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The problem setter has made a mistake. Cosine of an angle must be within -1 and 1. -15/3 is -5, which is outside of the range. (Unless the angle is a complex number, of course, but that is out of consideration for, supposedly, a math problem in high school.)

You can use Chinese Remainder Theorem.

The problem translates to: \(\begin{cases}1000 \leq N \leq 9999\\N \equiv 5 \pmod 9\\N \equiv 3 \pmod 7\\ N \equiv 3 \pmod 5\end{cases}\)

Now, we let \(N = 9k + 5\) for some integer k. Since \(N \equiv 3 \pmod 7\),

\(\begin{array}{rclc}9k+5&\equiv&3&\pmod7\\ 2k&\equiv&-2&\pmod{7}\\ k&\equiv&6&\pmod7\end{array}\)

Then we let \(k = 7n + 6\) for some integer n. Then \(N = 9(7n + 6) + 5 = 63n + 59\).

Repeat this process one more time with N = 3 (mod 5), and you will get \(N = 315m + 248\) for some integer m.

We can try m = 0, 1, 2, 3, ... to see which one gives the smallest possible 4-digit number. You will get \(N = 315(3) + 248 = 1193\). You are more than welcome to check that this solution satisfies the conditions described in the problem.

The quadratic polynomial has two distinct real roots. That means the discriminant must be positive.

\(\Delta > 0\\ m^2 - 4(1)(m) > 0\\ m^2 - 4m > 0\\ m > 4\text{ or }m < 0\)

Can you write that inequality in interval notation?

Suppose the first two digits are 6's.

The only possibilities are 660, 661, ..., 666, 667, ..., 669.

You can count the other possibilities by hand. Just make sure that you don't count 666 three times and 0 cannot be the first digit, and you're good to go.

b divides 168 and c divides 490 by definition of gcd.

\(\gcd(b, c) \geq \gcd(168, 490) = 14\)

Therefore the smallest value is 14.

Using chain rule, 

\(\begin{array}{rcl}f(x) &=& (2x - 1)^5\\ f'(x) &=& 5(2x - 1)^4 \dfrac{d}{dx} (2x - 1)\\ &=& 5(2x - 1)^4 (2)\\ &=& 10(2x - 1)^4\\ f'(2) &=& 10(2(2) - 1)^4\\ &=& 810\end{array}\)

If f(x) has a real number value, 

\(\dfrac{x - 7}{4 - x} \geq 0\\ (x - 7)(4 - x) \geq 0\text{ and }x \neq 4\\ 4 < x \leq 7\)

The smallest possible integer in that range is x = 5. 

If you have a polynomial, substitute 1 into the polynomial and that will be the sum of the coefficients.

Therefore, sum of coefficients = -(3 - 1)(1 + 2(3 - 1) - 7(1 - 5)) = -66.

To check the answer, we can go the normal way and see if we really get this sum of coefficients.

-(3 - c)(c + 2(3 - c) - 7(c - 5))

= (c - 3)(c + (6 - 2c) - (7c - 35))

= (c - 3)(c + 6 - 2c - 7c + 35)

= (c - 3)(-8c + 41)

= -8c^2 + 24c + 41c - 123

= -8c^2 + 65c - 123

sum of coefficients = -8 + 65 - 123 = -66.

Therefore the answer is correct.

Let \(y = a\cdot b^x +c\) for some constants a, b, and c

Substitute (x, y) = (0, 900) (x, y) = (5, 200) and (x, y) = (4, 450) to get the system of simultaneous equations \(\begin{cases}a + c= 900\\a\cdot b^4 + c=450\\a\cdot b^5 + c=200\end{cases}\).

It is really hard to find the exact solutions of this system of equations, but the approximate solution is (a, b, c) = (-147.7, 1.418, 1047.75).

Here's the graph:

heureka posted a great solution, so I will post a different one.

\(\quad1 - 2 + 3 - 4 + \cdots + 2019 - 2020 + 2021\\ =(1 - 2) + (3 - 4) + \cdots + (2019 - 2020) + 2021\\ = \underbrace{(-1) + (-1) + \cdots + (-1)}_{1010\text{ times}} + 2021\\ =-1010 + 2021\\ = 1011\)

Hint: When x = 1, y = A + B + C. Try to find the y-coordinate of the point on the graph with x-coordinate = 1.

What's the question? 

We can simplify this equation to: \(x^2 + kx- 10x + 4\).

Factoring out x gives us: \(x^2 + x(k-10) +4\)

For the quadratic to have only 1 solution, the discriminant(\(b^2 - 4ac\)) must be equal to 0. 

