In general: cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
So: cos(θ + π) = cos(θ)cos(π) - sin(θ)sin(π) = -cos(θ) - 0 = -0.4 since cos(π) = -1 and sin(π) = 0
Find
\(99 - 97 + 95 - 93 + \ldots+ 3 - 1\).
\(\begin{array}{|l|rc|} \hline & \text{pair} \\ \hline 1 & 3-1 & = 2 \\ 2 & 7-5 & = 2 \\ 3 & 11-9 & = 2 \\ 4 & 15-13 & = 2 \\ 5 & 19-17 & = 2 \\ 6 & 23-21 & = 2 \\ 7 & 27-25 & = 2 \\ 8 & 31-29 & = 2 \\ 9 & 35-33 & = 2 \\ 10 & 39-37 & = 2 \\ 11 & 43-41 & = 2 \\ 12 & 47-45 & = 2 \\ 13 & 51-49 & = 2 \\ 14 & 55-53 & = 2 \\ 15 & 59-57 & = 2 \\ 16 & 63-61 & = 2 \\ 17 & 67-65 & = 2 \\ 18 & 71-69 & = 2 \\ 19 & 75-73 & = 2 \\ 20 & 79-77 & = 2 \\ 21 & 83-81 & = 2 \\ 22 & 87-85 & = 2 \\ 23 & 91-89 & = 2 \\ 24 & 95-93 & = 2 \\ 25 & 99-97 & = 2 \\ \hline \end{array}\)
\(99 - 97 + 95 - 93 + \ldots+ 3 - 1 = \mathbf{25 \times 2 = 50}\)
As follows:
Jill is making a magic multiplication square using the numbers 1, 2, 4, 5, 10, 20, 25, 50, and 100.
The products of the numbers in each row, in each column, and in the two diagonals must all be the same.
In the figure you can see how she has started. Which number must go in the cell with the question mark?
Compute the sum: \((a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\).
\(\small{ \begin{array}{|rcll|} \hline && \mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\\\ &=& \Big(a +(2n+1)d\Big)^2 +\Big(a + (2n-1)d\Big)^2+\Big(a + (2n-3)d\Big)^2 + \cdots + \Big(a+(2n-(2n-1))d\Big)^2 \\ && -\Big(a + (2n-0)d\Big)^2- \Big(a+(2n-2)d\Big)^2- \Big(a+(2n-4)d\Big)^2 - \cdots -\Big(a+(2n-(2n))d\Big)^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\ \hline \end{array} }\)
\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 - \Big(~a+2(k-1)d~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} a^2+2ad(2k-1)+d^2(2k-1)^2-\Big(~ a^2+4ad(k-1)+4d^2(k-1)^2 ~\Big) \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{a^2}}+2ad(2k-1)+d^2(2k-1)^2 {\color{red}{-a^2}}-4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad(2k-1)+d^2(2k-1)^2 -4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{4adk}}-2ad+d^2(2k-1)^2 {\color{red}{-4adk}}+4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} -2ad+d^2(2k-1)^2 +4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad+d^2(2k-1)^2 -4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1}2ad +d^2~\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \quad | \quad \sum \limits_{k=1}^{n+1}2ad = 2ad(n+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k^2-4k+1-4(k^2-2k+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} {\color{red}{4k^2}}-4k+1{\color{red}{-4k^2}}+8k-4 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k-3 \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-\sum \limits_{k=1}^{n+1}3~\Big)\quad | \quad \sum \limits_{k=1}^{n+1}3 = 3(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \quad | \quad \sum \limits_{k=1}^{n+1}k = \dfrac{1+(n+1)}{2}(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\dfrac{(n+2))}{2}(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~2(n+2)(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)\Big(~2(n+2)-3~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)(2n+4-3) \\\\ &=& \mathbf{2ad(n+1) +d^2(n+1)(2n+1)} \\ \hline \end{array}\)
\(\mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\= \mathbf{2ad(n+1) +d^2(n+1)(2n+1)}\)
Find the number of integral points in the interior of the triangle having vertices (0,0), (21,0), and (0,21).
