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3 years ago

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A powered by B equals B powered by A. A does not equal B. What's A and what's B?

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

Step-by-step explanation:

I know that there is a zero and a 1 involved in this somewhere. The only answer I can get which is a bit of a cheat, is

1^0! = 0!^1

These are not exactly the same thing. The trouble is that they are equal. There is a difference between "not the same thing" and "equal."

If I think of something else, I'll put it in a comment.

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Let f(x) = x+2/x+6 <br> f^-1(-6) =

GREYUIT [131]

We can explicitly find the inverse. If $f^{-1}(x)$ is the inverse of $f(x)$ , then

$f\left(f^{-1}(x)\right) = \dfrac{f^{-1}(x)+2}{f^{-1}(x)+6} = x$

Solve for the inverse :

$\dfrac{f^{-1}(x) + 2}{f^{-1}(x) + 6} = x$

$\dfrac{f^{-1}(x) + 6 - 4}{f^{-1}(x) + 6} = x$

$1 - \dfrac4{f^{-1}(x) + 6} = x$

$1 - x = \dfrac4{f^{-1}(x) + 6}$

$f^{-1}(x) + 6 = \dfrac4{1-x}$

$\implies f^{-1}(x) = \dfrac4{1-x} - 6$

Then when x = -6, we have

$f^{-1}(-6) = \dfrac4{1-(-6)} - 6 = \dfrac47-6 = \boxed{-\dfrac{38}7}$

Alternatively, we can first solve for x such that $f(x) = -6$ . Then taking the inverse of both sides, $x = f^{-1}(-6)$ . (The difference in this method is that we don't compute the inverse for all x.)

We have

$\dfrac{x+2}{x+6} = -6$

$x + 2 = -6 (x + 6)$

$x + 2 = -6x - 36$

$7x = -38$

$\implies x = \boxed{-\dfrac{38}7}$

8 0

2 years ago

Find the coordinates of point P along the directed line segment AB with endpoints from A(2, 7) to B(-1,1) with the given ratio o

Umnica [9.8K]

The answer is 5, 9 if its not than it is 3,4 that is the answer no problem

7 0

3 years ago

Write x as a logarithm in the following: 10^x=7

pochemuha

Answer:

x = log 7

Step-by-step explanation:

10^x=7

Take the log base 10 on each side

log10 (10^x)=log10 (7)

We dont need to write the 10, it is implied

log(10^x)=log (7)

The exponent goes out front since log a^b = b log a

x log 10 = log 7

log 10 = 1 since the base is 10 log10(10) =1

x = log 7

6 0

2 years ago

Read 2 more answers

Someone help me please

natali 33 [55]

Part A:
1) 16
2) -25
3) 9
4) -7

I'm too lazy to do the rest, just pay attention in class next time

5 0

3 years ago

Given cscx/cotx=sqrt2, find a numerical value of one trigonometric function of x.

Firlakuza [10]

$\frac{csc (x)}{cot (x)} = \sqrt{2}$

⇒ $\frac{csc (x)}{1} * \frac{1}{cot (x)} = \sqrt{2}$

⇒ $\frac{1}{sin (x)} * \frac{sin(x)}{cos (x)} = \sqrt{2}$

⇒ $\frac{1}{cos (x)} = \sqrt{2}$ ; sin(x) ≠ 0, cos(x) ≠ 0

⇒ $\frac{cos (x)}{1} = \frac{1}{\sqrt{2}}$ ; sin(x) ≠ 0, cos(x) ≠ 0

⇒ $cos (x) = \frac{\sqrt{2}}{2}$

Use the Unit Circle to determine when $cos (x) = \frac{\sqrt{2}}{2}$

Answer: 45° and 315° $(\frac{\pi}{4} and \frac{7\pi }{4} )$

8 0

2 years ago

Read 2 more answers