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mars1129 [50]
3 years ago
11

What side length would you specify if you were required to create a regular hexagonal plate that was composed of 33 cm2 of sheet

metal? Sketch the hexagon, dimension the side length on the sketch to 0.1 cm, and indicate your grid spacing. Justify your dimension.
Mathematics
1 answer:
OverLord2011 [107]3 years ago
5 0
5.3 is proble the answer
i think

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You visit the Grand Canyon and drop a penny off the edge of a cliff. The distance the penny will fall is 16 feet the first secon
fiasKO [112]
The penny will fall 112 feet the fourth second, 144 feet the fifth second, and 176 feet the sixth second. Therefore, the penny will be a total of 576 feet from the cliff at six seconds.
5 0
3 years ago
Given: f(x) = 2x + 5, match the inputs with their outputs.
bekas [8.4K]
Yeah the answer is number 5
3 0
3 years ago
An unconfined compression test was performed to determine the average strength of concrete cylinders (in psi). It is believed th
Ronch [10]

Answer:

Explained below.

Step-by-step explanation:

(1)

The confidence level is, 91%.

Compute the value of α as follows:

\text{Confidence level}\% =(100-\alpha \%)\\\\91% = 100-\alpha \%\\\\\alpha \%=100-91\%\\\\\alpha \%=9\%\\\\\alpha =0.09

(2)

As the population standard deviation is provided, i.e. <em>σ</em> = 256 psi, the <em>z</em> value would be appropriate.

The <em>z</em> value for α = 0.09 is,

<em>z</em> = 1.69

(3)

Compute the 91% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

     =3000\pm 1.69\cdot\frac{256}{\sqrt{70}}\\\\=3000\pm 51.71\\\\=(2942.29, 3057.71)

(4)

The 91% confidence interval for population mean implies that there is a 0.91 probability that the true value of the mean is included in the interval, (2942.29, 3057.71) psi.

8 0
2 years ago
Prove that it is impossible to dissect a cube into finitely many cubes, no two of which are the same size.
solniwko [45]

explanation:

The sides of a cube are squares, and they are covered by the respective sides of the cubes covering that side of the big cube. If we can show that a sqaure cannot be descomposed in squares of different sides, then we are done.

We cover the bottom side of that square with the bottom side of smaller squares. Above each square there is at least one square. Those squares have different heights, and they can have more or less (but not equal) height than the square they have below.

There is one square, lets call it A, that has minimum height among the squares that cover the bottom line, a bigger sqaure cannot fit above A because it would overlap with A's neighbours, so the selected square, lets call it B, should have less height than A itself.

There should be a 'hole' between B and at least one of A's neighbours, this hole is a rectangle with height equal to B's height. Since we cant use squares of similar sizes, we need at least 2 squares covering the 'hole', or a big sqaure that will form another hole above B, making this problem inifnite. If we use 2 or more squares, those sqaures height's combined should be at least equal than the height of B. Lets call C the small square that is next to B and above A in the 'hole'. C has even less height than B (otherwise, C would form the 'hole' above B as we described before). There are 2 possibilities:

  • C has similar size than the difference between A and B
  • C has smaller size than the difference between A and B

If the second case would be true, next to C and above A there should be another 'hole', making this problem infinite. Assuming the first case is true, then C would fit perfectly above A and between B and A's neighborhood.  Leaving a small rectangle above it that was part of the original hole.

That small rectangle has base length similar than the sides of C, so it cant be covered by a single square. The small sqaure you would use to cover that rectangle that is above to C and next to B, lets call it D, would leave another 'hole' above C and between D and A's neighborhood.

As you can see, this problem recursively forces you to use smaller and smaller squares, to a never end. You cant cover a sqaure with a finite number of squares and, as a result, you cant cover a cube with finite cubes.

3 0
3 years ago
Find the measure of XY. Assume that segments that appear to be tangent are tangent. Round your final answer to the nearest tenth
Slav-nsk [51]

Answer:

Solution given:

7x-29=2x+16[length from the circumference of circle is equal to same point]

7x-2x=16+29

5x=45

x=45/5=9

now.

XY=2×9+16=18+16=34 is your answer

5 0
3 years ago
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