Lecture notes
... be able to perform calculations and prove results. . . . With these in mind, my goals in my half of this lecture course will be roughly as follows: • to introduce some aspects (and methods) of pure mathematics so as to provoke interest and enthusiasm for future courses; • to introduce the concept of ...
... be able to perform calculations and prove results. . . . With these in mind, my goals in my half of this lecture course will be roughly as follows: • to introduce some aspects (and methods) of pure mathematics so as to provoke interest and enthusiasm for future courses; • to introduce the concept of ...
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... calculate since it is equal to the current gain (b) multiplied by the resistance gain (E = bR). And, the power gain is equal to the voltage gain multiplied by the current gain b (P = bE). ...
... calculate since it is equal to the current gain (b) multiplied by the resistance gain (E = bR). And, the power gain is equal to the voltage gain multiplied by the current gain b (P = bE). ...
Basic Algebra Review
... i. Solve one equation for either variable. ii. Substitute that variable into the other equation. iii. Solve the equation from step ii. iv. Substitute the result from step iii into the equation from step i to find the remaining value. ...
... i. Solve one equation for either variable. ii. Substitute that variable into the other equation. iii. Solve the equation from step ii. iv. Substitute the result from step iii into the equation from step i to find the remaining value. ...
2.3 Solving Quadratic Equations
... Write the left side as the square of a binomial: (x − 3)2 = 36 Take the square root of both sides: x − 3 = ±6 Solve for x: x − 3 = 6 or x − 3 = −6 x = 9 or x = −3 The roots are 9 and −3. The solution to Example 1 can be modelled graphically. The equation has two distinct real roots. The graph of the ...
... Write the left side as the square of a binomial: (x − 3)2 = 36 Take the square root of both sides: x − 3 = ±6 Solve for x: x − 3 = 6 or x − 3 = −6 x = 9 or x = −3 The roots are 9 and −3. The solution to Example 1 can be modelled graphically. The equation has two distinct real roots. The graph of the ...