answer:Let's denote the number of yellow tiles as Y. According to the problem, the number of blue tiles is one more than the number of yellow tiles, so there are Y + 1 blue tiles. We also know that there are 6 purple tiles and 7 white tiles. The total number of tiles is the sum of the yellow, blue, purple, and white tiles, which is 20. So we can write the equation: Y (yellow) + (Y + 1) (blue) + 6 (purple) + 7 (white) = 20 Combining like terms, we get: 2Y + 1 + 6 + 7 = 20 2Y + 14 = 20 Subtracting 14 from both sides gives us: 2Y = 20 - 14 2Y = 6 Dividing both sides by 2 gives us: Y = 6 / 2 Y = 3 So there are boxed{3} yellow tiles.

answer:First, we need to analyze the continuity and monotonicity of the function f(x) = ln x + x. Since ln x is defined for x > 0 and both ln x and x are continuous functions, their sum is also a continuous function. Additionally, ln x is monotonically increasing for x > 0, and so is the function x. Therefore, their sum, f(x) = ln x + x, is a monotonically increasing function. Now we can check the values of f(x) at the endpoints of the given intervals to find where the solution falls. For the interval (0, 1): - At the left endpoint, lim_{x to 0^+} f(x) = -infty. - At the right endpoint, f(1) = ln 1 + 1 = 0 + 1 = 1 < 3. So the solution cannot be in this interval as f(x) does not reach the value of 3. For the interval (1, 2): - We have already calculated f(1) = 1. - At the right endpoint, f(2) = ln 2 + 2. Since 0 < ln 2 < ln e = 1, we have 2 < ln 2 + 2 < 3. Thus, the function value is getting closer to 3, but it is not equal to 3 within this interval. For the interval (2, 3): - We know f(2) < 3. - At the right endpoint, f(3) = ln 3 + 3. Note that since ln 2 < 1 and ln 3 > 1, we have f(3) = ln 3 + 3 > 4. Because f(x) is monotonically increasing, there is one point within this interval where f(x) will precisely equal 3. For the interval (3, 4): - We have f(3) > 4 from the previous calculation. - Therefore, the solution to the equation cannot be in this interval either, as f(x) jumps from a value greater than 3 to an even larger number since the function is increasing. Thus, since the function f(x) has a value less than 3 at x = 2 and a value greater than 3 at x = 3, and it is monotonically increasing, the solution x_0 to the equation ln x + x = 3 must lie within the interval (2, 3). This means that the correct answer is boxed{C: (2, 3)}.

answer:Let's denote the original price of the suit as ( P ). After a 30% increase, the new price of the suit is ( P + 0.30P = 1.30P ). Then, a 30% off coupon is applied to this increased price, which means the consumers pay 70% of the increased price: ( 0.70 times 1.30P ). We know that after applying the 30% off coupon, the consumers paid 182 for the suit: ( 0.70 times 1.30P = 182 ). Now, we can solve for ( P ): ( 0.91P = 182 ), ( P = frac{182}{0.91} ), ( P = 200 ). So, the original price of the suit was boxed{200} .

answer:First, convert 543210_{6} to base 10: [ 543210_{6} = 5 cdot 6^5 + 4 cdot 6^4 + 3 cdot 6^3 + 2 cdot 6^2 + 1 cdot 6^1 + 0 cdot 6^0 ] Calculating each term: [ 5 cdot 7776 + 4 cdot 1296 + 3 cdot 216 + 2 cdot 36 + 1 cdot 6 + 0 cdot 1 = 38880 + 5184 + 648 + 72 + 6 + 0 = 44790 ] Next, convert 43210_{7} to base 10: [ 43210_{7} = 4 cdot 7^4 + 3 cdot 7^3 + 2 cdot 7^2 + 1 cdot 7^1 + 0 cdot 7^0 ] Calculating each term: [ 4 cdot 2401 + 3 cdot 343 + 2 cdot 49 + 1 cdot 7 + 0 cdot 1 = 9604 + 1029 + 98 + 7 + 0 = 10738 ] Subtract the two results: [ 44790 - 10738 = boxed{34052} ]