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2.5 x 100 x 10^3

=2.5 x 10^5

e^.5x=30   take the ln of both sides

ln e^.5x= ln 30   and by a log property, we can write

(.5x) lne = ln 30      remember, ln e is just 1    ... so we have....

.5x  = ln 30           

(1/2) x = ln 30     now, just multiply both sides by 2 and that will give you your answer...!!!

Thanks Chris,    

If there were less questions I would have drawn numberlines to demonstrate the answers.  

I suppose I still can if the asker request is.  

Nice presentation, Melody...!!!

The coefficient of x^2 is 1/2 whid is POSITIVE

So the parabola point UP

it is CONCAVE UP

3x>7 or (-7/8)x-5>7 [This will be a Union]

x>7/3 or (-7/8)x>12

x>7/3 or x<12*(-8/7)

$$x>2\frac{1}{3}\qquad or\qquad x<13\frac{5}{7}$$

This includes ALL real numbers       $$(-\infty,+\infty)$$

x>-5 or x<=8             [This will be a Union]

Same as above

This includes ALL real numbers       $$(-\infty,+\infty)$$

2x-8<-18 and -6x+6>-6

2x<-10 and -6x>-12

x<-5 and x<2

x<-5

$$(-\infty,-5)$$

If the first is meant to be 11*5 = 55 and the second is 0.0011*5 = 0.55 then the second is wrong!  It should be 0.0011*5 = 0.0055

.

Here's one way....

Note that  300 / 6 = 50

But 294 is one less 6 than 300

So 294 / 6   must just be 49 "6s"

Here's a slightly different approach to Melody's

Note that we can write .5  as 1/2  = 2^-1 

So we have

log₂(19)  + log_0.5 (19)   .....and...using the"change of base" rule, we can write

log(19) / log(2)  +  log(19) / log(.05) =

log(19) / log(2)  + log(19) / log(1/2) =

log(19) / log(2)  +  log(19) / log2^-1 =

log(19) / log(2)  +  log(19) / (-1) log2 =

log(19) / log(2)  +  [ (-1) log(19) /  log2 ] = 

log(19) / log(2)  -  log(19) /  log2 =   0

Let one boat be distance d1 from the observer, and the other be at a distance d2.

 We have two right-angled triangles for which we know the angle and the opposite side (the height), so we can use the tangent function to find the adjacent sides (the distances from the observer).

tan(31°) = 95/d1  or d1 = 95/tan(31°)

similarly d2 = 95/tan(42°)

so distance between boats is d1 - d2

$${\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{31}}^\circ\right)}}}{\mathtt{\,-\,}}{\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{42}}^\circ\right)}}} = {\mathtt{52.598\: \!361\: \!914\: \!428\: \!964\: \!3}}$$

The boats are ≈ 52.6 feet apart.

.

LOL, Alan!!!!...yeah.....Usain Bolt couldn't even beat this guy.......no matter what the distance....!!!!

And he would put the  Kenyans to shame in the marathon.....completing it in about 43 minutes....!!!

I'll do a couple for you:

A.  x + 2y = 17

     x - y    = 2

Subtract the second equation from the first, term by term

x - x + 2y - (-y) = 17 - 2

or 3y = 15

Divide both sides by 3

y = 5

Put this back into the first equation

x +2*5 = 17

x + 10 = 17

x = 17 - 10

x = 7

B. 4x + 5y = 11

    2x + 6y = 16

Double all the terms in the second equation to get the number of x's the same as in the first equation

   4x + 12y = 32

Subtract the first equation from the third equation term by term

   4x - 4x + 12y - 5y = 32 - 11

or 7y = 21

Divide by 7

y = 3

Put this back into the first equation

4x + 5*3 = 11

4x + 15 = 11

4x = -4

x = -1

.

$$\\-50<10x<50\\
so\\
-5

1/4 of a mile in 25 seconds is 1 mile in 100 seconds.  There are 3600 seconds in one hour so there are 36 lots of 100 seconds in one hour.  This means the guy can run 36 mph (must be the six-million dollar man!).

. 

I would have thought is was 

|2n-15|  which is the same as  |15-2n|

So EITHER  A or B    

Not both, Not neither     

$$(3x^2-4x+7)(-2x^2+5x-9)=-6x^4+23x^3-61x^2+71x-63$$

Think of this as:

$$3x^2(-2x^2+5x-9)-4x(-2x^2+5x-9)+7(-2x^2+5x-9)$$

Then multiply the 3x² by all the terms inside the brackets; do the same for the -4x and the +7; then collect like terms together.

.

Mmm interesting.

Let 

$$\\x=log_{0.5}(19)\qquad and \qquad y=log_2(19)\\
so\\
19=0.5^x\qquad\qquad and \qquad 19=2^y\\
19=(\frac{1}{2})^x\\
2^x=\frac{1}{19}\\
so\\
2^x\times 2^y=\frac{1}{19}\times 19\\
2^x\times 2^y=1\\
2^x\times 2^y=2^0\\
2^{x+y}=2^0\\
$therefore$\\
x+y=0\\
so\\
log_{0.5}(19)+log_2(19)=0$$

It located near my house! I mean u just have to cross ten houses then the sea , then another ten houses then 5 factories , 1 river and another sea and there is the island! If u reach there.....don't bring help my brother ....I mean don't disturb him when he's with his fellow monkeys ! Lol!

Awe ....what a sad story.....!words are always with me.......untill and unless I have a bad temper! Words just leave me alone becoz they don't want to feel embarrassed! Lol!

Nice answer, rosala....!!!

BTW.....where, exactly, is the "Island of Fractions" located????

x^3 = 2x+1

$$x^3 -2x-1 = 0 \qquad \Rightarrow x_1=-1 \\ \\
(x^3 -2x-1) : (x+1) = x^2-x-1 \\\\
(x+1)\underbrace{(x^2-x-1)}_{=0} = 0\\\\
x^2-x-1 = 0\\
x_{2,3}=\frac{
1\pm\sqrt{1-1*4*(-1)}
}
{2*1} = \frac{
1\pm\sqrt{5 }
}
{2} \\ \\
x_2 = \frac{1+\sqrt{5}}{2} =1.61803398875 \\\\
x_3 = \frac{1-\sqrt{5}}{2} =-0.61803398875$$

Yes, rosala.....words have failed me.......!!!

x^3 = 2x+1    subtract everything on the right side to make it 0....so we have....

x^3 - 2x - 1  =  0     and from the Facror Theorem, we have that -1 is a root (solution)

Therefore, using a little synthetic division, we have

-1  [  1    0   - 2    -1 ]

            - 1     1     1

       -----------------------

       1     -1   -1     0

This tells us that the polynomial remaining after we divide  x^3 - 2x -1  by (x + 1)  = x^2 - x - 1

And setting this = 0, we have that x =

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$

So these are the 3 solutions......BTW.....the last 2 solutions are better known as  "-phi" and "Phi"......

This is the magical map that will take you to the Island Of Fractionts! Here is your help!

Wooooaaahhh!!!!

Sir CPhill is having a loss of words! How SaaaaaDDDD!!!!!