All Answers 358156 Answers

oh and I for somereason cant eddit my post but after you found the area of the triangle you can take that number and divide by 100 to get the persentage! 

~ Wolf

Hi there!!! So I will give you hints to get to the right direction! After these hints and parts of the soulution if you still dont understand please ask!

hints

1. The blue region is a triangle and perfictly alinged to the corners

2. If each box is 1 then you have the length and width or height and base.

3. We know that the formula for the triangle is \(\frac{(b\cdot h)}{2}\)

I hope this helps!!!!

~ Wolf :D

thank you!!

wow thank you so much!!!

Approximately  93.74/140.73 = 66.6%

Find a base 7 three-digit number which has its digits reversed when expressed in base 9.

Let the digits of the number be a,b,c where 

a is between 1 and 6 inclusive and 

b is between 0 and 6 inclusive

and 

c is between 1 and 6 inclusive

a>c  

\(49a+7b+c=81c+9b+a\\ 0=80c+2b-48a\\ 0=40c+b-24a\\ 24a-b=40c\\ \)

24a could be    24,  48,  72, 96,  120,  144

40c could be     40,  80,  

I do not think I will finish it since it is obviously a challenge question for you. You need to examine all the possibilities in orger to find triplets that work. It is not as tiresome as it may seem.

When you finish it, it would be nice for you to display the rest, (or the alternate) logic that you use.

That is another challenge for you.

Please let Kitten continue this. Do not finish it for her. Thanks

LaTex:

49a+7b+c=81c+9b+a\\
0=80c+2b-48a\\
0=40c+b-24a

Compute

\(\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{9700}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{8554}+\dfrac3{9118}+\dfrac3{9700}} \\\\ &=& \sum \limits_{k=0}^{32} \dfrac{3}{(3k+1)(3k+4)} \\\\ &=& \sum \limits_{k=0}^{32}\left( \dfrac{1}{3k+1} -\dfrac{1}{3k+4} \right) \\\\ &=& \sum \limits_{k=0}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0}^{32}\left(\dfrac{1}{3k+4}\right) \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0}^{31}\left(\dfrac{1}{3k+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0+1}^{31+1}\left(\dfrac{1}{3(k-1)+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=1}^{32}\left(\dfrac{1}{3k-3+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \dfrac{1}{100} \\\\ &=& 1 - \dfrac{1}{100} \\\\ &=& \dfrac{100-1}{100} \\\\ &=& \mathbf{\dfrac{99}{100}} \\ \hline \end{array}\)

\(\lfloor t\rfloor = 2t + 3\\ t \le 2t + 3

also

2t+3 must be an integer

so t must be a multiple of a half

so maybe  t=-3   or t=-2.5

sub in and t=-3 works but t=-2.5 does not

so

I only get one answer  t=-3

------------------------------------

LaTex:

\lfloor t\rfloor = 2t + 3\\
t \le 2t + 3 0\le t + 3<1\\
-3\le t <-2\\

 What angles are coterminal to an angle measuring −5π/3 or 8π/7?

Hello ocmac!

Two angles are "coterminal" if they differ by an integral multiple of 360 °.

Zwei Winkel sind "coterminal", wenn sie sich um ein ganzzahliges Vielfaches von  360°  unterscheiden.

\(\frac{2\pi}{-\frac{5\pi}{3}}=\color{red}-\frac{6}{5}\)    \(The\ angle\ is\ not\ coterminal\ to\ 2\pi.\)

\(\frac{2\pi}{\frac{8\pi}{7}}=\color{red}-\frac{7}{4}\)        \(The\ angle\ is\ not\ coterminal\ to\ 2\pi.\)

  !

sum_(n=1)^33 3/((3 n - 2) (3 n + 1)) = 99/100

A = ab    

Find the shaded area.

  \(x:3=3:\sqrt{3^2+4^2}\\ x:3=3:5\\ 5x=9\\ x=\frac{9}{5}\)      \(the\ section\ at\ the\ bottom\ left\)

\(b:3=4:5\\ b=\frac{12}{5}\)                       \(width\)

\(a=5-x=5-\frac{9}{5}\\a=\frac{16}{5}\)          \(length\)

\(A=a\times b\\ A=\frac{16}{5}\cdot \frac{12}{5}\)

\(A=\frac{192}{25}=7.68\)

  !

