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The interest on this is given by

I = Prt     

Here  P = $4000  r = 5% = .05   and t = 2 years 3 months = 2.25 years

So we have

I = 4000(.05)(2.25) = $450

So the total value is  $4000 + $450 = $4450

4 lb 12 oz  + 7lb 11 oz

Adding the ounces we have 11 + 12 = 23 oz

And since there are 16 ounces in one pound, this equals 1 lb 7 oz

And adding back the rest, we have

4 lb + 7lb + 1 lb + 7 oz  =

12 lbs 7 oz

Melody actually didn't  quite carry her calculations far enough.

We have

(n-3)(n-2)(n-1)n ≥ 8760

n⁴ - 6n³ + 11n² - 6n  ≥ 8760

Using WolframAlpha  ...we get the following  (somewhat messy) solution....

n  ≥ (3/2)  + (1/2)* sqrt(5 + 4√8761) =

n  ≥  about 11.239

Which confirms that we need more than 11.... (which means 12)

Her way is actually superior to my "guess and check" approach....!!!

But.....as she always says..."All roads lead to Rome"

If he combined them all.....there would be 1....

{It's an old joke....}

The square has equal sides and all vertex angles are right angles.

The rhombus has equal sides, and pairs of consecutive vertex angles angles are supplementary, but not right angles. And each pair is equal to the other.

Let's look at this one, again....I assume it's really supposed to be

 4x^2-12x+9

This is a perfect square trinomial....the factorization is

(2x - 3) (2x - 3)  = (2x - 3)²

sin(54)=2.44/p find = p

Rearranging, we have

p = 2.44 / sin(54)  = about 3.016

$${\mathtt{968}} \,{mod}\, {\mathtt{44}} = {\mathtt{0}}$$

there is not remainder

Good one Chris  :))

you need to use brackets

sqrt(144/4)

$${\sqrt{{\frac{{\mathtt{144}}}{{\mathtt{4}}}}}} = {\mathtt{6}}$$

I want brackets here please.

I think this is it

1)  there are 4 choices for the first row, the X can go anywhere.    Just put your X in

2)  There are only 2 choices for the next row because it cannot go in the same box as the first one did.

3)  There are 2 choices for the next row because it can go in either box but it cannot go under either of the 2 x's that are already there.

4)  There is only one possibility left for the last row

so the number of possibilities is        4*2*2*1=16

Use the compoound interest formula but remember that

n is the number of compounding periods =  30

and

r is the interest rate as a decimal for each of these.        0.058/12 = 0.00483  repeater

Now just plug the numbers in

$$FV=PV(1+r)^n$$

I saw a different pattern again

8,16,20,.....

I saw 16 as  3*4+4     (I added the corner blocks seperately)

I saw 20 as   5*4+4    (again I added the corner blocks afterwards.)

Looking at the patter I would get  8=1*4+4   that's true, great,

so I have

n                                 1           2            3

number of squares    1*4+4,  3*4+4,   5*4+4,  ...... 

now lets look at a pattern for 1,3,5 etc ...    they are getting bigger by 2s ....    mmm,    (2n-1)

So   

$$\\T_n=(2n-1)*4+4\\
T_n=4(2n-1)+4\\
T_n=4[(2n-1)+1]\\
T_n=4[2n]\\
T_n=8n\\$$

so

$$T_{100}=8*100=800 squares$$

Yet another proof that

(Mine is the same as CPhill but I think he overlooked a couple of possibilities )

I will need to either choose 2 digits or 1 digit

there are 4C2 = 6 ways of choosing 2 digits and 4 ways of choosing just one digit

Say I chose 2 digits and they are A and B  the palindomes i can make are

AABB

ABBA

BBAA

BAAB          there are only 4

If i choose just one letter, sat A,  then there is only one way to make a palindome, i.e.  AAAA

So

there will be   6*4+4*1 = 24+4 = 28 ways.

I think it is 2 of the 10 boys OR 2 of the ten girls, and order counts so that is

10*9 + 10*9 = 180

or

10P2 + 10P2 = 180

I  used formula to get a vague starting point of 10 but then I would do it by trial and error after that the same as CPhill did. 

$$\\^nC_4\ge 365\\\\
\frac{n!}{4!(n-4)!}\ge 365\\\\
\frac{(n-4)!*(n-3)*(n-2)*(n-1)*n}{24(n-4)!}\ge 365\\\\
(n-3)*(n-2)*(n-1)*n\ge 365*24\\\\
n^4.....\ge 8760\\\\$$

so     $${\sqrt[{{\mathtt{{\mathtt{4}}}}}]{{\mathtt{8\,760}}}} = {\mathtt{9.674\: \!444\: \!255\: \!582\: \!887\: \!1}}$$

My first 'guess' will be 10.    Now I will jut try 1 by 1 till I hone in on the right answer.   :)

so we'll do this like.... -1 + (-2) so,

- + = - so in + (-2) we'll put -2 so

= -1 -2

that means add the twoo and put a minus sign!

= -3

sooo easy!

Oh I see the problem - I am glad that you have been persistant :)

this problems is

20÷2(5+5)

$$=20\div2\times(5+5)$$

parentheses first

$$=20\div2\times10$$

Now, division and multiplication are equally important. The division comes first so you do it first.

$$=10\times10$$

now multiply

$$=100$$

This is a good question.

$$\displaystyle\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=2.718281828.....\qquad=e\\$$

This limit is used all the time (for continual compounding) so it was decided to give this number a formal symbol, that symbol is e      

e is called Euler's number after the Swiss mathematician Euler,  it is also called the Napier constant.

Here is the graph to show you

just move the point over 8 places in the direction that will make the number bigger. 

You will need to add some zeros to jump over

asin(2/3)

now enter it on the site calc 

arc sine is the same as inverse sine

I expect the 4! is supposed to be on the bottom so you need more brackets.

There is a calc on the home page.  :)

Hi Mathcalc,

The question is very vague but

I divided year 2 total by year 1 and got 1.105     So Y1*1.105=Y2

 I divided year 3 total by year 2 and got 1.05      So Y2*1.05 = Y3

I need to estimate the number of yellow b***s for Y4 so I decided to mult Y3 by 1.1    (it is between 1.105 and 1.05)

237108*1.1= about 260800

So I am going to estimate the number of yellow b***s as 260800

There are 5 times more white b***s    5*260800 = 1304000 white b***s

260800+1304000= 1564800 total b***s

1564800/12 = 130,400 dozen b***s in year 4   (estimated)

130400*$23.94=$2,475,396

So that is about  2.5 million dollars