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nydimaria [60]
3 years ago
6

Find an expression for f−1(x), if f(x) = 5 − 2x

Mathematics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

f^{-1}(x) = \frac{5-x}{2}

Step-by-step explanation:

Let y = f(x) and rearrange making x the subject

y = 5 - 2x ( add 2x to both sides )

2x + y = 5 ( subtract y from both sides )

2x = 5 - y ( divide both sides by 2 )

x = \frac{5-y}{2}

Change y back into terms of x

f^{-1} (x) = \frac{5-x}{2}

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Can someone explain the meaning of the Spiral of Theodorus?
drek231 [11]

Answer:

The Spiral of Theodorus is a construction of continuous right triangles, and each triangle has a side length of one representing the of the Pythagorean theorem, with the other sides filling in the spaces for the and in the theorem. hope that helped :)

Step-by-step explanation:

8 0
3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
30% of what equals 60​
Elena-2011 [213]
Create an inequality
x/60 = 30/100
x = 200
CHECK ANSWER: 200 x .30 = 60
3 0
3 years ago
Read 2 more answers
Please answer please
loris [4]
4/5 + a = 1
=1/5

*try using cymath it helps!
6 0
3 years ago
Read 2 more answers
A van traveling at a speed of 32.0 mi/h needs a minimum of 42.0 ft to stop. If the same van is traveling 67.0 mi/h, determine it
makkiz [27]

Answer:

Δx= 184.12 ft

Step-by-step explanation:

The equation you need to use is velocity as a function of displacement.

v^2= u^2 +2a(\delta x)

v = the speed at which the car is travelling,

and

v_o is the original speed (in this case zero).

The change in x (displacement) is how far the car travels. You will be solving for a (acceleration).

32^2= 0^2+ 2a\times42

solving we get

a= 12.19

now put this acceleration value into the second case when v= 67mi/h

67^2= 0^2 + 2\times12.19(change in x)

⇒Δx= 184.12 ft

5 0
3 years ago
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