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2 years ago

Is this correct? Explain. 48 - 24k = 12 (4 - 2k)

Mathematics

2 answers:

VLD [36.1K]2 years ago

7 0

It is correct because if you do the math you will come up with 48 - 24k = 48 - 24k

Ilia_Sergeevich [38]2 years ago

6 0

It is correct because when you simplify the expression on the right side, you will end up with the same expression that is to the other side of it.

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Solve by completing the square and show steps <br> x^(2)+6x+5=0

Crazy boy [7]

X ^2 +6x+ 5
(x+2)(x+3)

2 + 3 = 5
While also 2•3= 6
The two numbers have to equal b when multiplied and c when added

4 0

3 years ago

Round fraction as a decimal round three decimals place 6/13

frez [133]

Round fraction as a decimal: 6/13=.462 that's the answer

6 0

3 years ago

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

4 0

3 years ago

Mason plays a game by flipping two fair coins.He wins the game if both coins land facing heads up. If Mason plays 200 times, how

blagie [28]

Answer:

Therefore Mason would be expected to win 50 times.

Step-by-step explanation:

i) probability of flipping a head for a fair coin is $\frac{1}{2}$

ii) probability of flipping two fair coins is = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

iii) Therefore if Mason plays 200 times he would be expected to win

= $\frac{1}{4} \times 200 = 50\hspace{0.1cm} times.$

7 0

3 years ago

Order 0.2, 1/4, 1/12 from least to greatest

olchik [2.2K]

0.2 = 2/10 = 1/5

0.2

1/4

1/2

7 0

3 years ago