IGCSE /Grade 9
Maths MCQ Based On Find constant rate of change
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IGCSE Grade 9 Maths Find constant rate of change
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Multiply the rate of production by the number of hours to find the total number of toys produced.
Taking two pairs from the table: (2,1) and (4,2) \\(Rate of change=\\frac{change in y}{change in x}=\\frac{2-1}{4-2}=\\frac{1}{2}=0.5\\) 0.5 ml per minute. Amount of water leaked out after 8 minutes = 0.5×8=4ml.
Two points on the line are (2004, 7) and (2007, 10). \\(Rate of change=\\frac{change in y}{change in x}=\\frac{10-7}{2007-2004}=\\frac{3}{3}=1\\) Therefore the constant rate of change is 1 game per year.
Two points on the line are (0 , 50 ) and (8, 300) \\(Slope=\\frac{change in y}{change in x}=\\frac{300-50}{8-0}=\\frac{250}{8}=31.25kg\\). The rate of change is 31.25 kg per months.
Two points on the line are (0 , 5) and (12,50) \\(Rate of change=\\frac{change in y}{change in x}=\\frac{50-5}{12-0}=\\frac{45}{12}=3.75\\) 3.75 points per minutes.
1) From the graph it is clear that the y-coordinate corresponding to x=1 is 20 and the y-coordinate corresponding to x=3 is 60. So, the amount spent by Sam while purchasing 1 and 3 chocolates is 20 pounds and 60 pounds respectively. 2) Find any two points on the line. Two points on the line are (1,20) and (2,40). Put those points into the gradient formula. \\(gradient=\\frac{change in y}{change in x}=\\frac{40-20}{2-1}=\\frac{20}{1}=20\\) Gradient = 20. The rate of change is 20 pounds per chocolate.
\\(Rate of change=\\frac{change in y}{change in x}=\\frac{4-2}{20-10}=\\frac{2}{10}=0.2\\)
To find the temperature after 10 hours, multiply the rate of change by the number of hours and subtract it from the initial temperature.
The line is going down constantly represents that rate of constant change.
To find the population after a certain number of years, multiply the current population by 1 plus the rate of change as a decimal.
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