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  • stemjock.com
    Solutions to Vector Calculus 4e by S. J. Colley - stemjock.com

    https://stemjock.com/colleyvc4e.htm

    Exercise 40. Exercise 10. Exercise 20. Exercise 30. Exercise 40. These are my solutions to the fourth edition of Vector Calculus by Colley.

  • stemjock.com
    Solutions to Differential Equations 10e by Boyce and DiPrima

    https://stemjock.com/boyceode10e.htm

    These are my solutions to the tenth edition of Elementary Differential Equations and Boundary Value Problems 10e by Boyce and DiPrima.

  • stemjock.com
    Ross 1st Course in Probability 10e: Chapter 1 - stemjock.com

    https://stemjock.com/STEM%20Books/Ross%20Probability%2010e/Chapter%201/RossProch1p12.pdf

    Ross 1st Course in Probability 10e: Chapter 1 - Problem 12 Page 1 of 1 Problem 12 How many 3 digit numbers xyz, with x;y;z all ranging from 0 to 9 have at least 2 of their digits

  • stemjock.com
    Hibbeler Statics 14e: Problem 3-31 Page 1 of 2 - stemjock.com

    https://stemjock.com/STEM%20Books/Hibbeler%20Statics%2014e/Chapter%203/HibbelerS14eCh3p31.pdf

    Hibbeler Statics 14e: Problem 3-31 Page 2 of 2 As a result, cos˚= 4 p s2 +42 and sin˚= s p s2 +42 and cos = 2 p s2 +22 and sin = s p s2 +22 Since T AD = m Dg= 4gand T AE = m Eg= 6g, the last two equations become 6g 2 p s2 …

  • stemjock.com
    Problem 1 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%206/BoyceODEch3s6p01.pdf

    www.stemjock.com. Boyce & DiPrima ODEs 10e: Section 3.6 - Problem 1 Page 3 of 3 Integrate both sides with respect to t, setting the integration constant to zero. e tC0 1(t) = e 2t Multiply both sides by et. C0 1(t) = e t Integrate both sides with respect to t once more. C 1(t) = e t Consequently, the particular solution is y p(t) = C

  • stemjock.com
    Problem 20 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%204/Section%201/BoyceODEch4s1p20.pdf

    www.stemjock.com. Boyce & DiPrima ODEs 10e: Section 4.1 - Problem 20 Page 3 of 4 Alternatively, we can proceed as we’re told. W 0= y 1 y 2 y 3 y 1 y 2 y 3 y000 1 y 000 2 y 000 3 = y 1 y 2 y 3 y0 1 y 0 2 y 0 3 p 1(t)y00 1 p 2(t)y 1 0 p 3(t)y 1 p 1(t)y002 p 2(t)y 2 0 p 3(t)y 2 p 1(t)y003 p 2(t)y 3 0 p 3(t)y 3 Multiply the first row by p

  • stemjock.com
    Problem 3-33 - stemjock.com

    https://stemjock.com/STEM%20Books/Hibbeler%20Statics%2014e/Chapter%203/HibbelerS14eCh3p33.pdf

    www.stemjock.com. Hibbeler Statics 14e: Problem 3-33 Page 2 of 2 Solve these seven equations for the six tensions and . Then use the fact that W= 15 lb. T AB = (p 3 1)Wˇ11:0 lb T AC = q 2 p 3Wˇ7:76 lb T BC = (p 3 1)Wˇ11:0 lb T CD = r 5 2 (p 3 1)Wˇ17:4 lb T BE = (3 p 3)Wˇ19:0 lb T AL = W= 15:0 lb = cos 1 3 p 10

  • stemjock.com
    Exercise 3.2 - stemjock.com

    https://stemjock.com/STEM%20Books/Haberman%20Applied%20PDEs%205e/Chapter%203/HabermanPDEch3s32e2.pdf

