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DENIUS [597]
3 years ago
12

What is the standard form of two hundred fifty three thousandths

Mathematics
1 answer:
lozanna [386]3 years ago
6 0

We are given number in words " two hundred fifty three thousandths".

Let us write it in number form: First we would write the number form of "two hundred fifty".

Two hundred fifty = 253.

Now, we need to write three thousandths.

Three thousandths =  .003

Now, we need to combine 253 and 0.003.

On combining we get 253.003.

<h3>Therefore, two hundred fifty three thousandths in standard form is 253.003.</h3>
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hus, D satisfying ​(ABC)DequalsI exists. Why does the expression for D found above also satisfy ​D(ABC)equalsI​, thereby showing
Assoli18 [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second  Question

Option C is correct

Third   Question

     D =  A^{-1}  *  B^{-1} *  C^{-1}

Fourth   Question

  So substituting for D in  (ABC) D =  I

                 (ABC) *  A^{-1}  *  B^{-1} *  C^{-1} =  I

                 I =  I

This proof that  ABC is invertible

Step-by-step explanation:

From the question we are told that

   A , B and  C are invertible which means that A^{-1} , B^{-1}, C^{-1} exist

Now

 From the question

          (ABC) D =  I

Where I is an identity matrix

   Now when we multiply both sides by  A^{-1}  we have

          A^{-1}  A BCD =  A^{-1} * I

          IBCD =  A^{-1}

Now when we multiply both sides by  B^{-1}  we have  

         B^{-1 } *I BCD =  A^{-1}  *  B^{-1}

         I CD =  A^{-1}  *  B^{-1}

Now when we multiply both sides by  C^{-1}  we have  

          C^{-1} * I CD =  A^{-1}  *  B^{-1} *  C^{-1}

              I D =  A^{-1}  *  B^{-1} *  C^{-1}

                 D =  A^{-1}  *  B^{-1} *  C^{-1}

So substituting for D in the above equation

                 (ABC) *  A^{-1}  *  B^{-1} *  C^{-1} =  I

                 I =  I

This proof that  ABC is invertible

 

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