All Answers 358087 Answers

Well...if James beat Robert by 10 m and Sam by 20m, then Robert will beat Sam by 10m

it is 10 metres

2 is the answer

The area of the rectangle is 9 x 4 = 36cm^2

So the side of the square = √36cm^2 = 6cm

And the perimeter of the square is  4 x 6cm = 24 cm

20/30 +18/30 =38/30 = 1 8/30 =1 4/15

2cos^2(x)-sin(x)-1=0

$$\small{\text{$
\begin{array}{rclr}
2\cos^2{(x)}-\sin{(x)}-1 &=& 0 \qquad | \qquad \cos^2{(x)}=1-\sin^2{(x)} \\
2 [ 1-\sin^2{(x)} ]-\sin{(x)}-1 &=& 0 \\
2 - 2 \sin^2{(x)} -\sin{(x)}-1 &=& 0 \\
- 2 \sin^2{(x)} -\sin{(x)}+1 &=& 0 \qquad | \qquad \cdot (-1)\\
2 \sin^2{(x)} +\sin{(x)}-1 &=& 0 \\
\left[ 2 \sin{(x)} -1 \right]\cdot \left[ \sin{(x)}+1 \right] &=& 0 \\\\
\underbrace{\left[
\textcolor[rgb]{1,0,0}{2 \sin{(x)} -1}\right]}_{=0}
\cdot
\underbrace{\left[
\sin{(x)}+1 \right]}_{=0}
&=& 0 \\\\
\sin{(x)}+1 &=& 0 \\
\sin{(x)} &=& -1\\
x &=& \arcsin(-1) \pm 2k\pi \\
\mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{\pi}{2} \pm 2k\pi } \quad \mathbf{k}\in Z\\
\\
\hline
\\
\textcolor[rgb]{1,0,0}{2 \sin{(x)} -1} &\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{0} \\
2 \sin{(x)} &=&1 \\\\
\sin{(x)} &=&\dfrac{1}{2} \\
x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\
\mathbf{x} & \mathbf{=} & \mathbf{\dfrac{\pi} {6} \pm 2k\pi} \mathbf \quad {k}\in Z\\\\
\sin{(\pi-x)} &=&\dfrac{1}{2} \\
\pi-x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\
\pi-x &=& \dfrac{\pi} {6} \pm 2k\pi \\\\
\mathbf{x} & \mathbf{=} & \mathbf{\dfrac{5} {6}\pi \pm 2k\pi} \quad \mathbf{k}\in Z
\end{array}
$}}$$

10000238682384117842336328714874663743361993.5

whats 9999 to the power of 555

you can copy the technique that i just used for this one

 11^500 plus 25^500 + 36^500

$$y=11^{500} \\\\
logy=log11^{500}\\\\
logy=500log11\\\\
logy=520.6963426\\\\
y=10^{520.6963426}\\\\
y=10^{0.6963426}*10^{520}\\\\
y=4.969842206\times 10^{520}\\\\
11^{500}=4.969842206\times 10^{520}$$

Now you can copy the technique and do the others then add them together :)

$$\small{\text{
If $b$ is positive, what is the value of $b$ in the geometric sequence $9, a , 4, b$?}}\\
\small{\text{
Express your answer as a common fraction.
}}$$

$$\small{\text{We have $ u_1 = 9,~ u_2 = a, ~ u_3 = 4, and ~ u_4 = b
$ ~~\boxed{ \dfrac{u_n}{u_{n-1}} = constant. }
}}\\\\
\small{\text{ $
\begin{array}{rcl}
\dfrac{u_2}{u_1} &=& \dfrac{u_3}{u_2} \\\\
\dfrac{a}{9} &=& \dfrac{4}{a} \\\\
a^2 &=& 4\cdot 9 = 36 \\\\
\mathbf{a} & \mathbf{=} & \mathbf{6}\\\\\\
\dfrac{u_3}{u_2} &=& \dfrac{u_4}{u_3} \\\\
\dfrac{4}{a} &=& \dfrac{b}{4} \\\\
\dfrac{4}{6} &=& \dfrac{b}{4} \\\\
b &=& \dfrac{16}{6} \\\\
\mathbf{b} & \mathbf{=} & \mathbf{\dfrac{8}{3}}
\end{array}
$}}\\\\$$

$$\small{\text{
What is the next number in this sequence: $\dfrac12, 2, 5, 11, 23, 47, 95, \ldots$?
}}$$

$$\small{\text{$
t_1 = \dfrac{1}{2}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
t_n &=& 2^{n-1}t_1 +\sum \limits_{k=0}^{n-2}2^k \qquad | \qquad \sum \limits_{k=0}^{n-2}2^k = 2^{n-1}-1 \\\\
t_n &=& 2^{n-1}t_1 + 2^{n-1}-1 \\\\
t_n &=& 2^{n-1}(1+t_1)-1 \\\\
t_n &=& 2^{n-1}\left(1+\dfrac{1}{2}\right)-1 \\\\
t_n &=& 2^{n-1}\left(\dfrac{3}{2}\right)-1 \\\\
\mathbf{t_n} & \mathbf{=} & \mathbf{3\cdot 2^{n-2}-1} \\\\
t_8 &=& 3\cdot 2^6-1\\\\
t_8 &=& 3\cdot 64 - 1 \\\\
\mathbf{t_8} & \mathbf{=} & \mathbf{191}
\end{array}
$}}$$