    Haberman Applied PDEs 5e: Section 3.2 - Exercise 3.2.2 Page 2 of 6 Part (b) For f(x) = e x, the coefficients are A 0 = 1 2L L L e xdx= sinhL L A n= 1 L L L e xcos nˇx L dx= 2( 1)nLsinhL n2ˇ2 +L2 B n= 1 L L L e xsin nˇx L dx= 2( 1)nnˇsinhL n2ˇ2 +L2 www.stemjock.com

  • stemjock.com
    Problem 2B - stemjock.com

    https://stemjock.com/STEM%20Books/BSL%20Transport%20Phenomena%202e%20Revised/Chapter%202/BSLTPCh2p2B4.pdf

    www.stemjock.com. BSL Transport Phenomena 2e Revised: Chapter 2 - Problem 2B.4 Page 2 of 5 Figure 2: This is the shell over which the momentum balance is made for the flow in a slit. Rate of z-momentum into the shell at z= 0: (W x)˚ ...

  • stemjock.com
    Problem 2.1 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20ED%205e/Chapter%202/GriffithsED5eCh2p1.pdf

    www.stemjock.com. Griffiths Electrodynamics 5e: Problem 2.1 Page 3 of 4 Part (c) Suppose that now there are 13 charges equally spaced on a circle with radius r, one of them being at the 6 o’clock position. The position vector of charge q k is r ...

  • stemjock.com
    Griffiths Electrodynamics 5e: Problem 1.35 Page 1 of 2

    https://stemjock.com/STEM%20Books/Griffiths%20ED%205e/Chapter%201/GriffithsED5eCh1p35.pdf

    www.stemjock.com. Griffiths Electrodynamics 5e: Problem 1.35 Page 2 of 2 As a result, the surface integral of ∇×v over the five faces is S (∇×v)·dS = 1 0 1 0 [xˆ(4z2 −2x)+ˆz(2z)]·(xˆdydz) x=1 + 1 0 1 0 [xˆ(4z2 −2x)+ˆz(2z)]·(−ˆzdxdy) z=0 + 1 0 1 0 [xˆ(4z2 −2x)+ˆz(2z)]·(yˆdxdz)

  • stemjock.com
    Problem 29 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%202/BoyceODEch3s2p29.pdf

    Boyce & DiPrima ODEs 10e: Section 3.2 - Problem 29 Page 1 of 1 Problem 29 In each of Problems 29 through 32, find the Wronskian of two solutions of the given differential

  • stemjock.com
    Solutions to Mathematical Methods for Physicists 7e by Arfken

    https://stemjock.com/arfkenmm7e.htm

    Solutions to Mathematical Methods for Physicists 7e by Arfken. Visit the Textbook's Page on Amazon.com. Ch. 1-7. Ch. 8-15. Ch. 16-23. These are my solutions to the seventh …

  • stemjock.com
    Exercise 9.4 - stemjock.com

    https://stemjock.com/STEM%20Books/Arfken%20Mathematical%20Methods%207e/Chapter%209/ArfkenMMch9s4e2.pdf

    Exercise 9.4.2. Show that the Helmholtz equation, r2+ k2= 0; is still separable in circular cylindrical coordinates if k2is generalized to k2+ f(ˆ) + (1=ˆ2)g(’) + h(z). Solution Replace k2with k2+ f(ˆ) + (1=ˆ2)g(’) + h(z) in the Helmholtz equation. r2+ k2+ f(ˆ) + 1 ˆ2. g(’) + h(z) = 0 Expand the Laplacian operator in circular ...

