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question:an equilateral triangle, the perimeter is 45, so the length of each side is 45/3 = 15. Now, consider one of the sides of the equilateral triangle as one of the sides of an isosceles triangle. Let the angle between the two equal sides of the isosceles triangle be θ. The perimeter of the isosceles triangle is 40. Let the equal sides of the isosceles triangle be x, and the base be b. Then, 2x + b = 40. Using the given information, x = 15, making the isosceles triangle have sides 15, 15, and b. Now, to find the length of base b, we can use the trigonometric functions. Since we know the angle θ and the two equal sides (15) of the isosceles triangle, we can find the height of the triangle, which will bisect the base b. Let's denote the height as h. We can use the sine function: sin(θ) = h/15. Once we find the value of h, we can use the Pythagorean theorem to find half the length of base b, and then double it to find the full length.
answer:To find the length of the base b, we need to know the value of the angle θ. However, since the angle θ is not provided, we cannot directly calculate the height h using the sine function. If we had the value of θ, we could proceed as follows: 1. Calculate the height h using the sine function: sin(θ) = h/15 h = 15 * sin(θ) 2. Use the Pythagorean theorem to find half the length of the base (let's call it b/2): (b/2)^2 + h^2 = 15^2 (b/2)^2 = 15^2 - h^2 b/2 = sqrt(15^2 - h^2) 3. Double the value of b/2 to find the full length of the base b: b = 2 * sqrt(15^2 - h^2) Since we know that 2x + b = 40 and x = 15, we can find b without knowing θ: 2(15) + b = 40 30 + b = 40 b = 40 - 30 b = 10 Therefore, the length of the base b of the isosceles triangle is boxed{10} units.
question:For what base7 digit d is 3dd6_7 divisible by the base 10 number 13? (Here 3dd6_7 represents a base7 number whose first digit is 3, whose last digit is 6, and whose middle two digits are both equal to d).
answer:First, convert 3dd6_7 to base-10: [ 3dd6_7 = 3 cdot 7^3 + d cdot 7^2 + d cdot 7 + 6 ] [ = 1029 + 49d + 7d + 6 ] [ = 1035 + 56d ] Now, simplify and check for divisibility by 13: [ 1035 + 56d ] Subtract 1035 - 79 cdot 13 = 1035 - 1027 = 8: [ 8 + 56d ] We then test each base-7 digit (0 to 6) to see if 8 + 56d is divisible by 13: - d = 0: 8 + 56 cdot 0 = 8 (not divisible by 13) - d = 1: 8 + 56 cdot 1 = 64 (not divisible by 13) - d = 2: 8 + 56 cdot 2 = 120 (not divisible by 13) - d = 3: 8 + 56 cdot 3 = 176 (not divisible by 13) - d = 4: 8 + 56 cdot 4 = 232 (divisible by 13) - d = 5: 8 + 56 cdot 5 = 288 (not divisible by 13) - d = 6: 8 + 56 cdot 6 = 344 (not divisible by 13) Only d = 4 results in a number divisible by 13. Therefore, d = boxed{4}.
question:Given the function f(x)= begin{cases} log_{2}x, & xgeqslant 1 x+c, & x < 1 end{cases}, then c=1 is a ______ condition for the function to be monotonically increasing on mathbb{R}.
answer:To solve this, we have f(x)= begin{cases} log_{2}x, & xgeqslant 1 x+c, & x < 1 end{cases}. For the function to be monotonically increasing on mathbb{R}, therefore log_{2}1geqslant 1+c, therefore cleqslant -1, therefore "c=-1" is a sufficient but not necessary condition for the function to be monotonically increasing on mathbb{R}. Therefore, the answer is: Sufficient but not necessary. By analyzing the range of c for which the function f(x)= begin{cases} log_{2}x, & xgeqslant 1 x+c, & x < 1 end{cases} is monotonically increasing on mathbb{R}, and then referring to the definitions of sufficient and necessary conditions, we can make a judgment. This question examines the monotonicity of functions and the determination of sufficient and necessary conditions, and is considered a basic question. Thus, the final answer is boxed{text{Sufficient but not necessary}}.
question:On a sheet of paper, two lines forming an angle are drawn, with the vertex of the angle lying outside of this sheet. Using a compass and a ruler, draw the part of the angle bisector that lies on the sheet of paper.
answer:1. **Select an arbitrary point O on the sheet of paper**: - This point will be used as the center for homothety transformations. 2. **Perform a homothety transformation**: - Define a homothety transformation with center ( O ) and a small coefficient ( k ), such that the image of the intersection point, where the given lines intersect, is moved onto the sheet of paper. 3. **Determine images of the lines under the homothety**: - The original lines will transform into new lines that still keep their angles but their intersection point now lies within the paper. 4. **Construct the angle bisector of the transformed lines**: - Using the standard method with compass and straightedge, construct the angle bisector of the angle formed between the images of the lines obtained from the homothety. This involves: - Drawing arcs with equal radii from the intersection point. - Intersecting these arcs with the transformed lines. - Drawing arcs from the intersection points of the former arcs. - Finding the intersection of these latter arcs and drawing a line from the intersection point of the original lines through this new intersection point. 5. **Perform the reverse homothety transformation**: - Apply the inverse homothety transformation with the same center ( O ) and a coefficient ( 1/k ). This will map the bisector constructed in the previous step back to the original diagram, giving you the required segment of the bisector of the angle formed by the initial lines on the paper. # Conclusion: [ boxed{text{Segment of the bisector is constructed}} ]