1.0<x<1. Which must be true?
I. x^9 < x^7
II.x^7+x^5 < x^6+x^4
III. x^7-x^5 < x^6-x^4
2. A football team average x points (the mean) in the first n games, scored y points in the next game, and z points in the one after that . The mean of the n+2 is
A.(x+y+z) / (n+2)
B. (nxyz) / (n+2)
C. (nx+y+z) / (n+2)
D.x + (y+x)/2
E.(nx+yz)/n+2
3.The postage in country X is 50 gruff for the first ounce and 22 gruff for each additional ounce. A half-pound letter mailed in country X cost how many gruff?
4.Two sides of a triangle are 4 and 7. If the sides are integers, the product of the smallest possible perimeter and largest possible perimeter is
5.J, K, M are three consecutive odd integers, with j < k < m. Which is true?
l : m - j =4
II: jk +jm+ km + 1 is even
III:(j + k + m) / 3 is an integer
6. The sum of the three smallest primes than 30 is
A. 99
B. 109
C. 111
D. 115
E. 131
1. To determine which inequalities are true, we need to compare the exponents of x in each expression.
I. x^9 < x^7
The exponent is higher in the denominator, meaning that the fraction x^9 / x^7 can be simplified to 1 / x^2. Since x is between 1 and 1.0, x^2 will be between 1 and 1.0. Therefore, x^9 will be less than x^7.
II. x^7 + x^5 < x^6 + x^4
Again, we can simplify the equation using exponents. x can be factored out, giving us x^5(x^2 + 1) < x^4(x^2 + 1). Since x is between 1 and 1.0, x^2 + 1 will be positive. Therefore, we can cancel out x^2 + 1 and x^4 from both sides of the equation, leaving us with x^5 < x^4. This inequality is true.
III. x^7 - x^5 < x^6 - x^4
Similar to the previous inequality, we can factor out x. This gives us x^5(x^2 - 1) < x^4(x^2 - 1). Since x is between 1 and 1.0, x^2 - 1 will be negative. This means that we need to reverse the inequality when we cancel out x^2 - 1. Doing so, we get x^5 > x^4. This inequality is false.
Therefore, the correct options are I and II.
2. The mean (average) of the n+2 games can be calculated by taking the sum of all the points and dividing it by the total number of games:
Mean of n games = (x * n) + y + z
Mean of n+2 games = (x * n) + y + z + 2
The correct option is A, (x+y+z) / (n+2).
3. In country X, the first ounce costs 50 gruff, and each additional ounce costs 22 gruff. Since a half-pound is equal to 8 ounces, we can calculate the cost as follows:
Cost = 50 gruff + 22 gruff * (8 - 1)
Simplifying the equation, we get:
Cost = 50 gruff + 22 gruff * 7
= 50 gruff + 154 gruff
= 204 gruff
Therefore, a half-pound letter mailed in country X would cost 204 gruff.
4. Let's consider the triangle's side lengths. We have two sides measuring 4 and 7. To find the possible range of the third side, we apply the triangle inequality theorem:
The sum of any two sides of a triangle must be greater than the length of the remaining side.
By applying this theorem: 4 + 7 > third side length and 4 + third side length > 7.
Combining these inequalities, we get 11 > third side length > 3.
The smallest possible perimeter would be when the third side length is 4, resulting in a perimeter of 4 + 4 + 7 = 15.
The largest possible perimeter would be when the third side length is 10, resulting in a perimeter of 4 + 7 + 10 = 21.
Therefore, the product of the smallest possible perimeter (15) and largest possible perimeter (21) is 15 * 21 = 315.
5. Let's consider the given information.
- J, K, M are three consecutive odd integers: J, K, M.
- j < k < m.
To determine which statements are true, we can analyze each option:
I. m - j = 4:
Since consecutive odd integers have a difference of 2, the difference between the largest and smallest integers cannot be 4. Therefore, this statement is false.
II. jk + jm + km + 1 is even:
Since j, k, and m are consecutive odd integers, their sum will always be even. Adding 1 to an even number will make it odd. Therefore, the statement is false.
III. (j + k + m) / 3 is an integer:
Since j, k, and m are consecutive odd integers, their sum will always be divisible by 3. Thus, the statement is true.
Therefore, statement III is the only true statement.
6. The three smallest primes less than 30 are 2, 3, and 5. To find the sum, we add them together:
2 + 3 + 5 = 10.
The sum of the three smallest primes less than 30 is 10, so the correct option is not given.