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YEAH!!!!!!!! SAO!!!!!!!!! BEST ANIME EVA!!!!!!!!!!!!!!

2, 3, 5, 7 whats the pattern?

I expect there was a picture to go with this. :)

40^11*4^-11/10^10

3 1/8 = 25/8

2 1/2 =5/2

25/8 - 5/2 = 25/8 - 20/8 = 5/8

Verify the following identity:
sin(θ)^2/(cos(θ)) = tan(θ) sin(θ)

Write tangent as sine/cosine:
sin(θ)^2/(cos(θ)) = ^?(sin(θ))/(cos(θ)) sin(θ)

sin(θ) ((sin(θ))/(cos(θ))) = sin(θ)^2/(cos(θ)):
sin(θ)^2/(cos(θ)) = ^?sin(θ)^2/(cos(θ))

The left hand side and right hand side are identical:
Answer: |(identity has been verified)

The latent heat of fusion of mercury is 11kJ/kg, so multiply this by 0.06kg to get the result in kJ, then multiply that by 1000 to get the result in joules.

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http://web2.0calc.com/

It all depends on how many hours you worked. Some people work 5 hours per day. Some work 8 hours per day. And some work maybe 10 hours per day. If you were to take the AVERAGE office worker, working from 9 to 5, that is 8 hours per day inculding their Lunch hour. So, if they were paid for 8 hours per day, that would come to:

8 hours per day x 5 days a week x 52 weeks in one year =2,080 hours per year

$69,280 / 2,080 =~$33.30 per hour.

You can use this type of calculation to figure out the rates per hour for any number of hours that people get paid for.

Let the cost per pound for Turkey=T

Let the cost oer pound for Ham =H

15.4T + 11.6H =38.51

10.2T + 7.3H =24.84, solve for T, H

T = $1.19 per pound for Turkey

H = $1.74 per pound for Ham.

5x^2

a) Who will arrive at C first?

b) How many minutes later will the second person arrive?

what is 7^-3/2 expressed in radical form?

\(7^{-3/2}=\frac{1}{7^{+3/2}}=\frac{1}{(7^{3})^{1/2}}=\frac{1}{\sqrt{7^{3}}}\\ or\\ =\frac{1}{\sqrt{7}^3}\\ or\\ =(\frac{1}{\sqrt{7}})^3\\\)

8/7

Harvey’s expresso express, a drive through coffer stop, is famous for its great house coffee,

a blend of Colombian and mocha java beans.

Their archrival, JoJo’s java, sent a spy to steal their ratio for blending beans.

The spy returned with a torn part of an old receipt that showed only the total number of pounds and the total cost, 18 pounds for $92.07.

At first JoJo was angry, but then he realized that he knew the price per pound of each kind of coffee

($4.89 for Colombian and $5.43 for mocha java).

Show how he could use equations to figure out how many pounds of each type of beans Harvey’s used.

Let x = pounds of Colombian coffee
Let y = pounds of mocha jave coffee

\(\begin{array}{|rcll|} \hline x + y &=& 18\ \text{pounds} \qquad \text{ or } \qquad x= 18\ \text{pounds} - y\\ x\cdot \dfrac{$4.89}{\text{1 pound}} + y\cdot \dfrac{$5.43}{\text{1 pound}} &=& $92.07 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 4.89x+5.43y&=& 92.07 \quad &| \quad x = 18 - y \\ 4.89\cdot (18 - y) + 5.43y&=& 92.07 \\ 4.89\cdot18 -4.89y + 5.43y&=& 92.07 \\ 88.02 + 0.54y&=& 92.07 \quad &| \quad -88.02 \\ 0.54y&=& 92.07 -88.02 \\ 0.54y&=& 4.05 \quad &| \quad : 0.54 \\ y&=& \frac{4.05}{ 0.54 } \\ \mathbf{y}&\mathbf{=}& \mathbf{7.5\ \text{pounds } } \\\\ x &=& 18-y \\ x &=& 18 - 7.5 \\ \mathbf{x} &\mathbf{=}& \mathbf{10.5\ \text{pounds }} \\ \hline \end{array} \)

Harvey’s used 10.5 pounds of Colombian coffee and 7.5 pounds of mocha jave coffee

x² + y² – 6x + 8y = 144

The equation of a circle in the xy-plane is shown above. 

What is the diameter of the circle?

