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cos^-1( -2·sqrt(2) )  does not exist.

c = cats                    money from = $ 5.5 c

d = dogs  = 8-c         money from = $ 8(8-c)       money summed = $ 54

5.5c   + 8 *(8-c)  = 54       solve for c = number of cats

This is a modified sin fxn...... the very LAST term shifts the midline from zero    down to - 4.2

 c + v +p+l=15

but  p+l = 2 (c+v)

 and  p = l+ 8              l+8    +   l   =  2 (c+v)           (2l+8) 1/2 = c+v

c+v+  p+l = 15

(2l+8)1/2    +   l+8   +  l = 15            l + 4    + l+8     +   l = 15      3l = 3     l = 1                               (p=9       c+v = 5  ) 

-5 sqrt (25*3) - sqrt3

-25 sqrt3 - sqrt3

-26 sqrt3

simplest radical form

sqrt180 = sqrt 5*36     =   6 sqrt 5

Presumably you mean 10ft and 3ft.

The ladder forms the hypotenuse of a right triangle, so use Pythagoras theorem to determine the third side.

height = √(10^2 - 3^2) ft

I don't think we are talking about the same things:

you have:

Here's my final equation:

465000 = 132/d (d- .04) + 6852300

This should be

465000 = 132 (d-.04) + 6.8523

solving for d n     = 67841.2    N   =   67.841 kn

then cc = .802 x 32 x 300 x .89 dn  / 1000  = 464.86 mm

Thanks for your help pavlov. I've just found my error.

Upon calculating the term 132/dn * (dn-40). I used E as 200MPa for steel instead of 200GPa.

I appreiate your time.

unfortunately dn cannot exceed 460mm in this problem

I think the problem is lies with finding the term 6.85dn or 6852300dn

6.8523  kn   dn      when dn is in mm

    when you convert to m   and N

       this just equals    6.8523 N  dn  

ahhh yes. you're right

yeah I don't understand how it's 6.85 either. I was hoping someone could shed some light.

The third image shows how the 6.85dn was calculated

I am referring to your first equation.....it starts as 6.85 kn   which is   6852.3 N   not    6852300 N

Wait....now I see you converted dn to m   from mm which would make it   6852300.....

    looking some more....

Wait...again.....when you converted dn to m  you MULTIPLIED by 1000.....you should DIVIDE by 1000

Then I get   - 6.8523 d^2 + 464868 d + 5.28  = 0     resulting in dn = 67841.2

sorry, I wrote in the *10^-3 at the last second.

I've used 6852300 as Newtons

Pretty good....except you marked amplitude incorrectly on your drawing....it should be 1/2 of the wave height.

How do you go from 6.85 dn   to   6852300 dn

    shouldn't it be    6852.3 dn ?

cosine function

cos x

amplitude (which is 1/2 the total height of the wave) =5

5 cos x

period 2 pi

5 cos x

shifted left  pi/4

5 cos (x + pi/4)

shifted UP 4

5 cos (x+pi/4) +4

The midline of 10 rules out all of the choices except the last two.....

   normal period of sin fxn is 2 pi      if you divide 'x' by pi/2 you stretch the wave out by a factor of 2 /pi     2 pi *  2/pi =4

           so the last one is the correct one

???   Correct your answer. 

1. trapezoid ABCD is inscribed in a semicircle, as shown below. the length of base CD is 7, and the lengths of legs AD and BC are 3. Find the radius of the smicircle.

Radius is  4.5 units     

Please help with this equation my 4th grade math teacher is giving extension

and i know quadratics and other stuff but i don't know how to solve this properly
The real numbers \(x\) and \(y\) satisfy
\(\begin{align*} x + y &= 4, \\ x^2 + y^2 &= 22, \\ x^4 &= y^4 - 176 \sqrt{7}. \end{align*}\)
Compute \(x - y\).

\(\begin{array}{|rcll|} \hline x^4 &=& y^4 - 176 \sqrt{7} \quad | \quad \times(-1) \\ -x^4 &=& -y^4 + 176 \sqrt{7} \quad | \quad +y^4 \\ y^4-x^4 &=& 176 \sqrt{7} \\ && \boxed{\text{Formula }~a^2-b^2 =(a-b)(a+b) }\\ (y^2-x^2)(y^2+x^2) &=& 176 \sqrt{7} \quad | \quad y^2-x^2=(y-x)(y+x) \\ (y-x)(y+x)(y^2+x^2) &=& 176 \sqrt{7} \quad | \quad y+x = 4 \\ 4(y-x)(y^2+x^2) &=& 176 \sqrt{7} \quad | \quad y^2+x^2 = 22 \\ 4*22(y-x)&=& 176 \sqrt{7} \\ 88(y-x)&=& 176 \sqrt{7} \quad | \quad : 88 \\ y-x &=& 2 \sqrt{7} \quad | \quad \times(-1) \\ \mathbf{ x-y } &=& \mathbf{-2 \sqrt{7}} \\ \hline \end{array}\)