answer:1. **Assertion for ( lim_{x to infty} log_a x = infty ) when ( a > 1 ):** To show this, consider the exponential form. Let ( M = a^A ) for some constant ( A ). [ text{As } x to infty, text{ we can find a value of } x text{ such that } x > a^A. ] Rewriting the inequality in terms of logarithm, we get: [ a^{log_a x} > a^A implies log_a x > A. ] Since ( A ) can be any large positive number, as ( x to infty ), ( log_a x ) will also go to ( infty ). Therefore, [ lim_{x to infty} log_a x = infty. ] 2. **Assertion for ( lim_{x to 0} log_a x = -infty ) when ( a > 1 ):** Using the property of logarithms, we can observe that: [ log_a x = -log_a left(frac{1}{x}right). ] As ( x to 0 ), [ frac{1}{x} to infty. ] Hence, [ lim_{x to 0} log_a x = -lim_{frac{1}{x} to infty} log_a left(frac{1}{x}right) = -infty. ] 3. **Assertions for ( lim_{x to infty} log_a x = -infty ) and ( lim_{x to 0} log_a x = infty ) when ( a < 1 ):** Let’s rewrite ( a ) as ( a = frac{1}{b} ) where ( b > 1 ). Then, [ log_{a} x = log_{frac{1}{b}} x = -log_b x. ] Thus: [ lim_{x to infty} log_a x = lim_{x to infty} (-log_b x) = -lim_{x to infty} log_b x = -infty. ] For ( lim_{x to 0} log_a x ): [ lim_{x to 0} log_a x = lim_{x to 0} (-log_b x) = -lim_{x to 0} log_b x = infty. ] Conclusion: [ boxed{lim_{x to infty} log_a x = infty text{ and } lim_{x to 0} log_a x = -infty text{ for } a > 1.} ] [ boxed{lim_{x to infty} log_a x = -infty text{ and } lim_{x to 0} log_a x = infty text{ for } a < 1.} ]

answer:1. **Identify the initial condition:** Today's date is 17.02.2008, and the sum of the first four digits of the date equals the sum of the last four digits. 2. **Write out the date in numerical form:** The date 17.02.2008 can be written numerically as 17022008. 3. **Calculate the sum of the first four and the last four digits:** - First four digits: (1 + 7 + 0 + 2 = 10) - Last four digits: (2 + 0 + 0 + 8 = 10) Clearly, (1 + 7 + 0 + 2 = 2 + 0 + 0 + 8 = 10). 4. **Determine if there are more dates with this property in 2008:** - Let's systematically check the remaining months from 17.02.2008 to the end of 2008. 5. **Check the following months:** - **March**: Dates range from 01.03.2008 to 31.03.2008. - Sum of the first four digits (0103) ranges from 4 to 7, which doesn't match any sums from the last four digits (2008, 2009, ..., 2038). - **April to November**: Similar checking confirms there isn't any accounting date matching sums. - **December**: - **Check from 01.12.2008 to 24.12.2008**: - Similar computations above show no dates matching sums. - **On 25.12.2008**: - Date in numerical form: 25122008. - Sum of the first four digits: (2 + 5 + 1 + 2 = 10) - Sum of the last four digits: (2 + 0 + 0 + 8 = 10). Hence, (2 + 5 + 1 + 2 = 2 + 0 + 0 + 8 = 10). 6. **Conclusion**: The last date in 2008 when the sum of the first four digits equals the sum of the last four digits is 25.12.2008. [ boxed{text{25 December 2008}} ]

answer:Let's denote the monthly incomes of A, B, and C as A_income, B_income, and C_income respectively. Given that the ratio of their incomes is 5 : 3 : 2, we can write: A_income / B_income / C_income = 5 / 3 / 2 We are also given that B's monthly income is 12% more than C's monthly income. So we can write: B_income = C_income + 0.12 * C_income B_income = 1.12 * C_income Since C's monthly income is Rs. 17000, we can substitute this value into the equation: B_income = 1.12 * 17000 B_income = Rs. 19040 Now, we can find A's monthly income using the ratio. Since A's income is 5 parts and C's income is 2 parts, we can write: A_income / C_income = 5 / 2 Substituting the value of C_income, we get: A_income / 17000 = 5 / 2 A_income = (5 / 2) * 17000 A_income = 5 * 8500 A_income = Rs. 42500 Now, let's find D's monthly income. D's monthly income is 15% less than A's monthly income, so: D_income = A_income - 0.15 * A_income D_income = 0.85 * A_income D_income = 0.85 * 42500 D_income = Rs. 36125 Now we have the monthly incomes of A, B, C, and D. To find their combined annual income, we need to sum their monthly incomes and multiply by 12 (since there are 12 months in a year): Annual_income = (A_income + B_income + C_income + D_income) * 12 Annual_income = (42500 + 19040 + 17000 + 36125) * 12 Now, let's calculate the total: Annual_income = (114665) * 12 Annual_income = Rs. 1375980 Therefore, the combined annual income of A, B, C, and D is Rs. boxed{1375980} .

answer:Let's denote the width of the wall as W, the height as H, and the length as L. According to the information given: W = 3 (width of the wall) H = 6 * W (height is 6 times the width) L = 7 * H (length is 7 times the height) First, we calculate the height: H = 6 * W = 6 * 3 = 18 Next, we calculate the length: L = 7 * H = 7 * 18 = 126 Now that we have the dimensions of the wall, we can calculate the volume (V) using the formula for the volume of a rectangular prism: V = W * H * L Substitute the values we have: V = 3 * 18 * 126 V = 54 * 126 V = 6804 The volume of the wall is boxed{6804} cubic units.