All Answers 358108 Answers

Well Nauseated ....

"I might have thought so, but this forum is one of the few where I am not banned. I know because I can log on. Which begs the question: why do I want to be a member of a forum that would allow me as a member?

It’s not troll-like; it antithetical to our character --It’s a paradox. :)"

Yes - it is definitely a paradox      

Mind you - if you have been thrown out of all the other forum maybe you only want to be here because this is the ONLY place that will have you  :/

Plus, of course, we are also lovable.

Rosala is positively dripping with lovable ....      :) 

I completely skimmed over the detail the first corridor is only 8m wide -_- I thought they were the same width. Alan has the right idea.

14a - 93 = 49 - 57a

First let's re-arrange the equation so all "a"s are on one side.

14a + 57a - 93 = 49

14a + 57a = 49 + 93

Simplifying it down

71a = 142

Divide both sides by 71 so it's just "a" as the subject.

a = 142 / 71

a = 2

Thank you Alan and Sir-Emo, 

I intend to examine your answers in detail a little later.  

I am very curious about this one :)

[(p+1)-3] = p+1-3 = p-2

What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98 ?

$$\small{
\begin{array}{rcl}
a_n &=& a_1\cdot r^{n-1} \\
a_6 &=& a_1\cdot r^5 = 4 \\
a_{100} &=& a_1\cdot r^{99} = 98\\
\\
\hline
\\
\dfrac{a_{100}}{a_6} &=&
\dfrac{a_1\cdot r^{99}}{a_1\cdot r^{5}} = \dfrac{98}{4} \\\\
\dfrac{a_1\cdot r^{99}}{a_1\cdot r^{5}} &=& \dfrac{98}{4} \\\\
\dfrac{r^{99}}{r^{5}} &=& 24.5 \\\\
r^{99-5} &=& 24.5 \\\\
r^{94} &=& 24.5 \\\\
r &=& \sqrt[94]{24.5} \\\\
r &=& 1.03461402806 \\
\\
\hline
\\
a_6 &=& a_1\cdot r^5 = 4 \\\\
a_1 \cdot r^{5} &=& 4 \\\\
a_1 &=& \dfrac{4} { r^5 } \\\\
a_1 &=& \dfrac{4} { 1.03461402806^5 } \\\\
a_1 &=& 3.37417946747\\
\\
\hline
\\
a_{55} &=& a_1\cdot r^{54} \\\\
a_{55} &=& 3.37417946747 \cdot 1.03461402806^{54}\\\\
a_{55} &=& 21.1933572378
\end{array}
}$$

If this is an arithmetic sequence then the n'th term is a + (n-1)d where a is the first and d is the difference.

a + 5d = 4

a + 99d = 98

so a = -1 and d = 1

the 55th term is -1 + 54*1 = 53

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If this is a geometric sequence then the n'th term is a*r^n-1 where r is the multiplier.

a*r⁵ = 4

a*r⁹⁹ = 98

r=(98/4)^1/94 ≈ 1.0346

a = 4/1.0346⁵ ≈ 3.3742

The 55th term is 3.3742*1.0346⁵⁴ ≈ 29.193

.

I am not sure what this question means   

$${\frac{\left({\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{29}}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{96}}}{{\mathtt{5}}}} = {\mathtt{19.2}}$$  

At the moment the average age is 19.2 years

Maybe it means that the average has to be between 13 and 25     

$$\\13<\frac{96+X}{6}<25\\\\
78<96+X<150\\\\
0

But really the person should be between  13 AND 25

I think that this is a dumb question  

tan(45°) = 8/p  Hence p = 8/tan(45°)  or p = 8 m

cos(45°) = (w+p)/27  or  1/√2 = (w + 8)/27 so  w = 27/√2 - 8 m

$${\mathtt{w}} = {\frac{{\mathtt{27}}}{{\sqrt{{\mathtt{2}}}}}}{\mathtt{\,-\,}}{\mathtt{8}} \Rightarrow {\mathtt{w}} = {\mathtt{11.091\: \!883\: \!092\: \!036\: \!783\: \!2}}$$

or w ≈ 11.1 m

.

Amount deposited on monday = x

The amounts he has left by the end of each day:

Monday: x

Tuesday: x + 700

Thursday: (x+700) - 300 - 200 = x + 200

Friday: (x + 200) / 2

So (x + 200)/2 = 1400 remaining in the account

Multiply both sides by 2

x + 200 = 2800

Subtract both sides by 200

x = 2600

The amount deposited on monday was 2600

For question 2, consider this:

The girder must rotate a full 90 degrees (Or pi/4 if you like your radians), so at one point or another it's going to be at 45° to the corridor walls. Seeing as we ignore width of the beam we can forget about it 'catching' on the walls when it rotates.

So we have a nice situation of all we need is the corridor corner to support a three-quarter circle of diameter 27 metres. This is a radius of a 27/2 = 13.5 metres. So the corridor needs to be 13.5 metres wide.

If my explanation is a bit confusing, just say and I will try to mock up some diagrams to better explain it.

That's a lot of 0s. It's much faster to use scientific notation for it:

2 + 4 * 10⁴⁵

Now we have a dilemma. There are two ways of reading this.

