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5 integer solutions. (1,5) (2,6) (5,1) (6,2) (7,7)

it wrong :(

You forgot to include 100

1 = 2
2 = 4
3 = 9
4 = 11
5 = 16
6 = 18
7 = 23
8 = 25
9 = 30
10 = 32
11 = 37
12 = 39
13 = 44
14 = 46
15 = 51
16 = 53
17 = 58
18 = 60
19 = 65
20 = 67
21 = 72
22 = 74
23 = 79
24 = 81
25 = 86
26 = 88
27 = 93
28 = 95
Total== 28 integers

We can factor N^2+8N+15 into (N+3)(N+5). So if we want N^2+8N+15 to be divisible by 7, we need either n+3 or N+5 divisible by 7. The list of N such that N+3 is divisible by 7 is: 4,11,18,25,...,95, which is (95-4)/7+1=14 numbers. The list of N such that N=5 is divisible by 7 is 2, 9, 16, 23,...,100 which is (100-2)/7+1=15 numbers. 14+15=\(\boxed{29}\)

40 when x = -3

I believe you have missed some cases where all of the exponents are 1.

(x + 2)(3x^2 + x + 5)

= (x)(3x^2 + x + 5) + 2(3x^2 + x + 5)

= (3x^3 + x^2 + 5x) + (6x^2 + 2x + 10)

= 3x^3 + 7x^2 + 7x + 10

I'm assuming the question is a typo and you want to find Ax^3 + Bx^2 + Cx + D. It shouldn't be too hard to match up the coefficients to find the answer.

Let the original number be called x. We can rewrite the original problem as:

(five times a number) is divided by (7 more than that number) = -8

5x/(x + 7) = -8

5x = -8(x + 7)

5x = -8x - 56

13x = -56

x = -56/13

The minimum height is 39/2.

\((r_1, r_2, s_1, s_2) = (-20/29, 21/29, -35/37, 12/37)\)

The answer is 320.

Yeah, it's 5-3i.

I think you might mean  this  ???

(a + 6i) ( 5 - 3i)

5a + 30i - 3ai  - 18i^2

5a + 30i - 3ai + 18

a =  10

Because

5 (10)  + 30i  -3(10)i + 18  =

50 + 30i - 30i + 18  =

68

A graph might work best :

AB ≈   9.22 units

196  =  1 + 6 + 9  =  16

sqrt (196) = 14     

(1 + 4)^2  =  25

The 10th term  =   

18 * r^6 =  162

r^6  = 162 / 18  =  9                take the 9th root

r=  9^(1/6)

The 7th term is

18 * r^3  =

18 * (9^(1/6))^3  = 

18  *  (9)^(1/2)  =

18 * 3  =  

54

We have a lower power polynomial over a higher power polynomial

This produces a  horizontal asymptote of     y =  0

This  is the bottom of the range

To discover the max use some Calculus

Take the first derivative of the function and set =  0

0 - 2 * (4 + 6x)

____________  =   0

(2 + 4x + 3x^2)

This simplifies to

4 + 6x =  0

4 =  -6x

x =  -4/6 =  -2/3

The max occurs at  x  = -2/3

The max is           2 /  ( 2 + 4(-2/3) + 3(-2/3)^2 )  =    2 / [ 2 -8/3 + 4/3 ] =  2 / [  2/3] =  3

The range is   (0, 3 ]

a + b =   0 + 3  =  3

The answer is 676.

y^2 - x + 5y - 25 =  x^2 - 6x + 1  rewrite as

y^2 + 5y - x^2 + 5x  - 26 =  0 

-1x^2 + 0xy + 1y^2 + 5x + 5y  - 26 = 0

General (Standard Form) Equation Of A Conic Section

Ax^2+Bxy+Cy^2+Dx+Ey+F=0,where A,B,C,D,E,F are constants

From the standard equation, it is easy to determine the conic type eg

B^2−4AC<0 , if a conic exists, then it is a circle or ellipse

B^2−4AC=0, if a conic exists, then it is a parabola

B^2−4AC>0 , if a conic exists, it is a hyperbola

We have  :

0^2 - 4 (-1) (1)   =   4  >  0

This is a hyperbola that opens upward / downward

AC = sqrt(7).