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1/2*x^6*y^7 = 3/8

Note that \(3 - i\sqrt3 = 3 - \sqrt{-3}\).

Now, recall the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). Substituting \(a = 1\), we have \(3 - \sqrt {-3} = {-b - \sqrt{b^2-4c} \over 2}\) (take the minus root, not the plus)

Multiplying both sides by 2 gives us: \(6 - 2 \sqrt {-3} = -b - \sqrt{ b^2 - 4ac}\). Simplify the left-hand side: \(6 - \sqrt{-12} = -b - \sqrt {b^2 - 4ac}\).

Note that \(6 = -b\), meaning \(b = -6\). Plugging this into the discriminant, we have \({36 - 4c} = -12\) (disregard the square root because both have them)

Solving, we find that \(c = 12\).

Thanks for the feedback!!

Note that because all the bases in the term \(7^4 \times 2^4\) are raised to the 4th power, the 4th root of this is an integer. 

So, we basically have: \({7^4 \times 2^4 \over 7^x \times 2^y }= 7^2 \times 2^1\)

Start with \(7^4 \over 7^x\) . We want this to equal \(7^2\). 

Recall that when you are dividing exponents with the same base, you subtract the exponents (\(3^4 \div 3^2 = 3^{4 - 2} = 3^2\)).

Applying that here, we have \(7^4 \div 7^x = 7^2\), thus \(x = 2\). (\(7^4 \div 7^2 = 7 ^ {4 - 2} = 7^2\))

Now, doing the same thing to \(2^4 \over 2^y\). We want this to equal \(2^1\), so \(y = 3\). (\(2^4 \div 2^3 = 2^{4 - 3} = 2^1\)).

So, the answer we are looking for is \(7^2 \times 2^3 = \color{brown}\boxed{392}\)

Hope this helps!

Wait a minute... 

In \(\triangle BCD\), the hypotenuse is 17, while one of the legs is 18. 

That isn't a triangle...

Thanks but im still confused how you got 7^2 and 2^3  

Fill in the blanks to make the equation true.

(___)^3*(___)*(___)^2 = x^5*y^6

x y x^2 x*y y^2 x^3 x^2*y x*y^2 y^3 

The prime factorization of 98 is \(7^2 \times 2\). 

Because we are taking the fourth root, everything must be raised to a power of a multiple of 4, (4, 8, 12, etc.)

This means that the smallest value of n that works is \(7^2 \times 2^3 = \color{brown}\boxed{392}\) 

The minimum occurs at \(-{b \over 2a} = -{-8 \over 4} = 2\). 

Now, substitute this into the equation: \(\text{___} = 2\times 2^2 - 8 \times 2 + 19\) 

The other diagonal is 14 cm.

Simplify the left-hand side: \({3^2}^y = 3^{2y}\). 

Note that the exponents have the same bases, so we have \(2y = 6\). 

Now, just solve for y. 

Pythagorean Theorem

DG  = sqrt  [ DO^2 + OG^2 ]  = 

sqrt [ 24^2 + 12^2 ] =   

sqrt [ 12^2 ( 2^2 + 1) ] =

12 sqrt  [ 5 ] 

The common ratio of the series is \(b \over 7\), and the starting term is \(b \over 7\), so we have the equation: \(\large{{{b \over 7} \over 1 - {b \over 7}} = 7}\)

Multiplying the entire equation by \(1 - {b \over 7}\) gives us \({b \over 7} = 7(1 - {b \over 7})\), which simplifies to \({b \over 7} = 7 - b\). 

Multiplying the entire equation by 7 gives us \(b = 49 - 7b\), meaning \(b = \color{brown}\boxed{49 \over 8}\)

thank u

(N + 1)        (N + 2)       (N + 3)                     (N + 19)      (N + 20)

______ *   _______ * _______  *   ......  *    ________ * _______

     N             (N + 1)      (N + 2)                     ( N + 18)     (N + 19)

Note  that the product of all the  fractions will be

(N + 20) / N  =  3

N + 20  = 3N

20 =  2N

N = 10

The first fraction  was   (N + 1)  / N   =       11 /  10

Divide both sides by  13^5

13^17 / ( 13^11 *  13^5)  =    ____

13^17  /( 13^16)  =  ____

13^1  =     13  =    ____

As written, we have

mn^6 / (2mn)  -

n^5 / 2

The positive difference is 25.

There are 285 different dress designs.

The solution is (x,y) = (31/8, -21/8).

4*5=20. 2*5=10 

20+10+5=35

Kenny - 20

Jay - 10

Derek - 5

Derek collected 15 less flags than Kenny.

All integers from 1 to 81 inclusive.

I assume the dice are cubes, (6 sides)

The numbers on them are 1 to 6

1+2+3+4+5+6= 21     

so prob of rolling a 1 is  1/21

so prob of rolling a 2 is  2/21

etc

To roll a 3 you have to have (1,2)  ot (2,1)

P(rolling 1 followed by 2 )= 1/21 * 2/21 = 3/441

P(rolling 2 followed by 1)= 1/21 * 2/21 = 3/441

So prob of rolling a total of 3 is   6/441  = 2/147

WOW 

Thanks Alan,

That was a lot to display.  But I get it.  

Thanks!