All Answers

#1

+130466

+1

We're not solving anything ....we just want to factor this as the difference of cubes

So we have

y^3 / 27 - x^9 / 216

Difference of cubes formula

a^3 - b^3 = (a - b) (a^2 + ab + b^2)

let a = y / 3 b = x^3 / 6 a^2 = y^2 / 9 ab = [y x^3 ] / 18 and b^2 = x^6 / 36

So we have

( y / 3 - x^3 / 6) ( y^2 / 9 + [y x^3 ] / 18 + x^6/ 36 )

CPhillSep 16, 2017

#1

+130466

+1

3) 64j^15 + 125y^21

Notice that this is just the sum of two cubes....and we have

Remember that

a ^3 + b^3 = ( a + b) (a^2 - ab + b^2)

So...let a = 4j^5 b = 5y^7 a^2 = 16j^10 b ^2 = 25y^14 and -ab = - 4j^5*5y^7

And we have.....

( 4j^5 + 5y^7) ( 16j^10 - 4j^5*5y^7 + 25y^14) =

(4j^5 + 5y^7) ( 16j^10 - 20 j^5y^7 + 25y^14)

4) c^3d^3 + 125 = (cd)^3 + 125

This is similar......

a ^3 + b^3 = ( a + b) (a^2 - ab + b^2)

a = cd b = 5 a^2 = (cd)^2 - ab = - 5cd and b^2 = 25

So we have

( cd + 5) [ (cd)^2 - 5cd + 25]

CPhillSep 16, 2017

#2

+130466

+1

Thanks, Dr Dros!!!

Here's another method without resorting to logs

[(11)^ (1/4)] ^(3x -3) = 1/5 implies that

[ 11 ^(x - 1) ] ^ (3/4) = 1/5 take both sides to the 4/3 power

[11 ^(x - 1) ] = (1/5)^(4/3)

[11^x ] / 11 = (1/5)^(4/3)

11^x = 11* (1/5)^(4/3)

So..... 11^(6x) = (11^x)^6 = [ 11 * (1/5)^(4/3) ]^6 = [ 11^6] * [ 1/5]^8

So..... [11^(1/4)]^(6x + 2) = [ 11^(6x + 2) ] ^(1/4) = [ 11 ^(6x) * 11^2] ^ (1/4) =

[ 11^6 * (1/5)^8 ] ^(1/4) * 11^(1/2) =

[ 11^(6/4)] * [(1/5)^8]^(1/4) * 11^(1/2) =

[11 ^ (3/2) * [ (1/5) ^2] * 11^(1/2) =

[ 11 ^2] * (1 / 25) =

121 / 25

CPhillSep 16, 2017

#2

+130466

+1

Hey, rosala.....!!!

We always have time for your funny pictures and your stories!!!!

CPhillSep 16, 2017

#1

+130466

+2

Volume of the container = [10 in ]^3 = 1000 in^3

Volume of water = 500 in^3

Volume of 10 ice cubes = 10 * (2)^3 = 80 in^3

Unoccupied space = [ 1000 - 500 - 80 ] = 420 in^3

CPhillSep 16, 2017

#1

+130466

+1

Suppose that $f(x)$ is a polynomial that has degree $6$ and $g(x)$ is a polynomial that has degree $3$. If $h(x)$ is also a polynomial such that $f(g(x)) + g(h(x)) + h(f(x))$ is a polynomial of degree $36$, then what is the degree of the polynomial $h$?

If f is of degree 6 and g is of degree 3, then the greatest degree of either f(g(x)) or g(f(x)) wil be 18

But addding these together will only result in a polynomial of degree 18 [ at most ]

So...... h ( f(x)) must itself be of degree 36 which implies that h will be of degree 6

CPhillSep 16, 2017

#2

+130466

+1

Nice diagram, hectictar....!!!

CPhillSep 16, 2017

#1

+130466

+1

Number of full house hands

We want to choose any 1 of 13 ranks and and any 3 of 4 cards witihin the chosen rank

Then...we want to choose two cards from any one of the 12 remaining ranks

So.....full house hands =

C (13 , 1) * C(4,3) * C(12, 1) * C(4,2) = 3744

Three of a kind [not counting full house hands ]

We want to choose any 1 of 13 ranks and and any 3 of 4 cards witihin the chosen rank

Then...we want to choose any two of the 12 remaining ranks with one card from each rank

So...we have

C (13, 1) * C (4,3) * C (12,2) * [C(4,1)]^2 = 54,912

So....the total of three-of-a-kind hands [ including full house hands ] = 3744 + 54,912 = 58,656

CPhillSep 16, 2017

#1

+2446

0

True.

Both sides are the exact same, so this is a true equation.

TheXSquaredFactorSep 16, 2017

#2

+2446

+1

I'll show my method of proving. Of course, the standard form of a quadratic equation is in the following form of $ax^2+bx+c=0$ . If I start with the quadratic formula $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$ , then all I should have to do is utilize some algebraic manipulation to arrive to the standard form. Let's do that!