We have a 3-4-5 right triangle and a similar triangle with a hypotenuse of 3 units.

The longer leg of a smaller triangle:   3 / b = 5 / 4       b = (3 * 4) / 5  This is the width of the rectangle.

The length of the rectangle:        a = 5 - sqrt( 3² - b² )   

Find the number of real roots of  \(1 + x + x^{2} + x^{3} = x^{4} + x^{5}\)

\(\begin{array}{|lrcll|} \hline & 1 + x + x^2 + x^3 &=& x^4 + x^5 \\ & (1 + x)(x^2+1) &=& x^4(1+x) \\ & (1 + x)(x^2+1)-x^4(1+x) &=& 0 \\ & \mathbf{ (1 + x)(x^2+1-x^4) } &=& \mathbf{0} \\\\ 1) & x+1 &=& 0 \\ & \mathbf{x} &=& \mathbf{-1} \\\\ 2) & x^2+1-x^4 &=& 0 \\ & x^4-x^2-1 &=& 0 \\ & x^2 &=& \dfrac{1\pm \sqrt{1-4(-1)} }{2} \\ & \mathbf{x^2} &=& \mathbf{\dfrac{1\pm \sqrt{5} }{2}} \\\\ & x^2=\dfrac{1+\sqrt{5} }{2} &\text{or}& x^2=\dfrac{1-\sqrt{5} }{2} \quad < 0!\ \text{Complex solutions:} \\\\ & \mathbf{x=\sqrt{\dfrac{1+\sqrt{5}}{2}}} & \\\\ & \mathbf{x=-\sqrt{\dfrac{1+\sqrt{5} }{2}}} & \\\\ \hline \end{array}\)

Real solutions:
\(\mathbf{x=-1} \\ \mathbf{x=\sqrt{\dfrac{1+\sqrt{5}}{2}}} \\ \mathbf{x=-\sqrt{\dfrac{1+\sqrt{5} }{2}}}\)

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius \(1\).
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

A ball is thrown in air and it's height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t)=−4t2+16t. When is the ball higher than 12 feet off the ground?

Ein Ball wird in die Luft geworfen und seine Höhe h (t) in Fuß zu jeder Zeit, t in Sekunden, wird durch die Gleichung h (t) = - 4t2 + 16t dargestellt. Wann ist der Ball höher als 12 Fuß über dem Boden?

Hello Guest!

\(h (t) = - 4t^2 + 16t=12\ feets\)

\(- 4t^2 + 16t-12=0\)

\(t= {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(t= = {-16 \pm \sqrt{(16)^2-4(-4)(-12)} \over 2(-12)}\)

\(t=\frac{-16\pm 8}{-24}\)

\( t_1=\frac{1}{3}\ second\\ t_2=1\ second\\\)

The ball is higher than 12 feet off the ground

between  \( t_1=\frac{1}{3}\ second\) and \( t_2=1\ second\).

  !

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius \(1\).
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

If

\(x + y + z = 26\) and \(\dfrac1x + \dfrac1y + \dfrac1z = 31\), then find \(\dfrac xy + \dfrac yx + \dfrac xz + \dfrac zx + \dfrac yz + \dfrac zy\).

\(\begin{array}{|rcll|} \hline && \dfrac xy + \dfrac yx + \dfrac xz + \dfrac zx + \dfrac yz + \dfrac zy \\\\ &=& \dfrac xy +\dfrac xz + \dfrac yx + \dfrac yz + \dfrac zx + \dfrac zy \\\\ &=& x\left(\dfrac 1y +\dfrac 1z \right) + y\left(\dfrac 1x + \dfrac 1z\right) + z\left(\dfrac 1x + \dfrac 1y\right) \\\\ && \boxed{\dfrac1x + \dfrac1y + \dfrac1z = 31\\ \dfrac1x + \dfrac1y = 31- \dfrac1z \\ \dfrac1x + \dfrac1z = 31- \dfrac1y \\ \dfrac1y + \dfrac1z = 31-\dfrac1x } \\\\ &=& x\left(31-\dfrac1x\right) + y\left(31- \dfrac1y\right) + z\left(31- \dfrac1z\right) \\\\ &=& 31x-1 +31y-1 + 31z-1 \\\\ &=& 31(x+y+z)-3\quad | \quad \mathbf{x + y + z = 26} \\ &=& \mathbf{803} \\ \hline \end{array}\)

If

\(ab = 15\) and \(a^2 + b^2 = 40\), then find the value of \((a + b)^4 - (a - b)^4\).