  • stemjock.com
    Exercise 8.2 - stemjock.com

    https://stemjock.com/STEM%20Books/Arfken%20Mathematical%20Methods%207e/Chapter%208/ArfkenMMch8s2e1.pdf

    Arfken Mathematical Methods 7e: Section 8.2 - Exercise 8.2.1 Page 1 of 1 Exercise 8.2.1 Show that Laguerre’s ODE, Table 7.1, may be put into self-adjoint form by multiplying by e x and that w(x) = e x is the weighting function. Solution

  • stemjock.com
    Problem 4 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20QM%203e/Chapter%204/GriffithsQMCh4p19.pdf

    Problem 4. Griffiths Quantum Mechanics 3e: Problem 4.19 Page 1 of 6. Problem 4.19. A hydrogenic atom consists of a single electron orbiting a nucleus with Zprotons. (Z= 1 would be hydrogen itself, Z= 2 is ionized helium, Z= 3 is doubly ionized lithium, and so on.) Determine the Bohr energies E. n(Z), the binding energy E.

  • stemjock.com
    Problem 3-48 - stemjock.com

    https://stemjock.com/STEM%20Books/Hibbeler%20Statics%2014e/Chapter%203/HibbelerS14eCh3p48.pdf

    Hibbeler Statics 14e: Problem 3-48 Page 1 of 2 Problem 3-48 Determine the tension in the cables in order to support the 100-kg crate in the equilibrium

  • stemjock.com
    Problem 15 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%207/Section%205/BoyceODEch7s5p15.pdf

    www.stemjock.com. Boyce & DiPrima ODEs 10e: Section 7.5 - Problem 15 Page 2 of 2 Substitute these two eigenvalues into equation (1) to determine the corresponding eigenvectors. 5 1 1

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    Problem 3 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20QM%203e/Chapter%203/GriffithsQMCh3p26.pdf

    Griffiths Quantum Mechanics 3e: Problem 3.26 Page 2 of 2 A is not hermitian because A†= 1 −2i −1 0 0 0 −2i −4 2i ̸= 1 0 2i 2i 0 −4 −1 0 −2i

  • stemjock.com
    Griffiths Electrodynamics 5e: Problem 2 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20ED%205e/Chapter%202/GriffithsED5eCh2p16.pdf

    www.stemjock.com. Griffiths Electrodynamics 5e: Problem 2.16 Page 2 of 4 Integrate both sides of Gauss’s law over the volume of the spherical Gaussian surface. x2 0 +y 2 0 +z 0 ≤r 2 ∇·EdV 0 = x2 0 +y 0 +z 0

  • stemjock.com
    Problem 23 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%201/BoyceODEch3s1p23.pdf

    Boyce & DiPrima ODEs 10e: Section 3.1 - Problem 23 Page 1 of 1 Problem 23 In each of Problems 23 and 24, determine the values of , if any, for which all solutions tend to

  • stemjock.com
    Solutions to Engineering Mechanics Statics 14e by R. C. Hibbeler

    https://stemjock.com/hibbelers14e.htm

    These are my solutions to the fourteenth edition of Engineering Mechanics: Statics by R. C. Hibbeler.

  • stemjock.com
    Solutions to Fundamentals of Electric Circuits 6e by

    https://stemjock.com/asec6e.htm

    These are my solutions to the sixth edition of Fundamentals of Electric Circuits by Alexander and Sadiku.

  • stemjock.com
    Problem 12-25 - stemjock.com

    https://stemjock.com/STEM%20Books/Hibbeler%20Dynamics%2014e/Chapter%2012/HibbelerD14eCh12p25.pdf

    www.stemjock.com. Hibbeler Dynamics 14e: Problem 12-25 Page 2 of 2 Use the fact that the body is released from rest to determine C. 100ln secsin 1 0 100 +tansin 1 0 100 = 9:81(0)+C ! C= 100ln1 = 0 As a result, equation (2) becomes 100ln secsin 1 v 100 +tansin 1 v 100 = 9:81t: (3)

  • stemjock.com
    Griffiths Electrodynamics 5e: Problem 1 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20ED%205e/Chapter%201/GriffithsED5eCh1p20.pdf

    www.stemjock.com. Griffiths Electrodynamics 5e: Problem 1.20 Page 2 of 2 v = yxˆ+xyˆ is illustrated below in red. Some streamlines are drawn in blue. There’s no divergence because for every streamline pointing into the origin, another is pointing out. And there’s no curl because, if you imagine a circular wheel centered at the origin, the ...

