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 A rectangular box has interior dimensions 6-inches by 5-inches by 10-inches. The box is filled with as many solid 3-inch cubes as possible, with all of the cubes entirely inside the rectangular box. What percent of the volume of the box is taken up by the cubes?

Volume of the box:  V = 5*6*10 = 300 in³

Volume of the cube:  V_c= 3³ = 27 in³

300 / 27 = 11.1111

11.1111 : 100% = 11 : X            X = 99%

11 cubes take up 99% of the volume of the box.  

Nicely done heureka - a more elegant solution than mine!

For the first one, note that the term 3n⁵/(4n⁵ + 1)  gets very close to 3/4 by the time n has reached 10 or so.  What that means is that the sum consists of an infinite number of values that are all very nearly 3/4.  Adding an infinite number of 3/4 results in an infinite sum.

Thank you, CPhill !

Nice, heureka  !!!

The radius of circle 1 is 1.  What is the radius of circle 2?

Circles 1 and 2 have the same radii.  

The other question was different.

I answered the other one recently but obviously not on the thread that you just found.

In triangle \(ABC\),  let \(I\) be the intersection of angle bisectors \(\overline{AD}\) and  \(\overline{BE}\).
Prove that \( \angle DIB = 90^\circ - \dfrac{\angle BCA}{2}\).  

\(\begin{array}{rcll} \begin{array}{|rcll|} \hline \mathbf{\text{In AIB:}} \\ \hline 180^\circ-x + \alpha + \beta &=& 180^\circ \\ -x + \alpha + \beta &=& 0 \\ \mathbf{x} &=& \mathbf{\alpha + \beta} \qquad (1) \\ \hline \end{array} & \begin{array}{|rcll|} \hline \mathbf{\text{In BID:}} \\ \hline x + D + \beta &=& 180^\circ \\ \mathbf{180^\circ-D} &=& \mathbf{x+\beta} \qquad (2) \\ \hline \end{array} \\ & \begin{array}{|rcll|} \hline \mathbf{\text{In EIA:}} \\ \hline x + E + \alpha &=& 180^\circ \\ \mathbf{180^\circ-E} &=& \mathbf{x+\alpha} \qquad (3) \\ \hline \end{array} \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{In EIDC:}} \\ \hline (180^\circ-D)+(180^\circ-E)+(180^\circ-x) + C &=& 360^\circ\\ \boxed{ \mathbf{180^\circ-D=x+\beta}\\ \mathbf{180^\circ-E=x+\alpha}} \\ x+\beta+x+\alpha+(180^\circ-x) + C &=& 360^\circ \\ x+\beta+x+\alpha+180^\circ-x + C &=& 360^\circ \quad | \quad - 180^\circ \\ x+\beta+x+\alpha -x + C &=& 360^\circ - 180^\circ \\ x+\beta+x+\alpha -x + C &=& 180^\circ \\ x+\beta +\alpha + C &=& 180^\circ \\ x+\alpha+\beta + C &=& 180^\circ \quad | \quad \mathbf{\alpha+\beta=x} \\ x+x + C &=& 180^\circ \\ 2x + C &=& 180^\circ \\ 2x &=& 180^\circ -C \quad | \quad : 2 \\ \mathbf{x} &=& \mathbf{90^\circ -\dfrac{C}{2}} \\ \hline \end{array}\\\)

The desired point is  ( 60/80) = 6/8 = 3/4 the  distance between  M  and N

To  find this point, we have

[ 3  + (3/4)(9 - 3) ,  2 + (3/4) ( 13 - 2)  ]  =

[ 3 + (3/4)(6)  , 2 + (3/4)(11)  ] =

[ 3 + 18/4  , 2 + 33/4 ] = 

[ 12/4 + 18/4 , 8/4 + 33/4  ] =

[30/4 , 41/4 ] =

( 15/2, 41/4)

Since we know that a red disc has been selected......then  the  probability of selecting  the one with a "3"  is  just 1/5

Find the ordered pair \((a, b)\) of real numbers for which

\((ax + b)(x^5 + 1) - (5x + 1)\)

is divisible by \(x^{2}+1\).

