All Answers 358456 Answers

$${\mathtt{0.2}}{\mathtt{\,-\,}}{\mathtt{0.5}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{10}}}} = -{\mathtt{0.3}}$$

 That equals to, 93.25

7x-(x+4)=-7   I think you'd actually want to solve this !!!  ...... simplify first........

7x - x - 4 = -7

6x - 4 = -7      add 4 to both sides

6x = -3          divide both sides by 6

x = -3/6 = -1/2

We can model this with the following exponential function:

A = 100(.90)ⁿ    where A is the value of the account after n withdrawals

So, after 10 witdrawals,we have

A = 100(.90)¹⁰ = $34.87  left after 10 withdrawals

This seems counter-intuitive....but we are withdrawing 10% of a smaller and smaller balance  as n increases. Theoretically, since an exponential function is never 0, the account is never emptied - there always remains some fraction of one cent, even after infinite withdrawals !!!! (However, the limit of this fuction as n approaches infinity = 0)

6% written as a decimal is .06

To calculate 6% of 29, find .06 x 29.

There is a calculator at this site to help you.

It is worth knowing that 1/3 = .333333...

and                              2/3 = .666666...

So,  6.66666...  = 6 2/3  or  20/3.

20000  =  9330(1.10)^n

First, divide by 9330:

20000/9330  = (1.10)^n

To solve an equation that has the variable as an exponent, find the log of both sides:

log(20000/9330)  =  log(1.10^n)

The reason to find the log of both sides is that an exponent in the log will come out s a multiplier:

log(20000/9330)  =  n·log(1.1)

Divide both sides by log(1.1);

log(20000/9330) / log(1.1)  =  n

Enter this into a calculator to get your answer.

cot(x) = 1/tan(x)

The period of tan(x) and cot(x) is the same; it equals π.

If there is a multiplier of x, so that you have cot(ax), the period will be π/a; this means that the period of this function is π/(2π) = 1/2. The entire function will make one period between 0 and 1/2.

At 0, tan(x) = 0, so cot(x) is undefined; because the period is 1/2, it will also be undefined at 1/2.

Halfway between, at 1/4, 2cot(2π(1/4)  =  2cot(π/2) = 0.

The graph starts with a y-value of +∞ at x = 0, waves down through the point (1/4,0), and continues down to   -∞ at x = 1/2.

Graph it on a calculator, restrict the x-values to between -1/2 and +1/2.

STU, we tried several times to make a video conference call.

We have some fine talent on our end:

Chimp Stanley. A great engineering chimp, whose skill set includes vacuum tube theory.

Chimp Stuart. A lower level engineering student, who also dresses funny, but is a great translator of human and near human languages.

We made a successful video conference link for only a few moments.

We see you have a modern organic banana phone.

We use modern organic banana phones too.

After several more unsuccessful attempts, we had Chimp Loki activate your remote viewer.

We now understand. That is a problem with those modern, organic banana phones; they know they will be eaten after the call is completed, so they run for their lives!

Anyway, if you catch up to your banana phone, give us a call before you eat it.

Lancelot_Link Ph.M.E. (Philosophy of Monkey Engineering) and Top Banana of A.P.E.

Hi robtre

Q2 doesn't make sense ans can't you do 1 and 3 yourself?

There is a calculator on the home page for Q1.  

turn would be better than flip.  

on the web2 calc on the home page enter

nCr(49,6)

Hi anon,

let's look at your answers

6+6-9*9=12-81=-69

6-6-9-9=0-9-9=-18

So they do not work.   

a) The blocks form a banging noise when they are together.

b) it depends when i slopes onto a 1 degree angle and a 60 degree angle.

you measure it, duh.

the same number when you flip it upside-down.

Anon #23, it was even (although that some people think it is odd.).

Thumbs up to Chris for spotting an error

and thumbs up to Geno for all his effort.    :)

a) Graph it, either byhand or by calculator.

b) The h-intercept represent the height of the log.

c) The maximum height occurs at about 0.0765 secs. The maximum height is about 14.87 cm. These are the coordinates of the vertex of the funtion.

d) The spider lands after approximately 0.251 seconds.

e) Appropriate domain: [0, 0.251]    Appropriate range:  [0, 14.87]

f) Replace t with  0.05 to get 14.525  cm. 

 my school only does it only  for 1 month in summer.

The first anonymous is correct :)

you know, that it was $${\frac{{\mathtt{5}}}{{\mathtt{8}}}}$$.