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We will actually end up with this equation :

pi*x (10 (sqrt(225 -x^2) + 15) - x^2)  = 0

Divide by pi*x   ....this will drop one solution of x = 0   ...and we have remaining

10(sqrt(225 - x^2) + 15) - x^2 )  = 0

10sqrt(225 - x^2) + 150 - x^2  = 0

10sqrt(225-x^2)  = x^2 - 150      square both sides

100(225-x^2) =  x^4 - 300x^2 + 22500   simplify

22500 - 100x^2 = x^4 - 300x^2 + 22500

x^4 - 200x^2  = 0      factor

x^2(x^2 - 200) = 0     obviously x = 0  is one solution [ actually, x = 0 is a multiple solution]

The other positive solution is

x^2  = 200     take the positive sq root of both sides

x = sqrt(200)   = sqrt(100 * 2)  =  10*sqrt(2) 

Note.....revenue = quantity * price  =  x * p(x)     and the revenue function, R(x)  is :

R(x) =  x* p(x)  = x [75 - (1/4)x^2]  =  75x - (1/4)x^3   take the derivative

R'(x)  = 75 - (3/4)x^2      set this to 0

75 - (3/4)x^2  = 0

75 = (3/4)x^2      multiply both sides by 4/3

100  = x^2    take the positive root

10 =  x     so ....selling 10 cakes maximizes the revenuue

The second derivative = -(4/3)x

And subbing the critical point of x = 10  ito this produces a negative......which indicates a max

And the max revenue is shown by this graph : https://www.desmos.com/calculator/i1wj8zvt4k

The max revenue = $500  

So.....the cake price which produces the max revenue is : 75 - (1/4)(10)^2  =  75 - (1/4)(100)  =   75 - 25  = $50

Which makes sense....!!!

arc tan is the same as inverse tan

just hit [2nd] [atan] 

The answer on the answer sheet says \(x = {10\sqrt{2}}\).

Thanks for your input though, very much appreciated!

I got that x was equal to 0, obviously, but for the problem I needed the other answer which has to be positive because it's a length.

I will look into what you (Melody) did and I'll go on from there.

8/(4/5)=10

Flip the (4/5) to the top and it becomes (5/4). [looks like: (8/1)*(5/4)]

Then you multiply the top numbers and multiply the bottom numbers which comes to be (40/4).

Simplify and it becomes 10.

Made the whole calc in the image thing and it got to long, which resulted in that the "ok" button went too far to the right of the screen that I was unable to click it.. New to this site.

Anyway I got the answer,

X₁ = 0 X₂ ≈ +-12,247..

Although I might aswell be wrong..

Edit: forgot the +-

y = x + 7    (1)

y = 4    (2)

Put (2) into (1)

4 = x + 7     subtract 7 from both sides

-3  = x

So   {x,y}  = { -3, 4}

False.....a dilation does not change the shape....it increases or decreases the size of the shape.....however.....the shapes remain similar is all respects

Here's  a cool demo site :

http://www.mathwarehouse.com/transformations/dilations/dilations-in-math.php

A) is the answer 

I assume that IS the derivate and you just want the equation solved

\(0 = {{\pi x(10(\sqrt{225-x^2}+15)-x^2})\over{\sqrt{225-x^2}}}\\ 0 =\frac{{\pi x(\sqrt{10\sqrt{225-x^2}+15)}+x)(\sqrt{10\sqrt{225-x^2}15)}-x})}{{\sqrt{225-x^2}}}\\ x=0\:\;\:or\;\;(\sqrt{10\sqrt{225-x^2}+15)}+x)=0\;\;or\;\;(\sqrt{10\sqrt{225-x^2}+15)}-x)=0\\ x=0\:\;\:or\;\;(\sqrt{10\sqrt{225-x^2}+15)}=-x\;\;or\;\;(\sqrt{10\sqrt{225-x^2}+15)}=+x\\ x=0\:\;\:or\;\;10\sqrt{225-x^2}+15=x^2\\ x=0\:\;\:or\;\;\sqrt{225-x^2}=\frac{(x^2-15)}{10}\\ x=0\:\;\:or\;\;225-x^2=\frac{(x^4-30x^2+225)}{100}\\ x=0\:\;\:or\;\;225-x^2=\frac{(x^4-30x^2+225)}{100}\\ x=0\:\;\:or\;\;22500-100x^2=x^4-30x^2+225\\ x=0\:\;\:or\;\;x^4+70x^2-22275=0\\ \)

\(x=0\:\;\:or\;\;x^4+70x^2-22275=0\\ x ^2= {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x ^2= {-70 \pm \sqrt{4900+4*22275} \over 2}\\ x ^2= {-70 \pm \sqrt{94000} \over 2}\\ x ^2\approx {-70 \pm306.6 \over 2}\\ x ^2\approx {-70 \pm306.6 \over 2}\\ x ^2\approx 118.3\\ x\approx \pm10.9\\~\\ so\;I\; get\;\\~\\ x=0\;\;or\;\;x\approx \pm10.9\)

You probably should check those 2 approx answers though.   ://

Mmm I do not think it is right.  

