answer:Harry starts with 79 apples and buys 5 more. To find out how many apples he ends with, we add the 5 apples to the 79 he already has: 79 + 5 = 84 Harry ends with boxed{84} apples.

answer:To solve this problem, we need to understand that the graph of a quadratic function y = ax^2 + bx + c intersects the x-axis at points where y=0. For the given function y = x^2 - 2x + m, this means we are looking for the values of x that satisfy the equation x^2 - 2x + m = 0. Given that the graph intersects the x-axis at only one point, this implies that the quadratic equation has a single, repeated root. This condition is met when the discriminant Delta of the quadratic equation, given by Delta = b^2 - 4ac, equals zero. In our case, a=1, b=-2, and c=m. Substituting these values into the formula for the discriminant gives us: [ Delta = (-2)^2 - 4 cdot 1 cdot m = 4 - 4m = 0 ] Solving this equation for m involves isolating m on one side: [ 4 - 4m = 0 implies 4 = 4m implies m = 1 ] Therefore, the value of m that satisfies the condition is m = 1. So, the correct answer is boxed{C}.

answer:The original statement to examine is: "For a positive integer n, at least one of 6n - 1 and 6n + 1 is prime." We need to identify a value of n which serves as a counterexample to this statement, meaning we need to find an n such that both 6n - 1 and 6n + 1 are not prime. Let's evaluate the candidates: **Candidate: n = 10** 1. Calculate 6n - 1: [ 6 times 10 - 1 = 60 - 1 = 59 quad (text{Check if 59 is prime}) ] Since 59 is a prime number, n = 10 is not a counterexample. **Candidate: n = 19** 1. Calculate 6n - 1: [ 6 times 19 - 1 = 114 - 1 = 113 quad (text{Check if 113 is prime}) ] Since 113 is a prime number, n = 19 is not a counterexample. **Candidate: n = 20** 1. Calculate 6n - 1: [ 6 times 20 - 1 = 120 - 1 = 119 quad (text{Check if 119 is prime}) ] 119 is not a prime number (it is divisible by 7 and 17). 2. Calculate 6n + 1: [ 6 times 20 + 1 = 120 + 1 = 121 quad (text{Check if 121 is prime}) ] 121 is not a prime number (it is 11^2). Since both 119 and 121 are not prime, n = 20 is indeed a counterexample. **Candidate: n = 21** 1. Calculate 6n - 1: [ 6 times 21 - 1 = 126 - 1 = 125 quad (text{Check if 125 is prime}) ] 125 is not a prime number (it is 5^3). 2. Calculate 6n + 1: [ 6 times 21 + 1 = 126 + 1 = 127 quad (text{Check if 127 is prime}) ] Since 127 is a prime number, n = 21 is not a counterexample. **Candidate: n = 30** 1. Calculate 6n - 1: [ 6 times 30 - 1 = 180 - 1 = 179 quad (text{Check if 179 is prime}) ] Since 179 is a prime number, n = 30 is not a counterexample. Therefore, among the given choices, the value that acts as a counterexample is: [ boxed{text{C}} ]

answer:To find the degree of each exterior angle of a regular octagon, we follow these steps: 1. We know that the sum of the exterior angles of any polygon is 360^{circ}. 2. Since an octagon has 8 sides, it will have 8 exterior angles if it is regular. 3. To find the measure of each exterior angle, we divide the total sum of the exterior angles by the number of sides (or angles). Therefore, the calculation is as follows: [ text{Degree of each exterior angle} = frac{360^{circ}}{8} = 45^{circ} ] Hence, the degree of each exterior angle of a regular octagon is boxed{45^{circ}}.