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Step number 4 helps with that! I can describe it!

its 4 rays, X is the center point. it goes XP as one point, Below it XP, Below it XR, Below it XS. 

We need to see a diagram of what you're trying to prove.....!!!

The problem here is that we don't know which angle is opposite the side with the length of 10.....is it the 70° angle or the 20° angle ???

First, let us assume it is the 70° angle....by the Law of Sines, we have

x/sin90  = 10/sin70   and x = about 10.64  = the hypotenuse length

And the other leg (L) is just

10/sin70 = L / sin 20   →   L =   10sin20/sin70 = about 3.64

Check   (3.64)^2 + 10^2 =  10.64^2  ??   ...if you evalute this on a calculator, you will see that it is approximately true

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But.....what it the side of 10 is opposite the 20° angle??  .....again, using the Law of Sines we have

x/sin90 = 10/sin 20   and x = about 29.238 =  the hypotetnuse length

And the other leg L is given by:

10/sin20 = L/sin70   →  L = 10sin70/sin20 = about 27.475

Check  (27.475)^2 + 10^2 = 29.238^2 ????   and agian, this is approximately true........

So....we have two possible right triangles depending upon the orientation of the sides and angles

Thanks a bunch! Really helped me out.

A. 

Just divide 56 by 67. BAM

 I shall tell you how to do it!!

All you have to do is combine like terms, im guessing you are solving for A and B.

 so, by combining the "A" Terms, you will get 1014123a.

by combining the "B" Terms, you will get 221111b

by combining the number terms, you will get -721189

THEREFOR!

Your answer is....

1014123a+221111b-721189.

Shrek is love Shrek is Lyfe. Much like Chipotle..... CHIPOTLE IS LYYFE

V=B*h 

where B is the area of the bottom and h is the height. i assume you know how to calculate the volume. but maybe have some problems with the fractions. 

for example you can do like this. 

 2 and 2 over 3.= 2+ (2/3)  1=(3/3) -----> 2= (6/3) ------> 2+(2/3)= (6/3)+(2/3)= 8/3

try to do the other two sides by yourself.

Then you just have to multiply the three different numbers to get the volume.

an example of that. (8/3)*(3/5)*(1/2) = 24/30 = 4/5

i hope you follow 

good luck 

Thank you so much:)

when subtracting integers, change the operation to addition, then change the sign of the subtrahend. if negative, make it positive, if positive make it negative, follow the rules in addition.

1.5=1/cos(x) find x

$$\small{\text{$
\begin{array}{rcl}
\dfrac{ 1 } { \cos{(x)} } &=& 1.5 \\\\
\dfrac{ 1 } { \cos{(x)} } &=& \dfrac{3}{2} \\\\
\cos{(x)} &=& \dfrac{2}{3} \\\\
\cos{(x)} &=& \dfrac{2}{3} \qquad | \qquad \pm\arccos{()}\\\\
x &=& \pm \arccos{ \left(\dfrac{2}{3} \right) }\\\\
x_{1,2} &=& \pm 48.1896851042\ensurement{^{\circ}}\\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{48.1896851042\ensurement{^{\circ}} } \\\\
x_2 &=& -48.1896851042\ensurement{^{\circ}} \\\\
x_2 &=& -48.1896851042\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{311.810314896\ensurement{^{\circ}} }
\end{array}
$}}$$

What is an accumulator bet?

Do you need me to do it with the two points that you randomly chose?

Or can you work that out now.  

I used the proper regression line but the idea is the same and the answers should be more or less similar :)

Unless i have entered the values wrongly, Excel shows me that for 3 rebounds the expected height is approx 188.5cm

If i use the regression formula that Excel gave me

y=0.4063x-70.647

$${\mathtt{3}} = {\mathtt{0.406\: \!3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{70.647}} \Rightarrow {\mathtt{x}} = {\frac{{\mathtt{736\,470}}}{{\mathtt{4\,063}}}} \Rightarrow {\mathtt{x}} = {\mathtt{181.262\: \!613\: \!832\: \!143\: \!736\: \!2}}$$

It can't be 79, no one in the sample is that short :))

You can get 28200 if u dont reduce it with 5 % the last day.

