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Ok, I tried to do something, but I did not succeed in the end...

Derivative of  y = b^x .    b is a constant and x is the variable,so differentiate with respect to x,not b.

The result is dy/dx = b^x ln(b).   Here is a proof of the result.

Take base e logs of both sides to get

ln(y) = x ln(b)                             

differentiate both sides with respect to x

(1/y)dy/dx  = ln(b)                               {  using chain rule on left hand side   }

dy/dx  = y ln(b)   = b^x ln(b).

In general ,the derivative of a^x   = a^x ln(a).

a children's park is 350 m long and 200 m wide. It is surrounded by a pathway of uniform width. Suppose the total area of the park is 74464 m2. How wide is the pathway

\((2x+350)(2x+200)=74462\\ 2(2x+350)(x+100)=74462\\ (2x+350)(x+100)=37231\\ 2x^2+200x+350x+35000=37231\\ 2x^2+550x-2231=0\\\)

2x^2+550x-2231=0 = {x=-(((sqrt(80087)+275)/2)), x=((sqrt(80087)-275)/2)}

x is approx 4m

The path is 4 metres wide. That is one generous path :)

To get the answer that you are seeking, it should be written like this:

1,000 x (1 + 0.092512/12)^(12*42) =

1,000 x (1.007709333....)^504 =

1,000 x 47.971314.......=

=$47,971.31 - This is the FV of $1,000 earning 9.2512%compounded monthly for 42 years!!. Your answer, however, is still wrong, if that is what you are trying to calculate. We are only speculating on what you are trying to do!.

1000(1 +0.092512)(12*42)

Just type it in straight like that! Only don't leave any spaces

1000(1+0.092512)(12*42) = 550626.048

your answer is incorrect :/

Thank you so much

How many 1/2-inch cubes would fit into a 96inch cubed rectanular prism?

Do you mean that the volume is   0.5cm^3 or do you mean that the side length of the cube is 0.5cm?

Do you mean that the rectangular prism has a volume of 96cm^3?   I guess that is what you mean....

If 26 students each make 3 cards, there will be a total of  26 * 3 = 78  cards .

If there are 78 cards and 79 staff members, then 1 staff member won't get a card.

If there's something you don't understand about this answer, can you explain what it is?

Sagedragon has asked me how I factorised 

\(2cos^2(2x)+cos(2x)-1\)

in the denominator.

His actual question was why/how I changed  cos(2x) into 2cos(2x)-1cos(2x)

Hi Sagedragon I am really pleased that you have asked :)

It may be easier for you to see if I let     \(a=cos(2x)\)

then you have the expression    

\(\textcolor{blue}{2} a^2+1a \textcolor{blue}{-1}\ \)

To factorize this I looked for two numbers

that would multiply to 2*-1=-2        

[Since they multiply to a negative I know straight off that one is negative and one is positive]

and add to +1

That would have to be +2 and -1

So I will now split the coefficient for a into  2-1

1a becomes  2a-1a

\(\textcolor{blue}{2} a^2+1a \textcolor{blue}{-1}\\ =\textcolor{blue}{2} a^2+2a-1a \textcolor{blue}{-1}\\ \text{Now I factorise in pairs}\\ =\textcolor{blue}{2a} (a+1) \textcolor{blue}{-1}(a+1)\\ \text{Now I have 2a lots of (a+1)and -1 lots of (a+1)}\\ \text{which makes (2a-1) lot of (a+1)}\\ =\textcolor{blue}{(2a-1)}(a+1) \)

so

\(2a^2+1a -1=(2a-1)(a+1)\\ but\;\;a=cos(2x)\\ so\\ 2cos^2(2x)+1cos(2x) -1=[2cos(2x)-1][cos(2x)+1]\\ \)

Sagedragon,

Think about it and if you still have queries do not hesitate to ask.

You can ask on this thread if it is not locked but always send the link by private message as well becasue otherwise I may not see it. :)

Wow, this one was tricky!!! 

Thank you! This problem forced me to put in effort! I would consider this an elegant solution because I need not know the lengths of any of the sides.

(a + b)  / (a -b) +  (a - b) / (a + b)  →  [   (a + b)^2 +  ( a - b)^2 ]  / [ a^2 - b^2]  →

[ a ^ 2 + 2ab + b^2 + a^2 - 2ab + b^2) / [ a^2 - b^2]  →

[ 2a ^2 + 2 b^2]  / [ a^2  - b^2] →

2 [ a^2 + b^2] / [a^2 - b^2 ] → 

This expression will have the smallest possible value when the numerator of  a^2 + b^2  =  the denominator of a^2 - b^2

Note that when   a = any integer > 0 and  b  = 0   the numerator of  a^2 + b^2  =  the denominator of a^2 - b^2  = 1

So....the smallest positive value  is 2

Very nice, X²....!!!!!!!

