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Very Curious, George!  How is that you can post a moderately complex solution, but do not know how to post an image?

Why did you address this to Hectictar?

Here is my "guess" (if I understand your question) !.

I don't think there is any base between 7 and 36 where 663 is prime. I couldn't find a single base between 7 and 36 where 663 is a prime number. Sorry, that is the best I could do.

If I remember correctly, ALL numbers between 2 and 59 have an EVEN numbers of positive divisors, EXCEPT the Perfect Squares, or 4, 9, 16, 25, 36 and 49, which have 3 divisors each. So:

59 - 2 + 1 =58 - 6 perfect squares =52 such numbers that have EVEN divisors.

Hectictar: I can't and don't know how to draw a picture of a square inside a circle, but you most certainly can. Assume this circle contains the diameter of the pumpkin with a radius = 1. One side of the square can be the chord that will form the segment. Bisect the 4 right angle of the square, and where they intersect, is the center of the circle.
The segment should subtend an angle of 90 degrees at center of the circle. Find the area of the segment, which comes to =0.2853982....sq.units. Since this area is proportional to 5 kg weight and is proportional to pi, or the area of the entire circle, which is proportional to the weight "W" of the pumpkin.
In other words: W =pi / 0.2853982 * 5 =~55 kg - the weight of the pumpkin.

\(\text{Rational roots will occur when the discriminant is a non-negative perfect square}\\ D=b^2 - 4ac = 25-(4)(2)(b) = 25-8b\\ \text{$D$ is a perfect square for $b=0, 2, 3$}\\ \sum b = 5\)

\(\text{Each digit can be any of 1,3,5,7, 9}\\ N=5^4 = 625\)

The reason why your calculator evaluate  -7⁰  as  -1  is because, for the calculator, it is evaluated as  -(7⁰),  which is  -(1)  =  -1.

If you wanted it to be evaluated as  (-7)⁰,  which is  1,  you will need to include the parentheses.

You cant solve it without pictures

Both (-7)^0 and 7^0 =1. It is NOT true that -7^0 = -1.

Since both =1, then 1 - 1 = 0. And that is it.

Melody: I don't know much about Trig. But would this direct "solution" be valid? If not, why not? Just curious to know. Thanks.

Cos(x) = 1 /3

x = arccos(1/3)

x =70.529 - degrees

x / 2 = 70.529 / 2

x = 35. 2645 - degrees.

Sin(35.2645) =1 / sqrt(3)  ???.

If cos(x) = 1/3 and x is an acute angle, find the value of sin(x/2)

let 

\(\theta=\frac{x}{2}\qquad x=2\theta\\ \)

The question becomes.  

\(If\;\; cos(2\theta) = \frac{1}{3} \text{ and }x \text{ is less between 0 and 45 degrees, find the value of } sin(\theta)\\~\\ 1-sin^2\theta-sin^2\theta=\frac{1}{3}\\ 1-2sin^2\theta=\frac{1}{3}\\ -2sin^2\theta=\frac{-2}{3}\\ sin^2\theta=\frac{1}{3}\\ sin\theta=\frac{1}{\sqrt3}\\ sin(\frac{x}{2})=\frac{1}{\sqrt3}\\ \)

14  21  28  35  42  49  56  63  70  77  84  91  98  Total No =  13
Add all the numbers in the above list =728
728 / 13 = 56 - this is the average, or arithmetic mean, of all 2-digit multiples of 7

Well, Look at this list:

45  90  135  180  225  270  315  360  405  450  495  540  585  630  675  720  765  810  855  900  945  990  Total No =  22.

As you can see, there are 22 numbers between 45 and 1000 that are multiples of 45. But, only 2 of them

have 2 digits, 45 itself and 90. So, the probability would be:

2 / 22 =1 / 11 That a randomly picked number will be a 2-digit number.

As follows:

Just divide:

2,874 / 195,800 =0.0147 x 100 =~ 1.5%

The equation of a vertical parabola with a vertex of  (3, 2)  in vertex form is

y  =  A(x - 3)² + 2

We know that  (1, 0)  is a solution to the equation. We can use that to find  A

0  =  A(1 - 3)² + 2

0  =  A( -2 )² + 2

0  =  4A + 2

-4A  =  2

A  =  -\(\frac12\)

So the equation of the parabola is

y  =  -\(\frac12\)(x - 3)² + 2         Now we just have to get this equation into the form  ax² + bx + c

y  =  -\(\frac12\)(x - 3)(x - 3) + 2

y  =  -\(\frac12\)(x² - 6x + 9) + 2

y  =  -\(\frac12\)x² + 3x - \(\frac92\) + 2

y  =  -\(\frac12\)x² + 3x - \(\frac{5}{2}\)

Now we can see that     (a,  b,  c)  =  ( -\(\frac12\),  3,  -\(\frac{5}{2}\) )

(2x^2+3x+7)(x+1)-(x+1)(x^2+4x-63)+(3x-14)(x+1)(x+5).

\((2x^2+3x+7)(x+1)-(x+1)(x^2+4x-63)+(3x-14)(x+1)(x+5)\\ =(x+1)[(2x^2+3x+7)-(x^2+4x-63)+(3x-14)(x+5)]\\ =(x+1)[(2x^2+3x+7)-(x^2+4x-63)+(3x^2+x-70)]\\ =(x+1)[(2x^2+3x+7-x^2-4x+63+3x^2+x-70]\\ =(x+1)[4x^2+140]\\ =4(x+1)[x^2+35]\\ =4[x^3+35x+x^2+35]\\ =4x^3+140x+4x^2+140\\\)

Are you seriously asking for hw help here? 

If you know the textbook/teacher answer can you please share it with us.

In how many ways can I place 5 different beads on a bracelet if rotations and flips of the bracelet are not treated as different arrangements?

Without thinking to much I think it would be    4!/2 = 12 ways.

Thanks Sudsw12,

I will examine what you have said in detail when I have a little more time.

Thanks for answering and  doing this research for me.

It is much appreciated.

Do you understand what Rom wants you to do?

graph each side seperately, put y= out the front of each side and see where the graphs cross.

The x value of intersections will be your solutions.

\(A = x y = 2x + 4y\)

Now we just have to solve this for integer values of x and y.

There are 4 integer solutions

(x,y) = (5,10), (6,6), (8,4), (12,3)

Of these (5,10) yields an area of 50 which is the maximum