Simplify. -6+{14+2 [60 – 9(1+3)]} Enter your answer in the box.
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Answer:
1 - (π/6) - (√3/4)
Step-by-step explanation:
First, calculate the area of ABCD. That's just
Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.
Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.
Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = AD =
. MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.
Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.
The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.
The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.
Hope this helps, have a great day.
The 400th term is 425.
There are floor(√400) = 20 squares in the range 1..400, so the 400th term will be at least 420. There are floor(∛420) = 7 cubes in the range 1..400, so the 400th term may be as high as 427. However, there are![\lfloor\sqrt[6]{427}\rfloor=2](https://tex.z-dn.net/?f=%5Clfloor%5Csqrt%5B6%5D%7B427%7D%5Crfloor%3D2)
numbers that are both squares and cubes. Consequently, the 400th term will be 427-2 = 425.
There are floor(√400) = 20 squares in the range 1..400, so the 400th term will be at least 420. There are floor(∛420) = 7 cubes in the range 1..400, so the 400th term may be as high as 427. However, there are
Answer:
63°
Step-by-step explanation:
- ≈ 63°
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hope it helps...
have a great day!!