  • stemjock.com
    Griffiths Electrodynamics 5e: Problem 2.21 Page 1 of 3

    https://stemjock.com/STEM%20Books/Griffiths%20ED%205e/Chapter%202/GriffithsED5eCh2p21.pdf

    www.stemjock.com. Griffiths Electrodynamics 5e: Problem 2.21 Page 2 of 3 with position vectors, a and b, and use the fundamental theorem for gradients. b a

  • stemjock.com
    Stewart Calculus 8e: Chapter 9 Problems Plus - stemjock.com

    https://www.stemjock.com/STEM%20Books/Stewart%20Calculus%208e/Chapter%209/Problems%20Plus/StewartCalcch09ppp9.pdf

    www.stemjock.com. Stewart Calculus 8e: Chapter 9 Problems Plus - Problem 9 Page 2 of 5 Solution Part (a) There is a critical observation that needs to be made in order to derive this formula. The dog and the rabbit travel at the same speed. Therefore, the arc length of the dog’s path is equal to the

  • stemjock.com
    Exercise 1 - stemjock.com

    https://stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%205/Section%206/StraussPDEch5s6p01.pdf

    www.stemjock.com. Strauss PDEs 2e: Section 5.6 - Exercise 1 Page 6 of 6 Figure 1: This is a plot of the solution uas a function of xfor various times. The curves in red, orange, yellow, green, blue, and purple correspond to t= 0, t= 0:15, t= 0:3, t= 0:5, t= 0:8, and t= 10, respectively. Note that as t!1the graph approaches the equilibrium ...

  • stemjock.com
    Problem 14 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%201/BoyceODEch3s1p14.pdf

    www.stemjock.com. Boyce & DiPrima ODEs 10e: Section 3.1 - Problem 14 Page 2 of 2 www.stemjock.com. Created Date: 4/13/2019 7:19:54 PM ...

  • stemjock.com
    Problem 7 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%202/BoyceODEch3s2p07.pdf

    Boyce & DiPrima ODEs 10e: Section 3.2 - Problem 7 Page 1 of 1 Problem 7 In each of Problems 7 through 12, determine the longest interval in which the given initial value

  • stemjock.com
    Problem 2 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20QM%203e/Chapter%202/GriffithsQMCh2p7.pdf

    www.stemjock.com. Griffiths Quantum Mechanics 3e: Problem 2.7 Page 3 of 4 in this infinite series vanishes, then, except for the one corresponding to n= p. a 0 (x;0)sin nˇx a dx= r 2 a B n a 0 sin2 nˇx a dx = r 2 a B n a 2 = r a 2 B n Solve for B n. B n= r 2 a a 0 (x;0)sin nˇx a dx = r 2 a " a=2 0 2 r 3 a3 xsin nˇx a dx+ a a=2 2 r 3 a3 (a x ...

  • stemjock.com
    Problem 8 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%205/BoyceODEch3s5p08.pdf

    www.stemjock.com. Boyce & DiPrima ODEs 10e: Section 3.5 - Problem 8 Page 2 of 2 The complementary solution is then y c(t) = C 1te t +C 2e t: On the other hand, the particular solution satisfies y00 p +2y 0 p +y p = 2e t: We would use the trial solution, Ae t, but because e t is already a solution of equation (1), we

  • stemjock.com
    Problem 4B - stemjock.com

    https://stemjock.com/STEM%20Books/BSL%20Transport%20Phenomena%202e%20Revised/Chapter%204/BSLTPCh4p4B1.pdf

    www.stemjock.com. BSL Transport Phenomena 2e Revised: Chapter 4 - Problem 4B.1 Page 4 of 6 and substitute them into the PDE. y 4 p t3 f0= 1 4 t f00 1 4t f00+ y 4 p t3 f0= 0 f00+ y p t f0= 0 f00+2 y p 4 t f0= 0 f00+2 f0= 0 This is a first-order linear ODE for f, so it can be solved with an integrating factor I. I= exp

