All Answers 358095 Answers

Indeed, I do, Queen of the Forum.

If thou wilt knoweth, knoweth it as Andrew...

but I shalt be known as Anonymous4338 in thy kingdom.

P.S. I doth try hard to...

Hullo!

45,300 at 21% compounded monthly for 8 3/4 years

compounded monthly

\(\begin{array}{lcll} K_{n} &=& K_0 \cdot (1 + \frac{i}{12} )^n \qquad n = \text{Number of the months} \\\\ K_{n} &=& K_0 \cdot (1 + \frac{i}{12} )^n \qquad K_0 = $45300 \quad i=21\% \quad n = 8\cdot 12 + \frac34 \cdot 12 = 105 \\ K_{105} &=& 45300 \cdot (1 + \frac{\frac{21}{100}}{12} )^{105} \\ K_{105} &=& 45300 \cdot (1 + \frac{21}{100\cdot 12} )^{105} \\ K_{105} &=& 45300 \cdot (1 + 0.01750000000 )^{105} \\ K_{105} &=& 45300 \cdot (1.01750000000 )^{105} \\ K_{105} &=& 45300 \cdot 6.18178475531 \\ \mathbf{K_{105}} & \mathbf{=} & \mathbf{$280034.849416} \\ \end{array}\)

4^1000=1.148130 X 10^602. This calculator CANNOT handle big numbers like this one. So, it gives the answer as "infinity", which is meaningless.

45,300 at 21% compounded monthly for 8 3/4 years

The FV of 45,300 @21% for 8 3/4 years will be worth=$280,034.85.

The formula you use for this is FV=PV (1 + R)^N, where FV=Future value, PV=Present Value, R=Interest rate per period, N=Number of periods.

Alexis' niece is my guess. 😁😁

t = ????

Not sure what you mean BUT

(5*6)^2 means

\((5\times 6)^2 = 5^2\times 6^2 = 25\times 36 = 900\\ or\\ (5\times 6)^2 = 30^2=900\)

does that help?

The angle of elevation of the top of a tree is found to be 23° at one point and 76° at a point 49 feet nearer the tree. How high is the tree if both observation points and the base of the tree are on the same horizontal plane?

\(\begin{array}{rcll} \tan{(23^\circ)} &=& \frac{h}{x} \\ x &=& \frac{h}{ \tan{(23^\circ)}} \\\\ \tan{(76^\circ)} &=& \frac{h}{x-49} \\ \tan{(76^\circ)} &=& \frac{h}{\frac{h}{ \tan{(23^\circ)}}-49} \\ \left( \frac{h}{ \tan{(23^\circ)}}-49 \right)\cdot \tan{(76^\circ)} &=& h\\ h\cdot \frac{\tan{(76^\circ)}} { \tan{(23^\circ)}}-49\cdot \tan{(76^\circ)}&=& h\\ h\cdot \left( \frac{\tan{(76^\circ)}} { \tan{(23^\circ)}} - 1 \right) &=& 49\cdot \tan{(76^\circ)}\\ h\cdot \left( \frac{\tan{(76^\circ)}-\tan{(23^\circ)} } { \tan{(23^\circ)}} \right) &=& 49\cdot \tan{(76^\circ)}\\ h &=& 49\cdot \left( \frac{ \tan{(76^\circ)} \cdot \tan{(23^\circ)} } { \tan{(76^\circ)}-\tan{(23^\circ)} } \right) \\ h &=& 49\cdot \left( \frac{4.01078093354 \cdot 0.42447481621 } { 4.01078093354-0.42447481621 } \right) \\ h &=& 49\cdot \left( \frac{1.70247549962 } { 3.58630611733 } \right) \\ h &=& 49\cdot 0.47471561097 \\ h &=& 23.2610649376\ \text{feet} \\ \end{array}\)

The tree is 23.2610649376 feet high

whats a^-3*(b^4)-2

             b^5*a^4

\(\frac{a^{-3}*(b^4)^{-2}}{b^5*a^4}\\ =\frac{a^{-3}*b^{-8}}{b^5*a^4}\\\)

Now the way you deal with negativie indices is:

Is something is raised to a negaive indice then swap it to the other side of the fraction line and chang the negative power to a positive power.  

Anything NOT raised to a negative power stays right where it started!

(don't confuse negative coefficients with negative powers - but there are no negative coefficients here anyway)

\(=\frac{a^{-3}*b^{-8}}{b^5*a^4}\\ =\frac{1}{b^5*a^4*a^{+3}*b^{+8}}\\ \mbox{Since there was nothing left on the top I put a one there}\\ =\frac{1}{b^5*b^8*a^4*a^3}\\ =\frac{1}{b^{13}a^7}\\\)

A rectangle is 152 by 383 feet. Determine the length of the diagonal and the angle between the shorter side and the diagonal.

\(\begin{array}{rcll} \text{diagonal} &=& \sqrt{ 383^2+152^2} \\ &=& \sqrt{ 169793 } \\ &=& 412.059461729\ \text{feet} \\\\ \tan{(\varphi)} &=& \frac{383}{152} \\ &=& 2.51973684211 \\ \varphi &=& \arctan{(2.51973684211)}\\ \varphi &=& 68.3535133795^\circ \end{array}\)

Thanks Bertie, it is really good to see you. :)

I could see that there was no unique solution but I did not know how to work out the general solution.

Can you show me how to do it?

I am not at all good with hyperbolas  

By the way, do you use Firefox as your browser?

and

Have you tried logging on using a different browser?

THIS IS A VERY HARD QUESTION!!!!!!!! :-D

maybe 1+1=11      JK! 1+1=2      OBVI

how do i isolate x in y=b∗x^a?

\(\begin{array}{rcll} y &=& b\cdot x^a \qquad & | \qquad :b\\ \frac{y}{b} &=& x^a \qquad & | \qquad 1^{\frac{1}{a}}\\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x^{ a\cdot \frac{1}{a} } \\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x^{\frac{a}{a} } \\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x^{1 } \\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x \\ \end{array}\)

There is no unique solution to this.

The general solution is

65x^2 + 60y^2 - 128xy + 448x + 448y + C = 0,

where C is a constant.

Draw some graphs using desmos to check this out.

-Bertie

It is difficult, possibly if you are a baby. Try two...

Of course 2.........duh

There are an infinite number of solutions to the question as written!  

The four corners are triangles each of side 2.9 cm and diagonal sqrt(2)*2.9 cm

They can be put together to make two squares each of area 2.9^2 cm^2, so the area of the octagon is the area of the original  square less the area of the 4 triangles, namely:

Area = 10^2 - 2*2.9^2 cm^2

Perimeter = 4*sqrt(2)*2.9 + 4*(10 - 2*2.9) cm

I'll leave you to do the number crunching.

He earns a lot of money.

You'll need to specify the current, A, if you want a numerical answer!

Really nice diagram! I love visual aid...:)

If you are a new member, here is a little help to get you started...:)

Here is your answer.