Please help, even if you only know 1 answer.
1. The length of each side of rectangle R is an integer, and the area of R is 2009. What is the largest possible perimeter of R?
2. If i place all three operational symbols +, -, X in all possible way into the blanks if the expression 5_4_6_3, only one symbol blank, each resulting expression will have a unique value. What is the largest of these values?
3. How many integers n > 0 satisfy 3/65 < 1/n < 9/100
Thanks
Sure, I'd be happy to help! Here are the answers to your questions:
1. To find the largest possible perimeter of rectangle R, we need to first determine the dimensions of the rectangle. We know that the area of R is 2009. Since the length of each side of R is an integer, we can start by finding the factors of 2009.
The factors of 2009 are:
1, 7, 287, 2009
We can see that 287 is closest to being a square, so we can use that as the length of one side of the rectangle. The other side will be the quotient of 2009 divided by 287, which is approximately 6.99. Since the length of each side must be an integer, the other side of the rectangle will also be approximately 7.
Therefore, the dimensions of rectangle R are 287 x 7. The perimeter is calculated by adding all the sides, so P = 2(287 + 7) = 2(294) = 588.
Therefore, the largest possible perimeter of R is 588.
2. To find the largest value possible for the expression 5 _ 4 _ 6 _ 3, where only one symbol is missing, we need to consider all the possible combinations.
The possible combinations are:
5 + 4 - 6 x 3 = -7
5 - 4 + 6 x 3 = 23
5 x 4 + 6 - 3 = 23
5 + 4 x 6 - 3 = 23
5 - 4 - 6 + 3 = -2
5 - 4 - 6 x 3 = -27
5 x 4 - 6 + 3 = 17
The largest value among these combinations is 23. Therefore, the largest possible value for the expression is 23.
3. To find the number of integers n > 0 that satisfy the inequality 3/65 < 1/n < 9/100, we need to determine the range of values for n that satisfy both inequalities.
The first inequality implies that n > 65/3, which means n must be greater than approximately 21.67.
The second inequality implies that n < 100/9, which means n must be less than approximately 11.11.
Therefore, the values of n that satisfy both inequalities are 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99.
Therefore, there are 78 integers n > 0 that satisfy the given inequality.
I hope this helps! Let me know if you have any further questions.