Computer Simulation Lab
... Array Subscripts • r = rand(1,7) This gives you a row vector of seven random numbers. • r(3) This will display the third element of r. The number 3 is the subscript. • r(2:4) This should give you the second, third and fourth elements. • r(1:2:7) • r([1 7 2 6]) • r([1 7 2]) = [ ] will remove element ...
... Array Subscripts • r = rand(1,7) This gives you a row vector of seven random numbers. • r(3) This will display the third element of r. The number 3 is the subscript. • r(2:4) This should give you the second, third and fourth elements. • r(1:2:7) • r([1 7 2 6]) • r([1 7 2]) = [ ] will remove element ...
Divisibility tests be shared by .
... Divisibility tests These are simple tricks to test what a number can be shared by . We are going to learn tricks for testing if a number can be shared by 2, 3, 4, 5, 6, 8, 9, and multiples of 10. ...
... Divisibility tests These are simple tricks to test what a number can be shared by . We are going to learn tricks for testing if a number can be shared by 2, 3, 4, 5, 6, 8, 9, and multiples of 10. ...
Inductive Reasoning and Conjecture A conjecture is an educated
... Conjecture: The next number will increase by 6. So, it will be 15 + 6 or 21. Example 2 Geometric Conjecture For points P, Q, and R, PQ = 9, QR = 15, and PR = 12. Make a conjecture and draw a figure to illustrate your conjecture. Given: points P, Q, and R; PQ = 9, QR = 15, and PR = 12 Examine the mea ...
... Conjecture: The next number will increase by 6. So, it will be 15 + 6 or 21. Example 2 Geometric Conjecture For points P, Q, and R, PQ = 9, QR = 15, and PR = 12. Make a conjecture and draw a figure to illustrate your conjecture. Given: points P, Q, and R; PQ = 9, QR = 15, and PR = 12 Examine the mea ...
Section 4.3 - The Chinese Remainder Theorem
... One comes up with x ≡ 57 (mod 72). Thus since 12 divides 72, we must also have x ≡ 57 (mod 12). But 57 6≡ 2 (mod 12) thus there can be no solutions to this system of congruences. ...
... One comes up with x ≡ 57 (mod 72). Thus since 12 divides 72, we must also have x ≡ 57 (mod 12). But 57 6≡ 2 (mod 12) thus there can be no solutions to this system of congruences. ...
