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A/B =7/23 

7/(7+23) x 90 =21 degrees, angle A

23/(7+23) x 90 =69 degrees, angle B

B/A =69/21 =23/7

Listfor(n, 1, 10, (n+6) mod n) =(0, 0, 0, 2, 1, 0, 6, 6, 6, 6). Only 4 numbers that give "0" as the remainder satisfy the condition, as heureka found. They are: 1 + 2 + 3 + 6 =12. After 6, all other numbers will give a remainder of 6.

A "rational number" means: A number that can be expressed as a "Ratio" of two other numbers. For example:

3 can be expressed as a "Ratio" of two other number such as: 6/2, 9/3, 12/4.........and so on.

7/11 =0.6363636363..................etc, is "rational" number because it can easily be converted into a "Ratio" of two other numbers, which in this case are 7/11, On the other hand, some numbers are "irrational" because you cannot express them as a "ratio" of two other numbers. Example: Sqrt(2), Pi, e......etc.

Find the sum of all positive integral values of n for which

\(\mathbf{\large\dfrac{n+6}{n}}\)

is an integer.

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{n+6}{n}} \\\\ &=& \dfrac{n}{n} + \dfrac{6}{n}\\\\ &=& 1 + \dfrac{6}{n}\\ \hline \end{array}\)

So \(n\) are the divisors of \(6\).

The 4 divisors of 6 are: \(1,2,3,6\)
The sum is \(1+2+3+6 =\mathbf{ 12}\)

(C)
Each of the last two digits must be 5, 7, or 8 ?

\(\begin{array}{|lclc|c|} \hline & x&x &xx \\ \hline &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 5& 100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 7& +100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 8& +100\times 3 \\\\ \hline & & & & = 3\times 100\times 3 = {\color{red}900} \\ \hline \end{array}\)

There are 900 4-digit positive integers that satisfy the condition.

(B)
The last two digits cannot be the same digit ?

\(\begin{array}{|lcc|c|} \hline & xx &xx \\ \hline &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 00 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 11 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 22 & 90 \\\\ & \cdots \\ \\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 99 & 90 \\\\ \hline & & & = 90\times 10 = 900 \\ \hline \end{array}\)

1000 until 9999: There are 9000  four-digit integers.

9000 - 900 = 8100

There are 8100 4-digit positive integers that satisfy the condition.

How many 4-digit positive integers exist that satisfy the following conditions:

(A) Each of the first two digits must be 1, 4, or 5, and

(B) the last two digits cannot be the same digit, and

(C) each of the last two digits must be 5, 7, or 8?

(A)
Each of the first two digits must be 1, 4, or 5 ?

\(\begin{array}{|lllc|c|} \hline & x&x &xx \\ \hline &1 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & 3\times 100 \\\\ &4 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ &5 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ \hline & & & & = 3\times 3\times 100 = {\color{red}900} \\ \hline \end{array} \)

There are 900 4-digit positive integers that satisfy the condition.

(3x+2) / x I think

What is a rational number and why is it always terminating or a repeating decimal???

I don't think you're right, what about n=12?

Factor the following:
x^2 - 16 x + 60

The factors of 60 that sum to -16 are -6 and -10. So, x^2 - 16 x + 60 = (x - 6) (x - 10):
(x - 10)(x - 6)

a=10   and   b=6

3b - a =18 - 10=8

3x^2 = 4(16 - 4x), solve for x
Solve for x:
3 x^2 = 4 (16 - 4 x)

Expand out terms of the right hand side:
3 x^2 = 64 - 16 x

Subtract 64 - 16 x from both sides:
3 x^2 + 16 x - 64 = 0

The left hand side factors into a product with two terms:
(x + 8) (3 x - 8) = 0

Split into two equations:
x + 8 = 0 or 3 x - 8 = 0

Subtract 8 from both sides:
x = -8 or 3 x - 8 = 0

Add 8 to both sides:
x = -8 or 3 x = 8

Divide both sides by 3:
  x = 8/3 =2 2/3 miles per hour.

Solve for x:
x/(x + 4) = -9/(x + 3)

Cross multiply:
x (x + 3) = -9 (x + 4)

Expand out terms of the left hand side:
x^2 + 3 x = -9 (x + 4)

Expand out terms of the right hand side:
x^2 + 3 x = -9 x - 36

Add 9 x + 36 to both sides:
x^2 + 12 x + 36 = 0

Write the left hand side as a square:
(x + 6)^2 = 0

Take the square root of both sides:
x + 6 = 0

Subtract 6 from both sides:
x = -6

Solve for x:
6 = 35/x - 49/x^2

6 = 35/x - 49/x^2 is equivalent to 35/x - 49/x^2 = 6:
35/x - 49/x^2 = 6

Bring 35/x - 49/x^2 together using the common denominator x^2:
(7 (5 x - 7))/x^2 = 6

Multiply both sides by x^2:
7 (5 x - 7) = 6 x^2

Expand out terms of the left hand side:
35 x - 49 = 6 x^2

Subtract 6 x^2 from both sides:
-6 x^2 + 35 x - 49 = 0

The left hand side factors into a product with three terms:
-(2 x - 7) (3 x - 7) = 0

Multiply both sides by -1:
(2 x - 7) (3 x - 7) = 0

Split into two equations:
2 x - 7 = 0 or 3 x - 7 = 0

Add 7 to both sides:
2 x = 7 or 3 x - 7 = 0

Divide both sides by 2:
x = 7/2 or 3 x - 7 = 0

Add 7 to both sides:
x = 7/2 or 3 x = 7

Divide both sides by 3:
x = 7/2    or    x = 7/3

Solve for x:
2 x^2 + 9 x - 5 = 0

The left hand side factors into a product with two terms:
(x + 5) (2 x - 1) = 0

Split into two equations:
x + 5 = 0 or 2 x - 1 = 0

Subtract 5 from both sides:
x = -5 or 2 x - 1 = 0

Add 1 to both sides:
x = -5 or 2 x = 1

Divide both sides by 2:
 x = -5    or      x = 1/2

Here is the same answer as Rom's but simplified:

https://artofproblemsolving.com/community/c541205h1537311_daily_problem_19

If Joshua Smith were a single: 0.25(56,500)-3190=14125-3190=10933.

If Joshua Smith were the head of a household: 0.25E-4305=, 0.25(56,500)-4305, 14125-4305=9820.

So, the difference is 10933-9820=1113.

The formula for finding the slope: \(\frac{y_2-y_1}{x_2-x_1}=\frac{8-4}{0-2}=\frac{4}{-2}=\boxed{-2}\).

certainly looks to be

\(s_n = (-1)^{n-1} n^2,~n = 1, ~2,~\dots\)

can you predict \(s_5\) ?

This was pretty clever thinking actually.  If you had multiplied instead of added you would have gotten it right.  Well done!

−4x^2(2x^2+5)=−8x^4−20x^2

And

−5a^4(2a^2+4)=−10a^6−20a^4

Are correct

hmmmm

1 2 3 4 5 6 9 10 12 15 18 20 30 36 45 60 90   ?

Thanks!

Answer to

Which polynomial represents the area of the rectangle? http://prntscr.com/ku2v2m

2x2−9x−10

2x2−x−10

2x2+x−10

2x2−10

(2x+5)(x-2)=2x^2+x-10.