Welcome to IRSC’s LIVE Virtual Lesson on:
... • Why? To email to your instructor as proof of attendance. To get 1 hour of credit towards your 10 hours this week. • How? Place your cursor and left click your mouse on the participant window. On your keyboard, hold down the SHIFT and PRINT SCREEN keys. Then open a Word document and paste (Ctrl + V ...
... • Why? To email to your instructor as proof of attendance. To get 1 hour of credit towards your 10 hours this week. • How? Place your cursor and left click your mouse on the participant window. On your keyboard, hold down the SHIFT and PRINT SCREEN keys. Then open a Word document and paste (Ctrl + V ...
General School Presentation
... Adding Fractions If we were adding apples, 1 apple + 1 apple = 2 apples ...
... Adding Fractions If we were adding apples, 1 apple + 1 apple = 2 apples ...
Simple Block Code Parity Checks
... Multiplicative inverses may not exist for some numbers. Example: 2 × 5 ≡ 0 mod 10. Does 2 have a multiplicative inverse? Suppose it does, then 2 × 2−1 ≡ 1 mod 10. However, multiplying both sides by 5 yields 0 ≡ 5 mod 10, which is false. Note: If the modulus is a prime, p, then numbers not congruent ...
... Multiplicative inverses may not exist for some numbers. Example: 2 × 5 ≡ 0 mod 10. Does 2 have a multiplicative inverse? Suppose it does, then 2 × 2−1 ≡ 1 mod 10. However, multiplying both sides by 5 yields 0 ≡ 5 mod 10, which is false. Note: If the modulus is a prime, p, then numbers not congruent ...
7TH GRADE MATH MID YEAR STUDY GUIDE
... Rules of Fractions Ordering – change all fractions to decimals by dividing the top (numerator) by the denominator (bottom) and then multiply by 100 to see as a percent Adding and subtracting must have a common denominator which is the same as the least common multiple a number both denominators will ...
... Rules of Fractions Ordering – change all fractions to decimals by dividing the top (numerator) by the denominator (bottom) and then multiply by 100 to see as a percent Adding and subtracting must have a common denominator which is the same as the least common multiple a number both denominators will ...
Solutions
... If x leaves a remainder of 1 when divided by 2, 3 and 5, then x − 1 is divisible by 2, 3 and 5. Since 2, 3 and 5 have no common factors, x − 1 must be divisible by 2 × 3 × 5 = 30. So x is 1 more than a multiple of 30, i.e., x = 30k + 1 for some integer k. If x = 30k + 1, then x = (2 × 15k) + 1 = (3 ...
... If x leaves a remainder of 1 when divided by 2, 3 and 5, then x − 1 is divisible by 2, 3 and 5. Since 2, 3 and 5 have no common factors, x − 1 must be divisible by 2 × 3 × 5 = 30. So x is 1 more than a multiple of 30, i.e., x = 30k + 1 for some integer k. If x = 30k + 1, then x = (2 × 15k) + 1 = (3 ...
Revised Version 070511
... correspondence between the equivalence classes and the real numbers. Thus, the real numbers give us all possible slopes, except for the vertical line. When x = 0 , all the points in the equivalence class lie on the vertical line that is the y-axis. (Again the origin must be excluded from this equiva ...
... correspondence between the equivalence classes and the real numbers. Thus, the real numbers give us all possible slopes, except for the vertical line. When x = 0 , all the points in the equivalence class lie on the vertical line that is the y-axis. (Again the origin must be excluded from this equiva ...
