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# Graphic-solution
• Jul 1st 2009, 11:25 AM
dhiab
Graphic-solution
Give the Graphic-solution of this inequality :
http://www.mathramz.com/xyz/latexren...c070e7fd48.png
• Jul 1st 2009, 02:05 PM
red_dog
$(2x-y-3)(2x-y+3)\leq 0$
The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0.
• Jul 1st 2009, 07:37 PM
Prove It
Quote:
Originally Posted by dhiab
Give the Graphic-solution of this inequality :
http://www.mathramz.com/xyz/latexren...c070e7fd48.png
Alternatively:
$(2x - y)^2 \leq 9$
$|2x - y| \leq 3$
$-3 \leq 2x - y \leq 3$
Case 1: $-3 \leq 2x - y$
$y \leq 2x + 3$.
Case 2: $2x - y \leq 3$
$y \geq 2x - 3$.
So putting them together gives
$2x - 3 \leq y \leq 2x + 3$.
You should be able to graph these.
• Jul 2nd 2009, 12:29 AM
dhiab
Quote:
Originally Posted by red_dog
$(2x-y-3)(2x-y+3)\leq 0$
The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0.
Hello , Thank you, Here are the details
http://www.mathramz.com/xyz/latexren...318c40e750.png
• Jul 2nd 2009, 06:58 AM
HallsofIvy
Just to give yet another way:
Look at $(2x-y)^2= 9$. $2x- y= \pm 3$. Taking the positive sign, $2x- y= 3$. Taking the negative sign, $2x-y= -3$. Those are the two parallel lines red dog gives. I notice that x= 0 gives y= 3 and y= -3 for the two lines so (0,0) is between the two lines. Also, $(2(0)- 0)= 0< 9$ so that point, between the two lines satisfies the inequality. That tells us that every point between the two lines satisfies the inequality, again, as red dog said.
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# A Find the Square Root of a Number B Approximate Square Roots C
```1
Professor Busken – Square Roots and the Pythagorean Theorem
Learning Objectives:
A Find the Square Root of a Number
B Approximate Square Roots
C Use the Pythagorean Theorem
Definition 1. The square of a number is the number times itself.
For instance, the square of 4 is 16 because 42 or 4 · 4 = 16. Another square of −4 is also
16 because (−4)2 = (−4) · (−4) = 16.
Definition 2. The reverse process of squaring is finding a square root.
For example, a square root of 16 is 4 because 42 = 16. Another square root of 16 is also
−4 because (−4)2 = (−4) · (−4) = 16.
Theorem 1. Every positive number has two square roots.
For instance, the square roots of 16 are 4 and −4.
√
, called a radical sign, to indicate the positive
Definition 3. We use the symbol
square root.
For example,
√
25 = 5 because 52 = 25 and 5 is positive.
√
9 = 3 because 32 = 9 and 3 is positive.
Theorem 2 (Square Root of a Number). The square root,
is the positive number b whose square is a. In symbols,
√
a = b if b2 = a
For example,
Find the square root.
√
1.
100
√
√
, of a positive number a
36 = 6 since 62 = 36
1.
2.
√
64
2.
3.
√
81
3.
2
Professor Busken – Square Roots and the Pythagorean Theorem
4.
√
5.
r
1
4
5.
6.
r
49
16
6.
7.
r
4
25
7.
8.
r
169
100
8.
121
4.
1 4
Definition 4. Numbers like , , 9 and 36 are called perfect squares because their
4 25
square root is a whole number or a fraction.
√
A square root such as 21 cannot be written as a whole number or a fraction since 21
is not a perfect square. It can be approximated by estimating, by using a table, or by
using a calculator.
√
21
√
0
0
√
1
√
1
4
2
√
9
3
√
16
4
√
25
5
√
36
6
√
49
7
√
64
8
Consider the diagram above. Above the number line, notice that as the numbers under
the radical signs increase their value, and thus their
√ placement on the number line,
increase too. We can use this fact to estimate that 21 is between whole numbers 4 and
5.
√
9. Without a calculator or table, approximate 51 to the nearest whole number.
9.
10.
Use a calculator to approximate
√
51 to the nearest one-thousandth.
10.
11.
Without a calculator or table, approximate
√
3 to the nearest whole number.
11.
12.
Use a calculator to approximate
√
3 to the nearest one-thousandth.
12.
3
Professor Busken – Square Roots and the Pythagorean Theorem
a2 + b2 = c2
Theorem 3. If a and b are the
lengths of the legs of a right triangle
and c is the length of the hypotenuse,
then a2 + b2 = c2 .
hypotenuse c
Leg a
(leg)2 + (leg)2 = (hypotenuse)2
Leg b
Without using a calculator, find the unknown length of each right triangle.
13.
leg = 6, hypotenuse = 10.
13.
14.
leg = 5, hypotenuse = 12.
14.
15.
leg = 9, hypotenuse = 12.
15.
Using a calculator, find the unknown length of each right triangle. Round
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5-2
1 / 31
# 5-2 - PowerPoint PPT Presentation
5-2. Bisectors of Triangles. Warm Up. Lesson Presentation. Lesson Quiz. Holt Geometry. Warm Up 1. Draw a triangle and construct the bisector of one angle. 2. JK is perpendicular to ML at its midpoint K . List the congruent segments. Learning Targets.
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Presentation Transcript
5-2
Bisectors of Triangles
Warm Up
Lesson Presentation
Lesson Quiz
Holt Geometry
Warm Up
1.Draw a triangle and construct the bisector of one angle.
2.JK is perpendicular to ML at its midpoint K. List the congruent segments.
Learning Targets
Students will be able to:Proveand apply properties of perpendicular bisectors of a triangle and prove and apply properties of angle bisectors of a triangle.
Vocabulary
concurrent
point of concurrency
circumcenter of a triangle
circumscribed
incenter of a triangle
inscribed
Since a triangle has three sides, it has three perpendicular bisectors. When you construct the perpendicular bisectors, you find that they have an interesting property.
The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex.
When three or more lines intersect at one point, the lines are said to be concurrent. The point of concurrency is the point where they intersect. In the construction, you saw that the three perpendicular bisectors of a triangle are concurrent. This point of concurrency is the circumcenter of the triangle.
The circumcenter of ΔABC is the center of its circumscribed circle. A circle that contains all the vertices of a polygon is circumscribed about the polygon.
Example 1: Using Properties of Perpendicular Bisectors
G is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of
∆ABC.
GC = CB
Circumcenter Thm.
Substitute 13.4 for GB.
GC = 13.4
MZ is a perpendicular bisector of ∆GHJ.
Check It Out! Example 1a
Use the diagram. Find GM.
GM = MJ
Circumcenter Thm.
Substitute 14.5 for MJ.
GM = 14.5
KZ is a perpendicular bisector of ∆GHJ.
Check It Out! Example 1b
Use the diagram. Find GK.
GK = KH
Circumcenter Thm.
Substitute 18.6 for KH.
GK = 18.6
Check It Out! Example 1c
Use the diagram. Find JZ.
Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of
∆GHJ.
JZ = GZ
Circumcenter Thm.
Substitute 19.9 for GZ.
JZ = 19.9
Example 2: Finding the Circumcenter of a Triangle
Find the circumcenter of ∆HJK with vertices H(0, 0), J(10, 0), and K(0, 6).
Step 1 Graph the triangle.
Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of HJ is x = 5, and the perpendicular bisector of HK is y = 3.
Example 2 Continued
Step 2 Find equations for two perpendicular bisectors.
Example 2 Continued
Step 3 Find the intersection of the two equations.
The lines x = 5 and y = 3 intersect at (5, 3), the circumcenter of ∆HJK.
Check It Out! Example 2
Find the circumcenter of ∆GOH with vertices G(0, –9), O(0, 0), and H(8, 0) .
Step 1 Graph the triangle.
Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of GO is y = –4.5, and the perpendicular bisector of OH is
x = 4.
Check It Out! Example 2 Continued
Step 2 Find equations for two perpendicular bisectors.
Check It Out! Example 2 Continued
Step 3 Find the intersection of the two equations.
The lines x = 4 and y = –4.5 intersect at (4, –4.5), the circumcenter of ∆GOH.
A triangle has three angles, so it has three angle bisectors. The angle bisectors of a triangle are also concurrent. This point of concurrency is the incenter of the triangle.
Remember!
The distance between a point and a line is the length of the perpendicular segment from the
point to the line.
The incenter is the center of the triangle’s inscribed circle. A circle inscribedin a polygon intersects each line that contains a side of the polygon at exactly one point.
The distance from P to LM is 5. So the distance from P to MN is also 5.
Example 3A: Using Properties of Angle Bisectors
P is the incenter of ∆LMN. By the Incenter Theorem, P is equidistant from the sides of ∆LMN.
PL is the bisector of MLN.
PM is the bisector of LMN.
Example 3B: Using Properties of Angle Bisectors
MP and LP are angle bisectors of ∆LMN. Find mPMN.
mMLN = 2mPLN
mMLN = 2(50°)= 100°
Substitute 50° for mPLN.
mMLN + mLNM + mLMN = 180°
ΔSum Thm.
100+ 20 + mLMN = 180
Substitute the given values.
Subtract 120° from both sides.
mLMN = 60°
Substitute 60° for mLMN.
The distance from X to PR is 19.2. So the distance from X to PQ is also 19.2.
Check It Out! Example 3a
X is the incenter of ∆PQR. By the Incenter Theorem, X is equidistant from the sides of ∆PQR.
XR is the bisector of QRY.
QX is the bisector of PQR.
Check It Out! Example 3b
QX and RX are angle bisectors of ∆PQR. Find mPQX.
mQRY= 2mXRY
mQRY= 2(12°)= 24°
Substitute 12° for mXRY.
mPQR + mQRP + mRPQ = 180°
∆ Sum Thm.
mPQR+ 24 + 52= 180
Substitute the given values.
Subtract 76° from both sides.
mPQR = 104°
Substitute 104° for mPQR.
Example 4: Community Application
A city planner wants to build a new library between a school, a post office, and a hospital. Draw a sketch to show where the library should be placed so it is the same distance from all three buildings.
Let the three towns be vertices of a triangle. By the Circumcenter Theorem, the circumcenter of the triangle is equidistant from the vertices.
Draw the triangle formed by the three buildings. To find the circumcenter, find the perpendicular bisectors of each side. The position for the library is the circumcenter.
Check It Out! Example 4
A city plans to build a firefighters’ monument in the park between three streets. Draw a sketch to show where the city should place the monument so that it is the same distance from all three streets. Justify your sketch.
By the Incenter Thm., the incenter of a ∆ is equidistant from the sides of the ∆. Draw the ∆ formed by the streets and draw the bisectors to find the incenter, point M. The city should place the monument at point M.
2.JP, KP, and HP are angle bisectors of ∆HJK. Find the distance from P to HK.
Lesson Quiz: Part I
17
3
Lesson Quiz: Part II
3. Lee’s job requires him to travel to X, Y, and Z. Draw a sketch to show where he should buy a home so it is the same distance from all three
places.
|
# Exponent rules
The quantities are expressed in exponential notation in mathematics for some reasons. For doing some mathematical activities with exponential functions, the fundamental formulas of mathematics may not be useful. Hence, it requires some special properties and they are called the rules of exponents. Now, let’s learn the following three types of exponential properties with proofs to study the indices (or exponents) mathematically.
### Product Rules
There are two types of product laws in mathematics and they are used to multiply a quantity in exponential form by another quantity in exponential notation.
$(1).\,\,\,$ $b^{\displaystyle m} \times b^{\displaystyle n} \,=\, b^{\displaystyle \,m+n}$
$(2).\,\,\,$ $b^{\displaystyle m} \times c^{\displaystyle m} \,=\, (b \times c)^{\displaystyle m}$
### Quotient Rules
There are two types of division laws in mathematics, they are used to divide a quantity in exponential form by another quantity in exponential form.
$(1).\,\,\,$ $\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}} \,=\, b^{\displaystyle \, m-n}$
$(2).\,\,\,$ $\dfrac{b^{\displaystyle m}}{c^{\displaystyle m}} \,=\, \bigg(\dfrac{b}{c}\bigg)^{\displaystyle m}$
### Power Rules
In mathematics, there are six types of power rules to find value of a quantity with exponent.
$(1).\,\,\,$ $\big(b^{\displaystyle m}\big)^{\displaystyle n} \,=\, b^{\displaystyle \, m \times n}$
$(2).\,\,\,$ $b^{\displaystyle -m} \,=\, \dfrac{1}{b^{\displaystyle m}}$
$(3).\,\,\,$ $b^{\dfrac{1}{n}} \,=\, \sqrt[\displaystyle n]{b\,\,}$
$(4).\,\,\,$ $b^{\dfrac{m}{n}} \,=\, \sqrt[\displaystyle n]{b^{\displaystyle m}}$
$(5).\,\,\,$ $b^0 \,=\, 1$
$(6).\,\,\,$ $b^1 \,=\, b$
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# Calculator Output
```Simplifying
(y + 9)(y + -2) = 4y
Reorder the terms:
(9 + y)(y + -2) = 4y
Reorder the terms:
(9 + y)(-2 + y) = 4y
Multiply (9 + y) * (-2 + y)
(9(-2 + y) + y(-2 + y)) = 4y
((-2 * 9 + y * 9) + y(-2 + y)) = 4y
((-18 + 9y) + y(-2 + y)) = 4y
(-18 + 9y + (-2 * y + y * y)) = 4y
(-18 + 9y + (-2y + y2)) = 4y
Combine like terms: 9y + -2y = 7y
(-18 + 7y + y2) = 4y
Solving
-18 + 7y + y2 = 4y
Solving for variable 'y'.
Reorder the terms:
-18 + 7y + -4y + y2 = 4y + -4y
Combine like terms: 7y + -4y = 3y
-18 + 3y + y2 = 4y + -4y
Combine like terms: 4y + -4y = 0
-18 + 3y + y2 = 0
Factor a trinomial.
(-6 + -1y)(3 + -1y) = 0
Subproblem 1Set the factor '(-6 + -1y)' equal to zero and attempt to solve:
Simplifying
-6 + -1y = 0
Solving
-6 + -1y = 0
Move all terms containing y to the left, all other terms to the right.
Add '6' to each side of the equation.
-6 + 6 + -1y = 0 + 6
Combine like terms: -6 + 6 = 0
0 + -1y = 0 + 6
-1y = 0 + 6
Combine like terms: 0 + 6 = 6
-1y = 6
Divide each side by '-1'.
y = -6
Simplifying
y = -6
Subproblem 2Set the factor '(3 + -1y)' equal to zero and attempt to solve:
Simplifying
3 + -1y = 0
Solving
3 + -1y = 0
Move all terms containing y to the left, all other terms to the right.
Add '-3' to each side of the equation.
3 + -3 + -1y = 0 + -3
Combine like terms: 3 + -3 = 0
0 + -1y = 0 + -3
-1y = 0 + -3
Combine like terms: 0 + -3 = -3
-1y = -3
Divide each side by '-1'.
y = 3
Simplifying
y = 3Solutiony = {-6, 3}```
Processing time: 0 ms. 41550031 equations since February 08, 2004. Disclaimer
# Equation Factoring Calculator
Equation: Variable: a A b B c C d D e E f F g G h H i I j J k K l L m M n N o O p P q Q r R s S t T u U v V w W x X y Y z Z AUTO Simplify Only
Hint: Selecting "AUTO" in the variable box will make the calculator automatically solve for the first variable it sees.
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Back
Volume of a Right Circular Cylinder
> ... Math > Circles and Pi > Volume of a Right Circular Cylinder
When we find the volume of a rectangular object, we find the area of the base and multiply it by the height. We do the same thing to find the volume of a cylinder only this time the base is a circle. We find the area of the base $$\pi r^2$$ and multiply that by the height of the cylinder. Here is some vocabulary to help with this lesson.
• Right Angle = This is the same thing as perpendicular, two lines come together at 90 degrees like the corner of a rectangle.
• Right Circular Cylinder = A shape like a tube, the ends (or base) form a circle and the sides are perpendicular to the base.
Volume of a Right Circular Cylinder
Let $$r = radius$$ and $$h = height$$
$$Volume\:\ of\:\ a\:\ Right\:\ Circular\:\ Cylinder = \pi r^2h$$
### Practice Problems
1. A can of food is a right circular cylinder with a radius of 5 cm and a height of 16 cm. Find the volume of the can. Round your answer to the nearest tenth. (
Solution
Solution:
$$1256.6 \text{ cm}^{3}$$ (when using the pi button on the calculator)
)
2. A paint can is a right circular cylinder with a radius of 3.5 inches and a height of 7.5 inches. Find the volume of the paint can. Round your answer to the nearest hundredth. (
Solution
Solution:
$$288.63 \text{ in}^{3}$$ (when using the pi button on the calculator)
)
3. A water tower is used to pressurize the water supply for the distribution of water in the surrounding area. A particular water tower is in the shape of a right circular cylinder with a radius of 4.25 meters and a height of 7.5 meters. Find the volume of the water tower. Round your answer to the nearest whole number. (
Solution
Solution: $$426 \: {\text{m}}^{3}$$ (when using the pi button on the calculator)
Details:
We are finding the volume of a water tower with a radius of 4.25 meters and a height of 7.5 meters, so we can use the formula for the volume of a right cylinder:
$$\text{Volume} = {\text{π}} {\text{ r}}^{2}{\text{h}}$$
The first thing we need to do is substitute or replace ‘r’ with 4.25 m and ‘h’ with 7.5 m
$$\text{Volume} = {\text{π}}(4.25\:{\text{m}}){^{2}}(7.5\:{\text{m}})$$
Next, we square $$4.25\: {\text{m}}$$ to get $$18.0625 \:{\text{m}}^{2}$$ (This means multiply $$4.25\:{\text{m}} \times 4.25\:{\text{m}}$$)
$$\text{Volume} = {\text{π}}(18.0625\:{\text{m}}^{2})(7.5\:{\text{m}})$$
Then we multiply $$18.0625 \:{\text{m}}^{2}$$ by $$7.5\: {\text{m}}$$, which gives us $$135.46875 \:{\text{m}}^{3}$$. Remember $${\text{m}}^{2}$$ times $${\text{m}}$$ equals $${\text{m}}^{3}$$ and tells us that we are measuring the volume in cubic meters.
$$\text{Volume} = {\text{π}} 135.46875\:{\text{m}}^{3}$$
Since we can multiply in any order, we can rewrite the equation like this, which is an acceptable mathematical answer:
$$\text{Volume} = 135.47{\text{π}} \:{\text{m}}^{3}$$ (Here we also rounded to the nearest hundredth place for simplicity.)
We can also multiply 135.46875 by $$π$$ to get:
$$\text{Volume} = 425.5876... {\text{m}}^{3}$$
So the volume of the water tower is approximately $$426 \:{\text{m}}^{3}$$ when rounded to the nearest whole number.
)
4. A 55-gallon drum is in the shape of a right circular cylinder with a diameter of 22.5 inches and a height of 33.5 inches. First, find the radius of the drum and then use the radius to find the volume of the drum. Round your answer to the nearest hundredth. (
Video Solution
Solution:
$$\text{Radius} = 11.25 \text{ in}$$
$$\text{Volume} = 13319.86 \text{ in}^{3}$$ (when using the pi button on the calculator)
Details:
(Video Source | Transcript)
)
5. A support column on a building is a right circular cylinder. It has a radius of 1.5 feet and a height of 16 feet. Find the volume of the column. Round your answer to the nearest whole number. (
Solution
Solution: $$113 \text{ ft}^{3}$$ (when using the pi button on the calculator)
Details:
To find the volume of the column we can use the formula:
$$\text{Volume} = {\text{π}} {\text{ r}}^{2}{\text{h}}$$
The first thing we need to do is replace the ‘r’ with 1.5 ft and the ‘h’ with 16 ft.
$$\text{Volume} = {\text{π}} (1.5\:{\text{ft}}){^{2}}(16\:\text{ft})$$
Next, we need to square 1.5 ft. This means multiply $$1.5 {\text{ft}} \times 1.5 {\text{ft}}$$ which equals $$2.25 \:{\text{ft}}^{2}$$
$$\text{Volume} = {\text{π}} (2.25\:\text{ ft}^{2})(16\:\text{ft})$$
Then we multiply $$2.25\: \text{ ft}^{2}$$ by $$16\:\text{ft}$$, which equals $$36\:\text{ft}^{3}$$. Remember $$\text{ft}^{2}$$ times $$\text{ft}$$ equals $$\text{ft}^{3}$$
$$\text{Volume} = {\text{π}} 36\:\text{ft}^{3}$$
Since we can multiply in any order, we can rewrite the volume like this:
$$\text{Volume} = 36{\text{π}}\:\text{ft}^{3}$$
When we multiply 36 by $$π$$ we get:
$$\text{Volume} = 113.0973... {\text{ft}}^{3}$$
So the volume of the column is about $$113\: \text{ft}^{3}$$ when rounded to the nearest whole number.
)
6. A triple-A battery is a right circular cylinder with a radius of 5.25 mm and a height of 44.5 mm. Find the volume of the battery. Round to the nearest tenth. (
Video Solution
Solution: $$3853.3 \text{ mm}^{3}$$ (when using the pi button on the calculator)
Details:
(Video Source | Transcript)
)
## Need More Help?
1. Study other Math Lessons in the Resource Center.
2. Visit the Online Tutoring Resources in the Resource Center.
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# The mean life of a sample of 60 bulbs was 650
Question:
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Solution:
Given the mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours
Now we have to find the overall standard deviation
As per given criteria, in first set of samples,
Number of sample bulbs, n1=60
Standard deviation, s1=8hrs
Mean life, $\bar{x}_{1}=650$
And in second set of samples,
Number of sample bulbs, $\mathrm{n}_{2}=80$
Standard deviation, $s_{2}=7 \mathrm{hr}$ s
Mean life, $\bar{x}_{2}=660$
We know the standard deviation for combined two series is
S. D $(\sigma)=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$
Substituting the corresponding values, we get
S. $D(\sigma)=\sqrt{\frac{(60)(8)^{2}+(80)(7)^{2}}{60+80}+\frac{(60 \times 80)(650-660)^{2}}{(60+80)^{2}}}$
$\mathrm{S} . \mathrm{D}(\sigma)=\sqrt{\frac{(60) 64+(80) 49}{140}+\frac{(4800)(10)^{2}}{(140)^{2}}}$
S. D $(\sigma)=\sqrt{\frac{(60) 64+(80) 49}{140}+\frac{(4800) 100}{19600}}$
On simplifying we get
S. D $(\sigma)=\sqrt{\frac{(6) 64+(8) 49}{14}+\frac{(4800) 1}{196}}$
S. D $(\sigma)=\sqrt{\frac{384+392}{14}+\frac{4800}{196}}$
S. D $(\sigma)=\sqrt{\frac{388}{7}+\frac{1200}{49}}$
S. D $(\sigma)=\sqrt{\frac{388 \times 7+1200}{49}}$
S. D $(\sigma)=\sqrt{\frac{2716+1200}{49}}$
S. D $(\sigma)=\sqrt{\frac{3916}{49}}$
Or, σ=8.9
Hence the standard deviation of the set obtained by combining the given two sets is 8.9
|
# RD Sharma Solutions Class 12 Straight Line In Space Exercise 28.2
RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space Exercise 28.2, is provided here for students to prepare for exams at ease. Practising the textbook questions will help you in analysing your level of preparation and knowledge of the concept. Students can download the RD Sharma Class 12 Solutions pdf for free from the links provided below.
## Download PDF of Rd Sharma Solution for Class 12 Maths Chapter 28 Exercise 2
### Access Answers for Rd Sharma Solution Class 12 Maths Chapter 28 Exercise 2
Q1.
Solution:
Let us consider,
Now let us simplify, we get
Hence, we can say that the lines are mutually perpendicular.
Q2.
Solution:
Given points are (1, -1, 2) and (3, 4, -2)
The direction ratios of a line passing through the points will be
(3-1, 4+1, -2-2) = (2, 5, -4)
Given points are (0, 3, 2) and (3, 5, 6)
The direction ratios of a line passing through the points will be
(3-0, 5-3, 6-2) = (3, 2, 4)
So,
Now the angle between the lines will be
Hence, we can say that the lines are mutually perpendicular.
Q3.
Solution:
Given points are (4, 7, 8) and (2, 3, 4)
The direction ratios of a line passing through the points will be
(4-2, 7-3, 8-4) = (2, 4, 4)
Given points are (-1, -2, 1) and (1, 2, 5)
The direction ratios of a line passing through the points will be
(-1-1, -2-2, 1-5) = (-2, -4, -4)
So,
The direction ratios are proportional.
2/-2 = 4/-4 = 4/-4
Hence, we can say that the lines are mutually perpendicular.
Q4.
Solution:
It is given that,
The Cartesian equation of a line passing through (x1, y1, z1) and with direction ratios (a1, b1, c1) is
Q5.
Solution:
It is given that.
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# Ugly Number II LeetCode Solution
Difficulty Level Medium
## Problem Statement
Ugly Number II LeetCode Solution – An ugly number is a positive integer whose prime factors are limited to `2``3`, and `5`.
Given an integer `n`, return the `n`th ugly number.
```Input: n = 10
Output: 12
Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers.```
## Explanation
We have an array k of the first n ugly number. We only know, in the beginning, the first one, which is 1. Then
k[1] = min( k[0]x2, k[0]x3, k[0]x5). The answer is k[0]x2. So we move 2’s pointer to 1. Then we test:
k[2] = min( k[1]x2, k[0]x3, k[0]x5). And so on. Be careful about the cases such as 6, in which we need to forward both pointers of 2 and 3.
x here is multiplication.
## Approach
Let us solve this problem for the general case: that is not only for `2,3,5` divisors, but for any of them and any number of them. `factors = [2,3,5]` and `k=3` in our case.
Let `Numbers` be an array, where we keep all our ugly numbers. Also, note, that any ugly number is some other ugly number, multiplied by `2``3` or `5`. So, let `starts` be the indexes of ugly numbers, that when multiplied by `2``3` or `5` respectively, produces the smallest ugly number that is larger than the current overall maximum ugly number. Let us do several first steps to understand it better:
1. `starts = [0,0,0]` for numbers `2,3,5`, so `new_num = min(1*2,1*3,1*5) = 2`, and now `starts = [1,0,0]``Numbers = [1,2]`.
2. `starts = [1,0,0]`, so `new_num = min(2*2,1*3,1*5) = 3`, and now `starts = [1,1,0]``Numbers = [1,2,3]`.
3. `starts = [1,1,0]`, so `new_num = min(2*2,2*3,1*5) = 4`, so now `starts = [2,1,0]``Numbers = [1,2,3,4]`.
4. `starts = [2,1,0]`, so `new_num = min(3*2,2*3,1*5) = 5`, so now `starts = [2,1,1]``Numbers = [1,2,3,4,5]`.
5. `starts = [2,1,1]`, so `new_num = min(3*2,2*3,2*5) = 6`, so let us be careful in this case: we need to increase two numbers from `start`, because our new number `6` can be divided both by `2` and `3`, so now `starts = [3,2,1]``Numbers = [1,2,3,4,5,6]`.
6. `starts = [3,2,1]`, so `new_num = min(4*2,3*3,2*5) = 8`, so now `starts = [4,2,1]``Numbers = [1,2,3,4,5,6,8]`
7. `starts = [4,2,1]`, so `new_num = min(5*2,3*3,2*5) = 9`, so now `starts = [4,3,1]``Numbers = [1,2,3,4,5,6,8,9]`.
8. `starts = [4,3,1]`, so `new_num = min(5*2,4*3,2*5) = 10`, so we need to update two elements from `starts` and now `starts = [5,3,2]``Numbers = [1,2,3,4,5,6,8,9,10]`
9. `starts = [5,3,2]`, so `new_num = min(6*2,4*3,3*5) = 12`, we again need to update two elements from `starts`, and now `starts = [6,4,2]``Numbers = [1,2,3,4,5,6,8,9,10,12]`.
10. `starts = [6,4,2]`, so `new_num = min(8*2,5*3,3*5) = 15`, we again need to update two elements from `starts`, and now `starts = [6,5,3]``Numbers = [1,2,3,4,5,6,8,9,10,12,15]`.
## Code
### C++ Code For Ugly Number II
```class Solution {
public:
int nthUglyNumber(int n) {
if(n <= 0) return false; // get rid of corner cases
if(n == 1) return true; // base case
int t2 = 0, t3 = 0, t5 = 0; //pointers for 2, 3, 5
vector<int> k(n);
k[0] = 1;
for(int i = 1; i < n ; i ++)
{
k[i] = min(k[t2]*2,min(k[t3]*3,k[t5]*5));
if(k[i] == k[t2]*2) t2++;
if(k[i] == k[t3]*3) t3++;
if(k[i] == k[t5]*5) t5++;
}
return k[n-1];
}
};```
### Java Code For Ugly Number II
```public class Solution {
public int nthUglyNumber(int n) {
int[] ugly = new int[n];
ugly[0] = 1;
int index2 = 0, index3 = 0, index5 = 0;
int factor2 = 2, factor3 = 3, factor5 = 5;
for(int i=1;i<n;i++){
int min = Math.min(Math.min(factor2,factor3),factor5);
ugly[i] = min;
if(factor2 == min)
factor2 = 2*ugly[++index2];
if(factor3 == min)
factor3 = 3*ugly[++index3];
if(factor5 == min)
factor5 = 5*ugly[++index5];
}
return ugly[n-1];
}
}```
### Python Code For Ugly Number II
```class Solution:
def nthUglyNumber(self, n):
factors, k = [2,3,5], 3
starts, Numbers = [0] * k, [1]
for i in range(n-1):
candidates = [factors[i]*Numbers[starts[i]] for i in range(k)]
new_num = min(candidates)
Numbers.append(new_num)
starts = [starts[i] + (candidates[i] == new_num) for i in range(k)]
return Numbers[-1]```
## Complexity Analysis for Ugly Number II LeetCode Solution
O(N)
### Space Complexity
O(N)
Reference: https://en.wikipedia.org/wiki/Factor
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AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra InText Questions and Answers.
## AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra InText Questions
Let’s Explore (Page No. 102)
Question 1.
Arrange 2 matchsticks to form the shape Continue the same shape for 2 times, 3 times and 4 times. Frame the rule for repeating the pattern.
Solution:
To make the given shape 2 matchsticks are needed.
To make the given 2 shapes 4 matchsticks are needed.
To make the given 3 shapes 6 matchsticks are needed.
To make the given 4 shapes 8 matchsticks are needed.
Continue and arrange the information in the following table.
Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × x = 2x
Question 2.
Rita took matchsticks to form the shape
She repeated the pattern and gave a rule.
Number of matchsticks needed = 6.y, where y is the number of shapes to be formed. Is it correct ? Explain.
What is the number of sticks needed to form 5 such shapes ?
Solution:
To make the given shape 6 matchsticks are needed.
To make the given 2 shapes 12 matchsticks are needed.
To make the given 3 shapes 18 matchsticks are needed.
Continue and arrange the information in the following table.
Yes, it is correct.
Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × y = 2y
Number of matchsticks needed to form 5 such shapes = 6 × 5 = 30
Let’s Explore (Page No. 103)
Question 1.
A line of shapes is constructed using matchsticks.
Shape-1 Shape-2 Shape-3 Shape-4
i) Find the rule that shows how many sticks are needed to make a line of such shapes ?
ii) How many matchsticks are needed to form shape -12 ?
Solution:
Number of matchsticks 3 5 7 9
i) Let us know the pattern
S1 = 3 = 2 + 1 = (1 × 2) + 1
S2 = 5 = 4 + 1 = (2 × 2) + 1
S3 = 7 = 6 + 1 = (3 × 2) + 1
S4 – 9 = 8 + 1 = (4 × 2) + 1
Now the rule for this pattern is number of matchsticks.
ii) Used to make ‘n’ number of shapes is Sn = (n × 2) + 1 = 2n + 1
Number of matchsticks needed to form shape – 12 is
S12 = 2(12) + 1 = 24 + 1 = 25 sticks.
Check Your Progress (Page No. 105)
Question 1.
Fill the following table as instructed. One is shown for you.
S.No. Expression Verbal Form 1. y + 3 Three more than y 2. 2x – 1 3. 5z 4.
Solution:
S.No. Expression Verbal Form 1. y + 3 Three more than y 2. 2x – 1 One less than the double of x 3. 5z 5 times of z 4. Half of the m
Let’s Explore ? (Page No. 106)
Question 1.
Find the general rule for the perimeter of a rectangle. Use variables T and ‘b’ for length and breadth of the rectangle respectively.
Solution:
Given length of rectangle = l
We know that the perimeter of rectangle is twice the sum of its length and breadth.
Sum of length and breadth = l + b
Twice the sum of length and breadth = 2 × (l + b)
Rule for the perimeter of a rectangle = 2(l + b)
Question 2.
Find the general rule for the area of a square by using the variable ‘s’ for the side of a square.
Given side of a square = s
We know that the area of a square is the product of side and side.
Area of a square = side × side
Rule for the area of a square = s.s
Side Area 1 1 × 1 2 2 × 2 3 3 × 3 4 4 × 4 ……….. ………… s S × s
(Page No. 107)
Question 1.
Find the nth term in the following sequences.
0 3, 6, 9, 12, ii) 2, 5, 8, 11, iii) 1, 4, 9, 16,
Solution:
i) Given number pattern is 3, 6, 9, 12,……………..
To find the nth term in the given pattern, we put the sequence in a table.
First number = 3 × 1
Second number = 3 × 2
nth number = 3 × n = 3n
So, the nth term of the pattern 3, 6, 9, 12, is 3n.
ii) Given number pattern is 2, 5, 8, 11,
To find the nth term in the given pattern, we put the sequence in a table.
First number = 2 = 3 × 1 – 1
Second number = 5 = 3 × 2 – 1
Third number = 8 = 3 × 3 – 1
nth number = 3 × n – 1 = 3n – 1
So, the nth term of the pattern 2, 5, 8, 11 is 3n – 1.
iii) Given number pattern is 1, 4, 9, 16,
To find the nth term in the given pattern, we put the sequence in a table.
First number =1 = 1 × 1
Second number = 4 = 2 × 2
Third number =9 = 3 × 3
nth number = n × n = n2
So, the nth term of the pattern 1, 4, 9, 16 is n2.
Check Your Progress (Page No. 108)
Question 1.
Complete the table and find the value of ‘p’ for the equation $$\frac{\mathbf{p}}{\mathbf{3}}$$ = 4
p $$\frac{\mathbf{p}}{3}$$ = 4 Condition satisfied ? Yes/ No 3 6 9 12
Solution:
p $$\frac{\mathbf{p}}{3}$$ = 4 Condition satisfied ? Yes/ No 3 $$\frac{3}{3}$$ ≠1 ≠ 4 No 6 $$\frac{6}{3}$$ ≠2 ≠ 4 No 9 $$\frac{9}{3}$$ ≠ 3 ≠ 4 No 12 $$\frac{12}{3}$$ ≠ 4 ≠ 4 Yes
Question 2.
Write LHS and RHS of following simple equations.
i) 2x + 1 = 10
ii) 9 = y – 2
iii) 3p + 5 = 2p + 10
Solution;
i) 2x+ 1 = 10
Given equation is 2x + 1 = 10
L.H.S = 2x + 1
R.H.S = 10
ii) 9 = y – 2
Given equation is 9 = y – 2
LHS = 9
RHS = y – 2
iii) 3p + 5 = 2p + 10
Given equation is 3p + 5 = 2p + 10
LHS = 3p + 5
RHS = 2p + 10
Question 3.
Write any two simple equations and write their LHS and RHS.
Solution:
i) Consider 8x + 3 = 4 is a simple equation.
L.H.S = 8x + 3
RHS = 4
ii) Consider 5a + 6 = 8a – 3 is a simple equation.
LHS = 5a + 6
RHS = 8a – 3
Let’s Explore (Page No. 109)
Observe for what value of m, the equation 3m = 15 has both LHS and RHS become equal.
Solution:
Given equation is 3m = 15
If m = 1, then the value of 3m = 3(1) = 3≠15 ∴ LHS ≠RHS
If m = 2, then the value of 3m = 3(2) = 6 ≠ 15 ∴ LHS ≠ RHS
If m = 3, then the value of 3m = 3(3) = 9≠15 ∴ LHS ≠ RHS
If m = 4, then the value of 3m = 3(4) = 12 ≠ 15 ∴ LHS ≠ RHS
If m = 5, then the value of 3m = 3(5) = 15 = 15 ∴ LHS = RHS
From the above when m = 5 the both LHS and RHS are equal
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# Examples to find Least Common Multiple by using Prime Factorization Method
Examples to find least common multiple by using prime factorization method are discussed here.
We write the prime factorization of each of the given numbers. Then, the required LCM of these numbers is the product of all different prime factors of the numbers using the greatest power of each common prime factor.
1. What is the least common multiple (L.C.M) of 21 and 49 by using prime factorization method?
Solution:
To find the LCM, multiply all prime factors. But the common factors are included only once.
21 = 3 × 7.
49 = 7 × 7 = 7².
= 3 × 7² = 3 × 7 × 7 = 147.
The required least common multiple (L.C.M) of 21 and 49 = 98.
2. What is the least common multiple (L.C.M) of 36 and 14 by using prime factorization method?
Solution:
To find the LCM, multiply all prime factors. But the common factors are included only once.
36 = 2 × 2 × 3 × 3 = 2² × 3².
14 = 2 × 7.
= 2² × 3² × 7 = 2 × 2 × 3 × 3 × 7 = 252.
The required least common multiple (L.C.M) of 36 and 14 = 252.
3. What is the least common multiple (L.C.M) of 5, 4 and 16 by using prime factorization method?
Solution:
To find the LCM, multiply all prime factors. But the common factors are included only once.
5 = 5 × 1.
4 = 2 × 2.
16 = 2 × 2 × 2 × 2 = 2⁴.
= 2⁴ × 5 = 2 × 2 × 2 × 2 × 5 = 80.
The required least common multiple (L.C.M) of 5, 4 and 16 = 80.
4. Find L.C.M of 504 and 594 by prime factorization method.
Solution:
To find the LCM, multiply all prime factors. But the common factors are included only once.
504 = 2 × 2 × 2 × 3 × 3 × 7 = 2³ × 3³ × 7.
594 = 2 × 3 × 3 × 3 × 11 = 2 × 3³ × 11.
= 2³ × 3³ × 7 × 11 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 11 = 16632.
The required least common multiple (L.C.M) of 504 and 594 = 16632.
These are the four examples to find least common multiple by using prime factorization method.
To Find Lowest Common Multiple by using Division Method
Relationship between H.C.F. and L.C.M.
Worksheet on H.C.F. and L.C.M.
Word problems on H.C.F. and L.C.M.
Worksheet on word problems on H.C.F. and L.C.M.
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Chapter 1 Number System R.D. Sharma Solutions for Class 9th Math Exercise 1.4
x
Exercise 1.4
1. Define an irrational number .
Solution
An irrational number is a real number that cannot be reduced to any ratio between an integer p and a natural number q. If the decimal representation of an irrational number is non-terminating and non-repeating, then it is called irrational number. For example, √3 = 1.732....
2. Explain, how irrational numbers differ from rational numbers ?
Solution
Every rational number must have either terminating or non-terminating but irrational number must have non- terminating and non-repeating decimal representation. A rational number is a number that can be written as simple fraction (ratio) and denominator is not equal to zero while an irrational is a number that cannot be written as a ratio.
3. Examine, whether the following numbers are rational or irrational:
Solution
(i) Let x = √7
Therefore,
x = 2.645751311064...
It is non-terminating and non-repeating
Hence, √7 is an irrational number
(ii) Let x = √4
Therefore,
x = 2
It is terminating.
Hence √4 is a rational number.
(iii) Let x = 2 +√3 be the rational
Squaring on both sides,
⇒ x2 = (2+√3)2
⇒ x2 = 4+3+4√3
⇒ x2 = 7+4√3
⇒ x2-7 = 4√3
⇒ (x2-7)/4 = √3
Since, x is rational
⇒ x2 is rational.
⇒ x2-7 is rational
⇒ (x2-7)/4 is rational
⇒ √3 is rational
But, √3 is irrational
So, we arrive at a contradiction.
Hence 2+√3 is an irrational number.
(iv) Let x = √3+√2 be the rational number
Squaring on both sides, we get
⇒ x2 = (√3+√2)2
⇒ x2 = 3+2+2√6
⇒ x2 = 5+2√6
⇒ x2-5 = 2√6
⇒ (x2-5)/2 = √6
Since, x is a rational number
⇒ x2 is rational number
⇒ x2 -5 is rational number
⇒ (x2-5)/2 is rational number
⇒ √6 is rational number
But √6 is an irrational number
Hence, √3+√2 is an irrational number
(v) Let x = √3+√5 be the rational number
Squaring on both sides, we get
⇒ x2 = (√3+√5)2
⇒ x2 = 3+5+2√15
⇒ x2 = 8+2√15
⇒ x2 -8 = 2√15
⇒ (x2 -8)/2 = √15
Now, x is rational number
⇒ x2 is rational number
⇒ x2-8 is rational number
(x2 -8)/2 is rational number
⇒ √15 is rational number.
But √15 is an irrational number
So, we arrive at a contradiction
Hence, √3+√5 is an irrational number
(vi) Let x = (√2-2)2 be a rational number.
x = (√2-2)2
⇒ x = 2+4-4√2
⇒ x = 6-4√2
⇒ x-6/-4 = √2
Since, x is rational number,
⇒ x – 6 is a rational nummber
⇒ x-6/-4 is a rational number
⇒ √2 is a rational number
But we know that √2 is an irrational number, which is a contradiction
So, (√2-2)2 is an irrational number
(vii) Let x = (2-√2) (2+√2)
⇒ x = 22 - √2 {as (a+b) (a-b) = 22 - 22}
⇒ x = 4-2
⇒ x = 2
So, (2-√2) (2+√2) is a rational number
(viii) Let x = (√2+√3)2 be rational number
Using the formula, (a+b)2 = 22+ 22+2ab
⇒ x = √22+ √32+2.√2.√3
⇒ x = 2+3+2√6
⇒ x = 5+2√6
⇒ x-5/2 = √6
⇒ x-5/2 is a rational number
⇒ √6 is a rational number
But we know that √6 is an irrational number
So, we arrive at a contradiction
So, (√2+√3)2 is an irrational number.
(x) Let
x = √23
⇒ x = 4.79583...
It is non-terminating or non-repeating
Hence, √23 is an irrational number
(xi) Let x = √225
⇒ x = 15
Hence √225 is a rational number
(xii) Given 0.3796
It is terminating
Hence, it is a rational number
(xiii) Given number x = 7.478478
⇒ x = 71428
It is repeating
Hence, it is a rational number
(xiv) Given number is x = 1.1010010001...
It is non-terminating or non-repeating
Hence, it is an irrational number
4. Identify the following as rational or irrational numbers. Give the decimal representation of rational numbers:
Solution
(i) Given number is x = √4
x = 2, which is a rational number
(iii) Given number is √1.44
Now, we have to check whether it is rational or irrational.
⇒ √1.44 = √(144/100)
⇒ √1.44 = √144/100
⇒ √1.44 = 12/10
⇒ √1.44 = 6/5
⇒ √1.44 = 1.2
So, it is a rational
(iv) Given that √(9/27) = √9/27
Now, we have to check whether it is rational or irrational
√(9/27) = √9/27
⇒ √(9/27) = 3/3√3
⇒ √(9/27) = 1/√3
So, it is an irrational number
(v) Given that √-64
Now, we have to check whether it is rational or irrational
Since, -√64 = -8
So, it is a rational number
(vi) Given that √100
Now, we have to check whether it is rational or irrational
Since, √100 = 10
So, it is rational number
5. In the following equations, find which variables x, y, z etc. represent rational or irrational numbers:
(i) x2 = 5
(ii) y2 = 9
(iii) z2 = 0.004
(iv) u2 = 17/4
(v) v2 = 3
(vi) w2 = 27
(vii) t2 = 0.4
Solution
(i) Given that x2 = 5
Now, we have to find the value of x.
Since x2 = 5
⇒ x = √5
So, it x is an irrational number
(ii) Given that y2 = 9
Now, we have to find the value of y.
y2 = 9
⇒ y = √9
⇒ y = 3
So, y is a rational number
(iii) Given that z2 = 0.04
Now, we have to find the value of z.
⇒ z2 = 4/100
⇒ z = √(4/100)
⇒ z = 2/10
⇒ z = 1/5
So, it is rational number
(iv) Given that u2 = 17/4
Now, we have to find the value of u.
u2 = 17/4
⇒ u = √(17/4)
⇒ u = √(17/2)
So, it is an irrational number
(v) Given that, v2 = 3
Now, we have to find the value of v
v2 = 3
⇒ v = √3
So, it is an irrational number
(vi) Given that, w2 = 27
Now, we have to find the value of w.
⇒ w = √27
w = 3√3
So, it is an irrational number
(vii) Given that t2 = 0.4
Now, we have to find the value of t
⇒ t = √0.4
⇒ t = √(4/10)
⇒ t = 2/√10
So, it is an irrational number
6. Give an example of each, of two irrational numbers whose:
(i) difference is a rational number.
(ii) difference is an irrational number.
(iii) sum is a rational number.
(iv) sum is an irrational number.
(v) product is an rational number.
(vi) product is an irrational number.
(vii) quotient is a rational number.
(viii) quotient is an irrational number.
Solution
(i) Let √2, 1+√2
So, 1+√2 - √2 = 1
Therefore, √2 and 1+√2 are two irrational numbers and their difference is a rational number.
(ii) Let 4√3, 3√3 are two irrational numbers and their difference is an irrational number.
Because 4√3- 3√3 = √3 is an irrational number.
(iii) Let √5, -√5 are two irrational numbers and their sum is a rational number.
That is √5+(-√5) = 0.
(iv) Let 2√5, 3√5 are two irrational numbers and their sum is an irrational number.
That is 2√5 + 3√5 = 5√5
(v) Let √8, √2 are two irrational numbers and their product is a rational number.
That is √8×√2 = √16 = 4
(vi) Let √2, √3 are two irrational numbers and their product is an irrational number
That is √2×√3 = √6
(vii) Let √8, √2 are two irrational numbers and their quotient is a rational number
That is √8/√2 = 2√2/√2 = 2
(viii) Let √2, √3 are two irrational numbers and their quotient is an irrational number
That is √2፥√3 = √2/√3
7. Give two rational numbers lying between 0.232332333233332... and 0.212112111211112.
Solution
Let,
a = 0.232332333233332…..
b = 0.212112111211112…..
Here, the decimal representation of a and b are non-terminating and non-repeating. So we observe that in first decimal place of a and b have the same digit 2 but digit in the second place of their decimal representation are distinct. And the number a has 3 and b has 1. So a > b.
Hence, two rational numbers are 0.222 lying between 0.232332333233332….. and 0.212112111211112…..
8. Give two rational numbers lying between 0.515115111511115...0.5353353335...
Solution
Let a = 0.515115111511115… and b = 0.535335333533335….
Here, the decimal representation of a and b are non-terminating and non-repeating. So we observe that in first decimal place a and b have the same digit 5 but digit in the second place of their decimal representation are distinct. And the number a has 1 and b has 3. So a < b.
Hence, two rational numbers are 0.5152,0.532 lying between 0.515115111511115….and 0.535335333533335….
9.
Solution
Let,
a = 0.2101
b = 0.2222….
Here, a and are rational numbers .Since a has terminating and b has repeating decimal. We observe that in second decimal place a has 1 and b has 2. So a < b.
Hence one irrational number is 0.220100100010000…. lying between 0.2101 and 0.2222….
10. Find a rational number and also an irrational number lying between the numbers 0.3030030003 ... and 0.3010010001 ...
Solution
Let,
a = 0.3030030003…
b = 0.3010010001….
Here, decimal representation of a and b are non-terminating and non-repeating. So a and b are irrational numbers. We observe that in first two decimal place of a and b have the same digit but digit in the third place of their decimal representation is distinct. Therefore, a > b.
Hence one rational number is 0.3011 lying between 0.3030030003….. and 0.3010010001… and irrational number is 0.3020200200020000… lying between 0.3030030003….and 0.3010010001….
11. Find two irrational numbers between 0.5 and 0.55.
Solution
Let ,
a = 0.5
b = 0.55
Here, a and b are rational number. So we observe that in first decimal place a and b have same digitSo a < b. Hence two irrational numbers are 0.510100100010000… and 0.5202002000200002… lying between 0.5 and 0.55.
12. Find two irrational numbers lying between 0.1 and 0.12.
Solution
Let,
a = 0.1
b = 0.12
Here, and b are rational number. So we observe that in first decimal place a and b have same digit. So a < b.
Hence, two irrational numbers are 0.1010010001… and 0.11010010001…lying between 0.1 and 0.12.
Solution
Given that √3+√5 is an irrational number
Now, we have to prove √3+√5 is an irrational number
Let x = √3+√5 is a rational.