Ok, after some problem solving I have finally figured out the answer! The answer was right in front of me, too!

I will attempt to make a diagram, but I cannot guarantee that it will be clear.

   C        A                                                                      D

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   |                                                                                  |

   |                                                                                  | G

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  F                                                                          B     E

Ok, this is a diagram, and I have labeled the corners with letters so that it is clearer for you to understand!

If AB=12, that is the diagonal of the inner rectangle. Notice that AG=BH and BG=AH. We know that this by the properties of a rectangle; opposite sides of a rectangle are congruent. If all the triangles are right and isosceles, then we know that \(\triangle AGD \cong \triangle BHF\hspace{1mm}\text{and}\hspace{1mm}\triangle ACH \cong \triangle BEG\) because of angle-side-angle congruence theorem.

Let's remind ourselves, too, that these are, by the given information, right isosceles. This means that \(AD=DG=HF=FB\hspace{1mm}\text{and}\hspace{1mm}AC=CH=BE=EG\).

Let's label all the sides congruent to AD x and all sides congruent to AC y.

Let's find the length of the triangles' hypotenuses. Of course, we already know that \(AG=BH\hspace{1mm}\text{and}\hspace{1mm} BG=AH\), so we only need to calculate half of the diagonals. These are easier than they first appear, actually. A right isosceles triangle is 45-45-90 triangle, so the hypotenuses is the length of the leg times the square root of 2. Using this logic, \(AG=BH=x\sqrt{2}\hspace{1mm}\text{and}\hspace{1mm} BG=AH=y\sqrt{2}\).

Now that we have determined the lengths of the hypotenuses of the triangle. We can utilize the Pythagorean theorem, \(a^2+b^2=c^2\)

.

We are amazing close to the answer. To know the answer for sure, we must recognize something, first...

Well, let's find the area of all the triangles.

Add these areas together and see what you get.

Do you see the significance of this? I do. We had determined in the previous calculation that \(x^2+y^2=72units^2\). Oh look! That's our answer! The combined area of all the triangles is \(72units^2\). 

I sincerely hoped this helped you out.

Never mind solved it. 4

                                   5

help.

Thanks, heureka....here's another way using similar triangles...

Referring to heureka's pic....let the side of the square = s ...let WA  = x  and let BW  = 3 - x

Now....triangles ABC, AXW and WBZ are all similar

Using triangles ABC and AXW, we have that  XW / AW  = BC/ AC →  s / x  =  4 / 5  →  s  = (4/5)x  →  s  = .8x

And using triangles ABC  and WBZ, we have that  BW / ZW  = BA / CA  →  (3 - x ) / s  = 3 / 5  →  (3 - x) / (.8x)  = .6

Multiply both sides by  .8x  and we have that

3 - x  = .48x      add x to  both sides

3  = 1.48x       divide both sides by 1.48

3 / 1.48  = x   =  75 / 37    ..... and s  is .8   of  this  =  600 / 370  = 60/37 units 

This question doesn't quite make sense. Could you check the wording and maybe edit it a little?

"when P(x) is x+1 and x-2, the remainders are 9 and 12 respectively. What is the remainder when P(x) is divided by (x+1) and (x-2)"

Sorry, I did not respond sooner, but here is an explanation.

I also want to make sure that your question is clear. You want me to explain how to multiply \(\frac{4}{1}*-\frac{3}{2}\) in more detail, I think.

Hopefully, this cleared up any confusion you had earlier. If it did not, reply with a burning question!

How do I do partial fractions.

Hi Max :)

Here is an example.

