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Darn, waiting for moderation. Well, it is a 4 by 5 grid with an A in the bottom left hand side, and a B in the top right hand side.

thank you

Answered earlier today here :

https://web2.0calc.com/questions/need-help-plzzz

thank you

A quick google search says    trapezium (USA) fits the bill too

\(f(3) = 7 \neq 19 = g(3)\)

A "kite" is a named quadriateral with no parallel sides.  It has two pairs of adjacent, equal length, sides.

I don't know of any other named quadrilaterals w/o at least one pair of parallel sides.

All the values fall within 3 standard deviations of the mean so it is reasonable to use the Empirical Rule.

Ya got the first part correct.....BUT....

5.5x 10^5 / 1.1 x 10^3  =    5 x 10^2 = 500 times greater

f(x) = -2x^2 + 24x + 1280

This is a downward turning parabola........we can find the x coodinate of the vertex as

-24/ [ (2 (-2) ]  =    -24 / -4   =  6

This means that  the number of people that maximizes the revenue is  32 - 6   =  26

This is the answer to (2)

And for (1)....the max revenue is   

-2(6)^2 + 24(6) + 1280 =

-2(36) + 144 + 1280 =  $1352

reflect the graph over the x-axis                  due to the  negative sign

translate the graph down                             due to the    -1

stretch the graph away from the x-axis         due to the  2

For   f(x) =   1 /  [x - 3]

y intercept.....let x = 0     and we have   1 / [ 0 - 3]  =   -1/3

x intercept...let   y = 0 and we have that      0 = 1 / [ x - 3]    multiply both sides by x - 3   and we get that

0 = 1......so.....there is no x intercept

Vertical asymptote.....set  x - 3 =  0

Add 3 to both sides

x = 3   ⇒  the vertical asymptote

For  g(x)

y intercept =   0

x intercept = 0

Vertical asymptote ⇒  x = 4

LOL!!!

Yes....it was  "2"  complicated.....!!!!

Thanks, EP......I think either RainbowPanda or jennylove posted this before, but I couldn't find the question        

It is a sin function    so   start with   sin x

It is NOT a reflection so it is NOT -sin x        still sin x

Frequency = 1/(8pi)    period = 1/f =   8pi          2pi / (8pi) = 1/4     now we have    sin (1/4  x)

Amplitude = 6       Now we have     6 sin (1/4  x)

Midline 3                                          6 sin (1/4 x) + 3

Thanks, heureka.....that is a nice comprehensive answer !!!

Just raise this to the 6th power and we have

( 2 * 2^(1/3) )^6   =

2^6 *  (2^(1/3))^6 =     { remember  that  (a^m)^n  =  a^(m*n)  }

64  *  2^2 = 

64 * 4 =

256

I think this is the same question....

https://web2.0calc.com/questions/math_85961#r1

Looks good, cerenetie.....!!!

Ocean's depth describes the midline of the function is what we decided the last time this Q was posted.....    ~ EP

I'm sorry...I am still a little unclear.    From the digram you have drawn, is she saying the angle between the sliding cable and the first arm is 60 degrees   and the angle between the two arms is 60 degrees and the angle between the second arm and the cable is also 60 degrees? (meaning one arm is longer than the other)  Do I have that correct?    I'm pretty sure I still need the angle of the sliding cable.....is it 30 degrees as your diagram is labeled?

  The tensions in the two arms will be different from each other.    The magnitude of the  horizontal components must equal each other in opposite directions and the vertical components of the tension in the two arms must sum to the weight of the car (7600kg)....I do not think I can calc that until I have the angle of the sliding cable and clarification of  the  60 degree angles...

   Hope I am not making this muddier than it really is supposed to be! 

We have the form

y = Asin (Bt + C) + D

A is the amplitude.......it is the difference between the highest and lowest point divided by 2  = 20/2 = 10

B is the period......we can find this as follows

B =  2pi/period......the period is 8 seconds.....so we have    2pi / 8  =  pi/4  

C=  the phase  shift  ....since the first point is on the midline which is y = 0.....this is the normal starting point for the sine function....so.....there is no phase shift

D = shift up/down.......there is none since the first point is on the midline,  y = 0

Thus....the function is

y =  10 sin ( (pi/4) t)

Here is a graph :https://www.desmos.com/calculator/queeoexbii

th

We roll a fair 6-sided die 5 times.
What is the probability that exactly 3 of the 5 rolls are either a 1 or a 2?

\(\begin{array}{|rcll|} \hline && \dbinom{5}{3}\left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^2 \\\\ &=& \dbinom{5}{3}\left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right)^2 \\\\ &=& \dfrac{5}{3}\cdot\dfrac{4}{2}\cdot\dfrac{3}{1}\cdot \dfrac{1}{3^3}\cdot \dfrac{4}{3^2} \\\\ &=& 10\cdot \dfrac{4}{3^5} \\\\ &=& 10\cdot \dfrac{4}{243} \\\\ &=& \dfrac{40}{243} \\\\ &=& 0.16460905350\ (16.5 \%) \\ \hline \end{array} \)

Inverse

1.
\(\begin{array}{|rcll|} \hline g(f(x)) & \overset{?}{=} & x \quad | \quad f(x) = -\dfrac{1}{4}x + \dfrac{1}{4} \\ g(-\dfrac{1}{4}x + \dfrac{1}{4} ) & \overset{?}{=} & x \\ -4\left( -\dfrac{1}{4}x + \dfrac{1}{4} \right) +1 & \overset{?}{=} & x \\ x-1+1 & \overset{?}{=} & x \\ x & \overset{!}{=} & x \quad \text{the functions are inverses }\\ \hline \end{array}\)

2.
\(\begin{array}{|rcll|} \hline g(f(x)) & \overset{?}{=} & x \quad | \quad f(x) = \dfrac{5x+20}{8} \\ g(\dfrac{5x+20}{8}) & \overset{?}{=} & x \\ \dfrac{-20+8\left( \dfrac{5x+20}{8} \right) }{5}& \overset{?}{=} & x \\ -4 + \dfrac{5x+20}{5} & \overset{?}{=} & x \\ -4 + x + 4 & \overset{?}{=} & x \\ x & \overset{!}{=} & x \quad \text{the functions are inverses }\\ \hline \end{array}\)

But they are not frills,  they are important   !!!  :)