often you see $\mu$ as $\lambda$ in this formula.
This new formula is approximately the binomial probability mass function for a large number of trials and small probability. That’s basically what we did, right? Substitute an expression for p involving n and let n grow large. That means the probability got small while n got large. Which further means that before we had modern computers, the Poisson was very handy for approximating binomial probabilities since the binomial coefficient can be difficult to compute for large n.
For example, say we have 100 independent trials with probability of success 0.05. What is the probability of observing 7 successes?
Using the binomial:
$f(7) = \frac{100!}{7!(100-7)!}0.05^{7}(1-0.05)^{100-7} = 0.106026$
Using the Poisson as an approximation (where $\mu = np = 100(0.05) = 5$):
$f(7) = \frac{5^{7}e^{-5}}{7!} = 0.104445$
Very close indeed!
By the way, how did I calculate those probabilities? I used Excel functions:
Binomial: =BINOMDIST(7,100,0.05,0)
Poisson: =POISSON(7,5,0)
The numbers in the parentheses should be self-explanatory. The final 0 means “false, no cumulative”. If you set it to 1, you get the probability of 7 or fewer successes in 100 trials. You can also use a TI-83 Plus. Just press 2nd – DISTR and choose binompdf or poissonpdf. If want to do cumulative probabilities, select binomcdf or poissoncdf. The parameters are in a different order from Excel:
binomcdf/binompdf(n,p,x)
poissoncdf/poissonpdf(mu,x)
Of course “real statisticians” would use R:
> dbinom(7,100,0.05)
[1] 0.1060255
> dpois(7,5)
[1] 0.1044449
I would be careless not to mention that the Poisson distribution is not the same as the binomial. The binomial describes the distribution of events (or “successes”) over a fixed number of trials for a fixed probability of success on each trial. The Poisson describes the distribution of events in space or time for a given rate of occurrence. The binomial is specified by n (number of trails) and p (probability of success on each trial). The Poisson is specified by $\mu$, the rate at which events occur per unit of space or time.
And finally, and no doubt most important, the way to pronounce Poisson is \pwa-sawn\. Say it correctly and you earn instant credibility at cocktail parties.
## 2 thoughts on “Deriving the Poisson Distribution”
1. Jun Shen
This is very impressive. My question is how we know Poisson distribution can be derived in this way in the first place as Poisson and binomial distributions are two different processes? Thanks.
1. Clay Ford Post author
Well, the reason I know it is because I read about it in the book Principles of Statistics by M.G. Bulmer. This blog post was basically me working through his derivation to make sure I understood and filling in some of the steps. In his words: “The Poisson distribution is the limiting form of the binomial distribution where there is a large number of trials but only a small probability of success at each of them.”
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