  • stemjock.com
    Griffiths Electrodynamics 5e: Problem 1 - stemjock.com

    https://stemjock.com/STEM%20Books/Griffiths%20ED%205e/Chapter%201/GriffithsED5eCh1p61.pdf

    www.stemjock.com. Griffiths Electrodynamics 5e: Problem 1.61 Page 2 of 5 Therefore, D ∇TdV = bdy D TdS. Part (b) Begin with the divergence theorem, let v = u×c in which c is constant, use Identity 6 on the left, and use Identity 1 on the right. D ∇·vdV = bdy D v ·dS D ∇·(u×c)dV = bdy D (u×c)·dS D [c·(∇×u)−u·(∇×c)

  • stemjock.com
    Solutions to Introduction to Quantum Mechanics 3e by D. J. Griffiths

    https://stemjock.com/griffithsqm3e.htm

    These are my solutions to the third edition of Introduction to Quantum Mechanics by D. J. Griffiths.

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    Check if Stemjock.com is legit or scam, Stemjock.com reputation, customers reviews, website popularity, users comments and discussions.

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    Boyce & DiPrima ODEs 10e: Section 3.7 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%203/Section%207/BoyceODEch3s7p07.pdf

    www.stemjock.com. Boyce & DiPrima ODEs 10e: Section 3.7 - Problem 7 Page 2 of 4 This is a linear inhomogeneous ODE, so its general solution can be expressed a sum of the complementary solution and the particular solution. x(t) = x c(t) + x p(t) The complementary solution satisfies the associated homogeneous equation.

  • stemjock.com
    Problem 15 - stemjock.com

    https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%206/Section%203/BoyceODEch6s3p15.pdf

    Page 1 of 1. Problem 15. In each of Problems 13 through 18, find the Laplace transform of the given function. f(t) =. 8. 0; t .

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stemjock.comIN300Aip: 172.66.41.12
stemjock.comIN86400NStarget: carl.ns.cloudflare.com
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stemjock.comIN300AAAA
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stemjock.com Traffic Analysis

According to global rankings, stemjock.com holds the position of #283108. It attracts an approximate daily audience of 5.78K visitors, leading to a total of 5890 pageviews. On a monthly basis, the website garners around 173.33K visitors.

Daily Visitors5.78K
Monthly Visits173.33K
Pages per Visit3.06
Visit Duration0:04:5
Bounce Rate48.91%
Want complete report?Full SEMrush Report >>
Daily Unique Visitors:
5777
Monthly Visits:
173332
Pages per Visit:
3.06
Daily Pageviews:
5890
Avg. visit duration:
0:04:5
Bounce rate:
48.91%
Monthly Visits (SEMrush):
175788

Traffic Sources

SourcesTraffic Share
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0.42%
Paid Referrals:
0.00%
Mail:
0.01%
Search:
65.09%
Direct:
34.47%

Visitors by Country

CountryTraffic Share
United States:
53.82%
Korea, Republic of:
7.15%
Vietnam:
6.75%
Canada:
5.08%
India:
4.87%

SSL Checker - SSL Certificate Verify

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name
stemjock.com
hash
80d28021
issuer
Let's Encrypt
version
2
serialNumber
417856170951973980725519698217987567950619
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1713078913
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1720854912
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ecdsa-with-SHA384
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ecdsa-with-SHA384
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795
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Digital Signature
extendedKeyUsage
TLS Web Server Authentication, TLS Web Client Authentication
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CA:FALSE
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authorityInfoAccess
OCSP - URI:http://e1.o.lencr.org CA Issuers - URI:http://e1.i.lencr.org/
subjectAltName
DNS:*.stemjock.com, DNS:stemjock.com
certificatePolicies
Policy: 2.23.140.1.2.1

