All Answers 358516 Answers

I wish.

I'd shout you all an icecream if I was.

And I'd pay for you to have a holiday in  Australia so that you could collect your shout!

Does that sound good!

Kids can even bring their parents if the really want to!

Then I'd get to meet you all!

That is assuming that the fruit is all differerent and arranged in a straight line.  8!

If it was arranged around a circular platter then it would be 7!

If they were just randomly placed in the universe then the number of possibilities would be infinite!

I'm sorry - I just realised that Chris said some of this already.

Where abouts can you do this Rosala?

A parabola is defined by the equation . In which direction will the parabola open?

$$\boxed{y=ax^2+bx+c} \qquad \mbox{ parabola}$$

a > 0     opens upward 

a < 0     opens downward

$$\\x^2={3 \over 4}y\\\\
y=\underbrace{{4 \over 3}}_{ a > 0} x^2$$

a > 0 parabola opens upward

You got the right answer Rosala, but you made it look really hard.

it is just

22/88*100% = 25%

e^2x + 1 = 900   ?

$$\begin{array}{rcrc}
e^{2x+1} &=& 900 & \quad | \quad \ln{}\\
\ln{(e^{2x+1} )}&=&\ln{( 900 )} & \\
(2x+1)*\ln{( e )}&=&\ln{( 900 )} & \quad |\quad \ln{(e)} = 1 \quad !\\
2x+1 &=&\ln{( 900 )} & \\
2x&=&\ln{( 900 )} - 1& \\
x&=&{ \ln{(900)} -1 \over 2 } &\\
x&=&2.90119738166&\end{array}$$

9^−x/200 = 2  ?

$$\begin{array}{rcrc}
9^{
(-{x\over 200}) }
&=& 2 & \quad | \quad \ln{}\\

\ln{(
9^{
(-{x\over 200}) }
)
}&=&\ln{(2 )} & \\
(-{x\over 200}) }
*\ln{(9 )}&=&\ln{(2 )} & | \quad *\quad -({200\over \ln{(9)}})\\
x&=&-200*{\ln{(2)}\over\ln{(9)}} &\\
x&=&-63.0929753571&
\end{array}$$

There are still an infinite number of solutions.

7^0.9x = 6   ?

$$\begin{array}{rcrc}
7^{0.9x} &=& 6 & \quad | \quad \ln{}\\
\ln{(
7^{0.9x} )}
&=&
\ln{(
6 )} & \\
0.9*x*\ln{(
7 )}
&=&
\ln{(
6 )} & \\
x&=&{
\ln{(6)}
\over
0.9 *\ln{(7)}
}
&\\
x&=&1.02309135685&
\end{array}$$  

I think that this is what you mean

$$2x^2+4x-3=0$$

Hi Rosala,

The school system is different all over the world and the names and grades can be very confusing.

In most of Australia High School is from year 7 to year 12.   (Starts at 11 or 12 years old and finishes at 17 or 18 years old)

Junior High is years 7 to 10 and 

Senior high is years 11 and 12.

Just because a student can use a calculator in a test does not mean that the test has to be easy.  Sometimes it is harder because they teachers have to put in tricky questions. 

x ² 3 ^x −  4( 3 ^x) = 0    ?

$$\\x^2 3^x-4(3^x)=0 \\
3^x
(
x^2-4
)
=0 \\
3^x
(
x-2
)
(
x+2
)
=0\\\\
\underbrace{3^x}_{=0}
\times(
\underbrace{x-2}_{=0}
)
\times(
\underbrace{x+2}_{=0}
)
=0 \\\\
\boxed{3^x=0} \quad | \quad \ln{} \\\\
\ln{(3^x)} = \ln{(0)} \\
x\ln{(3)} = \ln{(0)} \\
x=
{
\ln{(0)}
\over \ln{(3)} }
} \quad | \quad \ln{(0)} \mbox{ no solution !}\\\\
\boxed{x-2=0} \quad \Rightarrow \quad \boxed{x=x_1=2}\\\\
\boxed{x+2=0} \quad \Rightarrow \quad \boxed{x=x_2=-2}$$

Take a look here

e^6x + 5e^3x − 14 = 0 ?