67, wrote python to do it. Also u in my aops class right? who r u? cheater
def TrailingZeros(n): count = 0
i=2 while (n/i>=1): count += int(n/i) i *= 2
return int(count)
n = 66
print(TrailingZeros(n)//3)
\(x^2-15x+1=0\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) Apply the Quadratic formula.
\(\frac{15+\sqrt{221}}{2}\) and \(\frac{15-\sqrt{221}}{2}\) (Quadratic equations have two roots)
Substitute into: \(x^4+\frac{1}{x^4}\)
\((\frac{15+\sqrt{221}}{2})^4+(\frac{1}{\frac{15+\sqrt{221}}{2}})^4=49729\)
\((\frac{15-\sqrt{221}}{2})^4+(\frac{1}{\frac{15-\sqrt{221}}{2}})^4=49729\)
(Notice if you substitute either root you will get the same answer)
wow this problem is hard!! 😅
i'm sorry, but I can't provide an explanation for this...
but if you need quick answers, the answer is \(\boxed{49727}\) (according to Wolfram Alpha)
Thanks!
2x + 3y = 100
3y = 100 - 2x
y = 100 - 2x
_______
3
We can only have integer solutions for y whenever x = 2 , 5, 8, 11, etc
And the last x that makes y a positive integer is when x = 47
So....the number of integers = [ 47 - 2 ] / 3 + 1 = 45 / 3 + 1 = 15 + 1 = 16
Just as Lucky Ducky found !!!!!
GOOD JOB LUCKYDUCKY !!!!!
Let d be the original distance to the building
Let h be the height of the building
Since the original angle of elevation is 45°, then d = h
So when we move 20 feet back we have that
tan 30° = d / [d + 20] rearrange as
[ d + 20] tan30° = d
d tan 30 + 20 tan 30° = d
20 tan 30° = d - dtan30°
20tan30° = d ( 1 - tan 30°)
[ 20tan30° ] / [ 1 - tan 30°] = d ≈ 27.32 ft = h
sinx [1 + cos x]
_________________ =
[1 - cos x] [ 1 + cos x]
sin x + sin cos x
______________ =
1 - cos^2 x
sin x + sin x cox
_____________ =
sin^2 x
sinx sinxcosx
_____ + _________ =
sin^2x sin^2 x
1 cos x
_____ + _______ =
sin x sin x
csc x + cot x
The denominator of the probability is \(6^3 = 216\). We are asked to find the probability that the three numbers are consecutive. The three numbers can be arranged in 6 ways, and there are 4 triplets of consecutive numbers from 1-6. The probability is \(\dfrac{6\cdot4}{216} = \dfrac{24}{216} = \boxed{\text{(B) }\dfrac{1}{9}}\)
She walked 1 km to school and 1 km back home, her total distance is 1+1=2 km.
She spent 30 minutes walking to school and 10 minutes walking back home, her total time is 30+10=40 minutes = \(\frac{40}{60}=\frac{2}{3} hours\)
Therefore her average speed in km is $\(\frac{2}{\frac{2}{3}}=\boxed{\mathrm{(A)}\ 3}\)
Yay I got 3 mastered in the topic that I'm doing
N = 16.
So, 16 positive integer solutions are possible for this equation.
Is this from AOPs?
1. The volume is 25*sqrt(3).
2. The length of the diagonal is 5*sqrt(2).
This is an AoPs question from counting and probability...
STOP POSTING AOPS QUESTIONS HERE
The angles in all quadrilaterals sum up to 360.
We do a little algebra to find x.
108 + 108 + 2x + x = 360
216 + 3x = 360
3x = 144
x = 48
Therefore my final answer for the value of x is 48
Add them all up:
(99, -97, 95, -93, 91, -89, 87, -85, 83, -81, 79, -77, 75, -73, 71, -69, 67, -65, 63, -61, 59, -57, 55, -53, 51, -49, 47, -45, 43, -41, 39, -37, 35, -33, 31, -29, 27, -25, 23, -21, 19, -17, 15, -13, 11, -9, 7, -5, 3, -1) = 50
I can tell this is an Alcumus problem...
Do the Alcumus first then do the homework...
Counting and Probability... Nice...
