CPhill

y= e^x sinx

Remember, Yura_chan.....we are using the Product Rule, here.....

y '  = e^xsinx + e^xcosx  =  e^x(sinx + cosx)

Use the Product and Chain Rules to find y "

y "   = e^x(sinx + cosx)   + e^x(sinx + cosx)^0 * ( cosx - sinx)   =

e^x(sinx + cosx) + e^x (cosx - sinx)  =

e^x [ sinx + cosx + cosx - sinx]  =

e^x [2 cosx] =

2e^xcosx

Nice, DS....

Call the amount in one acct  [ doesn't matter which one], x

Call the amount in the other acct   4500 - x    and we have

.10x  + .03(4500 - x)  = 205     simplify

.10x + 135  - .03.x   = 205

.07x + 135  = 205     subtract 135 from both sides

.07x =  70       divide boh sides by  .07

x = $1000  this was the amount in the 10% account

So  4500 - 1000  = $3500  = the amount in the 3% acct

Yep......maybe even more than one.....

Look at the following graph of the rational function,  [3 -x] / [x + 3]    ......  

https://www.desmos.com/calculator/9liadrn2if

We have a vertical asymptote at x = -3    and a horizontal asymptote at  y = -1

If the slope of  QR  =   x - 1/4    ....a line perpendicular to it will have a negative reciprocal slope....=

-1 / [ x - 1/4]

And since ST  is  perpendicular.....it's slope  8/3  must =  -1 / [x - 1/4]......cross-multiply and we have....

8 [x - 1/4]  = 3 * -1

8x - 2  = -3     add 2 to both sides

8x   = -1        divide both sides by  8

x = -1/8

cos(6x)=sqrt3sin(6x)    

sqaure both sides

cos^2(6x)  = 3sin^2(6x)

cos^2(6x)  = 3 [ 1 - cos^2(6x)]

cos^2(6x)  = 3  - 3cos^2(6x)     re-arrange as

4cos^2(6x) - 3  =  0      factor this as

[2cos(6x) + √3 ] [ 2 cos(6x) - √3] = 0

Setting the first factor to 0 we have

2cos(6x) + √3  = 0

cos(6x)  = -√3 / 2

cos(x)  = -√3 / 2   at      150° + n*360°      and  at  210° + n*360°    where n is an integer

Dividing each of these by 6, we have :

x = 25° + n* 60°      and  x = 35° + n* 60°

As will be shown shortly.....only the second solution is good......the first is an extraneous solution produced by squaring both sides of the original equation

Settintg the second factor to 0 , we have

2cos(6x) - √3  = 0

cos(6x)  = √3 / 2

cos(x)  = √3 / 2      at   30° + n * 360°     and  at 330°  + n*360°

Dividing each of these by 6 we have :

x = 5° + n* 60°     and  x = 55° + n* 60°

For the same reason as above, only the first solution is good.

Here's a graph of the solutions on the interval  [ 0, 360 ]°

https://www.desmos.com/calculator/ywvf0g7bxs

28x^2 / 2   =

(28/2) * x^2  =

14 x^2

You're very talented, DS.......!!!!!

(6/7)(3-5n) ≥ (3/5)(1-9n)

Let's start by multiplying both sides by the common denominator of 7 and 5, i.e., 35.....this will clear the fractions

(35) (6/7)(3-5n) ≥ (35) (3/5)(1-9n)  =

30 (3 - 5n)  ≥  21 (1 - 9n)       simplify

90 - 150n ≥  21 - 189n      add  150n  to both sides, subtract 21 from both sides

69 ≥  - 39n       divide both sides by -39   and reverse the inequality sign

-69/39 ≤ n         reduce the fraction by dividing top/bottom by 3

-23/13 ≤  n    or     written another way, 

n ≥  -23 / 13

Sorry, guest.....I couldn't see 20 clearly [or I mis-read it].....but.....if it says that   AD = DB and AE = EC.....then the sides are cut propotionally by DE. That is, each side is bisected by DE. So, triangle ABC ~  triangle ADE.

By Euclid.....if the sides of a triangle are cut proportionally, the line joining the points of intersection is parallel to the remaining side......thus DE is parallel to DC....and we proved in (18) that, if this was so, the triangles were similar

If you are  interested, here's Euclid's proof :

http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI2.html