\(\begin{array}{|rcll|} \hline \mathbf{(a + b)^4 - (a - b)^4} &=& \left((a + b)^2\right)^2 - \left((a - b)^2\right)^2 \\ &=& \left(a^2+2ab+b^2\right)^2 - \left(a^2-2ab+b^2\right)^2 \\ &=& \left(a^2+b^2+2ab\right)^2 - \left(a^2+b^2-2ab\right)^2 \\ &=& \left(40+2*15\right)^2 - \left(40-2*15\right)^2 \\ &=& 70^2 - 10^2 \\ &=& 4900-100 \\ &=& \mathbf{4800} \\ \hline \end{array}\)

Simplify 

\(\dfrac{\sin (A-B)}{\sin A \sin B} + \dfrac{\sin (B-C)}{\sin B \sin C} + \dfrac{\sin (C-A)}{\sin C \sin A}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\sin(A-B)}{\sin A \sin B} + \dfrac{\sin(B-C)}{\sin B \sin C} + \dfrac{\sin(C-A)}{\sin C \sin A} } \\\\ &=& \dfrac{\sin(A)\cos(B)-\cos(A)\sin(B)}{\sin A \sin B} + \dfrac{\sin(B)\cos(C)-\cos(B)\sin(C)}{\sin B \sin C} + \dfrac{\sin(C)\cos(A)-\cos(C)\sin(A)}{\sin C \sin A} \\\\ &=& \dfrac{\sin(A)\cos(B)}{\sin A \sin B}-\dfrac{\cos(A)\sin(B)}{\sin A \sin B} + \dfrac{\sin(B)\cos(C)}{\sin B \sin C}-\dfrac{\cos(B)\sin(C)}{\sin B \sin C} + \dfrac{\sin(C)\cos(A)}{\sin C \sin A}-\dfrac{\cos(C)\sin(A)}{\sin C \sin A} \\\\ &=& \dfrac{\cos(B)}{\sin B}-\dfrac{\cos(A)}{\sin A} + \dfrac{\cos(C)}{\sin C}-\dfrac{\cos(B)}{\sin B} + \dfrac{\cos(A)}{\sin A}-\dfrac{\cos(C)}{\sin C} \\\\ &=& \dfrac{\cos(B)}{\sin B}-\dfrac{\cos(B)}{\sin B} + \dfrac{\cos(C)}{\sin C}-\dfrac{\cos(C)}{\sin C} + \dfrac{\cos(A)}{\sin A}-\dfrac{\cos(A)}{\sin A} \\\\ &=& \mathbf{0} \\ \hline \end{array}\)

have got 100 lamp bulbs with 100 switches. Initially all switches are turned off.

After that I change the position of each switch. After that I change the position of each second, third, fourth... hundredth switch.

How many bulbs will light up?

I assume:

All lamps will light up on a perfect square position:  {1,4,9,16,25,36,49,64,81,100} = 10 lamps will light up.

What is the remainder when \(7^{42}\) is divided by \(16\)?

\(\begin{array}{|rcll|} \hline && 7^{42} \pmod{16} \quad | \quad 7^2 \pmod{16} = 1 \\ &\equiv& 7^{2*21} \pmod{16} \\ &\equiv& \left(7^{2}\right)^{21} \pmod{16} \\ &\equiv& \left(1\right)^{21} \pmod{16} \\ &\equiv& \mathbf{1} \pmod{16} \\ \hline \end{array} \)

The remainder is \(\mathbf{1}\)

The number of ways is 5^7*3 + 5^6*4*3 = 421875.