$$\\e^{6x}=e^{3x}e^{3x} \quad | \quad \mbox{ set }\quad \boxed{z=e^{3x}}\\
z^2 +5z -14=0\\
\underbrace{1}_{a=1}z^2 \underbrace{+5}_{b=5}z \underbrace{-14}_{c=-14} =0\\
\boxed{z_{1,2}=
{
-b\pm\sqrt{b^2-4ac}
\over
2a
}
}\\\\
z_{1,2}=
{ -5\pm\sqrt{25-4*1*(-14)}
\over
2*1 } \\\\
z_{1,2}=
{ -5\pm\sqrt{25+56}
\over
2} \\\\
z_{1,2}=
{ -5\pm9\over
2} \\\\
\boxed{z_1= 2 \quad z_2 = -7} \\\\
e^{3x}=z \quad | \quad \ln\\\\
\ln{(e^{3x})}=\ln{(z)}\\\\
3x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\
\boxed{x={1\over 3}\ln{(z)}} \\\\$$

$$\\x=x_1={
\ln{(2)}\over 3}
}
=0.23104906019\\ x=x_2={
\ln{(-7)}\over 3}
}
= \mbox{no solution !}\\\\
\boxed{x=0.23104906019}$$

I like it too - thanks anonymous!

Thanks Anonymous,

If I didn't look at every question when I write a wrap I would have missed this.

Thanks for sharing.

I will put it in with our puzzles too I think.   

2x+2/y=4w+2   what does X equal?

2x=4w+2-2/y

x=2W+1-1/y

e^2x − 9e^x + 8 = 0  ?

 $$\\e^{2x}=e^xe^x \quad | \quad \mbox{ set }\quad \boxed{z=e^x}\\
z^2 -9z +8=0\\
\underbrace{1}_{a=1}z^2 \underbrace{-9}_{b=-9}z \underbrace{+8}_{c=8} =0\\
\boxed{z_{1,2}=
{
-b\pm\sqrt{b^2-4ac}
\over
2a
}
}\\\\
z_{1,2}=
{ 9\pm\sqrt{81-4*1*8}
\over
2*1 } \\\\
z_{1,2}=
{ 9\pm\sqrt{81-32}
\over
2} \\\\
z_{1,2}=
{ 9\pm7\over
2} \\\\
\boxed{z_1= 8 \quad z_2 = 1} \\\\
e^x=z \quad | \quad \ln\\\\
\ln{(e^x)}=\ln{(z)}\\\\
x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\
\boxed{x=\ln{(z)}} \\\\$$

$$\\x=x_1=\ln{(8)}=2.07944154168\\
x=x_2=\ln{(1)}=0\\
\boxed{x_1=2.07944154168 \qquad x_2=0}$$

$${{\mathtt{23\,245}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{0.035\: \!039\: \!412\: \!554\: \!852\: \!7}}$$

You just do it straight on a calculator.

Indeed, I should have said spheres.

why is 1 not a prime number? why?

Prime numbers are never divisible by other prime numbers. If was 1 a prime number, would be valid this no more, because every number is divisible by 1

3.2=1.05^t  what does t equal?

$$\begin{array}{rcl}
3.2 &=& 1.05^{\textcolor[rgb]{1,0,0}{t}} \quad | \quad \ln{}\\ \\
\ln{(3.2)} &=& \ln{(1.05^{\textcolor[rgb]{1,0,0}{t}}} )} \\ \\
\ln{(3.2)} &=& \textcolor[rgb]{1,0,0}{t}} \times \ln{(1.05)} \\ \\
\textcolor[rgb]{1,0,0}{t}} &= &{\ln{(3.2)} \over \ln{(1.05)}
} \\ \\
\textcolor[rgb]{1,0,0}{t}} &= &{1.16315080981\over 0.04879016417
} \\ \\
\textcolor[rgb]{1,0,0}{t}} &= & 23.8398625954
\end{array}$$

e  ^x = 3   ?

$$\begin{array}{rcl}
\\e^x &=& 3 \quad | \quad \ln{} \\
\ln{(e^x)} &=& ln{(3)}\\
x\ln{(e)} &=& ln{(3)} \quad | \quad \ln{(e)} = 1 \;! \\
x &=& ln{(3)}\\
x &=& 1.09861228867
\end{array}$$

$$\boxed{x=\ln{(3)}}$$

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