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For the second method, you (correct me if I'm wrong) overlooked the fact that the probability of each of the possibilities may not be the same.

1) The function is

h(t)  = -16t^2  + vt  + s   =

h(t)  = -16t^2 + 63t  

2)  Here's the graph  (in x instead of t)  : https://www.desmos.com/calculator/96nhz6lqgq

3) Note that with the given function, the ball never rises above ≈  62 ft

4) If the punter's initial velocity is 68 ft/s we have the graph  of -16t^2 + 68t : https://www.desmos.com/calculator/2bn74gqnob

The ball will be > than the stadium height  between  1.75  and 2.125 sec

Let \(x\) be the number of sides in the polygon.

Using what we are given, we get the equation \(120+130(x-1)=180(x-2)\).

Expanding, we have \(120+130x-130=180x-360\). Then, by moving all the terms to the left and combining like terms, the equation turns into \(-50x+350=0\).

Therefore, the polygon has \(\fbox{7}\) sides.

Let \(N\) and \(O\) be functions such that \(N(x)=2\sqrt{x}\) and \(O(x)=x^2\). What is \(N(O(N(O(N(O(3))))\)?

\(O(3)=9 \rightarrow N(9)=6\rightarrow O(6)=36\rightarrow N(36)=12\rightarrow O(12)=144\rightarrow N(144)=\fbox{24}\)

By the property of geometric sequences, \(g_{43} = g_1 \times d^{42}\). 

Note that \(g_{28} = g_{23} \times d^5\). Substituting the given values, we have: \(28 = 16d^5\), meaning \(d = \sqrt[5]{7 \over 4}\).

Now, notice that \(g^{43} = g^{28} \times d^{15}\). 

We know that \(\sqrt[5]{7 \over 4} = {7 \over 4}^3 = {343 \over 64}\)

This means that \(d^{43} = 28 \times {343 \over 64} = \color{brown}\boxed{150 {1 \over 16}}\)

Because both equations are equal to y, we can set them equal to each other: \(3x^2 - 6x = 2x^2 + x - c\)

Bringing everything to the right-hand side gives us the quadratic: \(x^2 - 7x + c = 0\)

For the quadratic to have exactly 1 solution, the discriminant (\(b^2 - 4ac\)) must equal 0. This means \(49 - 4c = 0\)

For the quadratic to have more than 1 solution, the discriminant must be greater than 0. This means: \(49 - 4c >0\)

For the quadratic to have no real solutions, the discriminant must be less than 0. This means:\(49 - 4c <0\)

Can you take it from here?

We can view the problem as if there are 27 dots in a row split into three sections by two dividers. There would be C(26, 2) to choose where to place the dividers. That is equivilant to \(\frac{26!}{2!24!}=\frac{25\cdot26}{2}=325\), so that is our answer.

Steps:

\(\frac{1}{3}t-5

\(\frac{1}{3}t-5<\:t-2\quad \::\:t>\:-\frac{9}{2}\)

\(t-2\le -3t+7\:\:\:\:\:\::\:t\le \frac{9}{4} \)

\(​​t>-\frac{9}{2}\quad \mathrm{and}\quad \:t\le \frac{9}{4}\)

\(-\frac{9}{2}

\(\:(-\frac{9}{2},\:\frac{9}{4}]\)

Hope this helps!

Can you explain how you get the derivative? Please?

Take the derivative of the function

h' t =  -6.22t + 9.33

At 1 sec we have

h'(1)  =   -6.22 (1)  + 9.33  ≈   3.1  ft/s

f(-2)  = (-2)^2 - 2  =  2

f(2) =  2 (2)+ 20  =  24

So

2 +  24   + f(a)  =  0

26  + f(a)  = 0

f(a)  =  -26

So either                                       or                 

a^2 - 2 =  -26                             2a + 20  =  -26                                                                                                

a^2  =  -24                                   2a  = -46

not possible                                  a  = -23     { but a must be >=  0 }

No answer to this one

When it hits the surface, the height = 0...so

-t^2  + 8t + 10  = 0

Use the quad formula

-8  ±  sqrt [ 8^2 -  4 * -1 * 10 ]                   -8 ± sqrt [ 104]           -8 ± 2sqrt (26)

_______________________  =             _____________  =   _____________  = 

           2 (-1)                                                      -2                                -2

