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xy+4x=2880

xy-12y=2160

x and y in both equations is the same

\(\begin{array}{|lrcll|} \hline (1) & xy+4x &=& 2880 \\ (2) & xy-12y &=& 2160 \\ \hline \\ (1)-(2): & xy+4x -(xy-12y)&=& 2880-2160 \\ & xy+4x -xy+12y &=& 720 \\ & 4x+12y &=& 720 \quad & | \quad : 4 \\ & x+3y &=& 180 \\ (3) & x &=& 180-3y \\ \hline \end{array}\)

(3) in (2):

\(\begin{array}{|rcll|} \hline (2) & xy-12y &=& 2160 \\ & y\cdot (x-12) &=& 2160 \quad & | \quad x = 180-3y \\ & y\cdot (180-3y-12) &=& 2160 \\ & 180y-3y^2-12y &=& 2160 \\ & -3y^2+168y &=& 2160 \quad & | \quad : (-3) \\ & y^2-56y &=&-720 \\ & y^2-56y +720 &=& 0 \\ & (y-36)(y-20) &=& 0 \\\\ & y_1 = 36 &\text{or}& \quad y_2 = 20 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3) & x &=& 180-3y \\ & x_1 &=& 180-3\cdot 36 \\ & x_1 &=& 72 \\\\ & x_2 &=& 180-3\cdot 20 \\ & x_2 &=& 120 \\ \hline \end{array}\)

Solutions: (72,36) and (120,20)

x²+4x-5x²-1

= -4x² + 4x - 1

= \(-4(x^{2}-x+\frac{1}{4})\)

= \(-4(x - \frac{1}{2})(x - \frac{1}{2})\)

= \(-4(x - \frac{1}{2})^{2}\)

50! /(45! x5!)

\(\begin{array}{|rcll|} \hline && \frac{50!}{45!5!} \\ &=& \dbinom{50}{5} \\ &=& \frac{50}{5}\cdot \frac{49}{4}\cdot \frac{48}{3}\cdot \frac{47}{2}\cdot \frac{46}{1} \\ &=& \frac{50}{5}\cdot 49\cdot \frac{48}{12}\cdot \frac{47}{1}\cdot \frac{46}{2} \\ &=&10\cdot 49\cdot 4\cdot 47\cdot 23 \\ &=& \mathbf{2118760} \\ \hline \end{array}\)

The wall, ladder and ground form a right triangle. The hypotenuse is the ladder, which is 20 ft.

sin85°= x/20

20sin85°=x

x ≈ 19.924 ft.

arctan x = 90°

tangent both sides

tan(arctan x) = tan 90°

x = undefined, its 1/0

i want to k**l myself

it will be +10i or -10i

it is impossible to have a real number as a square root of a negative number

45/5=45

good work my dude

it is i dony know

Second Best = Best + 4 sec    (1)

(4/3)Best  =  Second Best  (2)

Sub (1) into (2)

(4/3)Best  =  Best + 4

(1/3)Best  = 4

Best  = 12  seconds

Second Best  = (4/3)Best  =  (4/3)12  =  48/3   = 16 seconds

x^2+4x-5x^2-1

You have to look for commmon terms

4x^2+4x-1

And you have a cuadratic equation

Yep...Happy BDay, Alan......!!!!!

lol ur gay

If you solve 6 = -4.9t^2 +14t -0.4 for t, you get that t = 4/7 and t = 16/7. The difference of these is 12/7, so the exact time is 12/7 seconds, as an improper fraction.

. A debt payment of $5500 is due in 27 months. If money is worth 8.4% p.a. compounded quarterly, what is the equivalent payment

( b) 15 months from now? 5 quarters  (c) 27 months from now? 9 quarters   (d) 36 months from now?   12 quarters

interest rate = 8.4/4 =2.1% = 0.021 every 3 months

I am assuming that the debt of $5500 is due in 27 months.  Its monetary value is not that great yet !

\(FV=PV(1+r)^n\\ PV=FV(1+r)^{-n}\\ PV=5500(1.021)^{-9}\)

5500*(1.021)^-9 = 4561.745063099648344

So if $5500 is due in 27 months its current value is  $4561.75

( b) 15 months from now? 5 quarters  

FV(15months)= 4561.75*1.021^5

4561.75*1.021^5 = $5061.28

(c) 27 months from now? 9 quarters

FV(15months)= $5500  that was given in the question

check:

FV(27months)=4561.75*1.021^9

4561.75*1.021^9 = $5500.0059523168345708    near enough :)

  (d) 36 months from now?   12 quarters

FV(36months)=4561.75*1.021^12

4561.75*1.021^12 = $5853.83

See the graph here :   https://www.desmos.com/calculator/3ceh7ii7po 

The  cannonball is above 6m   from ≈ .571s  to ≈ 2.286s

So    2.286 - .571   ≈   343 / 200  seconds    [1.715 sec ]

x = 1. 

No, x=3

3.96+.07=4.03

4.03:1.00= 4.03

The most important thing to understand is that log is a power (or an indice)

If you can remember this one it will help a lot.  You can substitute other values in and use it for other questions.

\(log_{10}100=2\\ because\\ 10^2=100\)

Other than that you just have the log laws to learn

\(log_bX+log_bY=log_b(XY)\\ log_bX-log_bY=log_b(\frac{X}{Y})\\ log(X^y)=ylog(X)\\ \text{And the change of base law:}\\ log_bX=\frac{log_aX}{log_ab} \)

[5 / (2-2)] ^ 1

\([\frac{5}{(2-2)}] ^ 1\\ \text{The value is undefined. You cannot divide by 0} \)

Limit x->infinity (sqrt 9x^2 + x -3x)

\(\displaystyle\lim_{ x\rightarrow \infty}(\sqrt 9x^2 + x -3x)\\ =\displaystyle\lim_{ x\rightarrow \infty}(3x^2 -2x)\\ =\displaystyle\lim_{ x\rightarrow \infty}(3x^2 -2x)\\ =\infty\)

Here is the graph, (y=3x^2-2x)you can see that as x grows larger so does y.

ax^2+bx+c=0      subtract c from both sides

ax^2 + bx  =  -c     divide through by a

x^2  + (b/a)x  =   -c/a    take  (1/2) of  (a/b)  =  b / [2a].....square this = b^2/[4a^2]...

add it  to both sides

x^2 + (b/a)x + b^2/[4a^2]  = -c/a  + b^2/[4a^2]

Factor the left side.....get a common denominator on the right

( x  + b/[2a] )^2  =   [ b^2 - 4ac] / [4a^2]

Take  both square roots

 x  + b/[2a]   =  ±  √  ( [ b^2 - 4ac] / [4a^2]  )

 x  + b/[2a]   =  ±  √  ( [ b^2 - 4ac] / [2a]  )     

Subtract  b/[2a]  from both sides

x  =  ±  √  ( [ b^2 - 4ac] / [2a]  )   -  b/[2a]

x =  ( -b  ±  √  ( [ b^2 - 4ac]  )  / [ 2a ]

Voila!!!!!......this is  the  Quadratic Formula.....

cos(41) = x/11   Multiply both sides by 11

11 cos(41) = x

cos(41)*11 = 8.301805382453 = x