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"It wasn't me Alan, I didn't put anything but I also noticed that error, I suppose I should have said so :))

the answer is 71

Why does Pe^(rt) work??

Look at 

lim n →∞   (1 + r/n)^n

Let  m = n/r    So we have

lim m →∞ ( 1 + 1/m)^(r*m)  = 

lim m →∞ [(1 + 1/m)^m ] ^r       but ...    lim m →∞ (1 + 1/m)^m   = e

So we have

e^r

And

Pe^[r(1)]  = P(e^r) = would represent continuous compounding for one year

So....

Pe^(rt)  =

P(e^r)(e^r)(e^r)...........   where (e^r) is multiplied "t" times ....{representing "t" years (or periods)......}

And that's it.......!!!

2/9 = 2 * 1/9 = 2 * 0.1(1) = 0.2(2)

At last, Nauseated

The die is cast

As Sisyphus hath said

And found

Such problems do not lend themselves

To Numbers

"Round"

One way to solve this problem, is to use the method of linear combinations:

4x  -  y  =  3     --->   multiply by -2     --->     -8x + 2y  =  -6

6x - 2y  =  5                                       --->      6x  - 2y  =  5

Add down the columns:                                  -2x          =  -1

Divide by -2:                                                            x  =  -1/-2   --->   x  =  1/2

Replace this value into the first equation:  4x - y  =  3   --->   4(1/2) - y  =  3  

                                                                               --->   2 - y  =  3   --->   -y  =  1   --->   y  =  -1  

Since the perimeter is 4 cm and all the sides of the square have the same length, each side must be 1 cm long.

The area of that square is 1 cm x 1 cm, which is 1 cm².

Since all sides are the same, the side length is just 1 cm.

And the area is just the side length squared = (1 cm)^2 = 1 cm^2   ....!!!!

The locus of points in a plane that are equally distant to two intersecting lines are lines which are the two bisectors of the given lines. (The distance from a point to a line is found by measuring along the perpendicular to the line.)

Since L1 is vertical and L2 are horizontal, the bisectors will be the two lines which have slopes of 45° (m = 1) and 135° (m = -1).

For this problem, you want the line that passes through the intersection point of the two lines (3, 3) with a slope of -1; defined by the equation y = -x + 6.

You can debate whether or not you want the point (3,3) included in the locus (Is a distance of 0 really a distance?); my preference is to include it.

Since velocity is a vector quantity the average velocity is just (-15 + 10) / 28  = (-5/28)m/s

Average speed is a scalar......it's just total distance/time = (15 + 10) /28m/s  = (25/28)m/s

I believe this is correct......

Well for speed he has traveled a total of 15m in 28 sec.

I know thats correct

now for velocity, direction counts. But you need magnitude so thats like absolute value.

well it  finishes 5m from the start point and it took 28sec to get there so the mag of the velocity is 5m/28sec.

I'm pretty sure that is correct ☺

Generally speaking, if the variable is in the exponent, use logs (either common logs or natural logs):

3^{x - 1}  =  24

Take the log of both sides:

log(3^{x - 1})  =  log(24)

The reason that logs are used is that the exponent within a logarithm comes out as a multiplier:

(x - 1)log(3)  =  log(24)

Now, divide both sides by the log(3):

x - 1  =  log(24) / log(3)                        If you want, simplify this by using a calculator ...

And, add 1 to both sides:

x  =  log(24) / log(3) + 1                       If you haven't used a calculator yet, use one now ...

very clear Thank you melody 

the radius of the hemisphere must be 15cm 

so all you have to do is change the R of the equation of what Melody had done from 9 to 15

1/4 *pi(6^2)= 28.2743338823081391

Hi 315

Average speed is 15m/ 28 sec

Magnitude of

Average vel = 5m/28sec

I think that's right :)

x²-2x- 8 = 0 solve by factorising quadratic equation

$$\\(-2)*(+4) = -8 \\
(-2) +4 = 2\\
\boxed{(x-4)(x+2) = x^2-2x-8=0 } \\
x_1 = -2\\
x_2 = 4$$

$${\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}^{{\mathtt{2}}} = {{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}$$

if you write your equation as this:$${\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}$$

you can use this rule:$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = \left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{b}}\right){\mathtt{\,\times\,}}\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)$$

finally you can write your  equation as:$$\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)\right)$$

so this means: 4 and -2 are your solutions. I hope I could explain

This is the cross section of the cone of solid X.    

The top bit is INSIDE the hemisphere.  So I am only interested in the bottom smaller cone.

Volume of little cone:

$$\\=\frac{1}{3}\pi r^2 h\\\\
=\frac{1}{3}\pi* 25* 15\\\\
=\frac{1}{3}\pi* 25* 15\\\\
=75\pi\\\\$$

Volume of hemisphere

$$\\=\frac{1}{2}*\frac{4}{3} \pi r^3\\\\
=\frac{4}{6}\pi* 9^3\\\\
=\frac{2}{3}\pi* 9^3\\\\
=486 \pi$$

total volume of  solid X = $$486\pi + 75\pi = 561 \pi\;\;cm^3$$

Now the ratio of surface areas of X:Y   =     25:36

so the ratio of lengths of   X:Y              =    5:6

and the ratio of volumes      X:Y           =   125: 216

$$\\\frac{216}{125}=\frac{Y}{561\pi}\\\\
\frac{216}{125}\times 561\pi=Y\\\\
Y=\frac{216}{125}\times 561\pi\\\\
Volume\;of\;Y=\frac{121176\pi}{125}\;\;cm^3\\\\$$

Thanks anon :))

I have taught in High Schools and in a uni college for many years but I am semi-retired now :))

As an addendum.....apparently......our calculator never heard of this, either...LOL!!!

If you key in  !N, where N is some positive integer........nothing.....!!!!

A long with being true that is quite funny, CPhill. You should get a professional troll’s license. It’s to die for!

Now, for a very important question: How could we do it with roman numerals?

I have just added this thread to our reference material sticky note.

I do not know if other people use this but there are some really good posts in there and I use them quite often. 

Actually Melody , as a secondary school's students , I really appreciate what you and the other helpers had done for those who are frustrated in maths , good job! Besides , are u a maths teacher in school?

Thanks Heureka  :)))