Find all real numbers t such that  $\frac{2}{3} t - 1 < t + 7 \le -2t + 15$. Give your answer as an interval.

Hello Guest!

$\frac{2}{3} t - 1 < t + 7 \le -2t + 15\\ -8 <\frac{1}{3}t\\ \color{blue}t<24$

$​ ​$

$t+7\le-2t+15\\ -8\le -3t\\\color{blue}t\le24$

$t\in \{ \mathbb{R}|t<24\}$

  !

You must have mistyped your question because the cosine of nothing is bigger than  1

Did you mean to say   cos θ = √3 / 3   ?     That is,   $\cos\theta\ =\ \frac{\sqrt3}{3}$   ?

ibelive there is 27 pairs sorry but thats what i got i may be wrong though because i did it off th example of cp's pic and tried my best

(c)

Let's define functions  a,  b,  and  c  like this:

a(x)  =  cot( arctan(x) )   =   $\frac1x$       where  x  is any real number except  0

b(x)  =  cos( arctan(x) )  =  $\sqrt{\frac{1}{x^2+1}}$       where  x  is any real number

c(x)  =  cos( arcsin(x) )   =   $\sqrt{1-x^2}$       where  x  is a real number the interval  [-1, 1]

We start with  0  on the display.

Then we can take the  cos  to get  1

Then we can do  b  to get  $\sqrt{\frac12}$

Then we can do  b  to get  $\sqrt{\frac23}$

Then we can do  c  to get  $\sqrt{\frac13}$

Then we can do  b  to get  $\sqrt{\frac34}$

Then we can do  c  to get  $\sqrt{\frac14}$

Then we can do  b  to get  $\sqrt{\frac45}$

Then we can do  b  to get  $\sqrt{\frac59}$

Then we can do  a  to get  $\frac{3}{\sqrt5}$

Suppose the price of the cellphone is reduced to $450-5x$ dollars; then it will sell $500x+10$ units, so the revenue will be$\begin{align*} (450 - 5x)(500 + 10x) &= 5(90 - x) 10(50 + x) \\ &= 50 (90 - x)(50 + x) \\ &= 50 (-x^2 + 40x + 4500), \end{align*}$

in dollars.

Completing the square on $-x^2 + 40x + 4500,$ we get

$\begin{align*} 50 (-x^2 + 40x + 4500) &= 50 (-(x - 20)^2 + 400 + 4500) \\ &= 50 (-(x - 20)^2 + 4900) \\ &= -50 (x - 20)^2 + 245000. \end{align*}$

This is maximized when $x=20,$ so the optimal price of the smartphone is $450 - 5(20) = \boxed{350}$ dollars.

The total possible committees = C(20,3)

The probability  is  P(2 girls) + P(3 girls)  =

[ C(8,2)* C(12,1)  + C(8,3)] / C(20,3)  =

[336 + 56 ] / 1140 =

392 / 1140  =  98/285 ≈ .349  ≈ 34.9%

Assuming that the balls are drawn sequentially........we can either draw two yelow balls or two blue balls

P( two yellow balls )  =  5/9 * 4/8  =  20/72  = 5/18

P(two blue balls) = 3/9 * 2/8  = 6/72  =  1/12

So.....the total probability is  5/18 + 1/12  =   [10 + 3 ] / 36  =  13 / 36

There's probably something quicker than this !

With thanks to asinis and hectictar, we have the two sequences

sequence [ 1 ] $\displaystyle \cot(\arctan(x))=1/x,$

and 

sequence [ 2 ] $\displaystyle \cot(\arctan(\cos(\arctan(\sqrt{x}))))=\sqrt{x+1}.$

Beginning with a zero on the display, selecting cos, will produce the number 1, which can be thought of the square root of 1.

So, then

sequence  [ 2 ] will get us sqrt(1 + 1 ) = sqrt( 2 ),

sequence  [ 2 ] will get us sqrt(2 + 1) = sqrt( 3 ),

sequence  [ 2 ] will get us sqrt(3 + 1) = sqrt( 4 ),

sequence  [ 1 ] will get us 1/sqrt(4) = sqrt(1/4),

sequence  [ 2 ] will get us sqrt(1/4 + 1) = sqrt(5/4),

sequence  [ 1 ] will get us 1/sqrt(5/4) = sqrt(4/5),

and finally,

sequence  [ 2 ] will get us sqrt(4/5 + 1) = sqrt( 9/5) = 3/sqrt(5).

I don't know why don't show the letters...

P ( 8 out of 10 flips )  = C(10,8) * ( 1/2)^10  =   45 /1024 ≈   4.4%

Note that the shaded area is composed of two triangles...

