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LUCKY_DIMON [66]
3 years ago
6

The function f(x)=(x-1)^4 is not one to one. However, if we restrict the domain to x greater than or equal to 0, we can find it'

s inverse. Which Graph is the graph of f^-1(x) ?

Mathematics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

Option A

Step-by-step explanation:

Domain should be greater than/equal to 1 for the function to be one-one.

f has vertex at (1,0)

f inverse has (0,1)

f(x) = 4

(x - 1)⁴ = 4

x = 2.414

On f: (2.414, 4)

On f inverse: (4, 2.414)

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Find the interior angle sum for the following polygon​
wolverine [178]

Answer:

1620°

Step-by-step explanation:

The figure is an 11-sided polygon and is called hendecagon

The interior angle sum:

180( n-2)   n=11

=180(11-2)

=180(9)

=1620°

I hope this help you

5 0
3 years ago
Anyone up for the question, prize awaits you
Vikki [24]

Answer:

Step-by-step explanation:

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8 0
3 years ago
B. Using your answer in letter a or write the equation of the line in the form Ax + By = C
Svetradugi [14.3K]

Answer:

Step-by-step explanation:

(1, 5.5) ; (2, 9)

Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\

         = \frac{9-5.5}{2-1}\\\\=\frac{3.5}{1}\\\\= 3.5

         = 7/2

m = 7/2 ; (2 , 9)

y - y1 =m(x -x1)

y - 9 = \frac{7}{2}(x- 2)\\\\y - 9 = \frac{7}{2}x - \frac{7}{2}*2\\\\y -9 = \frac{7}{2}x - 7\\\\y = \frac{7}{2}x - 7 + 9\\\\y = \frac{7}{2}x + 2

Multiply the equation  by 2

2y = 2*\frac{7}{2}x + 2 *2\\\\2y = 7x + 4\\\\7x - 2y = -4\\

         

4 0
3 years ago
Assuming that the equations define x and y implicitly as differentiable functions x=f(t),y=g(t) find the slope of the curve x=f(
Mumz [18]
The given equations are
x(t+1)-4t \sqrt{x} =9            (1)
2y+4y^{3/2}=t^{3}+t           (2)

When t=0, obtain
x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means \frac{dx}{dt}.

Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1 
y'(0) = 1/2.
Here, y' means \frac{dy}{dt}.

Because \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}, obtain
\frac{dy}{dx} |_{t=0}\, =  \frac{1/2}{3}= \frac{1}{6}

Answer:
The slope of the curve at t=0 is 1/6.



3 0
3 years ago
What is 5y=2x+3 in standard form ?
tatiyna

the answer in standard form is 2x -5y = -3

4 0
3 years ago
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