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=asodaksdopaksdopaksdpoasdk o0asod0- a=sd

ignore this message. 

sqrt  [  (4 -1)^2  +  (-10  - - 2)^2  ]  =  

sqrt   [  3^2   +  (-8)^2  ] =

sqrt [  73]

Perimeter =   4 sqrt  (73)

x   +   y  +   3z   =  10      ⇒    -3x   -  3y    -  9z   =   -30       (1)

-4x +  3y  +  5z   =  7            (2)

 kx            + z     =  3             (3)

Add (1)  and (2)   and  we get that

-7x  - 4z   =  -23     (4)

Multiply   (3)  by   -4  =    -4kx  -  4z  =   -12      (5)

Note  that     if      -4k = -7   ⇒    k   =  7/4.....we will  have the equations

-7x  - 4z =   -23

-7x  -  4z  =  -12         

And it is impossible  for  equal   values on the left side  to  produce unequal  values on  the right

So   no solutions if    k =  7/4

https://web2.0calc.com/

For a certain value of k, the system has no solutions.

\(\begin{align*} x + y + 3z &= 10, \\ -4x + 3y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)

Real numbers x and y satisfy
x + xy = 250y
x - xy = -240y

My attempt:

\(\begin{array}{|lrcll|} \hline (1): & x + xy &=& 250y \\ (2): & x - xy &=& -240y \\ \hline (1)+(2):& 2x &=& 10y \quad | \quad :2 \\ &x &=& 5y \quad \text{or} \quad y=\dfrac{x}{5} \\\\ & \dfrac{x}{y} &=& 5 \\ \hline (1):& x + xy &=& 250y \quad|\quad :y \\ & \dfrac{x}{y} + x &=& 250 \\\\ & 5+x &=& 250 \\ & \mathbf{x} &=& \mathbf{245} \\ \hline &y &=& \dfrac{x}{5} \\\\ &y &=& \dfrac{245}{5} \\ & \mathbf{y} &=& \mathbf{49} \\ \hline \end{array}\)

If   8 +  sqrt (3)  is a root, then so is its  conjugate,  8  -  sqrt (3)

6t^2 -47t + 30 = 0

Quadratic formula   a = 6   b = -47   c = 30

\(t = {-(-47) \pm \sqrt{(-47)^2-4(6)(30)} \over 2(6)}\)

Yields t = 47/12 +- sqrt (1489)/12       You can finish ...I hope  

Y

           28

24           

X                Z

XZ  =  sqrt   [  YZ^2  -  XY^2  ] =  sqrt [ 28^2  - 24^2  ]  =  sqrt (208)  =  sqrt (16 * 13)  =   4sqrt (13)

tan Y    =   XZ /  XY =    4sqrt(13) / 24   =      sqrt (13)  /  6

m = slope between the two points

   = ( -7 - -1) /(4-5) = -6/-1 = 6   

perpindicular slope = - 1/m =   - 1/6

150 = h

150 = -16t^2 + 28t + 144

-16t^2 + 28t - 6 = 0    Quadratic formula shows   t =    1/4  second  ( on the wy up)   and     1.5 seconds (on the way down)

The function \(f(n)\) satisfies \(f(1)=1\) and \(f(2n+1)=f(n)+2\) for \(n \ge 0\).
Find \(f(15)\).

\(\begin{array}{|lrcll|} \hline n=1:& f(2*1+1) &=& f(1) + 2 \quad | \quad f(1)=1 \\ &f(3) &=& 1+2 \\ &f(3)&=& 3 \\ \hline n=3:& f(2*3+1) &=& f(3) + 2 \quad | \quad f(3)=3 \\ &f(7) &=& 3+2 \\ &f(7)&=& 5 \\ \hline n=7:& f(2*7+1) &=& f(7) + 2 \quad | \quad f(7)=5 \\ &f(15) &=& 5+2 \\ &\mathbf{ f(15) } &=& \mathbf{7} \\ \hline \end{array}\)

Two ......

The square root of 2x is greater than 3 and less than 40. How many integer values of x satisfy this condition?

This question https://web2.0calc.com/questions/help_18510 has the same kind of problem.  You can use it as an example.  Solve this one the same way.  

Why is it that you call yourself 'thanks for the help' but you have never thanked anyone for the help?

Also, although you have only posted 4 question posts you have asked many more questions than that.

Please confine yourself to one question per post.  (like you have here)

60/250=0.24 = 24%

8!/(3!2!)

List them: 100008, 100017, ..., 100179 ==> $\boxed{19}$ multiples of 9.

bruv why was this blocked lmao

anyway it's kinda useless cuz you can just open it on another window

The problem can be easily solved using the law of sines:

\(\frac{\sin(120)}{2\sqrt{3}}=\frac{\sin(45)}{x}\\ \frac{\frac{\sqrt{3}}{2}}{2\sqrt{3}}=\frac{\frac{\sqrt{2}}{2}}{x}\\ \frac{1}{2}=\frac{\sqrt{2}}{x}\)

Should be trivial by now. Can you finish it?

sorry about that I don't think you can do anything about that I hope you get unblocked

https://web2.0calc.com/questions/please-help-asap-thanks_11

The length of side AB is 35/3.

The set of values of a is [28,inf).

The sum of the minimum values is -64.

By stars-and-bars, the number of ways is C(12,4)*C(11,4) = 163,350.