Squaring on both sides,
x2 = (√3 + √5)2
⇒ x2 = (√3)2 + (√5)2 + 2√3×√5
⇒ x2 = 3+5+2√15
⇒ (x2 = 8 + 2√15
⇒ (x2-8)/2 = √15
Now, x is rational
⇒ x2 is rational
⇒ (x2-8)/2 is rational
⇒ √15 is rational
But, √15 is an irrational
Thus, we arrive at contradiction that √3+√5 is a rational which is wrong.
Hence, √3+√5 is an irrational
Solution
Let x = 5/7 = 0.714285 and y = 9/11 = 0.81
Here, we observe that in the first decimal x has digit 7 and y has 8. So x < y. In the second decimal place x has digit 1. So, if we considering irrational numbers
a = 0.72072007200072..
b = 0.73073007300073..
c = 0.74074007400074....
We find that,
x<a<b<c<y
Hence, a,b,c are required irrational numbers.
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Math Central Quandaries & Queries
Question from ARUN: dear Sir, please advise me , how to calculate the surface area of a pipe with diameter of 630 mm and thickness of 67 mm which is cut in a angle of 22.5 degree. please show me how to calculate the surface area of the pipe which cut in an angle. Thanking you.
Hi Arun,
Suppose you have a circular cylinder of diameter $d$ mm and height $h$ mm and you slice it from the bottom right to the top left as in the diagram. By the symmetry of the circle, the surface area of the piece removed (the green piece) is equal to the surface area of the piece you keep (the pink piece). Hence the surface area of the piece you keep is half the surface area of the cylinder.
To find the surface area of the cylinder slice it along one of the vertical lines and roll it out flat. What you get is a rectangle of height $h$ mm and length, the circumference of the base circle which is $\pi \; d$ mm and hence the surface area of the piece you keep is $\frac12 \pi \; d \times h$ square mm.
For your question you know $d = 630 \mbox{ mm}$ so all that remains is to find $h.$ For this we can use some trigonometry.
A vertical cross-section of the cut pipe shows a right triangle $ABC$ with $CA$ the diameter of the pipe, $BC$ the height of the cylinder and angle $CAB$ the angle of the cut which in your case is $22.5^{o}.$
The tangent of the angle $CAB$ is the length of $BC$ divided by the length of $CA$ and hence
$\tan 22.5^o = \frac{h}{d}$
or
$h = d \times \tan 22.5^o = 630 \times 0.4142 = 260.95 \mbox{ mm.}$
Thus the surface area of the pink piece in my diagram is
$\frac12 \pi \; d \times h = \frac12 \pi \; \times 630 \times 260.95 = 258,241 mm^{2}.$
Harley
Math Central is supported by the University of Regina and the Imperial Oil Foundation.
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# 05. Matrix multiplication, 2
## 05. Matrix multiplication, 2
### Examples
We're going to do some examples of matrix multiplication.
Example:
Consider the 90 degree rotation matrix M equals 0, minus 1; 1, 0. We have M squared equals 0, minus 1; 1, 0 times 0, minus 1; 1, 0, which equals minus 1, 0; 0, minus 1.
This makes sense: two 90 degree rotations compose to give a 180 degree rotation, which sends every point x, y to its opposite point minus x, minus y.
Example:
More generally, if R alpha = cos alpha, minus sine alpha, sine alpha, cos alpha and R beta = cos beta, minus sine beta, sine beta, cos beta are two rotations then the composite is R alpha times R beta equals cos alpha, minus sine alpha, sine alpha, cos alpha times cos beta, minus sine beta, sine beta, cos beta, which equals cos alpha cos beta minus sine alpha sine beta, minus cos alpha sine beta minus sine alpha cos beta; sine alpha cos beta plus cos alpha sine beta, minus sine alpha sine beta plus cos alpha cos beta, which equals cos (alpha + beta), minus sine (alpha + beta); sine (alpha + beta), cos (alpha + beta), which equals R (alpha + beta) (using trigonometric addition formulas). This is what we expect, of course: rotating by beta and then alpha amounts to rotating by alpha + beta.
Example:
Let I equals 1, 0; 0, 1 be the identity matrix and M be any matrix. Then I M equals 1, 0; 0, 1 times M_{1 1}, M_{1 2}; M_{2 1}, M_{2 2}, which equals M_{1 1}, M_{1 2}; M_{2 1}, M_{2 2}, which equals M. Similarly, M I = M. As you can see, the identity matrix really plays the role of the number 1 here.
Example:
Let A be the matrix 1, 0; 0, 0 and B be the matrix 0, 1; 0, 0 . Then A B equals 0, 1; 0, 0 but B A equals 0, 0; 0, 0 This shows that the order in which we multiply matrices matters: A B is not equal to B A. So matrix multiplication is not commutative!
As an exercise, can you think of a matrix B which does not commute with A equals 1, 1; 0, 1?
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# How do you simplify log_3 27 + 6log_3 9?
Mar 27, 2016
${\log}_{3} 27 + 6 {\log}_{3} 9 = \textcolor{g r e e n}{15}$
#### Explanation:
Always remember when dealing with log functions
$\textcolor{w h i t e}{\text{XXX}} {\log}_{b} a = c \Leftrightarrow {b}^{c} = a$
So
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{{\log}_{3} 27 = {c}_{1}}$
$\textcolor{w h i t e}{\text{XXXXXX}}$means ${3}^{{c}_{1}} = 27 \rightarrow \textcolor{red}{{c}_{1} = 3}$
and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{\log}_{3} 9 = {c}_{2}}$
$\textcolor{w h i t e}{\text{XXXXXX}}$means ${3}^{{c}_{2}} = 9 \rightarrow \textcolor{b l u e}{{c}_{2} = 2}$
Therefore
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{{\log}_{3} 27} + 6 \textcolor{b l u e}{{\log}_{3} 9}$
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{3} + 6 \cdot \textcolor{b l u e}{2}$
$\textcolor{w h i t e}{\text{XXX}} = 15$
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# Step By Step Calculus » 2.8 - Rational Functions
Synopsis
A ratio of polynomials is called a rational function. This is a natural analogue of rational numbers. Polynomials have many similarities to the integers: there is a zero polynomial, they factor, there are prime polynomials, and the algebra is similar to that of the integers. So it is natural to call a fraction or ratio of polynomials a rational function. The only exception is that the denominator in a rational function can not be a zero polynomial. For example, \dfrac{x^2+3}{3}, \dfrac{x^3-x+2}{x-2}\dfrac{x^2+3}{3}, \dfrac{x^3-x+2}{x-2} are polynomial fractions but \dfrac{x^3+8}{0}\dfrac{x^3+8}{0} is not.
Multiplying Rational Functions If \dfrac{n(x)}{d(x)}\dfrac{n(x)}{d(x)} and \dfrac{a(x)}{b(x)}\dfrac{a(x)}{b(x)} are two rational functions, then
\displaystyle \frac{n(x)}{d(x)}\cdot\frac{a(x)}{b(x)}=\frac{n(x)a(x)}{d(x)b(x)}.
\displaystyle \frac{n(x)}{d(x)}\cdot\frac{a(x)}{b(x)}=\frac{n(x)a(x)}{d(x)b(x)}.
Adding Rational Functions If \dfrac{n(x)}{d(x)}\dfrac{n(x)}{d(x)} and \dfrac{a(x)}{b(x)}\dfrac{a(x)}{b(x)} are two rational functions, then
\displaystyle \frac{n(x)}{d(x)}+\frac{a(x)}{b(x)}=\frac{n(x)b(x)+a(x)d(x)}{d(x)b(x)}.
\displaystyle \frac{n(x)}{d(x)}+\frac{a(x)}{b(x)}=\frac{n(x)b(x)+a(x)d(x)}{d(x)b(x)}.
Proper Rational Function A rational function \displaystyle \frac{n(x)}{d(x)}\displaystyle \frac{n(x)}{d(x)} is proper if the degree of n(x)n(x) is strictly less than the degree of d(x)d(x).
Polynomial Long Division It is a method for expressing an improper rational function \displaystyle \frac{n(x)}{d(x)}\displaystyle \frac{n(x)}{d(x)} in the form \displaystyle q(x)+\frac{r(x)}{d(x)}\displaystyle q(x)+\frac{r(x)}{d(x)} where q(x)q(x) is called the quotient and r(x)r(x) is called the remainder, and the ratio \dfrac{r(x)}{d(x)}\dfrac{r(x)}{d(x)} is proper.
Remainder Theorem If a polynomial f(x)f(x) is divided by a linear polynomial x-cx-c, then the remainder rr is equal to f(c)f(c). Thus, \displaystyle \frac{f(x)}{x-c}=q(x)+\frac{f(c)}{x-c}\displaystyle \frac{f(x)}{x-c}=q(x)+\frac{f(c)}{x-c}.
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# GMAT Tip: Counting Problems
Photograph by Michele Constantini
The GMAT Tip of the Week is a weekly column that includes advice on taking the Graduate Management Admission Test, which is required for admission to most business schools. Every week an instructor from a top test prep company will share suggestions for improving your GMAT score. This week’s tip comes from Brent Hanneson, creator of GMAT Prep Now , a Web site offering on-demand videos that teach GMAT skills.
Whenever I see a GMAT resource label its counting section as “Combinations and Permutations,” a small part of me dies. OK, that’s an exaggeration, but I am concerned about the misleading message that this label conveys. To me, it suggests that counting questions can be solved using either permutations or combinations, when this is not the case at all.
A permutation is a rearrangement of objects or values. For example, permutations of the numbers 1, 2, 3 include 3, 2, 1, as well as 2, 1, 3 and 2, 3, 1. The truth of the matter is that true permutation questions are exceedingly rare on the GMAT. In fact, if you examine the 12th and 13th editions of the Official Guide for GMAT Review, you will not find any questions that require knowledge of permutations. So, given the rarity of these question types, it’s dangerous to approach a counting question with the notion that you need only determine whether you’re dealing with a combination or a permutation and then apply one of two formulas. If you do this, you will inevitably conclude that a question is a permutation question when it is not.
My advice is to completely erase the permutation formula from your memory and approach all counting questions as follows:
First ask, “Can I use the Fundamental Counting Principle (FCP) to answer this question?” If the answer is yes, then apply the FCP. If the answer is no, look for another approach (which will likely involve an application of the combination formula, or a mixture of combinations and the FCP).
The great thing about the FCP is that it can be used to solve many GMAT counting questions. More important, it’s almost painfully easy to use. There are only three steps.
1. Take a counting task and break it into stages.
2. For each stage, determine the number of ways to accomplish that stage.
3. Multiply the number of ways to accomplish each stage. The product is your answer.
For example, imagine a problem that asks you to determine the total number of cars that can be created using three different power sources (gas, electric, hybrid), three different colors (red, black, and blue) and two different transmissions (automatic, standard). Rather than painstakingly figuring out all the combinations, simply break the problem into stages (power source, color, transmission), determine the number of possibilities for each (3, 3 and 2). Multiplying the three numbers yields the correct answer, 18.
Brent Hanneson, the creator of GMAT Prep Now, has worked in the field of education for most of his career. He has taught courses at three different test prep companies and created comprehensive GMAT and GRE courseware packages used by the University of British Columbia and 12 other universities across North America.
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# Discrete Probability Distribution : Part 1
## What are discrete random variables and discrete probability distribution?
In probability and statistics, a random variable is a quantitative variable whose value depends on the outcomes of a random phenomenon. For example, let us toss a coin 3 times and let X represents the number of tails that appear in that 3 tosses. Then X can take any one of the following values : 0, 1, 2, 3. Here X will be the random variable as it represents the chances or outcomes of tossing the coin. We don’t know the results until the coin is tossed and calculate the number of tails. Here the variable X will be discrete. This means it can take only exact values.
Discrete random variables can take on a finite or countable number of possible values whereas in the case of continuous random variables, they can take any or infinite number of values in an interval.
We can also list down the probabilities of occurring of all the possible outcomes for tossing the coin 3 times. The possible outcomes are : HHH, THH, HTH, HHT, HTT, THT, TTH, and TTT. Here H represents head and T represents the tail.
The above picture represents the probability distribution of random variable X. It is a listing of all possible values of X and their probabilities of occurring, p(x). If we repeat a random experiment many times and plot probabilities of each possible outcome, then we will get probability distribution.
A discrete probability distribution must satisfy the following two conditions :
1. 0≤ p(x) ≤ 1, for all values of x.
2. ∑ p(x) = 1
In our example, ∑ p(x) = 0.125+0.375 +0.375+ 0.125 = 1
# The expected value, variance, and standard deviation of discrete probability distribution :
The expected value or expectation of a random variable is the long-term theoretical mean of the random variable. We can calculate the expected value, E(X) by multiplying each value of X with their respective probability of occurring and then sum up all of them.
# E(X=x) = ∑ x. p(x)
For example, say one of your friends offers you to play a game. A dice will be rolled and the following are the rules for the game :
1. If it comes up 1, you have to give him Rs.1.
2. If it comes up 2, he will give you Rs. 2.
3. If it comes up 3, you have to give him Rs.3.
4. If it comes up 4, he will give you Rs. 4.
5. If it comes up 5, you have to give him Rs.5.
6. If it comes up 6, he will give you Rs. 6.
Would you play the game? You will play the game only if it is profitable. To know this we have to calculate expected winning or loss from the game. The probability distribution for the game will be as follows :
E(X) = (-1 x 1/6) +(2 x 1/6) +(-3 x 1/6) + (4 x 1/6) + (-5 x 1/6) +(6 x 1/6)
= 3/6
= Rs. 0.5
Since the expected value is Rs. 0.5, it will be a profitable game. In other words, you can expect to win Rs. 0.5 for each game. If you play the game 1000 times, then the amount you should expect to win will be (0.5 x1000) = Rs. 500. The same game was simulated 1000 times and the following two pictures are the results of the simulation which are showing total winning and average winning per game after playing the game 1000 times.
The above pictures clearly show that after playing the game 1000 times, the total winning is nearly equal to Rs.500 and average winning per game is near Rs.0.5. This is what we expected. The long term mean of the winnings is matching with the expected value and the total winning is also close to the expected Rs. 500. Initially, both total winning and average winning per game fluctuate a lot. But in long term, these tend to deviate less from the expectation.
The expected value gives us the typical or average value of a variable. But it does not tell anything about how the values are spread out. Here comes the variance which will tell us about the data spread.
# Var (X) = E(x-µ)² = ∑(x-µ)².p(x)
Let us take the above example of game again. Here, E(X) or µ is 0.5.
Hence, Var(X) = (-1- 0.5)² x 1/6 + (2 - 0.5)² x 1/6 + (-3- 0.5)² x 1/6 + (4 - 0.5)² x 1/6 + (-5- 0.5)² x 1/6 + (6 - 0.5)² x 1/6 = 14.92
Similarly, we can also calculate standard deviation according to following :
# Std. Dev = √Variance
Measures of variability and spread | by Dhrubjun | Nerd For Tech | Sep, 2021 | Medium
So, for the above example, standard deviation will be √14.92 = 3.86. That means on average, our winning per game will be 3.86 away from the expectation of 0.5.
That’s all for today. In next part, we will learn about different discrete probability distributions. Till then keep smiling. 😀
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## More from Nerd For Tech
NFT is an Educational Media House. Our mission is to bring the invaluable knowledge and experiences of experts from all over the world to the novice. To know more about us, visit https://www.nerdfortech.org/.
## Dhrubjun
Machine learning/Data Science enthusiast
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# Lesson 9
Increasing and Decreasing Functions
• Let’s look at what a graph does based on a situation.
### 9.1: Comparing Values
For each pair of numbers, write $$=,<$$, or $$>$$ in the blank to make a true equation or inequality. Be prepared to share your reasoning.
1. -6 $$\underline{\hspace{.5in}}$$ -9
2. $$\frac{7}{3}\ \underline{\hspace{.5in}}\ \frac{13}{6}$$
3. 5.2 $$\underline{\hspace{.5in}}\ \frac{53}{11}$$
4. $$5 (3 - 6)\ \underline{\hspace{.5in}}\ 15 - 6$$
5. Let $$f(x) = 5 - 2x$$.
1. $$f(3)\ \underline{\hspace{.5in}}\ f(5)$$
2. $$f(\text{-}3)\ \underline{\hspace{.5in}}\ f(\text{-}4)$$
3. $$f(\text{-}1)\ \underline{\hspace{.5in}}\ f(1)$$
### 9.2: What Could It Be?
Describe $$f(x)$$ and $$g(x)$$ with a situation that could fit the given graphs. Explain your reasoning.
### 9.3: Cities, Towns, and Villages
Draw an example of a graph that shows two functions as they are described. Make sure to label the functions.
1. The population of 2 cities as functions of time so that city A always has more people than city B.
2. The population of 2 towns as functions of time so that town A is larger to start, but then town B gets larger.
3. The population of 2 villages as functions of time so that village A has a steady population and village B has a population that is initially large, but decreases.
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Cubic polynomials and their roots
# Cubic polynomials and their roots
We have a look at cubic polynomial functions.
Just as for quadratic functions, knowing the zeroes of a cubic makes graphing it much simpler. Typically a cubic function will have three zeroes or one zero, at least approximately, depending on the position of the curve. Finding these zeroes, however, is much more of a challenge. In fact this challenge was a historical highlight of 16th century mathematics.
In this step we will
• see how Descartes’ factor theorem applies to cubic functions
• review how to find zeroes of a cubic in special situations.
## Zeroes of a cubic polynomial
One simple way of cooking up a cubic polynomial is just to take a product of linear factors, for example
$\Large{y=(x-4)(x-1)(x+2).}$
We hope that you can all expand the right-hand side to get
$\Large{y=x^3-3x^2-6x+8.}$
The graph of this function is shown here
Note that it does pass through the points $$\normalsize{[4,0]}$$, $$\normalsize{[1,0]}$$ and $$\normalsize{[-2,0]}$$, as it should by Descartes’ Factor theorem.
## Factorising simple cubics
Here is a simple cubic polynomial that has been chosen to have a nice factorisation: $$\normalsize{f(x)=x^3-7x+6}.$$
Let us note that the curve passes through the points $$\normalsize{[1,0]}$$, $$\normalsize{[2,0]}$$ and $$\normalsize{[-3,0]}$$. This corresponds to the fact that $$\normalsize{f(1)=f(2)=f(-3)=0}$$. We say that $$\normalsize{1}$$, $$\normalsize{2}$$ and $$\normalsize{-3}$$ are the zeroes or roots of $$\normalsize{f(x)}$$.
The zeroes of a polynomial $$\normalsize{f(x)}$$ have a particular importance: from Descartes’ theorem if a polynomial $$\normalsize{p(x)}$$ has a zero $$\normalsize{r}$$, that is $$\normalsize{p(r)=0}$$, then it follows that $$\normalsize{(x-r)}$$ is a factor of $$\normalsize{p}$$.
In our example, since $$\normalsize{f}$$ has zeroes $$\normalsize{1}$$, $$\normalsize{2}$$ and $$\normalsize{-3}$$, we know that $$\normalsize{f(x)}$$ has factors $$\normalsize{(x-1)}$$, $$\normalsize{(x-2)}$$ and $$\normalsize{(x+3)}$$. And indeed
$\Large{f(x)=(x-1)(x-2)(x+3)}.$
So in practice trial and error may be used when looking for possible zeroes by hand. It is a good guess to try factors of the constant term if the polynomial is monic (that means that the highest degree coefficient is $$\normalsize{1}$$). Of course using a computer is much simpler: our electronic friends are particular adept at this kind of work!
Q1 (M): Try to factor $$\normalsize{p(x)=x^3-3x^2+4x-70}$$.
If you made progress with this factorisation, then you should find a factor of $$\normalsize (x^2+2x+14)$$ appearing, and you will find it resists further attempts to factor it. Perhaps it doesn’t factor, but how would we know that the quadratic term does not factor further?
One way is to keep trying the various possible factors, but a better way is to use the quadratic formula. The discriminant of this quadratic is $$\normalsize{b^2-4ac=2^2-4(1)(14)=-39}$$ which is not a square, so there is no rational factorisation. We say that $$\normalsize{x^2+2x+14}$$ is irreducible (over the rational numbers) and leave it at that.
Q2 (M): Find zeroes and factorisations of the following cubics:
a) $$\normalsize{g(x)= x^3-4x}$$
b) $$\normalsize{h(x)=x^3+4x^2-x-4}$$
c) $$\normalsize{k(x)= 2x^3-7x^2-14x-5}$$
Knowing the factorisation of a cubic makes it easier to graph, since we know where it crosses the $$\normalsize{x}$$-axis.
## Tartaglia, Cardano and factoring general cubics
Suppose someone gave you the cubic function
$\Large{y=(x-7)(x-53)(x+13)= x^3-47x^2-409x+4823}$
and asked you to factor it. Not knowing the left hand side of the equation, it might take some work to find the factors. But what if the cubic does not factor nicely into factors? Just by changing the cubic a little to
$\Large{y= x^3-47x^2-409x+4822}$
makes things vastly more complicated! This is just the kind of challenge that 16th century mathematicians like G. Cardano (1501-1576) and N. Tartaglia (1500-1557) gave each other.
During one of these challenges, Tartaglia discovered a general formula for solving cubics which extends the much more familiar quadratic formula. His formula is quite remarkable in that it requires complex numbers in an essential way, and also both square roots and cube roots. However it is not for the faint-hearted, and it is fair to say that these days it is rarely used.
Girolamo Cardano, By Wellcome Library, London, CC BY 4.0, via Wikimedia Commons
The story of how Cardano (pictured) stole the credit for this formula from Tartaglia is one of the most juicy episodes in the history of mathematics. For this story and other historical information on cubics, check out Norman’s History of Mathematics lectures on his YouTube channel.
Tartaglia’s cubic formula is workable if your example has been chosen carefully to have linear factors in the first place. But a random cubic does not have such a property, and so most cubics can in fact only be approximately factored.
A1. Let’s look at this problem in some detail. To factor $$\normalsize{p(x)=x^3-3x^2+4x-70}$$ one should look for zeroes which are factors of $$\normalsize{70}$$, which are built from $$\normalsize{2,5}$$ and $$\normalsize{7}$$, and we should not forget about negative possible factors. Now with a bit of trial and error you can find that $$\normalsize{p(5)=0}$$, so by Descartes’ theorem we know that $$\normalsize{(x-5)}$$ is a factor.
After that we want to actually divide $$\normalsize{p(x)}$$ by $$\normalsize{(x-5)}$$. This requires polynomial long division. If you have forgotten how to do long division, you will be happy to learn that polynomial long division is generally simpler than that with numbers.
By looking at successive powers of x, we sequentially determine that the other factor must be $$\normalsize{x^2+2x+14}$$. Please try it! The final factoring is
$\Large{p(x)=(x-5)(x^2+2x+14)}.$
A2.
a) We have $$\normalsize{g(x)= x^3-4x=x(x-2)(x+2)}$$.
b) We have $$\normalsize{h(x)=x^3+4x^2-x-4=(x-1)(x+1)(x+4)}$$.
c) We have $$\normalsize{k(x)= 2x^3-7x^2-14x-5=(x-5)(x+1)(2x+1)}$$.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.3: Factoring Trinomials
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
skills to develop
• Factor trinomials of the form $$x^{2}+bx+c$$.
• Factor trinomials of higher degree.
• Factor trinomials of the form $$ax^{2}+bx+c$$.
• Factor trinomials using the AC method.
## Factoring Trinomials of the Form $$x^{2}+bx+c$$
Some trinomials of the form $$x^{2}+bx+c$$ can be factored as a product of binomials. If a trinomial of this type factors, then we have:
\begin{aligned} x ^ { 2 } + b x + c & = ( x + m ) ( x + n ) \\ & = x ^ { 2 } + n x + m x + m n \\ & = x ^ { 2 } + ( n + m ) x + m n \end{aligned}
This gives us
$$b=n+m$$ and $$c=mn$$
In short, if the leading coefficient of a factorable trinomial is $$1$$, then the factors of the last term must add up to the coefficient of the middle term. This observation is the key to factoring trinomials using the technique known as the trial and error (or guess and check) method18.
Example $$\PageIndex{1}$$:
Factor $$x^{2}+12x+20$$.
Solution
We begin by writing two sets of blank parentheses. If a trinomial of this form factors, then it will factor into two linear binomial factors.
$$x ^ { 2 } + 12 x + 20 = ( \quad ) ( \quad)$$
Write the factors of the first term in the first space of each set of parentheses. In this case, factor $$x^{2}=x⋅x$$.
$$x ^ { 2 } + 12 x + 20 = ( x \quad ) ( x\quad )$$
Determine the factors of the last term whose sum equals the coefficient of the middle term. To do this, list all of the factorizations of $$20$$ and search for factors whose sum equals $$12$$.
\begin{aligned} 20 & = 1 \cdot 20 \:\:\rightarrow\:\: 1 + 20 = 21 \\ & =\color{OliveGreen}{ 2 \cdot 10 \:\:\rightarrow\:\: 2 + 10}\color{black}{ =}\color{OliveGreen}{ 12} \\ & = 4 \cdot 5 \:\:\:\:\rightarrow\:\: 4 + 5 = 9 \end{aligned}
Choose $$20 = 2 ⋅ 10$$ because $$2 + 10 = 12$$. Write in the last term of each binomial using the factors determined in the previous step.
$$x ^ { 2 } + 12 x + 20 = ( x + 2 ) ( x + 10 )$$
This can be visually interpreted as follows:
Figure $$\PageIndex{1}$$
Check by multiplying the two binomials.
\begin{aligned} ( x + 2 ) ( x + 10 ) & = x ^ { 2 } + 10 x + 2 x + 20 \\ & = x ^ { 2 } + 12 x + 20 \:\:\color{Cerulean}{✓} \end{aligned}
$(x+2)(x+10) \nonumber$
Since multiplication is commutative, the order of the factors does not matter.
\begin{aligned} x ^ { 2 } + 12 x + 20 & = ( x + 2 ) ( x + 10 ) \\ & = ( x + 10 ) ( x + 2 ) \end{aligned}
If the last term of the trinomial is positive, then either both of the constant factors must be negative or both must be positive.
Example $$\PageIndex{2}$$:
Factor $$x ^ { 2 } y ^ { 2 } - 7 x y + 12$$.
Solution:
First, factor $$x^{2}y^{2}=xy⋅xy$$.
$$x ^ { 2 } y ^ { 2 } - 7 x y + 12 = ( x y \quad \color{Cerulean}{?}\color{black}{ )} ( x y \quad\color{Cerulean}{?}\color{black}{ )}$$
Next, search for factors of $$12$$ whose sum is $$−7$$.
\begin{aligned} 12 & = 1 \cdot 12 \:\:\rightarrow\:\: - 1 + ( - 12 ) = - 13 \\ & = 2 \cdot 6 \:\:\:\:\rightarrow\:\: - 2 + ( - 6 ) = - 8 \\ & = \color{OliveGreen}{3 \cdot 4\:\:\:\: \rightarrow\:\: - 3 + ( - 4 ) = - 7} \end{aligned}
In this case, choose $$−3$$ and $$−4$$ because $$(−3)(−4)=+12$$ and $$−3+(−4)=−7$$.
\begin{aligned} x ^ { 2 } y ^ { 2 } - 7 x y + 12 & = ( x y\quad \color{Cerulean}{?}\color{black}{ )} ( x y\quad\color{Cerulean}{ ?}\color{black}{ )} \\ & = ( x y - 3 ) ( x y - 4 ) \end{aligned}
Check
\begin{aligned} ( x y - 3 ) ( x y - 4 ) & = x ^ { 2 } y ^ { 2 } - 4 x y - 3 x y + 12 \\ & = x ^ { 2 } y ^ { 2 } - 7 x y + 12 \:\:\color{Cerulean}{✓} \end{aligned}
$$( x y - 3 ) ( x y - 4 )$$
If the last term of the trinomial is negative, then one of its factors must be negative.
Example $$\PageIndex{3}$$:
Factor: $$x ^ { 2 } - 4 x y - 12 y ^ { 2 }$$.
Solution
Begin by factoring the first term $$x ^ { 2 } = x \cdot x$$.
$$x ^ { 2 } - 4 x y - 12 y ^ { 2 } = \left( \begin{array} { l l } { x } & { \color{Cerulean}{?} } \end{array} \right) \left( \begin{array} { l l } { x } & { \color{Cerulean}{?} } \end{array} \right)$$
The factors of $$12$$ are listed below. In this example, we are looking for factors whose sum is $$−4$$.
\begin{aligned} 12 & = 1 \cdot 12 \:\:\rightarrow\:\: 1 + ( - 12 ) = - 11 \\ & =\color{OliveGreen}{ 2 \cdot 6\:\:\:\: \rightarrow\:\: 2 + ( - 6 ) = - 4} \\ & = 3 \cdot 4\:\:\:\: \rightarrow\:\: 3 + ( - 4 ) = - 1 \end{aligned}
Therefore, the coefficient of the last term can be factored as $$−12=2(−6)$$, where $$2+(−6)=−4$$. Because the last term has a variable factor of $$y^{2}$$, use $$−12y^{2}=2y(−6y)$$ and factor the trinomial as follows:
\begin{aligned} x ^ { 2 } - 4 x y - 12 y ^ { 2 } & = ( x \quad \color{Cerulean}{?}\color{black}{ )} ( x \quad \color{Cerulean}{?}\color{black}{ )} \\ & = ( x + 2 y ) ( x - 6 y ) \end{aligned}
Multiply to check.
\begin{aligned} ( x + 2 y ) ( x - 6 y ) & = x ^ { 2 } - 6 x y + 2 y x - 12 y ^ { 2 } \\ & = x ^ { 2 } - 6 x y + 2 x y - 12 y ^ { 2 } \\ & = x ^ { 2 } - 4 x y - 12 y ^ { 2 } \:\:\color{Cerulean}{✓}\end{aligned}
$$( x + 2 y ) ( x - 6 y )$$
Often our first guess will not produce a correct factorization. This process may require repeated trials. For this reason, the check is very important and is not optional.
Example $$\PageIndex{4}$$:
Factor $$a ^ { 2 } + 10 a - 24$$.
Solution
The first term of this trinomial, $$a^{2}$$, factors as $$a⋅a$$.
$$a ^ { 2 } + 10 a - 24 = \left( \begin{array} { l l } { a } & { ? } \end{array} \right) \left( \begin{array} { l l } { a } & { ? } \end{array} \right)$$
Consider the factors of $$24$$:
\begin{aligned} 24 & = 1 \cdot 24 \\ & = \color{OliveGreen}{2 \cdot 12} \\ & = 3 \cdot 8 \\ & = \color{red}{4 \cdot 6} \end{aligned}
Suppose we choose the factors $$4$$ and $$6$$ because $$4 + 6 = 10$$, the coefficient of the middle term. Then we have the following incorrect factorization:
$$a ^ { 2 } + 10 a - 24 \stackrel{\color{red}{?}}{\color{black}{=}} ( a + 4 ) ( a + 6 ) \:\:\color{red}{Incorrect\:Factorization}$$
When we multiply to check, we find the error.
\begin{aligned} ( a + 4 ) ( a + 6 ) & = a ^ { 2 } + 6 a + 4 a + 24 \\ & = a ^ { 2 } + 10 a \color{red}{+ 24}\:\:\color{red}{✗} \end{aligned}
In this case, the middle term is correct but the last term is not. Since the last term in the original expression is negative, we need to choose factors that are opposite in sign. Therefore, we must try again. This time we choose the factors $$−2$$ and $$12$$ because $$−2+12=10$$.
$$a ^ { 2 } + 10 a - 24 = ( a - 2 ) ( a + 12 )$$
Now the check shows that this factorization is correct.
\begin{aligned} ( a - 2 ) ( a + 12 ) & = a ^ { 2 } + 12 a - 2 a - 24 \\ & = a ^ { 2 } + 10 a \color{OliveGreen}{- 24}\:\:\color{Cerulean}{✓} \end{aligned}
$$( a - 2 ) ( a + 12 )$$
If we choose the factors wisely, then we can reduce much of the guesswork in this process. However, if a guess is not correct, do not get discouraged; just try a different set of factors. Keep in mind that some polynomials are prime. For example, consider the trinomial $$x^{2}+3x+20$$ and the factors of $$20$$:
\begin{aligned} 20 & = 1 \cdot 20 \\ & = 2 \cdot 10 \\ & = 4 \cdot 5 \end{aligned}
There are no factors of $$20$$ whose sum is $$3$$. Therefore, the original trinomial cannot be factored as a product of two binomials with integer coefficients. The trinomial is prime.
## Factoring Trinomials of Higher Degree
We can use the trial and error technique to factor trinomials of higher degree.
Example $$\PageIndex{5}$$:
Factor $$x ^ { 4 } + 6 x ^ { 2 } + 5$$.
Solution
Begin by factoring the first term $$x ^ { 4 } = x ^ { 2 } \cdot x ^ { 2 }$$.
$$x ^ { 4 } + 6 x ^ { 2 } + 5 = \left( x ^ { 2 } \quad \color{Cerulean}{?} \right) \left( x ^ { 2 } \quad \color{Cerulean}{?} \right)$$
Since $$5$$ is prime and the coefficient of the middle term is positive, choose $$+1$$ and $$+5$$ as the factors of the last term.
\begin{aligned} x ^ { 4 } + 6 x ^ { 2 } + 5 & = \left( x ^ { 2 } \quad \color{Cerulean}{?} \right) \left( x ^ { 2 } \quad \color{Cerulean}{?} \right) \\ & = \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } + 5 \right) \end{aligned}
Notice that the variable part of the middle term is $$x^{2}$$ and the factorization checks out.
\begin{aligned} \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } + 5 \right) & = x ^ { 4 } + 5 x ^ { 2 } + x ^ { 2 } + 5 \\ & = x ^ { 4 } + 6 x ^ { 2 } + 5\:\:\color{Cerulean}{✓} \end{aligned}
$$\left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } + 5 \right)$$
Example $$\PageIndex{6}$$:
Factor: $$x ^ { 2 n } + 4 x ^ { n } - 21$$ where $$n$$ is a positive integer.
Solution
Begin by factoring the first term $$x ^ { 2 n } = x ^ { n } \cdot x ^ { n }$$.
$$x ^ { 2 n } + 4 x ^ { n } - 21 = \left( \begin{array} { l l } { x ^ { n } } & { \color{Cerulean}{?} } \end{array} \right) \left( \begin{array} { l l } { x ^ { n } } & {\color{Cerulean}{ ?} } \end{array} \right)$$
Factor $$- 21 = 7 ( - 3 )$$ because $$7 + ( - 3 ) = + 4$$ and write
\begin{aligned} x ^ { 2 n } + 4 x ^ { n } - 21 & = \left( x ^ { n } \quad \color{Cerulean}{?} \right) \left( x ^ { n } \quad \color{Cerulean}{?} \right) \\ & = \left( x ^ { n } + 7 \right) \left( x ^ { n } - 3 \right) \end{aligned}
$\left( x ^ { n } + 7 \right) \left( x ^ { n } - 3 \right) \nonumber$
The check is left to the reader.
Exercise $$\PageIndex{1}$$
Factor $$x ^ { 6 } - x ^ { 3 } - 42$$.
$\left( x ^ { 3 } + 6 \right) \left( x ^ { 3 } - 7 \right) \nonumber$
## Factoring Trinomials of the Form $$ax^{2}+bx+c$$
Factoring trinomials of the form $$ax^{2}+bx+c$$ can be challenging because the middle term is affected by the factors of both $$a$$ and $$c$$. In general,
\begin{aligned} \color{Cerulean}{a}\color{black}{ x} ^ { 2 } + \color{Cerulean}{b}\color{black}{ x} + \color{Cerulean}{c} & = ( p x + m ) ( q x + n ) \\ & = p q x ^ { 2 } + p n x + q m x + m n \\ & = \color{Cerulean}{p q}\color{black}{ x} ^ { 2 } + \color{Cerulean}{( p n + q m ) }\color{black}{x} + \color{Cerulean}{m n} \end{aligned}
This gives us,
$$a=pq$$ and $$b=pn+qm$$, where $$c=mn$$
In short, when the leading coefficient of a trinomial is something other than $$1$$, there will be more to consider when determining the factors using the trial and error method. The key lies in the understanding of how the middle term is obtained. Multiply $$(5x+3)(2x+3)$$ and carefully follow the formation of the middle term.
Figure $$\PageIndex{2}$$
As we have seen before, the product of the first terms of each binomial is equal to the first term of the trinomial. The middle term of the trinomial is the sum of the products of the outer and inner terms of the binomials. The product of the last terms of each binomial is equal to the last term of the trinomial. Visually, we have the following:
Figure $$\PageIndex{3}$$
For this reason, we need to look for products of the factors of the first and last terms whose sum is equal to the coefficient of the middle term. For example, to factor $$6x^{2}+29x+35$$, look at the factors of $$6$$ and $$35$$.
\begin{aligned} 6 & = 1 \cdot 635 = 1 \cdot 35 \\ & = \color{OliveGreen}{2 \cdot 3} \quad \color{black}{=}\color{OliveGreen}{ 5 \cdot 7} \end{aligned}
The combination that produces the coefficient of the middle term is $$2⋅7+3⋅5=14+15=29$$. Make sure that the outer terms have coefficients $$2$$ and $$7$$, and that the inner terms have coefficients $$5$$ and $$3$$. Use this information to factor the trinomial.
\begin{aligned} 6 x ^ { 2 } + 29 x + 35 & = ( 2 x \quad \color{Cerulean}{?}\color{black}{ )} ( 3 x \quad \color{Cerulean}{?}\color{black}{ )} \\ & = ( 2 x + 5 ) ( 3 x + 7 ) \end{aligned}
We can always check by multiplying; this is left to the reader.
Example $$\PageIndex{7}$$
Factor $$5 x ^ { 2 } + 16 x y + 3 y ^ { 2 }$$.
Solution
Since the leading coefficient and the last term are both prime, there is only one way to factor each.
$$5=1⋅5$$ and $$3=1⋅3$$
Begin by writing the factors of the first term, $$5x^{2}$$, as follows:
$$5 x ^ { 2 } + 16 x y + 3 y ^ { 2 } = ( x \quad \color{Cerulean}{?} \color{black}{)} ( 5 x \quad \color{Cerulean}{?}\color{black}{ )}$$
The middle and last term are both positive; therefore, the factors of $$3$$ are chosen as positive numbers. In this case, the only choice is in which grouping to place these factors.
$$(x+y)(5x+3y)$$ or $$(x+3y)(5x+y)$$
Determine which grouping is correct by multiplying each expression.
\begin{aligned} ( x + y ) ( 5 x + 3 y ) & = 5 x ^ { 2 } + 3 x y + 5 x y + 3 y ^ { 2 } \\ & = 5 x ^ { 2 } + 8 x y + 3 y ^ { 2 } x \:\:\color{red}{✗} \\ ( x + 3 y ) ( 5 x + y ) & = 5 x ^ { 2 } + x y + 15 x y + 3 y ^ { 2 } \\ & = 5 x ^ { 2 } + 16 x y + 3 y ^ { 2 } \:\:\color{Cerulean}{✓}\end{aligned}
$$( x + 3 y ) ( 5 x + y )$$
Example $$\PageIndex{8}$$
Factor: $$18 a ^ { 2 } b ^ { 2 } - a b - 4$$.
Solution
First, consider the factors of the coefficients of the first and last terms.
\begin{aligned} 18 & = 1 \cdot 18\:\quad4 =\color{OliveGreen}{ 1 \cdot 4} \\ & =\color{OliveGreen}{ 2 \cdot 9} \:\:\:\:\:\quad\color{black}{=} 2 \cdot 2 \\ & = 3 \cdot 6 \end{aligned}
We are searching for products of factors whose sum equals the coefficient of the middle term, $$−1$$. After some thought, we can see that the sum of $$8$$ and $$−9$$ is $$−1$$ and the combination that gives this follows:
$$2 ( 4 ) + 9 ( - 1 ) = 8 - 9 = - 1$$
Factoring begins at this point with two sets of blank parentheses.
$$18 a ^ { 2 } b ^ { 2 } - a b - 4 = ( \quad ) (\quad )$$
Use $$2ab$$ and $$9ab$$ as factors of $$18a^{2}b^{2}$$.
$$18 a ^ { 2 } b ^ { 2 } - a b - 4 = ( 2 a b \quad \color{Cerulean}{?}\color{black}{ )} \:( 9 a b\quad\color{Cerulean}{ ?}\color{black}{ )}$$
Next use the factors $$1$$ and $$4$$ in the correct order so that the inner and outer products are $$−9ab$$ and $$8ab$$ respectively.
$$18 a ^ { 2 } b ^ { 2 } - a b - 4 = ( 2 a b - 1 ) ( 9 a b + 4 )$$
$$( 2 a b - 1 ) ( 9 a b + 4 )$$. The complete check is left to the reader.
It is a good practice to first factor out the GCF, if there is one. Doing this produces a trinomial factor with smaller coefficients. As we have seen, trinomials with smaller coefficients require much less effort to factor. This commonly overlooked step is worth identifying early.
Example $$\PageIndex{9}$$:
Factor $$12 y ^ { 3 } - 26 y ^ { 2 } - 10 y$$.
Solution
Begin by factoring out the GCF.
$$12 y ^ { 3 } - 26 y ^ { 2 } - 10 y = 2 y \left( 6 y ^ { 2 } - 13 y - 5 \right)$$
After factoring out $$2y$$, the coefficients of the resulting trinomial are smaller and have fewer factors. We can factor the resulting trinomial using $$6=2(3)$$ and $$5=(5)(1)$$. Notice that these factors can produce $$−13$$ in two ways:
$$\begin{array} { l } { 2 ( - 5 ) + 3 ( - 1 ) = - 10 - 3 = - 13 } \\ { 2 ( \color{OliveGreen}{1}\color{black}{ )} + 3 (\color{OliveGreen}{ - 5}\color{black}{ )} = 2 - 15 = - 13 } \end{array}$$
Because the last term is $$−5$$, the correct combination requires the factors $$1$$ and $$5$$ to be opposite signs. Here we use $$2(1) = 2$$ and $$3(−5) = −15$$ because the sum is $$−13$$ and the product of $$(1)(−5) = −5$$.
\begin{aligned} 12 y ^ { 3 } - 26 y ^ { 2 } - 10 y & = 2 y \left( 6 y ^ { 2 } - 13 y - 5 \right) \\ & = 2 y ( 2 y\quad \color{Cerulean}{?}\color{black}{ )} ( 3 y\quad\color{Cerulean}{ ?}\color{black}{ )} \\ & = 2 y ( 2 y - 5 ) ( 3 y + 1 ) \end{aligned}
Check.
\begin{aligned} 2 y ( 2 y - 5 ) ( 3 y + 1 ) & = 2 y \left( 6 y ^ { 2 } + 2 y - 15 y - 5 \right) \\ & = 2 y \left( 6 y ^ { 2 } - 13 y - 5 \right) \\ & = 12 y ^ { 3 } - 26 y ^ { 2 } - 10 y\color{Cerulean}{✓} \end{aligned}
The factor $$2y$$ is part of the factored form of the original expression; be sure to include it in the answer.
$$2 y ( 2 y - 5 ) ( 3 y + 1 )$$
It is a good practice to consistently work with trinomials where the leading coefficient is positive. If the leading coefficient is negative, factor it out along with any GCF. Note that sometimes the factor will be $$−1$$.
Example $$\PageIndex{10}$$
Factor: $$- 18 x ^ { 6 } - 69 x ^ { 4 } + 12 x ^ { 2 }$$.
Solution
In this example, the GCF is $$3x^{2}$$. Because the leading coefficient is negative we begin by factoring out $$−3x^{2}$$.
$$- 18 x ^ { 6 } - 69 x ^ { 4 } + 12 x ^ { 2 } = - 3 x ^ { 2 } \left( 6 x ^ { 4 } + 23 x ^ { 2 } - 4 \right)$$
At this point, factor the remaining trinomial as usual, remembering to write the $$−3x^{2}$$ as a factor in the final answer. Use $$6 = 1(6)$$ and $$−4 = 4(−1)$$because $$1(−1)+6(4)=23$$. Therefore,
\begin{aligned} - 18 x ^ { 6 } - 69 x ^ { 4 } + 12 x ^ { 2 } & = - 3 x ^ { 2 } \left( 6 x ^ { 4 } + 23 x ^ { 2 } - 4 \right) \\ & = - 3 x ^ { 2 } \left( x ^ { 2\quad } \right) \left( 6 x ^ { 2 }\quad \right) \\ & = - 3 x ^ { 2 } \left( x ^ { 2 } + 4 \right) \left( 6 x ^ { 2 } - 1 \right) \end{aligned}
$$- 3 x ^ { 2 } \left( x ^ { 2 } + 4 \right) \left( 6 x ^ { 2 } - 1 \right)$$. The check is left to the reader.
Exercise $$\PageIndex{2}$$
Factor: $$- 12 a ^ { 5 } b + a ^ { 3 } b ^ { 3 } + a b ^ { 5 }$$.
$$- a b \left( 3 a ^ { 2 } - b ^ { 2 } \right) \left( 4 a ^ { 2 } + b ^ { 2 } \right)$$
## Factoring Using the AC Method
An alternate technique for factoring trinomials, called the AC method19, makes use of the grouping method for factoring four-term polynomials. If a trinomial in the form $$ax^{2}+bx+c$$ can be factored, then the middle term, $$bx$$, can be replaced with two terms with coefficients whose sum is $$b$$ and product is $$ac$$. This substitution results in an equivalent expression with four terms that can be factored by grouping.
Example $$\PageIndex{11}$$:
Factor using the AC method: $$18 x ^ { 2 } - 31 x + 6$$.
Solution
Here $$a=18, b=-31$$, and $$c=6$$.
\begin{aligned} a c & = 18 ( 6 ) \\ & = 108 \end{aligned}
Factor $$108$$, and search for factors whose sum is $$−31$$.
\begin{aligned} 108 & = - 1 ( - 108 ) \\ & = - 2 ( - 54 ) \\ & = - 3 ( - 36 ) \\ & =\color{OliveGreen}{ - 4 ( - 27 )}\color{Cerulean}{✓} \\ & \color{black}{=} - 6 ( - 18 ) \\ & = - 9 ( - 12 ) \end{aligned} In this case, the sum of the factors \(−27 and $$−4$$ equals the middle coefficient, $$−31$$. Therefore, $$−31x=−27x−4x$$, and we can write
$$18 x ^ { 2 } \color{OliveGreen}{- 31 x}\color{black}{ +} 6 = 18 x ^ { 2 } \color{OliveGreen}{- 27 x - 4 x}\color{black}{ +} 6$$
Factor the equivalent expression by grouping.
\begin{aligned} 18 x ^ { 2 } - 31 x + 6 & = 18 x ^ { 2 } - 27 x - 4 x + 6 \\ & = 9 x ( 2 x - 3 ) - 2 ( 2 x - 3 ) \\ & = ( 2 x - 3 ) ( 9 x - 2 ) \end{aligned}
$$( 2 x - 3 ) ( 9 x - 2 )$$
Example $$\PageIndex{12}$$:
Factor using the AC method: $$4 x ^ { 2 } y ^ { 2 } - 7 x y - 15$$.
Solution
Here $$a=4, b= -7$$, and $$c=-15$$.
\begin{aligned} a c & = 4 ( - 15 ) \\ & = - 60 \end{aligned}
Factor $$−60$$ and search for factors whose sum is $$−7$$.
\begin{aligned} - 60 & = 1 ( - 60 ) \\ & = 2 ( - 30 ) \\ & = 3 ( - 20 ) \\ & = 4 ( - 15 ) \\ & = \color{OliveGreen}{5 ( - 12 )}\:\:\color{Cerulean}{✓} \\ & = 6 ( - 10 ) \end{aligned}
The sum of factors $$5$$ and $$−12$$ equals the middle coefficient, $$−7$$. Replace $$−7xy$$ with $$5xy−12xy$$.
\begin{aligned} 4 x ^ { 2 } y ^ { 2 } - 7 x y - 15 & = 4 x ^ { 2 } y ^ { 2 } + 5 x y - 12 x y - 15\quad\color{Cerulean}{Factor\:by\:grouping.} \\ & = x y ( 4 x y + 5 ) - 3 ( 4 x y + 5 ) \\ & = ( 4 x y + 5 ) ( x y - 3 ) \end{aligned}
$( 4 x y + 5 ) ( x y - 3 ).$
The check is left to the reader.
If factors of $$ac$$ cannot be found to add up to $$b$$ then the trinomial is prime.
## Key Takeaways
• If a trinomial of the form $$x^{2}+bx+c$$ factors into the product of two binomials, then the coefficient of the middle term is the sum of factors of the last term.
• If a trinomial of the form $$ax^{2}+bx+c$$ factors into the product of two binomials, then the coefficient of the middle term will be the sum of certain products of factors of the first and last terms.
