answer:1. **Define the digits of a four-digit number:** Let the digits of the four-digit number be (a), (a+1), (a+2), and (a+3). 2. **Form the four-digit number:** The original number can be expressed as: [ 1000a + 100(a+1) + 10(a+2) + (a+3) ] Simplifying the expression, we get: [ 1000a + 100a + 100 + 10a + 20 + a + 3 ] [ 1111a + 123 ] 3. **Form the number with digits in reverse order:** The reversed number is written as: [ 1000(a+3) + 100(a+2) + 10(a+1) + a ] Simplifying the expression, we get: [ 1000a + 3000 + 100a + 200 + 10a + 10 + a ] [ 1111a + 3210 ] 4. **Find the difference between the original and reversed number:** Subtract the original number from the reversed number: [ (1111a + 3210) - (1111a + 123) ] [ 1111a + 3210 - 1111a - 123 ] [ 3210 - 123 ] [ 3087 ] 5. **Conclusion:** The difference between the four-digit number formed by digits (a), (a+1), (a+2), (a+3) and its reverse is always: [ boxed{3087} ]

answer:First, let's calculate the compound interest on Rs. 4000 for 2 years at 10% per annum. The formula for compound interest is: A = P(1 + r/n)^(nt) Where: A = the future value of the investment/loan, including interest P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate (decimal) n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for In this case, P = 4000, r = 10/100 = 0.10, n = 1 (compounded annually), and t = 2. A = 4000(1 + 0.10/1)^(1*2) A = 4000(1 + 0.10)^2 A = 4000(1.10)^2 A = 4000 * 1.21 A = 4840 The compound interest earned is A - P: CI = 4840 - 4000 CI = 840 Now, we know that the simple interest (SI) on the certain sum of money is half of this compound interest (CI). SI = CI / 2 SI = 840 / 2 SI = 420 The formula for simple interest is: SI = P * r * t Where: SI = simple interest P = principal amount r = annual interest rate (as a decimal) t = time in years We are given that the sum placed on simple interest is Rs. 1750 and the annual interest rate is 8% (or 0.08 as a decimal). 420 = 1750 * 0.08 * t Now, we solve for t: t = 420 / (1750 * 0.08) t = 420 / 140 t = 3 Therefore, the money was placed at simple interest for boxed{3} years.

answer:Given the circle M:(x-sqrt{a})^2+(y-sqrt{a})^2=9, we can deduce several things: 1. The center of circle M is M(sqrt{a}, sqrt{a}), and it has a radius r=3. 2. Points A(-2,0) and B(2,0) are given, and there exists a point P on circle M such that angle APB=frac{pi}{2}. 3. The circle with diameter AB that passes through points A and B intersects circle M. This circle can be described by the equation x^{2}+y^{2}=4, which means its center is O(0,0) and its radius r_{1}=2. 4. To find the range of values for a, we consider the distance |OM| between the centers of the two circles, which is calculated as |OM|=sqrt{(sqrt{a}-0)^{2}+(sqrt{a}-0)^{2}}=sqrt{2a}. 5. For the circles to intersect in such a way that angle APB=frac{pi}{2}, the distance between their centers must satisfy the inequality |3-2|leqslant sqrt{2a}leqslant |3+2|. 6. Solving the inequality for a, we get frac{1}{2}leqslant a leqslant frac{25}{2}. Therefore, the range of values for a is boxed{C: left[frac{1}{2},frac{25}{2}right]}.

answer:There are nine numbers with one digit: 1,2,...,9, so the probability of selecting a one-digit number is frac{9}{20}. There are eleven numbers with two digits: 10,11,...,20, so the probability of selecting a two-digit number is frac{11}{20}. The expected number of digits is therefore: [ E = left(frac{9}{20} times 1right) + left(frac{11}{20} times 2right) = frac{9}{20} + frac{22}{20} = frac{31}{20} = 1.55 ] Thus, the expected number of digits Brent obtains when rolling the 20-sided die is boxed{1.55}.