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The answer is 1. Any number to the power of 0 will always be 1

1.06^x=3.4(1.02)^x    divide both sides by (1.02)^x

[ 1.06 ] ^x  / [ 1.02]^x    = 3.4      and we can write this as

[ 1.06 / 1.04 ] ^x   =  3.4           simplify

[ 53 / 51 ] ^x   = 3.4      take the log of both sides

log [ 53/51 ]^x  = log  [ 3.4]      divide both sides by log [53/51]

x =  log [ 3.4] / log [ 53 / 51]  ≈  31.814

2/12 + 2/4   =

1/6 + 1/2          get a common denominator  = 12

2/12 + 6/12   =

8/12      reduce

2/3

(5/6)  / (1/5)      invert the second fraction and multiply

(5/6) * (5/1)  =

25 / 6    =     4 + 1/6    servings

8c^7  = -0.72     divide both sides by 8

c^7   = -0.09       take the 7th root of both sides

c  = (-0.09)^1/7

c  ≈  -0.70893    [ real solution ]

50^325  ≈   1.463023860841311977236202941830836428098444793373182... × 10^552

Thank you!

Note that "Q"   is  "a" units from "O"   and since   OQ  = PQ,  then  P  must lie the same  "a" units from Q

So....the coordinates of "P"   must be  ( -a, a )

A   2x - 3y = 12       y = -(2/3)x + 5

Solving for y in the first equation, we have,  y = (2/3)x - 4........the slope would have to be the same as in the second equation to be parallel and it would have to be the negative reciprocal of the second to be perpendicular......since it's neither, these lines are not parallel or perpendicular

B   3x + 2y = 6      2x - 3y = 7     rearrange both  in terms of y

y = -(3/2)x  +  3    y =  (2/3)x + 7/3   the solpes are negative reciprocals.......these lines are perpendicular

C   y = 4x + 13       y = (1/4)x - 13   ......neither

D   x + y  = 1    2y = -2x + 2      dividing the second equation through by 2, we have   y  = -x + 1   and adding x to both sides we have     x + y = 1    ...........these lines llie on top of each other  [ they are the same line ]........so they must have the same slope, but they are  not "parallel" in the true sense of the word since parallel lines never intersect

1  /  [r^3 + 7]   - 7   =  -r^3 / [ r^3 + 7 ]    rearrange

1 / [ r^3 + 7 ] + r^3 / [ r^3 + 7]    =  7

[ r^3 + 1 ] / [ r^3 + 7 ]   = 7       cross-multiply

r^3 + 1    =   7 [ r^3 + 7 ]

r^3 + 1 =  7r^3  + 49

-6r^3   =    48     divide both sides by -6

r^3  =  -8       →   r  =  -2

Here's the way I see this one.......

Let x be the fraction......and we have

 [(1/4) + (1/6) - x ] / 3  =  1/12    simplify

[ 10/24 - x] / 3  =  = 1/12   multiply both sides by 3

10/24 - x =  3/12      simplify

5/12  - x =  1/4         rearrange

5/12 - 1/4  = x      simplify again

5/12  -  3/12   = x

2/12  = x    

1/6 = x

We have, for the first question :

y = log (15) + 7.9  = 9.07  →  9.1

For the second, we have :

8.6 = lox x + 7.9        subtract 7.9 from each side

8.6 - 7.9  = log x

0.7 = log x    this says that x  =    10^0.7  ≈   about 5 times

The correct answer is  D

Solve for x:

1/4 + 1/6 - 1/x =1/12 , solve for x
5/12-1/x = 1/12

Bring 5/12-1/x together using the common denominator 12 x:
(5 x-12)/(12 x) = 1/12

Cross multiply:
12 (5 x-12) = 12 x

Expand out terms of the left hand side:
60 x-144 = 12 x

Subtract 12 x-144 from both sides:
48 x = 144

Divide both sides by 48:
Answer: |  x = 3  OR THE FRACTION IS: 1/3

Zero.

Go online to this link and see the accurate solution to your inequality:

https://www.symbolab.com/solver/inequalities-calculator/2000%5Cleft(1.020%5Cright)N-1%3E2000%2B%5Cleft(N-1%5Cright)50

https://www.desmos.com/calculator/lnr8fgakmj

Hi :D How can I help you?

Yes and just type 10+(-2) in your calculator

Logarithmic functions takes the form \(f(x)=log_ax\), and if y=\(log_ax\), \(a^y=x\), and that's why some people say that logarithm is the reverse of exponentiation.

The question didn't mention x, y, and z should be different integers. It is correct that it's the form of Fermat's Last Theorem but the conditions aren't the same or it would be so hard for a student. Therefore geno3141's answer is correct.

\(\color{blue} 52-\frac{-80}{-40}+ \frac{-5}{-5}\)

\(\color{blue}= 52-2+1\)

\(\color{blue}=50+1\)

\(\color{blue}=51\)

\(\color{blue}x^{\frac{a}{b}}= \sqrt[b]{x^a}\)

The solution sets are {-1,-1} and {2,5}

Graphical Representation:

https://www.desmos.com/calculator/jft8sfnkxr

\(\color{aqua} \frac{21}{4}\)

\(\color{aqua}= \frac{20}{4}+\frac{1}{4}\)

\(\color{aqua} = 5 + \frac{1}{4}\)

\(\color{aqua}=5\frac{1}{4}\)

what do you mean by that? That isn't a question!