\(\begin{array}{|lrcll|} \hline & P(x) &=& Q(x) (x^{2}+1) \\ & \mathbf{(ax + b)(x^5 + 1) - (5x + 1)} &=& \mathbf{Q(x) (x^{2}+1)} \\ & && \boxed{x^2+1 = 0\\ x^2 = -1\\ x=\pm i } \\ \hline x=i: & (ai + b)(i^5 + 1) - (5i + 1) &=& Q(i)\times 0 \quad | \quad i^5 = i \\ & (ai + b)(i + 1) - (5i + 1) &=& 0 \\ & (ai + b)(i + 1) - 5x - 1 &=& 0 \\ & ai^2+i(a+b-5) +b-1 &=& 0 \quad | \quad i^2 = -1 \\ & -a+i(a+b-5) +b-1 &=& 0 \\ (1) & \mathbf{-a+b-1+i(a+b-5)} &=& \mathbf{0} \\ \hline x=-i: & \left(a(-i) + b\right)\left((-i)^5 + 1\right) - \left(5(-i) + 1\right) &=& Q(-i) \times 0 \quad | \quad (-i)^5= -i \\ & (-ai + b)(-i + 1) - (-5i + 1) &=& 0 \\ & (-ai + b)(-i + 1) 5i-1 &=& 0 \\ & ai^2-i(a+b-5) +b-1 &=& 0 \quad | \quad i^2 = -1 \\ & -a-i(a+b-5) +b-1 &=& 0 \\ (2) & \mathbf{-a+b-1-i(a+b-5)} &=& \mathbf{0} \\ \hline (1)+(2):& (-a+b-1)+i(a+b-5) \\ & + (-a+b-1)-i(a+b-5)&=& 0 \\ & 2(-a+b-1) &=& 0 \\ & -a+b-1 &=& 0 \\ &\mathbf{b} &=& \mathbf{1+a} \\ \hline (1)-(2):& (-a+b-1)+i(a+b-5) \\ &- \Big( (-a+b-1)-i(a+b-5) \Big) &=& 0 \\ & (-a+b-1)+i(a+b-5) \\ & -(-a+b-1)+i(a+b-5) &=& 0 \\ & 2i(a+b-5) &=& 0 \\ & a+b-5 &=& 0 \\ &\mathbf{b} &=& \mathbf{5-a} \\ \hline &b=1+a &=& 5-a \\ & 1+a &=& 5-a \\ & 2a &=& 4 \\ & \mathbf{a} &=& \mathbf{2} \\\\ & b&=&1+a \\ & b&=&1+2 \\ & \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}\)

The sum of the first n terms of a certain sequence is n(n+1)(n+2). Find the tenth term of the sequence.

The first term   is    1 ( 2) (3)  =  6  =  6(1)

The sum of the first two terms  =  2 (3) ( 4)  = 24...so   the second term  =  18  =  6(3)

The sum of the first three terms =  3 * 4 * 5  = 60.....so the third term =  36 = 6(6)

The sum of the first four terms =  4 * 5 *6  =  120....so....the fourth term = 60  = 6(10)

So....it appears that the  nth term  =   6 * nth triangular number  =  6 * n (n + 1) / 2

So....the 10th term is   6 * (10) ( 11)  / 2  =   30 * 11  =  330

Find the area of triangle PQR if PQ = QR = 12 and

It,s impossible to solve! Not enough information. 

a.   

Let  u(x)  =   ∛ x

Let v (x) =  1 - 4e^x

Then   u ( v(x) )   means that we are subbing  v  into the "x" in   u    =    ∛[ 1 - 4e^x ]   =  f(x) 

b. This one is a little tricky   .....

Let  u(x)  =    1 + x

                     ______

                      2 + x

Let v (x)  =   x^2

                                    1  + x^2

So... u(v (x))   =           _________  =   f(x) 

                                    2  +  x^2

c.

Let u(x)  =  ln (x)

Let v(x)  =  2 -  1/x

Then   u (v(x))  =  ln ( 2 - 1/x)  =  f(x)

Unit semicircle, with AB = 2.  Angle alpha = 30.  Find the area of the green region.

There are 2 sectors: ADG and BCG, and 2 right angled triangles: DFG, and CFE.

Area of both sectors is:  [(AB/2)² * pi] /3 = 1.04719755 u²

Area of triangle DFG =(DF * GF) /2 = 0.21650635 u²

Area of triangle CFE = (CF * EF) /2 = 0.072168783 u²

Area of semicircle is:  pi / 2 = 1.570796327 u²

Area of the green (yellow) region is:  A = 0.234923643 u²  

Let N  be the number of scarves she can make

Let  k  be the proportionality constant

Let A  be the area of a scarf   .....so we have that...

N  =  k / A

And we know that

15  =  k / 4            multiply both sides by 4

60  = k

So

N  = 60 /  5

12   =  number of scarves she can make when each has an area of 5 ft^2

ok boomer

I care that you cared that he cared that they said nobody cares.

Its not cheating if they understand the process and don't just copy the answer, Guest.

[previous deleted] ok imma gonna do dis tommorow because it is 10:30 for  me now

Drawing a picture can HELPPPPPPP

The Gecko is 6 feet above the fly (vertical), 8 feet horizontally forward from it, and 12 feet horizontally left from it.

We first use pythagorean theorem.

8² + 12² = 208

\(\sqrt{208}=4\sqrt{13}\)

Ok then we use the pythagorean theorem again.

6² + \((4\sqrt{13}^2)\) = 244

\(\sqrt{244}=\boxed{2\sqrt{61}}\)

PLEASE CHECK THIS IS PROBABLY SUPER DUPER BUPER MOOPER LOOPER WRONG!!!!

In triangle $ABC$, we have $\angle A = 30^\circ$, $\angle B = 60^\circ$, and $\angle C=90^\circ$. Point $X$ is on $\overline{AC}$ such that $\overline{BX}$ bisects $\angle ABC$. If $AB = 12$, then what is the area of triangle $ABX$?

 Area of the triangle  ABX = ( AB/2) * [( AB/2) * tan(A)] 

                                    Area   ABX ≈ 20.785 u²