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The sum of two consecutive integers is 41. What are the integers?
Two consecutive integers are numbers in sequence like 4 and 5 or -30 and -29, that are each 1 number apart. Let x = the ?rst consecutive integer. Let x + 1 = the second consecutive integer. Sum is a key word for addition so the equation becomes: (x )+ (x + 1) = 41. Combine like terms on the left side of the equation: 2x + 1 = 41. Subtract 1 from both sides of the equation: 2x + 1 - 1 = 41 - 1. Simplify: 2x = 40. Divide each side of the equation by 2: 2x/2 = 40/2. The variable is now alone: x = 20. Therefore the larger integer is: x + 1 = 21. The two integers are 20 and 21.
7(y + 3) - 2(x + 2) = 14, 4 (y - 2) + 3(x - 3) = 2 Ans: 7(y + 3) - 2 (x+ 2) = 14 --------- (1) 4(y- 2) + 3(x - 3) = 2 ----------(2) From (1) 7y +21 -
Solve -10 cos(3t )= 7 on [-2,5]. Solution Let's first get the inverse cosine portion of this problem taken care of. cos(3 t )= - 7/10 ⇒ 3t = cos -1 ( - 7
Find the values of x and y
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Provide me some Example of in-equations.
The value of y that minimizes the sum of the two distances from (3,5) to (1,y) and from (1,y) to (4,9) can be written as a/b where a and b are coprime positive integers. Find a+b.
1. (‡) Prove asymptotic bounds for the following recursion relations. Tighter bounds will receive more marks. You may use the Master Theorem if it applies. 1. C(n) = 3C(n/2) + n
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