4 -sqrt (26) sec  (reject)  or

4  + sqrt (26)sec ≈   9.1 sec

We have  the form

y = a ( x - h)^2  + k

(h,k)   = the vertex =  ( -1, 7)   

So we have

y  = a( x  - -1)^2  + 7

y = a ( x + 1)^2  + 7

And the point (0,20) is on the graph so

20  = a ( 0 +1)^2 + 7

20 - 7  =   a * 1^2

13  = a

The equation is

y =  13 ( x + 1)^2  + 7

Here's a graph :  https://www.desmos.com/calculator/hnmgnpsmel

f(f(x))  + 2x - 1 = (x^2 -2x)^2 - 2 (x^2 - 2x) + 2x - 1 =   [ x^4 - 4x^3 + 4x^2 ] - 2x^2 + 4x + 2x- 1 =

x^4 - 4x^3 + 2x^2 + 6x - 1

So

x^2 - 2x =  x^4 - 4x^3 + 2x^2 + 6x - 1            rearrange as

x^4 - 4x^3 + x^2 + 8x - 1   =  0

See the graph here :  https://www.desmos.com/calculator/os2uyqpfze

The real numbers that make this true   are     x ≈ -1.209   and x ≈ .124

Area of figure = Area of 1st rectangle + Area of 2nd rectangle + Area of semicircle  = (8 * 3) + (6 * 5) + 1/2 (π * 7 * 7) = 24 + 30 + 76.93 = 130.93

Area of quarter circle = 1/4πr²

                                  = 1/4 * π * 3² = 7.065 km²
Area of rectangle = Length * width 

                            = 5 * 8 = 40 km²

Area = 1 * 6 = 6 km²

Total area = (7.065 + 40 + 6) km²

                 = 53.065 km²

For rectangle 1st, length(1) = 13 yard, base(1) = 5 yd

For rectangle 2nd, length(2) = 8 yard, base(2) = 4 yd

Area = area of triangle + area of A rectangle = area of 1st rectangle + area of 2nd rectangle

        = 1/2 * base * height + length(1) * base(1) + length(2) * base(2)

        = 1/2 * 3 * 3 + 5 * 13 + 8 * 4 yd²

        = 9/2 + 65 + 32 yd²

        = 4.5 + 65 + 32 yd²

        = 101.5 yd²

1)  Quarter circle at the top with a radius   =  3.....area =  pi (3)^2 / 4  =  (9/4)pi

2)  A    (4 + 2) x 4  = 6 x 4  rectangle in the middle =  24

3) A 1 x 4  rectangle at the bottom

Again....add these together....answer is in km^2

h(t)  = -(t -14)^2 + 74

The first statement is true

To see why, whenever   t is anything but 14, the first part of the function is negative.....adding this to 74  gives us something < 74

But when t =14, the first part of the function =  0^2....adding this to 74    =   74

Break it up into pieces

1) A 5 x 5  square on the left  = 30

2) A 4 x 8  rectangle at the bottom =  32

3) A (8+1) x 5 rectangle in the middle  = 9 x 5 =  45

4) A triangle at top with a base of   (8 + 5) - (6 + 4)  =  3   and a height of 3 =  (3 * 3) / 2  = 4.5

Now....just add the areas together

Let the amount that Joseph paid =  N

The amount that Tony paid =  3N

The amount that Andy paid  =  2 [3n]  = 6n

N + 3N + 6N  = 126

10N = 126

N = 126/10 =  $12.60

Sharing the cost equally each would pay   126/3 =  $42

So    Joseph would have to pay Andy    42 -12.60  = $29.40

And Tony would have to pay Andy    42 - 3(12.60)  = 42 -  37.80  =  $4.20

Normal  price of the towel set  =   15.96 / (4/5)   =  19.95

floor  [ 200 / 19.95 ]  =  10  sets

floor [ 200 /15.96]  =  12 sets

All we need to do is to look  at 8^(2/3)  = [ (8)^(1/3) ] ^2   =  [cube root of 8] ^2  =   [2] ^2 =  4

The last answer is the only one that works

sqrt (144)  = 12    (1 + 2)^2     = 9  = 1 + 4 + 4

sqrt (169) = 13     ( 1 + 3)^2  = 16  =  1 + 6 + 9

sqrt ( 196)  = 14    ( 1 + 4)^2 =  25  but 1 + 9 + 6  =   16

100°

To convert from degrees to radians use  ( pi / 180)

So

100  * (p /180)  =   (10/18) pi   =   (5/9) pi  (rads)