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Could also do this as follows.

Notice that 4096 = 2^12

Let x = 2^p

Then we have:  2^12 = 2^(180p)

180p = 12, so p = 12/180

x = 2^(12/180) ≈ 1.04729

2 to the power of any real number will always be positive, so won't give -70000.

tan(51°) = (h + a)/12       (1)   where h is height of house and a is length of antenna

tan(42°) = h/12               (2)

Subtract (2) from (1):    tan(51°) - tan(42°) = a/12

a = 12*(tan(51°) - tan(42°))

I'll leave you to crunch the numbers.

Like this perhaps:

.

Hmm looks like I was originally correct, just made some calculation errors...

This is my working, I'm sure there is a more efficient way to do it however...

\(\frac{x^2}{16}+\frac{(5-x)^2}{9}=1\)

\(\frac{9x^2+16(5-x)^2}{144}=1\)

\(9x^2+16(5-x)^2=144\)

\(9x^2+16(25-10x+x^2)=144\)

\(9x^2+400-160x+16x^2=144\)

\(25x^2-160x=-256\)

\(25x^2-160x+256=0\)

And then just solve for x using the quadratic formula...

\(x=\frac{160+-\sqrt{25600-25600}}{50}\)

x = 3.2

y = 1.8

Thus point of intersection = (3.2,1.8)

The answer in the book however is (doesn't show any working):

(16/5, 9/5) which of course is the same answer, but there must be a more efficient way to get to this answer if the answer is a fraction.

Suppose I have a bag with 12 slips of paper in it. Some of the slips have a 2 on them, and the rest have a 7 on them. If the expected value of the number shown on a slip randomly drawn from the bag is X , then how many slips have a 2?

Let A be the number of slips with an A  :)

ElectricPavlov is right of course :)

\(\frac{2A+7(12-A)}{12}=X\\ \frac{2A+84-7A}{12}=X\\ \frac{84-5A}{12}=X\\ 84-5A=12X\\ A=\frac{12X-84}{-5}\\ A=\frac{84-12X}{5}\\\)

How to solve 4096=x^180

log 4096=log x^180

log 4096=180log x

(log 4096)/180 =log x

\(10^{log (4096)\div 180} = x\)

10^(log(4096)/180) = 1.0472941228206267

\(Let A(t) = 3- 2t^2 + 4^t. Find A(2) - A(1).\\ A(2)-A(1)=3-2*4+4^2-(3-2*1+4)\\ A(2)-A(1)=3-8+16-(3-2+4)\\ A(2)-A(1)=11-(5)\\ A(2)-A(1)=6\\\)

\(\mbox{The function f(x) satisfies } f(\sqrt{x+1}) = \frac{1}{x}\quad for \;\;all \;\;x ≥ 1, \;\;x≠0. \quad Find \;\;f(2)\\ \sqrt{x+1}=2\\ x+1=4\\ x=3\\~\\ f(2)=\frac{1}{3}\)

next time please just ask one question per post. :)

This cannot be factorised

Hi Bobo2k16, it is very nice to meet you :)

1. How ways are there to put 6 b***s in 3 boxes if the b***s are not distinguishable but the boxes are?

ok boxes are different but ball are the same.  Mmm do I have to put at least one ball in every box.  I guess some boxes can be empty...

So the only relevent thing is how many b***s are in each individual box.

I think I wil use the stars and bars method.

I'l put the boxes in order   A,B,C

now I have the b***s

1,2,3,4,5,6    they are the stars, now I wil use 2 bars to divide those b***s into three groups.

So now there are 6 stars and 2 bars and I want to know how many places in the line I can put the 2 bars.

That will be 8C2=28

So that is 28 ways :)

Here is a clip to explain the stars and bars method to you

"If the expected value of the number shown on a slip randomly drawn from the bag is_____ , then.."

Ya left off the most important part of the question.......shucks.

What is 0.5401 expressed in scientific notation?

3 6 12 24 48   = 93   ?

5 terms = 93     yah?   Did I do this correctly?

How many terms are there in a geometric series if the first term is 3, the common ratio is 2, and the sum of the series is 93?

Sum  = a₁ [ 1 - r^n] / [ 1 -r ]      where a₁ is the first term, n is the number of terms and r is the common ratio

93 = 3 [ 1 - 2^n ] / [ 1 - 2]     simplify.......divide both sides by 3

31 =  [1-2^n] / -1         multiply the fraction on the right by  -1 on top/bottom

31 = 2^n - 1         add 1 to both sides

32 = 2^n        and we can write  32 as 2^5

2^5  = 2^n

So......n = 5  terms

Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that one of the cards is a 7 and the other is a 3, and are these events independent?

No they are not independent because the first draw effects the probablility of the second draw.

If the first card WAS replaced then they would be independent.

P(a seven and a 3) = P(7,3) + P(3,7)

=

 \(\frac{4}{52}*\frac{4}{51}+\frac{4}{52}*\frac{4}{51}\\ =2*\frac{4}{52}*\frac{4}{51}\\ =\frac{32}{2652}\\ =\frac{8}{663}\)

If P(A and B) = P(A)*P(B), then they are independent....so....

(3/10)  = (1/2) * (3/5)   

(3/10)  =  (3/10)      ...so......they are independent

The table indicates that 4/90   of the TVs sold are 54-inch Projection models....so

4/90  ≈   .0444  ≈   4.44%   ≈  4%   [rounded]

I gave you five stars Chris - you deserved them for all that effort :)

Eh....I'll probably get over it........

Is our guest calling my solution an elephant ??    LOL     

Thanks guest      

Melody's solution is the only ELEGANT solution!!. Eat your heart out, CPhill!!.