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Using the Pythagorean theorem...

1.8^2 (That is PQ) + (QR)^2 = 8.2^2 (That is PR)

Subtract 1.8^2 to both sides to get QR^2 = 8.2^2-1.8^2

QR^2 = 64

QR=8, and QR is the base. PQ is the height.

Get the triangle's area.

8*1.8*1/2=7.2

Is this correct???

2b^2 + b  = 6      

step 1....  divide both sides by 2

b^2  + (1/2)b   = 3        

step 2    .....take (1/2)  of (1/2)  = 1/4....square this = 1/16  add to both sides

b^2 + 1/2b + 1/16  = 3 + 1/16

step 3  ......factor the left side.....simplify the right side

(b + 1/4)^2  = 49/16

step 4  ......take positive/negative roots of both sides

b + 1/4  = ±√[ 49/16]

b + 1/4  =   ±7/4

step 5......subtract  1/4 from both sides

b = 7/4 - 1/4  = 6/4  = 3/2        or    b =  -7/4 - 1/4  = -8/4   = -2

The mistake was in step 2.....1/4 was added to both sides, but it should have been (1/4)^2 = 1/16 !!

Substituting x for t, this function is not one-to-one....thus....it has no inverse unless we restrict its domain

y = 100 ( 1 - t/40 )^2   divide both sides by 100

y / 100  = ( 1 - t/40)^2       take both roots of  sides

 ±√ (y/100)  = 1 - t/40       multiply both sides by -1

 ±√ ( y / 100 )  = t / 40 - 1     add 1 to both sides

1  ±  √ ( y / 100 )  = t/40        multiply both sides by 40

40  ± 40  √ ( y / 100 )  = t       swap t and y

40  ± 40  √ ( t / 100 )  = y    for  y, write  f^-1(t)

40  ± 40 √ ( t / 100 )  =   f^-1(t)

Here is the graph : https://www.desmos.com/calculator/lf5zoe7lei

If we restrict the original domain to ( -infinity, 40),  the inverse is represented by the orange graph

If we restrict the original domain to (40, infinity), the inverse is represented by the red graph

Thanks guest and Melody. I'll try to find the big dipper the next time I go under a clear sky. I wonder when that will be...

Well, you are multiplying -9   180 times. So the result will be even. Unfortunately, you will have to plug this in a calculator, or your hand might fall off from calculating. 

I am a reasonably sensible person (or am I? ), so I used a calculator.

The answer is 5802988355301064263572941269018812264886757498342118039827603956893368389206154994705426429662904979537365662938773593537918329060451697709838267789886857631025156060199201.

I am fine, thank you. And you?

The most important thing is that you know what it looks like.

Here are some pictures. You certainly will not be able to see it if you do not know what you are supposed to be lookng for. :)

yea, thanks Ghosty :)

The challenge question which is a few posts below this one.   

Sorry...I'm not CPhil. Just ignore me... :)

Well, let's say y is v(t). It'll make it simpler, or at least... it will for me.  :)

Then the equation is now y=100(1-t/40)^2

To find the inverse you need to switch the y and x, or in this case, y and t.

That would make t=100(1-y/40)^2

Solve for y

First divide both sides by 100 to make t/100=(1-y/40)^2

Then square root both sides to get √(t/100)=1-y/40, which simplifies to √(t)/10=1-y/40.

Then subtract 1 to both sides to get  -1+√(t)/10=-y/40.

Next, multiply -1 to both sides to get 1-√(t)/10=y/40

Then multiply 40 to both sides to get 40-4√(t)=y

That is y=40-4√(t)

Which is v(t)=40-4√(t)  because I had used y instead of v(t).

And that's the answer

v(t)=40-4√(t)

Correct me if I'm wrong. 

1.483 g•mL^-1 x 10.00 mL =14.83 g. 

[ml/ml] cancels.

He will probably put them up after a few hours/minutes. He has done that 2 or 3 times already...

Or maybe it's a lag...

Buy yourself the "Sky and Telescope" magazine, or subscribe to it. They will teach how to recognize all constellations.

In the left column, you are adding Phi n times. In the right column, you are multiplying Phi by itself n times. That is it.

500+50+25? I don't have to use all the numbers, right? And it doesn't say that I can't use '+' more than once, right? 

Explanation: I used 50 and 25 and 500. I added them. I got the answer. 

Am I wrong? I think I am...

Sorry, but what are all the {nI}'s?

 0.000 000 000 154 m =1.54 x 10^-10 m

There are no questions here!

Use the  ^y√x key (2nd row, 4'th column), or type in sqrt(x,y) for y'th root of x.

Geometric series

Hi, I start doing the task, but then I dont know what to do next. Here is:

a₁+a₂+a₃=62

a₁*a₂*a₃=1000

\(\begin{array}{|rcll|} \hline a_1*a_2*a_3 &=& 1000 & | \quad a_1 = a \qquad a_2 = a*q \qquad a_3 = a*q^2 \\ a*(a*q)*(a*q^2) &=& 1000 \\ a^3*q^3 &=& 1000 \\ a^3*q^3 &=& 10^3 & | \quad \sqrt[3]{} \\ a*q &=& 10 \\ \mathbf{ a} &\mathbf{ =}&\mathbf{ \frac{10}{q} }\\\\ a_1+a_2+a_3 &=& 62 & | \quad a_1 = a \qquad a_2 = a*q \qquad a_3 = a*q^2 \\ a +a*q + a*q^2 &=& 62 \\ a *(1 + q + q^2) &=& 62 & | \quad \mathbf{a= \frac{10}{q}} \\ \frac{10}{q} *(1 + q + q^2) &=& 62 \\ \frac{1}{q} *(1 + q + q^2) &=& 6.2 \\ \frac{1}{q} +1+q &=& 6.2 \\ \frac{1+q^2}{q} &=& 5.2 \\ 1+q^2 &=& 5.2*q \\ q^2-5.2*q +1 &=& 0 \\ (q-5)*(q-0.2) &=& 0 \\\\ \mathbf{q = 5} &\mathbf{ \text{ or } }& \mathbf{q = 0.2 }\\\\ a =\frac{10}{5} && a =\frac{10}{0.2} \\ \mathbf{a = 2} &\mathbf{ \text{ or } }& \mathbf{a = 50} \\ \hline \end{array} \)

Geometric series:

1.

