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[9,999,999,998 - 1,000,000,000] / 2  + 1 =4,500,000,000 such numbers

I'm assuming that "·" means "multiplication", not "composition".

f(x)  =  x - 1          g(x)  =  4x² + 3x - 5

(f · g)(x)  =  f(x) · g(x)  =  (x - 1) · (4x² + 3x - 5)

                                   =  (x)(4x²) + (x)(3x) - (x)(5) + (-1)(4x²) + (-1)(3x) - (-1)(5)

                                   =  4x³ + 3x² - 5x - 4x² - 3x + 5

                                          =   4x³ - x² - 8x + 5

(f · g)(-2)  =  4(-2)³ - (-2)² - 8(-2) + 5  =  -15

The next problem is like this problem

(1023 , 1024 , 1032 , 1034 , 1042 , 1043 , 1203 , 1204 , 1230 , 1234 , 1240 , 1243 , 1302 , 1304 , 1320 , 1324 , 1340 , 1342 , 1402 , 1403 , 1420 , 1423 , 1430 , 1432 , 2013 , 2014 , 2031 , 2034 , 2041 , 2043 , 2103 , 2104 , 2130 , 2134 , 2140 , 2143 , 2301 , 2304 , 2310 , 2314 , 2340 , 2341 , 2401 , 2403 , 2410 , 2413 , 2430 , 2431 , 3012 , 3014 , 3021 , 3024 , 3041 , 3042 , 3102 , 3104 , 3120 , 3124 , 3140 , 3142 , 3201 , 3204 , 3210 , 3214 , 3240 , 3241 , 3401 , 3402 , 3410 , 3412 , 3420 , 3421 , 4012 , 4013 , 4021 , 4023 , 4031 , 4032 , 4102 , 4103 , 4120 , 4123 , 4130 , 4132 , 4201 , 4203 , 4210 , 4213 , 4230 , 4231 , 4301 , 4302 , 4310 , 4312 , 4320 , 4321) = 96 such numbers =259,980 - their sum

26640

Find the area

Hi, I can't seem to see the picture of the triangle...

A formula to find the sum of an arithmetic sequence is:  Sum  =  N(F + L)/2

   N = number of numbers = 2197

   F = first number

   L = last number = F + 2196

   Sum = 65⁴ = 17,850,625

--->     17,850,625  =  2197( F + (F + 2196) ) / 2

           35,701,250  =  2197( 2F + 2196)

                   16250  =  2F + 2196

                   14054  =  F

                     7027  =  F

You can find the last term ...

pls guys

H  x  L  = 28 ft²  

H = L-3     subst this in

(L-3)    x  L = 28

L^2 -3L -28 = 0      use Quadratic Formula to find   L  =   7ft    or  -4 (throw out)

The   H = 7-3 = 4 ft

When f(x) = -5

-5 = x^2 - 8x + 7

0 = x^2 - 8x + 12

0 = (x-6)(x-2)             so a value of   x = 6  or  2    will result in f(x) = -5

Wellllll   now you have two points     (6, -5)    and   (2,-5)  

Leading coefficient needs to be  1    so filter out 2

2  (x^2 -3/2 x )     = 5     divied both sides by 2

*x^2 - 3/2 x    = 5/2

    Now take 1/2 of the x coeeficient   3/4     square it    and add it to the  equation.....both sides

    (x^2 -3/2 x + 9/16 )      = 5/2 + 9/16

(x- 3/4)^2   =  49/16       or   (x-3/4)^2 - 49/16 = 0 

     Now you can SOLVE it  (though Q did not ask this)    take the sqrt of both sides

x-3/4  = +- 7/4

x = -1   and  10/4

To have a vertical asymptote of x = 7, the function needs a factor of (x - 7) in the denominator.

The function then can look like this:  f(x)  =  a / (x - 7)  for any non-zero value of a.

This function will have a horizontal asymptote at y = 0.

For this function to have a horizontal asymptote of y = -3, the function can be modified to be:  f(x)  =  a / (x - 7) - 3.

Rewriting the function, using the common denominator:  f(x)  =  [ -3x + (21 + a) ] / (x - 7)

Since the y-intercept is (0,6):                                              6    =   (21 + a) / -7

Solving for a:                                                                         a  =  -63

The function is now:  f(x)  =  (-3x - 42) /(x - 7)

Thank you for your time, seriously. With the fraction part, I'm just as confused as you are. It'd just be easier to enter in a decimal in my opinion, but it is what it is lol. I appreciate you helping me :)

any tips guys?

No worries.....   I make a lot of mistakes !

Mine is supposed to be 4π/15. That's my bad, omigosh. I'm sorry, I didn't even realize that until you said something. 

Rounded up, right? 

Hmmmm     2/15  * 49 pi = 20.53   in²       via my calculator......

Also I used  4pi/15         your question   had   4 pi 15       ....  Which is it supposed to be?

   since it asks for a fraction answer , I'm thinking it might be something different.....

    Maybe in terms of pi?

  2/15  x 49 pi =   98/15  pi     ( 6  8/15) pi

                                                  3sqrt(x)  =  x + k

Squaring both sides:                          9x  =  x² + 2xk + k²

Rewriting:                   x² - 9x + 2kx + k²  =  0

                                   x² + (2k - 9)x + k²  =  0

This is a quadratic; it will have a unique solution when the discriminant of the quadratic formula is zero.

For the quadratic formula:  a = 1     b = 2k - 9     c = k²

Discriminant:                  b² - 4·a·c  =  0

                       (2k - 9)² - 4·(1)·(k²)  =  0       

                       4k² - 36k + 81 - 4k²  =  0

                                       -36k + 81  =  0

                                               -36k  = -81

                                                    k  =  9/4

volume of a cone   =    1/3  pi r^2 h      r = 2.3   h=15 m       Have at it !  

Alright, so I attempted it for myself & I'm still very confused. When I enter what you gave me for the total area, I got 1.315947252 = .133333333.

I'm not sure what I'm doing wrong.. And when I tried to solve for the last part, I got 20.525. But is that the area of the whole circle or the sector? 

Guest is correct -- but there is an easier way ...

Using the Law of Cosines:

[You can't use the Pythagorean Theorem because it isn't a right triangle.]

Use the Law of Cosines to find the third side of the triangle:  c²  =  a² + b² - 2·a·b·cos(C)

   --->   c²  =  (2.6)² + (5.6)² - 2·(2.6)·(5.6)·cos(70)

   --->   c²  =  28.160

   --->    c  =  5.3

Now, use Heron's formula:

s  =  (2.6 + 5.6 + 5.3) / 2  =  6.75

Area  =  sqrt( 6.75 · (6.75 - 2.6) · (6.75 - 5.6) · (6.75 - 5.3) )  =  6.83

But, there is an easier way:  use this formula for area:  Area  =  ½ · a · b · sin(C)

                                                                                        Area  =  ½ · (2.6) · (5.6) · sin(70)

                                                                                        Area  =  6.84

The difference between the two answers is due to a rounding error in my using Heron's formula.

Pretty good try....but....Pythagorean Theorem only applies in RIGHT triangles.....

Thank you so much! I wasn't even given that formula.. It's so confusing lol. Thank you again.