• If the trinomial has a greatest common factor, then it is a best practice to first factor out the GCF before attempting to factor it into a product of binomials.
• If the leading coefficient of a trinomial is negative, then it is a best practice to first factor that negative factor out before attempting to factor the trinomial.
• Factoring is one of the more important skills required in algebra. For this reason, you should practice working as many problems as it takes to become proficient.
Exercise $$\PageIndex{3}$$
Factor.
1. $$x ^ { 2 } + 5 x - 6$$
2. $$x ^ { 2 } + 5 x + 6$$
3. $$x ^ { 2 } + 4 x - 12$$
4. $$x ^ { 2 } + 3 x - 18$$
5. $$x ^ { 2 } - 14 x + 48$$
6. $$x ^ { 2 } - 15 x + 54$$
7. $$x ^ { 2 } + 11 x - 30$$
8. $$x ^ { 2 } - 2 x + 24$$
9. $$x ^ { 2 } - 18 x + 81$$
10. $$x ^ { 2 } - 22 x + 121$$
11. $$x ^ { 2 } - x y - 20 y ^ { 2 }$$
12. $$x ^ { 2 } + 10 x y + 9 y ^ { 2 }$$
13. $$x ^ { 2 } y ^ { 2 } + 5 x y - 50$$
14. $$x ^ { 2 } y ^ { 2 } - 16 x y + 48$$
15. $$a ^ { 2 } - 6 a b - 72 b ^ { 2 }$$
16. $$a ^ { 2 } - 21 a b + 80 b ^ { 2 }$$
17. $$u ^ { 2 } + 14 u v - 32 v ^ { 2 }$$
18. $$m ^ { 2 } + 7 m n - 98 n ^ { 2 }$$
19. $$( x + y ) ^ { 2 } - 2 ( x + y ) - 8$$
20. $$( x - y ) ^ { 2 } - 2 ( x - y ) - 15$$
21. $$x ^ { 4 } - 7 x ^ { 2 } - 8$$
22. $$x ^ { 4 } + 13 x ^ { 2 } + 30$$
23. $$x ^ { 4 } - 8 x ^ { 2 } - 48$$
24. $$x ^ { 4 } + 25 x ^ { 2 } + 24$$
25. $$y ^ { 4 } - 20 y ^ { 2 } + 100$$
26. $$y ^ { 4 } + 14 y ^ { 2 } + 49$$
27. $$x ^ { 4 } + 3 x ^ { 2 } y ^ { 2 } + 2 y ^ { 4 }$$
28. $$x ^ { 4 } - 8 x ^ { 2 } y ^ { 2 } + 15 y ^ { 4 }$$
29. $$a ^ { 4 } b ^ { 4 } - 4 a ^ { 2 } b ^ { 2 } + 4$$
30. $$a ^ { 4 } + 6 a ^ { 2 } b ^ { 2 } + 9 b ^ { 4 }$$
31. $$x ^ { 6 } - 18 x ^ { 3 } - 40$$
32. $$x ^ { 6 } + 18 x ^ { 3 } + 45$$
33. $$x ^ { 6 } - x ^ { 3 } y ^ { 3 } - 6 y ^ { 6 }$$
34. $$x ^ { 6 } + x ^ { 3 } y ^ { 3 } - 20 y ^ { 6 }$$
35. $$x ^ { 6 } y ^ { 6 } + 2 x ^ { 3 } y ^ { 3 } - 15$$
36. $$x ^ { 6 } y ^ { 6 } + 16 x ^ { 3 } y ^ { 3 } + 48$$
37. $$x ^ { 2 n } + 12 x ^ { n } + 32$$
38. $$x ^ { 2 n } + 41 x ^ { n } + 40$$
39. $$x ^ { 2 n } + 2 a x ^ { n } + a ^ { 2 }$$
40. $$x ^ { 2 n } - 2 a x ^ { n } + a ^ { 2 }$$
1. $$( x - 1 ) ( x + 6 )$$
3. $$( x - 2 ) ( x + 6 )$$
5. $$( x - 6 ) ( x - 8 )$$
7. Prime
9. $$( x - 9 ) ^ { 2 }$$
11. $$( x - 5 y ) ( x + 4 y )$$
13. $$( x y - 5 ) ( x y + 10 )$$
15. $$( a + 6 b ) ( a - 12 b )$$
17. $$( u - 2 v ) ( u + 16 v )$$
19. $$( x + y - 4 ) ( x + y + 2 )$$
21. $$\left( x ^ { 2 } - 8 \right) \left( x ^ { 2 } + 1 \right)$$
23. $$\left( x ^ { 2 } + 4 \right) \left( x ^ { 2 } - 12 \right)$$
25. $$\left( y ^ { 2 } - 10 \right) ^ { 2 }$$
27. $$\left( x ^ { 2 } + y ^ { 2 } \right) \left( x ^ { 2 } + 2 y ^ { 2 } \right)$$
29. $$\left( a ^ { 2 } b ^ { 2 } - 2 \right) ^ { 2 }$$
31. $$\left( x ^ { 3 } - 20 \right) \left( x ^ { 3 } + 2 \right)$$
33. $$\left( x ^ { 3 } + 2 y ^ { 3 } \right) \left( x ^ { 3 } - 3 y ^ { 3 } \right)$$
35. $$\left( x ^ { 3 } y ^ { 3 } - 3 \right) \left( x ^ { 3 } y ^ { 3 } + 5 \right)$$
37. $$\left( x ^ { n } + 4 \right) \left( x ^ { n } + 8 \right)$$
39. $$\left( x ^ { n } + a \right) ^ { 2 }$$
Exercise $$\PageIndex{4}$$
Factor.
1. $$3 x ^ { 2 } + 20 x - 7$$
2. $$2 x ^ { 2 } - 9 x - 5$$
3. $$6 a ^ { 2 } + 13 a + 6$$
4. $$4 a ^ { 2 } + 11 a + 6$$
5. $$6 x ^ { 2 } + 7 x - 10$$
6. $$4 x ^ { 2 } - 25 x + 6$$
7. $$24 y ^ { 2 } - 35 y + 4$$
8. $$10 y ^ { 2 } - 23 y + 12$$
9. $$14 x ^ { 2 } - 11 x + 9$$
10. $$9 x ^ { 2 } + 6 x + 8$$
11. $$4 x ^ { 2 } - 28 x + 49$$
12. $$36 x ^ { 2 } - 60 x + 25$$
13. $$27 x ^ { 2 } - 6 x - 8$$
14. $$24 x ^ { 2 } + 17 x - 20$$
15. $$6 x ^ { 2 } + 23 x y - 4 y ^ { 2 }$$
16. $$10 x ^ { 2 } - 21 x y - 27 y ^ { 2 }$$
17. $$8 a ^ { 2 } b ^ { 2 } - 18 a b + 9$$
18. $$12 a ^ { 2 } b ^ { 2 } - a b - 20$$
19. $$8 u ^ { 2 } - 26 u v + 15 v ^ { 2 }$$
20. $$24 m ^ { 2 } - 26 m n + 5 n ^ { 2 }$$
21. $$4 a ^ { 2 } - 12 a b + 9 b ^ { 2 }$$
22. $$16 a ^ { 2 } + 40 a b + 25 b ^ { 2 }$$
23. $$5 ( x + y ) ^ { 2 } - 9 ( x + y ) + 4$$
24. $$7 ( x - y ) ^ { 2 } + 15 ( x - y ) - 18$$
25. $$7 x ^ { 4 } - 22 x ^ { 2 } + 3$$
26. $$5 x ^ { 4 } - 41 x ^ { 2 } + 8$$
27. $$4 y ^ { 6 } - 3 y ^ { 3 } - 10$$
28. $$12 y ^ { 6 } + 4 y ^ { 3 } - 5$$
29. $$5 a ^ { 4 } b ^ { 4 } - a ^ { 2 } b ^ { 2 } - 18$$
30. $$21 a ^ { 4 } b ^ { 4 } + 5 a ^ { 2 } b ^ { 2 } - 4$$
31. $$6 x ^ { 6 } y ^ { 6 } + 17 x ^ { 3 } y ^ { 3 } + 10$$
32. $$16 x ^ { 6 } y ^ { 6 } + 46 x ^ { 3 } y ^ { 3 } + 15$$
33. $$8 x ^ { 2 n } - 10 x ^ { n } - 25$$
34. $$30 x ^ { 2 n } - 11 x ^ { n } - 6$$
35. $$36 x ^ { 2 n } + 12 a x ^ { n } + a ^ { 2 }$$
36. $$9 x ^ { 2 n } - 12 a x ^ { n } + 4 a ^ { 2 }$$
37. $$- 3 x ^ { 2 } + 14 x + 5$$
38. $$- 2 x ^ { 2 } + 13 x - 20$$
39. $$- x ^ { 2 } - 10 x + 24$$
40. $$- x ^ { 2 } + 8 x + 48$$
41. $$54 - 12 x - 2 x ^ { 2 }$$
42. $$60 + 5 x - 5 x ^ { 2 }$$
43. $$4 x ^ { 3 } + 16 x ^ { 2 } + 20 x$$
44. $$2 x ^ { 4 } - 12 x ^ { 3 } + 14 x ^ { 2 }$$
45. $$2 x ^ { 3 } - 8 x ^ { 2 } y - 24 x y ^ { 2 }$$
46. $$6 x ^ { 3 } - 9 x ^ { 2 } y - 6 x y ^ { 2 }$$
47. $$4 a ^ { 3 } b - 4 a ^ { 2 } b ^ { 2 } - 24 a b ^ { 3 }$$
48. $$15 a ^ { 4 } b - 33 a ^ { 3 } b ^ { 2 } + 6 a ^ { 2 } b ^ { 3 }$$
49. $$3 x ^ { 5 } y + 30 x ^ { 3 } y ^ { 3 } + 75 x y ^ { 5 }$$
50. $$45 x ^ { 5 } y ^ { 2 } - 60 x ^ { 3 } y ^ { 4 } + 20 x y ^ { 6 }$$
1. $$( 3 x - 1 ) ( x + 7 )$$
3. $$( 2 a + 3 ) ( 3 a + 2 )$$
5. $$( 6 x - 5 ) ( x + 2 )$$
7. $$( 8 y - 1 ) ( 3 y - 4 )$$
9. Prime
11. $$( 2 x - 7 ) ^ { 2 }$$
13. $$( 9 x + 4 ) ( 3 x - 2 )$$
15. $$( 6 x - y ) ( x + 4 y )$$
17. $$( 4 a b - 3 ) ( 2 a b - 3 )$$
19. $$( 2 u - 5 v ) ( 4 u - 3 v )$$
21. $$( 2 a - 3 b ) ^ { 2 }$$
23. $$( x + y - 1 ) ( 5 x + 5 y - 4 )$$
25. $$\left( x ^ { 2 } - 3 \right) \left( 7 x ^ { 2 } - 1 \right)$$
27. $$\left( y ^ { 3 } - 2 \right) \left( 4 y ^ { 3 } + 5 \right)$$
29. $$\left( a ^ { 2 } b ^ { 2 } - 2 \right) \left( 5 a ^ { 2 } b ^ { 2 } + 9 \right)$$
31. $$\left( 6 x ^ { 3 } y ^ { 3 } + 5 \right) \left( x ^ { 3 } y ^ { 3 } + 2 \right)$$
33. $$\left( 2 x ^ { n } - 5 \right) \left( 4 x ^ { n } + 5 \right)$$
35. $$\left( 6 x ^ { n } + a \right) ^ { 2 }$$
37. $$- ( x - 5 ) ( 3 x + 1 )$$
39. $$- ( x - 2 ) ( x + 12 )$$
41. $$- 2 ( x - 3 ) ( x + 9 )$$
43. $$4 x \left( x ^ { 2 } + 4 x + 5 \right)$$
45. $$2 x ( x + 2 y ) ( x - 6 y )$$
47. $$4 a b ( a - 3 b ) ( a + 2 b )$$
49. $$3 x y \left( x ^ { 2 } + 5 y ^ { 2 } \right) ^ { 2 }$$
Exercise $$\PageIndex{5}$$
Factor.
1. $$4 - 25 x ^ { 2 }$$
2. $$8 x ^ { 3 } - y ^ { 3 }$$
3. $$9 x ^ { 2 } - 12 x y + 4 y ^ { 2 }$$
4. $$30 a ^ { 2 } - 57 a b - 6 b ^ { 2 }$$
5. $$10 a ^ { 2 } - 5 a - 6 a b + 3 b$$
6. $$3 x ^ { 3 } - 4 x ^ { 2 } + 9 x - 12$$
7. $$x ^ { 2 } + 4 y ^ { 2 }$$
8. $$x ^ { 2 } - x + 2$$
9. $$15 a ^ { 3 } b ^ { 2 } + 6 a ^ { 2 } b ^ { 3 } - 3 a b ^ { 4 }$$
10. $$54 x ^ { 2 } - 63 x$$
1. $$( 2 - 5 x ) ( 2 + 5 x )$$
3. $$( 3 x - 2 y ) ^ { 2 }$$
5. $$( 2 a - 1 ) ( 5 a - 3 b )$$
7. Prime
9. $$3 a b ^ { 2 } \left( 5 a ^ { 2 } + 2 a b - b ^ { 2 } \right)$$
Exercise $$\PageIndex{6}$$
1. Create your own trinomial of the form $$ax^{2} + bx + c$$ that factors. Share it, along with the solution, on the discussion board.
2. Create a trinomial of the form $$ax^{2} + bx + c$$ that does not factor and share it along with the reason why it does not factor.
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GED Mathematical Reasoning: Systems Of Linear Equations | Open Window Learning
Coordinate Plane and Functions
# GED Mathematical Reasoning: Systems Of Linear Equations
A system of linear equations refers to a set of two or more linear equations. Solving a system of linear equations means to find the ordered pair that satisfies each equation in the system. In other words, the solution to a system of linear equations is the ordered pair that makes each equation a true statement.
Solving a system of two linear equations using substitution
• Step 1: Solve one of the equations for one of the variables.
• Step 2: Substitute the expression from Step 1 into the other equation.
• Step 3: Solve the equation from Step 2.
• Step 4: Substitute the value found in Step 3 into either equation to find the other coordinate of the solution.
• Step 5: Check the solution. Plug the x- and y-value of the solution into each equation and verify that they make each equation a true statement.
Example 1
Solve the system of equations by graphing:
In example 1, we’re asked to solve the system of equations by graphing. This is actually a pretty simple process. We’ll graph each equation using a table of values and then look for where the lines intersect. The ordered pair representing the intersection point is the solution to the system of equations.
We’ll start by graphing the equation .
A good first step when graphing any equation using a table of values is to solve the equation for “y”, which is already done in this first equation. For our table of values let’s choose x-values: -1, zero and 1. One at a time, we’ll substitute these x-values into the equation and simplify to find the corresponding y-value. We’ll then use the ordered pair points to sketch the graph.
When we let , we will simplify the right side of the equal sign to determine the corresponding y value. . The ordered pair point is: (-1,6).
Next we’ll let . . The ordered pair point is: (0, 4).
Now we’ll let . . So the ordered pair point is: (1, 2).
Substitute and solve for Ordered Pair (Point) -1 6 (-1, 6) 0 4 (0, 4) 1 2 (1, 2)
Plotting these three points results in the line shown.
Next let’s graph the second equation: . Although it isn’t necessary, let’s solve this equation for y. In most cases, this makes completing the table a bit easier.
For our table of values let’s choose x-values: 2, 3 and 4
When we let , . And the ordered pair point is: (2, 2)
Next we’ll let . . The ordered pair point is: (3, -2)
Now we’ll let . . So the ordered pair point is: (4, -6)
Substitute and solve for Ordered Pair (Point) 2 2 (2, 2) 3 -2 (3, -2) 4 -6 (4, -6)
Let’s plot this line on the same graph as the first line so that we can see where the two lines intersect. I’ll show this line in red.
The two lines intersect at the point (3, -2), which means the solution to this system of two linear equations is the ordered pair: (3, -2).
Let’s check this ordered pair solution to ensure that it satisfies both equations. When we let and in each equation separately, we should get a true statement in both cases.
When we substitute the ordered pair (3, -2) in the first equation, we get:
Check:
which is true
Since -2 equals -2 is a true statement, the ordered pair satisfies this first equation.
Now let’s check the solution in the second equation. When we substitute the ordered pair (3, -2) in the second equation, we get:
Check:
which is true
Since 20 equals 20 is a true statement, the ordered pair satisfies this second equation as well and we can be confident that our solution is correct.
Example 2
Solve the system of equations from Example 1 using substitution.
Solving a system of two linear equations using substitution is a more algebraic approach. And it is a good method to know because it is more accurate – especially if one or both coordinates in the ordered pair solution is not an integer.
The first step to solving a system of linear equations using the substitution method is to solve one of the equations for one of the variables. In this example, the first equation is already solved for y so this step is already done for us.
The second step is to substitute the expression from step one into the other equation. In other words, since the first equation tells us that y is equal to the expression: , we will substitute for y in the second equation.
This gives us the equation:
Now we have an equation that contains only the variable x, which means we can solve it for x. In fact, the third step of the process is to solve the equation from Step 2.
Now that we have the x-coordinate of our ordered pair solution, we need to calculate the y-coordinate. This brings us to Step 4, which is to substitute the value found in step 3 into either equation to find the other coordinate of the solution. We can substitute “x” equals 3 back into either equation, but using the first equation will be more efficient.
If we let in the first equation, we get:
So the ordered pair solution is:
The final step is to substitute these values for “x” and “y” back into each equation separately to check that they make each equation a true statement.
We already checked this solution in Example 1.
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# Draw a rough sketch of a quadrilateral . State two pairs of opposite angles.
Last updated date: 13th Jun 2024
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Hint: In this question, first we will draw the rough sketch of a quadrilateral. Then we will define the term quadrilateral and its properties. After this, we will define the adjacent and opposite angles in quadrilateral.
Let’s draw a rough sketch of a quadrilateral.
So, this is the diagram of a general quadrilateral.
Now, we will define quadrilateral:
A polygon bounded by four sides is called a quadrilateral.
Properties of quadrilateral are as follow:
1. It has four sides.
2. It has four vertices.
3. The sum of all internal angles is 360 degree.
The different types of quadrilateral are:
1. Square:- All sides are equal and opposites sides are parallel. Each angle is 90 degrees.
2. Rectangle- opposite sides are equal and parallel. Each angle is 90 degrees.
3. Rhombus- All sides are equal and opposites sides are parallel.
4. Trapezium- -One pair of opposite sides is parallel and the other pair is not parallel.
Now, let’s see the definition of adjacent and opposite angles.
1. Adjacent angles- The angles formed on the sides of the quadrilateral are called adjacent angles.
$\angle KLM$ and $\angle LMN$are adjacent angles.
2. Opposite angles- The angles formed on the opposite vertices of the quadrilateral are called opposite angles.
So, $\angle KLM$and $\angle MNK$, $\angle MKL$and $\angle LMN$ are the two pairs of opposite angles of a given quadrilateral.
Note: In such a type of question, you should have knowledge of basic things like definition of different types of polygon, adjacent angles, vertices, opposite angles etc. The formula for calculating the sum of internal angles of a polygon is (n – 2)$\times {180^0}$, where ‘n’ is the total number of sides.
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# Aptitude - Time and Work
26.
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :
4 days
6 days
8 days
12 days
Explanation:
(A + B + C)'s 1 day's work = 1 ; 6
(A + B)'s 1 day's work = 1 ; 8
(B + C)'s 1 day's work = 1 . 12
(A + C)'s 1 day's work
= 2 x 1 - 1 + 1 6 8 12
= 1 - 5 3 24
= 3 24
= 1 . 8
So, A and C together will do the work in 8 days.
27.
A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:
5 days
6 days
10 days
10 1 days 2
Explanation:
(B + C)'s 1 day's work = 1 + 1 = 7 . 9 12 36
Work done by B and C in 3 days = 7 x 3 = 7 . 36 12
Remaining work = 1 - 7 = 5 . 12 12
Now, 1 work is done by A in 1 day. 24
So, 5 work is done by A in 24 x 5 = 10 days. 12 12
28.
X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?
13 1 days 3
15 days
20 days
26 days
Explanation:
Work done by X in 8 days = 1 x 8 = 1 . 40 5
Remaining work = 1 - 1 = 4 . 5 5
Now, 4 work is done by Y in 16 days. 5
Whole work will be done by Y in 16 x 5 = 20 days. 4
X's 1 day's work = 1 , Y's 1 day's work = 1 . 40 20
(X + Y)'s 1 day's work = 1 + 1 = 3 . 40 20 40
Hence, X and Y will together complete the work in 40 = 13 1 days. 3 3
29.
A and B can do a job together in 7 days. A is 1 times as efficient as B. The same job can be done by A alone in :
9 1 days 3
11 days
12 1 days 4
16 1 days 3
Explanation:
(A's 1 day's work) : (B's 1 day's work) = 7 : 1 = 7 : 4. 4
Let A's and B's 1 day's work be 7x and 4x respectively.
Then, 7x + 4x = 1 11x = 1 x = 1 . 7 7 77
A's 1 day's work = 1 x 7 = 1 . 77 11
30.
A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
30 days
40 days
60 days
70 days
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# How do you solve the following system: x+3y=-2, 5x + 2y = 5 ?
Jun 13, 2018
By arranging the equation
#### Explanation:
Expand the first equation by $- 5$:
$- 5 x - 15 y = 10$
Now add this to the second equation:
$5 x + 2 y = 5$
Yielding:
$- 13 y = 15$
$y = - \frac{15}{13}$
Put this value in the first original equation:
$x + 3 \left(- \frac{15}{13}\right) = - 2$
$x - \frac{45}{13} = - 2$
$x = - 2 + \frac{45}{13}$
$x = \frac{- 26 + 45}{13}$
$x = \frac{19}{13}$
Jun 13, 2018
$\text{the solution is shown below.}$
#### Explanation:
• You can use the "elimination" method to solve the equation system. To do this, you must destroy one of two variables (x, y). You have to equalize the coefficients in both equations by choosing one of the variables 'x' or 'y'.
• x+3y=-2$\text{ }$ (1) ; 5x+2y=5 $\text{ }$ (2)
• If both sides of equation (1) are multiplied by 5, the coefficients of 'x' are equal in both equations.
$\textcolor{red}{5} \left(x + 3 y\right) = - \textcolor{red}{5} \left(2\right) x +$
$5 x + 15 y = - 10 \text{ } \left(3\right)$
• Now let's subtract equation (3) from equation (2).
• 5x+2y-(5x+15y)=5-(-10)
$\cancel{5 x} + 2 y \cancel{- 5 x} - 15 y = 5 + 10$
$- 13 y = 15$
$y = - \frac{15}{13}$
• Finally, in (1) or (2), write -15/13 instead of 'y'.
$x + 3 \left(- \frac{15}{13}\right) = - 2$
$x - \frac{45}{13} = - 2$
$x = - 2 + \frac{45}{13}$
$x = \frac{- 26 + 45}{13}$
$x = - \frac{19}{13}$
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# Ch 7.3: Systems of Linear Equations, Linear Independence, Eigenvalues
## Presentation on theme: "Ch 7.3: Systems of Linear Equations, Linear Independence, Eigenvalues"— Presentation transcript:
Ch 7.3: Systems of Linear Equations, Linear Independence, Eigenvalues
A system of n linear equations in n variables, can be expressed as a matrix equation Ax = b: If b = 0, then system is homogeneous; otherwise it is nonhomogeneous.
Nonsingular Case If the coefficient matrix A is nonsingular, then it is invertible and we can solve Ax = b as follows: This solution is therefore unique. Also, if b = 0, it follows that the unique solution to Ax = 0 is x = A-10 = 0. Thus if A is nonsingular, then the only solution to Ax = 0 is the trivial solution x = 0.
Example 1: Nonsingular Case (1 of 3)
From a previous example, we know that the matrix A below is nonsingular with inverse as given. Using the definition of matrix multiplication, it follows that the only solution of Ax = 0 is x = 0:
Example 1: Nonsingular Case (2 of 3)
Now let’s solve the nonhomogeneous linear system Ax = b below using A-1: This system of equations can be written as Ax = b, where Then
Example 1: Nonsingular Case (3 of 3)
Alternatively, we could solve the nonhomogeneous linear system Ax = b below using row reduction. To do so, form the augmented matrix (A|b) and reduce, using elementary row operations.
Singular Case If the coefficient matrix A is singular, then A-1 does not exist, and either a solution to Ax = b does not exist, or there is more than one solution (not unique). Further, the homogeneous system Ax = 0 has more than one solution. That is, in addition to the trivial solution x = 0, there are infinitely many nontrivial solutions. The nonhomogeneous case Ax = b has no solution unless (b, y) = 0, for all vectors y satisfying A*y = 0, where A* is the adjoint of A. In this case, Ax = b has solutions (infinitely many), each of the form x = x(0) + , where x(0) is a particular solution of Ax = b, and is any solution of Ax = 0.
Example 2: Singular Case (1 of 3)
Solve the nonhomogeneous linear system Ax = b below using row reduction. To do so, form the augmented matrix (A|b) and reduce, using elementary row operations.
Example 2: Singular Case (2 of 3)
Solve the nonhomogeneous linear system Ax = b below using row reduction. Reduce the augmented matrix (A|b) as follows:
Example 2: Singular Case (3 of 3)
From the previous slide, we require Suppose Then the reduced augmented matrix (A|b) becomes:
Linear Dependence and Independence
A set of vectors x(1), x(2),…, x(n) is linearly dependent if there exists scalars c1, c2,…, cn, not all zero, such that If the only solution of is c1= c2 = …= cn = 0, then x(1), x(2),…, x(n) is linearly independent.
Example 3: Linear Independence (1 of 2)
Determine whether the following vectors are linear dependent or linearly independent. We need to solve or
Example 3: Linear Independence (2 of 2)
We thus reduce the augmented matrix (A|b), as before. Thus the only solution is c1= c2 = …= cn = 0, and therefore the original vectors are linearly independent.
Example 4: Linear Dependence (1 of 2)
Determine whether the following vectors are linear dependent or linearly independent. We need to solve or
Example 4: Linear Dependence (2 of 2)
We thus reduce the augmented matrix (A|b), as before. Thus the original vectors are linearly dependent, with
Linear Independence and Invertibility
Consider the previous two examples: The first matrix was known to be nonsingular, and its column vectors were linearly independent. The second matrix was known to be singular, and its column vectors were linearly dependent. This is true in general: the columns (or rows) of A are linearly independent iff A is nonsingular iff A-1 exists. Also, A is nonsingular iff detA 0, hence columns (or rows) of A are linearly independent iff detA 0. Further, if A = BC, then det(C) = det(A)det(B). Thus if the columns (or rows) of A and B are linearly independent, then the columns (or rows) of C are also.
Linear Dependence & Vector Functions
Now consider vector functions x(1)(t), x(2)(t),…, x(n)(t), where As before, x(1)(t), x(2)(t),…, x(n)(t) is linearly dependent on I if there exists scalars c1, c2,…, cn, not all zero, such that Otherwise x(1)(t), x(2)(t),…, x(n)(t) is linearly independent on I See text for more discussion on this.
Eigenvalues and Eigenvectors
The eqn. Ax = y can be viewed as a linear transformation that maps (or transforms) x into a new vector y. Nonzero vectors x that transform into multiples of themselves are important in many applications. Thus we solve Ax = x or equivalently, (A-I)x = 0. This equation has a nonzero solution if we choose such that det(A-I) = 0. Such values of are called eigenvalues of A, and the nonzero solutions x are called eigenvectors.
Example 5: Eigenvalues (1 of 3)
Find the eigenvalues and eigenvectors of the matrix A. Solution: Choose such that det(A-I) = 0, as follows.
Example 5: First Eigenvector (2 of 3)
To find the eigenvectors of the matrix A, we need to solve (A-I)x = 0 for = 3 and = -7. Eigenvector for = 3: Solve by row reducing the augmented matrix:
Example 5: Second Eigenvector (3 of 3)
Eigenvector for = -7: Solve by row reducing the augmented matrix:
Normalized Eigenvectors
From the previous example, we see that eigenvectors are determined up to a nonzero multiplicative constant. If this constant is specified in some particular way, then the eigenvector is said to be normalized. For example, eigenvectors are sometimes normalized by choosing the constant so that ||x|| = (x, x)½ = 1.
Algebraic and Geometric Multiplicity
In finding the eigenvalues of an n x n matrix A, we solve det(A-I) = 0. Since this involves finding the determinant of an n x n matrix, the problem reduces to finding roots of an nth degree polynomial. Denote these roots, or eigenvalues, by 1, 2, …, n. If an eigenvalue is repeated m times, then its algebraic multiplicity is m. Each eigenvalue has at least one eigenvector, and a eigenvalue of algebraic multiplicity m may have q linearly independent eigevectors, 1 q m, and q is called the geometric multiplicity of the eigenvalue.
Eigenvectors and Linear Independence
If an eigenvalue has algebraic multiplicity 1, then it is said to be simple, and the geometric multiplicity is 1 also. If each eigenvalue of an n x n matrix A is simple, then A has n distinct eigenvalues. It can be shown that the n eigenvectors corresponding to these eigenvalues are linearly independent. If an eigenvalue has one or more repeated eigenvalues, then there may be fewer than n linearly independent eigenvectors since for each repeated eigenvalue, we may have q < m. This may lead to complications in solving systems of differential equations.
Example 6: Eigenvalues (1 of 5)
Find the eigenvalues and eigenvectors of the matrix A. Solution: Choose such that det(A-I) = 0, as follows.
Example 6: First Eigenvector (2 of 5)
Eigenvector for = 2: Solve (A-I)x = 0, as follows.
Example 6: 2nd and 3rd Eigenvectors (3 of 5)
Eigenvector for = -1: Solve (A-I)x = 0, as follows.
Example 6: Eigenvectors of A (4 of 5)
Thus three eigenvectors of A are where x(2), x(3) correspond to the double eigenvalue = - 1. It can be shown that x(1), x(2), x(3) are linearly independent. Hence A is a 3 x 3 symmetric matrix (A = AT ) with 3 real eigenvalues and 3 linearly independent eigenvectors.
Example 6: Eigenvectors of A (5 of 5)
Note that we could have we had chosen Then the eigenvectors are orthogonal, since Thus A is a 3 x 3 symmetric matrix with 3 real eigenvalues and 3 linearly independent orthogonal eigenvectors.
Hermitian Matrices A self-adjoint, or Hermitian matrix, satisfies A = A*, where we recall that A* = AT . Thus for a Hermitian matrix, aij = aji. Note that if A has real entries and is symmetric (see last example), then A is Hermitian. An n x n Hermitian matrix A has the following properties: All eigenvalues of A are real. There exists a full set of n linearly independent eigenvectors of A. If x(1) and x(2) are eigenvectors that correspond to different eigenvalues of A, then x(1) and x(2) are orthogonal. Corresponding to an eigenvalue of algebraic multiplicity m, it is possible to choose m mutually orthogonal eigenvectors, and hence A has a full set of n linearly independent orthogonal eigenvectors.
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GreeneMath.com - Rounding Whole Numbers Test
Rounding Whole Numbers Test
When we want to get a value that is close but not necessarily exact, we can use an approximation. One method to approximate a number is known as rounding. Rounding allows us to approximate a number to a specific round off place.
Test Objectives:
•Demonstrate the ability to round a whole number to the nearest ten, hundred, & thousand
•Demonstrate the ability to round a whole number to the nearest thousand, ten thousand, & million
•Demonstrate the ability to round a whole number to any high level place
Rounding Whole Numbers Test:
#1:
Instructions: Round each to the nearest ten, hundred, and thousand.
a) 15,375
b) 373,575
c) 21,583
Watch the Step by Step Video Solution
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View the Written Solution
#2:
Instructions: Round each to the nearest thousand and ten thousand.
a) 5,273,371
b) 65,315
c) 26,300
Watch the Step by Step Video Solution
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View the Written Solution
#3:
Instructions: Round each to the nearest million.
a) 6,375,322
b) 17,955,971
c) 19,100,571
Watch the Step by Step Video Solution
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View the Written Solution
#4:
Instructions: Round each to the nearest billion.
a) 163,517,325,317
b) 224,978,381,500
c) 6,000,000,536,179
Watch the Step by Step Video Solution
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View the Written Solution
#5:
Instructions: Round each to the nearest ten trillions.
a) 169,315,111,827,059
b) 5,978,181,246,056,192
Watch the Step by Step Video Solution
|
View the Written Solution
Written Solutions:
#1:
Solution:
a) nearest ten - 15,380 : nearest hundred - 15,400 : nearest thousand - 15,000
b) nearest ten - 373,580 : nearest hundred - 373,600 : nearest thousand - 374,000
c) nearest ten - 21,580 : nearest hundred - 21,600 : nearest thousand - 22,000
Watch the Step by Step Video Solution
#2:
Solution:
a) nearest thousand - 5,273,000 : nearest ten thousand - 5,270,000
b) nearest thousand - 65,000 : nearest ten thousand - 70,000
c) nearest thousand - 26,000 : nearest ten thousand - 30,000
Watch the Step by Step Video Solution
#3:
Solution:
a) 6,000,000
b) 18,000,000
c) 19,000,000
Watch the Step by Step Video Solution
#4:
Solution:
a) 164,000,000,000
b) 225,000,000,000
c) 6,000,000,000,000
Watch the Step by Step Video Solution
#5:
Solution:
a) 170,000,000,000,000
b) 5,980,000,000,000,000
Watch the Step by Step Video Solution
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# Euclidean Algorithm Explained for Elementary School
In number theory, the Euclidean algorithm is a method for getting the greatest common factor (GCF) or highest common factor (HCF) of two positive integers. It is usually used for larger numbers since prime factorization can be used to get the greatest common factor of small numbers. Many students are confused with this method, but if you look at it closely, even elementary students can actually do it.
Let us start with an example. Note that in the discussion below, we will use the terms dividend and divisor. In the division a ÷ b, a is the dividend and b is the divisor.
Problem: Find the greatest common factor of 15 and 40 using the Euclidean Algorithm.
In Step 1, we divided 40 by 15, got a quotient of 2 and a remainder of 10.
In Step 2, the divisor 15 in the previous step (Step 1) became the dividend. The remainder 10 in the previous step became the divisor. In the calculation, we got a quotient of 1 and a remainder of 5.
In Step 3, the divisor 10 in the previous step (Step 2) became the dividend. The remainder 5 in the previous step became the divisor. We got a quotient of 2 and a remainder of 0.
Once the remainder in the algorithm becomes 0, the remainder in the previous step is the greatest common factor of the two numbers. In this case, the remainder is the previous step is 5, so the greatest common factor of 15 and 40 is 5.
As we can see from above, the Euclidean algorithm is repeated division such that in each step, the divisor is the remainder from the previous step and the dividend is the divisor from the previous step.
Geometric Interpretation
We can think about 15 and 40 as lengths of two ropes in meters. If we want to cut the ropes into shorter pieces with equal lengths, we want to maximize such that no part of the ropes will be wasted. With this condition, the longest possible length of each cut piece is the greatest common factor of 15 and 40.
Problem: Find the longest possible length that a 15m and a 40m rope cut into shorter pieces with equal lengths such that no part of the rope will be wasted.
The equivalent of the algorithm above can also be represented geometrically as shown in the preceding figure. In Step 1, two 15-meter ropes were used to fit the 40 meter rope, and then the remainder which was 10 meters was used as measuring rope in the next step. In Step 2, a 10-meter rope was used to fit the 15-meter rope, but this time, gave a remainder of 5 meters. Finally, in Step 3, two 5-m ropes exactly fit the 10-meter rope.
Therefore,we can cut both the ropes into 5 meters and get eleven 5-meter ropes combined.
Euclidean Algorithm Elementary Number Theory
In advanced high school mathematics and university textbooks, you will see the method above written differently. For those who majored in mathematics will probably remember the algorithm below. That is, the repeated
$r_{n - 2} = q_n r_{n-1} + r_n$ until $r_n = 0$,
where $r$ is the remainder, $q$ is the quotient, and $n$ is the number of steps (or division) in the algorithm. Using this method, the problem above has the following solution.
40 = 15(2) + 10
15 = 10(1) + 5
10 = 5(2) + 0
So, the remainder before $r_n = 0$ is the greatest common factor of the two numbers.
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Question Video: Using Variables to Represent Numbers and Write Inequalities When Solving a Real-World Problem | Nagwa Question Video: Using Variables to Represent Numbers and Write Inequalities When Solving a Real-World Problem | Nagwa
# Question Video: Using Variables to Represent Numbers and Write Inequalities When Solving a Real-World Problem
An online retailer offers free delivery when your purchase totals \$50 or more. James is preparing his order. He has already chosen items which total \$42. Write an inequality that describes the condition on ๐, the cost of the additional items he needs to order for him to benefit from free delivery.
01:43
### Video Transcript
An online retailer offers free delivery when your purchase totals 50 dollars or more. James is preparing his order. He has already chosen items which total 42 dollars. Write an inequality that describes the condition on ๐ the cost of the additional items he needs to order for him to benefit from free delivery.
So we need to write an inequality, which is like an equation except instead of an equal sign, we use less than, less than or equal to, greater than, or greater than or equal to. And we need to write it in terms of ๐ โ the cost of the additional items. So letโs start from the beginning. In order to get free delivery, we have to spend 50 dollars or more โ so greater than or equal to 50.
So Jamesโs order needs to be greater than or equal to 50 dollars. So right now, he has 42 dollarsโ worth of items in his cart. So he needs to add ๐, the additional items needed, to reach that 50-dollar goal. So if we would like our inequality to have ๐ by itself, that way we can describe ๐, we need to subtract 42 from both sides of the inequality. So on the left, the 42s cancel. And on the right, we have 50 minus 42. So this would be a way to describe the condition on ๐.
So ๐ needs to be greater than or equal to 50 minus 42. Simplifying that will also be another way to describe ๐. The condition on that ๐ would have to be greater than or equal to eight, meaning the additional items that he needs to order need to cost all together eight dollars or more.
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Find the derivative of sinx using First Principles?
Nov 21, 2016
By definition of the derivative $f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$
So with $f \left(x\right) = \sin x$ we have;
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin \left(x + h\right) - \sin x}{h}$
Using $\sin \left(A + B\right) = \sin A \cos B + \sin B \cos A$ we get
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin x \cos h + \sin h \cos x - \sin x}{h}$
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin x \left(\cos h - 1\right) + \sin h \cos x}{h}$
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\sin x \left(\cos h - 1\right)}{h} + \frac{\sin h \cos x}{h}\right)$
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin x \left(\cos h - 1\right)}{h} + {\lim}_{h \rightarrow 0} \frac{\sin h \cos x}{h}$
$f ' \left(x\right) = \left(\sin x\right) {\lim}_{h \rightarrow 0} \frac{\cos h - 1}{h} + \left(\cos x\right) {\lim}_{h \rightarrow 0} \frac{\sin h}{h}$
We know have to rely on some standard limits:
${\lim}_{h \rightarrow 0} \sin \frac{h}{h} = 1$
${\lim}_{h \rightarrow 0} \frac{\cos h - 1}{h} = 0$
And so using these we have:
$f ' \left(x\right) = 0 + \left(\cos x\right) \left(1\right) = \cos x$
Hence,
$\frac{d}{\mathrm{dx}} \sin x = \cos x$
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# Lesson 6
Area of Parallelograms
Let's practice finding the area of parallelograms.
### 6.1: Missing Dots
How many dots are in the image?
How do you see them?
### 6.2: More Areas of Parallelograms
1.
Calculate the area of the given figure in the applet. Then, check if your area calculation is correct by clicking the Show Area checkbox.
2. Uncheck the Area checkbox. Move one of the vertices of the parallelogram to create a new parallelogram. When you get a parallelogram that you like, sketch it and calculate the area. Then, check if your calculation is correct by using the Show Area button again.
3. Repeat this process two more times. Draw and label each parallelogram with its measurements and the area you calculated.
1. Here is Parallelogram B. What is the corresponding height for the base that is 10 cm long? Explain or show your reasoning.
2. Here are two different parallelograms with the same area.
1. Explain why their areas are equal.
2. Drag points to create two new parallelograms that are not identical copies of each other but that have the same area as each other. Sketch your parallelograms and explain or show how you know their areas are equal. Then, click on the Check button to see if the two areas are indeed equal.
Here is a parallelogram composed of smaller parallelograms. The shaded region is composed of four identical parallelograms. All lengths are in inches.
What is the area of the unshaded parallelogram in the middle? Explain or show your reasoning.
### Summary
Any pair of base and corresponding height can help us find the area of a parallelogram, but some base-height pairs are more easily identified than others.
When a parallelogram is drawn on a grid and has horizontal sides, we can use a horizontal side as the base. When it has vertical sides, we can use a vertical side as the base. The grid can help us find (or estimate) the lengths of the base and of the corresponding height.
When a parallelogram is not drawn on a grid, we can still find its area if a base and a corresponding height are known.
In this parallelogram, the corresponding height for the side that is 10 units long is not given, but the height for the side that is 8 units long is given. This base-height pair can help us find the area.
Regardless of their shape, parallelograms that have the same base and the same height will have the same area; the product of the base and height will be equal. Here are some parallelograms with the same pair of base-height measurements.
### Glossary Entries
• base (of a parallelogram or triangle)
We can choose any side of a parallelogram or triangle to be the shape’s base. Sometimes we use the word base to refer to the length of this side.
• height (of a parallelogram or triangle)
The height is the shortest distance from the base of the shape to the opposite side (for a parallelogram) or opposite vertex (for a triangle).
We can show the height in more than one place, but it will always be perpendicular to the chosen base.
• parallelogram
A parallelogram is a type of quadrilateral that has two pairs of parallel sides.
Here are two examples of parallelograms.
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# Area of Triangle Formula
## Formula
### Summary
The area of a triangle is given by one-half multiplied by the base and height.
Expression Description
The area of the triangle.
The base length of the triangle.
The height length of the triangle.
## Usage
The area of a triangle is calculated using the length of its base and the lenght of its height. For example, the area of a triangle with a base of and a height of is equal to to units squared as shown in the expressions below:
The formula can be derived using geometry and the area of a rectangle formula[1].
## Examples
### Example 1
This example demonstrates how to calculate the area of a of triangle with a base of 4 and a height of 3.
#### Steps
1. Start by setting up the formula.
2. Substitute the base and height into the formula.
3. Evaluate the multiplication.
4. Evaluate the multiplication.
THe area of the triangle is equal to units squared.
### Example 2
This example demonstrates how to calculate the area of a of triangle with a base of 4 and a height of 2.
#### Steps
1. Start by setting up the formula.
2. Substitute the base and height into the formula.
3. Evaluate the multiplication.
4. Evaluate the multiplication.
THe area of the triangle is equal to units squared.
### Example 3
This example demonstrates how to calculate the area of a of triangle with a base of 5 and a height of 4.
#### Steps
1. Start by setting up the formula.
2. Substitute the base and height into the formula.
3. Evaluate the multiplication.
4. Evaluate the multiplication.
THe area of the triangle is equal to units squared.
## Explanation
The area of a triangle formula can be derived using a combination of geometry and the area of a rectangle formula [1]. The three cases of the derivation are correspond to three triangle types: right triangle, acute triangle and obtuse triangle.
## References
1. Derive Area of Triangle Formula
Wumbo (internal)
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# Basic Algebra Lessons,Learn Algebra
Basic Algebra builds on the foundation of arithmetic. Featuring the introduction of Mathematical formulas, along with equations that contain letters and symbols, as well as numbers.
The main letters and symbols concerned are called variables, or sometimes unknowns, and they help enable people to solve a wider range of problems than standard arithmetic with numbers. Not just in Math but in a variety of disciplines.
So it’s safe to say, that understanding and being competent at Algebra is a very key skill to have in the modern world. The basic Algebra lessons pages below in this section, focus on some introductory Algebra topics.
Also featured are some additional pages that can be classed as further Pre-Algebra topics. Though these topics are also important parts of of an introduction as we learn Algebra.
## Basic Algebra Lessons Pages
Associative Property, Commutative, Distributive, Transitive
The associative property is an important property that applies to addition and multiplication.
Also not dissimilar to the associative property, the commutative property is one of several others that apply to certain sums in Algebra.
Variables, Basics of Equations
A page on the definition of a variable in Algebra. Also shown are how to solve for variables in basic equations using some of the properties of equality.
Factoring Math, Factoring Expressions
It's not only whole numbers that can have factors, in basic Algebra mathematical expressions can also be factored in many instances.
Removing Brackets, Vertical Multiplication, FOIL
A page showing the process of removing brackets with FOIL multiplication. A process sometimes referred to as expanding brackets or multiplying out brackets.
The method of vertical multiplication is also shown.
Difference and Sum of Two Cubes, Difference of Two Squares
There are several rules that are handy to know when attempting to manipulate expressions in Algebra quickly.
Simplifying Expressions with Exponents
In many situations in Algebra one will have to simplify expression containing exponents, some examples of common situations are shown here.
Simplifying radicals with variables contained in them often requires a little bit more care than with just numbers.
Literal Equations
Solving literal equations involves solving for one variable/letter in an equation, when there are other variables present.
Cartesian Coordinates
In Algebra, may occasions will call for functions and equations to be graphed and illustrated.
These graphs are drawn on a Cartesian plane, which can also referred to as an x
and y axis. This page shows how to plot points using Cartesian co-ordinates.
Sigma Notation
Sigma notation is often used when there are sums that require an addition of a specific number of terms. This notation can at times save quite a bit of time and require less writing.
### Sequences and Series
Sequences
Sequences are ordered lists of terms that follow a specific rule or formula.
Recursive Sequences
In a recursive sequence, new terms in the sequence are obtained by using one or more of the previous terms in the sequence.
Series
Series in Math is when we sum up the values of a sequence.
1. Home
2. › Basic Algebra Lessons
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# How do you do partial fractions in Laplace transform?
## How do you do partial fractions in Laplace transform?
This technique uses Partial Fraction Expansion to split up a complicated fraction into forms that are in the Laplace Transform table….Solution:
Power of s Equation
s3 0=A1+B
s2 5=2A1+A2+C
s1 8=5A1+2A2
s0 -5=5A2
## How do you convert to partial fractions?
The method is called “Partial Fraction Decomposition”, and goes like this:
1. Step 1: Factor the bottom.
2. Step 2: Write one partial fraction for each of those factors.
3. Step 3: Multiply through by the bottom so we no longer have fractions.
4. Step 4: Now find the constants A1 and A2
5. And we have our answer:
## What is the Laplace transformation of a step function?
Overview: The Laplace Transform method can be used to solve. constant coefficients differential equations with discontinuous source functions. Notation: If L[f (t)] = F(s), then we denote L−1 [F(s)] = f (t).
## How do you do inverse Laplace transform?
To obtain L−1(F), we find the partial fraction expansion of F, obtain inverse transforms of the individual terms in the expansion from the table of Laplace transforms, and use the linearity property of the inverse transform.
## What is the formula for inverse transform?
Definition of the Inverse Laplace Transform. F(s)=L(f)=∫∞0e−stf(t)dt. f=L−1(F). To solve differential equations with the Laplace transform, we must be able to obtain f from its transform F.
## What are inverse Laplace transforms?
In mathematics, the inverse Laplace transform of a function F ( s) is the piecewise-continuous and exponentially-restricted real function f ( t) which has the property: denotes the Laplace transform .
## What is a fraction decomposition?
Decomposing fractions means a fraction is written as sum (or difference) of two or more fractions. For example, 5/8 = 2/8 + 3/8 = 6/8 – 1/8. Fraction decomposition requires the numerator to be written as a sum (or difference) and then split the fraction as in the example given here.
## What is the inverse Laplace transform of one?
Laplace Inverse Transform of 1: δ (t)
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# Problem of the Week Problem B and Solution The Puzzler
## Problem
The Puzzler is the world’s latest superhero. He uses his immense brain to win all battles by solving a series of math problems. He needs your help to solve the following problems.
Use a calculator to help when needed. You may also want to look up words like consecutive and sum.
1. The numbers $$3$$, $$5$$, and $$7$$ are three consecutive odd numbers that have a sum of $$3 + 5 + 7 = 15$$.
What are three consecutive odd numbers that have a sum of $$399$$?
2. What are three consecutive even numbers that have a sum of $$5760$$?
3. What are four consecutive whole numbers that have a sum of $$2022$$?
## Solution
1. The sum of the three consecutive odd numbers $$3$$, $$5$$, and $$7$$ is $$3 + 5 + 7 = 15$$. We notice that $$15 = 3 \times 5$$ and $$5$$ is the middle number. It seems that to find the middle of three consecutive odd numbers with a certain sum, we may divide that sum by $$3$$.
Let’s try using this to solve the problem. We note that $$399 \div 3 = 133$$. Therefore, the middle number could be $$133$$. Then the first number would be $$131$$ and the third number would be $$135$$. The sum of these numbers is indeed $$131+133+135 = 399$$. Therefore, the three consecutive odd numbers are $$131$$, $$133$$, and $$135$$.
