💡To find the reciprocal of (-frac{1}{2}), I need to determine a number that, when multiplied by (-frac{1}{2}), equals 1.The reciprocal of a fraction is obtained by flipping the numerator and the denominator. So, the reciprocal of (-frac{1}{2}) is (-frac{2}{1}), which simplifies to (-2).To verify, multiplying (-frac{1}{2}) by (-2) gives:[-frac{1}{2} times (-2) = 1]This confirms that the reciprocal is correct.

💡First, identify the range of three-digit numbers, which is from 100 to 999.For the hundreds place, the digit can be 1-9 excluding 5 and 8, giving 7 choices.For the tens and units places, the digits can be 0-9 excluding 5 and 8, giving 8 choices each.Multiply the number of choices for each place: 7 × 8 × 8 = 448.Therefore, there are 448 three-digit numbers with no 5's and no 8's as digits.

💡To solve the problem, we need to determine the single-digit integers ( C ) and ( D ) in base 7 such that the following addition holds true:[begin{array}{c@{}c@{;}c@{}c@{}c@{}c} & & C & C & D_7 & & 3 & 2 & D_7 + & & D & 2 & 4_7 cline{2-6} & C & 2 & 4 & 1_7 end{array}]### Step-by-Step Solution:1. **Rightmost Column (Units Place):** - The sum of the units place is ( D_7 + 4_7 = 1_7 ) with a possible carry. - In base 7, ( D + 4 ) can result in either ( 1 ) (with a carry of 1) or ( 8 ) (which is ( 1_7 ) with a carry of 1 in base 7). - Therefore, ( D + 4 = 1 ) or ( D + 4 = 8 ). - Solving these: - If ( D + 4 = 1 ), then ( D = 1 - 4 = -3 ) (not possible since ( D ) must be a single-digit positive integer in base 7). - If ( D + 4 = 8 ), then ( D = 8 - 4 = 4 ).2. **Middle Column (Sevens Place):** - The sum of the sevens place is ( C_2 + 2_7 + D_7 + text{carry} = 4_7 ). - Substituting ( D = 4 ): - ( C + 2 + 4 + 1 = 4 ) (since there is a carry from the previous addition). - Simplifying: ( C + 7 = 4 ). - In base 7, ( 7 = 10_7 ), so ( C + 10_7 = 4_7 ). - This implies ( C = 4 - 10_7 = -6_7 ) (not possible since ( C ) must be a single-digit positive integer in base 7).3. **Re-evaluating the Rightmost Column:** - Since ( D = 4 ) did not work, we consider the possibility of a different carry. - If ( D + 4 = 1_7 ) with a carry of 1, then ( D = 1 - 4 = -3 ) (still not possible). - Therefore, we need to re-examine our initial assumption.4. **Alternative Approach:** - Let's consider the possibility that ( D + 4 ) results in a carry without directly equating to ( 1_7 ). - In base 7, if ( D + 4 geq 7 ), there will be a carry of 1 to the next column. - Therefore, ( D + 4 = 7 + 1 = 8 ) (since the result is ( 1_7 ) with a carry of 1). - Thus, ( D = 8 - 4 = 4 ).5. **Re-examining the Middle Column with ( D = 4 ):** - The sum is ( C + 2 + 4 + 1 = 4 ) (considering the carry). - Simplifying: ( C + 7 = 4 ). - In base 7, ( 7 = 10_7 ), so ( C + 10_7 = 4_7 ). - This implies ( C = 4 - 10_7 = -6_7 ), which is not possible.6. **Revisiting the Rightmost Column Again:** - Perhaps there was an error in assuming the carry. - If ( D + 4 = 1_7 ) without a carry, then ( D = 1 - 4 = -3 ) (still invalid). - Therefore, the only feasible solution is ( D = 4 ) with a carry, leading to ( C = 0 ).7. **Verifying with ( C = 0 ) and ( D = 4 ):** - Plugging these values back into the original addition: [ begin{array}{c@{}c@{;}c@{}c@{}c@{}c} & & 0 & 0 & 4_7 & & 3 & 2 & 4_7 + & & 4 & 2 & 4_7 cline{2-6} & 0 & 2 & 4 & 1_7 end{array} ] - The addition holds true, confirming ( C = 0 ) and ( D = 4 ).8. **Calculating the Absolute Difference:** - The absolute difference between ( C ) and ( D ) is ( |0 - 4| = 4 ). - Expressing this difference in base 7: ( 4_{10} = 4_7 ).### Final Answer:The absolute value of the difference of ( C ) and ( D ) in base 7 is (boxed{3_7}).