\(a=2 \quad q = 5 : \\ 2+2*5+2*5^2 = 62\ \checkmark \\ 2*(2*5)*(2*5^2) = 1000\ \checkmark \)

2

.\(a=50 \quad q = 0.2 : \\ 50 + 50*0.2 + 50*0.2^2 = 62 \ \checkmark \\ 50*(50*0.2)*(50*0.2^2) = 1000\ \checkmark \)

Each product of n x Phi is the product of n - 1 plus Phi. And each product of Phi^n is the product of n - 1 times Phi. So the product of n =10 is the product n=10 - 1 + Phi. The product of Phi^10 is the product Phi^(10-1)=Phi^9 x Phi. 

Heureka, how did you get that ϕ^n=Fnϕ+Fn-1 ?

Formula:  \( \phi^2 = \phi +1\)

mathematical proof:

\(\begin{array}{|rcll|} \hline \phi = \frac{1+\sqrt{5}} {2} \\\\ \phi^2 &=& \left( \frac{1+\sqrt{5}} {2} \right)^2 \\ &=& \frac{\left( 1+\sqrt{5}\right)^2 } {4} \\ &=& \frac{ 1+2\sqrt{5}+5 } {4} \\ &=& \frac{ 6+2\sqrt{5} } {4} \\ &=& \frac{ 3+\sqrt{5} } {2} \\ &=& \frac{ 1+2+\sqrt{5} } {2} \\ &=& \frac{ 1+\sqrt{5}+2 } {2} \\ &=& \frac{ 1+\sqrt{5} } {2} +\frac{2}{2} \\ &=& \frac{ 1+\sqrt{5} } {2} + 1 \\ &=& \mathbf{ \phi + 1 } \\ \hline \end{array}\)

\(\small{ \begin{array}{|lclclclcl|} \hline \phi^1&&&& &=& 1\phi+0 &=& F_1\phi + F_0 \\ \phi^2&&&& &=& 1\phi+1 &=& F_2\phi + F_1 \\ \phi^3 = \phi\phi^2 = \phi(\phi+1) &=&1\phi^2 + 1\phi &=&1(\phi+1)+1\phi &=& 2\phi+1 &=& F_3\phi + F_2 \\ \phi^4 = \phi\phi^3 = \phi(2\phi+1) &=&2\phi^2 + 1\phi &=&2(\phi+1)+1\phi &=& 3\phi+2 &=& F_4\phi + F_3 \\ \phi^5 = \phi\phi^4 = \phi(3\phi+2) &=&3\phi^2 + 2\phi &=&3(\phi+1)+2\phi &=& 5\phi+3 &=& F_5\phi + F_4 \\ \phi^6 = \phi\phi^5 = \phi(5\phi+3) &=&5\phi^2 + 3\phi &=&5(\phi+1)+3\phi &=& 8\phi+5 &=& F_6\phi + F_5 \\ \phi^7 = \phi\phi^6 = \phi(8\phi+5) &=&8\phi^2 + 5\phi &=&8(\phi+1)+5\phi &=& 13\phi+8 &=& F_7\phi + F_6 \\ \phi^8 = \phi\phi^7 = \phi(13\phi+8) &=&13\phi^2 + 8\phi &=&13(\phi+1)+8\phi &=& 21\phi+13 &=& F_8\phi + F_7 \\ \phi^9 = \phi\phi^8 = \phi(21\phi+13) &=&21\phi^2 + 13\phi &=&21(\phi+1)+13\phi &=& 34\phi+21 &=& F_9\phi + F_8 \\ \phi^{10} = \phi\phi^9 = \phi(34\phi+21) &=&34\phi^2 + 21\phi &=&34(\phi+1)+21\phi &=& 55\phi+34 &=& F_{10}\phi + F_9 \\ \cdots \\ \mathbf{ \phi^n } &&&& && &\mathbf{ =}& \mathbf{ F_n\phi + F_{n-1}} \\ \hline \end{array} }\)

Me neither...

Assuming the bullet points are not negative signs....

Look at the first part of the phrase:

"Five more than three times the number"

The variable 'n' would represent "the number."

Three times the number would be 3n. 

Five more than that would be 3n+5.

Now look at the second part of the phrase: 

"one third more than the sum of the number and itself."

This is a bit tricky. (Didn't quite understand this, so I had to look at the four possible answers.)

I assume "the sum of the number and itself" is n+n.

And 1/3 more than that would simply be (n+n) + 1/3.

Now using the 'equal' sign let's put these two parts together. (No, we are not adding)

3n+5=(n+n)+1/3

And that is the fourth answer.

So the fourth answer is correct.

(This was a tricky question...especially because I spent a lot of time wondering if those bullet points were negative signs)

Assuming the bullet points are not negative signs,

let's look at "the sum of a number and three." 

That would obviously be n+3, where n is the 'number.'

Look at the phrase again.

"Twice the sum of a number and three,"

That would be twice of n+3

2*(n+3), which is 2(n+3)

So it is the fourth answer.