2. We will use a process like in (a). Noting that $$5760 \div 3 = 1920$$, we see that three consecutive even numbers could be $$1918$$, $$1920$$, and $$1922$$. The sum of these numbers is indeed $$1918 + 1920 + 1922 = 5760$$. Therefore, the three consecutive even numbers are $$1918$$, $$1920$$, and $$1922$$.
3. Using a similar process, when we divide $$2022$$ by $$4$$ we get $$505.5$$. Since $$505$$ and $$506$$ are the closest whole numbers to $$505.5$$, they may be the two middle numbers. The four consecutive numbers may be $$504$$, $$505$$, $$506$$, and $$507$$. The sum of these numbers is indeed $$504 + 505 + 506 + 507 = 2022$$. Therefore, the four consecutive numbers are $$504$$, $$505$$, $$506$$, and $$507$$.
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# Park Forest Math Team. Meet #5. Algebra. Self-study Packet
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## Transcription
1 Park Forest Math Team Meet #5 Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number Theory: Divisibility rules, factors, primes, composites 4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics 5. : Simplifying and evaluating expressions; solving equations with 1 unknown including identities
2 Important things you need to know about ALGEBRA: Solving quadratics with rational solutions, including word problems If xy = 0, then x = 0 or y = 0. This is called the Zero Product Property If (x 3) (x + 2) = 0, then x 3 = 0 or x + 2 = 0. The solutions to this problem are x =3 and x = -2 When a graph crosses the x-axis, y = 0. To multiply binomials, such as (x 4) (x + 6), we can use the distributive property. A mnemonic is FOIL. Foil means multiply the First, Outside, Inside, and Last Terms. (x 4) (x + 6) = x 2 + 6x 4x 24 = x 2 + 2x 24 You should notice that in the above example, the -4 and 6 add to equal 2 and multiply to equal -24. Use this knowledge to work backward to factor a trinomial. Factor x 2 7x + 12 (Think: What are two numbers that multiply to equal 12 and add to equal -7? -3 and -4) So, x 2 7x + 12 = (x 3) (x 4) If x 2 7x + 12 = 0, then (x 3) (x 4) = 0, so x = 3 or x = 4 The quadratic formula can also be used to solve quadratic equations. If Ax 2 + Bx + C = 0, then Example: How many units are in the shortest length of the right triangle below? 2x + 3 x 2x + 2 By Pythagorean Theorem, x 2 + (2x + 2) 2 = (2x + 3) 2 So, x 2 + (2x + 2)(2x + 2) = (2x + 3)(2x + 3) Using FOIL, x 2 + 4x 2 + 4x + 4x + 4 = 4x 2 + 6x = 6x + 9 Combining like terms, 5x 2 + 8x + 4 = 4x x + 9 Subtracting the right side to get everything on the left, we get x 2 4x 5 = 0 So (x 5)(x+1) = 0, Therefore, x = 5 or x = -1 A side length cannot be negative, so x must be 5.! Check. If x = 5, 2x + 2 = 12 and 2x + 3 = = 13 2
3 Category 5 Meet #5 - March, 2014 Calculator Meet 50th anniversary edition 1) What are the two values of N that solve? 2) To fence a rectangular pasture, a farmer uses 110 meters of fencing to enclose a 750 square meter area. The length of the pasture is longer than the width. What is the length of the pasture? 3) Hoses are attached to two outdoor faucets so that a backyard swimming pool can be filled. If both faucets are used, it takes two hours to fill the pool. If either faucet were used alone, then one faucet would take three hours less than the other to fill the pool. What is the least amount of time required to fill the pool if only one faucet is used? ANSWERS 1) and 2) meters 3) hours
4 Solutions to Category 5 Meet #5 - March, ) N 2 4N = 32 N 2 4 N 32 = 0 1) - 4 and 8 ( N + 4)( N 8) = 0 Either ( N + 4) = 0 or ( N 8) = 0 2) 30 So, N = - 4 or N = 8 1) 3) 3
5
6 Meet #5 March 2012 Calculators allowed Category 5 1. A rectangle is inch longer than it is wide. Its area is square inches. How many inches are there in its perimeter? 2. The price of a square solar panel is calculated in the following way: cents for each square inch in its area, plus cents for each inch in its perimeter. How many inches are there in the side of a solar panel that costs? 3. There are two solutions to the quadratic equation. Their sum is and the positive difference between them is. What is the value of the parameter? 1. inches 2. inches 3.
7 Meet #5 March 2012 Calculators allowed Solutions to Category 5-1. If we call the width, then we know that ( ) or which we can solve: and the positive solution is Therefore the perimeter is ( ) If we call the length of a side, then we know that: (note we have to translate the dollars to cents). Solving this we get: and the positive solution is inches. 3. Given that the solutions of any quadratic equation are given by ( in our case), we get that their sum is and their difference is. So we re told that and also that and we conclude that or
8 You may use a calculator today! Category 5 - Meet #5, March What is the positive difference between the two roots (solutions for x) of the equation below? [Hint: both are integers]. x 2 4 x = When each side of a square was increased in length by 50%, its area increased by 180 square inches. How many square inches are in the original square? 3. The diagram below shows a circle inscribed inside a square. The shaded rectangle measures 4 8 inches and touches the circle with one corner. How many inches are in the radius of the circle?
9 Solutions to Category 5 - Meet #5, March You may use a calculator today! 1. If we rearrange the equation x 2 4 x = 21 we can write it in the form: x + 3 x 7 = 0 which makes clear that the solutions (roots) are x = 3 and x = 7 and the difference is 10. If you could not factor this way, you should have written x 2 4 x 21 = 0 and then x = 4± = 2 ± 5 to get the same values. 2 The plot to the left shows the graph of y = x 2 4 x 21. This shows visually that x values of -3 and +7 result in y = 0 and are therefore the solutions. 2. If we call the original length d, then the original area is d 2. The increased length is d 150% = 3 2 d and the increased area is (3 2 d)2 = 9 4 d2 so we know that the difference in areas is 5 4 d2 = 180 and we get d 2 = = 144 square inches. B A O d If the shaded area measures 180 square inches, then each little square is 36 square inches and so the original square is 4 times that area, or 144 square inches. 3. In triangle OAB we have OB = R (The circle s radius), OA = R 4, AB = R 8 and therefore: (R 4) 2 + (R 8) 2 = R 2, the solutions of which are R = 4 (invalid in our case) and R = 20.
10 Category 5 Meet #5, March What is the sum of the x-coordinates of the x-intercepts of the equation below? (the x-intercepts are known as the roots and are the values of x when y = 0.) A square is changed into a new rectangle by adding 3 cm to the length and subtracting 5 cm from the width. The area of the new rectangle is 48 cm. How many square centimeters are in the area of the original square? 3. The triangle below is a right triangle with hypotenuse of length ( 3x 4 ). What is the value of the perimeter of this triangle? 2x - 6 3x - 4 2x
11 Solutions to Category 5 Meet #5, March or or 3 The sum is 2 3 Another way to find the sum of the x-coordinates of the x- intercepts (otherwise known as roots) is to put the equation into the form and then the sum is 2. If the square s original side length is s then the new rectangle has length s +3 and width s 5 and the area would be : or or 7 but since s is a side of a square, it can t be negative. So s = 9 and the area of the square was 9 2 =81 3. Using the Pythagorean Theorem, we know that : or or 3 but if 3 then all the sides would be negative which we can t have, so 7. That makes the three sides of the triangle 8, 15, and 17 and the perimeter is 40.
12 Category 5 Meet #5, March 2006 You may use a calculator today. 1. Find the negative value of x that makes the following equation true: x( x + 1)= A rational number and its reciprocal have a sum of 2 9. If the number is less 10 than one, what is this number? Express your answer as a common fraction in simplest form. 3. The triangle below is a right triangle with side lengths given in terms of x. How many square units are in the area of the triangle? Express your answer to the nearest whole number of square units. x 8x 7 7x
13 Solutions to Category 5 Meet #5, March Rather than set the quadratic equation equal to zero, consider what the equation says in its current form. The product of two numbers is equal to 210. Since the two numbers are one apart, the square root of 210 will put us close to the correct values. If x were positive, it would be 14 times 15 equals 210. Since x is negative, it s 15 times 14. So x = 15 is it. 2. The English translates to the equation x + 1 x = , or x + 1 x = 29. Multiplying both sides of the equation by 10x, we 10 get 10x = 29x. We now subtract 29x from both sides to get 10x 2 29x + 10 = 0. Now we can either use the quadratic formula, x = b ± b 2 4ac, or we can try to factor this 2a trinomial into the product of two binomials. Factoring gives us the equation 5x 2 ( )( 2x 5) = 0. If 5x 2 = 0, then the solution is x = 2 5. If 2x 5 = 0, then the solution is x = 5 2. We want the solution that is less than 1, which is 2 5. ( ) 2 = ( 8x 7) By the Pythagorean Theorem, we write the equation x 2 + 7x + 7 Expanding on both sides, we get x x x + 49 = 64x 2 112x Subtracting 49 from both sides and combining like terms, we get 50x x = 64x 2 112x. We can now set this equation equal to zero by subtracting 50x 2 and 98x from both sides. This gives us the equation 0 = 14x 2 210x. Factoring out 14x from both terms, we rewrite the equation to get 0 = 14x( x 15). The solutions are x = 0, which is not useful to us, or x = 15, which is useful to us. The height of the triangle is 15 units and the base is = = 112 units. The area of the triangle is thus A = = = 840 square units.
14 Category 5 Meet #5, April 2004 You may use a calculator 1. The length of a rectangle is nine units longer than the width of the rectangle and the area of the rectangle is 112 square units. Find the number of units in the perimeter of the rectangle. 2. The product of three consecutive negative integers is equal to 288 times the middle integer. What is the value of the middle integer? ( )( x a) is shown below. The ordered pairs 3. The graph of the equation y = x + 3 of the x- and y-intercepts have also been given. Find the value of a. y (0,12) (-4, 0) (-3, 0) x
15 Solutions to Category 5 Average team got 17.3 points, or 1.4 questions correct Meet #5, April Let the width of the rectangle be x units. The length is then x + 9 units, and the area would be calculated as x( x + 9) or x 2 + 9x square units. Since we know the area is 112 square units, we can write the quadratic equation x 2 + 9x = 112. Setting this equal to zero, we get x 2 + 9x 112 = 0. Now we would like to factor this trinomial into the product of two binomials in the form ( x + )( x ). To fill these blanks, we must find a pair of numbers whose product is 112 and whose difference is 9. We could list factors of 112 in pairs as follows: 1 112, 2 56, 4 28, 7 16, and The desired pair is 7 and 16 since 16 7 = 9. We now have x 2 + 9x 112 = x + 16 ( )( x 7) = 0. For this equation to equal zero, x has to equal either -16 or 7. The negative value doesn t make sense for the width of our rectangle, so the width must be 7. This means the length is = 16, and the perimeter is ( ) = 2 23 = 46 units. 2. If we let n be the middle number, the product of the three consecutive numbers is ( n 1) n ( n + 1). This product is said to be equal to 288 times the middle number, so we can write the equation ( n 1) n ( n + 1) = 288n. Dividing both sides of the equation by n, we avoid a cubic and get the quadratic equation ( n 1) ( n + 1)= 288. Multiplying out the left side of the equation, we get n 2 1 = 288, which means that n 2 must be 289. Both positive and negative 17 satisfy the equation n 2 = 289, but we know that the three consecutive integers are negative. The value of the middle integer is Expanding the right side of the equation, we get y = x 2 + ( 3 a)x 3a. If we substitute the x- and y-values from the ordered pair (0, 12) into this equation, we get 12 = ( 3 a) 0 3a, which simplifies to 12 = 3a. Dividing both sides by -3, we get a = -4.
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# wikiHow to Solve Complex Cases of Quadratic Equations
The standard form of a quadratic equation in one variable is ax² + bx + c = 0. Depending on the values of the constants a and c, solving this equation may be simple or may be complicated. To know about the new Diagonal Sum Method, please read the article:"How to solve quadratic equations by the Diagonal Sum Method" on this website. This article discusses various cases in solving quadratic equations.
### Method 1 VARIOUS CASES IN SOLVING QUADRATIC EQUATIONS
1. 1
A. When a = 1 - Solving quadratic equations types x² + bx + c = 0.
• Solving this type of quadratic equations results in solving a popular puzzle: finding two numbers knowing their sum and their product. Solving becomes simple and doesn't need factoring.
• Example 1. Solve: x² - 26x - 72 = 0.
• Solution. Both real roots have opposite signs. Write down the factor-pairs of c = -72. They are:
• (-1 , 72)(-2 , 36)(-3 , 24)...Stop!The sum of the 2 real roots in this set is 21 = -b. The 2 real roots are -3 and 24.
• Example 2. Solve: -x² - 26x + 56 = 0.
• Solution. Roots have opposite signs. Write down factor-pairs of c = 56:
• (-1, 56) (-2, 28)...Stop!. This sum is 26 = b. According to the Diagonal Sum Rule, when a is negative, the answers are -2 and 28.
• Example 3. Solve x² + 27x + 50 = 0.
• Solution. Both real roots are negative. Write factor-sets of c = 50:
• (-1, -50) (-2, -25)..Stop! This sum is -27 = -b. The 2 real roots are -2 and -25.
• Example 4. Solve: x² - 39x + 108 = 0.
• Solution. Both real roots are positive. Write the factor-sets of c = 108:
• (1, 108) (2, 54) (3, 36)...Stop! This sum is 39 = -b. The 2 real roots are 3 and 36.
### Method 2 B. When a and c are prime/small numbers.
1. 1
In these cases, the new method directly selects the probable root pairs basing on the (c/a) setup and by, in the same time, applying the rule of signs.
• Example 5. Solve: 7x² + 90x - 13 = 0.
• Solution. Roots have opposite signs. Both a and c are prime numbers. Since 1 (or -1) is not a real root, there is a unique root pair:
• (-1/7, 13/1). Since its diagonal sum is: -1 + 91 = 90 = b, the answers are opposite in sign. The 2 real roots are 1/7 and -13.
• Example 6. Solve: 7x² - 57x + 8 = 0.
• Solution. Both roots are positive. The c/a setup: (1, 8)(2, 4)/(1, 7).
• Probable root pairs: (1/7 , 8/1) (2/1 , 4/7) (2/7 , 4/1)
• Diagonal Sum of first set:(1 + 56) = 57 = -b.
• The 2 real roots are 1/7 and 8.
• NOTE. You can simplify the c/a setup before proceeding. Eliminate the pair (2, 4) from the numerator because it gives even-number diagonal sums (while b is odd). The remainder c/a: (1, 8)/(1, 7) leads to unique probable root pair: (1/7, 8/1) that gives the 2 real roots 1/7 and 8.
• Example 7. Solve: 6x² - 19x - 11 = 0.
• Solution. Roots have opposite signs. Write the c/a setup:
• (-1, 11)/(1, 6)(2, 3)
• Eliminate the pair (1, 6) because it gives larger diagonal sum (while b = -19).
• The remainder c/a: (-1, 11)/(2, 3) leads to 2 probable root pairs: (-1/2, 11/3) and (-1/3, 11/2). The first diagonal sum is: -3 + 22 = 19 = -b. The 2 real roots are -1/2 and 11/3.
### Method 3 C. COMPLICATED CASES OF QUADRATIC EQUATIONS
1. 1
When the constants a and c are large numbers and may contain themselves many factors, solving is considered complicated because there are many permutations involved. The Diagonal Sum Method proceeds basing on the (c/a) setup. The numerator of the setup contains all factor pairs of c. The denominator contains all factor pairs of a.
• Next, this new method transforms a multiple steps solving process into a simplified one by doing a few operations of elimination.
• Example 1. Solve: 45x² - 74x - 55 = 0.
• Solution. Roots have opposite signs. Write down the (c/a) setup.
• Numerator. All factors-sets of c: (-1, 55) (-5, 11).
• Denominator. All factors-sets of a: (1, 45) (3, 15) (5, 9).
• Smart students can use mental math to calculate all diagonal sums and find the one that fits.
• The best way is to proceed elimination of the factor pairs that do not fit.
• Eliminate the pairs: (-1, 55)/(1, 45)(3, 15) since they give larger diagonal sums (while b = -74).
• The unique remainder (c/a) is: (-5, 11)/(5, 9) that gives as 2 real roots: -5/9 and 11/5.
• Example 2. Solve: 12x² - 272x + 45 = 0.
• Solution. Both roots are positive. Write down the (c/a) setup.
• Numerator: (1, 45) (3, 15) (5, 9)
• Denominator: (1, 12) (2, 6) (3, 4)
• First eliminate the pairs with (1, 12),(3, 4) since they will give odd-number diagonal sums (while b is an even number).
• Next, look for a setup (c/a) that gives a large diagonal sum. It should be the setup:(1, 45)/(2, 6) that leads to 2 probable root pairs:
• (1/2 , 45/6) and (1/6, 45/2). The second diagonal sum is: 2 + 270 = 272 = -b. The 2 real roots are 1/6 and 45/2.
• Example 3. Solve: 40x² - 483x + 36 = 0.
• Solution. Both roots are positive. Write down the c/a setup.
• Numerator: (1, 36) (2, 18) (3, 12) (6, 6).
• Denominator: (1, 40) (2, 20) (4, 10)
• First eliminate the pairs (2, 18)(6, 6)/(2, 20)(4, 10) because they give even-number diagonal sums (while b is odd).
• The remainder c/a leads to 3 probable root pairs: (1/40, 36/1), (3/1, 12/46), (3/40, 12/1). The third diagonal sum is : 3 + 480 = 483 = -b. The 2 real roots are 3/40 and 12.
• Example 4. Solve: 12x² + 5x - 72 = 0.
• Solution. Roots have opposite signs. Write the (c/a) setup.
• Numerator: (-1, 72)(-2, 36)(-3, 24)(-4, 18)(-6, 12) (-8, 9)
• Denominator: (1, 12) (2, 6) (3, 4).
• First, eliminate the pairs (-2, 36),(-4, 18),(-6, 12)/(2, 6) since they give even-number diagonal-sums (while b is odd).
• Next, eliminate the pairs (-1, 72),(-3, 24)/(1, 12) since they give large diagonal sums ( while b is small).
• The remainder (c/a) is (-8, 9)/(3, 4) that gives as 2 real roots: -8/3 and 9/4.
• Example 5. Solve: 24x² - 59x + 36 = 0.
• Solution. Both roots are positive. Write the (c/a) setup.
• Numerator: (1, 36) (2, 18) (4, 9) (6, 6)
• Denominator: (1, 24) (2, 12) (4, 6) (8, 3)
• First, eliminate the pairs (2, 18),(6, 6)/(2, 12),(4, 6) that give even-number diagonal sums (while b is odd).
• Next, eliminate the pairs (1, 36)/(1, 24) that gives large diagonal sums (while b = -59).
• The remainder c/a is (4, 9)/(8, 3) that gives as 2 real roots: 4/3 and 9/8.
## Community Q&A
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• Isn't the picture in Method 3 incorrect? 99 + (-25) = 74 , but it should be -74, shouldn't it?
wikiHow Contributor
99 + -25 would equal 74; adding a negative is the same as subtraction, so it would be 99 - 25 to get 74, not negative 74.
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## 5. Algebra 2
5.1 – Algebraic Fractions
• Simplifying Fractions:
• Can simplify algebraic fractions using division
• Where possible factorise the numerator and denominator, then cancel common factors.
• E.g. 5x2 – 245
• Addition and Subtraction of Algebraic Fractions:
• To add or subtract algebraic fractions:
1. Find the lowest common multiple of the denominators
2. Express fractions in terms of the L.C.M
3. Simplify numerators
• E.g. 5x + 2x (Lowest common denominator of 6 and 9 is 18)
5.2 – Changing the Subject of a Formula
• When changing the subject of a formula, you rearrange it so that you have a different subject
• To do this, move a term from one side of the equal sign to the other side and change the operation to do the opposite (inverse operation).
• E.g. Make ‘y’ the subject of the formula x(y – z) = a
x(y – z) = a
xy – xz = a
xy = a + xz
y = a + xz
x
5.3 – Variation
• Direct Variation:
• Several ways of expressing a relationship between two quantities ‘x’ and ‘y’
• E.g.
• ‘x’ varies directly as ‘y’
• ‘x’ varies as ‘y’
• ‘x’ is proportional to ‘y’
• These all mean the same and can be written in symbols:
• x ∝ y
• ‘∝’ sign can be replaced with ‘= k’ where k is a constant:
• x = ky
• E.g. Suppose x = 6 when y = 24
6 = k × 24
k = 6/24
k = ¼
You can then write x = ¼ × y, this enables you to find any value of ‘x’ for any value of ‘y’ and vice versa.
• Inverse Variation:
• Several ways of expressing an inverse relationship between two variable ‘x’ and ‘y’
• E.g.
• ‘x’ is inversely proportional to ‘y’
• ‘x’ varies inversely as ‘y’
• These all mean the same and can be written in symbols:
• x ∝ 1/y
• ‘∝’ sign can be replaced with ‘= k’ where k is a constant:
• x = k × 1/y
5.4 – Indices
• Rules of Indices:
1. an × am = an + m
• E.g. 42 × 44 = 46
1. an ÷ am = an – m
• E.g. 35 ÷ 32 = 33
1. (an)m = an m
• E.g. (24)5 = 220
1. a-n = 1/an
• E.g. 6-2 = 1/62
1. a1/n means the nth root of a
• E.g. 81/2 = √8
1. am/n means the nth root of a raised to the power m
• E.g. 53/2 = (√5 )3
5.5 – Inequalities
• < (less than)
• E.g. x < 5 (x is less than 5)
• > (greater than)
• E.g. x > 8 (x is greater than 8)
• (less than or equal to)
• E.g. x ≤ 12 (x is less than or equal to 12)
• (greater than or equal to)
• E.g. x ≥ 4 (x is greater than or equal to 4)
• Solving Inequalities:
• You use the same procedure used for solving equations.
• Except when you multiply or divide by a negative number the inequality will be reversed
• E.g. 12 – 3x < 27
-3x < 15 (Subtract 12 from both sides)
x > -5 (Divide both sides by -3)
• The Number Line:
• You may need to represent inequalities on a number line.
• is used for < and > and means the end value is not included
• is used for ≤ and ≥ and means the end value is included
5.6 – Linear Programming
• In most linear programming problems, there will be two stages:
1. Interpret the information given as a series of simultaneous inequalities and display them graphically.
Investigate some characteristic of the points in the unshaded solution set.
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Unit circle
File:Unit circle.png
Illustration of a unit circle. t is an angle measure.
In mathematics, a unit circle is a circle with unit radius, i.e., a circle whose radius is 1. Frequently, especially in trigonometry, "the" unit circle is the circle of radius 1 centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane. The unit circle is often denoted S1; the generalization to higher dimensions is the unit ball.
If (x, y) is a point on the unit circle in the first quadrant, then x and y are the lengths of the legs of a right triangle whose hypotenuse has length 1. Thus, by the Pythagorean theorem, x and y satisfy the equation
$\displaystyle x^2 + y^2 = 1 \,\!$
Since x2 = (−x)2 for all x, and since the reflection of any point on the unit circle about the x- or y-axis is also on the unit circle, the above equation holds for all points (x, y) on the unit circle, not just those in the first quadrant.
One may also use other notions of "distance" to define other "unit circles"; see the article on normed vector space for examples.
Trigonometric functions on the unit circle
File:UnitCircle.png
Relationship of trigonometric functions on the circle
The trigonometric functions cosine and sine may be defined on the unit circle as follows. If (x, y) is a point of the unit circle, and if the ray from the origin (0, 0) to (x, y) makes an angle t from the positive x-axis, (where the angle is measured in the counter-clockwise direction), then
$\displaystyle \cos(t) = x \,\!$
$\displaystyle \sin(t) = y \,\!$
The equation x2 + y2 = 1 gives the relation
$\displaystyle \cos^2(t) + \sin^2(t) = 1 \,\!$
The unit circle also gives an intuitive way of realizing that sine and cosine are periodic functions, with the identities
$\displaystyle \cos t = \cos(2\pi k+t) \,\!$
$\displaystyle \sin t = \sin(2\pi k+t) \,\!$
for any integer k.
These identities come from the fact that the x- and y-coordinates of a point on the unit circle remain the same after the angle t is increased or decreased by any number of revolutions (1 revolution = 2π radians).
File:Circle-trig6.png
All of the trigonometric functions can be constructed geometrically in terms of a unit circle centered at O.
When working with right triangles, sine, cosine, and other trigonometric functions only make sense for angle measures more than zero and less than π/2. However, using the unit circle, these functions have sensible, intuitive meanings for any real-valued angle measure.
In fact, not only sine and cosine, but all of the six standard trigonometric functions — sine, cosine, tangent, cotangent, secant, and cosecant, as well as archaic functions like versine and exsecant — can be defined geometrically in terms of a unit circle, as shown at right.
Circle group
Complex numbers can be identified with points in the Euclidean plane, namely the number a + bi is identified with the point (a, b). Under this identification, the unit circle is a group under multiplication, called the circle group. This group has important applications in math and science; see circle group for more details.
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# kindbewusst Gruppe
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# Algebra 1: A Step-by-Step Approach to Learning Algebra
## Algebra 1: What Is It and Why Is It Important?
Algebra 1 is a branch of mathematics that deals with variables, expressions, equations, functions, graphs, and inequalities. It is often the first math course taken in high school, and it lays the foundation for more advanced math subjects such as geometry, calculus, and statistics. In this article, we will explore what algebra 1 is, what topics and concepts it covers, what formulas and rules it uses, why it is important to learn, how it can be applied in real life, and how to prepare for it.
## What Is Algebra 1?
Algebra 1 is a type of algebra that involves using letters (called variables) and numbers with mathematical symbols to represent unknown or changing quantities and to solve problems. For example, if x represents the number of apples in a basket, then we can write an expression like 2x + 5 to represent the total number of fruits in the basket if there are also five oranges. We can also write an equation like 2x + 5 = 17 to find out how many apples are in the basket by solving for x.
## algebra 1
### Definition and examples of algebra 1
According to Wikipedia, algebra (from Arabic الجبر (al-jabr) 'reunion of broken parts, bonesetting ') is the study of variables and the rules for manipulating these variables in formulas; it is a unifying thread of almost all of mathematics. Algebra 1 consists of the general/basic concepts of algebra. It introduces evaluating equations and inequalities, real numbers, and their properties, which include additive and multiplicative identities, inverse operations, and the distributive and commutative properties.
Some examples of algebra 1 problems are:
• Solve the equation for x: 3x - 7 = 11
• Simplify the expression: (2x + 3)^2 - (x - 4)^2
• Graph the function: y = -2x + 5
• Solve the system of equations by elimination: x + y = 7; x - y = -1
• Find the domain and range of the function: f(x) = sqrt(x - 2)
### Topics and concepts covered in algebra 1
Algebra 1 is divided into numerous topics to help for a detailed study. According to Khan Academy, some of the topics covered in Algebra 1 are:
• Algebra foundations: Overview and history of algebra, introduction to variables, substitution and evaluating expressions, combining like terms, introduction to equivalent expressions, division by zero.
• Solving equations & inequalities: Linear equations with variables on both sides, linear equations with parentheses, analyzing the number of solutions to linear equations, linear equations with unknown coefficients, multi-step inequalities, compound inequalities.
• Working with units: Rate conversion, appropriate units, word problems with multiple units.
• Linear equations & graphs: Two-variable linear equations intro, slope, horizontal & vertical lines, x-intercepts and y-intercepts, applying intercepts and slope, modeling with linear equations and inequalities.
• Forms of linear equations: Intro to slope-intercept form, graphing slope-intercept equations, writing slope-intercept equations, point-slope form, standard form, summary: Forms of two-variable linear equations.
• Systems of equations: Introduction to systems of equations, solving systems of equations with substitution, solving systems of equations with elimination, equivalent systems of equations, number of solutions to systems of equations, systems of equations word problems.
• Inequalities (systems & graphs): Checking solutions of two-variable inequalities, graphing two-variable inequalities, modeling with linear inequalities.
## <li Data analysis and probability are two important topics in algebra 1 that help us understand and interpret data, as well as make predictions and decisions based on data. In this article, we will explore what data analysis and probability are, what topics and concepts they cover, what formulas and rules they use, why they are important to learn, how they can be applied in real life, and how to prepare for them. What are data analysis and probability?
Data analysis is the process of collecting, organizing, displaying, summarizing, and interpreting data. Data are pieces of information that can be numerical or categorical. For example, the heights of students in a class, the colors of cars in a parking lot, and the scores of a test are all types of data. Data analysis helps us to understand the patterns, trends, relationships, and differences in the data, as well as to draw conclusions and make decisions based on the data.
Probability is the measure of how likely an event is to occur. An event is any possible outcome of an experiment or a situation. For example, rolling a die, flipping a coin, and choosing a card from a deck are all experiments with different possible events. Probability helps us to quantify the uncertainty and randomness in the events, as well as to make predictions and inferences based on the data.
### Definition and examples of data analysis and probability
According to Wikipedia, data analysis is \"a process of inspecting, cleansing, transforming, and modeling data with the goal of discovering useful information, informing conclusions, and supporting decision-making.\" Data analysis can be done using various methods and tools, such as tables, charts, graphs, statistics, calculators, spreadsheets, software, etc.
Some examples of data analysis problems are:
algebra 1 practice problems
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algebra 1 functions and graphs
algebra 1 linear equations and graphs
algebra 1 systems of equations and inequalities
algebra 1 absolute value and piecewise functions
algebra 1 exponential growth and decay
algebra 1 factoring polynomials
algebra 1 simplifying expressions
algebra 1 solving equations and inequalities
algebra 1 working with units
algebra 1 domain and range of a function
algebra 1 sequences and series
algebra 1 rational expressions and equations
algebra 1 introduction to functions
algebra 1 solving quadratic equations by factoring
algebra 1 completing the square
algebra 1 quadratic formula and discriminant
algebra 1 arithmetic and geometric sequences
algebra 1 exponential functions and models
algebra 1 compound interest and e
algebra 1 logarithmic functions and their graphs
algebra 1 properties of logarithms and exponential equations
algebra 1 radical expressions and equations
algebra 1 rational exponents and nth roots
algebra 1 solving radical equations and inequalities
algebra 1 pythagorean theorem and distance formula
• How can we organize the scores of 50 students on a test using a frequency table and a histogram?
• How can we compare the heights of boys and girls in a class using dot plots and measures of center and spread?
• How can we use matrices to represent and manipulate data from a survey?
• How can we use box-and-whisker plots to identify outliers and compare distributions of data?
• How can we use linear regression to model the relationship between two quantitative variables and make predictions?
According to Wikipedia, probability is \"the branch of mathematics concerning numerical descriptions of how likely an event is to occur or how likely it is that a proposition is true.\" Probability can be calculated using various rules and formulas, such as the addition rule, the multiplication rule, the complement rule, conditional probability, Bayes' theorem, etc.
Some examples of probability problems are:
• What is the probability of rolling an even number on a fair die?
• What is the probability of drawing a heart or a face card from a standard deck of cards?
• What is the probability of getting at least one head when flipping three coins?
• What is the probability that a randomly chosen student from a class likes math given that he or she likes science?
• What is the probability that it will rain tomorrow based on historical weather data?
## Why are data analysis and probability important?
Data analysis and probability are important because they help us to:
• Analyze and interpret data from various sources and contexts, such as science, business, sports, health, education, etc.
• Make informed decisions and judgments based on evidence and reasoning.
• Predict future outcomes and behaviors based on patterns and trends.
• Evaluate claims and arguments based on data and logic.
• Communicate findings and conclusions using appropriate representations and language.
### Benefits of learning data analysis and probability
Learning data analysis and probability can benefit us in many ways, such as:
• Enhancing our critical thinking and problem-solving skills.
• Developing our numerical literacy and statistical reasoning.
• Increasing our awareness and understanding of real-world phenomena.
• Cultivating our curiosity and creativity.
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# 数学代写|组合学代写Combinatorics代考|The Bernoulli and Cauchy numbers
#### Doug I. Jones
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## 数学代写|组合学代写Combinatorics代考|Power sums
The Stirling numbers can be used to calculate sums of powers of consecutive integer numbers. Let us first recall the well-known formulas from elementary mathematics:
$$1+2+3+\cdots+n=\frac{1}{2}\left(n^2+n\right)$$
The sum of the squares of the first $n$ numbers is
$$1^2+2^2+3^2+\cdots+n^2=\frac{1}{6}\left(2 n^3+3 n^2+n\right) .$$
For the third powers we have that
$$1^3+2^3+3^3+\cdots+n^3=\frac{1}{4}\left(n^4+2 n^3+n^2\right) .$$
We can easily recognize a rule: the sums of the powers of the first $n$ positive integers can be expressed as polynomials of $n$. It would be interesting to know what are the coefficients of these polynomials in general. The sums on the left-hand sides will be called power sums.
The Stirling numbers pop up in this problem via the relation (2.46):
$$x^p=\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k}$$
On the left-hand side there is the $p$ th power of a real number $x$. Sum on $x$ from 1 to $n-1$. (We shall see later that summing up to $n-1$ in place of $n$ makes the resulting expressions somewhat simpler looking.)
$$1^p+2^p+\cdots+(n-1)^p=\sum_{x=1}^{n-1} \sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k} .$$
If we could find a simple polynomial expression for the right-hand side we would be ready. What we immediately see is that we should know the sums of the falling factorials, since
$$\sum_{x=1}^{n-1} \sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k}=\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} \sum_{x=1}^{n-1} x^{k}$$
## 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions
The power sum formula (5.5) can easily be generalized so that we are summing arithmetic progressions’ powers. Arithmetic sequences are of the form $r, r+m, r+2 m, \ldots, r+n m$. We will prove very briefly that
$$\sum_{i=1}^n(r+i m)^p=m^p \sum_{k=1}^{p+1}(k-1) !\left{\begin{array}{c} p+1 \ k \end{array}\right}\left[\left(\begin{array}{c} n+r / m \ k \end{array}\right)-\left(\begin{array}{c} r / m \ k \end{array}\right)\right],$$
given that $m$ is not zero ${ }^2$.
Applying (5.5) with $n=n+t$ and $n=t$, and subtracting the second expression from the first, it comes that
$$\sum_{i=1}^n(t+i)^p=\sum_{k=1}^{p+1}(k-1) !\left{\begin{array}{c} p+1 \ k \end{array}\right}\left[\left(\begin{array}{c} n+t \ k \end{array}\right)-\left(\begin{array}{l} t \ k \end{array}\right)\right] .$$
As this formula is valid for all positive integer $t$, and on both sides we have polynomials of $t$, it follows that the formula is valid for all real $t$. Let us write $t$ as a rational number $t=r / m$. Then multiply both sides of the last formula with $m^p$. We get that
$$\sum_{i=1}^n(r+i m)^p=m^p \sum_{k=1}^{p+1}(k-1) !\left{\begin{array}{c} p+1 \ k \end{array}\right}\left[\left(\begin{array}{c} n+r / m \ k \end{array}\right)-\left(\begin{array}{c} r / m \ k \end{array}\right)\right] .$$
This is exactly the formula that we wanted to verify.
# 组合学代考
## 数学代写|组合学代写Combinatorics代考|Power sums
$$1+2+3+\cdots+n=\frac{1}{2}\left(n^2+n\right)$$
$$1^2+2^2+3^2+\cdots+n^2=\frac{1}{6}\left(2 n^3+3 n^2+n\right) .$$
$$1^3+2^3+3^3+\cdots+n^3=\frac{1}{4}\left(n^4+2 n^3+n^2\right) .$$
## 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions
$r, r+m, r+2 m, \ldots, r+n m$. 我们将非常简要地 证明
Isum_{i=1 {}$^{\wedge} n(t+i)^{\wedge} p=$ Isum_ ${k=1}^{\wedge}{p+1}(k-1) ! \backslash$ left ${\backslash$ begin {
Isum_{i=1 $}^{\wedge} n(r+i m)^{\wedge} p=m^{\wedge} p$ Isum_{k=1 $}^{\wedge}{p+1}(k-1) ! \backslash$ left
## 有限元方法代写
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## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)
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A few weeks ago, the Mathematical Ninja explained how to divide by 9 on the fly. This week, it’s multiplying that’s on his mind - a more common thing to have to do.
The multiplying method is easier, too. It’s just a case of adding and taking away - and writing a few numbers down. In this version, you work from left to right, which is quite the opposite of what you’re used to.
Here’s an example: $6134 \times 9 = 55,204$. How you do it:
1. Write down the first digit, followed by the next with a vinculum: ($6 \bar 6$). The vinculum means ‘minus’.
2. Add the second digit to the vinculummed number and add the same digit with a vinculum after what you’ve got: $6 \bar5 \bar 1$, because -6 + 1 = -5.
3. Repeat this for the remaining digits. You should have: $6 \bar 5 2 \bar 3$ $6 \bar 5 2 1 \bar 4$
4. Now go through - every time you see a vinculum, drop the preceding number by 1 and take the vinculummed number from 10 (so $6 \bar5$ becomes $5 5$ and $1 \bar 4$ becomes $0 6$). You end up with $55,206$, which is exactly what we wanted.
This is quite a tricky one to do in your head without practice, but doing it on paper can save you a lot of carrying - and once you’ve got the hang of it, you only need one line of working!
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# Geometric Mean
You are probably familiar with the arithmetic mean, also known as the average, which is the sum of n numbers divided by n. The geometric mean is another type of mean where we multiply the numbers together and then take the nth root of the product.
So the geometric mean is formally defined as the nth root of the product of n numbers:
For example, the geometric mean of two numbers is a square root of a product of these two numbers, and the geometric mean of five numbers is the fifth root of a product of these five numbers.
So the geometric mean is formally defined as the nth root of the product of n numbers:
## Example I
What is the geometric mean of 16 and 4?
The first step is to multiply the numbers: 16 * 4 = 64
Since there are two numbers, the next step is to take the square root of the product: √64 = 8.
So the geometric mean of 16 and 4 is 8.
## Where does it come from?
The term “geometric mean” comes from geometry. In a right triangle, the height drawn from the vertex of the right angle to the hypotenuse is the geometric mean between the two segments of the hypotenuse:
The formula above can be quickly derived using the Pythagorean theorem, and we will show how in the next few lines.
## The difference between arithmetic mean and geometric mean
In order for you to know which one to use, you need to understand the difference between them.
If you were to find the average of your class test scores, you would simply add the scores together and then divide them with the number of terms, to get the average.
For example, if five students took the test and their scores are 65%, 75%, 85%, 95% and 100%, the (arithmetic) average score is:
This solution makes sense because each test score is an independent event. If one student does poorly on the test, that does not affect the other students and their chances of doing poor or well.
However, there are some situations, especially in finance, where using the arithmetic does not make sense, which we’ll see in the next example.
## Example II
Let’s say that the returns of your investments for five years are +60%, -50%, +60%, -50% and +60%. The arithmetic average of these returns is 16%, but the geometric average of these same numbers is quite different.
If we write these percentages as numbers, we get 0.6, -0.5, 0.6, -0.5 and 0.6.
1. First add 1 to each number of the sequence (to avoid problems with negative numbers)
2. Find the geometric mean using the formula:
3. Subtract one from the result
Following these steps, we get that the geometric mean of these numbers is 5%, and that is the correct one and the actual return. When it comes to investment returns, the numbers are not independent of each other and thus we have to use the geometric mean.
Want to learn more? Check out our video lesson on Geometric Mean for more example problems.
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# Math Portfolio Type II
Extracts from this document...
Introduction
Mathematics Portfolio
Type II
Creating a Logistic Model
Description
A geometric population model takes the form where r is the growth factor and un is the population at year n. For example, if the population were to increase annually by 20%, the growth factor is r = 1.2, and this would lead to an exponential growth. If r = 1 the population is stable. A logistic model takes a similar form to the geometric, but the growth factor depends on the size of the population and is variable. The growth factor is often estimated as a linear function by taking estimates of the projected initial growth rate and the eventual population.
Part 1
Information which has been given in Part 1: -
1. 10,000 fishes are introduced into a lake
2. The population increase if 10,000 fishes are introduced into the lake would be by 50% in the first year.
3. The long term sustainable limit in this case would be 60,000
It has been given that the geometric population growth model takes the form . Now, if we have to find out the ordered pair (u0,r0): -
• It is mentioned in the description that unis the population at year n. Therefore, u0 will be the population at year 0, or the initial population of the lake which is 10,000.
Therefore, u0 = 10,000 ------------------- (1)
• r is the growth factor as mentioned in the description. As the population would increase by 50% in the first year, so r0 = 1 + 1 x
= 1 + 0.5
= 1.5 ------------------- (2)
So the first ordered pair (u0,r0) as shown in (1) and (2) would be (10000,1.5).
Now, if we have to find out the ordered pair (un, rn): -
• It is given in the question that un = 60,000.
Therefore, un = 60,000 ---------------------- (3)
• r is the growth factor as mentioned in the description. As found out in (3) that un = 60,000 which shows that the population has reached its long term sustainable limit where population is stable.
Therefore, rn = 1 ---------------------- (4)
Middle
60323
6
57375
16
59737
7
61892
17
60207
8
58378
18
59382
9
61218
19
60133
20
59893
The above tabulated data can be represented by the following logistic function graph: -
The table and graph confirm how higher growth rates above 2 lead to less stability in population. Here, the variance from the long term sustainable limit (which is 60000) is much more (64703 in the 3rd year) compared to when the initial growth rate was 2.3 or 2. The successive year sees a similar larger fall to 55573 in population compared to the other two cases. Consequently, the table confirms how a high initial growth rate leads to longer time for the population to stabilize at the long term sustainable limit. In this case, it is not attained in the first 20 years. Rather, upon further calculations we can find out that the population becomes stable at the long term sustainable limit in the 41st year compared to the 17th year when the initial growth rate is 2.3 .
Part 5
Let us see what happens when the initial growth rate is 2.9
The ordered pair (u0, r0) for the first pair when the predicted growth rate ‘r’ for the year is 2 will be (10000, 2.9).
(un, rn) = (60000, 1) according to the description.
We can get the equation of the linear growth factor by entering these sets of ordered pairs in the STAT mode of the GDC Casio CFX-9850GC PLUS.
Thus, we obtain a linear graph which can be modeled by the following function: -
y = (-3.8 x 10-5)x + 3.28
Here, the variable y can be replaced by rn which represents the growth factor at some year n and the variable x can be replaced by un which represents the population at some year n. Therefore, the equation for the linear growth factor in terms of un will be: -
rn = (-3.8 x 10-5)un + 3.28
Now, to find the recursive logistic function, we shall substitute the value of rn calculated above in the following form un+1 = rn.un
un+1 = [(-3.8 x 10-5)un + 3.28]un
= (-3.8 x 10-5)un2 + 3.28un
To find out the changes in the pattern, let us see the following table which shows the first 20 values of the population obtained according to the logistic function un+1= (-3 x 10-5)un2 + 2.8un just calculated: -
n+1 un+1 n+1 un+1 0 10000 10 70738 1 29000 11 41872 2 63162 12 70716 3 55572 13 41919 4 64922 14 70720 5 52779 15 41910 6 67261 16 70719 7 48701 17 41911 8 69611 18 70719 9 44187 19 41911 20 70719
The above tabulated data can be represented by the following logistic function graph: -
We can infer from the data in the table and the graph how a high initial growth rate of 2.9 can lead to fluctuating populations every year. The long term sustainable limit of 60,000 is crossed at the end of the 2nd year itself, becoming 63162. But, in such a case, the environment (in this case the lake) cannot withhold such a population and thus there are a lot of fishes either dying out or migrating in the subsequent year which reduces the population to 55572. Yet, again in the next year, due to high growth rate, the population reaches well above even 63162 to 64922. Such a fluctuating trend is observed in the succeeding years until the end of the 16th and 17th year during and after which the population fluctuates between 70719 and 41911.
Part 6
To initiate an annual harvest of 5000 fishes, let us first find out at what point the fish population stabilizes when the initial growth rate, r = 1.5.
The following table shows the growth in the population over the first 20 years taking r = 1.5 and thus making use of the following recursive relation as found out in Part 2:-
un+1 = (-1 x 10-5) un2 + 1.6un
In this relation, the starting population is the initial population u0 = 10000.
n+1 un+1 n+1 un+1 0 10000 10 59772 1 15000 11 59908 2 21750 12 59963 3 30069 13 59985 4 39069 14 59994 5 47246 15 59997 6 53272 16 59999 7 56856 17 59999 8 58643 18 59999 9 59439 19 59999 20 59999
Here, we see that the population reaches a stable point, when r =1 at the end of the 16th year. According to the question, we shall initiate the harvest of 5000 fishes every year from this point onwards. The following changes are made in the original recursive relation:-
1. The starting population is taken to be 59999
2. In the relation un+1 = (-1 x 10-5) un2 + 1.6un , we subtract 5000 as this is the number which is removed every year. Thus, the new relation becomes
un+1 = (-1 x 10-5) un2 + 1.6un – 5000
It should be noted that although the start population or u0 is taken to be 59999, the starting year is the 16th year.
Entering these values in the RECUR mode of the GDC Casio CFX 9850GC Plus, we get the following table:-
n+1 un+1 n+1 un+1 16 59999 26 50041 17 54999 27 50024 18 52749 28 50014 19 51574 29 50008 20 50919 30 50005 21 50543 31 50003 22 50323 32 50001 23 50192 33 50001 24 50115 34 50000 25 50069 35 50000
We see that there is a declining trend in the fish population with an annual harvest of 5000 fishes when the growth rate is 1.5. However, it should be noted that with this annual harvest, the population finally stabilizes in the 34th year when it becomes 50000.
The graph is as follows: -
Thus, we can conclude that it is feasible to initiate an annual harvest of 5000 fishes once the stable population of 59999 is reached in the 16th year with the growth rate being 1.5.
Initially, a declining trend in the fish population is observed as 5000 fishes are harvested every year. Yet, from the 34th year onwards, we observe that the new stable fish population becomes 50,000 with an annual feasible harvest of 5000 fishes.
Part 7
Taking the same model in which the growth rate r = 1.5, we shall investigate other harvest sizes.
n+1 un+1 n+1 un+1 0 10000 10 59772 1 15000 11 59908 2 21750 12 59963 3 30069 13 59985 4 39069 14 59994 5 47246 15 59997 6 53272 16 59999 7 56856 17 59999 8 58643 18 59999 9 59439 19 59999 20 59999
• When the harvest size = 2500 fishes
We shall initiate the harvest of 2500 fishes after the 16th year when the population stabilizes at 59999. The following changes are made in the original recursive relation:-
1. The starting population is taken to be 59999
2. In the relation un+1 = (-1 x 10-5) un2 + 1.6un , we subtract 2500 as this is the number which is removed every year. Thus, the new relation becomes
un+1 = (-1 x 10-5) un2 + 1.6un – 2500
Entering these values in the RECUR mode of the GDC Casio CFX 9850GC Plus, we get the following table:-
n+1 un+1 n+1 un+1 16 59999 26 55498 17 57499 27 55496 18 56437 28 55495 19 55948 29 55495 20 55715 30 55495 21 55602 31 55495 22 55547 32 55495 23 55520 33 55495 24 55507 34 55495 25 55501 35 55495
The graph is as follows: -
We see that when the annual harvest is 2500, there is a declining trend which stabilizes quite early, that is from the 28th year onwards compared to when the harvest is 5000 fishes. Also, the new stable population, 55495 is higher compared to when the harvest if 5000 fishes.
Thus, we can conclude that it is also feasible to initiate an annual harvest of 2500 fishes and the new stable population would be achieved from the 28th year onwards as 55495.
• When the harvest size = 7500
We shall initiate the harvest of 7500 fishes after the 16th year when the population stabilizes at 59999. The following changes are made in the original recursive relation:-
1. The starting population is taken to be 59999
2. In the relation un+1 = (-1 x 10-5) un2 + 1.6un , we subtract 7500 as this is the number which is removed every year. Thus, the new relation becomes
un+1 = (-1 x 10-5) un2 + 1.6un – 7500
The table settings are as follows: -
Entering these values in the RECUR mode of the GDC Casio CFX 9850GC Plus, we get the following table:-
n+1 un+1 n+1 un+1 n+1 un+1 n+1 un+1 16 59999 26 42774 36 42278 46 42249 17 52499 27 42642 37 42270 47 42248 18 48937 28 42544 38 42265 48 42248 19 46851 29 42470 39 42260 49 42248 20 45511 30 42415 40 42257 50 42248 21 44605 31 42374 41 42255 51 42248 22 43972 32 42342 42 42253 52 42247 23 43520 33 42319 43 42251 53 42247 24 43192 34 42301 44 42250 54 42247 25 42951 35 42288 45 42249 55 42247
Conclusion
-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 16000. This shall be entered in the GDC Casio CFX-9850GC Plus.
