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I can only think of 9 such numbers:

1234  1357  2345  2468  3456  3579  4567  5678  6789  Total =  9

$ 12.10  /  $ 2.20/gal  =  12.10/2.20 gallons = .........gallons

speech-less    

speechless      Answer B

Convert 152,100,000 km  x 1,000 meters =152,100,000,000 meters.

152,100,000,000 x 10^24 =1.521  x  10^35 - yoctometers - distance to the Sun.

152,100,000,000 x 10^-24=1.521  x  10^-13 - yottameters - distance to the Sun.

I’m kinda partial to Alexandra Airhead, and I used Alexandra Ignoramus as a pseudonym during an online debate of religious philosophy.

I have to find angles x and y in these problems.  Please help!

You have No Idea how to solve this?  Why?  Are you too fucking lazy to pay attention in class, or to read your text examples? 

These are very easy questions! 

Guess what?  I’m too fucking lazy to do your homework for you. 

--. .-

GA

The remainder is 2^2 = 4.

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Extra! Extra! Mathematician tosses firecracker... Projectile Equations explode in his face...

thank u much

If the x value is 3 find the y coordinate of the point when the x coordinate is 3.

m<2=180°-88°-58°=34°

m<5=m<3+m<2

m<2=m<5-m<3

m<2=113°-82°=31°

The answer is approx 52241.06414.

Unfortunately I have already thrown out the working that I did for your question....

oh well..

Part c.  I think your answer is a bit wrong.

the points are   (3, 39) and  (3.001, 39.042)               rounded more

the gradient between these points will be    \(\frac{39.042-39}{3.001-3}\)

Can you see what you have done wrong?

It is an approximation because they actually asked you for the gradient of the tangent at (3, 39) 

And you have found the gradient of the secant between (3,39) and (3.001, 39.042)

You could just as easily found the gradient of the secant between the points where x=3 and x=2.999.  That would have been slightly different but also an acceptable estimation.

Part b

b. Given is that the differential quotient of f on the interval [0, a] equals 0. How do I calculate a?

This just means that the gradient of the line joining the 2 points, one where x=0 and the other point where x=a has a gradient of 0

  \(f(0)=12\\ f(a)=2a^3-a^2-6a+12\\ so\\ \frac{f(a)-f(0)}{a-0}=0\\ \frac{2a^3-a^2-6a+12-12}{a-0}=0\\ \frac{2a^3-a^2-6a}{a}=0\\\)

Can you take it from there?

The other side of that coin is that some people give too little explanation, or none at all. 

_.

REMEMBER:  in a right triangle, sin = opposite / hypotenuse

                                      sin x = 52/81

                                          x = arcsin (52/81) = .........

2 = e^{.075 t}          '2' represents 'double'

t = 9.24 years

x component of speed =  x cos 60 = 30      reveals  x = initial speed = 60 m/s

now the path the firecracker follows is given by

    x = 60 cos 60   t  = 30 t

    y = 60 sin 60  t        (this ignores the effect of gravity.....for this short period  t  .....to make much simpler calculations )

Using distance formula      15²   = (30t)2  + (30 sqrt3 t)²

                                                  15^2 = 900t^2 + 2700 t^2

                                                    t = 1/4 sec                  in 1/4 second the firecracker will get this high:    60 sin 60  (1/4) = ~ 13 meters

x+y = 7      or      y = 7-x

xy = 4       x(7-x) = 4

                7x - x^2 -4 = 0     use Quadratic Formula to find       7/2  +- sqrt (33) /2

test 1234

test 123

Your answer is incorrect!

1/       Rotation of 90º counterclockwise about the origin 

2/      Translation 2 units up

" Find the area inside the square but outside the 4 small circles."

What about the big circle???