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Drupady [299]
3 years ago
10

An electronic devise beeps every 4 seconds. Another devise beeps every 6 seconds if they both start beeping at the same time how

many seconds will it be before the next time they beep together
Mathematics
1 answer:
Nat2105 [25]3 years ago
4 0
Answer:

3

Explanation:

The smallest whole number that 4 and 6 share when multiplied is 12
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If frank has 380 condoms and he uses 180 in two weeks and by the 4th week he has 2 left how many did he use in 2 weeks (btws thi
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4 years ago
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I need question 3 only
dsp73

Answer:

  • table: 14, 16, 18
  • equation: P = 2n +12

Step-by-step explanation:

Perimeter values will be ...

rectangles . . . perimeter

  1 . . . 14

  2 . . . 16

  3 . . . 18

__

The perimeter of a figure is twice the sum of the length and width. Here, the length is a constant 6. The width is n, the number of rectangles. So, the perimeter is ...

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Your equation is ...

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<em>Additional comment</em>

You may be expected to write the equation using y and x for the perimeter and the number of rectangles. That would be ...

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4 0
3 years ago
Use the following matrices, A, B, C and D to perform each operation.
Vinvika [58]

Step-by-step explanation:

A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]

B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]

C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]

D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]

1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]

2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]

3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]

5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}

6 0
3 years ago
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