\(\frac{4}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\\ \frac{4}{(x+2)(x-1)}=\frac{Ax-A}{(x+2)(x-1)}+\frac{Bx+2B}{(x+2)(x-1)}\\ \frac{4}{(x+2)(x-1)}=\frac{Ax-A+Bx+2B}{(x+2)(x-1)}\\ Ax-A+Bx+2B=4\\ (A+B)x+2B-A=4\\ A+B=0 \qquad and \qquad 2B-A=4\\ A=-B \qquad and \qquad 2B-A=4\\ 2B--B=4\\ 3B=4\\ B=\frac{4}{3}\qquad and \qquad A=-\frac{4}{3}\\ so\\ \frac{4}{(x+2)(x-1)}=\frac{-4/3}{x+2}+\frac{4/3}{x-1}\\ \)

There are a number of you tube clips on this topic that will probably help more :)

It is discriminant  Not discrimante   :)

\(y = x^2-4x+5\\ \triangle=b^2-4ac\\ \triangle=(-4)^2-4*1*5=16-20=-4<0\\\)

The discriminant is -4 which is negative so there are no real roots :)

 y = Root(x) * (3-x)2

Find the x-coordinates of all points on y = Root(x) * (3-x)2  where the tangent is horizontal.

The first derivative GIVES the grandient of the tangent at any specific or general point. It is very important that you understand and remember this!

If the tangent is horizontal then the gradient of the tangent is 0. 

So yes, you are being asked where dy/dx=0

\( y = \sqrt x * (3-x)^2\\ y = x ^{0.5}* (3-x)^2\\ \frac{dy}{dx}=0.5x^{-0.5}(3-x)^2\quad+\quad x^{0.5}*2(3-x)*-1\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad 2(3-x)x^{0.5}\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad 2(3-x)x^{0.5}*\frac{2x^{0.5}}{2x^{0.5}}\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad \frac{4(3-x)x}{2x^{0.5}}\\ \frac{dy}{dx}=\frac{(3-x)^2-4(3-x)x}{2x^{0.5}}\quad\\ \frac{dy}{dx}=\frac{(3-x)[(3-x)-4x]}{2x^{0.5}}\quad\\ \frac{dy}{dx}=\frac{(3-x)(3-5x)}{2x^{0.5}}\quad\\\)

This will equal zero when  3-x=0   and when  3-5x=0

That is when x=3 and when x= 0.6

Substitute these points back inot the original equation to find the vaues of y.

\(When\;\; x=3\\ y=\sqrt3(0)=0\\ When\;\; x=0.6\\ y=\sqrt{0.6}(3-0.6)^2\\ y=\frac{\sqrt{3}}{\sqrt5}\frac{12^2}{5^2}\\ y=\frac{144\sqrt{3}}{25\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}\\ y=\frac{144\sqrt{15}}{125}\\~\\ \text{The tangent to the curve will be horizontal}\\ \text{ at the points} (3,\frac{3}{5})\;\;and\;\;(\frac{3}{5},\frac{144\sqrt{15}}{125})\)

You should check my working. I have not done so there could be errors.

Your derivative is correct...here are the steps...

\(\begin{array}\ y&=&x^{\frac12} \cdot(3-x)^2 \\ \frac{dy}{dx}&=&(\frac{d}{dx} x^{\frac12})(3-x)^2+(\,\frac{d}{dx}(3-x)^2\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(\,\frac{d}{dx}(3-x)\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(-1)(x^{\frac12}) \\ \frac{dy}{dx}&=&(3-x)(\quad(\frac12)(x^{-\frac12})(3-x)+(2)(-1)(x^{\frac12})\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(\quad(x^{-\frac12})(3-x)-4x^{\frac12}\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(\quad(3-x)-4x\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(3-5x) \end{array}\)

An isosceles right triangle is removed from each corner of a square piece of paper, as shown, to create a rectangle. If  AB=12 units, what is the combined area of the four removed triangles, in square units?

Hello MIRB15!

The sides of the square are a.
Let the distance to A be x.

Then

\(a=2x +\sqrt{12^2-a^2}\) 

\(x=\frac{a-\sqrt{12^2-a^2}}{2}\)

Small triangles

\(A_{small}=x^2\)
Large triangles

\(A_{large}=(a-x)^2=a^2-2ax+x^2\)

All triangels

\(A_{all}=x^2+a^2-2ax+x^2\\ A_{all}=a^2-2x(a-x)\\ \color{blue}A_{all}=a^2-2\times \frac{a-\sqrt{12^2-a^2}}{2}\times (a-\frac{a-\sqrt{12^2-a^2}}{2})\)

\(\)\(A_{all}=a^2-2\times \frac{a-\sqrt{12^2-a^2}}{2}\times \frac{a-\sqrt{12^2-a^2}}{2}\) 

\(A_{all}=a^2-\frac{1}{2}\times (a-\sqrt{12^2-a^2})^2\)

\(A_{all}=-\frac{1}{2}\times (a^2-2\sqrt{12^2-a^2}+12^2)\)

Later it goes on. There should be a number coming out.

  !