I am not sure I am answering the right question anyway.://

x=0 is definitely a solu​tion to the question that I answered.  ://

I'm not sure if this is what you want, or not....

Since "c" = 0  in the quadratic formula....the discriminant is just b^2  = 20^2

The roots of this function will be   [ -20 ± 20] / [2*-16)     t =   0   and   t = -40/-32   = 5/4    [seconds]

Due to the parabola's symmetry, the max height will be reached half-way between these two t values =  at 5/8 seconds......

So....the max height is

-16(5/8)^2 + 20(5/8)   =  25/4 ft =  6.25 ft

[Maybe some other mathematician sees something else here that I don't  ???]

p^(2/5) - 5p^(1/5) + 6  = 0     factor as

[p^(1/5]  - 3)  [ p^(1/5) - 2]  = 0

Set each factor to 0

p^(1/5)  - 3  = 0      add 3 to both sides

p^(1/5)  = 3       take each side to the 5th power

p = 3^5   = 243

p^(1/5) - 2  =   0      add 2 to both sides

p^(1/5)  = 2    take each side to the 5th power

p  = 2^5   = 32

I'm assuming that you want to solve this system :

(1/2)x-1=(2/3)y

(20/9)=(1/3)x+(1/3)y

Multiply the top equation through by 6   and the bottom one through by 9

3x - 6   =  4y     →  3x - 4y   = 6      (1)

20 = 3x +3y    →   3x  +3y  = 20    (2)

Multiply (2)   by -1 all the way through

-3x - 3y  = -20      add this to (1)

-7y = -14      divide both sides by - 7

y = 2

And using (1) to   find x, we have

3x - 4(2)  = 6

3x - 8 =  6     add 8 to both sides

3x = 14    divide both sides by 3

x = 14/3

don't we have to multiply probablities?

Just like this

6*6^2*tan(pi/6)

But make sure you set the units to radians.  There is a toggle button for this on the bottom left of the calculator.

4 l x - 3 l  - 15 ≤  -7     add 15 to both sides

4 l x - 3 l ≤  8        divide   both sides by 4

l x - 3 l  ≤  2

We have this inequality

-2 ≤ x - 3 ≤ 2        add 3 to everything

1 ≤ x ≤ 5

Here's the number line solution :

1-489,657,454,172.74719491603734695166- This is the correct answer.

2-5,660,939,593.3647550067912213-This is the correct answer.

3 gals = 12 qt

The area to be planted =  pi* [275 yds]^2 .....however....we need to convert the yards to ft....so we have......275 yds = 825 ft

Area  = pi* [ 825ft]^2   = 680,625pi ft^2

And the cost  =  $.09 * 680,625pi  =  $192,442.18......!!!!!

(-1)^(1/2)

-14 - 63 = -(14+ 63 ) = ????

How would I prove that "tan(x+1)2 = (1+sin(2x))/cos2x" ?

It is not clear what you want me to prove....

\(tan(x+1)^2 = \frac{1+sin(2x)}{cos2x}\\ or\\ tan(x+1)^2 = \frac{1+sin(2x)}{cos^2x}\\\)

The symbol for power on the net is ^

ex       3^2=9

Could you please rewrite your question clearly.

0<x<1, 2x<y<3x, xy/5<z<5xy

You are told x is between 0 and 1

I seperated the rest into 4 different constraints

2x<y

y>2x

The smallest 2x can be (not quite) is 0 so y>0

y<3x

The limit for 3x is 3 so y has to be smaller than 3

0<y<3

xy/5<z

z>xy/5

the lower limit for x is 0 and the lower limit for y is 0 so 

xy>0

xy/5>0

z>0

z<5xy

The upper limit of xy is 1*3=3

so the upper limit of 5xy is 15

z<15

0<z<15

limit constraints

0<x<1

0<y<3

0<z<15

(20^2) + (20^3) - 6663 = 1738