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Charra [1.4K]
3 years ago
5

Which two numbers on the number line have an absolute value of 1.75? Select the location of both numbers on the number line

Mathematics
2 answers:
hram777 [196]3 years ago
8 0

Solution

             To    find   Which two numbers on the number line have  sn  

            absolute value of 1.75  .

            Absolute   value   means    " how  far   a  number    is   from  

            zero "  .

           "1.75"    is    1.75   away   from   zero  .

           and  " -1.75"   is  also   1.75    away   from  zero  .

          i.e

               |  0  - 1.75 |   =   1.75

              |  1.75 - 0 |  =  1.75  

           So ,  the  absolute   value  of   "1.75"    is  1.75  

          similarly  , absolute  value  of  "1.75"   is   1.75  .

          Hence  , the  two  numbers   are   -1.75  and  1.75  .

                 

           

svp [43]3 years ago
3 0

Answer:

Absolute value is the same answer but on different sides, so the absolute value of 1.75 is -1.75

Step-by-step explanation:

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Answer:

1250 m²

Step-by-step explanation:

Let x and y denote the sides of the rectangular research plot.

Thus, area is;

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Thus, if y is the erected wall, and we are using 100m wire for the remaining sides, it means;

2x + y = 100

Thus, y = 100 - 2x

Since A = xy

We have; A = x(100 - 2x)

A = 100x - 2x²

At maximum area, dA/dx = 0.thus;

dA/dx = 100 - 4x

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3 years ago
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Let v⃗ 1=⎡⎣⎢033⎤⎦⎥,v⃗ 2=⎡⎣⎢1−10⎤⎦⎥,v⃗ 3=⎡⎣⎢30−3⎤⎦⎥ be eigenvectors of the matrix A which correspond to the eigenvalues λ1=−1, λ2
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Answer:

- x as a linear combination :

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Step-by-step explanation:

Given v1 = (0 3 3),v2 = (1 −1 0), v3 = (3 0 −3) be eigenvectors of the matrix A which correspond to the eigenvalues λ1 = −1, λ2 = 0, and λ3 = 1, respectively, and let x = (−2 −4 0). Express x as a linear combination of v1, v2, and v3, and find Ax .

To write x as a linear combination of v1, v2, and v3

x = -1 v1+ 0 v2+ 1 v3.

To find Ax

Write A = (0 ......3 ......3 )

...................(1 ......-1 ......0)

...................(3 ......0......-3)

Since transpose x = (-2, 4, 0)

Ax =......... (0 ......3......3 )(-2)

...................(1 ......-1 ......0)(4)

...................(3 ......0......-3)(0)

= (0×-2 + 3×4 + 3×0)

...(1×-2 + -1×4 + 0×0)

.. (3×-2 + 0×4 + -3×0)

As = (12)

....(-6)

....(-6)

Transpose Ax = (12, -6, -6)

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