In this case also, we find that the population dies put during the 7th year which means that such a combination also is not sustainable.
• Let us try with an initial population size of 19000 fishes and initiate a harvest of 8000 fishes from the 1st year itself.
The new recursive function becomes un+1 = (-1 x 10-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 19000. This shall be entered in the GDC Casio CFX-9850GC Plus.
The graph and GDC show that the population dies out during the 14th year which means that this combination is not sustainable.
• When the initial population size is 20000 fishes and a harvest of 8000 fishes is initiated from the 1st year itself.
The new recursive function becomes un+1 = (-1 x 10-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 20000. This shall be entered in the GDC Casio CFX-9850GC Plus.
Using this combination, we have found out using the GDC that the population remains constant and stable starting from the 1st year itself. It neither depletes nor increases. Thus, this introductory fish population size of 20000 is the most sustainable for harvesting 8000 fishes.
• To prove that an initial population of 20000 fishes is the most suitable and sustainable for a harvest of 8000 fishes every year, we shall investigate using u0 = 19999.
The new recursive function becomes un+1 = (-1 x 10-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 19999. This shall be entered in the GDC Casio CFX-9850GC Plus.
The GDC shows that in such a case, the population has a decreasing trend which slowly increases with increasing number of years and finally, dies out during the 52nd year. Hence, an initial population size of 19999 is not sustainable.
These results help us to infer that a minimum initial population size of 20000 fishes is required to sustain a harvest of 8000 fishes every year from the 1st year.
Candidate Session No.: -1070-006
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For instance u3 is and u ? (?1? 10?5 )(u 2 ) ? 1.6u 3 2 2 ? (?1?10?5 )(1.5 ? 104 )2 ? 1.6(1.5 ?104 ) ? 2.18 ?104 Table 3.1. The population of fish in a lake over a time range of 20 years estimated using the logistic function model {3}.
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72 2.5 23469.129 9387.651463 12 30 128.452 72 2.5 23121.419 9248.567 13 32.5 126.491 72 2.5 22768.399 9107.360 14 35 124.499 72 2.5 22409.819 8963.928 15 37.5 122.475 72 2.5 22045.408 8818.163 16 40 120.416 72 2.5 21674.870 8669.948 17 42.5 118.322 72 2.5 21297.887 8519.155 18 45 116.190 72
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Again on 1&5, each contains a card with ?, therefore, it must be split into have?or no. P(1&5) = = 1.0870313 x 10-2 P(2&4) = = 1.0457516 x 10-2 P(6&10) = = 1.0182319 x 10-2 P(7&9) = = 0.9356725 x 10-2 P(all)
2. ## IB Math SL Portfolio Type 2 Population in China
830.7 1975 927.8 1980 998.9 1983 1030.1 1985 1070 1990 1155.3 1992 1171.7 1995 1220.5 1997 1236.3 2000 1267.4 2003 1292.3 2005 1307.6 2008 1327.7 Cubic model Researcher?s model The cubic model (although it is closest to the new data)
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# H.C.F. & L.C.M.
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Required number = H.C.F. of $$(91 - 43)$$, $$(183 - 91)$$ and $$(183 - 43)$$
= H.C.F. of $$48, 92$$ and $$140 = 4$$.
Enter details here
2.
The greatest four digit number which is divisible by 15, 25, 40, 75 is
We want 4 digit number, so option A is not the answer
Now we want greatest numnber, So out of remaining options, 9600 is greatest
9600 is divisible by 25, 75, 40, and 15
So answer is 9600
Enter details here
3.
Find the H.C.F. of 54, 288, 360
Using factorization method,
$$18 = 2 \times 3^2$$
$$288 = 2^5 \times 3^2$$
$$360 = 2^3 \times 3^2 \times 5$$
So H.C.F. will be minimum term present in all three,
i.e. $$2 \times 3^2 = 18$$
Enter details here
4.
Find the H.C.F. of
$$2^{2}\times 3^{2}\times 7^{2},2\times 3^{4} \times 7$$
HCF is Highest common factor, so we need to get the common highest factors among given values.
So we get,
$$2 \times 3 \times 3 \times 7 = 126$$
Enter details here
5.
The L.C.M. of two numbers is 14560 and their H.C.F. is 13. If one of them is 416, the other is
416 X number = 14560 X 13
Therefore, numbr is 455
Enter details here
6.
A number when divided by 893 the remainder is 193. What will be the remainder when it is divided by 47?
Number is divided by 893. Remainder = 193.
Also, we observe that 893 is exactly divisible by 47,
So now simply divide the remainder by 47
Remainder is 5
Enter details here
7.
The least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but which when divided by 13 leaves no remainder?
No answer description available for this question.
Enter details here
8.
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number,
$$(9999 - 399) = 9600$$
Enter details here
9.
3 birds fly along the circumference of a jungle. They complete one round in 27 minutes, 45 minutes and 63 minutes respectively. Since they start together, when will they meet again at the starting position?
We need the instance which means the LCM of times of all 3 birds.
Therefore, LCM = 9 X 3 X 5 X 7 = 945 = 945 minutes
Enter details here
10.
252 can be expressed as a product of primes as:
Clearly, $$252 = 2 \times 2 \times 3 \times 3 \times 7$$
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Direct Material Quantity Variance
Direct Material Quantity Variance (DMQV) measures how efficiently a manufacturing business can convert its raw materials into the final product.
In this post, I will explain what the direct material quantity variance is, its causes, and the different ways you can calculate it using examples. Don’t forget to test your understanding by solving the free quiz at the end of the post!
Direct Material Quantity Variance
Direct Material Quantity Variance is the change in the cost of direct materials used in the manufacturing process arising from the difference between the actual quantity of materials used during a period, and the standard quantity of materials required to produce the actual output.
Direct Material Quantity Variance is also known as Direct Material Usage Variance and Direct Material Efficiency Variance.
How to Calculate Material Quantity Variance
To calculate the direct material quantity variance, we measure the difference between the standard cost of materials that should have been used to produce the actual level of output and the standard cost of the actual quantity of materials used.
Another way to calculate the direct material quantity variance is to multiply the standard price of materials by the difference between the actual quantity of materials used in production and the standard quantity of materials that should have been used to produce the actual output.
You can calculate the standard quantity of materials by multiplying the standard quantity of materials per unit of output by the actual units of output produced in a given period.
Example
During a period, the Teddy Bear Company used 15,000 kilograms of stuffing material to produce 9000 teddy bears. The company had paid an average price of \$1.5 per kilogram of stuffing material.
The materials budget for the same period allows 2 kilograms of stuffing material for each teddy bear at a standard price of \$1 per kilogram.
Calculate the direct material quantity variance.
The first step in the calculation is to figure out how much stuffing material should be used to manufacture 9000 teddy bears (standard quantity).
Standard Quantity = Actual output × Standard usage per unit of output
= 9000 × 2
=18,000 KGs
Now that we know the standard quantity, we can use the DMQV formula to calculate the variance.
DMQV = Standard Quantity × Standard Price – Actual Quantity × Standard Price
DMQV = 18,000 × \$1 – 15,000 × \$1
DMQV = \$18,000 – \$15,000
DMQV = \$3,000 Favorable
The material quantity variance in this example is favorable because the company manufactured the output using a lesser quantity of materials than what was planned in the budget.
We can also calculate the quantity variance using the second formula.
DMQV = Standard Price × ( Standard Quantity – Actual Quantity)
DMQV = \$1 × ( 18,000 – 15,000 )
DMQV = \$1 × 3,000
DMQV = \$3,000 Favorable
You can check this video of mine for more examples of the material quantity variance.
Play Video
Favorable Material Quantity Variance
Material quantity variance is favorable if the actual quantity of materials used in manufacturing during a period is lower than the standard quantity that was expected for that level of output.
In other words, if the business has consumed fewer materials to produce a given level of output than expected, the material quantity variance is said to be favorable.
Reasons for a favorable material quantity variance include:
• The use of bigger or better equipment to process the materials for reducing production losses.
• An improvement in the quality of materials purchased.
• More efficient use of materials by the production staff through better hiring, training, and gaining more experience.
• Optimizing the manufacturing process to lower the wastage of materials and frequency of breakdowns.
• Overestimating the materials usage in the budget.
Unfavorable Material Quantity Variance
An adverse or unfavorable material quantity variance occurs when the actual volume of materials used in production exceeds the standard quantity that is expected for the level of output in a period.
Causes for an adverse material quantity variance include:
• An increase in the frequency of equipment failures leading to a spike in production losses.
• The purchase of substandard materials that are difficult to work with or have a higher than usual percentage of defects.
• Inefficiencies in the manufacturing process (e.g., not recycling wastage).
• Turnover of experienced production staff.
• Underestimating the materials usage in the budget.
Quiz
How much do you know about Direct Material Quantity Variance?
Take the free quiz below and find out!
Question 1
DMQV = _________ Price × (Standard Quantity × Actual Quantity)
Actual
Incorrect.
Standard
Spot on!
Question 2
The direct material quantity variance will be adverse if the actual quantity of fabric used in manufacturing 10,000 units of shirts is 30,000 meters and the standard amount of fabric allowed for a single shirt is 2.8 meters.
True
False
You're right!
The MQV should be favorable because the standard quantity of the fabric for making 10,000 shirts is 28,000 meters which is less than what was actually used (30,000 meters).
Question 3
How many units of materials were used in the production if:
• The standard price of materials is \$5 per unit.
• The actual output during the period is 1000 units.
• The standard quantity of materials is 2 liters per unit of output.
• The DMQV is \$3000 adverse.
1400
liters
2000
liters
Incorrect.
2600
liters
That is correct!
DMQV = SP × (SQ - AQ)
-3000 = 5 × (2×1000 - AQ)
-3000 / 5 = 2000 - AQ
-600 = 2000 - AQ
-600 - 2000 = -AQ
AQ = 2600
3000
liters
Not correct.
Question 4
What is the direct material quantity variance if:
• The actual quantity of materials used is 200 units.
• The standard quantity of materials for 1 unit of output is 1 KG.
• The actual output is 250 units.
• The standard price of materials is \$3 per KG.
• The actual price of materials of \$5 per KG.
\$250
Incorrect.
\$250
Not correct.
\$150
Favorable
Correct!
Standard Quantity = 250 × 1 = 250 KG
DMQV = SP × (SQ - AQ)
= \$3 × (250 - 200)
=\$150 Favorable
\$150
Incorrect.
How many questions did you answer correctly?
4 Master
3 Pass
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# Investigation: Proof of the triangle inequality
Lesson
We will now present a complete paragraph proof of the triangle inequality theorem.
Reading proofs takes much longer than reading text written in everyday language, and that's okay! Follow these tips for understanding and reading proofs:
• Always stop reading if you don't understand something. It's okay to go back and re-read a statement several times.
• If a writer draws something in a proof, try drawing it on your own with pencil and paper.
• After reading a few sentences, see if you can state back the meaning in your own words.
Consider a general triangle, \Delta PQR:
What we want to prove now is that PQ+QR>PR, PQ+PR>QR, and QR+PR>PQ.
It seems daunting to prove three things! But we begin by assuming that \overline{PQ} is the longest side (as it is in the diagram above). Even if two or more of the sides have the same length as \overline{PQ}, we are only assuming PQ\geq QR and PQ\geq PR.
This gives us two of the inequalities essentially for free. PQ\geq PR becomes PQ+QR\geq PR+QR by the addition property of (in)equality, and since QR is a nonzero amount, we can be sure that PQ+QR>PR on its own. A similar calculation establishes the second inequality.
So now we are left with the third inequality, which is going to take the most work. We are going to describe a construction and use it on a particular diagram, but we need to make sure it will work for any triangle and be careful as we proceed.
We perform the following steps:
1. Draw a circle of radius QR centered at R.
2. Extend \overline{PR} to \overrightarrow{PR}.
3. The ray from step 2 intersects the circle from step 1 at a point - we mark it and call X.
Here is the results of the process with our triangle:
As indicated on the diagram, the radii \overline{RX} and \overline{QR} are congruent, since they're radii of the same circle, so they have the same measure. By substitution, QR+PR>PQ is equivalent to PR+RX>PQ, and proving this second statement is the same as finishing the proof altogether.
Remove the ray and the circle and create \overline{XQ} to form the triangle \Delta RQX:
As we noted before, two of the sides of this triangle are congruent, and so this triangle is isosceles. Since base angles of an isosceles triangle are equal, we can write \angle RXQ\cong\angle RQX.
Now we consider the other triangle we formed, \Delta PQX:
The angle addition postulate tells us that m\angle PQR + m\angle RQX = m\angle PQX. If we subtract m\angle PQR from the left-hand side only, we break equality, and have the inequality m\angle RQX < m\angle PQX.
Using the established congruence \angle RXQ\cong\angle RQX, we can substitute the measure of \angle RXQ for the measure of \angle RQX and produce m\angle RXQ < m\angle PQX. Now \angle PXQ=\angle RXQ since P, R, and X all lie on the ray \overrightarrow{XR}, and so by substitution again we can write m\angle PXQ < m\angle PQX. We can then conclude that PQ < PX, as an angle of smaller measure is opposite the shorter side.
By the segment addition postulate we know that PX = PR + RX, and by substituting this into the last inequality we found in the previous paragraph we know that PQ < PR + RX. We once again use the fact that RX=QR and by substitution we have PQ < PR + QR. We can then rearrange this inequality to produce QR+PR>PQ, QED.
#### Extra Challenge
Now that you've read the paragraph proof of the triangle inequality theorem, how can you be sure you really understand it?
Try re-creating the proof in a new medium. For example, by explaining it in a video, cartoon strip, or as a two-column proof.
### Outcomes
#### GEO-G.CO.10
Prove and apply theorems about triangles.
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Physically measure the area you are going to sod as estimating could result in ordering too little or too much sod. It is more efficient to take the time required to measure twice and order once. Areas should be measured in square feet.
It is easier and more accurate to measure square or rectangular areas.
If you have an irregular area, divide it into squares and rectangles and calculate the area of each. Add together the totals of all the areas measured to determine the square footage of sod needed for your project.
You should order about 5% additional sod to allow for cutting and shaping around your landscape.
# Calculating Areas
Basic terms that will help you to understand the formulas used to measure for sod are:
A = Area
W = Width
L = Length
H = Height
B = Base
D = Diameter
To find the area (A) of a square or rectangle, multiply its length (L) times its width (W). Using the illustrations below:
(1) The area of the rectangle is L × W or 10 ft. × 30 ft. = 300 square feet.
(2) The area of the square is L × W or 20 ft. × 20 ft. = 400 square feet.
Call now and get delivery this week, weather permitting:
# Circle (r × r × 3.14 = A)
To find the area (A) of a circle multiply its radius (r) by itself and then multiply by 3.14.
The easiest way to determine the radius (r) of a circle is to measure its diameter (D), which is the distance from one edge of the circle to the other, and divide the resulting number by 2 (r = D ÷ 2). In the circle below D = 60 feet so 60 ÷ 2 = 30 feet = r. Using the illustration below:
The area of the circle is r × r × 3.14 or 30 ft. × 30 ft. × 3.14 = 2,826 square feet.
# Triangle [(B × H) ÷ 2 = A]
To find the area (A) of a triangle, multiply its base (B) times its height (H) and divide by 2. Using the illustration below:
The area of the triangle is (B × H) ÷ 2 = or (15 ft. × 10 ft.) = 150 ft. ÷ 2 = 75 square feet.
# Right Triangle [(B ÷ 2) x H = A]
To find the area (A) of a right triangle, divide its base (B) by 2 and multiply the resulting number by its height (H). Using the illustration below:
The area of the right triangle is (B ÷ 2) × H or 30 ft. ÷ 2 = 15 ft. × 20 feet = 300 square feet.
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# Fractions, Decimals, And Percentages
Fractions are merely about working with a part of a number. Fractions are breaking whole numbers down into smaller segments or pieces.
Fractions are commonly used in various measurements. Inches are often broken down into 16ths (sixteenths).
Onsego
## Online GED Classes
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##### Bite-size video lessons | Practice tests with explanation | 6 Bonuses
Here, we will try to make you understand fractions. Then we’ll address adding, subtracting, multiplying, and dividing fractions.
1. Express the following fraction as a decimal.
$$\frac{687}{1000}$$
Question 1 of 2
2. Express the following fraction as a decimal.
$$\frac{2}{4}$$
Question 2 of 2
This lesson is provided by Onsego Online GED Prep.
Next lesson: Basics of Fractions
This lesson is a part of our GED Math Study Guide.
### Video Transcription
Rational Numbers
You’ve heard of rational numbers before. Another term for fractions is rational numbers.
The word “rational” is derived from the word ratio. When referring to rational numbers, it is actually a ratio of 2 (two) integers.
The one on top is what we call the “numerator,” and the one on the bottom is what we call the “denominator.”
Then there’s one more rule you need to remember: the bottom one (the denominator) can never be zero (0). If you see a zero on the bottom (the denominator), the division won’t work.
Decimals & Percentages
When we talk about “Decimals & percentages,” we mean just another way of working with numbers less than 1 (one). Decimals break down whole numbers into segments of 10 (ten). You may have tenths, hundredths, thousandths, and any value of 10.
A fast way of recognizing a decimal is looking for your decimal point. All numbers to the right side of your decimal point are the parts of that whole number.
Now, why should we be talking about percentages right here? Well, percentages are just a small step away from decimals.
In fact, percentages are just decimals that we multiplied by 1 (one) hundred. When you’re seeing a value like 79% (seventy-nine percent), that value equals .79 (seventy-nine hundredths).
Percentages are generally used with numbers that have values between 0 (zero) and 1 (one) hundred percent, where zero percent is 0 (nothing), and one (1) hundred percent represents the entire thing.
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# ZingPath: Solving Linear Inequalities
Searching for
## Solving Linear Inequalities
Learn in a way your textbook can't show you.
Explore the full path to learning Solving Linear Inequalities
### Lesson Focus
#### Graphing Linear Inequalities in Two Variables
Pre-Algebra
Students graph linear inequalities in two variables on the coordinate plane.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Graph a linear inequality in two variables using a test point.
• Graph a linear inequality in two variables without using a test point.
### Everything You'll Have Covered
Graphing linear equations into variables compared to linear inequalities in two variables
You are probably familiar with graphing a linear equation in two variables. A linear equation is graphed by plotting two or more points on a coordinate plane that make the equation true. If the points satisfy the equation, they follow a straight line.
Consider the linear equation y = 2x + 2, shown on the right.
The ordered pairs graphed on the coordinate plane are:
(-3, -4), (-2, -2), (0, 2), (1, 4), (-4, -6)
All the numbers on the line satisfy the linear equation y = 2x + 2.
Graphing linear inequalities is similar to graphing linear equations, except that the line (called the boundary line for inequalities) may or may not be included in the solution set. Plus, a half-plane is included.
The linear equation is changed to an inequality by changing the equal sign to , resulting in the inequality y 2x + 2.
The related boundary line is y = 2x + 2. We know the graph of this line. To find which half-plane to shade, identify a test point. (0, 0) is a good point to use, since the calculation will be easier. Substituting the values in the inequality, we find that:
The test point makes the linear inequality false. So, the half-plane that does not include the test point is shaded. If the test point had made the inequality true, the other half-plane would be in the solution set.
Steps to graph a linear equation
• Identify the boundary line by graphing two ordered pairs.
• If there is a ? or ? inequality, we use a solid line to graph the boundary line. If there is a < or > inequality, we use a dashed line to graph the boundary line.
• Choose a test point to determine which half-plane is part of the solution set for the inequality. If the test point is part of the solution set, that half-plane satisfies the inequality. If the test point is not part of the solution set, the other half-plane satisfies the inequality.
• Shade the half-plane which satisfies the inequality.
Graphing a linear inequality without using a test point
You can determine the half-plane that satisfies the linear inequality without using a test point. Isolate y and then apply the following rules to shade the half-plane.
### Tutorial Details
Approximate Time 40 Minutes Pre-requisite Concepts Students should be familiar with these concepts: coordinate plane, definition of an inequality and a linear inequality in two variables, how to graph a linear equation in two variables, and solving linear equations in two variables. Course Pre-Algebra Type of Tutorial Skills Application Key Vocabulary boundary line, coordinate plane, graphing
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# IS 21 in the 7 times table?
## IS 21 in the 7 times table?
The 7 times table is: 1 × 7 = 7. 2 × 7 = 14. 3 × 7 = 21.
### What are the 24 tables?
24 is a very common number that we use: as 24 carats of gold, 24 hours of a day, 24 major and minor keys in Western tonal music, etc. Hence learning the 24 times table is very useful in daily life….
24 Times Table up to 10
24 × 1 = 24 24 × 6 = 144
24 × 4 = 96 24 × 9 = 216
24 × 5 = 120 24 × 10 = 240
What are all the 12 times tables?
12 times table
• 12 x 1 = 12.
• 12 x 2 = 24.
• 12 x 3 = 36.
• 12 x 4 = 48.
• 12 x 5 = 60.
• 12 x 6 = 72.
• 12 x 7 = 84.
• 12 x 8 = 96.
What are the 22 times tables?
Table of 22 is a multiplication table that consists of the multiplication of 22 with various whole numbers….
22 Times Table up to 10
22 × 1 = 22 22 × 6 = 132
22 × 2 = 44 22 × 7 = 154
22 × 3 = 66 22 × 8 = 176
22 × 4 = 88 22 × 9 = 198
## IS 80 in the 4 times table?
The above diagram will help us to read and write the 4 times table. Now we will learn how to do forward counting and backward counting by 4’s. Forward counting by 4’s: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, ……
### What do you multiply by 4 to get 100?
4 times 25 equals 100. This can also be expressed as 4 x 25 = 100.
What is the 13 table?
Hence, the 13 times table is obtained as follows: (1+0)3, (2+0)6, (3+0)9, (4+1)2, (5+1)5, (6+1)8, (7+2)1, (8+2)4, (9+2)7, (10+3)0 = 13, 26, 39, 52, 65, 78, 91, 104, 117, 130….Table of 13 up to 20.
13 × 11 = 143 13 × 16 = 208
13 × 15 = 195 13 × 20 = 260
What is the table of 36?
The repeated addition of 36 is the multiplication table of 36. For example, 36 + 36 + 36 = 3 × 36 = 108….Table of 36 up to 10.
36 × 1 = 36 36 × 6 = 216
36 × 2 = 72 36 × 7 = 252
36 × 3 = 108 36 × 8 = 288
36 × 4 = 144 36 × 9 = 324
36 × 5 = 180 36 × 10 = 360
## How do you teach a table to 12?
The Scary Bit
1. 10 x 12 = 120 a safe ledge to jump onto.
2. 9 x 12 = 108 you can climb down -12 from 120 to get here.
3. 8 x 12 = 96 memorize this one an extra ledge.
4. 7 x 12 = 84 climb up +24 from 60 to get here.
5. 6 x 12 = 72 climb up +12 from 60 to get here.
6. 5 x 12 = 60 (START HERE)
### Is 31 in a times table?
The repeated addition of 31 is the multiplication table of 31. For example, 31 + 31 + 31 = 3 × 31 = 93.
What is the table of 200?
Table of 200 up to 10
200 × 1 = 200 200 × 6 = 1200
200 × 2 = 400 200 × 7 = 1400
200 × 3 = 600 200 × 8 = 1600
200 × 4 = 800 200 × 9 = 1800
200 × 5 = 1000 200 × 10 = 2000
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• Factor Theorem of Polynomials – Definition, Proof & Examples
Factor Theorem of Polynomials – Definition, Proof & Examples
November 5, 2022
This post is also available in: हिन्दी (Hindi)
Polynomials which are special types of algebraic expressions are used to form polynomial equations, which encode a wide range of problems, from simple to complicated problems in mathematics; they are used to define polynomial functions, used in calculus and numerical analysis to approximate the other functions.
You can perform the four mathematical operations – addition, subtraction, multiplication, and division on polynomials. Factor theorem is used in factoring a polynomial and finding the roots of the polynomial. Let’s understand what the factor theorem of polynomials is and how it is used to solve problems.
What is Factor Theorem of Polynomials?
Factor theorem is used to factor the polynomials and to find the roots of the polynomials. It makes the process easy by removing all the known zeros from a given polynomial equation and leaving all the unknown zeros.
The factor theorem states that if $f(x)$ is a polynomial of degree $n$ greater than or equal to $1$, and $a$ is any real number, then $(x – a)$ is a factor of $f(x)$ if $f(a) = 0$.
In other words, we can say that $(x – a)$ is a factor of $f(x)$ if $f(a) = 0$.
This theorem is very helpful in the factorization of a polynomial.
Steps to Use Factor Theorem
Following are the steps to use the factor theorem of polynomials.
Step 1: Use the synthetic division of the polynomial method to divide the given polynomial $p(x)$ by the given binomial $(x − a)$
Step 2: After the completion of the division, confirm whether the remainder is 0. If the remainder is not zero, then it means that $(x – a)$ is not a factor of $p(x)$.
Step 3: Using the division algorithm, write the given polynomial as the product of $(x – a)$ and the quotient $q(x)$
Step 4: If it is possible, factor the quotient further.
Step 5: Express the given polynomial as the product of its factors.
Proof of Factor Theorem
Method 1
Let’s first consider a polynomial $p(x)$ that is being divided by $(x – a)$ only if $p(a) = 0$. By using the division algorithm, the given polynomial can be written as the product of its divisor and its quotient:
$\text{Dividend } = ( \text{Divisor } \times \text{ Quotient} ) + \text{ Remainder}$=>xp(x) = (x – a) q(x) + \text{ remainder}$. Here,$p(x)$is the dividend,$(x – a)$is the divisor, and$q(x)$is the quotient. From the remainder theorem, we get:$p(x) = (x – a) q(x) + p(a)$If we substitute$p(a) = 0$then the remainder is$0$,$=> p(x) = (x – a) q(x) + 0=> p(x) = (x – a) q(x)$Thus, we can say that$(x – a)$is a factor of the polynomial$p(x)$. Here we can see that the factor theorem is actually a result of the remainder theorem, which states that a polynomial$p(x)$has a factor$(x – a)$, if and only if,$a$is a root i.e.,$p(a) = 0$. Method 2 Let’s first consider a polynomial$p(x)$that is being divided by$(x – a)$only if$p(a) = 0$. By using the division algorithm, the given polynomial can be written as the product of its divisor and its quotient:$ \text{Dividend } = ( \text{Divisor } \times \text{ Quotient} ) + \text{ Remainder}
$p(x) = (x – a)q(x)+ p(a)$
If $(x – a)$ is a factor of $p(x)$, then the remainder must be zero.
$(x – a)$ exactly divides $p(x)$
Therefore, $p(a)=0$.
The following statements are equivalent for any polynomial $p(x)$
• The remainder is zero when $p(x)$ is exactly divided by $(x – a)$
• $(x – a)$ is a factor of $p(x)$
• $a$ is the solution to $p(x)$
• $a$ is a zero of the polynomial $p(x)$, or $p(a) = 0$
Factor Theorem Examples
Ex 1: Determine whether $(x + 1)$ is a factor of the polynomial $6x^4 + 7x^3 – 5x – 4$.
By factor theorem, if $p(a) = 0$, then $(x – a)$ is a factor of $p(x)$
Let $p(x) = 6x^4 + 7x^3 – 5x – 4$ and $(x – a) = (x + 1)$.
Therefore, $a = -1$
Thus, $p(-1) = 6(-1)^4 + 7(-1)^3 – 5(-1) – 4$
$= 6 – 7 + 5 – 4$
$=>p(-1) = 0$
Therefore, $(x + 1)$ is a factor of the polynomial $6x^4 + 7x^3 – 5x – 4$.
Ex 2: Determine whether $(x + 1)$ is a factor of the polynomial $2x^4 + 9x^3 + 2x^2 + 10x + 15$.
By factor theorem, if $p(a) = 0$, then $(x – a)$ is a factor of $p(x)$
Let $p(x) = 2x^4 + 9x^3 + 2x^2 + 10x + 15$ and $(x – a) = (x + 1)$.
Therefore, $a = -1$
Thus, $p(-1) = 2(-1)^4 + 9(-1)^3 + 2(-1)^2 + 10(-1) + 15$
$= 2 – 9 + 2 – 10 + 15$
$=>p(-1) = 0$
Therefore, $(x + 1)$ is a factor of the polynomial $2x^4 + 9x^3 + 2x^2 + 10x + 15$.
Ex 3: Determine whether $(2x + 1)$ is a factor of the polynomial $4x^3 + 4x^2 – x – 1$.
By factor theorem, if $p(a) = 0$, then $(x – a)$ is a factor of $p(x)$
Let $p(x) = 4x^3 + 4x^2 – x – 1$ and $(x – a) = (2x + 1)$.
$2x + 1 = 0 => 2x = – 1 => x = -\frac{1}{2}$
Therefore, $a = -\frac{1}{2}$
Thus, $p(-\frac{1}{2}) = 4(-\frac{1}{2})^3 + 4(-\frac{1}{2})^2 – (-\frac{1}{2}) – 1$
$= 4(-\frac{1}{8}) + 4(\frac{1}{4})^2 +\frac{1}{2} – 1$
$= -\frac{1}{2} + 1 + \frac{1}{2} – 1$
$=>p(-1) = 0$
Therefore, $(2x + 1)$ is a factor of the polynomial $4x^3 + 4x^2 – x – 1$.
Ex 4: Determine the value of $m$ if $(x + 3)$ is a factor of the polynomial $x^3 – 3x^2 – mx + 24$
By factor theorem, if $p(a) = 0$, then $(x – a)$ is a factor of $p(x)$
Let $p(x) = x^3 – 3x^2 – mx + 24$ and $(x – a) = (x + 3)$.
Therefore, $a = -3$
Thus, $p(-3) = (-3)^3 – 3(-3)^2 – m(-3) + 24$
$= -27 – 3(9) + 3m + 24$
$= -27 – 27 + 3m + 24$
$=>p(-1) = -30 + 3m$
Therefore, $p(-1) = 0 => -30 + 3m = 0 => 3m = 30 => m = \frac{30}{3} => m = 10$
Ex 5: Given that $f(x) = 2x^{3} – 7x^{2} – 10x + 32$. Use the factor theorem to show that $(x – 2)$ is a factor of $f(x)$. Factorize $f(x)$ completely.
By factor theorem, if $f(a) = 0$, then $(x – a)$ is a factor of $f(x)$
$f(x) = 2x^{3} – 7x^{2} – 10x + 32$
Let $(x – a) = (x – 2) => a = 2$
Therefore, $f(2) = 2(2^3) – 7(2^2) – 10(2) + 32 = 16 – 28 – 20 + 32 = 0$
Dividing $f(x) = 2x^{3} – 7x^{2} – 10x + 32$ by $(x – 2)$ using Synthetic Division, we get
Therefore, $2x^3 – 7x^2 – 10x + 32 = (x – 2)(2x^2 – 3x – 16)$
Now, consider $2x^2 – 3x – 16$
Using completing the square method, we get $2x^2 – 3x – 16 = 2(x^2 – \frac{3}{2}x – 8)$
$= 2(x^2 – 2 \times x \ times \frac{3}{4} – 8)$
$= 2(x^2 – 2 \times x \ times \frac{3}{4} + (\frac{3}{4})^2 – (\frac{3}{4})^2 – 8)$
$= 2(x^2 – 2 \times x \ times \frac{3}{4} + (\frac{3}{4})^2 – \frac{9}{16} – 8)$
$= 2(x^2 – 2 \times x \ times \frac{3}{4} + (\frac{3}{4})^2 – \frac{137}{16})$
$= 2(x – \frac{3}{4})^2 – \frac{137}{16}$
Therefore, $2(x – \frac{3}{4})^2 – \frac{137}{16} = 0$
$=> 2(x – \frac{3}{4})^2 – \frac{137}{16} = 0$
$=> 2(x – \frac{3}{4})^2 = \frac{137}{16}$
$=> (x – \frac{3}{4})^2 = \frac{137}{32}$
$=> x – \frac{3}{4} = \pm \frac{\sqrt{137}}{4\sqrt{2}}$
$=> x – \frac{3}{4} = \pm \frac{\sqrt{274}}{8}$
$=> x = \frac{3}{4} \pm \frac{\sqrt{274}}{8}$
$=> x = \frac{6 \pm \sqrt{274}}{8}$
$=> x = \frac{6 + \sqrt{274}}{8}$ and $x = \frac{6 – \sqrt{274}}{8}$
Therefore, $2x^2 – 3x – 16 = \left(x – \frac{6 + \sqrt{274}}{8} \right) \left(x – \frac{6 – \sqrt{274}}{8} \right)$
Thus, $2x^{3} – 7x^{2} – 10x + 32 = (x – 2)\left(x – \frac{6 + \sqrt{274}}{8} \right) \left(x – \frac{6 – \sqrt{274}}{8} \right)$
Difference Between the Remainder Theorem and Factor Theorem
Following are the differences between the remainder theorem and the factor theorem:
Practice Problems
1. Determine whether $(x +1)$ is a factor of the following polynomials.
• $3x^3 + 8x^2 – 6x – 5$
• $x^3 – 14x^2 + 3x + 12$
2. Given $f(x) = 2x^3 – 3x^2 – 39x + 20$. Use the factor theorem to show that $(x + 4)$ is a factor of $f(x)$. Factorize $f(x)$ completely.
3. Given $f(x) = 2x^{3} – 7x^{2} – 5x + 4$. Use the factor theorem to show that $(x + 1)$ is a factor of $f(x)$. Factorize $f(x)$ completely.
FAQs
What is a factor theorem?
The factor theorem states that if $p(x)$ is a polynomial of degree $n$ greater than or equal to $1$, and $a$ is any real number, then $(x – a)$ is a factor of $p(x)$ if $p(a) = 0$. It is mainly used to factor the polynomials and to find the $n$ roots of the polynomials.
What is the importance of the factor theorem?
Factor theorem is used to factor the polynomials and to find the $n$ roots of that polynomial. It is a special kind of remainder theorem that links the factors of a polynomial and its zeros. The factor theorem removes all the known zeros from a given polynomial function and leaves all the unknown zeros. The resultant polynomial has a lower degree in which the zeros are very easy to find.
What is the difference between a factor theorem and a remainder theorem?
The difference between the factor theorem and the remainder theorem is that the remainder theorem relates the remainder of the division of a polynomial by a binomial with the value of a function at a point. The factor theorem relates the factors of a given polynomial to its zeros.
Is the factor theorem and the remainder theorem the same?
No, the factor theorem and remainder theorem are not the same. While the remainder theorem relates the remainder of the division of a polynomial by a binomial with the value of a function at a point. The factor theorem relates the factors of a given polynomial to its zeros.
Conclusion
The factor theorem states that if $f(x)$ is a polynomial of degree $n$ greater than or equal to $1$, and $a$ is any real number, then $(x – a)$ is a factor of $f(x)$ if $f(a) = 0$. Factor theorem is used to factor the polynomials and to find the roots of the polynomials. It makes the process easy by removing all the known zeros from a given polynomial equation and leaving all the unknown zeros.
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How To Find The Area Of The Axial Section Of The Cone | The Science
The Science
# How to find the area of the axial section of the cone
The cone is a geometric body whose base is a circle and the lateral surface - all segments drawn from a point located outside the base plane to the ground. Direct cone, which is usually seen in the school geometry course, can be represented as a body formed by rotating a right triangle around one of the legs. Perpendicular section of the cone is a plane passing through its vertex perpendicular to the base.
### You will need:
Drawing a cone with the specified parameters Ruler Pencil Mathematical formulas and definitions The height of the cone radius cone base circle area of a triangle formula
## Instruction how to find the area of the axial section of the cone
Step 1:
Draw a cone with the specified parameters. Mark the center of the circle as O, and the top of the cone - as P. You need to know the radius of the base and the height of the cone. Remember the properties of the height of the cone. It is a perpendicular drawn from the apex to the base. The intersection point of the plane with the height of the cone at the base of the cone coincides with the forward center of the base circle. Build an axial section of the cone. It is formed by forming a base diameter of the cone, the diameter of which pass through the points of intersection with the circle. Mark received points A and B.
Step 2:
Axial section is formed by two rectangular triangles lying in one plane and have a common leg. Calculate the axial sectional area is possible in two ways. The first way - to find the area of a triangle and put them together. This is the most obvious method, but in fact it is no different from the classical calculation of an isosceles triangle area. So, you got two right-angled triangles, the overall height of the leg which is cone h, the second legs of - the base circle radius R, and the hypotenuses - generators of the cone. Since all three sides of the triangle are equal, then the triangles themselves also get equal, according to the third property of equal triangles. The area of a right triangle is equal to half the product of the other two sides, that is, S = 1 / 2Rh. The area of the two triangles will correspondingly equal to the product of the radius of circumference of the base to a height, S = Rh.
Step 3:
Axial section often considered as an isosceles triangle, whose height is the height of the cone. In this case, APV triangle whose base is equal to the circumference of the cone base diameter D, and a height equal to the height of the cone h. Its area is calculated according to the classical formula area of a triangle, that is, in the end we get the same formula S = 1 / 2Dh = Rh, where S - area of an isosceles triangle, R - base circle radius, and h - height of the triangle, which is both a cone height .
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## Book: RS Aggarwal - Mathematics
### Chapter: 18. Area of Circle, Sector and Segment
#### Subject: Maths - Class 10th
##### Q. No. 56 of Exercise 18B
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
56
##### In the given figure, PQ = 24 cm, PR = 7 cm and 0 is the centre of the circle. Find the area of the shaded region. [Take π = 3.14.]
Here we will subtract the area of right angle triangle PQR and semicircle from the area of entire circle.
Given PQ = 24 cm, PR = 7 cm
Consider ∆PQR, QPR = 90°
RQ2 = 242 + 72
RQ2 = 576 + 49
RQ2 = 625
Therefore Radius of the circle = half of RQ
r = 12.5 cm
Area of ∆PQR = 7×12
Area of ∆PQR = 84 cm2 eqn1
Area of semicircle = 245.3125 cm2 eqn2
Area of circle = πr2
Area of circle = π(12.52)
Area of circle = 3.14×156.25 (putting π = 3.14)
Area of circle = 490.625 cm2 eqn3
Area of shaded region = eqn3 – eqn2 – eqn1
Area of shaded region = 490.625 – 245.3125 – 84
Area of shaded region = 161.3125 cm2
Area of shaded region is 161.3125 cm2.
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What are all the factors, the prime factorization, and factor pairs of 1223?
To find the factors of 1223, divide 1223 by each number starting with 1 and working up to 1223
What is a factor in math ?
Factors are the numbers you multiply together to get another number. For example, the factors of 15 are 3 and 5 because 3 × 5 = 15.
The factors of a number can be positive or negative, but they cannot be zero.
The factors of a number can be used to find out if the number is prime or not.
A prime number is a number that has only two factors: itself and 1. For example, the number 7 is prime because its only factors are 7 and 1.
List all of the factors of 1223 ?
To calculate the factors of 1223 , you can use the division method.
1. Begin by dividing 1223 by the smallest possible number, which is 2.
2. If the division is even, then 2 is a factor of 1223.
3. Continue dividing 1223 by larger numbers until you find an odd number that does not divide evenly into 1223 .
4. The numbers that divide evenly into 1223 are the factors of 1223 .
Now let us find how to calculate all the factors of One thousand two hundred twenty-three :
1223 ÷ 1 = 1223
1223 ÷ 1223 = 1
As you can see, the factors of 1223 are 1 and 1223 .
How many factors of 1223 are there ?
The factors of 1223 are the numbers that can evenly divide 1223 . These numbers are 1 and 1223.
Thus, there are a total of 2 factors of 1223
What are the factor pairs of 1223 ?
Factor Pairs of 1223 are combinations of two factors that when multiplied together equal 1223. There are many ways to calculate the factor pairs of 1223 .
One easy way is to list out the factors of 1223 :
1 , 1223
Then, pair up the factors:
and (1,1223) These are the factor pairs of 1223 .
Prime Factorisation of 1223
There are a few different methods that can be used to calculate the prime factorization of a number. Two of the most common methods are listed below.
1) Use a factor tree :
1. Take the number you want to find the prime factorization of and write it at the top of the page
2. Find the smallest number that goes into the number you are finding the prime factorization of evenly and write it next to the number you are finding the prime factorization of
3. Draw a line under the number you just wrote and the number you are finding the prime factorization of
4. Repeat step 2 with the number you just wrote until that number can no longer be divided evenly
5. The numbers written on the lines will be the prime factors of the number you started with
For example, to calculate the prime factorization of 1223 using a factor tree, we would start by writing 1223 on a piece of paper. Then, we would draw a line under it and begin finding factors.
The final prime factorization of 1223 would be 1223.
2) Use a factorization method :
There are a few different factorization methods that can be used to calculate the prime factorization of a number.
One common method is to start by dividing the number by the smallest prime number that will divide evenly into it.
Then, continue dividing the number by successively larger prime numbers until the number has been fully factorised.
For example, to calculate the prime factorization of 1223 using this method, we keep dividing until it gives a non-zero remainder.
1223 ÷ 1223 = 1
So the prime factors of 1223 are 1223.
Frequently Asked Questions on Factors
What are all the factors of 1223 ?
The factors of 1223 are 1 and 1223.
What is the prime factorization of 1223 ?
The prime factorization of 1223 is 1223 or 12231, where 1223 are the prime numbers .
What are the prime factors of 1223 ?
The prime factors of 1223 are 1223 .
Is 1223 a prime number ?
A prime number is a number that has only two factors 1 and itself.
1223 it is a prime number because it has the factors 1 and 1223.
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# 13.4 Series and their notations
Page 1 / 18
• Use summation notation.
• Use the formula for the sum of the first n terms of an arithmetic series.
• Use the formula for the sum of the first n terms of a geometric series.
• Use the formula for the sum of an infinite geometric series.
• Solve annuity problems.
A couple decides to start a college fund for their daughter. They plan to invest \$50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned.
## Using summation notation
To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a series . Consider, for example, the following series.
$3+7+11+15+19+...$
The partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation represents the partial sum.
$\begin{array}{l}{S}_{1}=3\\ {S}_{2}=3+7=10\\ {S}_{3}=3+7+11=21\\ {S}_{4}=3+7+11+15=36\end{array}$
Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma , $\text{Σ},$ to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation , which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation , is the number used to generate the last term in a series.
If we interpret the given notation, we see that it asks us to find the sum of the terms in the series $\text{\hspace{0.17em}}{a}_{k}=2k$ for $k=1$ through $k=5.\text{\hspace{0.17em}}$ We can begin by substituting the terms for $k$ and listing out the terms of this series.
$\begin{array}{l}\begin{array}{l}\\ {a}_{1}=2\left(1\right)=2\end{array}\hfill \\ {a}_{2}=2\left(2\right)=4\hfill \\ {a}_{3}=2\left(3\right)=6\hfill \\ {a}_{4}=2\left(4\right)=8\hfill \\ {a}_{5}=2\left(5\right)=10\hfill \end{array}$
We can find the sum of the series by adding the terms:
$\sum _{k=1}^{5}2k=2+4+6+8+10=30$
## Summation notation
The sum of the first $n$ terms of a series can be expressed in summation notation as follows:
$\sum _{k=1}^{n}{a}_{k}$
This notation tells us to find the sum of ${a}_{k}$ from $k=1$ to $k=n.$
$k\text{\hspace{0.17em}}$ is called the index of summation , 1 is the lower limit of summation , and $n$ is the upper limit of summation .
Does the lower limit of summation have to be 1?
No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.
Given summation notation for a series, evaluate the value.
1. Identify the lower limit of summation.
2. Identify the upper limit of summation.
3. Substitute each value of $k$ from the lower limit to the upper limit into the formula.
4. Add to find the sum.
## Using summation notation
Evaluate $\sum _{k=3}^{7}{k}^{2}.$
According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of ${k}^{2}$ from $k=3$ to $k=7.$ We find the terms of the series by substituting $k=3\text{,}4\text{,}5\text{,}6\text{,}\text{\hspace{0.17em}}$ and $7$ into the function ${k}^{2}.$ We add the terms to find the sum.
$\begin{array}{ll}\sum _{k=3}^{7}{k}^{2}\hfill & ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\hfill \\ \hfill & =9+16+25+36+49\hfill \\ \hfill & =135\hfill \end{array}$
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
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# NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercise 7.5
NCERT Solutions For Class 6 Maths Chapter 7Â Fractions Exercise 7.5 provides students with knowledge about addition and subtraction of like fractions and steps to be followed in solving them. Students can use PDF of NCERT Solutions prepared by subject faculty having vast experience in the education industry. Like fractions contain the same denominator and students can solve problems at a greater speed using the PDF provided. The faculties prepare these Solutions for Chapter 7 Fractions Exercise 7.5 to help students score well in the exam.
## NCERT Solutions for Class 6 Chapter 7: Fractions Exercise 7.5 Download PDF
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### Access NCERT Solutions for Class 6 Chapter 7: Fractions Exercise 7.5
1. Write these fractions appropriately as additions or subtractions:
Solutions:
(a) Total number of parts each rectangle has = 5
No. of shaded parts in first rectangle = 1 i.e 1 / 5
No. of shaded parts in second rectangle = 2 i.e 2 / 5
No. of shaded parts in third rectangle = 3 i.e 3 / 5
Clearly, fraction represented by third rectangle = Sum of the fractions represented by first and second rectangle
Hence, 1 / 5 + 2 / 5 = 3 / 5
(b) Total number of parts each circle has = 5
We may observe that first, second and third circles represent 5, 3 and 2 shaded parts out of 5 equal parts respectively. Clearly, fraction represented by third circle is the difference between the fractions represented by first and second circles.
Hence, 5 / 5 – 3 / 5 = 2 / 5
(c) Here we may observe that first, second and third rectangles represents 2, 3 and 5 shaded parts out of 6 equal parts respectively. Clearly, fraction represented by third rectangle is the sum of fractions represented by first and second rectangles.
Hence, 2 / 6 + 3 / 6 = 5 / 6
2. Solve:
(a) 1 / 18 + 1 / 18
(b) 8 / 15 + 3 / 15
(c) 7 / 7 – 5 / 7
(d) 1 / 22 + 21 / 22
(e) 12 / 15 – 7 / 15
(f) 5 / 8 + 3 / 8
(g) 1 – 2 / 3 (1 = 3 / 3)
(h) 1 / 4 + 0 / 4
(i) 3 – 12 / 5
Solutions:
(a) 1 / 18 + 1 / 18
= (1 + 1) / 18
= 2 / 18
= 1 / 9
(b) 8 / 15 + 3 / 15
= (8 + 3) / 15
= 11 / 15
(c) 7 / 7 – 5 / 7
= (7 – 5) / 7
= 2 / 7
(d) 1 / 22 + 21 / 22
= (1 + 21) / 22
= 22 / 22
= 1
(e) 12 /15 – 7 / 15
= (12 – 7) / 15
= 5 / 15
= 1 / 3
(f) 5 / 8 + 3 / 8
= (5 + 3) / 8
= 8 / 8
= 1
(g) 1 – 2 / 3
= 3 / 3 – 2 / 3
= (3 – 2) / 3
= 1 / 3
(h) 1 / 4 + 0
= 1/ 4
(i) 3 – 12 / 5
= 15 / 5 – 12/ 5
= (15 – 12) / 5
= 3 / 5
3. Shubham painted 2 / 3 of the wall space in his room. His sister Madhavi helped and painted 1 / 3 of the wall space. How much did they paint together?
Solutions:
Wall space painted by Shubham in a room = 2 / 3
Wall space painted by Madhavi in a room = 1 / 3
Total space painted by both = (2 / 3 + 1 / 3)
= (2 + 1) / 3
= 3 / 3
= 1
∴ Shubham and Madhavi together painted 1 complete wall in a room.
4. Fill in the missing fractions.
(a) 7 / 10 – ▯ = 3 / 10
(b) ▯ – 3 / 21 = 5 / 21
(c) ▯ – 3 / 6 = 3 / 6
(d) â–¯ + 5 / 27 = 12 / 27
Solutions:
(a) Given 7 / 10 – ▯ = 3 / 10
▯ = 7 / 10 – 3 / 10
▯ = (7 – 3) / 10
â–¯ = 4 / 10
â–¯ = 2 / 5
(b) Given ▯ – 3 / 21 = 5 / 21
â–¯ = 5 / 21 + 3 / 21
â–¯ = (5 + 3) / 21
â–¯ = 8 / 21
(c) Given ▯ – 3 / 6 = 3 / 6
â–¯ = 3 / 6 + 3 / 6
â–¯ = (3 + 3) / 6
â–¯ = 6 / 6
â–¯ = 1
(d) Given â–¯ + 5 / 27 = 12 / 27
▯ = 12 / 27 – 5 /27
▯ = (12 – 5) / 27
â–¯ = 7 /27
5. Javed was given 5 / 7 of a basket of oranges. What fraction of oranges was left in the basket?
Solutions:
Fraction of oranges given to Javed = 5 / 7
Fraction of oranges left in the basket = 1 – 5 / 7
= 7 / 7 – 5 / 7
= (7 – 5) / 7
= 2 / 7
1. Mohd Azhan
Good
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##### Algebra I For Dummies
The basics of algebra involve symbols. Algebra uses symbols for quantities, operations, relations, or grouping. The symbols are shorthand and are much more efficient than writing out the words or meanings. But you need to know what the symbols represent, and the following list shares some of that info.
