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3 years ago

I need help with this problem! Please show work so i can get credit for my work

Mathematics

1 answer:

ser-zykov [4K]3 years ago

4 0

Answer:

x= -43 or x=57

Step-by-step explanation:

Apply absolute value rule: If |u|=a, a>0 then u=a or u=-a

So applying that to our equation:

|x-7|=50, x-7 can either equal +50 or -50 so we solve for both

x-7= -50 ---> x= -43

x-7=50 ---> x=57

I hope this helps!

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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

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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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