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The optimal point is (x,y) = (20,50).

f2=5

(a,b,c,d) = (5,6,2,4)

This is a very easy problem.  Just use the parametric form of the cycloid.

(a^2)(b^3) = k

(49)(27) = k 

(a^2)(6^3) = 49 * 27

a^2(8) = 49

a^2 = 49/8. 

20! + 12! = 12!(x + 1). Since we know that x is an even number, (x + 1) is an odd number, and therefore cannot "produce" any more factors of 12.

Therefore, the largest integer n for which 12^n divides it would be how much 12^n divides 12!.

12 = 2^2 * 3

We have to see how many factors of 2 are there in 12!, divide that by 2 and set that equal to y.

In addition, we can see how many factors of 3 there are in 12! and set that equal to z. 

1: 2, 4, 6, 8, 10, 12

2: 4, 8, 12

3: 8

Adding up gives us: 10

Therefore, there are 5 2^2 terms.

1: 3, 6, 9, 12

2: 9

There are also 5 factors of 3 in the expression.

Therefore, 5 is the largest integer for n that divides 20! + 12!

THANKS A LOT DUDE!

x^2 - 5x + 3 = 0 

The polynomial that has the recirpocal of the roots of the equation above is:

3x^2 - 5x + 1 = 0 

We just wnat the polynomial with the square of its roots.

Suppose, that polynomial above is P(x) and has roots a and b. And let Q(x) be the polynomial that has the square of its roots. 

P(x) = (x - a)(x - b)

Q(x) = (x - a^2)(x - b^2) 

Q(x^2) = (x-a)(x-b) * (x + a)(x + b)

= P(x)P(-x)

= (3x^2 - 5x + 1)(3x^2 + 5x + 1) 

= (3x^2 + 1)^2 - 25x^2

= 9x^4 + 6x^2 + 1 - 25x^2

= 9x^4 - 19x^2 + 1
Q(x) = 9x^2 - 19x + 1
The sum of the roots of this equation is 19/9
 

Bricklayer Ben placed a total of 462 bricks.

How to use equation to find the number of bricks placed?

Bricklayer Ben places 42 bricks per hour.

Bricklayer Bob places 36 bricks per hour.

Let

x = number of hours worked by Bricklayer Bob

y = number of hours worked  by Bricklayer Ben

Therefore,

x = 2y

42y + 36x = 1254

42y + 36(2y) = 1254

42y + 72y = 1254

114y = 1254

y = 1254 / 114

y = 11

Hence,  Bricklayer Ben placed a total of 42 × 11 = 462 bricks.

Draw a line from point O to point C.

Because OB and OC are radii, OB = BC = OC = 4. Therefore, we know that triangle OBC is an equilateral triangle. Solving for the area of OBC gives us 4sqrt(3). 

In addition, since we know that Angle COA is 120 degrees(as angle BOC is 60), the rest of the area would be 1/3 of the area of the whole circle. This is 16/3pi. 

Hence, the area of the shaded region is: 4sqrt(3) + 16/3pi

ab = 64sqrt(3)/3

OOOH. What a hard question. I think you need to use your fingers and count

1

2

YAAAAY

I got the answer.

 Is it that I'm blind or that there is nothing shown

OBVIOUSLY I CAN'T ANSWER IF THERE IS NO QUESTION

QR is the hypotenuse, so:

tan R = PQ/PR      sin R = PQ/20

so  PQ/PR = 2*PQ/20   hence PR = 10  

Thanks Voldemort

Yes of course, there was not too little info given at all.  Sometimes I have rocks in my head.

Thanks Guest.

There is a way to find the exact value! :D

x^3 + 3x^2 + 3x + 1 = 2

(x+1)^3 = 2

x = \(\sqrt[3]{2}-1\)

There is three odd floors. 1, 3, and 5. There are three rooms per floor that are divisable by 3: 3, 6, and 9. The total number of rooms that Hans can choose from is: 3 possible floors times 3 possible rooms equals 9 choices. The probability that Hans picked the correct room is \(p=\frac{1}{9}\).

The difference in tax is 1.5 %. This equals to a difference in price of \(x = \frac{5.00}{100}\cdot 1.5 = 0.075 dollars = 7.5 cents\).

I do not see why it should be 300 miles or why there should be too little information given. As the rockets fly towards each other, they approach each other with a speed of \(v_{approach}=v_{rocket,1}+v_{rocket,2}=8500 mph+3500 mph=12000 mph\). Convert mph (miles per hour) to mpm (miles per minute): \(v_{mpm}=\frac{v_{mph}}{60min}=\frac{12000mph}{60min}=200mpm\). The distance the rockets are apart one minute prior to collision is then \(d=\frac{200 mpm}{1 min}=200 miles.\)

\(x^3+3x^2+3x+1=2\\ x^3+3x^2+3x-1=0\\ x\approx 0.26 \)

Not enough info given.

1 * 0.5 * 0.25 = 0.125

Check:

Number of paths = 4^3 = 64

Number of favourable pths = 4*2*1 = 8

8/64 = 1/8  = 0.125

Your right Builderboi, I was thinking of an octagon.