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We can divide it like this: o | o|  o | o o

We have 5 apples and 3 dividers.

We then use the stars and bars formula. So we have (5+3)!/5!3!= 8C3=56 ways.

We have that the long leg of a 30-60-90 trianlge is sqrt3*[short leg] so it is sqrt3*sqrt3*sqrt3*sqrt3 since the *2 and the /2 cancel out so we have (sqrt3)^4/sqrt3 so we have sqrt3^3=3sqrt3 and our final answer is $\boxed{3\sqrt3}$

128 y^7 - 2240 y^6 + 16800 y^5 - 70,000 y^4 + 175000 y^3 - 262500 y^2 + 218750 y - 78125
(8 terms)

plz help!

what is 5th root of 0.071?            

Using the scientific calculator on my iPhone, I obtain   

that the 5^th root of 0.071 is the same as EP's answer.   

_.

 39 sin 43  is the dotted line

      39 sin 43   * tan 32 = x     = =   16.6

sqrt(.071, 5 = 0.5891849838433908

Determine f(6)

Just substitute 6 for x in the equations

f (x) = - 2/3 x - 1

f (6) = - 2/3 *6 - 1 = -4-1 =-5    So one point on the graph is (6,-5)

Now find f(-9)

Determine x if f(x) = –7

this time it is the other way around

You have to solve

-7=-2/3x-1      

That is a good start for you.

(5x^4 -8x^3 +2x^2 +4x+7)/(x+2)^4 = a+b/(x+2) +c/(x+2)^2 +d/(x+2)^3 +e/(x+2)^4

$\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =a+\frac{ b}{(x+2)} +\frac{c}{(x+2)^2} +\frac{d}{(x+2)^3} +\frac{e}{(x+2)^4}\\ \\~\\ \frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =\frac{a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e}{(x+2)^4}\\~\\ (5x^4 -8x^3 +2x^2 +4x+7) =a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\ (5x^4 -8x^3 +2x^2 +4x+7) =a(x^4+8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\ a=5\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+3*2*x^2+3*4x+8)+c(x+2)^2+d(x+2) +e\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\ -8=40+b\\ b=-48\\~\\ ( 2x^2 +4x+7) =5(24x^2+32x+16)+ -48(6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\ $

Just keep going till you have them all.

There is probably a better way though.

Coding:

\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =a+\frac{ b}{(x+2)} +\frac{c}{(x+2)^2} +\frac{d}{(x+2)^3} +\frac{e}{(x+2)^4}\\
\\~\\
\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =\frac{a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e}{(x+2)^4}\\~\\
(5x^4 -8x^3 +2x^2 +4x+7) =a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\
(5x^4 -8x^3 +2x^2 +4x+7) =a(x^4+8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\
a=5\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+3*2*x^2+3*4x+8)+c(x+2)^2+d(x+2) +e\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\
-8=40+b\\
b=-48\\~\\
( 2x^2 +4x+7) =5(24x^2+32x+16)+ -48(6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\

Here's a bigger hint:

Find the coordinates of P so that P partitions the segment AB in the ratio 3:1 if A(0,7) and B(−4,2).

A(0,7)      B(-4,2)

-4 / 4 = -1            -1 * 3 = -3

7 - 2 / 4 = 1.25         7 - ( 3 * 1.25 ) = 3.25

Coordinates of   P( -3, 3.25 )  

Hint:    ∠ABD = 74 degrees 

5y - 3 = 3y + 27

2y = 30

y = 15

Angle A = 5*15-3 = 72º

Angle B = 180 - 72 = 108º

Since the digits can be repeated, you will simply have: 4^4 / 2^4 = 16 2-digit permutations as follows:

{2, 2} | {2, 4} | {2, 6} | {2, 7} | {4, 2} | {4, 4} | {4, 6} | {4, 7} | {6, 2} | {6, 4} | {6, 6} | {6, 7} | {7, 2} | {7, 4} | {7, 6} | {7, 7} (total: 16)

As follows:

(The other two solutions are ≈ -0.975 and 3.203)

How long  I will live..... will I fall off of the mountain I am climbing tomorrow?  (cloud peak WY USA)  Hope not.....but at least I will be smiling !!

     I make a lot of mistakes...I learn from them all !  (If I survive)

Please help ASAP

$\begin{array}{|rcll|} \hline AB=AC \\ \hline \angle ABC &=& \angle ACB \quad | \quad \angle ABC =2x \\ 2x &=& \angle ACB \\ \mathbf{\angle ACB} &=& \mathbf{2x} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD=BC \\ \hline \angle BDC &=& \angle ACB \quad | \quad \angle ACB =2x \\ \mathbf{\angle BDC} &=& \mathbf{2x} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \angle ADB &=& 180^\circ - \angle BDC \quad | \quad \mathbf{\angle BDC=2x} \\ \mathbf{\angle ADB} &=& \mathbf{180^\circ - 2x} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ \text{In $\triangle$ BCD }: } \\ \hline 2x+2x+x &=& 180^\circ \\ 5x &=& 180^\circ \\\\ x &=& \dfrac{180^\circ}{5} \\\\ \mathbf{x} &=& \mathbf{36^\circ} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ \text{In $\triangle$ ADB }: } \\ \hline A+ (180^\circ-2x) +x &=& 180^\circ \\ A+ 180^\circ-2x +x &=& 180^\circ \\ A+ 180^\circ-x &=& 180^\circ \\ A -x &=& 0 \\ A &=& x \quad | \quad \mathbf{x=36^\circ} \\ \mathbf{A} &=& \mathbf{36^\circ} \\ \hline \end{array}$

The graph would make this much easier !

First way: [5/9 * 4/8  + 4//9 * 3/8] = 4/9

2nd way: [9C2 ] = 36 number of ways you can 2 balls

Number of ways of drawing 2 balls of the same color =5C2 + 4C2 =16

Probability =16 / 36 = 4/9

a + b + c + d + e = 1 + (-4) + 8 + 7 + (-3) = 9.

You subtract C(7,2)*C(5,2), which gives you C(12,5) - C(7,2)*C(5,2) = 582.

What about   828 ?    880?      Hmmmmm....

   ( i do not think '0' counts)   ......

Area of the triangular ends   1/2   b * h =   1/2  * 12 * 5.8 = 34.8 km²

                                                              Multiply by the height 10     volume   =   348 km³

The solution to the system of equations ...graphed... is where the two lines intersect (where they cross) ...

   yo should be able to identify that point easily o the graph