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

$$\small{\text{We have $t_x=t_5=9$ and $t_y=t_{32}=-84$ and we want $t_z=t_{23}=?$
}}\\\\
\small{\text{$\boxed{~~t_z = t_x\cdot \left( \dfrac{y-z}{y-x}\right)
+t_y\cdot \left(\dfrac{z-x}{y-x}\right) ~~}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
t_{23} &=& t_5\cdot \left( \dfrac{32-23}{32-5}\right)
+t_{32}\cdot \left(\dfrac{23-5}{32-5}\right) \\\\
t_{23} &=& 9\cdot \left( \dfrac{9}{27}\right)
-84\cdot \left(\dfrac{18}{27}\right) \\\\
t_{23} &=& 9\cdot \left( \dfrac{1}{3}\right)
-84\cdot \left(\dfrac{2}{3}\right) \\\\
t_{23} &=& -\dfrac{159}{3}\\\\
\mathbf{t_{23}} & \mathbf{=} & \mathbf{-53}\\
\hline
\\
\end{array}
$}}\\\\$$

The 23rd term is -53

3.2808399  feet in a metre

so

$${{\mathtt{3.280\: \!839\: \!9}}}^{{\mathtt{2}}} = {\mathtt{10.763\: \!910\: \!449\: \!432\: \!01}}$$    feet squared in a metre squared

$${\frac{{\mathtt{490}}}{{\mathtt{10.763\: \!910\: \!449\: \!432\: \!01}}}} = {\mathtt{45.522\: \!489\: \!461\: \!611\: \!631\: \!9}}$$      metres squared in a 490 feet squared

$${{\mathtt{3\,125}}}^{{\mathtt{0.2}}} = {\mathtt{5}}$$

so

(-3125)^0.2 = -5 

2logx+3logx=10

$$\\logx^2+logx^3=10\\\\
log(x^2*x^3)=10\\\\
log(x^5)=10\\\\
10^{log(x^5)}=10^{10}\\\\
x^5=10^{10}\\\\
(x^5)^{1/5}=(10^{10})^{1/5}\\\\
x=10^2\\\\
x=100$$

OR BETTER STILL

2logx+3logx=10

$$\\5logx=10\\\\
logx=2\\\\
x=10^2\\\\
x=100$$

BUT why do it the easy way when there is a perfectally good difficult route that can be traversed?   :))

enter

atan(324/54)     into the web2.0calc

atan stands for arc tan which is the same as inverse tan :)

Where do you want the brackets to go?

Yes I will go for that

g(x)>0,    f(x)>0   and     f(x)-g(x)>0

Almost Geno :)

But there are 2 R's so I think that the answer is 

$$\frac{6!}{2!}=\frac{6!}{2}$$

$${\frac{{\mathtt{6}}{!}}{{\mathtt{2}}}} = {\mathtt{360}}$$   ways

The sum of 40 consecutive integers is 100. What is the smallest of these 40 integers?

$$\small{\text{The sum of a arithmetic sequence is $
\boxed{~~S_n = t_1\cdot \binom{n}{1}+d\cdot \binom{n}{2}~~}
$}}\\\\
\small{\text{We have the sum of 40 consecutive integers is 100 so $d=1$ and $ n = 40 $ and $S_{40}=100$ we have :
}}\\
\small{\text{$
\begin{array}{rcl}
S_{40} = 100 &=& t_1\cdot \binom{40}{1} + \binom{40}{2}\\\\
t_1\cdot \binom{40}{1} &=& 100 - \binom{40}{2}\\\\
t_1 &=& \dfrac{100 - \binom{40}{2} }{\binom{40}{1}} \qquad | \qquad \binom{40}{2} = 39\cdot 20 = 780 \quad \binom{40}{1} = 40 \\\\
t_1 &=& \dfrac{100-780}{40} \\\\
t_1 &=& \dfrac{-680}{40}\\\\
t_1 &=& \dfrac{-68}{4}\\\\
t_1 &=& -\dfrac{68}{4}\\\\
\mathbf{t_1} &\mathbf{=}& \mathbf{-17}
\end{array}
$}}$$

The smallest of these 40 integers is -17

4.5

*insert imaginary details*

the asnwer is 10, because 0.1(1/10) times ten is 1, and if you do 1 times ten again you get ten. it is reallyn quite simple

Put it into the onsite calculator like this: (.1065)^(90/365) ≈ .576