2017 - Problems and Solutions
... divisor of 30 because 30 ÷ 5 = 6; but 5 is not a divisor of 47 because 47 ÷ 5 = 9 with remainder 2. In this problem we consider only positive integer numbers n and positive integer divisors of n. Thus, for example, if we multiply all divisors of 6 we will obtain 36.) D/2 The area of square ABCD is 1 ...
... divisor of 30 because 30 ÷ 5 = 6; but 5 is not a divisor of 47 because 47 ÷ 5 = 9 with remainder 2. In this problem we consider only positive integer numbers n and positive integer divisors of n. Thus, for example, if we multiply all divisors of 6 we will obtain 36.) D/2 The area of square ABCD is 1 ...
Converting mixed numbers and improper fractions
... 3. Add fractions with common denominators: Add the numerator, denominator stays the same. 4. Add fractions with different denominators: Find a common denominator by multiplying them together. Use a “magic 1” to find an equivalent fraction for each fraction in the problem that uses the new common den ...
... 3. Add fractions with common denominators: Add the numerator, denominator stays the same. 4. Add fractions with different denominators: Find a common denominator by multiplying them together. Use a “magic 1” to find an equivalent fraction for each fraction in the problem that uses the new common den ...
X and div by powers of 10
... As we move down the list -- as we push the digits one place left -- the number has been multiplied by 10, because each next place is worth 10 times more. (As we move from 2.658 to 26.58, we go from 2 ones to 2 tens.) It appears, though, as if the decimal point has shifted one place right, or, with w ...
... As we move down the list -- as we push the digits one place left -- the number has been multiplied by 10, because each next place is worth 10 times more. (As we move from 2.658 to 26.58, we go from 2 ones to 2 tens.) It appears, though, as if the decimal point has shifted one place right, or, with w ...
Chapter 2
... Zeros after a decimal point are significant Zeros between any other digit are significant Initial zeros are not significant Zeros at the end of a whole number may or may not be significant Depends on if you place a decimal after the zero ...
... Zeros after a decimal point are significant Zeros between any other digit are significant Initial zeros are not significant Zeros at the end of a whole number may or may not be significant Depends on if you place a decimal after the zero ...
Exponential and Logarithmic Functions
... ● Populations → Animals making more animals. ● Investment → Money making more money ● Kinematics → Motion making more motion ● Radiology → Radioactive decay is negative ...
... ● Populations → Animals making more animals. ● Investment → Money making more money ● Kinematics → Motion making more motion ● Radiology → Radioactive decay is negative ...
Curriculum Map
... 1. Make sense of problems and persevere in solving them. In fourth grade, students know that doing mathematics involves solving problems and discussing how they solved them. Students explain to themselves the meaning of a problem and look for ways to solve it. Fourth graders may use concrete objects ...
... 1. Make sense of problems and persevere in solving them. In fourth grade, students know that doing mathematics involves solving problems and discussing how they solved them. Students explain to themselves the meaning of a problem and look for ways to solve it. Fourth graders may use concrete objects ...
UNIT-I - WordPress.com
... omega notation represent the lower bound of an algorithm’s running time, i.e. we can give smallest amount of time taken by the algorithm to complete. Let f(n) and g(n) be the two non-negative functions. We say that f(n) is said to be O(g(n)) if and only if there exists a positive constant ‘c’ and ...
... omega notation represent the lower bound of an algorithm’s running time, i.e. we can give smallest amount of time taken by the algorithm to complete. Let f(n) and g(n) be the two non-negative functions. We say that f(n) is said to be O(g(n)) if and only if there exists a positive constant ‘c’ and ...
