a. 6.5 and 7.8

b. 2.3 and 3.2

c. 1/7 and 2/7

d. root 3/2 and root 5/2

a. 6.5 and 7.8

b. 2.3 and 3.2

c. 1/7 and 2/7

d. root 3/2 and root 5/2

a) First, the rational numbers. Average of 6.5 and 7.8=(6.5+7.8)/2=7.15. Now take two more averages: (6.5+7.15)/2=6.825 and (7.15+7.8)/2=7.475. So 3 rational numbers could be 6.825, 7.15 and 7.475.

Now the irrationals. Square both numbers: 42.25 and 60.84. If we take any three numbers between these limits (excluding perfect squares) and then take the square root of the numbers we will get 3 irrational numbers. Lets’s pick 43, 51 and 59. The square roots are √43, √51 and √59. These all lie between 6.5 and 7.8.

b) Using the techniques from (a), we have rationals: 2.75, 2.525, 2.975. And irrationals √6, √10, √7.

c) Rationals: 3/14, 5/28, 1/4; irrationals: (√3)/10, (√5)/10, (√7)/10. (1/7)²=1/49=0.02 approx, (2/7)²=4/49=0.08 approx. The numbers 0.03=3/100, 0.05=5/100, 0.07=7/100 lie between these squares. So the square roots must be n the range 1/7 to 2/7: (√3)/10, (√5)/10, (√7)/10.

d) We need 3 rationals whose square lies between 1.5 and 2.5. Multiply these by 100: 150 and 250. Now it’s easier to find perfect squares between these: 169=13², 196=14², 225=15². Divide each of these candidates by 10 and we get 1.3, 1.4 and 1.5. These are rational examples that can also be expressed as fractions: 13/10, 7/5, 3/2. Right between 3/2 and 5/2 lies 4/2=2, so √2 lies between √(3/2) and √5/2). We need two more irrationals. √(3/2)=√(6/4)=(√6)/2 and √(5/2)=√(10/4)=(√10)/2. Can we pick an integer between 6 and 10? We can’t pick 8 or 9 because we end up with √2 (which we’ve already chosen) and 9 is a perfect square. But √7/2 we can pick. The average of the two square roots is also irrational, so we have ((√6)/2+(√10)/2)/2=((√6)+(√10))/4 as our third irrational.

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