Substituting what we know, we have: \((k-10)^2 - 16 = 0\)

Adding 16 to both sides gives us: \((k-10)^2 = 16\)

Taking the square root of both sides gives us: \(k - 10 = \pm 4\).

Now, we can find the 2 values of \(k\) and add them up. 

Can you take it from here?

Dividing is the same as multiplying a value by its reciprocal. The reciprocal is the value when you swap the numerator and the denominator. For example, the reciprocal of \(7 \over 9 \) is \(9 \over 7\).

This means that \({5 \over 8} \div {7 \over 9} = {5 \over 8 } \times {9 \over 7}\). Notice how the first fraction is the same, and ONLY the second fraction was converted into a reciprocal. 

Can you take it from here?

Compute the value of
\(1 - 2 + 3 - 4 + \dots + 2019 - 2020 + 2021\).

\(\begin{array}{|rcll|} \hline s &=& 1 - 2 + 3 - 4 + \dots + 2019 - 2020 + 2021 \\ s&=& 2021 - 2020 +2019 - \dots -4+3-2+1 \\ \hline 2s &=& 2022-2022+2022-2022 + \dots + 2022-2022+2022 \\ 2s &=& 0+0+ \dots +0+2022 \\ 2s &=& 2022 \\ s &=& 1011 \\ \hline \end{array}\)

\(1 - 2 + 3 - 4 + \dots + 2019 - 2020 + 2021 = \mathbf{1011} \)

1  = (450, 50, 9)
2  = (270, 54, 9)
3  = (180, 60, 9)
4  = (126, 70, 9)
5  = (120, 72, 9)
6  = (90, 90, 9)
7  = (480, 32, 10)
8  = (330, 33, 10)
9  = (255, 34, 10)
10  = (210, 35, 10)
11  = (180, 36, 10)
12  = (130, 39, 10)
13  = (120, 40, 10)
14  = (105, 42, 10)
15  = (90, 45, 10)
16  = (80, 48, 10)
17  = (75, 50, 10)
18  = (66, 55, 10)
19  = (60, 60, 10)
20  = (110, 30, 11)
21  = (420, 21, 12)
22  = (220, 22, 12)
23  = (120, 24, 12)
24  = (100, 25, 12)
25  = (70, 28, 12)
26  = (60, 30, 12)
27  = (45, 36, 12)
28  = (40, 40, 12)
29  = (156, 20, 13)
30  = (84, 20, 14)
31  = (70, 21, 14)
32  = (35, 30, 14)
33  = (240, 16, 15)
34  = (90, 18, 15)
35  = (60, 20, 15)
36  = (40, 24, 15)
37  = (30, 30, 15)
38  = (120, 16, 16)
39  = (48, 20, 16)
40  = (45, 18, 18)
41  = (36, 20, 18)
42  = (30, 20, 20)
43  = (28, 21, 20)
44  = (24, 24, 20)
 

find angle CED.

Hello Guest!

AEO = 41°   AOE = 98°   BOE = 82°

ECD = 82°   ADB = 49°   CED = 49° 

  !

The apothem is not 3.6 it is approximately 3.44  

So the question is nonsense.

I suppose you can ignore this and take it at face value.

Have you done anything yourself towards answering this question?  What?

Hi Heureka,

It is really nice to see you here again :)

Oh , so sorry but i m not sure about that !!

Find the value of series
\(1 + 4 + 2 + 8 + 3 + 12 + 4 + 16 + \cdots + 24 + 96 + 25 + 100.\)

\(\begin{array}{|rcll|} \hline 1+4 &=& 5 \\ 2+8 &=& 10 \\ 3+12 &=& 15 \\ 4+16 &=& 20 \\ \ldots \\ 24 + 96 &=& 120 \\ 25 + 100 &=& 125 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && 1 + 4 + 2 + 8 + 3 + 12 + 4 + 16 + \cdots + 24 + 96 + 25 + 100\\ &=& 5+10+15+20 + \cdots +120+125 \\ &=& 5*(1+2+3+4+\cdots +24+25) \\ && \boxed{ 1+2+3+4+\cdots +24+25 = \frac{1+25}{2}*25 \\ =13*25 \\ = 325 } \\ &=& 5*325 \\ &=& \mathbf{1625} \\ \hline \end{array}\)

Yes I get that, and it is good logic,  but how did you know that k was defined at all. 

OR maybe k is 12 and then there would not be a second k value.

....

I guess it was a reasonable assumtion that k was defined....and that it was not 12.

And yes if any such k exists (which I know it does becasue I worked it out ) and k is not 12, then there will be 2  k values, and their sum is 24 just as you have said.    

The answer was -6.

Thank you!