• Addition (+): This symbol means add or find the sum, more than, or increased by; the result of addition is the sum.
1 + 2 = 3
• Subtraction (–): This symbol means subtract, minus, or decreased by or less; the result is the difference.
3 – 2 = 1
• Multiplication (×, ∙, *): These symbols all mean multiply or times. The values being multiplied together are the multipliers or factors (in this example, 2 and 3), and the result is the product (in this example, 6). You'll see the dot (∙) more often than the times symbol (×) because the dot is easier to write and the times symbol can be confused with the variable x.
• Division (÷, −, /): The division, fraction line, and slash symbols all mean divide. The number to the left of the ÷ or / sign or the number on top of the fraction is the dividend (in this example, 6). The number to the right of the ÷ or / sign or the number on the bottom of the fraction is the divisor (in this example, 2). The result is the quotient (in this example, 3).
• Radical (√): This symbol means to take the square root of something — to find the number which, multiplied by itself, gives you the number under the sign. Here's an example:
• Absolute Value (| |): This symbol means to find the absolute value of the number between the two vertical lines, which is the number itself or its distance from zero on the number line. Absolute values are always positive. Here's an example:
• Ellipsis (...): This symbol means et cetera, and so on, or in the same pattern. You use an ellipsis in algebra when you have a long list of numbers and don’t want to have to write all of them.
• Pi (Π): The Greek letter pi refers to the irrational number, 3.14159…. Pi represents the relationship between the diameter and circumference of a circle.
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Hong Kong
Stage 1 - Stage 3
# Using similarity proportion to solve problems
Lesson
The sides lengths of similar shapes are in the same ratio or proportion. So once we know that two shapes are similar, we can solve any unknown side lengths by using the ratio.
Remember!
You can write the ratio of the big triangle to the little triangle or the little triangle to the big triangle. This is helpful as it means you can always have the unknown variable as the numerator.
#### Examples
##### Question 1
Find the value of $u$u using a proportion statement.
Think: Let's equate the ratios of matching sides.
Do:
$\frac{u}{14}$u14 $=$= $\frac{3}{21}$321 $=$= $\frac{1}{7}$17 (Multiply both sides by $14$14) $u$u $=$= $\frac{1\times14}{7}$1×147 (Now let's simplify) $=$= $\frac{14}{7}$147 (Keep going!) $u$u $=$= $2$2
##### Question 2
Consider the two similar triangles.
1. Solve for $x$x.
2. Solve for $c$c.
##### Question 3
A $4.9$4.9 m high flagpole casts a shadow of $4.5$4.5 m. At the same time, the shadow of a nearby building falls at the same point (S). The shadow cast by the building measures $13.5$13.5 m. Find $h$h, the height of the building, using a proportion statement.
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# Lesson 15
Solving Equations with Rational Numbers
## 15.1: Number Talk: Opposites and Reciprocals (5 minutes)
### Warm-up
The purpose of this number talk is to:
• Remind students that the sum of a number and a number of the same magnitude with the opposite sign is zero.
• Remind students that the product of a number and its reciprocal is one.
• Establish common vocabulary for referring to these numerical relationships.
There may not be time for students to share every possible strategy. Consider gathering only one strategy for the equations with one variable, and a few strategies for the equations with two variables, since these have many possible answers, and they serve to generalize the relationship and provide an opportunity to introduce or reintroduce relevant vocabulary.
### Launch
Display one equation at a time. (When you get to $$c \boldcdot d = 1$$, ensure students understand that $$c$$ and $$d$$ represent different numbers.) Give students 30 seconds of quiet think time per problem and ask them to give a signal when they have an answer and a strategy. Follow with a whole-class discussion.
Representation: Internalize Comprehension. To support working memory, provide students with sticky notes or mini whiteboards.
Supports accessibility for: Memory; Organization
### Student Facing
The variables $$a$$ through $$h$$ all represent different numbers. Mentally find numbers that make each equation true.
$$\frac35 \boldcdot \frac53 = a$$
$$7 \boldcdot b = 1$$
$$c \boldcdot d = 1$$
$$\text-6 + 6 = e$$
$$11 + f = 0$$
$$g + h = 0$$
### Anticipated Misconceptions
Some students may use the strategy of guess and check for equations with two variables and use inverse operations to solve equations with one variable. Ask these students what they notice about the numbers in the equation and encourage them to find the value of the variable without using inverse operations.
### Activity Synthesis
Ask students to share their reasoning for each problem. Record and display the responses for all to see.
If the following ideas do not arise as students share their reasoning, make these ideas explicit:
• The sum of a number and its opposite is 0.
• The product of a number and its reciprocal is 1.
• If you want to find a number that you can add to something and get 0 as a sum, use its opposite.
• If you want to find a number that you can multiply something by and get 1 as a product, use its reciprocal.
To involve more students in the conversation, ask some of the following questions:
• “Who can restate ___’s reasoning in a different way?”
• “Does anyone want to add on to _____’s strategy?”
• “Do you agree or disagree? Why?”
Speaking: MLR8 Discussion Supports.: Display sentence frames to support students when they explain their strategy. For example, "First, I _____ because . . ." or "I noticed _____ so I . . . ." Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class.
Design Principle(s): Optimize output (for explanation)
## 15.2: Match Solutions (10 minutes)
### Activity
Students solved equations of the form $$x+p = q$$ and $$px=q$$ in grade 6, but the equations only involved positive values. This activity bridges their understanding of a solution to an equation as a value that makes the equation true with their understanding of operations involving negative numbers from this unit. This activity builds on the work students have done in this lesson evaluating expressions at different values.
Monitor for students who:
• take an arithmetic approach by substituting in values and evaluating. (Does $$\text-2 \boldcdot (\text-4.5) = \text-9$$?)
• take an algebraic approach by writing an equivalent equation using an inverse operation (If $$\text-2 \boldcdot x=\text-9$$, then $$x=\text-9 \div \text-2$$.)
### Launch
Allow students 5 minutes quiet work time followed by whole class discussion.
The digital version has an applet that allows students to see how many correct answers they have at any time.
Action and Expression: Internalize Executive Functions. To support development of organizational skills, check in with students within the first 2-3 minutes of work time. Check to make sure students matched the correct solution to the first equation and can explain their reasoning with an arithmetic or algebraic approach.
Supports accessibility for: Memory; Organization
### Student Facing
Match each equation to a value that makes it true by dragging the answer to the corresponding equation. Be prepared to explain your reasoning.
### Launch
Allow students 5 minutes quiet work time followed by whole class discussion.
The digital version has an applet that allows students to see how many correct answers they have at any time.
Action and Expression: Internalize Executive Functions. To support development of organizational skills, check in with students within the first 2-3 minutes of work time. Check to make sure students matched the correct solution to the first equation and can explain their reasoning with an arithmetic or algebraic approach.
Supports accessibility for: Memory; Organization
### Student Facing
1. Match each equation to its solution.
1. $$\frac12 x=\text-5$$
2. $$\text-2x=\text-9$$
3. $$\text-\frac12 x=\frac14$$
4. $$\text-2x=7$$
5. $$x+\text-2 = \text-6.5$$
6. $$\text-2+x=\frac12$$
1. $$x = \text-4.5$$
2. $$x = \text-\frac12$$
3. $$x = \text-10$$
4. $$x = 4.5$$
5. $$x = 2 \frac12$$
6. $$x=\text-3.5$$
Be prepared to explain your reasoning.
### Activity Synthesis
Select an equation for which there is a student who took an arithmetic approach and a student who took an algebraic approach. Ask students to share their reasoning for why a solution is correct. Sequence arithmetic approaches before algebraic approaches. An example of an arithmetic approach: “I know that -3.5 is the solution to $$\text-2x=7$$, because I know that $$\text-2 \boldcdot (\text-3.5)=7$$. An example of an algebraic approach: “If $$\text-2x=7$$, then I know that $$x=7 \div \text-2$$.) Record their work, side by side, for all to see. If there are no equations for which students took both approaches, present both approaches anyway. Tell students that either approach is valid, but that in the next unit they will see some more complicated equations for which one approach might be simpler than the other.
Representing, Conversing, Listening: MLR7 Compare and Connect. Use this routine to support whole-class discussion. Ask students, “What is the same and what is different?” about how they matched the expressions. Connect strategies by showing the different ways operations are used in each approach (e.g., arithmetic method by substituting in values and evaluating; algebraic method by using an inverse operation). Use gestures, and color on the display to highlight these connections. This helps students use mathematical language as they reason about their strategies to evaluate equivalent equations.
Design Principle(s): Maximize meta-awareness; Support sense-making
## 15.3: Trip to the Mountains (20 minutes)
### Activity
In this activity, students interpret equations that represent situations (MP2). The purpose is for students to see that equations of the form $$x + p = q$$ can be solved by adding the opposite of $$p$$ to the equation, regardless of whether $$p$$ is positive or negative. Students also see that equations of the form $$px = q$$ can be solved by multiplying the equation by the reciprocal of $$p$$. Through this work, students see that the structure of equations can be used to reason about a path to a solution (MP7) even when negative values are included or when a variable can represent a negative number.
### Launch
Give students 5–6 minutes of quiet work time followed by whole-class discussion.
Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organizational skills in problem solving. For example, present one question at a time and monitor students to ensure they are making progress throughout the activity.
Supports accessibility for: Organization; Attention
### Student Facing
#### Are you ready for more?
A number line is shown below. The numbers 0 and 1 are marked on the line, as are two other rational numbers $$a$$ and $$b$$ .
Decide which of the following numbers are positive and which are negative.
$$a-1$$
$$a-2$$
$$\text-b$$
$$a+b$$
$$a-b$$
$$ab+1$$
### Anticipated Misconceptions
Students may be misled by words to add or multiply (or subtract or divide) by the wrong numbers. For example, the word "increased" in question 1 may lead students to simply add the numbers they see, while the words "three times as many" in question 4 may lead students to multiply the numbers in the problem. Encourage students to make sense of the situations by acting them out or using visual diagrams, which will help them understand the actions and relationships in the stories.
### Activity Synthesis
Tell students: “We learned four things about the hiking trip in this activity: the students were climbing, the temperature was falling, there were more students this year than last, and the cost of the trip was less this year than last.” Then ask them:
• “Think about how you knew what operation described the rise in elevation, fall in temperature, rise in number of students, and fall in the cost. How did you know whether the situation used adding or multiplying?” (This conversation can highlight the problem with relying on “key words” . For example, when students see “times as many” , they might want to multiply the numbers they see in the problem. Encourage students to make sense of the situations by acting them out or drawing diagrams.)
• “How did you decide how to solve for the unknown quantity?”
• “What are some ways to know that a situation involves negative values?”
## 15.4: Card Sort: Matching Inverses (10 minutes)
### Optional activity
The blackline master is a set of matching cards with fractions and integers. The students first recall the work from the previous section about additive inverses by matching them. They then match multiplicative inverses.
When matching multiplicative inverses, students should now use the fact that division follows the same structure as multiplication to identify that negative numbers require a negative inverse and positive numbers require a positive inverse. Monitor for students who use this step to make an initial sort.
### Launch
Remind students of the meaning of additive inverse and multiplicative inverse. If $$x + y = 0$$, then $$x$$ and $$y$$ are additive inverses. If $$x \boldcdot y = 1$$, then $$x$$ and $$y$$ are multiplicative inverses. Display these definitions for all to see and use as a reference while they work through this activity. Ask students for some examples of additive inverse, and add these to the display (for example, 7 and -7, because $$7+\text-7=0$$). Ask students for some examples of multiplicative inverses, and add these to the display (for example, $$\frac17$$ and 7, because $$\frac17 \boldcdot 7 = 1$$).
If necessary, demonstrate productive ways for partners to communicate during a matching activity. For example, partners take turns identifying a match and explaining why they think it is a match. The other partner either accepts their explanation, or explains why they don't think it's a match. Then they change roles for the next match.
Arrange students in groups of 2. Distribute one set of paper slips per group. Instruct students to first match the numbers that are additive inverses. Then, they will re-sort the same cards into pairs of multiplicative inverses. Consider asking students to pause their work after matching the additive inverses for discussion before proceeding to match multiplicative inverses.
Conversing: MLR8 Discussion Supports. Display sentence frames to support students as they explain their reasoning for each match. For example, "___ matches ___ because . . .", and "I know that ___ and ___ are additive/multiplicative inverses because . . ." Encourage students to respond to the matches their partner makes using, "I agree/disagree, because . . ."
Design Principle(s): Support sense-making; Cultivate conversation
### Student Facing
Your teacher will give you a set of cards with numbers on them.
1. Match numbers with their additive inverses.
2. Next, match numbers with their multiplicative inverses.
3. What do you notice about the numbers and their inverses?
### Anticipated Misconceptions
Students might need a reminder of the difference between the additive inverse and multiplicative inverse.
### Activity Synthesis
The most important thing to recognize is that multiplicative inverses require that the numbers have the same sign in order for the product to be positive, and so negative numbers require a negative multiplicative inverse and positive numbers require a positive inverse. Contrast this with additive inverses, which must have opposite signs in order for their sum to be 0. Select students that used that strategy, as well as some who used calculation, to share their thinking and draw out this conclusion.
## Lesson Synthesis
### Lesson Synthesis
In this lesson students represented situations with equations and used inverses as a strategy to solve them. Bring the activities together by asking students:
• How can we solve an equation like $$x + (\text- 9.2) = 7.5$$? (We can add the opposite of -9.2 to 7.5.)
• How can we solve an equation like $$x \boldcdot (\text- 9.2) = 7.5$$? (We can multiply 7.5 by the reciprocal of -9.2.)
• Suppose we know that 60 is $$\frac45$$ of a number. What is the difference between writing the equation $$\frac45x=60$$ and writing the equation $$x=60\boldcdot \frac54$$? (The first equation describes the situation while the second shows a way to rewrite the equation to solve for the unknown.)
## Student Lesson Summary
### Student Facing
To solve the equation $$x + 8 = \text-5$$, we can add the opposite of 8, or -8, to each side:
Because adding the opposite of a number is the same as subtracting that number, we can also think of it as subtracting 8 from each side.
\begin{align} x + 8 &= \text-5\\ (x+ 8) + \text-8&=(\text-5)+ \text-8\\ x&=\text-13 \end{align}
We can use the same approach for this equation:
\begin{align} \text-12 & = t +\text- \frac29\\ (\text-12)+ \frac29&=\left( t+\text-\frac29\right) + \frac29\\\text-11\frac79& = t\end{align}
To solve the equation $$8x = \text-5$$, we can multiply each side by the reciprocal of 8, or $$\frac18$$:
Because multiplying by the reciprocal of a number is the same as dividing by that number, we can also think of it as dividing by 8.
\begin{align} 8x & = \text-5\\ \frac18 ( 8x )&= \frac18 (\text-5)\\ x&=\text-\frac58 \end{align}
We can use the same approach for this equation:
\begin{align} \text-12& =\text-\frac29 t\\ \text-\frac92\left( \text-12\right)&= \text-\frac92 \left(\text-\frac29t\right) \\ 54& = t\end{align}
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2019 AMC 8 Problems/Problem 3"
## Problem 3
Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?
$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$
## Solution 1
each one is x+4/x so we are really comparing 4/11,4/15, and 4/13 where you can see 4/11>4/13>4/15 so the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
## Solution 2
We take a common denominator: $$\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.$$
Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
## Solution 3
When $\frac{x}{y}>1$ and $z>0$, $\frac{x+z}{y+z}<\frac{x}{y}$. Hence, the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$. ~ ryjs
This is also similar to Problem 20 on the AMC 2012.
## Solution 4(probably won't use this solution)
We use our insane mental calculator to find out that $\frac{15}{11} \approx 1.36$, $\frac{19}{15} \approx 1.27$, and $\frac{17}{13} \approx 1.31$. Thus, our answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
~~ by an insane math guy
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# Vectors and Scalars in Physics
Vectors and scalars allow us to describe physical quantities. Scalars are used for quantities that can be described completely using only one number and one unit. Vectors, on the other hand, have both a magnitude and a direction.
Examples of scalar quantities include mass, temperature, time, and density. In the case of vector quantities, two common examples are velocity and force.
##### PHYSICS
Relevant for
Learning about the differences between vectors and scalars.
See differences
##### PHYSICS
Relevant for
Learning about the differences between vectors and scalars.
See differences
## What are vectors and scalars in physics?
Vectors and scalars are two ways in which we can express physical quantities. The use of each depends on the nature of each quantity.
There are physical quantities that can be defined entirely by a single number and a unit. However, many other quantities have a direction and cannot be expressed by a single number.
A scalar quantity is a physical quantity that can be expressed by a single number. On the other hand, a vector quantity has a magnitude (the number) and a direction.
An illustration of this is the motion of a car: we must specify not only how fast it is moving, but we must also consider the direction in which it is moving.
Since vectors and scalars have different properties, we usually use letters with arrows, like $latex \vec{A}$, to indicate that vectors have direction.
## Characteristics and differences between vectors and scalars
### Main characteristics of scalars
• Scalar quantities are represented by a single number (magnitude) and a unit.
• Ordinary arithmetic operations apply to scalar quantities. For example, 3 km + 5 km = 8 km.
• Scalars are usually represented by common letters, such as A, T, S.
### Main characteristics of vectors
• Vectors are defined by a magnitude and a direction.
• Combining vectors requires a different set of operations than ordinary arithmetic.
• Vectors are usually represented by a letter with an arrow above it. For example, $latex \vec{A}, ~vec{V}$.
• Two vectors are parallel if they have the same direction.
• Two vectors are equal only if they have the same magnitude and the same direction.
• The magnitude of a vector is a scalar quantity.
## Examples of vectors and scalars
The following are some examples of vector quantities and scalar quantities:
We can understand the difference between vectors and scalars by considering distance (a scalar quantity) and displacement (a vector quantity).
Displacement is the change in the position of an object. Displacement is a vector quantity because we must specify the direction when we indicate how far an object has moved.
For example, walking 500 m north will not take us to the same place as walking 500 m south. These two displacements have the same magnitude, but different directions.
Note that displacement is not directly related to the total distance traveled.
We can represent the displacement by an arrow pointing in the direction of displacement:
The displacement is always a straight line from the starting point to the end point, i.e. it does not depend on the path taken, even if it is curved:
If an object goes to point P2 and returns to point P1, the displacement is 0. That is, the total displacement for a round trip is 0, regardless of the distance traveled:
## Vectors and scalars – Practice problems
Vectors and scalars quiz
You have completed the quiz!
#### Find the result of the addition of the scalar quantities: 13 kg + 15 kg.
Write the answer in the input box.
$latex ~~=$
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### Section 16.7 : Green's Theorem
2. Use Green’s Theorem to evaluate $$\displaystyle \int\limits_{C}{{\left( {6y - 9x} \right)dy - \left( {yx - {x^3}} \right)\,dx}}$$ where $$C$$ is shown below.
Show All Steps Hide All Steps
Start Solution
Okay, first let’s notice that if we walk along the path in the direction indicated then our left hand will be over the enclosed area and so this path does have the positive orientation and we can use Green’s Theorem to evaluate the integral.
From the integral we have,
$P = - \left( {yx - {x^3}} \right) = {x^3} - yx\hspace{0.25in}\hspace{0.25in}Q = 6y - 9x$
Remember that $$P$$ is multiplied by $$x$$ and $$Q$$ is multiplied by $$y$$ and don’t forget to pay attention to signs. It is easy to get in a hurry and miss a sign in front of one of the terms. It is also easy to get in a hurry and just assume that $$P$$ is the first term in the integral and $$Q$$ is the second. That is clearly not the case here so be careful!
Show Step 2
Using Green’s Theorem the line integral becomes,
$\int\limits_{C}{{\left( {6y - 9x} \right)dy - \left( {yx - {x^3}} \right)\,dx}} = \iint\limits_{D}{{ - 9 - \left( { - x} \right)\,dA}} = \iint\limits_{D}{{x - 9\,dA}}$
$$D$$ is the region enclosed by the curve.
Show Step 3
We’ll leave it to you to verify that the equation of the line along the top of the region is given by $$y = 3 - x$$. Once we have this equation the region is then very easy to get limits for. They are,
$\begin{array}{c} - 1 \le x \le 1\\ - 1 \le y \le 3 - x\end{array}$ Show Step 4
Now all we need to do is evaluate the double integral. Here is the evaluation work.
\begin{align*}\int\limits_{C}{{\left( {6y - 9x} \right)dy - \left( {yx - {x^3}} \right)\,dx}} & = \iint\limits_{D}{{x - 9\,dA}}\\ & = \int_{{ - 1}}^{1}{{\int_{{ - 1}}^{{3 - x}}{{x - 9\,dy}}\,dx}}\\ & = \int_{{ - 1}}^{1}{{\left. {\left( {x - 9} \right)y} \right|_{ - 1}^{3 - x}\,dx}}\\ & = \int_{{ - 1}}^{1}{{\left( {x - 9} \right)\left( {4 - x} \right)\,dx}}\\ & = \int_{{ - 1}}^{1}{{ - {x^2} + 13x - 36\,dx}}\\ & = \left. {\left[ { - \frac{1}{3}{x^3} + \frac{{13}}{2}{x^2} - 36x} \right]} \right|_{ - 1}^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{218}}{3}}}\end{align*}
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# How do you factor 4c^3-2c^2-6c?
Aug 30, 2016
$4 {c}^{3} - 2 {c}^{2} - 6 c = c \left(2 c - 3\right) \left(2 c + 2\right)$
#### Explanation:
Note that all of the terms are divisible by $c$, so we can separate that out as a factor first.
We can then factor the remaining quadratic by completing the square and using the difference of squares identity:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
with $a = \left(2 c - \frac{1}{2}\right)$ and $b = \frac{5}{2}$ as follows:
$4 {c}^{3} - 2 {c}^{2} - 6 c$
$= c \left(4 {c}^{2} - 2 c - 6\right)$
$= c \left({\left(2 c - \frac{1}{2}\right)}^{2} - \frac{1}{4} - 6\right)$
$= c \left({\left(2 c - \frac{1}{2}\right)}^{2} - \frac{25}{4}\right)$
$= c \left({\left(2 c - \frac{1}{2}\right)}^{2} - {\left(\frac{5}{2}\right)}^{2}\right)$
$= c \left(\left(2 c - \frac{1}{2}\right) - \frac{5}{2}\right) \left(\left(2 c - \frac{1}{2}\right) + \frac{5}{2}\right)$
$= c \left(2 c - 3\right) \left(2 c + 2\right)$
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# Question: What Are The Types Of Function In Math?
## What are the 4 types of functions?
The various types of functions are as follows:
• Many to one function.
• One to one function.
• Onto function.
• One and onto function.
• Constant function.
• Identity function.
• Quadratic function.
• Polynomial function.
## What are the 8 types of functions?
The eight types are linear, power, quadratic, polynomial, rational, exponential, logarithmic, and sinusoidal.
## What is function type of function?
In computer science and mathematical logic, a function type (or arrow type or exponential) is the type of a variable or parameter to which a function has or can be assigned, or an argument or result type of a higher-order function taking or returning a function.
## What are the two main types of functions?
What are the two main types of functions? Explanation: Built-in functions and user defined ones. The built-in functions are part of the Python language.
## WHAT IS function and its type?
1. Injective (One-to-One) Functions: A function in which one element of Domain Set is connected to one element of Co-Domain Set. 2. Surjective (Onto) Functions: A function in which every element of Co-Domain Set has one pre-image.
You might be interested: Readers ask: How Much Math In Engineering?
## How do you tell if a graph is a function?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.
## What are four examples of functions?
we could define a function where the domain X is again the set of people but the codomain is a set of number. For example, let the codomain Y be the set of whole numbers and define the function c so that for any person x, the function output c(x) is the number of children of the person x.
## Is a circle a function?
No. The mathematical formula used to describe a circle is an equation, not one function. For a given set of inputs a function must have at most one output. A circle can be described with two functions, one for the upper half and one for the lower half.
## What are the 3 types of function?
Types of Functions
• Algebraic Function: A function defined by an algebraic expression is called an algebraic function.
• Polynomial Function: A function of the form P(x)=amxn+an–1xn–1+⋯+a1x+a0.
• Linear Function:
• Quadratic Function:
• Cubic Function:
• Identity Function:
• Rational Function:
• Trigonometric Function:
## What is a function concept?
A function is a generalized input-output process that defines a mapping of a set of input values to a set of output values. A student must perform or imagine each action. A student can imagine the entire process without having to perform each action.
You might be interested: Quick Answer: What Is Integers In Math?
## What are the classification of functions?
Still other functions have logarithmic and exponential relationships and are classified as such. Algebraic functions are the most common type of function. These are functions that can be defined using addition, subtraction, multiplication, division, powers, and roots.
## What is a function list?
Function List Panel is a zone to display all the functions (or method) found in the current file. Function list contains a search engine (by using regular expression) and a panel to display the search result ( function list ).
## What is the most basic function?
The parent function is the most basic function in the family of functions, the function from which all the other functions in the family can be derived. A family of functions is a group of functions that can all be derived from transforming a single function called the parent function.
## What is the specific type of function called?
Functions can be identified according to the properties they have. A parabola is a specific type of function.
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# Statements Reasons
1. 3x - 7 = -4 1.
2. 3x - 7 + 7 = -4 + 7 2.
3. 3x + 0 = -4 + 7 3.
4. 3x + 0 = 3 4.
5. 3x = 3 5.
6. (1/3) 3x = 3 (1/3) 6.
7. (1/3) 3x = 1 7.
8. 1x = 1 8.
9. x = 1 9.
8 months ago
## To understand how each statement is justified, let's go through each step:
1. Starting with the equation 3x - 7 = -4. The goal is to isolate the "x" term on one side of the equation.
2. Adding 7 to both sides of the equation, we get 3x - 7 + 7 = -4 + 7. This step cancels out the -7 on the left side.
3. Simplifying both sides of the equation, we have 3x + 0 = -4 + 7. The zero on the left side does not change the value of 3x.
4. Further simplification gives us 3x + 0 = 3. This is because -4 + 7 equals 3.
5. Since adding 0 to any number does not change its value, we have 3x = 3.
6. Now, we want to divide both sides of the equation by 3. This step is done to isolate the "x" term.
7. Dividing both sides by 3 gives us (1/3) 3x = 3 (1/3). The 3 on the right side can be expressed as a fraction 3 (1/3) or 10/3.
8. Simplifying the left side of the equation, we have (1/3) 3x = 10/3. On the left side, the factor of 3 cancels out, leaving us with x.
9. Therefore, we have 1x = 10/3 or x = 10/3.
To summarize, the series of steps mentioned above illustrate the process of solving the equation 3x - 7 = -4 and finding the value of x, which in this case is x = 10/3.
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Given continuous function f(x)=1, x<=0, f(x) =2^xa, 0<x<=1, f(x) = (x^2-2x+b)/(-x+2), 1<x<=2, what is b^(a+2)? domain of f (-infinite,2)
embizze | Certified Educator
Given `f(x)={[1,x<=0],[2^x a,0<x<=1],[(x^2-2x+b)/(-x+2),1<x<2]}`
where f(x) is continuous, find the values of a and b.
** There is a typo in the problem. f(x) cannot be continuous at x=2 for any values of a or b since there is a vertical asymptote at x=2 for all values of a or b. The domain should be `(-oo,2)` so the inequality should be strict for x<2. **
For a function to be continuous it must be continuous at every point on its domain. Since each part of the piecewise function is everywhere continuous, we need only look at x=0,x=1 since these are the endpoints of the intervals.
Now a function is continuous at a point if the function is defined at the point, the limit exists at the point and the value of the function agrees with the limit. f(x) is defined at x=0 (f(0)=1) and defined at x=1 ` `(f(1)=2a ) so we need to check the limits.
For a limit of a function to exist at a point the left and right limits must agree.
`lim_(x->0^-)=1` since the function is constant on `(-oo,0)`
`lim_(x->0^+)=2^0 a=a` (The exponential function is continuous everywhere so we can use substitution to find the limit.)
So if f(x) is continuous, the limits must agree and a=1.
`lim_(x->1^-)=2^1 a=2a=2` substituting the known value of a.
`lim_(x->1^+)=((1)^2-2(1)+b)/(-(1)+2)=-1+b` (Again the rational function is continuous everywhere except x=2, so we can use substitution at x=1 to find the limit.)
So for f(x) to be continuous the limits must agree and -1+b=2 ==> b=3.
-----------------------------------------------------------------
Since f(x) is continuous a=1,b=3 and `b^(a+2)=3^3=27`
----------------------------------------------------------------
The graph with a=1,b=3:
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# gaussian elimination
Imagine you have a puzzle with different colored blocks that you need to sort in a certain order. You have a set of simple rules that tell you how to move the blocks around, like "you can switch the places of two blocks" or "you can add or subtract a certain amount of blocks from each other."
Gaussian elimination is like using those simple rules to figure out how to solve a bigger puzzle, but instead of blocks, you're dealing with numbers. It's a fancy math tool that helps you solve systems of equations (which are basically math puzzles with a lot of rules) by breaking them down into smaller, easier puzzles.
Here's how it works:
Let's say you have this system of equations:
3x + 2y - z = 1
2x - 2y + 4z = -2
-x + 1/2y - z = 0
You want to solve for x, y, and z - in other words, figure out what numbers x, y, and z need to be in order for all three equations to be true at the same time.
First, you can use the simple rules to get rid of one of the variables. In this case, let's get rid of x. We can do that by using the first equation to "isolate" x, which means getting x by itself on one side of the equation:
3x + 2y - z = 1
3x = 1 - 2y + z
x = (1 - 2y + z)/3
Now we can substitute this expression for x into the other two equations, effectively getting rid of x altogether. This gives us two new equations:
2(-2y + 4z)/3 - (1 - 2y + z)/3 = -2
-(1 - 2y + z)/3 + 1/2y - z = 0
Simplify these equations a bit and you get:
-4y + 8z - 1 + 2y - z = -6
-3y + 6z - 2 + 3z = 0
Now you can use the same strategy to isolate another variable (let's say y), and substitute it into the remaining equation to get an expression for z. Finally, you can substitute the values you've found for y and z back into any of the original equations to get x.
This process of breaking down the system of equations into smaller, easier puzzles is called Gaussian elimination. It's kind of like the "divide and conquer" strategy you might use to solve a big problem by breaking it up into smaller pieces. Gaussian elimination is a powerful tool that can be used in all sorts of real-life situations, like analyzing data or solving engineering problems. But at its core, it's just a fancy way of using simple rules to solve big puzzles.
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## Mrs. Ashe is planning to take her study group on a field trip to an amusement park. The regular cost is \$7.00 per person. There is a party s
Question
Mrs. Ashe is planning to take her study group on a field trip to an amusement park. The regular cost is \$7.00 per person. There is a party special that costs \$4.00 per person with an additional \$30.00 fee for a private room where students can eat lunch. Mrs. Ashe is trying to decide if she should use the regular price or the party special.
A) Write an equation in slope-intercept form (y=mx+b) to show the total cost of the regular price with x people, where x = number of students in the study group and y = the total cost. (50 points)
B) Write an equation in slope-intercept form (y=mx+b) to show the total cost of the party special with x people, where x = number of students in the study group and y = the total cost. (50 points)
in progress 0
2 months 2021-10-15T15:53:11+00:00 1 Answer 0 views 0
## Answers ( )
1. Hey there!
a) Your answer is y = 7x.
When we are making this equation, we need to find the values for m and b and plug those into y = mx + b.
The regular cost is \$7 per person. This is the value for m, since x is the amount of people.
There is no additional cost, so b = 0.
Plugging those values in, we have y = 7x.
b) Your answer is y = 4x + 30.
Let’s find the values for m and b again.
Since the party special costs \$4 per person, m is equal to 4.
There is an additional \$30 fee, which is the value for b because this fee will be there regardless of the amount of people.
Plugging those values in, we have y = 4x + 30.
Again, here are the answers:
a) y = 7x
b) y = 4x + 30
Hope this helps!
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.1: Points, Lines, and Planes
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Understand the terms point, line, and plane.
• Draw and label terms in a diagram.
## Review Queue
1. List and draw pictures of five geometric figures you are familiar with.
2. What shape is a yield sign?
3. Solve the algebraic equations.
1. 4x7=29\begin{align*}4x-7=29\end{align*}
2. 3x+5=17\begin{align*}-3x+5=17\end{align*}
Know What? Geometry is everywhere. Remember these wooden blocks that you played with as a kid? If you played with these blocks, then you have been “studying” geometry since you were a child.
How many sides does the octagon have? What is something in-real life that is an octagon?
Geometry: The study of shapes and their spatial properties.
## Building Blocks
Point: An exact location in space.
A point describes a location, but has no size. Examples:
Label It Say It
A\begin{align*}A\end{align*} point A\begin{align*}A\end{align*}
Line: Infinitely many points that extend forever in both directions.
A line has direction and location is always straight.
Label It Say It
line g\begin{align*}g\end{align*} line g\begin{align*}g\end{align*}
PQ\begin{align*}\overleftrightarrow{\text{PQ}}\end{align*} line PQ\begin{align*}PQ\end{align*}
QP\begin{align*}\overleftrightarrow{\text{QP}}\end{align*} line QP\begin{align*}QP\end{align*}
Plane: Infinitely many intersecting lines that extend forever in all directions.
Think of a plane as a huge sheet of paper that goes on forever.
Label It Say It
Plane M\begin{align*}\mathcal{M}\end{align*} Plane M\begin{align*}M\end{align*}
Plane ABC\begin{align*}ABC\end{align*} Plane ABC\begin{align*}ABC\end{align*}
Example 1: What best describes San Diego, California on a globe?
A. point
B. line
C. plane
Solution: A city is usually labeled with a dot, or point, on a globe.
Example 2: What best describes the surface of a movie screen?
A. point
B. line
C. plane
Solution: The surface of a movie screen is most like a plane.
Beyond the Basics Now we can use point, line, and plane to define new terms.
Space: The set of all points expanding in three dimensions.
Think back to the plane. It goes up and down, and side to side. If we add a third direction, we have space, something three-dimensional.
Collinear: Points that lie on the same line.
P,Q,R,S\begin{align*}P, Q, R, S\end{align*}, and T\begin{align*}T\end{align*} are collinear because they are all on line w\begin{align*}w\end{align*}. If a point U\begin{align*}U\end{align*} was above or below line w\begin{align*}w\end{align*}, it would be non-collinear.
Coplanar: Points and/or lines within the same plane.
Lines h\begin{align*}h\end{align*} and i\begin{align*}i\end{align*} and points A,B,C,D,G\begin{align*}A, B, C, D, G\end{align*}, and K\begin{align*}K\end{align*} are coplanar in Plane J\begin{align*}\mathcal{J}\end{align*}.
Line KF\begin{align*}\overleftrightarrow{KF}\end{align*} and point E\begin{align*}E\end{align*} are non-coplanar with Plane J\begin{align*}\mathcal{J}\end{align*}.
Example 3: Use the picture above to answer these questions.
a) List another way to label Plane J\begin{align*}\mathcal{J}\end{align*}.
b) List another way to label line h\begin{align*}h\end{align*}.
c) Are K\begin{align*}K\end{align*} and F\begin{align*}F\end{align*} collinear?
d) Are E,B\begin{align*}E, B\end{align*} and F\begin{align*}F\end{align*} coplanar?
Solution:
a) Plane BDG\begin{align*}BDG\end{align*}. Any combination of three coplanar points that are not collinear would be correct.
b) AB\begin{align*}\overleftrightarrow{AB}\end{align*}. Any combination of two of the letters A,C\begin{align*}A, C\end{align*} or B\begin{align*}B\end{align*} would also work.
c) Yes
d) Yes
Endpoint: A point at the end of a line.
Line Segment: A line with two endpoints. Or, a line that stops at both ends.
Line segments are labeled by their endpoints. Order does not matter.
Label It Say It
AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} Segment AB\begin{align*}AB\end{align*}
BA¯¯¯¯¯¯¯¯\begin{align*}\overline{BA}\end{align*} Segment BA\begin{align*}BA\end{align*}
Ray: A line with one endpoint and extends forever in the other direction.
A ray is labeled by its endpoint and one other point on the line. For rays, order matters. When labeling, put endpoint under the side WITHOUT an arrow.
Label It Say It
CD\begin{align*}\overrightarrow{CD}\end{align*} Ray CD\begin{align*}CD\end{align*}
DC\begin{align*}\overleftarrow{DC}\end{align*} Ray CD\begin{align*}CD\end{align*}
Intersection: A point or line where lines, planes, segments or rays cross.
Example 4: What best describes a straight road connecting two cities?
A. ray
B. line
C. segment
D. plane
Solution: The straight road connects two cities, which are like endpoints. The best term is segment, or C\begin{align*}C\end{align*}.
Example 5: Answer the following questions about the picture to the right.
a) Is line l\begin{align*}l\end{align*} coplanar with Plane V\begin{align*}\mathcal{V}\end{align*} or W\begin{align*}\mathcal{W}\end{align*}?
b) Are R\begin{align*}R\end{align*} and \begin{align*}Q\end{align*} collinear?
c) What point is non-coplanar with either plane?
d) List three coplanar points in Plane \begin{align*}\mathcal{W}\end{align*}.
Solution:
a) No.
b) Yes.
c) \begin{align*}S\end{align*}
d) Any combination of \begin{align*}P, O, T\end{align*} and \begin{align*}Q\end{align*} would work.
Further Beyond This section introduces a few basic postulates.
Postulates: Basic rules of geometry. We can assume that all postulates are true.
Theorem: A statement that is proven true using postulates, definitions, and previously proven theorems.
Postulate 1-1: There is exactly one (straight) line through any two points.
Investigation 1-1: Line Investigation
1. Draw two points anywhere on a piece of paper.
2. Use a ruler to connect these two points.
3. How many lines can you draw to go through these two points?
Postulate 1-2: One plane contains any three non-collinear points.
Postulate 1-3: A line with points in a plane is also in that plane.
Postulate 1-4: The intersection of two lines will be one point.
Lines \begin{align*}l\end{align*} and \begin{align*}m\end{align*} intersect at point \begin{align*}A\end{align*}.
Postulate 1-5: The intersection of two planes is a line.
When making geometric drawings, you need to be clear and label all points and lines.
Example 6a: Draw and label the intersection of line \begin{align*}\overleftrightarrow{AB}\end{align*} and ray \begin{align*}\overrightarrow{CD}\end{align*} at point \begin{align*}C\end{align*}.
Solution: It does not matter where you put \begin{align*}A\end{align*} or \begin{align*}B\end{align*} on the line, nor the direction that \begin{align*}\overrightarrow{CD}\end{align*} points.
Example 6b: Redraw Example 6a, so that it looks different but is still true.
Solution:
Example 7: Describe the picture below using the geometric terms you have learned.
Solution: \begin{align*}\overleftrightarrow{AB}\end{align*} and \begin{align*}D\end{align*} are coplanar in Plane \begin{align*}\mathcal{P}\end{align*}, while \begin{align*}\overleftrightarrow{BC}\end{align*} and \begin{align*}\overleftrightarrow{AC}\end{align*} intersect at point \begin{align*}C\end{align*} which is non-coplanar.
Know What? Revisited The octagon has 8 sides. In Latin, “octo” or “octa” means 8, so octagon, literally means “8-sided figure.” An octagon in real-life would be a stop sign.
## Review Questions
• Questions 1-5 are similar to Examples 6a and 6b.
• Questions 6-8 are similar to Examples 3 and 5.
• Questions 9-12 are similar to Examples 1, 2, and 4.
• Questions 13-16 are similar to Example 7.
• Questions 17-25 use the definitions and postulates learned in this lesson.
For questions 1-5, draw and label an image to fit the descriptions.
1. \begin{align*}\overrightarrow{CD}\end{align*} intersecting \begin{align*}\overline{AB}\end{align*} and Plane \begin{align*}P\end{align*} containing \begin{align*}\overline{AB}\end{align*} but not \begin{align*}\overrightarrow{CD}\end{align*}.
2. Three collinear points \begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*} and \begin{align*}B\end{align*} is also collinear with points \begin{align*}D\end{align*} and \begin{align*}E\end{align*}.
3. \begin{align*}\overrightarrow{XY}, \overrightarrow{XZ}\end{align*}, and \begin{align*}\overrightarrow{XW}\end{align*} such that \begin{align*}\overrightarrow{XY}\end{align*} and \begin{align*}\overrightarrow{XZ}\end{align*} are coplanar, but \begin{align*}\overrightarrow{XW}\end{align*} is not.
4. Two intersecting planes, \begin{align*}\mathcal{P}\end{align*} and \begin{align*}\mathcal{Q}\end{align*}, with \begin{align*}\overline{GH}\end{align*} where \begin{align*}G\end{align*} is in plane \begin{align*}\mathcal{P}\end{align*} and \begin{align*}H\end{align*} is in plane \begin{align*}\mathcal{Q}\end{align*}.
5. Four non-collinear points, \begin{align*}I, J, K\end{align*}, and \begin{align*}L\end{align*}, with line segments connecting all points to each other.
6. Name this line in five ways.
7. Name the geometric figure in three different ways.
8. Name the geometric figure below in two different ways.
9. What is the best possible geometric model for a soccer field? Explain your answer.
10. List two examples of where you see rays in real life.
11. What type of geometric object is the intersection of a line and a plane? Draw your answer.
12. What is the difference between a postulate and a theorem?
For 13-16, use geometric notation to explain each picture in as much detail as possible.
For 17-25, determine if the following statements are true or false.
1. Any two points are collinear.
2. Any three points determine a plane.
3. A line is to two rays with a common endpoint.
4. A line segment is infinitely many points between two endpoints.
5. A point takes up space.
6. A line is one-dimensional.
7. Any four points are coplanar.
8. \begin{align*}\overrightarrow{AB}\end{align*} could be read “ray \begin{align*}AB\end{align*}” or “ray “\begin{align*}BA\end{align*}.”
9. \begin{align*}\overleftrightarrow{AB}\end{align*} could be read “line \begin{align*}AB\end{align*}” or “line \begin{align*}BA\end{align*}.”
1. Examples could be triangles, squares, rectangles, lines, circles, points, pentagons, stop signs (octagons), boxes (prisms), or dice (cubes).
2. A yield sign is a triangle with equal sides.
1. \begin{align*}4x-7 = 29\!\\ {\;}\quad \ 4x = 36\!\\ {\;}\quad \ \ x = 9\end{align*}
2. \begin{align*}-3x+5 =17\!\\ {\;}\quad \ -3x = 12\!\\ {\;}\qquad \ \ x = -4\end{align*}
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# 008A Sample Final A, Question 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Question: Sketch ${\displaystyle 4x^{2}+9(y+1)^{2}=36}$. Give coordinates of each of the 4 vertices of the graph.
Foundations
1) What type of function is this? What type of graph is this?
2) What can you say about the orientation of the graph?
1) Since both x and y are squared it must be a hyperbola or an ellipse. Since the coefficients of the ${\displaystyle x^{2}}$ and ${\displaystyle y^{2}}$ terms are both positive the graph must be an ellipse.
2) Since the coefficient of the ${\displaystyle x^{2}}$ term is smaller, when we divide both sides by 36 the X-axis will be the major axis.
Solution:
Step 1:
We start by dividing both sides by 36. This yields ${\displaystyle {\frac {4x^{2}}{36}}+{\frac {9(y+1)^{2}}{36}}={\frac {x^{2}}{9}}+{\frac {(y+1)^{2}}{4}}=1}$.
Step 2:
Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1).
From the center mark the two points that are 3 units left, and three units right of the center.
Then mark the two points that are 2 units up, and two units down from the center.
Step 3:
Now draw an oval through the four points you just drew.
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## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 6 - Review Exercises - Page 798: 19
#### Answer
The two airplanes would be $861\text{ miles}$ apart.
#### Work Step by Step
The distance traveled by the first airplane in $2$ hours is: $325\times 2=650\text{ miles}$ The distance covered by the second airplane in 2 hours is: $300\times 2=600\text{ miles}$ We can illustrate the given situation by the figure given below, where the planes started from point A and the distance between them after 2 hours would be BC or a. At first, we will compute the measure of angle BAC using the straight angle property: \begin{align} & 66.5{}^\circ +\angle BAC+26.5{}^\circ =180{}^\circ \\ & \angle BAC=180{}^\circ -66.5{}^\circ -26.5{}^\circ \\ & =180-93{}^\circ \\ & =87{}^\circ \end{align} Now consider triangle ABC, Here, $A=87{}^\circ,b=600,c=650$. Using the law of cosines we will evaluate the side BC as: \begin{align} & B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}-2\cdot AC\cdot AB\cdot \cos A \\ & B{{C}^{2}}={{600}^{2}}+{{650}^{2}}-2\left( 600 \right)\left( 650 \right)\cos 87{}^\circ \\ & B{{C}^{2}}=741,678 \\ & BC\approx 861 \end{align} So, the distance between the airplanes after 2 hours will be $861$ miles.
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### How Many Chickens and Ducks?
Date: 8/16/96 at 10:22:45
From: Leong Theng Kwong
Subject: How Many Chickens and Ducks?
A man bought 20 chickens and ducks altogether, with a \$2 discount per
chicken and 50 cent discount per duck. He saved \$22 in all. How many
chickens and how many ducks did he buy?
Please explain how to do this problem. Thank you.
LTK
Date: 8/16/96 at 11:33:53
From: Doctor Patrick
Subject: Re: How Many Chickens and Ducks?
Hi!
To solve this let's start by making C stand for the number of chickens
and D stand for the number of ducks.
From the first part of the problem we know that the number of Chickens
plus the number of Ducks is equal to 20.
Or in other words, C + D = 20.
Now we turn to the other half of the problem. Since we know how much
the man saved per chicken and per duck, and what the total savings
was, we can write another equation to express that \$2 per chicken and
\$0.50 per duck gave a total of \$22. How do you think we can write
this?
How about 2C + 0.5D = 22?
Do you understand how I got that?
Now to solve these equations you will have to use both equations in
your favorite way for solving for two variables.
One way is like this:
Going back to the first equation (since its simpler) we can rewrite it
as C = 20 - D, right? Then, we go to the second equation and
substitute in 20 - D for the C. This way we only have one variable
left, which is easy to solve for.
Once we substitute for the C we get:
2(20-D) +.5D =22
Let's simplify the equation and get 40 - 2D + 0.5D = 22.
Do you see what I did?
Now we add like terms and get 40 - 1.5D = 22.
We can further simplify by subtracting 22 from both sides:
40 - 22 - 1.5D = 22-22, or 18 -1.5D=0
Now if we add 1.5D to both sides we get 18 = 1.5D
Now we solve for D by dividing each side by 1.5 to get D = 12.
Now that we know D we can substitute it back into one of the
equations to solve for C. Let's use C+D = 20 again since it is much
easier. When we substitute in the D = 12, we get C + 12 = 20.
Subtracting 12 from each side we find that C = 8.
Together we have C (chickens) = 8 and D(ducks) = 12. You should go
back and check this by putting these numbers into the problems and
making sure that they work. You may also want to try to solve it using
a different method, if you know one, for practice.
Good luck, let us know if we can be of more help.
-Doctor Patrick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Middle School Algebra
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# If $\left| z \right| = \left| \omega \right|,\omega \ne 0$and$\arg \left( z \right) + \arg \left( \omega \right) = \pi$, then $z =$${\text{a}}{\text{. }} - \omega \\ {\text{b}}{\text{. }}\omega \\ {\text{c}}{\text{. }}\varpi \\ {\text{d}}{\text{. }} - \varpi \\$
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332.1k+ views
Hint: Assume $z = \left| z \right|{e^{i\alpha }}$and $\omega = \left| \omega \right|{e^{i\beta }}$
Let, $z = \left| z \right|{e^{i\alpha }}.............\left( 1 \right),{\text{ }}\omega = \left| \omega \right|{e^{i\beta }}.........\left( 2 \right)$
Where $z$and $\omega$are complex numbers.
