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maks197457 [2]
4 years ago
14

What is 58.14 kg into a lb

Mathematics
2 answers:
madam [21]4 years ago
7 0

Answer:

There is 127.91lbs in that many kilograms.

Step-by-step explanation:

In order to find this, you need to know the unit rate. There is 2.2lbs in every kilogram. Therefore, we can multiply the number of kilograms that we have by 2.2 to get the correct answer.

58.14*2.2 = 127.91

hjlf4 years ago
6 0

It is 128.1 pounds.....


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Answer:

Your answer is absolutely correct

Step-by-step explanation:

The work would be as follows:

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\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4

Hence proved that your solution is accurate.

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What are the domain and range of f(x) = 2|x – 4|?
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Step-by-step explanation:

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4 years ago
#24, i am getting <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%26%5Cfrac%7B18%7D%7B7%7D%20
sveticcg [70]

It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.

The given system translates to

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 1 & -1 & 2 & 2 \\ 1 & 2 & -3 & 4 \end{array} \right]

Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,

(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)

2x - 3y + z - 2x + 2y - 4z = 2 - 4

-y - 3z = -2

and

(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)

2x - 3y + z - 2x - 4y + 6z = 2 - 8

-7y + 7z = -6

In augmented matrix form, this step yields

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & -7 & 7 & -6 \end{array} \right]

I'll omit these details in the remaining steps.

Eliminate y from the last row by combining -7 (row 2) and row 3 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 28 & 8 \end{array} \right]

Multiply the last row by 1/28 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Eliminate z from the second row by combining 3 (row 3) and row 2 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & 0 & -8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Multiply the second row by -1 :

\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :

\left[ \begin{array}{ccc|c} 2 & 0 & 0 & 36/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

Multiply the first row by 1/2 :

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 18/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]

8 0
3 years ago
geraldo is going to paint the interor walls of the lobby of the local history museum.the room is 40 ft by 60 ft.its height is 16
maks197457 [2]
Since the dimensions are 40 ft by 60 ft, and 16 ft high, I assume the floor of the room is shaped like a rectangle.
Each wall is a rectangle. Two walls measure 40 ft by 16 ft, and two walls measure 60 ft by 16 ft.

lateral area = 40 ft * 16 ft * 2 + 60 ft * 16 ft * 2

lateral area = 1280 ft^2 + 1920 ft^2

lateral area = 3200 ft^2
3 0
4 years ago
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