(22, 24, 26, 33, 36, 42, 62, 63)>>Total = 8
The probability is = 8 / 6^2 =8 / 36 = 2 / 9
What are the solutions of the equation x^4 – 5x^2 – 36 = 0?
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\( x^4 – 5x^2 – 36 = 0\)
\(u=x^2\)
\( u^2 – 5u – 36 = 0 \)
p q
\(u=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ u=2.5\pm\sqrt{6.25+36}\\ u=2.5\pm 6.5\)
\(u_1=9\\ u_2=-4\)
\((x_{1,2})^2=9\\ (x_{3,4})^2=-4\)
\(x_1=3\\ x_2=-3\)
\(x_3=+\sqrt{-4}\ complex\ number\\ x_4=-\sqrt{-4}\ complex\ number\)
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\(8\sqrt2-\sqrt{24}-\sqrt3+\sqrt{128}=8\sqrt2-2\sqrt6-\sqrt3+8\sqrt{2}=16\sqrt{2} - \sqrt{3} - 2 \sqrt{6}\)
Nice.
\(\frac{a}{1-r}=3 \rightarrow a=3-3r\) by geometric sum formula, where r is the common ratio. Our sequence would be \(a,ar,ar^2,ar^3,...\). When squared, it would be \(a^2, a^2r^2, a^2r^4, a^2r^6\). This sequence has first term a^2 and common ratio r^2, so by the geometric sum formula this is \(\frac{a^2}{1-r^2}=3 \rightarrow \frac{(3-3r)^2}{1-r^2}=3\rightarrow 9r^2-12r+9=3-3r^2 \rightarrow r=\frac{1\pm i}{2}\). So also \(a=3-\frac{3\pm 3i}{2}=\frac{3\pm 3i}{2}\).
Let the number of gallons of the 60% alcohol solution =G
0.60G + 0.25 * 90 = 0.50 * [G + 90], solve for G
G = 225 gallons of 60% alcohol solution will be needed.
Let z=x^2 --> z^2-5z-36=0 --> (z+4)(z-9)=0 --> z=-4,9 --> x^2=-4 (no real solutions), x^2=9 --> x=3,-3
If you wanted imaginary solutions as well --> x^2=-4 --> x=2i,-2i --> x=3,-3,2i,-2i
\(2\pi r*h=2\pi *5*2=20\pi\). Now for top and bottom. Top is \(5^2\pi-2^2\pi=25\pi-4\pi=21\pi\), and so is bottom. So ans is \(20\pi+21\pi+21\pi=\boxed{62\pi}\).
I don't like point-slope so I will convert.
3x-2y=-9 --> 2y=3x+9 --> y=3/2x+9/2
Perpendicular has slope -2/3. So y=-2/3x+b. Plug (-4,-2) --> -2=(-2/3)*(-4)+b --> -2=8/3+b --> b=-14/3
So eqn. is y=-2/3x-14/3 --> 3y=-2x-14 --> 3y+2x+14=0
x
216 27 8 1
- 216
- 27
- 8
- 1
x = 216 + 27 + 8 + 1 - 216 - 27 - 8 - 1= 0
Case one:
4(x+7)+3x+2=32 --> 4x+28+3x+2=32 --> 7x=2 --> x=7/2
Case two:
-4(x+7)+3x+2=32 --> -4x-28+3x+2=32 --> -x=58 --> x=-58
x=7/2, -58
Note that \((x+y)^2-2xy=100-2xy=x^2+y^2=56\rightarrow 100-2xy=56\rightarrow xy=22, x+y=10\). So x and y are the roots to this quadratic: \(z^2-10z+22=0\rightarrow z^2-10z+25=3\rightarrow (z-5)^2=3\rightarrow z=5\pm \sqrt{3}\rightarrow\boxed{(x,y)=(5+\sqrt3,5-\sqrt3), (5-\sqrt3, 5+\sqrt3)}\).
n=1010101010101010101010101010..........;s=0;p=0;cycle: s=(n%10);if(s==1,p=p+1, 0);n=(n div 10);if(n!=0, goto cycle,0);print"Total Num =",p
There are 2005 digits in the quotient of N/11.
1002 of them are zeros, and:
1003 of them are "1's"
There are 506 zeros in N/11.
The sum of all possible values of x is 736.
that is wrong..... :(
can you explain what e is?
The sum of all the different possible areas of cool right triangles is 46.
The maximum value is sqrt(10).
since it was 18 more birds, 131-113 which is 18, and there were 3 extra days, you do 18/3 which is 6
then it is 131 mod 6 = 5
so 5 is the answer
Check the numbers 18 and 128 in your question carefully. Can't find a right triangle with perimeter of 18 and their squares summing up to 128. Maybe 122??
By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4). Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i. Then the other roots work out as
4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,
4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and
4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.
a) 4 red balls must be added
b) 11 black balls mus be added.
The score 72 has been removed.
a=listforeach(n, (-1, 1, 2), listforeach(m,(-2, -1, 0, 1, 2),(n^m));print sort(a),"Total =>>", count a
(-1, -1, 0.25, 0.5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4) Total =>> 15
I count 5 distinct positive values [0.25, 0.5, 1, 2, 4]
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