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\(\frac{3}{8}-6=\frac{3}{8}-\frac{48}{8}=-\frac{45}{8}=-5\frac{5}{8}\)

expand and simplify

The circular table cloth had to be designed so that it completely covers a square table with sides 2 metres. What is the diameter of the smallest possible tablecloth which the designer could make?

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20 mins ago     

m(t)=85e^(-t/29) describes the amount of a radioactive isotope as a function of time t [ years]. After how long the amount has dropped to 1/10 of the original ?

 (e^x+3)/(4e^-x+5)=2

e^x + 3   =  2 (4e^(-x) + 5)

e^x + 3  = 8e^(-x) + 10      

e^x - 8e^(-x) - 7  = 0       multiply through by  e^x

e^(2x) - 7e^x - 8  = 0       factor

(e^x + 1)  (e^x - 8)  = 0           setting both factors to  0

e^x + 1   = 0    has no real solutions

e^x - 8 = 0     add 8 to both sides

e^x = 8         take the ln of both sides

ln e ^x  = ln 8

x * ln e   = ln 8       [ ln e = 1, so we can disregard this ]

x = ln 8 

f(x)=(x+1)²

Thanks, Omi......!!!!

If the exponent is even greater ........

m(t)=85e^(-t/409) describes the amount of a radioactive isotope as a function of time t [ years]. After how long the amount has dropped to 1/6 of the original ? The response can be written as t=alnb where a and b are integers and a=1

We are solving this :

85(1/6)  = 85e^(-t/409)    divide through by 85

(1/6)  = e^(-t/409)       take the natural log of each side

ln (1/60  = ln e^(-t/409)   and we can wriite

ln (1/6) = (-t/409)*ln e         and ln e = 1, so we can eliminate this

ln (1/6)  = -t/ 409      multiply both sides by -409

ln(1/6) * ( -409)   =  t  

(-409)[ ln 1 - ln 6]  = t

 (-409)(-ln 6)  =  t

(409)(ln 6)   = t  =    about 732.83 years

(x+1)^3

We can use the binomial theorem to expand this....so we have

C(3,0)*x^3*1^0 + C(3,1)*x^2*1^1  + C(3,2)*x*1^2 + C(3,3)*x^0*1^3  =

1*x^3 + 3*x^2 + 3*x + 1*1^3  =

x^3  + 3x^2  + 3x  + 1

Another method is to expand the expression as follows :

(x + 1) (x + 1) (x + 1)  =

(x^2 + 2x + 1) (x + 1) =

x^3 + 2x^2 + x + x^2 + 2x + 1  =   [combine like terms]

x^3 + 3x^2 + 3x + 1

Since this is a parabola that the turns upwards, the turning point occurs at the vertex. The x coordinate of the vertex can be found as : 

-b/ [ 2a] =   -14 / [2 (1)]  =  -14 / 2  = -7

The y coordinate of the vertex can be found by putting the x coordinate value back into the function

(-7)^2 + 14 (-7) + 33  =  

49 - 98 + 33  =

-49 + 33 =

-16

So.......the turning point is ( -7, -16)

hello

thank you

I have made a mistake. It should read as follows:

Sorry

You HAVE been busy!  Way to go...

Hello.....   you can graph it on this calculator (or others)

to find turning point ~  (-7, -16)

or you can take the derivative (which results in the equation of the SLOPE of the fxn)  and set it = 0 to get

2x + 14 = 0

2x=-14

x= -7       then substitute into the ORIGINAL equation to get 'y' at that point:

y= x^2 + 14x +33  =  49 - 98 +33 = -16

ANOTHER way is to find the 'zeroes' using the quadratic formula or factor the equation into

(x+11)(x+3)       to find the ZEROES as  -11 and -3    the turning point is exactly between these two at  x=-7

then substitute  - 7 into the equation to get y = -16

write the following numbers in the form m/n where m,n element of z. i)0.666666666666

Rational numbers

Agree with Melody EXCEPT there is a small error in her last line calculation:

-4      *3.0612^2+30*3.0612 = 54.35221824     should be

-4.9   *3.0612^2+30*3.0612 = 54.35221824    resulting in max height of 45.92 m

but how do u get th 9? thats is what i dont get

write the following numbers in the form m/n where m,n element of z. i)0.666666666666

Rational numbers

Find the coordinates of the turning point for the equation of the graph below. y=x2+14x+33

Max's equation.

\(\begin{align*}\sqrt[3]{(8^{2x})}&=\sqrt{2^{(4x-1)}}\\ (8^{2x})^{1/3}&=(2^{(4x-1)})^{1/2}\\ ((8^{2x})^{1/3})^6&=((2^{(4x-1)})^{1/2})^6\\ (8^{2x})^2&=(2^{(4x-1)})^3\\ ((2^3)^{2x})^2&=(2^{3(4x-1)})\\ (2^3)^{4x}&=2^{3(4x-1)}\\ (2)^{12x}&=2^{3(4x-1)}\\ 12x&=3(4x-1)\\ 12x&=12x-3\\ 0&=-3\\ &\mbox{No solutions}\end{align*}\)

OR

\(\begin{align*}\sqrt[3]{(8^{2x})}&=\sqrt{2^{(4x-1)}}\\ 8^{2x/3}&=2^{((4x-1)/2)}\\ (2^{2x})&=2^{((4x-1)/2)}\\ 2x&=2x-0.5\\ 0&=-0.5\\ &no \;solution\end{align*}\\ \)

Thanks Alan.  ://

Only true in the limit as x tends to minus infinity!

Thx Melody :D

CONGRATULATIONS   MAX !!!