7 + 3b ≥ 12 Subtract 7 from both sides: 3b ≥ 5 Divide both sides by 3: b ≥ 5/3 or 1.66...
It is: 5*(b+2) which is the same as 5b+10
1.Start 2. Input a,b,c 3. Sum = a+b+c 4. Average = sum/3 5. Output - Sum,Average 6. Stop
a + b = 14, so a = 14 - b; 6a - 3b = 3;substitute: 6(14 - b) - 3b = 3, ie 84 - 6b - 3b = 3, ie 81 = 9b so b = 9 and a = 5
We are looking for A B and C if they are consecutive odd integers B=A+2 and C=A+4 We are told 3B=5+A+C If we substitue the first two equations into the third 3(A+2)=5+A+(A+4) so 3A+6=2A+9 A=3 so B=5 and C=7
3(b5)
3a+b
It is the same as: 3*(a+b) or simply 3(a+b)
3*(a + b) or 3a + 3b
3(a + b) + a = 3a + 3b + a = 4a + 3b
3(a+b)+a
7 + 3b ≥ 12 Subtract 7 from both sides: 3b ≥ 5 Divide both sides by 3: b ≥ 5/3 or 1.66...
To find the sum of two vectors, you add their corresponding components together. For example, if you have two vectors A = (3, 5) and B = (2, -1), the sum would be A + B = (3 + 2, 5 + (-1)) = (5, 4).
(b+5)
Is your question 3 * ( 5 - b ) ? If so, 3*(5 - b) = 15 - 3b
It is: 5*(b+2) which is the same as 5b+10
5(a+b)-7