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3 years ago

13

Find the inverse of the function. f(x)=3 root x/9 -4

1 answer:

Anastasy [175]3 years ago

5 0

Step-by-step explanation:

Explanation:

y

=

√

x

+

1

+

3

x

=

√

y

+

1

+

3

x

−

3

=

√

y

+

1

(

x

−

3

)

2

=

(

√

y

+

1

)

2

x

2

−

6

x

+

9

=

y

+

1

x

2

−

6

x

+

8

=

y

This is a quadratic, therefore it is a function.

Hopefully this helps!

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Did I do this right? (Accounting)

levacccp [35]

Yes you did, congratulations ;-)

4 0

3 years ago

2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

mel-nik [20]

Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

4 0

2 years ago

Read 2 more answers

Please Help! The graph of y=f(x) is shown below. What are all of the real solutions of f(x)=0? (Picture of graph included)

qwelly [4]

Answer:

(7, 0), (-8, 0), (0, 0)

Step-by-step explanation:

y = f(x) = 0

the real solutions are the ones that cross the x-axis because when a point is on the x-axis, its y coordinate will be 0.

=> (7, 0), (-8, 0), (0, 0)

6 0

3 years ago

1) 7+a=15<br><br><br><br> 2) 23=d+44<br><br><br><br> 3) 28=n-14<br><br><br><br> 4) T-22=-31

Romashka [77]

Hi, I'd be glad to help.

1) a = 8

2) d = (-21)

3) n = 42

4) t = (-9)

6 0

3 years ago

Read 2 more answers

338.08=8(0.07m+28.90+.30(28.90)

AlladinOne [14]

Answer:

m= - 396.253089

Step-by-step explanation:

hope this helps

5 0

2 years ago