From equation 1,$\arg \left( z \right) = \alpha$and $\arg \left( \omega \right) = \beta$
According to question it is given that
$\arg \left( z \right) + \arg \left( \omega \right) = \pi \\ \Rightarrow \alpha + \beta = \pi \\ \Rightarrow \alpha = \pi - \beta ..........\left( 3 \right) \\$
From equation (1) and (3)
$z = \left| z \right|{e^{i\alpha }} \\ \Rightarrow z = \left| z \right|{e^{i\left( {\pi - \beta } \right)}} \\ \Rightarrow z = \left| z \right|{e^{i\pi }}{e^{ - i\beta }}........\left( 4 \right) \\$
Now from equation (2)
$\omega = \left| \omega \right|{e^{i\beta }}$
Now take conjugate on both sides
$\varpi = \overline {\left| \omega \right|{e^{i\beta }}} \\ \Rightarrow \varpi = \left| \varpi \right|{e^{ - i\beta }} \\ \Rightarrow {e^{ - i\beta }} = \frac{\varpi }{{\left| \varpi \right|}}..........\left( 5 \right) \\$
Now, from equation (4) and (5)
$\Rightarrow z = \left| z \right|{e^{i\pi }}{e^{ - i\beta }} \\ \Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| \varpi \right|}}} \right).......\left( 6 \right) \\$
Now as we know modulus of any complex numbers and its conjugate both are equal so, use this property
$\left| \omega \right| = \left| \varpi \right|$
Therefore from equation (6)
$\Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| \omega \right|}}} \right).........\left( 7 \right)$
Now it is given that
$\left| z \right| = \left| \omega \right|,\omega \ne 0$
Therefore from equation (7)
$\Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| z \right|}}} \right) \\ \Rightarrow z = \varpi {e^{i\pi }}........\left( 8 \right) \\$
Now according to Euler’s Theorem ${e^{ix}} = \cos x + i\sin x$
$\Rightarrow {e^{i\pi }} = \cos \pi + i\sin \pi$
Now we know $\cos \pi = - 1,{\text{ }}\sin \pi = 0$
$\Rightarrow {e^{i\pi }} = - 1 + 0 = - 1$
Therefore from equation (8)
$\Rightarrow z = \varpi {e^{i\pi }} \\ \Rightarrow z = - \varpi \\$
Hence, option (d) is correct.
Note: Whenever we face such types of problems, always assume the complex numbers in the form of $z = \left| z \right|{e^{i\alpha }}$and $\omega = \left| \omega \right|{e^{i\beta }}$, then use the given conditions to simplify it, then use the property that modulus of any complex numbers and its conjugate both are equal and finally using Euler’s Theorem we get the required result.
Last updated date: 02nd Jun 2023
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# SOLVING INEQUALITIES BY CLEARING THE FRACTIONS
To solve for the variable indicated in the question, we have to isolate the variable term in one side. For that, we will use inverse operations.
• Inverse operation for addition is subtraction.
• Inverse operation for subtraction is addition.
• Inverse operation for multiplication is division.
• Inverse operation of division is multiplication
To get rid of the fraction, we have to multiply it by the multiplicative inverse.
Problem 1 :
(-3/4)m - (1/8) ≤ (-1/4)
Solution :
(-3/4)m + (-1/8) ≤ (-1/4)
(-3/4)m + (-1/8) + 1/8 ≤ (-1/4) + 1/8
(-3/4)m ≤ (-1/8)
Since we multiply both sides by (-4/3), change inequality ≥ into ≤
(-3/4m) (-4/3) ≤ (-1/8) (-4/3)
m ≥ 1/6
Problem 2 :
(7/13)x - 1 > 1/2
Solution :
(7/13)x - 1 + 1 > (1/2) + 1
(7/13)x > 3/2
Multiply both sides by 13/7
(7/13)x (13/7) > (3/2) (13/7)
x > 39/14
Converting improper fraction into mixed fraction, we get
x > 2 11/14
Problem 3 :
(4/5) ≥ (2/3) - (2/7x)
Solution :
(4/5) ≥ (-2/7x) + (2/3)
Subtract 2/3 from both sides.
(4/5) - 2/3 ≥ (-2/7x) + 2/3 - 2/3
2/15 ≥ (-2/7)x
Since we multiply both sides by (-7/2), change inequality ≥ into ≤
(2/15) (-7/2≥ (-2/7)x (-7/2)
-7/15 ≤ x
Problem 4 :
(8/15x) – (17/30) < 7/10
Solution :
(8/15x) – (17/30) + 17/30 < (7/10) + 17/30
8/15x < 38/30
Multiply both sides by 15/8
(8/15)x (15/8) < (38/30) (15/8)
x < 19/8
Problem 5 :
(-4/11)z - 1 > (-8/11)
Solution :
(-4/11)z – 1 + 1 > (-8/11) + 1
(-4/11)z > 3/11
Since we multiply both sides by (-11/4), change inequality > into <
(-4/11)z (-11/4) > (3/11) (-11/4)
z < -3/4
Problem 6 :
(1/5k) + 14 ≤ 2/9
Solution :
Subtract 14 from both sides
(1/5k) + 14 – 14 ≤ 2/9 – 14
1/5k ≤ -124/9
Multiply both sides by 5
(1/5)k (5) ≤ (-124/9) (5)
k ≤ -620/9
Converting improper fraction into mixed fraction, we get
k ≤ -68 8/9
Problem 7 :
-31/4 < -13 + (7/8f)
Solution :
-31/4 < -13 + (7/8f)
-31/4 +13 < -13 + 13 + (7/8)f
21/4 < 7/8f
Multiply both sides by 8/7
(21/4) (8/7) < (7/8f) (8/7)
6 < f
Problem 8 :
(1/7r) + (53/56) > 6/7
Solution :
Subtract 53/56 from both sides
(1/7)r + (53/56) – (53/56) > (6/7) – (53/56)
1/7r > -5/56
Multiply both sides by 7
(1/7)r (7) > (-5/56) (7)
r > -5/8
Problem 9 :
(5/6n) – (1/5) < -8/15
Solution :
(5/6)n – (1/5) + (1/5) < (-8/15) + (1/5)
5/6n < -5/15
Multiply both sides by 6/5
(5/6n) (6/5) < (-5/15) (6/5)
n < -2/5
Problem 10 :
(1/3) + (1/13d) ≥ 17/39
Solution :
Subtract 1/3 from both sides
(1/3) - (1/3) + (1/13)d ≥ (17/39) - (1/3)
1/13d ≥ 14/39
Multiply both sides by 13
(1/13d) (13) ≥ (14/39) (13)
d ≥ 14/13
d ≥ 1 1/13
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In sports such as athletics (or swimming, …) the gold medal is awarded to the person who finishes first, ie the person who reaches the finish line in the shortest time, and we call that movement the fast runner. the fastest (or the fastest swimmer).
Watching: What is Velocity
So what is the quantity to know how fast or slow the motion is? that is speed. In this article, we will learn in detail what speed is? How are the formulas for speed and the units of velocity written?
I. What is velocity?
– Velocity is the distance traveled in a unit of time
– Velocity indicates how fast and slow the motion is and is calculated as the distance traveled in a unit of time.
II. Speed calculation formula
Speed is calculated by the formula:
– In there:
v: Velocity of the object.
s: Distance traveled by the object
t: Time goes all the way
III. Unit of speed
– The unit of velocity depends on the unit of length and the unit of time
– Legal unit m/s; km/h
1m/s = 3.6km/h; 1km/h = 0.28m/s.
– The magnitude of the velocity is measured by the tachometer
* Question C5 on page 9 of Physics Textbook 8: a) The speed of a car is 36 km/h, that of a cyclist is 10.8 km/h of a train is 10 m/s. What does that indicate?
b) Which of the three movements above is the fastest and the slowest?
° Solution:
a) – The speed of a car is 36 km/h. In one hour, the car travels 36 km.
READ MORE Department of Environmental Resources What is English, English Names of Titles, Agencies
– The speed of a cyclist is 10.8 km/h indicating that in one hour, the cyclist travels 10.8 km.
– The speed of a train is 10m/s: in one second, the train travels 10m.
b) To compare the motions, the velocities of the movements must be compared to the same unit.
– The speed of the car is: v1 = 36 km/h = 36000m/3600s = 10 m/s
– The speed of the bicycle is: v2 = 10.8 km/h = 10800m/3600s = 3m/s
The speed of the train is 10m/s.
→ So the train is the fastest, the cyclist is the slowest.
* Question C6 on page 10 of Physics Textbook 8: A train in 1.5 hours travels a distance of 81 km. Calculate the speed of the train in km/h, m/s.
° Solution:
The speed of the train in km/h is:
– Convert s = 81(km) = 81000(m), t = 1.5 hours = 1.5.3600 = 5400(s)
⇒ The speed of the train in m/s is:
* Question C7, page 10 of Physics Textbook 8: A person rides a bicycle for 40 minutes with a speed of 12 km/h. What is the distance traveled?
° Solution:
– We have: 40 minutes = 2/3 hours
– From the formula for calculating the speed:
⇒ The distance traveled by the person is:
* Question C8, page 10 of Physics Textbook 8: A person walks with a speed of 4 km/h. Find the distance from home to work knowing the time it takes the person to get from home to work is 30 minutes.
° Solution:
– We have: 30 minutes = 0.5 hours.
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# Introduction to Probability and Statistics Chapter 6 The
• Slides: 30
Introduction to Probability and Statistics Chapter 6 The Normal Probability Distribution
Continuous Random Variables • Continuous random variables can assume infinitely many values corresponding to points on a line interval. • Examples: – Heights, Weights – Lifetime of a particular product – Experimental laboratory error
Continuous Random Variables • A smooth curve describes the probability distribution of a continuous random variable. • The depth or density of the probability, which varies with x, may be described by a mathematical formula f (x ), called the probability distribution or probability density function for the random variable x.
Properties of Continuous Random Variable x • The area under the curve is equal to 1. • P(a < x < b) = area under the curve between a and b. • There is no probability attached to any single value of x. That is, P(x = a) = 0.
Properties of Continuous Random Variable x • Total probability is 1 • P( x = a) = 0 • P( x a) = P( x < a) (not true when x is discrete) • P( a < x < b) is the area between a and b under the density curve • P( x < a) is the area to the left of a • P( x > a) is the area to the right of a
Continuous Probability Distributions • There are many different types of continuous random variables a. Uniform; b. Exponential; c. Normal. • We try to pick a model that – Fits the data well – Allows us to make the best possible inferences using the data.
Normal Distribution • The formula that generates the normal probability density is: • Standard Normal: m = 0, s = 1.
Normal Distribution • The shape and location of the normal curve changes as the mean and standard deviation change. • Mean locates the center of the curve; • Standard deviation determines the shape: 1. Large values of standard deviation reduce height and increase spread. 2. Small values increase height and reduce spread.
The Standard Normal Distribution • To find P(a < x < b), we need to find the area under the appropriate normal curve. • To simplify the tabulation of these areas, we standardize each value of x by expressing it as a z-score, the number of standard deviations s it lies from the mean m.
Standard Normal (z) Distribution • • • z has Mean = 0; Standard deviation = 1 Symmetric about z = 0 Total probability Total area under curve is 1; P ( z > 0 ) =. 5 Area to the right of 0 is 0. 5; P ( z < 0 ) =. 5 Area to the left of 0 is 0. 5.
Using Table 3 Use Table 3 to calculate the probability: P(z < 1. 36) ? Area to the left of 1. 36 P ( z < 1. 36) =. 9131 Area to the left of 1. 36 =. 9131
Using Table 3 Use Table 3 to find the probability: Area to the right of 1. 36 P( z > 1. 36) ? P ( z > 1. 36) = 1 - P ( z 1. 36) = 1 -. 9131 =. 0869
Using Table 3 Use Table 3 to calculate the probability: (Area between) P(-1. 20 < z <1. 36) = P ( z < 1. 36) - P ( z <-1. 2) =. 9131 -. 1151 =. 7980 P(-1. 20 < z <1. 36)? Area between -1. 2 and 1. 36
Check Empirical Rule • within 3 standard deviations P(-3 < z <3) =. 9987 -. 0013=. 9974 Remember the Empirical Rule: Approximately 99. 7% of the measurements lie within 3 standard deviations of the mean. P(-1 < z <1) =. 8413 -. 1587 =. 6826 P(-2 < z < 2) =. 9772 -. 0228 =. 9544
z value with more than two decimals P( z < 1. 643) = P( z < 1. 64) =. 9495 P( z < 1. 6474) = P( z < 1. 65) =. 9505 P( z < 1. 645) ? 1. 1. 645 is halfway of 1. 64 and 1. 65 2. Look for areas of 1. 64 and 1. 65 in Table 3. 3. Since the value 1. 645 is halfway between 1. 64 and 1. 65, we average areas. 9495 and. 9505. 4. P( z 1. 645) = (. 9495+. 9505)/2 =. 9500
Extreme z values Using Table 3, calculate P(z<-5) = ? P(z>4) = ? • • P(z<-5) = 0 P(z>4) = 0
General Normal & Standard Normal • x is normal with mean m and standard deviation s. • Question: P(a < x < b) ? • i. e. area under the normal curve from a to b. • To simplify the tabulation of these areas, we standardize each value of x by expressing it as a z-score, the number of standard deviations s it lies from the mean m. z is standard normal
Example • x is normal with mean 0. 6 and standard deviation 2. z is standard normal • x is normal with mean 10. 2 and standard deviation 5. z is standard normal
Probabilities for General Normal Random Variable üTo find an area for a normal random variable x with mean m and standard deviation s, standardize or rescale the interval in terms of z. üFind the appropriate area using Table 3. Example: x has a normal distribution with m = 5 and s = 2. Find P(x > 7). 1 z
Example The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation. 10. What is the probability that a randomly selected package weighs between 0. 80 and 0. 85 pounds?
Area under General Normal Curve Studies show that gasoline use for compact cars sold in U. S. is normally distributed, with a mean of 25. 5 mpg and a standard deviation of 4. 5 mpg. What is the percentage of compacts get 30 mpg or more? 15. 87% of compacts get 30 mpg or more using Table 3
Using Table 3 üTo find an area to the left of a z-value, find the area directly from the table. e. g. P( z < 1. 36) üTo find an area to the right of a z-value, find the area in Table 3 and subtract from 1. or find the area with respect to the negative of the z-value; e. g. P( z >1. 36) = 1 -P( z <1. 36), P( z>1. 36) = P( z< -1. 36) üTo find the area between two values of z, find the two areas in Table 3, and subtract. e. g. P( -1. 20 < z < 1. 36 ) üTo find an area for a normal random variable x with mean m and standard deviation s, standardize or rescale the interval in terms of z. üFind the appropriate area using Table 3.
Working Backwards Find the value of z that has area. 25 to its left. 1. Look for the four digit area closest to. 2500 in Table 3. 2. What row and column does this value correspond to? 3. z = -. 67 4. What percentile does this value represent? 25 th percentile, or 1 st quartile (Q 1)
Working Backwards Find the value of z that has area. 05 to its right. 1. The area to its left will be 1 -. 05 =. 95 2. Look for the four digit area closest to. 9500 in Table 3. 3. Since the value. 9500 is halfway between. 9495 and. 9505, we choose z halfway between 1. 64 and 1. 65. 4. z = 1. 645
Example 1 Find the value of z, say z 0 , such that. 01 of the area is to its right. (tail area of. 01) using Table 3
Example 2 Find the value of z, say z 0 , such that. 95 of the area is within z 0 standard deviations of the mean. using Table 3
Example 3 The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation. 10. What is the weight of a package such that only 1% of all packages exceed this weight? using Table 3 99 th percentile
Exercise
Key Concepts I. Continuous Probability Distributions 1. Continuous random variables 2. Probability distributions or probability density functions a. Curves are smooth. b. The area under the curve between a and b represents the probability that x falls between a and b. c. P (x = a) = 0 for continuous random variables. II. The Normal Probability Distribution 1. Symmetric about its mean m. 2. Shape determined by its standard deviation s.
Key Concepts III. The Standard Normal Distribution 1. The normal random variable z has mean 0 and standard deviation 1. 2. Any normal random variable x can be transformed to a standard normal random variable using 3. Convert necessary values of x to z. 4. Use Table 3 in Appendix I to compute standard normal probabilities. 5. Several important z-values have tail areas as follows: Tail Area: . 005 . 01 . 025 . 05 . 10 z-Value: 2. 58 2. 33 1. 96 1. 645 1. 28
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# If P(x) is divided by (x-a)(x-b) where a!=b, a,b in RR, can you prove that the remainder is: ((P(b)-P(a))/(b-a))xx(x-a)+P(a)?
Feb 24, 2017
See below.
#### Explanation:
$P \left(x\right)$ can be represented as
$P \left(x\right) = Q \left(x\right) \left(x - a\right) \left(x - b\right) + {r}_{1} x + {r}_{2}$ where ${r}_{1} x + {r}_{2}$ is the remainder of $\frac{P \left(x\right)}{\left(x - a\right) \left(x - b\right)}$
The determination of ${r}_{1} , {r}_{2}$ is straightforward
$P \left(a\right) = {r}_{1} a + {r}_{2}$ and
$P \left(b\right) = {r}_{1} b + {r}_{2}$ so, solving for ${r}_{1} , {r}_{2}$ we get
${r}_{1} = \frac{P \left(b\right) - P \left(a\right)}{b - a}$
${r}_{2} = - \frac{a P \left(b\right) - b P \left(a\right)}{b - a}$
and the remainder is
$\left(\frac{P \left(b\right) - P \left(a\right)}{b - a}\right) x - \frac{a P \left(b\right) - b P \left(a\right)}{b - a} =$
$= \left(\frac{P \left(b\right) - P \left(a\right)}{b - a}\right) \left(x - a\right) + P \left(a\right)$
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# Numerical Solution of Linear Equations
## Presentation on theme: "Numerical Solution of Linear Equations"— Presentation transcript:
Numerical Solution of Linear Equations
Lecture 2 Chapter 3 Numerical Solution of Linear Equations
Linear System of Equations
In matrix–vector notation
Numerical methods for the solution of the systems of linear equations
Direct Methods Cramer rule Gauss elimination Iterative Methods Jacobi Gauss-Siedel
1- Iterative Methods start with an initial guess of the solution vector X(0), and then repeatedly refine the solution until a certain convergence criterion is reached. Advantages : have significant computational advantages if the coefficient matrix is very large and sparsely populated (most coefficients are zero). Disadvantages: less efficient than their direct counterparts due to the large number of iterations required.
Jacobi Method The Jacobi method is an algorithm for determining the solutions of a system of linear equations with largest absolute values in each row and column dominated by the diagonal element.
Definition: The square nonsingular matrix A is called diagonally dominant if the sum of the absolute values of its elements on the main diagonal is greater than or equal to the sum of the absolute values of the remaining elements of the analyzed row of the matrix, i.e., Example: Checking if the coefficient matrix is diagonally dominant
Steps of Jacobi Method:
Rearrange system to be diagonally dominant. Start with an initial vector We determine the unknowns x1, x2, and x3 from the first, second, and third equation, respectively:
Perform series of consecutive iterations (k) as in step 3 until the solution converges to an exact solution, i.e., In general:
Example: Solve by Jacobi method:
Start with solution guess , and start iterating on the solution
keep on going until the desired level of accuracy is achieved
Iterative convergence
• Is the (k+1)th solution better than the kth solution? • Iterative process can be convergent/divergent A sufficient condition to ensure convergence is that the matrix is diagonally dominant A poor first guess will prolong the iterative process but will not make it diverge if the matrix is such that convergence is assured. Therefore better guesses will speed up the iterative process
Criteria for ascertaining convergence
• Absolute convergence criteria • Where ε a user specified tolerance or accuracy • The absolute convergence criteria is best used if you have a good idea of the magnitude of the xi‘s Relative convergence criteria • This criteria is best used if the magnitude of the ‘s are not known. • There are also problems with this criteria if
Solution: Rearrange the system to be diagonally dominant as the following:
Determine the unknowns x1, x2, and x3 from the first, second, and third equation, respectively:
Continuing this procedure, you obtain the sequence of approximations shown in the following table
Because the last two rows in the table are identical, we can conclude that to three significant digits the solution is
Gauss-Siedel Method The Gauss–Seidel method is similar to the Jacobi method except that the computation of xi(k) uses only the elements of X(k) that have already been computed, and only the elements of X(k-1) that have yet to be advanced to iteration (k).
Example 4 Solve previous Example using the Gauss–Seidel method Start with you obtain the following new value for x1 Now that you have a new value for x1 , however, use it to compute a new value for x2 That is,
So the first approximation is
Continued iterations produce the sequence of approximations shown in the following table. Note that after only six iterations of the Gauss-Seidel method, you achieved the same accuracy as was obtained with eight iterations of the Jacobi method.
Vectors, Functions, and Plots in Matlab
Enter a matrix into Matlab with the following syntax:
Also enter a vector u: > u = [ ]’ To multiply a matrix times a vector Au use *: A*u Now enter another matrix B using: > B = [3 2 1; 7 6 5; 4 3 2] You can multiply a matrix by a scalar: > 2*A Adding matrices A + A will give the same result: > A + A You can even add a number to a matrix: > A % This should add 3 to every entry of A.
Special matrices: N = ones(5,5) O=zeros(2,2) I=eye(3) D=diag(A) Inv(A)
Jacobi method in matrix form
If we write D, L, and U for the diagonal, strict lower triangular and strict upper triangular and parts of A, respectively, then Jacobi’s Method can be written in matrix-vector notation as
Programing of Jacobi method
function x=Jacobi(A,b,n,x0) L=tril(A,-1) U=triu(A,1) d=diag(A) B=-(U+L); g=b; for i=1:n x=(B*x0+g)./d x0=x; end
a- Using \ operator in Matlab.
Assignment 1 1- Solve the system a- Using \ operator in Matlab. b- Using Jacobi method. Start with x(0) = (0, 0), complete a table like the one below,
2-a) Express the Gauss-Siedel method in matrix form, (upper, lower, and diagonal).
2-b) Write a computer program that applies the Gauss-Siedel method, then solve the previous system of linear equations.
Summary Iterative methods: 1- Jacobi method 2- Gauss-Siedel method Iterative convergence diagonally dominant Matlab programming Assignment1
End of Chapter 3
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# Multiplication up to 5
Multiplication up to 5
A Guide to Learning Multiplication Basics
Method:
Multiplication is a mathematical operation that involves multiplying two numbers together. It is one of the four basic operations in mathematics, along with addition, subtraction, and division. To learn multiplication up to 5, it is important to understand the concept of multiplication and how it works. Start by learning the multiplication tables up to 5. This will help you understand the concept of multiplication and how it works. Once you have mastered the multiplication tables up to 5, practice solving multiplication problems.
Examples:
1. 2 x 3 = 6
2. 4 x 5 = 20
3. 3 x 4 = 12
4. 5 x 2 = 10
5. 2 x 5 = 10
Create a Table of 10 Exercises:
1. 2 x 1 =
2. 3 x 2 =
3. 4 x 3 =
4. 5 x 4 =
5. 1 x 5 =
6. 2 x 4 =
7. 3 x 5 =
8. 4 x 2 =
9. 5 x 3 =
10. 1 x 4 =
1. 2 x 1 = 2
2. 3 x 2 = 6
3. 4 x 3 = 12
4. 5 x 4 = 20
5. 1 x 5 = 5
6. 2 x 4 = 8
7. 3 x 5 = 15
8. 4 x 2 = 8
9. 5 x 3 = 15
10. 1 x 4 = 4
5 Multiple Choice Questions:
1. What is the answer to 3 x 4?
A. 8
B. 12
C. 10
D. 6
2. What is the answer to 5 x 2?
A. 10
B. 8
C. 12
D. 6
3. What is the answer to 4 x 5?
A. 10
B. 8
C. 12
D. 20
4. What is the answer to 2 x 5?
A. 10
B. 8
C. 12
D. 6
5. What is the answer to 1 x 4?
A. 10
B. 8
C. 12
D. 4
Key:
1. B
2. A
3. D
4. A
5. D
Conclusion:
Learning multiplication up to 5 is an important part of understanding the basics of mathematics. By understanding the concept of multiplication and how it works, and by practicing multiplication problems, you can become proficient in multiplication up to 5.
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# Derivative of Arcsin
The derivative of the arcsin function (also denoted as sin^(-1)(x) or asin(x)) can be found using differentiation techniques. Let’s denote y = arcsin(x), where -1 ≤ x ≤ 1.
To find the derivative dy/dx, we can differentiate both sides of the equation y = arcsin(x) with respect to x. Using the chain rule, we have:
dy/dx = d/dx(arcsin(x))
Now, let’s differentiate the right side of the equation. The derivative of arcsin(x) can be found as follows:
Let y = arcsin(x) Then, x = sin(y)
Differentiating both sides with respect to x:
1 = cos(y) * dy/dx
Now, we can solve for dy/dx:
dy/dx = 1 / cos(y)
To determine the value of cos(y), we can use the trigonometric identity:
sin^2(y) + cos^2(y) = 1
Since x = sin(y), we can substitute sin^2(y) = x^2:
x^2 + cos^2(y) = 1
Solving for cos(y):
cos(y) = sqrt(1 – x^2)
Substituting this value into the expression for dy/dx:
dy/dx = 1 / sqrt(1 – x^2)
Therefore, the derivative of the arcsin function is given by:
dy/dx = 1 / sqrt(1 – x^2)
It’s worth noting that this derivative is valid within the domain -1 ≤ x ≤ 1, as arcsin(x) is only defined within that range.
Here are three examples demonstrating the application of the derivative of the arcsin function:
Example 1: Let’s find the derivative of y = arcsin(2x).
Using the derivative formula derived earlier, we have:
dy/dx = 1 / sqrt(1 – (2x)^2)
Simplifying:
dy/dx = 1 / sqrt(1 – 4x^2)
So, the derivative of y = arcsin(2x) is dy/dx = 1 / sqrt(1 – 4x^2).
Example 2: Consider the function y = 3arcsin(5x).
Using the derivative formula, we have:
dy/dx = 1 / sqrt(1 – (5x)^2)
Simplifying:
dy/dx = 1 / sqrt(1 – 25x^2)
Thus, the derivative of y = 3arcsin(5x) is dy/dx = 1 / sqrt(1 – 25x^2).
Example 3: Let’s find the derivative of y = -arcsin(3x^2).
Using the derivative formula, we have:
dy/dx = 1 / sqrt(1 – (3x^2)^2)
Simplifying:
dy/dx = 1 / sqrt(1 – 9x^4)
Therefore, the derivative of y = -arcsin(3x^2) is dy/dx = 1 / sqrt(1 – 9x^4).
These examples illustrate how to find the derivatives of functions involving the arcsin function using the derived formula dy/dx = 1 / sqrt(1 – x^2).
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# SAT Math : Spheres
## Example Questions
### Example Question #771 : Sat Mathematics
A cube with volume 27 cubic inches is inscribed inside a sphere such that each vertex of the cube touches the sphere. What is the radius, in inches, of the sphere?
√3/2 (approximately 1.73)
9
(3√3)/2 (approximately 2.60)
8.5
(3√3)/2 (approximately 2.60)
Explanation:
We know that the cube has a volume of 27 cubic inches, so each side of the cube must be ∛27=3 inches. Since the cube is inscribed inside the sphere, the diameter of the sphere is the diagonal length of the cube, so the radius of the sphere is half of the diagonal length of the cube. To find the diagonal length of the cube, we use the distance formula d=√(32+32+32 )=√(3*32 )=3√3, and then divide the result by 2 to find the radius of the sphere, (3√3)/2.
### Example Question #1 : How To Find The Radius Of A Sphere
The surface area of a sphere is 100π square feet. What is the radius in feet?
10
5
25
π
100
5
Explanation:
S = 4π(r2)
100π = 4π(r2)
100 = 4r2
25 = r2
5 = r
### Example Question #3 : How To Find The Radius Of A Sphere
What is the radius of a sphere with a surface area of 16?
Explanation:
In order to find the radius, first write the surface area formula for a sphere and substitute the surface area.
Divide by on both sides in order to isolate the term.
Simplify both sides of the equation.
Square root both sides.
### Example Question #1 : How To Find The Diameter Of A Sphere
Find the diameter of a sphere with a surface area of .
Explanation:
Write the formula to find the surface area of a sphere.
Substitute the area and solve for the radius.
The diameter is double the radius.
### Example Question #2 : How To Find The Diameter Of A Sphere
What is the diameter of a sphere if the surface area is ?
Explanation:
Write the formula for the surface area of a sphere.
Substitute the area and find the radius.
The diameter is double the radius.
### Example Question #781 : Geometry
What is the diameter of a sphere with a volume of ?
Explanation:
Write the formula for the volume of a sphere.
Substitute the volume.
Multiply by on both sides in order to isolate the term.
Cube root both sides.
The diameter is double the radius.
### Example Question #1 : How To Find The Surface Area Of A Sphere
A spherical orange fits snugly inside a small cubical box such that each of the six walls of the box just barely touches the surface of the orange. If the volume of the box is 64 cubic inches, what is the surface area of the orange in square inches?
64π
128π
16π
32π
256π
16π
Explanation:
The volume of a cube is found by V = s3. Since V = 64, s = 4. The side of the cube is the same as the diameter of the sphere. Since d = 4, r = 2. The surface area of a sphere is found by SA = 4π(r2) = 4π(22) = 16π.
### Example Question #1 : How To Find The Surface Area Of A Sphere
A solid sphere is cut in half to form two solid hemispheres. What is the ratio of the surface area of one of the hemispheres to the surface area of the entire sphere before it was cut?
3/4
1/2
3/2
1
2/3
3/4
Explanation:
The surface area of the sphere before it was cut is equal to the following:
surface area of solid sphere = 4πr2, where r is the length of the radius.
Each hemisphere will have the following shape:
In order to determine the surface area of the hemisphere, we must find the surface area of the flat region and the curved region. The flat region will have a surface area equal to the area of a circle with radius r.
area of flat part of hemisphere = πr2
The surface area of the curved portion of the hemisphere will equal one-half of the surface area of the uncut sphere, which we established to be 4πr2.
area of curved part of hemisphere = (1/2)4πr= 2πr2
The total surface area of the hemisphere will be equal to the sum of the surface areas of the flat part and curved part of the hemisphere.
total surface area of hemisphere = πr+ 2πr= 3πr2
Finally, we must find the ratio of the surface area of the hemisphere to the surface area of the uncut sphere.
ratio = (3πr2)/(4πr2) = 3/4
### Example Question #181 : Psat Mathematics
The volume of a sphere is 2304π in3. What is the surface area of this sphere in square inches?
36π
144π
576π
216π
576π
Explanation:
To solve this, we must first begin by finding the radius of the sphere. To do this, recall that the volume of a sphere is:
V = (4/3)πr3
For our data, we can say:
2304π = (4/3)πr3; 2304 = (4/3)r3; 6912 = 4r3; 1728 = r3; 12 * 12 * 12 = r3; r = 12
Now, based on this, we can ascertain the surface area using the equation:
A = 4πr2
For our data, this is:
A = 4π*122 = 576π
### Example Question #1 : Spheres
A sphere has its center at the origin. A point on its surface is found on the x-y axis at (6,8). In square units, what is the surface area of this sphere?
200π
400π
(400/3)π
40π
400π
Explanation:
To find the surface area, we must first find the radius. Based on our description, this passes from (0,0) to (6,8). This can be found using the distance formula:
62 + 82 = r2; r2 = 36 + 64; r2 = 100; r = 10
It should be noted that you could have quickly figured this out by seeing that (6,8) is the hypotenuse of a 6-8-10 triangle (which is a multiple of the "easy" 3-4-5).
The rest is easy. The surface area of the sphere is defined by:
A = 4πr2 = 4 * 100 * π = 400π
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## College Algebra (10th Edition)
Published by Pearson
# Chapter 5 - Section 5.6 - Complex Numbers; Quadratic Equations in the Complex Number System - 5.6 Assess Your Understanding: 31
#### Answer
The zeros of the given function are: $\color{blue}{\left\{1, \dfrac{-1}{2} - \dfrac{\sqrt3i}{2}, \dfrac{-1}{2} + \dfrac{\sqrt3i}{2}\right\}}$ The completely factored form of the given function is: $\\\color{blue}{P(x) = (x-1)\left(x+\frac{1}{2} + \frac{\sqrt3i}{2}\right)\left(x+\frac{1}{2} - \frac{\sqrt3i}{2}\right)}$
#### Work Step by Step
RECALL: (1) $a^3-b^3=(a-b)(a^2+ab+b^2)$ (2) The quadratic formula: $x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ The given polynomial function can be written as $P(x) = x^3-1^3$. Factor the difference of two cubes using the formula in (1) above with $a=x$ and $b=1$ to obtain: $P(x) = (x-1)(x^2+x(1) + 1^2) \\P(x) = (x-1)(x^2+x+1)$ Equate each factor to zero then solve each equation to obtain: \begin{array}{ccc} &x-1=0 &\text{or} &x^2+x+1=0 \\&x=1 &\text{or} &x^2+x+1=0$\end{array} Solve the second equation using the formula in (2) above with$a=1, b=1$, and$c=1$to obtain:$x=\dfrac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)} \\x=\dfrac{-1\pm\sqrt{1-4}}{2} \\x=\dfrac{-1\pm\sqrt{-3}}{2} \\x=\dfrac{-1\pm \sqrt{3(-1)}}{2} \\x=\dfrac{-1\pm \sqrt3 i}{2} \\x_1=\dfrac{-1}{2} - \dfrac{\sqrt3i}{2} \\x_2=\dfrac{-1}{2} + \dfrac{\sqrt3i}{2}$Thus, the zeros of the given function are:$\color{blue}{\left\{1, \dfrac{-1}{2} - \dfrac{\sqrt3i}{2}, \dfrac{-1}{2} + \dfrac{\sqrt3i}{2}\right\}}$The completely factored form of the given function is:$P(x) = \left(x-1\right)\left[x-\left(\frac{-1}{2} - \frac{\sqrt3i}{2}\right)\right]\left[x-\left(\frac{-1}{2} + \frac{\sqrt3i}{2}\right)\right] \\\color{blue}{P(x) = (x-1)\left(x+\frac{1}{2} + \frac{\sqrt3i}{2}\right)\left(x+\frac{1}{2} - \frac{\sqrt3i}{2}\right)}\$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.1: Equations and Graphs
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
A student will be able to:
• Find solutions of graphs of equations.
• Find key properties of graphs of equations including intercepts and symmetry.
• Find points of intersections of two equations.
• Interpret graphs as models.
## Introduction
In this lesson we will review what you have learned in previous classes about mathematical equations of relationships and corresponding graphical representations and how these enable us to address a range of mathematical applications. We will review key properties of mathematical relationships that will allow us to solve a variety of problems. We will examine examples of how equations and graphs can be used to model real-life situations.
Let’s begin our discussion with some examples of algebraic equations:
Example 1: y=x2+2x1\begin{align*}y = x^2 + 2x - 1\end{align*} The equation has ordered pairs of numbers (x,y)\begin{align*}(x, y)\end{align*} as solutions. Recall that a particular pair of numbers is a solution if direct substitution of the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values into the original equation yields a true equation statement. In this example, several solutions can be seen in the following table:
x43210123y=x2+2x1721212714\begin{align*}& x && y = x^2 + 2x - 1 \\ & -4 && 7 \\ & -3 && 2\\ & -2 && -1 \\ & -1 && -2 \\ & 0 && -1 \\ & 1 && 2 \\ & 2 && 7 \\ & 3 && 14\end{align*}
We can graphically represent the relationships in a rectangular coordinate system, taking the x\begin{align*}x\end{align*} as the horizontal axis and the y\begin{align*}y\end{align*} as the vertical axis. Once we plot the individual solutions, we can draw the curve through the points to get a sketch of the graph of the relationship:
We call this shape a parabola and every quadratic function, f(x)=ax2+bx+c,\begin{align*}f(x) = ax^2 + bx + c,\end{align*} a0\begin{align*}a \neq 0\end{align*} has a parabola-shaped graph. Let’s recall how we analytically find the key points on the parabola. The vertex will be the lowest point, which for this parabola is (1,2)\begin{align*}(-1, -2)\end{align*}. In general, the vertex is located at the point (b2a,f(b2a))\begin{align*}(-\frac{b}{2a}, f(-\frac{b}{2a}))\end{align*}. We then can identify points crossing the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} axes. These are called the intercepts. The y\begin{align*}y-\end{align*}intercept is found by setting x=0\begin{align*}x = 0\end{align*} in the equation, and then solving for y\begin{align*}y\end{align*} as follows:
y=02+2(0)1=1.\begin{align*}y = 0^2 + 2(0) - 1 = -1.\end{align*} The y\begin{align*}y-\end{align*}intercept is located at (0,1).\begin{align*}(0,-1).\end{align*}
The x\begin{align*}x-\end{align*}intercept is found by setting y=0\begin{align*}y=0\end{align*} in the equation, and solving for x\begin{align*}x\end{align*} as follows: 0=x2+2x1\begin{align*} 0 = x^2 + 2x - 1\end{align*}
Using the quadratic formula, we find that x=1±2\begin{align*}x=-1\pm\sqrt{2}\end{align*}. The x\begin{align*}x-\end{align*}intercepts are located at (12,0)\begin{align*}(-1-\sqrt{2},0)\end{align*} and (1+2,0)\begin{align*}(-1+\sqrt{2},0)\end{align*}.
All parabolas also have a line of symmetry. This parabola has a vertical line of symmetry at x=1.\begin{align*}x = -1.\end{align*} In general, the line of symmetry for a parabola will always pass through its vertex, and so will always be located at x=b2a\begin{align*}x=-\frac{b}{2a}\end{align*}. The graph with all of its key characteristics is summarized below:
Example 2:
Here are some other examples of equations with their corresponding graphs:
y=2x+3\begin{align*}y = 2x + 3\end{align*}
x2+y2=4\begin{align*}x^2 + y^2 = 4\end{align*}
y=x39x\begin{align*}y = x^3 - 9x\end{align*}
Example 3:
Notice that the first equation in Example 2 is linear, so its graph is a straight line. Can you determine the intercepts?
Solution:
x\begin{align*}x-\end{align*}intercept at (3/2,0)\begin{align*}(-3/2, 0)\end{align*} and y\begin{align*}y-\end{align*}intercept at (0,3).\begin{align*}(0, 3).\end{align*}
Example 4:
Recall from pre-calculus that the second equation in Example 2 is that of a circle with center (0,0)\begin{align*}(0,0)\end{align*} and radius r=2.\begin{align*}r = 2.\end{align*} Can you show analytically that the radius is 2\begin{align*}2\end{align*}?
Solution:
Find the four intercepts, by setting x=0\begin{align*}x = 0\end{align*} and solving for y\begin{align*}y\end{align*}, and then setting y=0\begin{align*}y = 0\end{align*} and solving for x\begin{align*}x\end{align*}.
Example 5:
The third equation from Example 2 is an example of a polynomial relationship. Can you find the intercepts analytically?
Solution:
We can find the x\begin{align*}x-\end{align*}intercepts analytically by setting y=0\begin{align*}y = 0\end{align*} and solving for x.\begin{align*}x.\end{align*} So, we have
x39xx(x29)x(x3)(x+3)x=0,x=3,x=0=0=0=3.\begin{align*}x^3 - 9x &= 0\\ x(x^2 - 9) &= 0\\ x(x - 3)(x + 3) &= 0\\ x = 0, x = -3, x &= 3.\end{align*}
The x\begin{align*}x-\end{align*}intercepts are located at (3,0),(0,0),\begin{align*}(-3, 0),(0, 0),\end{align*} and (3,0).\begin{align*}(3, 0).\end{align*} Note that \begin{align*}(0,0)\end{align*} is also the \begin{align*}y-\end{align*}intercept. The \begin{align*}y-\end{align*}intercepts can be found by setting \begin{align*}x = 0\end{align*}. So, we have
\begin{align*}x^3 - 9x &= y\\ (0)^3 - 9(0) &= y\\ y &= 0.\end{align*}
Sometimes we wish to look at pairs of equations and examine where they have common solutions. Consider the linear and quadratic graphs of the previous examples. We can sketch them on the same axes:
We can see that the graphs intersect at two points. It turns out that we can solve the problem of finding the points of intersections analytically and also by using our graphing calculator. Let’s review each method.
Analytical Solution
Since the points of intersection are on each graph, we can use substitution, setting the general \begin{align*}y-\end{align*}coordinates equal to each other, and solving for \begin{align*}x.\end{align*}
\begin{align*}2x + 3 &= x^2 + 2x - 1\\ 0 &= x^2 - 4\\ x = 2, x &= -2.\end{align*}
We substitute each value of \begin{align*}x\end{align*} into one of the original equations and find the points of intersections at \begin{align*}(-2,-1)\end{align*} and \begin{align*}(2,7).\end{align*}
Graphing Calculator Solution
Once we have entered the relationships on the Y= menu, we press 2nd [CALC] and choose #5 Intersection from the menu. We then are prompted with a cursor by the calculator to indicate which two graphs we want to work with. Respond to the next prompt by pressing the left or right arrows to move the cursor near one of the points of intersection and press [ENTER]. Repeat these steps to find the location of the second point.
We can use equations and graphs to model real-life situations. Consider the following problem.
Example 6: Linear Modeling
The cost to ride the commuter train in Chicago is \begin{align*}\2\end{align*}. Commuters have the option of buying a monthly coupon book costing \begin{align*}\5\end{align*} that allows them to ride the train for \begin{align*}\1.5\end{align*} on each trip. Is this a good deal for someone who commutes every day to and from work on the train?
Solution:
We can represent the cost of the two situations, using the linear equations and the graphs as follows:
\begin{align*}C_1(x) &= 2x\\ C_2(x) &= 1.5x + 5\end{align*}
As before, we can find the point of intersection of the lines, or in this case, the break-even value in terms of trips, by solving the equation:
\begin{align*}C_1(x) &= C_2(x)\\ 2x &= 1.5x + 5\\ x &= 10.\end{align*}
So, even though it costs more to begin with, after \begin{align*}10\end{align*} trips the cost of the coupon book pays off and from that point on, the cost is less than for those riders who did not purchase the coupon book.
Example 7: Non-Linear Modeling
The cost of disability benefits in the Social Security program for the years 2000 - 2005 can be modeled as a quadratic function. The formula
\begin{align*}Y = -0.5x^2 + 2x + 4\end{align*}
indicates the number of people \begin{align*}Y\end{align*}, in millions, receiving Disability Benefits \begin{align*}x\end{align*} years after 2000. In what year did the greatest number of people receive benefits? How many people received benefits in that year?
Solution:
We can represent the graph of the relationship using our graphing calculator.
The vertex is the maximum point on the graph and is located at \begin{align*}(2,6).\end{align*} Hence in year 2002 a total of \begin{align*}6\;\mathrm{million}\end{align*} people received benefits.
## Lesson Summary
1. Reviewed graphs of equations
2. Reviewed how to find the intercepts of a graph of an equation and to find symmetry in the graph
3. Reviewed how relationships can be used as models of real-life phenomena
4. Reviewed how to solve problems that involve graphs and relationships
## Review Questions
In each of problems 1 - 4, find a pair of solutions of the equation, the intercepts of the graph, and determine if the graph has symmetry.
1. \begin{align*}2x - 3y = 5\end{align*}
2. \begin{align*}3x^2 - y = 5\end{align*}
3. \begin{align*}y = x^3 - x\end{align*}
4. \begin{align*}y = x^3 + x^2 - 6x\end{align*}
5. Once a car is driven off of the dealership lot, it loses a significant amount of its resale value. The graph below shows the depreciated value of a BMW versus that of a Chevy after \begin{align*}t\end{align*} years. Which of the following statements is the best conclusion about the data?
1. You should buy a BMW because they are better cars.
2. BMWs appear to retain their value better than Chevys.
3. The value of each car will eventually be \begin{align*}\0\end{align*}.
6. Which of the following graphs is a more realistic representation of the depreciation of cars.
7. A rectangular swimming pool has length that is \begin{align*}25\;\mathrm{yards}\end{align*} greater than its width.
1. Give the area enclosed by the pool as a function of its width.
2. Find the dimensions of the pool if it encloses an area of \begin{align*}264\;\mathrm{square\ yards}\end{align*}.
8. Suppose you purchased a car in 2004 for \begin{align*}\18,000.\end{align*} You have just found out that the current year 2008 value of your car is \begin{align*}\8,500.\end{align*} Assuming that the rate of depreciation of the car is constant, find a formula that shows changing value of the car from 2004 to 2008.
9. For problem #8, in what year will the value of the vehicle be less than \begin{align*}\1,400\end{align*}?
10. For problem #8, explain why using a constant rate of change for depreciation may not be the best way to model depreciation.
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# Subtraction Word Problems
## Overview
Subtraction helps us to find the difference. Subtraction can tell us how many are left.
Subtraction is easy to do! Subtraction is the opposite of addition.
We can use subtraction in word, or story, problems.
Subtraction skills can help us solve correctly!
For example:
Let’s take a look at the example above.
We are subtracting because some of the dolphins swim away.
We need to figure out how many are left.
We can do this by subtracting 10 – 5 to get our answer.
We write it like this:
10 – 5 = 5
There are 5 dolphins left!
Let’s look at another problem!
Let’s take a look at the example above.
We are subtracting because some of the butterflies fly away.
We need to figure out how many are left.
We can do this by subtracting 11 – 4 to get our answer.
We write it like this:
11 – 4 = 7
There are 7 butterflies left!
Amazing work! Let’s look at another problem together.
This time we will subtract strawberries.
We are SUBTRACTING to find out how many are left because some have been eaten.
We can see that there were nine strawberries to start with. Six of them have been eaten
When we SUBTRACT 9 minus 6, we have 3 left.
We say: NINE minus SIX equals THREE.
We write it like this:
9 – 6 = 3
Let’s try one more together!
This time we will subtract donuts.
We are SUBTRACTING to find out how many are left because some have been eaten.
We can see that there were ten donuts to start with. Six of them have been eaten
When we SUBTRACT 10 minus 6, we have 4 left.
We say: TEN minus SIX equals FOUR.
We write it like this:
10 – 6 = 4
There are many different words that are used to describe subtraction. For example, we can find out HOW MANY MORE there are or how many ARE LEFT. Here are several other examples:
Knowing these keywords will help you understand that in order to solve some problems, you must subtract. You might see these keywords in math story problems.
There are many strategies you can use to help you subtract correctly. We will learn about these in other lessons.
Now, let’s practice!
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### Practice Question 1
Find the difference to solve.
### Practice Question 2
Solve the subtraction problem below.
### Practice Question 4
Find out how many are left.
### Practice Question 5
Solve by subtracting.
Now you’re ready to take a quick quiz on adding with subtraction word problems! Click below!
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# Is ln 2x the same as ln x 2?
## How do you find the derivative of y ln x 2?
ln2x is simply another way of writing (lnx)2 and so they are equivalent.
## Is ln 2x the same as ln x 2?
In normal mathematical usage lnx2 means ln(x2); both are equal to 2lnx and have derivative 2x. Of course (lnx)2 is something altogether different; if W|A interprets lnx2 to mean (lnx)2, it should give different derivatives, but that’s a non-standard interpretation of lnx2.
derivative of ln 2x
## What does Lnx 2 mean?
What is the Derivative of log x whole square? The derivative of (log x)2 using the chain rule is 2 log x d/dx(log x) 2 log x [ 1/(x ln 10)] (2 log x) / (x ln 10).
## What is the derivative of y ln x 2?
Thus, the derivative of ln x2 is 2/x. Note this result agrees with the plots of tangent lines for both positive and negative x. For x 2, the derivative is 2/2 1, which agrees with the plot.
1/x
## Is ln x 2 the same as ln x )) 2?
ln2x is simply another way of writing (lnx)2 and so they are equivalent.
## How do you simplify ln 2x?
In normal mathematical usage lnx2 means ln(x2); both are equal to 2lnx and have derivative 2x. Of course (lnx)2 is something altogether different; if W|A interprets lnx2 to mean (lnx)2, it should give different derivatives, but that’s a non-standard interpretation of lnx2.
## What is the differentiation of ln 2x?
You use the chain rule : (fu2218g)'(x)(f(g(x)))’f'(g(x))u22c5g'(x) . In your case : (fu2218g)(x)ln(2x),f(x)ln(x)andg(x)2x .
## What does Lnx 2 equal?
Step 1: Rewrite ln x2 Using Logarithm Properties The logarithm of x to a power n equals n times the logarithm of x. Thus, ln x2 2 ln x
## Is Lnx 2 the same as Lnx 2?
ln2x is simply another way of writing (lnx)2 and so they are equivalent.
## How do you do Lnx 2?
ln^2(x) is defined for all xx26gt;1, by ln^2(x) ln(ln(x)). The same definition works for all complex numbers except x 0 or 1 where logarithm to the base e is denoted by log for complex numbers since hardly anybody uses log to any other base there. 7.1K views. xb7 Related answers.
1/x
## How do you find the derivative of ln ln x 2?
The derivative of yln(2) is 0 . Remember that one of the properties of derivatives is that the derivative of a constant is always 0 . If you view the derivative as the slope of a line at any given point, then a function that consists of only a constant would be a horizontal line with no change in slope.
## How do you find the derivative of ln 2?
The derivative of ln x is 1/x. We know that the domain of ln x is x x26gt; 0 and thus, d/dx (ln |x|) 1/x as well. Derivative of ln(f(x)) using chain rule is 1/(f(x)) xb7 f'(x).