Let d = diameter of the circle

Let r = radius of the circle

Let d = 2r

\(\begin{array}{|rcll|} \hline x^2 + y^2 -6x+8y &=& 144\\ (x^2-6x) + (y^2+8y) &=& 144 \\ \left(x-\frac{6}{2}\right)^2 -\left(\frac{6}{2}\right)^2 +\left(y+\frac{8}{2}\right)^2 -\left(\frac{8}{2}\right)^2 &=& 144 \\ \left(x-3\right)^2 - 3^2 +\left(y+4\right)^2 -4^2 &=& 144 \\ \left(x-3\right)^2 - 9 +\left(y+4\right)^2 -16 &=& 144 \\ \left(x-3\right)^2 +\left(y+4\right)^2 &=& \underbrace{144+16 + 9}_{=r^2}\\\\ r^2 &=& 144+16 + 9 \\ r^2 &=& 169 \\ r &=& 13 \quad & | \quad \cdot 2 \\ 2r &=& 26 \quad & | \quad d = 2r \\ \mathbf{d} &\mathbf{=}& \mathbf{26} \\ \hline \end{array}\)

The diameter of the circle is 26.

The center of the circle is ( 3, -4 ).

What is the rule if x is 6 and y is 7. Then x is 5 and y is 5.

Table

IN(X)      6    5    4    3      2      1        0

OUT(Y)  7    5    3     1     -1     -3      -5

arithmetical sequence:

\(\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)\cdot d \quad & | \quad d = +2 \qquad a_1 = -3 \\ a_n &=& -3 + (n-1)\cdot 2 \\ a_n &=& -3 + 2n-2 \\ \mathbf{a_n} &\mathbf{=}& \mathbf{-5 + 2n} \quad & | \quad n = IN(X) \qquad a_n = OUT(Y) \\\\ \mathbf{OUT(Y)} &\mathbf{=}& \mathbf{-5 + 2\cdot IN(X)} \\ \hline \end{array} \)

b)  We begin by noting the rules for an integer to have more than four divisors.
The smallest number that has 4 divisors is 6. So, all of the divisors of 8400 that have at least 4 divisors must be greater than or equal to 6. 
Also, the divisor cannot be a squared prime, as these numbers have only 3 divisors. 
Finally, the divisor cannot be prime, since all prime numbers have only 2 divisors. 

 Now, we apply our rules to the divisors of 8400.
The prime factorization of 8400 is 2^4*3*5^2*7. 
Since the divisor must be greater than or equal to 6, it cannot be 2, 2^2, 3, or 5. 
Since the divisor cannot be a squared prime, it cannot be 5^2.
Since the divisor cannot be prime, it cannot be 7, the only prime factor we haven't already cancelled out. 
These are six unwanted divisors. 
Finally, the divisor cannot be 1, since 1 is a divisor of every integer and has less than 4 divisors. 
So, these 7 divisors have less than 4 positive divisors. 

Now we must find the total number of divisors of 8400. To do this, we add 1 to each exponent: 2^4 becomes 2^5, 3^1 becomes 3^2, 5^2 becomes 5^3, and 7^1 becomes 7^2. Now, we multiply all these new exponents: 5*2*3*2 = 60. The total number of divisors of 8400 is 60. 

Subtracting our 7 unwanted divisors from the 60 total divisors yields \(\fbox{53}\) divisors that have at least 4 positive divisors.

 -2x^3+x+7x^3+10

\( -2x^3+x+7x^3+10\\ = -2x^3+7x^3\quad +x+10\\ =5x^3\quad +x+10\\\)

\((2x-5)\times(3x+1)\)

\({6x}^{2}+2x-15x-5\)

\({6x}^{2}-13x-5\)

Thanks for telling us this MathPeacock.

CPhill will be very pleased   

\(a=22\times cot(30)\)

\(a=22\times \frac{cos(30)}{sin(30)}\)

\(a≈22\times \frac{0.154251449888}{sin(30)}\)

\(a≈22\times \frac{0.154251449888}{-0.988031624093}\)

\(a≈22\times-0.1561199521620584\)

\(a≈-3.4346389475652842\)

You need to be more explicit :)

Thanks Goldfish  :)

I definitely will not fit in either of the first 2 rooms, it MIGHT fit in the last room but only if it is nearly square in shape.

If it is longer than 13 foot it will not fit in any of the rooms