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2 years ago

What is the y-intercept of y = 3 - 4z?

Mathematics

1 answer:

Neporo4naja [7]2 years ago

3 0

Answer:

pretty sure its 3

Step-by-step explanation:

just fix it: y = -4z+3

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If f x equals x -3 + g x equals 1 - x 2 - 9 find ( f og) (x and give any restrictions on its domain

evablogger [386]

$\begin{gathered} f(x)=x-3 \\ g(x)=\frac{1}{x^2-9} \\ (\text{fog)(x)}=\frac{1}{x^2-9}-3 \end{gathered}$

the denominator cannot be zero, because the division by zero is not defined, therefore:

$\begin{gathered} x^2-9=0 \\ \text{Solving for x:} \\ x^2=9 \\ \sqrt[]{x^2}=\sqrt[]{9} \\ x=\pm3 \end{gathered}$

Therefore the domain of (f o g)(x) is:

$D\colon(-\infty,-3)\cup(-3,3)\cup(3,\infty)$

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9 months ago

What is the lenght in miles of the field

maw [93]

(Perimeter/6)2= Length

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3 years ago

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If Roberto does a job in 11 hours and with the help of Sarah they can do it together in 2 hours, how long would it take Sarah to

Levart [38]

I think it is seven but im probaly wrong

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3 years ago

Which polynomial correctly combines the like terms and expresses the given polynomial in standard form? 8mn5 – 2m6 + 5m2n4 – m3n

photoshop1234 [79]

So it's multiple choice?
whichever looks most like this is the answer:
30mn+6n-36m

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2 years ago

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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

4 0

2 years ago