I think it's likely you meant this:

(2 + 4) * 10⁴⁵ = 6.0 * 10⁴⁵

But in the standard order of mathmatics, you would have this:

2 + (4 * 10⁴⁵) = 4[fourty four 0's...]2

You use inverse trig        OR   arc trig      (these are different terms for the same thing)

$$\\sin\theta =0.125\\\\
then\\\\
\theta = sin^{-1}(0.125) = asin(0.125)$$

On the web2 calc it is asin         

right so what you do is you get a calculator and you look for it your self btw its 47.8

Firstly and most importantly, do you understand what a "Mobile app" is?

It's short for "Mobile Phone Application" and refers to small program you can download and use on a mobile phone.

an "Application interface" is what users of the app will see. This is the layout and design of buttons to push and whatever the application is meant to give back. For example the interface on a calculator is built up of a large area that shows the currently input number or the resulting number when you click "=". More advanced calculators also likely have small things to tell you if you have it set to "radians" or "degrees", whether something is stored on "Memory", etc.

"MS- Word '07" is just shorthand for "Microsoft Word 2007", which is a very common word processor.

So basically it wants you to use MS Word 2007 to design the "look" of an application, including all relevant details the user would be given on-screen.

Add 15 to both sides

then

divide both sides by 5    

Vectors A and B are at angles α = 44.9° and β = 26.2° up from the x-axis respectively. If the vector sum A B C = 0, what are the magnitudes of A and B?

a = magnitude of A,   b = magnitude of B

$$\small{
\vec{A} = a\cdot \begin{pmatrix}\cos{\alpha}\\\sin{\alpha}\end{pmatrix}\quad
\vec{B} = b\cdot \begin{pmatrix}\cos{\beta}\\\sin{\beta}\end{pmatrix}
\qquad
\vec{A}_{\bot }= \begin{pmatrix}-\sin{\alpha}\\\cos{\alpha}\end{pmatrix}\quad
\vec{B}_{\bot } =\begin{pmatrix}-\sin{\beta}\\\cos{\beta}\end{pmatrix}
}\\\\
\small{
\begin{array}{rcl}
a\vec{A}+b\vec{B}+\vec{c}&=&\vec{0}\quad | \quad \cdot \vec{B}_{\bot }\\
a\vec{A}\vec{B}_{\bot }+b\vec{B}\vec{B}_{\bot }+\vec{c}\vec{B}_{\bot }&=&\vec{0}\quad | \quad \vec{B}\cdot {\vec{B}_{\bot }=0 \\
a\vec{A}\vec{B}_{\bot }+\vec{c}\vec{B}_{\bot }&=&\vec{0} \\
a &=&\dfrac { -\vec{c}\vec{B}_{\bot } } {\vec{A}\vec{B}_{\bot }}\\
a &=& \vec{c}\cdot \begin{pmatrix}\frac{ \sin{\beta} }{ \sin{ (\alpha-\beta) }} \\\\ \frac{-\cos{\beta}}{\sin{ (\alpha-\beta) }} \end{pmatrix}
= \vec{c}\cdot \begin{pmatrix}1.37706788482\\ -2.79857147409\end{pmatrix}\\
\end{array}
}\\\\\\
\small{
\begin{array}{rcl}
a\vec{A}+b\vec{B}+\vec{c}&=&\vec{0}\quad | \quad \cdot \vec{A}_{\bot }\\
a\vec{A}\vec{A}_{\bot }+b\vec{B}\vec{A}_{\bot }+\vec{c}\vec{A}_{\bot }&=&\vec{0}\quad | \quad \vec{A}\cdot {\vec{A}_{\bot }=0 \\
b\vec{B}\vec{A}_{\bot }+\vec{c}\vec{A}_{\bot }&=&\vec{0} \\
b &=&\dfrac { -\vec{c}\vec{A}_{\bot } } {\vec{B}\vec{A}_{\bot }}\\
b &=& \vec{c}\cdot \begin{pmatrix}\frac{ -\sin{\alpha} }{ \sin{(\alpha-\beta)}} \\\\ \frac{\cos{\alpha}}{\sin{(\alpha-\beta)}} \end{pmatrix}
= \vec{c}\cdot \begin{pmatrix}-2.20163122333\\ 2.20932981047\end{pmatrix}\\
\end{array}
}$$

If the square root sign is written into the question then you only take the positive answer.

Plus,

sqrt(2) is an irrational number so the answers above are approximations :)

12 7/28 - 12/28 = 343/28 - 12/28 = 331/28 = 11 23/28

Hallo anonymous !

if 5/7 = 448 , what is 7/7 for

(7/7) : (5/7) = x : 448

(5/7) * x = (7/7) * 448

x = 448 * (7/5)

x = 627,2

if 5/7 = 448 then is 7/7 = 627,2

regards  :- )

What is the square root of 2 ?

The root has 2 results .

√2 = ± 1,41421356237...

Gruß  :- )

Why not show us what you did for problem 2 and then we can comment on it?

.

This requires use of that well known equation  E = mc²

Rearrange to get m = E/c²

m=7.84*10⁷/(3*10⁸)²  kg = (7.84/9)*10^-9 kg. ≈ 8.7*10^-10 kg

.