answer:To determine the range of the independent variable x for the function y=frac{{sqrt{x+3}}}{x-1}, we need to consider the conditions under which the function is defined. 1. The numerator sqrt{x+3} implies that x+3 must be greater than or equal to 0 for the square root to be real. This gives us the inequality: [x+3 geqslant 0] Solving for x, we find: [x geqslant -3] 2. The denominator x-1 implies that x-1 cannot equal 0 because division by zero is undefined. Therefore, we have: [x-1 neq 0] Solving for x, we get: [x neq 1] Combining these two conditions, we conclude that the range of the independent variable x is such that x is greater than or equal to -3 and x is not equal to 1. Therefore, the range of x is: [x geqslant -3 quad text{and} quad x neq 1] Encapsulating the final answer: [boxed{x geqslant -3 quad text{and} quad x neq 1}]

answer:To find the probability of getting at least one contract, we can use the principle of Inclusion-Exclusion. The formula for two events A and B is: P(A or B) = P(A) + P(B) - P(A and B) Where: P(A or B) is the probability of getting at least one contract. P(A) is the probability of getting the hardware contract. P(B) is the probability of not getting the software contract. P(A and B) is the probability of getting both contracts. Given: P(A) = 3/4 P(B) = 5/9 (since this is the probability of not getting the software contract, the probability of getting the software contract is 1 - 5/9 = 4/9) P(A and B) = 0.3944444444444444 Now we can plug these values into the formula: P(A or B) = P(A) + (1 - P(B)) - P(A and B) P(A or B) = 3/4 + (1 - 5/9) - 0.3944444444444444 First, calculate 1 - P(B): 1 - 5/9 = 9/9 - 5/9 = 4/9 Now, plug this into the formula: P(A or B) = 3/4 + 4/9 - 0.3944444444444444 To add the fractions, we need a common denominator. The least common multiple of 4 and 9 is 36. So we convert the fractions: P(A) = 3/4 = 27/36 1 - P(B) = 4/9 = 16/36 Now we can add them: P(A or B) = 27/36 + 16/36 - 0.3944444444444444 P(A or B) = 43/36 - 0.3944444444444444 Now we need to convert 0.3944444444444444 to a fraction with a denominator of 36 to subtract it from 43/36. Since the decimal is repeating, we can round it to a close fraction for simplicity. The closest fraction to 0.3944444444444444 with a denominator of 36 is 14/36 (since 0.3944444444444444 ≈ 14/36). P(A or B) = 43/36 - 14/36 P(A or B) = (43 - 14)/36 P(A or B) = 29/36 So the probability of getting at least one contract is boxed{29/36} .

answer:Since ( a^2, b^2, c^2 ) form a geometric sequence, we have ( b^2 = sqrt{a^2 c^2} = ac ). Thus, ( b = sqrt{ac} ). The three logs convert to: [ log_c a, log_{sqrt{ac}} c, log_a sqrt{ac} ]. Let ( x = log_c a ). Using change-of-base, [ log_{sqrt{ac}} c = frac{log_c c}{log_c sqrt{ac}} = frac{1}{frac{1}{2} log_c ac} = frac{2}{log_c a + log_c c} = frac{2}{x + 1}, ] and [ log_a sqrt{ac} = frac{1}{2} log_a ac = frac{log_c ac}{2 log_c a} = frac{log_c a + log_c c}{2 log_c a} = frac{x + 1}{2x}. ] Let ( d ) be the common difference, so [ d = frac{2}{x + 1} - x = frac{x + 1}{2x} - frac{2}{x + 1}. ] Solving, [ 4x - 2x^2 (x + 1) = (x + 1)^2 - 4x, ] which simplifies to ( 2x^3 + 3x^2 - 6x + 1 = 0 ). This factors as ( (x - 1)(2x^2 + 5x - 1) = 0 ). Not using ( x = 1 ) as ( a neq c ), so from ( 2x^2 + 5x - 1 = 0 ), [ x = frac{-5 pm sqrt{25 + 8}}{4} = frac{-5 pm sqrt{33}}{4}. ] Using the positive root (as ( x ) represents a log of positives), [ x = frac{-5 + sqrt{33}}{4}, ] remaining calculations imply, [ d = frac{3}{2}. ] Conclusion: Common difference ( d = boxed{frac{3}{2}} ).

answer:1. **Initial Setup**: On Day 1, Elsie puts 2 quarts of seeds into the feeder, 40% of which are sunflower seeds. This means 0.4 times 2 = 0.8 quarts of sunflower seeds and 1.2 quarts of other seeds are placed. 2. **Daily Seed Addition and Consumption**: - Each day, Elsie adds another 2 quarts of seed mix with the same ratio: 0.8 quarts of sunflower seeds and 1.2 quarts of other seeds. - Each day, the birds eat 30% of the sunflower seeds and all other seeds. 3. **Sunflower Seed Amount Calculation**: - On Day 1, the total amount of sunflower seeds is 0.8 quarts. - On Day 2, the remaining sunflower seeds from Day 1 are 0.7 times 0.8 = 0.56 quarts. With the addition of 0.8 quarts, the total becomes 0.56 + 0.8 = 1.36 quarts. - On Day 3, the remaining from previous days are 0.7 times 1.36 = 0.952 quarts. Adding another 0.8 quarts, the total is 0.952 + 0.8 = 1.752 quarts. 4. **General Formula for Sunflower Seeds**: - The amount of sunflower seeds on day ( n ) is given by: [ 0.8 (1 + 0.7 + 0.7^2 + cdots + 0.7^{n-1}) ] - This simplifies to (using the sum of a geometric series): [ 0.8 frac{1 - 0.7^n}{1 - 0.7} = frac{8}{3}(1 - 0.7^n) ] 5. **Condition for Sunflower Seeds Being More Than Half**: - Need frac{8}{3}(1 - 0.7^n) > 2, simplifying to 1 - 0.7^n > frac{3}{4} or 0.7^n < frac{1}{4}. 6. **Finding the Day**: - Analyze 0.7^3 = 0.343 and 0.7^4 = 0.2401. - Therefore, 0.7^n < frac{1}{4} occurs first on day 4. text{Thursday} The final answer is boxed{C) Thursday}