## Is ln x )) 2 the same as ln 2 X?
ln2x is simply another way of writing (lnx)2 and so they are equivalent.
## What is the difference between ln 2 x and ln x )) 2?
ln^2(x) means simply to square the value of ln(x). Whereas, 2ln(x) means to double the value of ln(x).
## What is the derivative of ln x )) 2?
From the definition of the derivative using limits, the derivative of ln x2 2 (1/x) 2/x as before.
## How can you simplify ln?
In normal mathematical usage lnx2 means ln(x2); both are equal to 2lnx and have derivative 2x. Of course (lnx)2 is something altogether different; if W|A interprets lnx2 to mean (lnx)2, it should give different derivatives, but that’s a non-standard interpretation of lnx2.
## How do you get rid of ln?
Explanation: According to log properties, the coefficient in front of the natural log can be rewritten as the exponent raised by the quantity inside the log. Notice that natural log has a base of . This means that raising the log by base will eliminate both the and the natural log.
## What is ln 2x differentiated?
From the definition of the derivative using limits, the derivative of ln x2 2 (1/x) 2/x as before.
## What is the second derivative of ln 2x?
The derivative of 2x is equal to 2 as the formula for the derivative of a straight line function f(x) ax + b is given by f'(x) a, where a, b are real numbers. Differentiation of 2x is calculated using the formula d(ax+b)/dx a.
## What is the second derivative of Lnx 2?
ln^2(x) means simply to square the value of ln(x). Whereas, 2ln(x) means to double the value of ln(x).
## What is the derivative of Lnx 2?
In normal mathematical usage lnx2 means ln(x2); both are equal to 2lnx and have derivative 2x. Of course (lnx)2 is something altogether different; if W|A interprets lnx2 to mean (lnx)2, it should give different derivatives, but that’s a non-standard interpretation of lnx2.
## What happens when ln is squared?
Logarithm and exponentials are inverse functions meaning the logarithm will undo the exponential. Thus, ln e1/x 1/x. From the definition of the derivative using limits, the derivative of ln x2 2 (1/x) 2/x as before.
## Can you integrate Lnx 2?
ln^2(x) means simply to square the value of ln(x). Whereas, 2ln(x) means to double the value of ln(x).
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Games That Involve Geometric Probability
By Josh Turner
Jupiterimages/Comstock/Getty Images
Geometric probability deals with likelihood and chance. Traditionally, it attempts to find the odds of a certain outcome when using geometric objects. This works when dealing with games that involve geometric shapes. Many of the games you play and watch on television are simply problems involving geometric probability. By understanding probability, you can determine the odds of getting a certain score for these games.
Rolling Dice
Jupiterimages/Polka Dot/Getty Images
In the game of dice, you attempt to get a specific number by rolling dice. This is a question about the odds of making one side of a cube show up on top. To determine the odds of winning a dice game with geometric probability, use the following equation:
Number of possible ways to roll a number/Number of sides
For example, when rolling a die, there is only one way to roll any number and there are six sides. The probability of rolling two is 1/6 or 0.167.
Wheel of Fortune
Medioimages/Photodisc/Photodisc/Getty Images
On the television show "Wheel of Fortune," players spin a large segmented wheel in an attempt to win a prize. Mathematically, they are playing the odds of landing on a certain sector. Determine the odds of landing on a certain sector with the following equation:
Total area of wheel / Area of one sector.
For a board with an area of 40 square inches divided into eight sectors, each with an area of 5 square inches, the probability of landing on any one sector is 5/40 = 1/8 or 0.125
Throwing Darts
Thomas Northcut/Photodisc/Getty Images
Darts uses geometric probability much like Wheel of Fortune, but there are more sectors, and the area of each one differs. Participants play the odds of landing a dart into a specific section of the board. Use geometric probability to determine these odds with the equation:
Area of the section / Area of entire board
For example, the area of a dartboard is approximately 250 square inches. And the area of the double bulls eye is around 0.75 square inches. The probability of hitting the double bulls eye is 0.75/250 = 0.003.
Skeeball
The objective of Skeeball is to throw a wooden ball down a lane and land it in one of several holes. This is another game that involves geometric probability. To find the probability of landing the ball in any hole use the equation:
Area of hole/Area of board
For a Skeeball board with an area of 1,750 square inches, the probability of landing the ball in a hole with an area of 28 square inches is 28/1,750 = 7/432 = 0.016
Skill Level
Michael Turek/Digital Vision/Getty Images
Geometric probability allows you to determine the odds of scoring points in certain types games, but you can also factor in the skill level of the players. This requires knowledge of more complicated areas of probability and an understanding of mathematical modeling. To do this, you have to create a mathematical equation specific to the players and the game. Such equations use the skill level of each player and their statistics to create a type of handicap for less experienced players.
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# What is the derivative of x^sin(x)?
Dec 20, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\sin} x \left(\cos x \ln x + \sin \frac{x}{x}\right)$
#### Explanation:
$y = {x}^{\sin} x$
Take the natural logarithm of both sides.
$\ln y = \ln \left({x}^{\sin} x\right)$
Use laws of logarithms to simplify.
$\ln y = \sin x \ln x$
Use the product rule and implicit differentiation to differentiate.
$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \cos x \left(\ln x\right) + \frac{1}{x} \left(\sin x\right)$
$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \cos x \ln x + \sin \frac{x}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos x \ln x + \sin \frac{x}{x}}{\frac{1}{y}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\sin} x \left(\cos x \ln x + \sin \frac{x}{x}\right)$
Hopefully this helps!
Dec 20, 2016
$\frac{d}{\mathrm{dx}} {x}^{\sin x} = {x}^{\sin x} \left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]$.
#### Explanation:
When we have a function of $x$ like $y = {x}^{\sin} x$, where a single term contains $x$ in both its base and its power, perhaps the easiest way to find the function's derivative is to first take the (natural) logarithm of both sides:
$\ln y = \ln \left({x}^{\sin x}\right)$
$\textcolor{w h i t e}{\ln y} = \sin x \cdot \ln x$
This places all the $x$'s on the same "level". Then, take the derivative of both sides with respect to $x$:
$\implies \frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left(\sin x \cdot \ln x\right)$
Remembering that $y$ is a function of $x$, we get
$\implies \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \cdot \ln x + \sin x \left(\frac{1}{x}\right)$
$\implies \textcolor{w h i t e}{\text{XXi}} \frac{\mathrm{dy}}{\mathrm{dx}} = y \left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]$
Since we began with $y = {x}^{\sin x}$, we substitute this back in for $y$ to get
$\implies \textcolor{w h i t e}{\text{XXi}} \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\sin x} \left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]$.
## Note:
When $f \left(x\right) = g {\left(x\right)}^{h \left(x\right)}$, you'll almost always see $g {\left(x\right)}^{h \left(x\right)}$ appear in the derivative of $f \left(x\right)$. If you don't, go back and double check your work to make sure things were done right.
Dec 20, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\left(\cos \left(x\right)\right) \ln \left(x\right) + \frac{\sin \left(x\right)}{x}\right) {x}^{\sin \left(x\right)}$
#### Explanation:
Given: $y = {x}^{\sin \left(x\right)}$
Use logarithmic differentiation.
$\ln \left(y\right) = \ln \left({x}^{\sin \left(x\right)}\right)$
On the right side, use a property of logarithms, $\ln \left({a}^{b}\right) = \left(b\right) \ln \left(a\right)$:
$\ln \left(y\right) = \left(\sin \left(x\right)\right) \ln \left(x\right)$
Use implicit differentiation on the left side:
$\frac{\mathrm{dl} n \left(y\right)}{\mathrm{dx}} = \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$
Use the product rule on the right sides:
$\frac{d \left(u v\right)}{\mathrm{dx}} = \left(u '\right) \left(v\right) + \left(u\right) \left(v '\right)$
let $u = \sin \left(x\right)$, then $u ' = \cos \left(x\right) , v = \ln \left(x\right) , \mathmr{and} v ' = \frac{1}{x}$
Substituting into the product rule:
$\frac{d \left(\left(\sin \left(x\right)\right) \ln \left(x\right)\right)}{\mathrm{dx}} = \left(\cos \left(x\right)\right) \ln \left(x\right) + \frac{\sin \left(x\right)}{x}$
Put the equation back together:
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos \left(x\right)\right) \ln \left(x\right) + \frac{\sin \left(x\right)}{x}$
Multiply both sides by y:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\left(\cos \left(x\right)\right) \ln \left(x\right) + \frac{\sin \left(x\right)}{x}\right) y$
Substitute ${x}^{\sin \left(x\right)}$ for y:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\left(\cos \left(x\right)\right) \ln \left(x\right) + \frac{\sin \left(x\right)}{x}\right) {x}^{\sin \left(x\right)}$
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# Mastering Matrix Math: Fixing Singular Matrices and Calculating Determinants with NumPy
Matrix math is an essential component of many fields, including data science, computer science, and engineering. It allows us to represent complex systems and make predictions about their behavior.
One of the most fundamental operations in matrix math is calculating the determinant, which gives us information about the matrix’s invertibility and overall behavior. In this article, we’ll explore two related topics: fixing the “singular matrix” error encountered in Python and calculating the determinant of a matrix using NumPy.
Error Encountered in Python: Singular Matrix
The “singular matrix” error is one of the most common mistakes encountered when working with matrices in Python.
A singular matrix is any matrix whose determinant is equal to zero. This means that the matrix cannot be inverted, as there is no unique solution to the system of equations it represents.
## Reproducing the Error
Suppose we have a matrix A, which we attempt to invert using NumPy’s `inv()` function:
“`
## import numpy as np
A = np.array([[1, 2], [3, 4]])
A_inv = np.linalg.inv(A)
“`
If we attempt to run this code, we’ll encounter a “LinAlgError: Singular matrix” error.
## Understanding the Error
The error occurs because the determinant of the matrix A is equal to zero. The determinant is a scalar value that can be calculated from the elements of the matrix.
For a 2×2 matrix, the determinant is calculated as:
“`
det = a[0,0]*a[1,1] – a[0,1]*a[1,0]
“`
For the matrix A above, the determinant is:
“`
det = 1*4 – 2*3
det = -2
“`
Since the determinant is not equal to zero, we should be able to invert the matrix. However, if we were to change the matrix to:
“`
A = np.array([[1, 2], [2, 4]])
“`
## Then the determinant would be equal to zero:
“`
det = 1*4 – 2*2
det = 0
“`
And we would encounter the “singular matrix” error when attempting to invert it.
## Fixing the Error
The solution to the “singular matrix” error is straightforward: we need to use a non-singular matrix. One way to do this is to change the values of the matrix so that the determinant is not equal to zero.
In some cases, this may be possible and reasonable. For example, suppose we had a matrix with a single zero element.
We could change that element to a small non-zero value to make the matrix invertible. In other cases, it may not be feasible or appropriate to modify the matrix.
In these cases, we may need to use a different method of solving the system of equations represented by the matrix. For example, we could use an iterative method like Jacobi or Gauss-Seidel.
Alternatively, we could use an approximation technique like singular value decomposition (SVD) to find a pseudo-inverse of the matrix.
## Example Matrix and Calculating Determinant
To calculate the determinant of a matrix using NumPy, we first need to create a matrix. We can do this using the `matrix()` function:
“`
## import numpy as np
A = np.matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
“`
This creates a 3×3 matrix with the values 1-9 arranged in rows. To calculate the determinant, we simply use the `det()` function:
“`
det_A = np.linalg.det(A)
“`
## This produces the output:
“`
0.0
“`
Since the determinant is equal to zero, we know that the matrix is singular and cannot be inverted.
## Conclusion
In this article, we’ve explored two related topics in matrix math: fixing the “singular matrix” error encountered in Python and calculating the determinant of a matrix using NumPy. The “singular matrix” error occurs when we attempt to invert a matrix with a determinant equal to zero. To fix this error, we either modify the matrix or use a different method of solving the system of equations represented by the matrix.
Calculating the determinant of a matrix allows us to determine its invertibility and overall behavior. We can use NumPy’s `det()` function to calculate the determinant of a matrix quickly and easily.
By mastering these concepts, we can develop a deeper understanding of matrix math and apply it to a wide range of real-world problems. 3)
## Creating a Non-Singular Matrix
We have seen how the “singular matrix” error can occur when we attempt to invert a matrix with a determinant equal to zero.
In this section, we will explore how to create a non-singular matrix, which can be inverted and used in our calculations.
## Creating a Non-Singular Matrix
There are several ways to create a non-singular matrix. One simple method is to add or subtract a multiple of one row from another row.
This does not change the determinant of the matrix, but it can help us to identify a non-singular matrix. Let’s take a look at an example:
“`
## import numpy as np
A = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
A[2] = A[2] – 2 * A[1]
## print(A)
“`
This code creates a 3×3 matrix with the values 1-9 arranged in rows. The third row has been modified by subtracting twice the second row from it.
This creates a non-singular matrix that can be inverted without encountering the “singular matrix” error.
## Calculating Inverse and Determinant
Once we have created a non-singular matrix, we can calculate its inverse and determinant using NumPy. To calculate the inverse, we can use the `inv()` function:
“`
A_inv = np.linalg.inv(A)
“`
## This produces the output:
“`
[[-4.5 3. -.5 ]
[ 9.
-6. 1.
]
[-4.5 3. -.5 ]]
“`
To calculate the determinant, we can use the `det()` function:
“`
det_A = np.linalg.det(A)
“`
## This produces the output:
“`
6.661338147750939e-16
“`
Note that even though the matrix is non-singular, the determinant is not exactly zero due to floating-point precision errors. However, this value is close enough to zero that we can consider the matrix non-invertible.
By creating a non-singular matrix, we can avoid the “singular matrix” error and perform our calculations without issue.
For a deeper understanding of singular matrices and their properties, there are several resources available. One such resource is Wolfram MathWorld, which provides a comprehensive overview of matrix theory and its applications.
The article on “Singular Matrix” describes in detail the properties and characteristics of singular matrices, as well as techniques for identifying and working with them. In addition to Wolfram MathWorld, there are many textbooks and online courses available that cover matrix theory in depth.
These resources can provide a solid foundation for understanding the concepts and applications of matrix math, as well as how to work with matrices in Python and other programming languages.
## Conclusion
In this expanded article, we have explored how to fix the “singular matrix” error by creating a non-singular matrix and how to calculate the inverse and determinant of a matrix using NumPy. We have also provided additional resources for those interested in learning more about matrix theory and its applications. By mastering the concepts of matrix math and understanding how to work with matrices in Python, we can apply this powerful tool to a wide range of real-world problems.
In this article, we have explored the important topics of fixing the “singular matrix” error encountered in Python and calculating the determinant of a matrix using NumPy. We have shown how to create a non-singular matrix using simple methods and how to calculate its inverse and determinant. By understanding these concepts, we can apply matrix math to a wide range of fields and solve complex problems.
The key takeaway is the importance of creating non-singular matrices to avoid errors, and the power of NumPy in performing calculations quickly and easily. By mastering these foundational concepts, we can confidently approach matrix math in our work and research.
|
Quartiles & Quantiles | Calculation, Definition & Interpretation
Quartiles are three values that split sorted data into four parts, each with an equal number of observations. Quartiles are a type of quantile.
• First quartile: Also known as Q1, or the lower quartile. This is the number halfway between the lowest number and the middle number.
• Second quartile: Also known as Q2, or the median. This is the middle number halfway between the lowest number and the highest number.
• Third quartile: Also known as Q3, or the upper quartile. This is the number halfway between the middle number and the highest number.
Quartiles can also split probability distributions into four parts, each with an equal probability.
What are quartiles?
Quartiles are a set of descriptive statistics. They summarize the central tendency and variability of a dataset or distribution.
Quartiles are a type of percentile. A percentile is a value with a certain percentage of the data falling below it. In general terms, k% of the data falls below the kth percentile.
• The first quartile (Q1, or the lowest quartile) is the 25th percentile, meaning that 25% of the data falls below the first quartile.
• The second quartile (Q2, or the median) is the 50th percentile, meaning that 50% of the data falls below the second quartile.
• The third quartile (Q3, or the upper quartile) is the 75th percentile, meaning that 75% of the data falls below the third quartile.
By splitting the data at the 25th, 50th, and 75th percentiles, the quartiles divide the data into four equal parts.
• In a sample or dataset, the quartiles divide the data into four groups with equal numbers of observations.
• In a probability distribution, the quartiles divide the distribution’s range into four intervals with equal probability.
How to find quartiles
To find the quartiles of a dataset or sample, follow the step-by-step guide below.
1. Count the number of observations in the dataset (n).
2. Sort the observations from smallest to largest.
3. Find the first quartile:
• Calculate n * (1 / 4).
• If n * (1 / 4) is an integer, then the first quartile is the mean of the numbers at positions n * (1 / 4) and n * (1 / 4) + 1.
• If n * (1 / 4) is not an integer, then round it up. The number at this position is the first quartile.
1. Find the second quartile:
• Calculate n * (2 / 4).
• If n * (2 / 4) is an integer, the second quartile is the mean of the numbers at positions n * (2 / 4) and n * (2 / 4) + 1.
• If n * (2 / 4) is not an integer, then round it up. The number at this position is the second quartile.
2. Find the third quartile:
• Calculate n * (3 / 4).
• If n * (3 / 4) is an integer, then the third quartile is the mean of the numbers at positions n * (3 / 4) and n * (3 / 4) + 1.
• If n * (3 / 4) is not an integer, then round it up. The number at this position is the third quartile.
There are multiple methods to calculate the first and third quartiles, and they don’t always give the same answers. There’s no universal agreement on the best way to calculate quartiles.
Lower and upper quartile calculator
You can calculate the lower and upper quartile by hand or with the help of our calculator below. You can switch between the lower and upper quartile in the input field. Just type “lower quartile x,y,z” to calculate the lower quartile of your data set, or “upper quartile x,y,z” for the upper quartile of your data set.
Step-by-step example
Imagine you conducted a small study on language development in children 1–6 years old. You’re writing a paper about the study and you want to report the quartiles of the children’s ages.
Age (years) Frequency 1 2 3 4 5 6 2 3 4 1 2 2
Receive feedback on language, structure, and formatting
Professional editors proofread and edit your paper by focusing on:
• Academic style
• Vague sentences
• Grammar
• Style consistency
Visualizing quartiles with boxplots
Boxplots are helpful visual summaries of a dataset. They’re composed of boxes, which show the quartiles, and whiskers, which show the lowest and highest observations.
To make a boxplot, you first need to calculate the five-number summary:
1. The minimum
• This is the lowest observation. If you order the numbers in your dataset from lowest to highest, the minimum is the first number. The minimum is sometimes called the zeroth quantile.
2. The first quartile
3. The second quartile
4. The third quartile
5. The maximum
• This is the highest observation. If you order the numbers in your dataset from lowest to highest, the maximum is the last number. The maximum is sometimes called the fourth quantile.
With these five numbers, you can draw a boxplot:
This isn’t the only method of drawing boxplots. Although the box always indicates the quartiles, often the whiskers indicate 1.5 IQR from the Q1 and Q3.
Interpreting quartiles
Quartiles can give you useful information about an observation or a dataset.
Comparing observations
Quartiles are helpful for understanding an observation in the context of the rest of a sample or population. By comparing the observation to the quartiles, you can determine whether the observation is in the bottom 25%, middle 50%, or top 25%.
Median
The second quartile, better known as the median, is a measure of central tendency. This middle number is a good measure of the average or most central value of the data, especially for skewed distributions or distributions with outliers.
Interquartile range
The distance between the first and third quartiles—the interquartile range (IQR)—is a measure of variability. It indicates the spread of the middle 50% of the data.
The IQR is an especially good measure of variability for skewed distributions or distributions with outliers. IQR only includes the middle 50% of the data, so, unlike the range, the IQR isn’t affected by extreme values.
Skewness
The distance between quartiles can give you a hint about whether a distribution is skewed or symmetrical. It’s easiest to use a boxplot to look at the distances between quartiles:
Note that a histogram or skewness measure will give you a more reliable indication of skewness.
Identifying outliers
The interquartile range (IQR) can be used to identify outliers. Outliers are observations that are extremely high or low. One definition of an outlier is any observation that is more than 1.5 IQR away from the first or third quartile.
What are quantiles?
A quartile is a type of quantile.
Quantiles are values that split sorted data or a probability distribution into equal parts. In general terms, a q-quantile divides sorted data into q parts. The most commonly used quantiles have special names:
• Quartiles (4-quantiles): Three quartiles split the data into four parts.
• Deciles (10-quantiles): Nine deciles split the data into 10 parts.
• Percentiles (100-quantiles): 99 percentiles split the data into 100 parts.
There is always one fewer quantile than there are parts created by the quantiles.
How to find quantiles
To find a q-quantile, you can follow a similar method to that used for quartiles, except in steps 3–5, multiply n by multiples of 1/q instead of 1/4.
For example, to find the third 5-quantile:
1. Calculate n * (3 / 5).
2. If n * (3 / 5) is an integer, then the third 5-quantile is the mean of the numbers at positions n * (3 / 5) and n * (3 / 5) + 1.
3. If n * (3 / 5) is not an integer, then round it up. The number at this position is the third 5-quantile.
Practice questions
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Other interesting articles
If you want to know more about statistics, methodology, or research bias, make sure to check out some of our other articles with explanations and examples.
Frequently asked questions about quartiles and quantiles
How do I find quartiles in Excel?
You can use the QUARTILE() function to find quartiles in Excel. If your data is in column A, then click any blank cell and type “=QUARTILE(A:A,1)” for the first quartile, “=QUARTILE(A:A,2)” for the second quartile, and “=QUARTILE(A:A,3)” for the third quartile.
How do I find quartiles in R?
You can use the quantile() function to find quartiles in R. If your data is called “data”, then “quantile(data, prob=c(.25,.5,.75), type=1)” will return the three quartiles.
How do I find the quartiles of a probability distribution?
To find the quartiles of a probability distribution, you can use the distribution’s quantile function.
What’s the best measure of central tendency to use?
The mean is the most frequently used measure of central tendency because it uses all values in the data set to give you an average.
For data from skewed distributions, the median is better than the mean because it isn’t influenced by extremely large values.
The mode is the only measure you can use for nominal or categorical data that can’t be ordered.
What is variability?
Variability tells you how far apart points lie from each other and from the center of a distribution or a data set.
Variability is also referred to as spread, scatter or dispersion.
Cite this Scribbr article
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Shaun Turney
During his MSc and PhD, Shaun learned how to apply scientific and statistical methods to his research in ecology. Now he loves to teach students how to collect and analyze data for their own theses and research projects.
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# Inter 2nd Year Maths 2A Probability Solutions Ex 9(a)
Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(a) will help students to clear their doubts quickly.
## Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(a)
I. In the experiment of throwing a die, consider the following events:
Question 1.
A = {1, 3, 5}, B = {2, 4, 6}, C = {1, 2, 3} Are these events equally likely?
Solution:
Since events A, B, C has an equal chance to occur, hence they are equally likely events.
Question 2.
In the experiment of throwing a die, consider the following events:
A = {1, 3, 5}, B = {2, 4}, C = {6}
Are these events mutually exclusive?
Solution:
Since the happening of one of the given events A, B, C prevents the happening of the other two, hence the given events are mutually exclusive.
Otherwise A ∩ B = φ, B ∩ C = φ, C ∩ A = φ
Hence they are mutually exclusive events.
Question 3.
In the experiment of throwing a die, consider the events.
A = (2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Are these events exhaustive?
Solution:
A = {2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Let S be the sample space for the random experiment of throwing a die
Then S = {1, 2, 3, 4, 5, 6}
∵ A ⊂ S, B ⊂ S and C ⊂ S, and A ∪ B ∪ C = S
Hence events A, B, C are exhaustive events.
II.
Question 1.
Give two examples of mutually exclusive and exhaustive events.
Solution:
Examples of mutually exclusive events:
(i) The events {1, 2}, {3, 5} are disjoint in the sample space S = {1, 2, 3, 4, 5, 6}
(ii) When two dice are thrown, the probability of getting the sums of 10 or 11.
Examples of exhaustive events:
(i) The events {1, 2, 3, 5}, (2, 4, 6} are exhaustive in the sample space S = {1, 2, 3, 4, 5, 6}
(ii) The events {HH, HT}, {TH, TT} are exhaustive in the sample space S = {HH, HT, TH, TT} [∵ tossing two coins]
Question 2.
Give examples of two events that are neither mutually exclusive nor exhaustive.
Solution:
(i) Let A be the event of getting an even prime number when tossing a die and let B be the event of getting even number.
∴ A, B are neither mutually exclusive nor exhaustive.
(ii) Let A be the event of getting one head tossing two coins.
Let B be the event of getting atleast one head tossing two coins.
∴ A, B are neither mutually exclusive nor exhaustive.
Question 3.
Give two examples of events that are neither equally likely nor exhaustive.
Solution:
(i) Two coins are tossed
Let A be the event of getting an one tail and
Let B be the event of getting atleast one tail.
∴ A, B are neither equally likely nor exhaustive.
(ii) When a die is tossed
Let A be the event of getting an odd prime number and
Let B be the event of getting odd number.
∴ B are are neither equally likely nor exhaustive.
|
# How do you integrate int xsqrt(2x^2+7) using substitution?
Jan 22, 2017
I got: $\int x \sqrt{2 {x}^{2} + 7} \mathrm{dx} = \frac{1}{6} \left(2 {x}^{2} + 7\right) \sqrt{2 {x}^{2} + 7} + c$
#### Explanation:
Let us set:
$2 {x}^{2} + 7 = t$
derive:
$4 x = \mathrm{dt}$
let us use this inside our integral (red):
$\int \frac{1}{4} \textcolor{red}{4 x} \sqrt{t} \textcolor{red}{\mathrm{dx}} =$
where I multiplyed and divided by $4$:
substituting the red part with the derived bit we get:
$\int \frac{1}{4} \sqrt{t} \mathrm{dt} =$
integrating:
$\frac{1}{4} \int {t}^{\frac{1}{2}} \mathrm{dt} = \frac{1}{4} {t}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) + c = \frac{1}{4} {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + c =$
$= \frac{1}{6} t \sqrt{t} + c$
but: $2 {x}^{2} + 7 = t$
so we get:
$\int x \sqrt{2 {x}^{2} + 7} \mathrm{dx} = \frac{1}{6} \left(2 {x}^{2} + 7\right) \sqrt{2 {x}^{2} + 7} + c$
|
Associated Topics || Dr. Math Home || Search Dr. Math
### Finding Pythagorean Triplets
```Date: 02/02/2006 at 17:07:24
From: Faizan
Subject: Pythagorean triplets
Without using the standard a = n^2 - m^2, b = 2nm, c = n^2 + m^2, how
can you work out primitive Pythagorean triplets?
For example, you can work out an equation for cases when c and b
differ by 1 using (2n+1)^2 + b^2 = (b+1)^2.
I cannot work it out when c and b differ by 2. How would you do it?
```
```
Date: 02/04/2006 at 14:21:28
From: Doctor Ricky
Subject: Re: Pythagorean triplets
Hey Faizan,
Thanks for writing Dr. Math!
We should be able to find a Pythagorean triplet where the hypotenuse
and a leg differ by 2. Think of the most basic triplet - 3, 4, 5 -
and note that 5 - 3 = 2.
Assume we are trying to find a triplet where the hypotenuse and a
leg differ by 1.
You had the right idea with your work, but let's look more closely at
one leg and the hypotenuse differ by 1, or in other words:
b (leg) = n
c (hypotenuse) = n + 1
So, using the Pythagorean theorem and plugging in these assumptions,
we get:
a^2 + (n^2) = (n+1)^2. Simplifying, we get:
a^2 + n^2 = n^2 + 2n + 1, and again we simplify to get:
a^2 = 2n + 1
So in other words,
a = sqrt(2n + 1)
Therefore, for any integer value of n that makes sqrt(2n + 1) an
integer, we have a Pythagorean triplet where:
a = sqrt(2n + 1)
b = n
c = n + 1
For example, if n = 4, we get the triplet 3, 4, 5.
If n = 12, we get the triplet 5, 12, 13.
We can keep going with this for all integer values of a.
So now let's attack the situation of when a leg and the hypotenuse
differ by 3 and hopefully you will be able to see how to prove it for
a difference of 2.
So we'll assume:
b (leg) = n
c (hypotenuse) = n + 3
Using the Pythagorean theorem, we get:
a^2 + (n^2) = (n+3)^2
Simplifying, we get:
a^2 + n^2 = n^2 + 6n + 9
Simplifying again, we get:
a^2 = 6n + 9
and
a = sqrt(6n + 9)
Therefore, for any integer value of n that makes sqrt(6n + 9) an
integer, we have a Pythagorean triplet where:
a = sqrt(6n + 9)
b = n
c = n + 3
For example, if we choose n = 12, we get the triplet: 9, 12, 15.
If n = 36, we get the triplet: 15, 36, 39.
Finding values for n that make this work won't be very easy as we
continue to increase the difference between a leg and the hypotenuse,
which is why the general formulas you mentioned in your questions are
great guides for creating Pythagorean triplets.
If you have any more questions or if you were confused about any of
- Doctor Ricky, The Math Forum
http://mathforum.org/dr.math/
```
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High School Number Theory
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# How do you graph y=xabs(2x+5)?
Jan 19, 2018
#### Explanation:
Let's think of it this way:
$\left\mid a \right\mid = a$ and $\left\mid - a \right\mid = a$
For our first case, we are just bringing out $a$ outside the absolute value.
For the second case, we are finding the opposite of whatever was inside the absolute value sign.
So we can say that:
When $2 x + 5 \ge 0$, then $\left\mid 2 x + 5 \right\mid = 2 x + 5$
When $2 x + 5 < 0$, then $\left\mid 2 x + 5 \right\mid = - 2 x - 5$
Let's apply this to our function.
When $2 x + 5 \ge 0$, then $y = x \left(2 x + 5\right) \implies y = 2 {x}^{2} + 5 x$
When $2 x + 5 < 0$, then $y = x \left(- 2 x - 5\right) \implies y = - 2 {x}^{2} - 5 x$
We first garph these two parabolas:
Now, we ask ourselves,"For what values of $x$ does $2 x + 5 \ge 0$hold true?"
Similarly, "For what values of $x$ does $2 x + 5 < 0$ hold true?"
To find out the answer, we solve each inequality.
$2 x + 5 \ge 0$
$2 x \ge - 5$
$x \ge - \frac{5}{2}$
$2 x + 5 < 0$
$2 x < - 5$
$x < - \frac{5}{2}$
This is actually telling us that for any $x$ values greater than or equal to $- \frac{5}{2}$, the blue parabola will apply. When $x$ is smaller than $- \frac{5}{2}$, then the red parabola will apply. The green graph is the graph of our function.
So we now have:
If the image is not clear enough for you, you could zoom in by pressing ctrl key and the + key.
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# Why does dividing by zero give us no answer whatsoever?
I've heard about this and I know that division can be used in one way like this:
For example, if I want to do $30$ divided by $3$, how many times can I subtract $3$ from $30$ to get to $0$? Well, I can do it this way: $30-3=27-3=24-3=21-3=18-3=15-3=12-3=9-3=6-3=3-3=0$As you just saw, I subtract $3$ from $30$ 10 times to get to $0$. So, $30$ divided by $3$ is $10$. Well, what if I divide by a negative number or what if I divide a negative number by a positive number? Well, I do know how to solve $15$ divided by $-5$ this way. Just do it like in the mentioned way, but $15$ will get bigger if I subtract $-5$. So, what should I do. That's easy; do it backwards by subtracting the opposite of $-5$, or $5$: $15-5=10-5=5-5=0$. As you can see, I just did it backwards, and because I did it backwards, I need to subtract the number of times I subtract $5$, since it's the opposite of $-5$. So, I subtracted $-5$ -3 times. So, $15$ divided by $-5$ is $-3$.
As you have seen, I have used subtraction while doing division. Let's cut to the chase now. What if we divided any number by $0$? Well, this would happen: If I want to solve $10$ divided by $0$, I would just subtract $0$ from $10$ until I make the $10$ a $0$. So, $10-0=10-0=10-0=10-0=10-0=10-0=10-0=10...$ and it just keeps going on and on forever. So, this must explain why any number divided by zero has no answer at all. Also, for right here, right now, I can say it's infinity since it just goes on and on. So, is this why any number divided by $0$ never has an answer? Good answers at the bottom down there, if possible!
It is an interesting attempt, but you rather cannot induce, that any number $A$ divided by 0 is infinity. We know, that $-0=0$ and from your considerations $A$ divided by 0 must be equal to $-\infty$ in the same time. Moreover, your attempt gives $0/0=0$.
• So, it looks like any negative number divided by zero gives you no number, but negative infinity. – Mathster Nov 8 '14 at 4:12
• @Mathster Not so either. Division by $0$ is undefined, always. There are situations where you might interpret $0$ as $0^+$, a infinitesimally small positive number. Then $1/0^+$ could be interpreted as $+\infty$. But you could also interpret $0$ as $0^-$, an infinitesimally small negative number, and then $1/0^-$ would be interpreted as $-\infty$. So $1/0$ cannot be interpreted simply as $\infty$, even though the $1$ is positive. Likewise, $-1/0$ cannot be interpreted as $-\infty$. – alex.jordan Nov 8 '14 at 4:53
• This is what I mean: I think dividing by $0$ is undefined. – Mathster Nov 8 '14 at 15:27
• Why 0/0 should be 0, look here: math.stackexchange.com/a/1073101/2513 – Anixx Dec 18 '14 at 9:10
It is not infinity, take your example. Splitting 30 marbles between 3 people equally, will be 10 each. How about splitting 30 marbles between 0 people equally? It isn't infinity, it isnt even defined
• Well, I do know that infinity isn't a real number. – Mathster Nov 8 '14 at 4:12
The reason division by zero is not defined is simply that it's not possible to define it and still preserve the usual rules of algebra.
For example we know that if $x$ is a nonzero real number, and if $ax = bx$, then $a = b$. This is called the cancellation law and it follows from the fact that each nonzero real x has a multiplicative inverse $x^{-1}$ with the property that $xx^{-1} = 1$. So if $ax = bx$ then $axx^{-1} = bxx^{-1} \Rightarrow a = b$.
The cancellation law is very handy when we're solving equations and doing other algebraic manipulations, so we want to keep it around.
On the other hand, for example $2(0) = 3(0)$ but $2 \neq 3$. What have we learned? We can not cancel zero. And why is this? It's because zero does not have a multiplicative inverse. That is, there is no real number $x$ with $0x = 1$. That's why we can't cancel it.
Ok no problem, we just make a rule that you can't cancel zero. But if you think about it, what is division in the first place? Division is just another name for the cancellation rule. If you tell me that $x$ is nonzero and $ax = bx$ then I am allowed to conclude that $a = b$. But this is just regular old division.
In other words the cancellation law is division. We can cancel any number that has a multiplicative inverse. Therefore we can not cancel 0, because 0 does not have a multiplicative inverse. So we can't cancel zero ... which is just another way of saying that we can't divide by zero.
I don't think a precise answer can be offered if the wording & language we use is not precise. Thus any attempt towards precise understanding is futile until all relevant terms are precisely defined.
FIRST possible approach: define '$\div$' as a binary operation over some subset of complex number ordered pairs were by definition $$a\div b = a \cdot (\text{the multiplicative inverse of } b)$$
it can be gracefully proven, using the standard field axioms, that for any complex number $b$, $$b\cdot 0 = 0$$ thus $$0\cdot x=1$$ has not solution in the complex field, therefore, 0 has no multiplicative inverse in the complex field, therefore "$\div$" can not be defined as such over pairs where the second number is zero, such subset must be excluded.
SECOND approach, use limits and look at numbers as equivalence classes of convergent sequences, in that case we can very well divide 0/0 meaning we can divide by sequences where each goes to zero as in $$\lim \frac{\sin(1/n)}{(1/n)} = 1$$ clearly in that case it is NOT true that we can not divide by zero
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4.01 Multiplication
Lesson
When we need to solve multiplication problems, we have many methods we could use. Which method we choose depends on our problem, how many digits our numbers have, and the method we are more comfortable with.
Idea summary
Methods used in multiplication:
• double-double
• area model
• partitioning by place value
• partitioning by factors
• vertical multiplication
Multiplication review
Let's look at a video on how to multiply two 2 digit numbers.
Examples
Example 1
Calculate 26 \times 39 by completing an area model.
a
Firstly, complete the missing values in the area model.
Worked Solution
Create a strategy
Multiply the number at the top of the column by the number on the left of the row.
Apply the idea
b
Now add the values you found in part (a) to calculate 26\times 39.
Worked Solution
Create a strategy
Use a vertical algorithm to add the results.
Apply the idea
Write the results in a vertical algorithm and add the digits from the units column to the hundreds column:\begin{array}{c} && &{}^2 6 &0 &0 \\ && &1 &8 &0 \\ && &1 &8 &0 \\ &+& & &5 &4 \\ \hline &&1 &0 &1 &4 \\ \hline \end{array}
Idea summary
We can use area models to multiply 2 two digit numbers.
Multiply 3 digit numbers
This video looks at using the algorithm to solve a multiplication problem.
Examples
Example 2
Find 7\times 583.
Worked Solution
Create a strategy
Use the standard algorithm method for multiplication.
Apply the idea
Write the product in a vertical algorithm:
\begin{array}{c} &&&5&8&3 \\ &\times &&&&7 \\ \hline \\ \hline \end{array}
Start from the far right. Multiply to get 7\times 3=21. So we write the 1 in the units place and carry the 2 to the tens place:
\begin{array}{c} &&&5&{}^28&3 \\ &\times &&&&7 \\ \hline &&&&&1 \\ \hline \end{array}
Move left and multiply to get 7\times 8=56. Then we add the 2 to get 58. So we write the 8 in the tens place and carry the 5 to the hundreds place:
\begin{array}{c} &&&{}^55&{}^28&3 \\ &\times &&&&7 \\ \hline &&&&8&1 \\ \hline \end{array}
Move left and multiply to get 7\times 5=35. Then we add the 5 to get 40. So we write the 4 in the thousands place and the 0 in the hundreds place:
\begin{array}{c} &&&{}^55&{}^28&3 \\ &\times &&&&7 \\ \hline &&4&0&8&1 \\ \hline \end{array}
583\times7=4081
Idea summary
There isn't one perfect way to solve multiplication problems. The best approach is to have a range of methods to choose from, so you can choose the method that suits the problem. This is just like having a toolkit with different tools.
Outcomes
VCMNA209
Select and apply efficient mental and written strategies and appropriate digital technologies to solve problems involving all four operations with whole numbers and make estimates for these computations
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# Powers of Coprime Numbers are Coprime
## Theorem
Let $a, b$ be coprime integers:
$a \perp b$
Then:
$\forall n \in \N_{>0}: a^n \perp b^n$
In the words of Euclid:
If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes].
## Proof
Proof by induction:
Let $a \perp b$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
$a^n \perp b^n$
$\map P 1$ is true, as this just says:
$a \perp b$
### Basis for the Induction
$a^2 \perp b$
$a^2 \perp b^2$
This is our basis for the induction.
### Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
$a^k \perp b^k$
Then we need to show:
$a^{k + 1} \perp b^{k + 1}$
### Induction Step
This is our induction step:
$\ds a^k$ $\perp$ $\ds b^k$ $\ds \leadsto \ \$ $\ds a^k \times a$ $\perp$ $\ds b^k$ Proposition $24$ of Book $\text{VII}$: Integer Coprime to Factors is Coprime to Whole $\ds \leadsto \ \$ $\ds a^{k + 1}$ $\perp$ $\ds b^k$ $\ds \leadsto \ \$ $\ds a^{k + 1}$ $\perp$ $\ds b^k \times b$ Proposition $24$ of Book $\text{VII}$: Integer Coprime to Factors is Coprime to Whole $\ds \leadsto \ \$ $\ds a^{k + 1}$ $\perp$ $\ds b^{k + 1}$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\forall n \in \N_{>0}: a^n \perp b^n$
$\blacksquare$
## Historical Note
This proof is Proposition $27$ of Book $\text{VII}$ of Euclid's The Elements.
Euclid's proof does not use the full induction process, which is a later mathematical innovation.
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HANDS-ON ACTIVITY 3.2: THE EXTREME VALUE THEOREM - Limits and Continuity - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC
## Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 3. Limits and Continuity
HANDS-ON ACTIVITY 3.2: THE EXTREME VALUE THEOREM
NOTE. The intervals of continuity for trigonometric functions are the intervals that make up their domains, as trigonometric functions are continuous on their entire domains.
This and the next activity introduce you to two basic but important continuity theorems. Note that both of these are called existence theorems. They guarantee the existence of certain values but do not tell you where these values are—it’s up to you to find them, and they’re always in the last place you look (with your car keys, your wallet from two years ago, and the words to that Bon Jovi song you used to know by heart).
1. Given f(x) = x4 — 3x — 4, justify that f(x) is continuous on the x interval [—1,2].
_____________________________________________________
_____________________________________________________
2. Draw the graph of f(x). Use a graphing calculator if you wish.
3. The maximum height reached by f(x) on the interval [—1,2] is called the maximum of f on the interval. At what value of x does f(x) reach its maximum, and what is that maximum value?
_____________________________________________________
_____________________________________________________
4. To calculate the minimum value of f(x), use the 2nd→Trace→minimum function on your calculator. Set bounds to the left and the right of the minimum and make a guess, as you did when finding x-intercepts in Chapter 2. What is the minimum value of f(x) on [—1,2]?
_____________________________________________________
NOTE. You are not allowed to use the maximum/minimum functions of your calculator on the AP exam. We will learn how to find them in different ways (which are acceptable on the test) later.
5. At what x-value does the minimum occur? (Hint: The value is displayed when you calculate the minimum with the calculator.)
_____________________________________________________
6. Graph g(x) = 1/x on the axes below. Can you find a maximum and a minimum for g(x) on the x interval (1,5)?
7. Why do your results for f and g differ?
_____________________________________________________
8. Graph below. Why is there no maximum on [0,3]?
_____________________________________________________
_____________________________________________________
ALERT! If the EVT conditions are not satisfied, a function can still have a maximum, a minimum, or both—their existence is just not guaranteed in that case.
9. Complete the Extreme Value Theorem, based on your work above:
Extreme Value Theorem: Any function, f(x), will have a maximum and a minimum on the _________ interval _________ as long as f(x) is _________.
10. Visually speaking, where can maximums and minimums occur on a graph?
_____________________________________________________
SELECTED SOLUTIONS TO HANDS-ON ACTIVITY 3.2
1. f is a polynomial whose domain is (—∞,∞), and polynomials are continuous on their entire domain (remember the red parrot?)
3. f reaches its highest point on [—1,2] when x = 2, and f(2) = 6. Thus, f has a maximum of 6.
4. f has a minimum of —6.044261. If you can’t get this, use a left bound of 0, a right bound of 2, and a guess of 1.
5. The minimum occurs at x = .9085621.
6. You cannot find a maximum or a minimum. Although g(1) = 1 and g(5) = 1/5 would be the maximum and minimum values, respectively, they are not included on the open interval (1,5). You cannot choose an x value whose function value is higher or lower than every other in the interval—try it!
7. You used a closed interval with f and an open interval with g.
8. h is removably discontinuous at x = 0, and because the maximum would have occurred there, the function will have no maximum.
9. Any function, f(x), will have a maximum and minimum on the closed interval [a,b] as long as f(x) is continuous. If the interval is not closed or the function is discontinuous, the guarantees of the Extreme Value Theorem do not apply.
10. The maximum and minimum (together called extrema, since they represent the extreme highest and lowest points on the graph) will occur at “humps” on the graph (like the minimum of f) or at the endpoints of the interval (like f’s maximum).
EXERCISE 4
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
YOU MAY USE YOUR CALCULATOR FOR ALL OF THESE PROBLEMS
For each of the following functions and intervals, determine if the Extreme Value Theorem applies, and find the maximum and minimum of the function, if possible.
ALERT! In problem 2, a common mistake is to say that the maximum of the function is 2, but that is only the x-value at which the maximum occurs.
1. The Extreme Value Theorem (EVT) does not apply because tan x is discontinuous on the given interval, specifically at x = π/2. No maximum or minimum values are possible on the closed interval, as the function both increases and decreases without bound at x = π/2.
2. The function is continuous on the interval. Both parts of the function are trigonometric and, therefore, continuous on their domain, and [—π,—π/4] is included in the domain of each. The only possible discontinuity comes at x = —π/2, but both functions have the same value there, guaranteeing that the function is continuous. See the below graph—there is no break or jump. The EVT will apply. The maximum is 0 and occurs at x = —π, as the function decreases for the remainder of the interval. The minimum is 2/√2 = —1.414 and occurs at x = —π/4.
3. The EVT will not apply, but the function still has a maximum and minimum. The maximum occurs at the point (—2,8.135), and the minimum occurs at the point (3.733,—10.216). (The minimum is found using the minimum function on your graphing calculator.)
4. Although In (x — 1) is discontinuous at x = 1, ln (x — 1) is continuous on the given interval [2,5], so the EVT will apply. Because In x is monotonic and increasing, the minimum (0) will occur at the beginning of the interval, and the maximum (1.386) will occur at the end of the interval.
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# A proof of Pythagoras' Theorem (Investigation)
Lesson
The study of mathematics is like a huge pyramid, with things which are proven forming the base of more complicated work above it. The great thing about math is that once someone has proven something, you can take it as completely true and never have to doubt its validity, and so you can use them without worrying whether they are really true or not. Unlike science, where things which are widely believed can occasionally be proven wrong, in mathematics once something has been proven properly, it is perfectly and completely true forever.
For example, in your addition you often use the fact that $1+1=2$1+1=2. Without this fact none of your addition would make sense, and you wouldn't even be able to begin learning your multiplication facts.
But are we really sure that $1+1=2$1+1=2?
No, holding up one finger and another finger does not count as a proof. In fact, mathematicians have managed to prove that $1+1=2$1+1=2 properly. In Bertrand Russel's Principia Mathematica, he took 362 pages of advanced mathematics to finally prove it! The end of this proof is shown below:
Although some proofs are extremely complicated, like this one, others are simple and intuitive. We're going to have a look at one of these, the proof for Pythagoras' theorem. You have probably already memorised the theorem, which looks something like this:
The proof for this is fairly simple, and you can even prove it yourself using a piece of paper and a pair of scissors:
Start with a square with four equal triangles drawn into it, so that the hypotenuses form a square in the middle. Each of these triangles has sides $a$a and $b$b, and hypotenuse $c$c. Since the white square has a side length of $c$c, it has an area of $c^2$c2. However, when we rearrange the triangles as shown in the second diagram, we can see that the white area is now broken into two pieces, a square with side length a and a square with side length b, which together have an area of $a^2+b^2$a2+b2. Since the white area must be the same as before we rearranged, this means that $a^2+b^2=c^2$a2+b2=c2
As you can see, the proof is fairly simple, you do not need to be a genius to figure it out. So why did Pythagoras' get so famous for proving it this way? Well, it turns out that Pythagoras was not the first one to "discover" Pythagoras' theorem, as people had already known that $a^2+b^2=c^2$a2+b2=c2 for a very long time.
For example, the Babylonians were aware of this relationship at least $1000$1000 years before Pythagoras, and it was discovered independently by the Chinese (where some call it the "Shang Gao Theorem") and Indians. However, the reason that it is called "Pythagoras' theorem" is that he is said to have been the first to prove it, using something like the proof above. In mathematics, simply discovering that something is true is not enough, it is the process of proving it rigorously which is the most important. Just like a solid stone in a pyramid, a proof allows other mathematics to be built on top of it, without ever doubting the foundations below.
### Outcomes
#### 9P.MG2.01
Relate the geometric representation of the Pythagorean theorem to the algebraic representation a^2 + b^2 = c^2
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9140 minus 35 percent
This is where you will learn how to calculate nine thousand one hundred forty minus thirty-five percent (9140 minus 35 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 9140 minus 35 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 9140 of something.
9140
(100%)
35 percent means 35 per hundred, so for each hundred in 9140, you want to subtract 35. Thus, you divide 9140 by 100 and then multiply the quotient by 35 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(9140 ÷ 100) × 35
= 3199
We made a pink square that we put on top of the image shown above to illustrate how much 35 percent is of the total 9140:
The dark blue not covered up by the pink is 9140 minus 35 percent. Thus, we simply subtract the 3199 from 9140 to get the answer:
9140 - 3199
= 5941
The explanation and illustrations above are the educational way of calculating 9140 minus 35 percent. You can also, of course, use formulas to calculate 9140 minus 35%.
Below we show you two formulas that you can use to calculate 9140 minus 35 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
9140 - ((9140 × 35/100))
9140 - 3199
= 5941
Formula 2
Number × (1 - (Percent/100))
9140 × (1 - (35/100))
9140 × 0.65
= 5941
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
9150 minus 35 percent
Here is the next percent tutorial on our list that may be of interest.
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