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$	First, let's eliminate the fraction on the right hand side by multiplying by 2a on both sides.
$2ax=-b\pm\sqrt{b^2-4ac}$	Add b to both sides of the equation.
$2ax+b=\pm\sqrt{b^2-4ac}$	Square both sides. Normally, you would have to consider both cases, but squaring always results in the positive answer--no matter if the original number is positive or negative. Therefore, both will result to the same thing.
$(2ax+b)^2=b^2-4ac$	Expand the left hand side of the equation by knowing that $(x+y)^2=x^2+2xy+y^2$ .
$(2ax)^2+2(2ax)(b)+b^2=b^2-4ac$	Subtract b^2 from both sides.
$(2ax)^2+2(2ax)(b)=-4ac$	Simplify the left hand side by dealing with the multiplication and the exponent.
$4a^2x^2+4axb=-4ac$	Divide by the GCF of both sides, which is 4a.
$ax^2+bx=-c$	And finally, subtract c on both sides.
$ax^2+bx+c=0$

I have now proven that the quadratic formula works for all numbers for A,B, and C.

TheXSquaredFactorSep 16, 2017

#1

+9488

+3

A monic polynomial is a polynomial where the cofficient of the highest order term is 1. (I didn't know this until now, I had to look it up here.)

f(x) is a monic polynomial with a degree of 2. So we can say...

f(x) = 1x² + bx + c = x² + bx + c

The problem says f(0) = 4 . So...

f(0) = 0² + b(0) + c

4 = 0 + 0 + c

4 = c

Now that we know c = 4 , we know that f(x) = x² + bx + 4 .

The problem says f(1) = 10 . So...

f(1) = 1² + b(1) + 4

10 = 1 + b + 4

10 = 5 + b

5 = b

Now we know b = 5 and c = 4 , so f(x) = x² + 5x + 4 .

hectictarSep 16, 2017

#2

+109

+1

I'm assuming you meant the tenth's place, so the answer is 2.

Jeff123Sep 16, 2017

#1

+9488

+1

Since triangle BCD is isosceles with BC = DC , ∠DBC = ∠BDC = 70°

Since ∠BDC and ∠CDA form a straight line...

∠BDC + ∠CDA = 180°

Subtract ∠BDC from both sides of the equation.

∠CDA = 180° - ∠BDC

∠BDC = 70°

∠CDA = 180° - 70°

∠CDA = 110°

Since there are 180° in every triangle...

∠CDA + ∠DAC + ∠ACD = 180°

∠CDA = 110°

110° + ∠DAC + ∠ACD = 180°

Since triangle ACD is isosceles with CD = AD, ∠ACD = ∠DAC.

110° + ∠DAC + ∠DAC = 180°

110° + 2∠DAC = 180°

2∠DAC = 70°

∠DAC = 35° = ∠BAC

hectictarSep 16, 2017

#1

+4

I hope this helps ( i took a snapshot from word, cause LaTeX gave me cancer ).

DrDrosSep 16, 2017

#1

+9488

+1

The longest diagonal will be from one corner to the corner completely opposite it (top and bottom, front and back, left and right). Like the black line in this picture:

We can find it's length by making it into a triangle like this:

Let's call this new gray line " b ".

Now we can use the Pythagorean theorem to find b and the length of the prism's diagonal.

9² + 8² = b²

81 + 64 = b²

145 = b²

b² + 12² = (diagonal)² Substitute 145 in for b² .

145 + 12² = (diagonal)²

145 + 144 = (diagonal)²

289 = (diagonal)² Take the positive square root of both sides.

√289 = diagonal

17 = diagonal

hectictarSep 16, 2017

#1

+1

Go online to this link and see how to prove the "Quadratic Formula":

https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/v/proof-of-quadratic-formula

GuestSep 16, 2017

#1

0

which way do you mean like right or left

be more pacific.

GuestSep 16, 2017

#1

0

Factor the following:
a^3 + 64 b^3

Hint: | Write 64 b^3 as a cube in order to express a^3 + 64 b^3 as a sum of cubes.
a^3 + 64 b^3 = a^3 + (4 b)^3:
a^3 + (4 b)^3

Hint: | Factor the sum of two cubes.
Factor the sum of two cubes. a^3 + (4 b)^3 = (a + 4 b) (a^2 - a×4 b + (4 b)^2):
(a + 4 b) (a^2 - 4 a b + (4 b)^2)

Hint: | Distribute exponents over products in (4 b)^2.
Multiply each exponent in 4 b by 2:
(a + 4 b) (a^2 - 4 a b + 4^2 b^2)

Hint: | Evaluate 4^2.
4^2 = 16:
(a + 4 b) (a^2 - 4 a b + 16 b^2)

Factor the following:
27 - y^12

Hint: | Factor a minus sign out of 27 - y^12.
Factor -1 out of 27 - y^12:
-(y^12 - 27)

Hint: | Express y^12 - 27 as a difference of cubes.
y^12 - 27 = (y^4)^3 - 3^3:
-(y^4)^3 - 3^3

Hint: | Factor the difference of two cubes.
Factor the difference of two cubes. (y^4)^3 - 3^3 = (y^4 - 3) ((y^4)^2 + y^4 3 + 3^2):
-(y^4 - 3) ((y^4)^2 + 3 y^4 + 3^2)

Hint: | For all positive integer exponents (a^n)^m = a^(m n). Apply this to (y^4)^2.
Multiply exponents. (y^4)^2 = y^(4×2):
-(y^4 - 3) (y^(4×2) + 3 y^4 + 3^2)

Hint: | Multiply 4 and 2 together.
4×2 = 8:
-(y^4 - 3) (y^8 + 3 y^4 + 3^2)

Hint: | Evaluate 3^2.
3^2 = 9:
-(y^4 - 3) (y^8 + 3 y^4 + 9)

GuestSep 16, 2017

#1