The one on the left has a height of 4 and a base of 2.....so its area  = (1/2)(2)(4) =  4 units^2

The one on the right has a base of 3 and a height of 4....so its area  = (1/2)(3)(4)  = 6 units ^2

So.....the total shaded area  is  [ 4 + 6 ] iunits^2  = 10 units^2

Juriemagic, it is good to see you here   

$S=2\pi x*h+2*\pi x^2\\ S=2\pi x*\frac{2000}{\pi x^2}+2*\pi x^2\\ S= \frac{4000}{x}+2\pi x^2\\ S= 4000x^{-1}+2\pi x^2\\ \frac{dS}{dx}= -4000x^{-2}+4\pi x\\ \frac{d^2S}{dx^2}= 8000x^{-3}+4\pi\\ \frac{d^2S}{dx^2}>0 \qquad \text{since x>0, concave up} \\ \text{So any turning point where x>0 will be a minimum}\\ \text{find minimum}\\ \frac{dS}{dx}= -4000x^{-2}+4\pi x=0\qquad x>0\\ -4000+4\pi x^3=0\\ \pi x^3=1000\\ x=\frac{10}{\sqrt[3]{\pi}}$

so for minimum surface area  the radius is    $\frac{10}{\sqrt[3]{\pi}}\;\;cm$        and the height is   

$h=  \frac{2000}{\pi x^2}\\ h=  2000 \div( \pi x^2)\\ h=  2000 \div[ \pi (\frac{10}{\pi ^{1/3}})^2]\\ h=  2000 \div[ \frac{100\pi}{\pi ^{2/3}}]\\ h=  2000 \div[ 100\pi^{1/3}]\\ h=\frac{20}{\sqrt[3]{\pi}}$

$min SA=4000x^{-1}+2\pi x^2\\ min SA=4000(\frac{10}{\pi^{1/3}})^{-1}+2\pi (\frac{10}{\pi^{1/3}})^2\\ min SA=4000(\frac{\pi^{1/3}}{10})+2\pi (\frac{100}{\pi^{2/3}})\\ min SA=400\pi^{1/3}+(\frac{200\pi}{\pi^{2/3}})\\ min SA=600\pi^{1/3}\\$

solve the equation. 

$6 = -4.9t^2 + 14t - 0.4$

Hint :  it is a quadratic.  Take it all to one side and use the quadratic formula.

The height will be above 6m between the time periods.

And for peats sake no one rush in a give a full answer. 

This forum is supposed to help people learn!

Asker if you have problems you can show what you have done and maybe we will help you more.

Thanks!

(b)

We start with  sqrt(x)  on the display.

Then take the  arctan  of what is on the display.

Then take the  cos  of what is on the display.  This gives us  $\frac{1}{\sqrt{x+1}}$

Then...just like from part (a).....we can take the  arctan  of what is on the display.

Then take the  cot  of what is on the display.

Altogether:

$\begin{array}{ccc} &\cot( \arctan( &\cos( \arctan(\sqrt{x}\,)\,) & ) \quad ) \\~\\ =&\cot( \arctan( &\frac{1}{\sqrt{x+1}}& ) \quad ) \\~\\ =&\sqrt{x+1} \end{array}$

I re-wrote the proof by myself to make sure I got it:

Prove of $\frac{2\cdot\sin2^{\circ}+4\cdot\sin4^{\circ}+\cdots+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}}{90} = \cot 1^{\circ}$:

L.H.S.= $\frac{2\cdot\sin2^{\circ}+4\cdot\sin4^{\circ}+\cdots+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}}{90} = \cot 1^{\circ}$
 
We can use the identity $\sin x = \sin (180^{\circ} - x)$ to get:

L.H.S.= $\frac{2\cdot\sin2^{\circ}+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}+4\cdot\sin4^{\circ}+\cdots}{90} = \cot 1^{\circ}$

We can simplify to get:

L.H.S.= $2\cdot(\sin 2^{\circ}+\sin 4^{\circ}+\cdots+\sin 88^{\circ})+\sin 90^{\circ}$

We can multiply by $\frac{\sin1^{\circ}}{\sin1^{\circ}}$ to get:

L.H.S.= $\frac{2\cdot(\sin 2^{\circ}\cdot\sin 1^{\circ}+\sin 4^{\circ}\cdot\sin 1^{\circ}+\cdots+\sin 88^{\circ}\cdot\sin 1^{\circ})+\sin 90^{\circ}\cdot\sin 1^{\circ}}{\sin1^{\circ}}$

Then, we can use the sum-to-product (or Prosthaphaeresis :)) identities to get:

L.H.S.= $\frac{\cos1^{\circ}-\cos3^{\circ}+\cos3^{\circ}-\cos5^{\circ}+\cdots+\cos87^{\circ}-\cos89^{\circ}+\sin90^{\circ}\cdot\sin1^{\circ}}{\sin1^{\circ}}$