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Status
HTTP/1.1 403 Forbidden
Date
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Content-Type
text/html
Connection
keep-alive
strict-transport-security
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x-content-type-options
nosniff
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SAMEORIGIN
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"1; mode=block"
referrer-policy
no-referrer-when-downgrade
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DYNAMIC
Report-To
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cloudflare
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Domain Updated Date:2023-11-09
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Registrar WHOIS Server:whois.cloudflare.com
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SEO Analysis

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Title Tag:
Textbook Index

Length: 14 characters

Title tags are usually best kept short, within 50-70 characters. It's important to note that search engines will typically read the entire title tag even if it exceeds 70 characters, but there is a chance they may cut it off or disregard it.

Meta Description:
This is a website where solutions to textbooks in mathematics, science, and engineering are posted. It is dedicated to the future generations of students.

Length: 154 characters

When crafting website descriptions, keep in mind that search engines only show the first 150-160 characters in search results. To ensure your entire description is visible, aim for a length of 25-160 characters. If your description is too long, it may get cut off. Conversely, if it's too short, search engines may add text from elsewhere on your page. Additionally, search engines may modify the description you provide to better match the user's search intent. It's best to strike a balance between brevity and relevance for optimal visibility.

Meta Keywords:
  • stemjock
  • textbook
  • index
  • solutions
  • answers
  • textbooks

In the realm of search engine optimization, the meta keywords tag has become a relic of the past due to its potential for misuse, ultimately leading major search engines to disregard it in their ranking algorithms.

Keywords Cloud:
Term Count Density
physics 21 3.95%
openstax 18 3.39%
chemistry 15 2.82%
mechanics 14 2.64%
calculus 12 2.26%
algebra 12 2.26%
applications 11 2.07%
equations 11 2.07%
differential 11 2.07%
fundamentals 11 2.07%
introduction 10 1.88%
engineers 10 1.88%
volume 8 1.51%
engineering 8 1.51%
scientists 7 1.32%
college 7 1.32%
methods 6 1.13%
elementary 6 1.13%

A crucial factor in search engine optimization is keyword density, which refers to the proportion of a particular keyword present in the text of a webpage. In order to achieve high rankings on search engine results pages, it is essential to maintain the appropriate keyword density for your primary keyword.

Headings:
<H1>
1
<H2>
0
<H3>
0
<H4>
0
<H5>
0
<H6>
0
<h1>Textbook Index</h1>

In SEO, the primary focus is placed on keywords within the content. The title of the page holds the highest importance, followed by heading tags such as h1, h2, and h3. The h1 heading should be the largest on the page, while the h2 heading should be slightly smaller, and the h3 heading even smaller. This hierarchical structure is crucial for optimizing search engine rankings.

Image Alt Attribute:
4 images found in your page, and 4 images are without "ALT" text.

What is the issue about?
The tag does not have an ALT attribute defined. As a general rule, search engines do not interpret the content of image files. The text provided in the attribute enables the site owner to provide relevant information to the search engine and to the end user. Alt text is helpful to end users if they have images disabled or if the image does not properly load. In addition, the Alt text is utilized by screen readers. Make sure that your Alt text is descriptive and accurately reflects what the image represents and supports the content on the page.

How to fix?
Use the <img alt> attribute to write descriptive content for the image: <img source='pic.gif' alt='Accurate and descriptive keyword text that represents the image.' />.

Website Speed Test (Desktop):
0.02 seconds

Website speed is a measurement of how fast the content on your page loads. Website speed is one of many factors involved in the discipline of search engine optimization (SEO), but it is not the only one. In a recent study, the average load time for a web page was 3.21s.

Top Organic Search Terms:
Term Search Volume Traffic Traffic (%)
stemjock 210 2 0.01%
stem jock 70 0 0%
solutions to stem textbooks 30 0 0%

CO-Hosted

CoHosted refers to a situation where multiple domain names (websites) are using the same IP address to point to their respective web servers. They could be owned by different individuals or organizations and may serve entirely different purposes.

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rating 5

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