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Gennadij [26K]
3 years ago
9

Which lengths would form a right triangle? 9, 12, and 14 24, 7, and 26 21, 16, and 12 30, 24, and 18

Mathematics
2 answers:
Kamila [148]3 years ago
6 0

right triangle
    c^2 = a^2+b^2
30^2  = 24^2+18^2
   900 = 576 +324
   900 = 900

answer
<span> 30, 24, and 18</span>

Sati [7]3 years ago
4 0

Answer:

D

Step-by-step explanation:

Which lengths would form a right triangle?

A.)9, 12, and 14

B.)24, 7, and 26

C.)21, 16, and 12

D.)30, 24, and 18

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Volume with cubes help pls 2
il63 [147K]

Answer:

  • 192

Step-by-step explanation:

The volume of the prism is 3 cube units, which means that the prism can be filled with 3 1*1*1.

Here, there are 64 cubes with 1/4 units side lengths in 1 cubic unit.

Since the rectangular prism is made up of 3 cubic units, we would have

3 * 64  total cubes with 1/4 units side lengths.

3*64=192

Therefore, it takes 192 of the 1/4 unit cubes to fill the prism.

~

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2 years ago
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Premier Bank is planning to have a new building constructed. They would like the length of the building to be 40 feet longer tha
MaRussiya [10]

Let L, W, F, H, P represent, respectively, the length, width, number of floors, height, and perimeter of the building. Here are some equations that model the given information.

  • L = W + 40
  • H = 12×F
  • foundation cost = 10×L×W
  • cost of floors = 35000×F
  • total cost = foundation cost + cost of floors
  • total cost = 1,180,000
  • P = 2×(L+W)
  • P = 3×H

We can rewrite these into a system of equations involving L, W, F.

Let's look first at the relation between the perimeter and height. The two versions of perimeter are equal to each other, so we have (after substituting for H) ...

... 3H = 2(L+W)

... 3(12F) = 2((W +40) +W) . . . . substitute for H and L

... 36F = 4W +80 . . . . . . . . . . . simplify

... 9F = W +20 . . . . . . . . . . . . . divide by 4

Now, we can look at the cost equations.

... 1180000 = 35000F + 10W(W+40) . . . . . substitute for the various cost terms and for L

... 118000 = 3500F +W² +40W . . . . . . . . . divide by 10

The two simplified equations in W and F are ...

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You can solve the first for F and substitute for F in the second. Then you have a quadratic equation in W that can be solved by the usual methods. (Likely, the quadratic formula would be the best choice.) Here, we have elected to let a graphing calculator show the two solutions. One solution has negative dimensions, so is clearly infeasible.

There is one (nearly feasible) solution:

... W ≈ 180.788

... F ≈ 22.31

There is no combination of floor dimensions and integral numbers of floors that will match <em>all</em> of the problem requirements.

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A solution that comes about as close as possible is ...

  • length = 221 ft
  • width = 181 ft
  • perimeter = 804 ft
  • # of floors = 22
  • height = 264 ft . . . . . (3×height = 792 ft vs. 804 ft)
  • cost = $1,170,010 . . . (vs. 1,180,000)

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Rectangle rhombus is and square

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<span>9 + 4p

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