We can cancel the terms to get:

L.H.S.= $\frac{\cos1^{\circ}-\cos89^{\circ}+\frac{\cos89^{\circ}-\cos91^{\circ}}{2}}{\sin1^{\circ}}$

We can simplify to get:

L.H.S.= $\frac{\cos1^{\circ}-\frac{\cos89^{\circ}+\cos91^{\circ}}{2}}{\sin1^{\circ}}$

We can use the identity $\cos{x}=\cos{(180-x)}$ to get:

L.H.S.= $\frac{\cos1^{\circ}-\frac{\cos89^{\circ}-\cos89^{\circ}}{2}}{\sin1^{\circ}}$

L.H.S.= $\frac{\cos1^{\circ}}{\sin1^{\circ}}=\cot1^{\circ}$ R.H.S.= $\cot1^{\circ}$

$L.H.S.=R.H.S.$

If $\frac{2\cdot\sin2^{\circ}+4\cdot\sin4^{\circ}+\cdots+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}}{90} = \cot 1^{\circ}$, that means the average of $n \sin n^\circ$($n = 2, 4, 6, \ldots, 180$) is $\cot 1^\circ$.

Therefore, the average of $n \sin n^\circ$($n = 2, 4, 6, \ldots, 180$)is $\cot 1^\circ$.

Twelve 1 by 1 squares form a rectangle, as shown. What is the total area of the shaded region?

$\text{Let the area of the lower triangle is $\dfrac{2\cdot 4}{2} = 4 $ }\\ \text{Let the area of the upper triangle is $\dfrac{3\cdot 4}{2} = 6 $ }\\ \text{The area of the shaded region is $4+6=\mathbf{10} $ }$

A square and a triangle have the same area.

If the square has a side length of 6 units and the triangle has a base of 8 units, what is the length, in units, of the altitude to that base of the triangle?

$\begin{array}{|rcll|} \hline \dfrac{8h}{2} &=& 6^2 \\ 4h &=& 36 \\ h &=& \dfrac{36}{4} \\ \mathbf{h} &=& \mathbf{9} \\ \hline \end{array}$

The altitude  of the triangle is 9 units

A square is divided, as shown. What fraction of the area of the square is shaded? Express your answer as a fraction.

$\text{Let $s$ is the side of the square }\\ \text{Let $s^2$ is the area of the square }\\ \text{Let $s\sqrt{2}$ is the diagonal of the square } \\ \text{Let $s\dfrac{\sqrt{2}}{2}$ is the parallel diagonal of the square }$

$\begin{array}{|rcll|} \hline \mathbf{A_{\text{trapezoid}}} &=& \left(\dfrac{s\sqrt{2}+s\dfrac{\sqrt{2}}{2}}{2}\right)\cdot s\dfrac{\sqrt{2}}{4} \quad &| \quad s\dfrac{\sqrt{2}}{4} \text{ is the height of the trapezoid } \\ &=& \left(\dfrac{1+ \dfrac{1}{2} }{2}\right)\cdot s\sqrt{2} s\dfrac{\sqrt{2}}{4} \\ &=& \left(\dfrac{\dfrac{3}{2} }{2}\right)\cdot s^2 \dfrac{2}{4} \\ &=& \left(\dfrac{3}{4}\right)\cdot s^2 \dfrac{1}{2} \\ &=& \mathbf{ \dfrac{3}{8} s^2} \\\\ \mathbf{A_{\text{shaded}}} &=& \dfrac{A_{\text{trapezoid}}}{2} \\ &=& \dfrac{\dfrac{3}{8} s^2}{2} \\ \mathbf{A_{\text{shaded}}} &=& \mathbf{ \dfrac{3}{16} s^2} \\ \hline \end{array}$

The fraction of the area of the square shaded is $\dfrac{\dfrac{3}{16} s^2}{s^2} = \dfrac{3}{16}$

OK, I wont delete my stuff next time

Thanks!

Thank You, CPhill !

yes theres some to spear 

🎂 

Call the side of the square , S

And its area  = S^2

We can see that  3/4   of the area of the square is unshaded

So...we just need to  find the area shaded in the remaining  1/4 th

Note that at the bottom right of the figure we have an isosceles right triangle  with leg lengths of S / 2

So....the area of this triangle  is   (1/2) (product of the leg lengths)  = (1/2) (S/2) (S/2) =  S^2/8

But  1/2 of  the area of this triangle  plus the shaded area  comprise the other 1/4th area of the square

So 1/2 of the area of the right triangle  = (1/2)(S^2)/8  = S^2/16  = (1/16)S^2

So.....the shaded area  =   (1/4)S^2  - S^2/16   =   (4/16)S^2 - (1/16)S^2  =  (3/16)S^2

So....the shaded part  is just 3/16  of the area of the square