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it says that is incorrect? thats strange. ok i got it thx using what you told me in was able to do the math and i got 1998.

Since the positive slope of the line appears greatest between  1997  and 1998, the year 1997 had the greatest increase....

19th....

No you aren't....I should pay more attention to my answers.....LOL!!!

Oh ok thanks! I'm stupid...

See my corrected solution....

Sorry, ACG....only x =   6 is correct.....x  = -1 as an answer to the first one isn't in the domain, either....

Thanks, Guest.....a slight correction....only x  = 6  is a solution...x  = 3  makes an original denominator 0, so that is not allowed

Sorry, but that would make the sum 5 and 5 isn't right... Am I missing something?

Let the cost without the tax  = C

So

C(1.06)  = 7.95       divide both sides by  1.06

C  = 7.95/1.06 ≈    7.5   =    $7.50

If f(x)  = 10....we have that

x^2  +  9  =  10       subtract  10 from both sides

x^2  -1  = 0       factor

(x + 1) (x - 1)  = 0      set each  factor to 0 and solve for x and we get that

x  = 1   or   x  = -1

But only the second answer is in the domain

Also

3x  -  8  =  10   subtract  10 from both sides

3x  - 18  = 0     factor

3(x  - 6)  = 0       divide both sides by 3

 x - 6  = 0       add 6 to both sides

x  = 6       and this is in the domain

So....possible  values of x are   -1   and   6

x+6/x-3=2x/x-3   Cross multiply

(x + 6) (x - 3) = 2 x (x - 3) 

 x (x + 3) - 18 =2x(x - 3)

x^2 + 3x -18 =2x^2 - 6x

x^2 + 3x - 18 - 2x^2 + 6x =0

-x^2 + 9x - 18 = 0, solve 

Factor the left-hand side.
The left-hand side factors into a product with three terms:
-((x - 6) (x - 3)) = 0

Multiply both sides by a constant to simplify the equation.
Multiply both sides by -1:
(x - 6) (x - 3) = 0

Find the roots of each term in the product separately.
Split into two equations:
x - 6 = 0 or x - 3 = 0

Look at the first equation: Solve for x.
Add 6 to both sides:
x = 6             or             x = 3

Go online to this page and read it carefully:

http://gbhsweb.glenbrook225.org/gbs/science/phys/class/sound/u11l2b.html

Here is a general solution. You translate it into the language they want!!.

FV=PV x [1 + R]^N, where FV=Future value, PV=Present value, R=Interest rate per period, N=Number of periods:

350 = 250 x [1 + 0.04/12]^N

350 = 250 x [1 + 0.003333333333]^N

350 = 250 x           1.003333333333^N   Divide both sides by 250

1.4 = 1.003333333333^N                        Take the log of both sides

N =Log(1.4) / Log(1.003333333333)

N =~101.1099 months, or about 8 years and 5 months.

Let the price of the Cooler before tax = C

C x [1 + 6%] =$7.95

C x 1.06 =$7.95   Divide both sides 1.06

C = $7.95 / 1.06

C =$7.50 - Price of the Cooler before 6% tax.

I'm a bit confused, with your formatting. . . 

When you say what is the exact value of x and you post a bunch of log equations are those the options? 

As for the second part 

f(x)= log x AND f(x)=log(14x) 

I know this is some sort of Horizontal stretch/compression. Horizontal stuff are always opposite. 

So 14 is actually 1/14=0.07 <---- This is obviously less than zero so it's a compression. 

So there is a horizontal compression by 1/14. Not sure about those options though... 

y=log(7x+3)-2 

For the x-intercept, you plug in 0 for y and solve. You get: (13.86,0) 

For the y-intercept: you plug in 0 for x and solve. You get: (0,-1.52) 

Hope that helps! 

Oh! Also, I accidentally clicked out of your message and couldn't figure out how to re-open it but, you were right about my calculator oops!

Thanks so much!

f(2)   =   2(2)²- 3   =   5

The first part of the function ends at the point  (2,5) .  So the second part of the function must go right up to that point. The line  f(x)  =  ax + 4  would pass through  (2, 5)  if  x  had no restrictions. So...

a(2) + 4  =  5

2a  =  1

a  =  1/2

12 - 8 = 4 m/s - difference in boat's speed due to the current

4 / 2  = 2 m/s - the speed of  the current.

CPhill's birthday messages 2017 

Happy birthday once again Chris !!!

Ah yes, in my own birthday messages from 2015.  

What a super sleuth you are Hectictar.

And how very lucky we all are that you found our little forum and made it your home  

What is the expression in radical form?

(3p^3q)^(3/4)

\(\begin{array}{|rcll|} \hline && (3p^3q)^{\frac34} \\ &=&\sqrt[4] {(3p^3q)^{3} } \\ &=&\sqrt[4] {3^3(p^3)^3q^3 } \\ &=&\sqrt[4] {27p^{3\cdot3}q^3 } \\ &=&\sqrt[4] {27p^9q^3 } \\ \hline \end{array}\)

Simplify.

cuberoot(27x^6y^3)

Assume all variables are nonnegative.

\(\begin{array}{|rcll|} \hline && \sqrt[3]{27x^6y^3} \\ &=& \sqrt[3]{3^3x^3x^3y^3} \\ &=& \sqrt[3]{3^3}\sqrt[3]{x^3}\sqrt[3]{x^3}\sqrt[3]{y^3} \\ &=& 3xxy \\ &=& 3x^2y \\ \hline \end{array}\)