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Презентация была опубликована 2 года назад пользователемНаталия Милорадович

2 Here are multiplication tables written in a code. The tables are not in the correct order. Find the digit, represented by each letter.

3 a x f = jc b x f = he g x f =f h x f = ai c x f = ge f x f = ea j x f = ca d x f = bc e x f = af

4 i x g = bf f x g = de a x g = ec d x g = ei c x g = g h x g = fa g x g = dh e x g = cd b x g = ab

5 s x q = lp p x q = no k x q = om n x q = os m x q = q r x q = pk q x q = nr o x q = mn l x q = kl

6 You can order fractions by converting them to decimals first, by dividing the numerator by the denominator. For example, 3/8 = and 6/13 =0.462, so 6/13 is greater than 3/8. You can also order fractions by converting them to equivalent fractions with a common denominator. When you multiply or divide both sides of an inequality by a negative number, you need to reverse the inequality sign for the inequality to remain true. For example, 2.5 > but – 2.5 < 3.9 You can use inequality signs to describe intervals.

7 To add or subtract fractions with different denominators, find equivalent fractions with a common denominator. Then add or subtract them. Answers that are improper fractions should be converted to mixed numbers. To add mixed numbers, add the whole-number parts and then add the fractional parts. You can use the same method for subtraction in the fraction part of the second fraction is smaller than the fractional part of the first. To subtract mixed numbers where the fractional part of the second fraction is larger than the fractional part of the first, convert them both into improper fractions before subtracting. To add or subtract more complex fractions, you may need to identify the multiples of each denominator to find the lowest common denominator.

8 An expression has no equals sign. It is a way of expressing something general. Equations are true for particular values. For example, 3x + 7 = 13 is only true for x = 2. An identity shows the same thing in two different ways. An identity is true for all values of the variable. To expand a bracket, multiply every term inside the bracket by the term on the outside. For example, 4(3x - 2) = 4 x 3x + 4 x (-2) = 12x – 8.

9 You can construct an equation for a problem from the facts that you are given. Some facts are given explicitly, others implicitly. For example, find each angle of a triangle, which has angles (2x + 13), (3x + 4) and (4x + 19). The angles are given explicitly. The fact that angles add up to 180 degrees is given implicitly, by saying that they are the angles of a triangle. You can use the facts to form an equation.

10 2x x x + 19 = 180. Now simplify the equation and solve it to find the unknown. 9x + 36 = 180 Subtract 36 from both sides 9x = 144 Divide both sides by 9: x = 16. Use the solution to the equation to solve the original problem. When x = 16: 2x + 13 = 2 x = 45, 3x + 4 = 3 x = 52, 4x + 19 = 4 x = 83. So, the angles are 45, 52 and 83.

11 When you have an equation with x on both sides, you need to add or subtract the same number of xs from both sides in order to get x on one side only.

12 When you have an equation with x on both sides, you need to add or subtract the same number of xs from both sides in order to get x on one side only. For example, x – 6 = 29 – 4x. The left-hand side has higher number of xs. Add 4x to both sides to eliminate the xs on the right-hand side. 5x – 6 = 29, 5x = 35, x = 7. Equivalent equations have the same solution for x. For example, subtracting x from both sides of 3x + 10 = x + 4 gives the equivalent equation 2x + 10 = 4. The solution to both equations is x = - 3. When an equation includes brackets, first expand the brackets and collect like terms.

13 Trial and improvement involves making an initial guess at a solution to an equation, trying it in the equation to find out whether the guess is too big or too small, and then improving the guess as required. You may need to form an equation, and then solve it using trial and improvement.

14 For example, the length of a rectangle is 4cm more than its width. Its area is 70. Call the width w, so the length is w + 4. This gives the equation w (w + 4) = 70. To solve this, try different values of w on the left-hand side. Record in a table whether each area is too high or too low. When you have the solution sandwiched between two values of the required level of accuracy, you need to do the halfway test. In the example w(w + 4) = 70, w = 6.6 is too low and w = 6.7 is too high. Next check w = Here w(w + 4) = 6.65 x = This is too high, so the correct value of w is bellow 6.65 and the answer to one decimal place is 6.6cm.

15 You can construct a table of values to plot a graph. Simultaneous equations are true at the same time, that is for the same values of x and y. You need two equations if there are two unknowns, three equations if there are three unknowns and so on. For two simultaneous equations, the point on a graph where the two lines cross is their solution.

16 There is no solution to the simultaneous equations of parallel lines because they do not cross. If two simultaneous equations are equivalent equations, there is an infinite number of solutions to them because the lines coincide at every point. For example, 2y = 6x + 12 and y – 3x = 6 are both equivalent to y = 3x + 6.

17 If you multiply every term on both sides of an equation by the same number you get an equivalent equation. You can add or subtract one equation to (or from) another, because what you are adding (or subtracting) is the same on both sides. Simultaneous equations are true at the same time. Each equation has two unknowns, such as x and y. The solution is the value of x and y that satisfy both equations.

18 To solve simultaneous equations: If necessary, multiply one or both equations so that they have either the same number (ignoring the sign in front) of xs or the same number of ys. Add or subtract the equations to eliminate one of the variables. Then solve the resulting equation to find the other variable. Use substitution to find the remaining variable.

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HPC Pipelining Parallelism is achieved by starting to execute one instruction before the previous one is finished. The simplest kind overlaps the execution.

HPC Pipelining Parallelism is achieved by starting to execute one instruction before the previous one is finished. The simplest kind overlaps the execution.

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