You get max 20 * 1410 = 28200.

$$\small{\text{$
\boxed{~~t_n=0.95^n\cdot \left( \dfrac{25547}{0.95}-20\cdot 1410 \right)+20\cdot 1410~~}
$}}\\\\
\small{\text{We have $t_n = 27924$ and get n by rearrange:
}}\\\\
\small{\text{$
\boxed{
n=\dfrac
{\ln{\left( \dfrac{t_n -20\cdot 1410}{ \dfrac{25547}{0.95} -20\cdot 1410} \right)}}
{\ln{(0.95)}}
}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
n &=& \dfrac
{\ln{\left( \dfrac{27924 -20\cdot 1410}{\dfrac{25547}{0.95}-20\cdot 1410} \right)}}
{\ln{(0.95)}}\\\\
n &=& \dfrac
{\ln{\left( \dfrac{-276}{ -1308.42105263} \right)}}
{\ln{(0.95)}}\\\\
n &=& \dfrac
{\ln{\left( 0.21094127112\right)}}
{\ln{(0.95)}}\\\\
n &=& \dfrac
{-1.55617552018}
{-0.05129329439}\\\\
\mathbf{n} & \mathbf{=} & \mathbf{30.3387711544}
\end{array}
$}}$$

It will take to get 27924 about 30.3387711544 days

day 30 = 27919.16 

day 31 = 27933.20


$$\small{\text{$ 
 \begin{array}{lr}
 \small{\text{day }} 1 & 26957.00 \\
 \small{\text{day }} 2 & 27019.15 \\
 \small{\text{day }} 3 & 27078.19 \\
 \small{\text{day }} 4 & 27134.28 \\
 \small{\text{day }} 5 & 27187.57 \\
 \small{\text{day }} 6 & 27238.19 \\
 \small{\text{day }} 7 & 27286.28 \\
 \small{\text{day }} 8 & 27331.97 \\ 
 \small{\text{day }} 9 & 27375.37 \\
 \small{\text{day }} 10 & 27416.60 \\
 \small{\text{day }} 11 & 27455.77 \\
 \small{\text{day }} 12 & 27492.98 \\
 \small{\text{day }} 13 & 27528.33 \\
 \small{\text{day }} 14 & 27561.92 \\
 \small{\text{day }} 15 & 27593.82 \\
 \small{\text{day }} 16 & 27624.13 \\
 \small{\text{day }} 17 & 27652.92 \\
 \small{\text{day }} 18 & 27680.28 \\
 \small{\text{day }} 19 & 27706.26 \\
 \small{\text{day }} 20 & 27730.95 \\
 \small{\text{day }} 21 & 27754.40 \\
 \small{\text{day }} 22 & 27776.68 \\
 \small{\text{day }} 23 & 27797.85 \\
 \small{\text{day }} 24 & 27817.96 \\
 \small{\text{day }} 25 & 27837.06 \\
 \small{\text{day }} 26 & 27855.20 \\
 \small{\text{day }} 27 & 27872.44 \\
 \small{\text{day }} 28 & 27888.82 \\
 \small{\text{day }} 29 & 27904.38 \\
 \small{\text{day }} 30 & 27919.16 \\
 \small{\text{day }} 31 & 27933.20
 \end{array}
 $}}$$

Good reasoning Melody.

Here's an alternative approach as a check:

.

Thanks Chris,     

It was a great little problem :)

12 cm

14400/(40*30) = 12

I should have clarifyed.

You can get 28200 if u dont reduce it with 5 % the last day.

So how many days will it take to reach 27924 if u start at 25547, add 1410 each day and reduse that new totalsum with 5%?

On the day u reach 27924 u dont reduce the sum.

Because math problem will not solve itself.

baby don't hurt me

don't hurt me

no more...

$${\frac{{\mathtt{48}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{958.41}} = {\mathtt{460.036\: \!8}}$$

$460.04

180% of 820%

$$\left({\frac{{\mathtt{180}}}{{\mathtt{100}}}}\right){\mathtt{\,\times\,}}{\mathtt{820}} = {\mathtt{1\,476}}$$

1476%

if i have 25547 and add 1410 to that sum i get = 26957. if i reduce that new sum with 5 % i get = 25609,15. if i repeat this process each day, how many days will it take to get to 27924?

If I have properly understood: You get max 19 * 1410 = 26790

(19 = 0.95/0.05)

You never reach 27924!

262.5 = 250 + 250*(1/2)*(1/10) so

5% of 262.5 = 12.5 + 12.5*(1/2)*(1/10) = 12.5 + 6.25*(1/10) = 12.5 + 0.625 = 13.125

Is that what you are looking for?

.