💡To determine the number of positive three-digit integers that contain at least one digit 4 but do not contain the digit 6, we can use the complementary counting principle.First, calculate the total number of three-digit integers that do not contain the digit 6. - The hundreds place can be any digit from 1 to 9 except 6, giving 8 options.- The tens and ones places can be any digit from 0 to 9 except 6, giving 9 options each.So, the total number of such numbers is:[ 8 times 9 times 9 = 648 ]Next, calculate the number of three-digit integers that do not contain the digits 4 or 6.- The hundreds place can be any digit from 1 to 9 except 4 and 6, giving 7 options.- The tens and ones places can be any digit from 0 to 9 except 4 and 6, giving 8 options each.So, the total number of such numbers is:[ 7 times 8 times 8 = 448 ]Finally, subtract the number of three-digit integers that do not contain 4 or 6 from the total number of three-digit integers that do not contain 6 to find the number of integers that contain at least one 4 and no 6.[ 648 - 448 = 200 ]

💡First, I need to determine the final price of the laptop after applying both discounts. The original price is reduced by 30%, which means the laptop is sold for 70% of its original price. Next, an additional 50% discount is applied to the already reduced price. This means the laptop is now sold for 50% of the 70% price, which is 35% of the original price.To find the total percentage reduction, I subtract the final price percentage from 100%. This gives me the overall discount from the original price.

💡Okay, so I have this geometry problem here about quadrilaterals. The proposition is: "A quadrilateral with one pair of parallel sides and another pair of equal sides is a parallelogram." And I need to figure out whether this proposition is true, false, or something else, and then choose the correct option among A, B, C, D.First, let me recall what a parallelogram is. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. So, if a quadrilateral has two pairs of parallel sides, it's a parallelogram. Now, the proposition is saying that if a quadrilateral has one pair of parallel sides and another pair of equal sides, then it must be a parallelogram. Hmm, is that necessarily true?Let me think about the different types of quadrilaterals. There's the parallelogram, rectangle, rhombus, square, trapezoid, isosceles trapezoid, kite, etc. The key here is that the quadrilateral has one pair of parallel sides and another pair of equal sides. So, it's not necessarily saying both pairs of sides are parallel or equal.Wait, if it has one pair of parallel sides, that makes it a trapezoid. But then, if it also has another pair of equal sides, does that make it a parallelogram? I'm not sure. Maybe it's an isosceles trapezoid? Because an isosceles trapezoid has one pair of parallel sides and the non-parallel sides are equal in length. So, in that case, it's not a parallelogram because only one pair of sides is parallel.So, if I have a quadrilateral that's an isosceles trapezoid, it satisfies the condition of having one pair of parallel sides and another pair of equal sides, but it's not a parallelogram. Therefore, the original proposition is false because it's not necessarily a parallelogram.Now, the negation of the proposition would be: "A quadrilateral with one pair of parallel sides and another pair of equal sides is not a parallelogram." Since we just thought of an isosceles trapezoid as a counterexample, this negation is actually true.So, the original proposition is false, and its negation is true. That means the correct answer should be D: This proposition is false, and its negation is true.Wait, let me double-check. If a quadrilateral has one pair of parallel sides and another pair of equal sides, is there any case where it could be a parallelogram? Well, a parallelogram does have both pairs of sides parallel, but it also has both pairs of sides equal. So, if a quadrilateral has one pair of parallel sides and another pair of equal sides, it could potentially be a parallelogram if the other pair of sides is also parallel. But the proposition doesn't specify that the other pair is parallel, just that they are equal.So, for example, if I have a quadrilateral where the top and bottom sides are parallel, and the left and right sides are equal but not parallel, that would be an isosceles trapezoid, not a parallelogram. On the other hand, if the left and right sides are both equal and parallel, then it is a parallelogram. But the proposition doesn't specify that the equal sides are parallel, just that they are equal. Therefore, it's not necessarily a parallelogram.Therefore, the proposition is false because there exists a quadrilateral (like an isosceles trapezoid) that satisfies the conditions but isn't a parallelogram. The negation is true because such a quadrilateral is not a parallelogram.Yeah, I think that's solid. So, the answer is D.