All Answers 358517 Answers

Thank you, you two. Thank you.

Okay, I solved one this morning before I went to school:

$$\frac{(b-5)^{3}}{15+7b-2b^{2}}$$

$$\frac{(b-5)(b-5)(b-5)}{(-2b-3)(b-5)}$$

Cancel like terms (the (b-5) at the top and bottom)

$$-\frac{(b-5)^{2}}{(2b-3)}$$   This is your final answer :D

8 choose 3 + 8 choose 5

$${ 8 \choose 3 } + { 8 \choose 5 } = 2 { 8 \choose 3 } = 2{ 8 \choose 5 } = 112$$

$$2 { 8 \choose 3 }
=2*\left(
\frac{8!}{3!(8-3)!}
\right)
=2*\left(
\frac{8!}{3!5!}
\right)
=2*\left(
\frac{5!*6*7*8}{3!5!}
\right)
=
2*\left(
\frac{6*7*8}{3!}
\right)\\\\
=
2*\left(
\frac{336}{6}
\right)
=
\frac{336}{3} =112$$

no your beautiful <3

9*(x - 9) = -11

Divide both sides by 9

x - 9 = -11/9

Add 9 to both sides

x = 9 - 11/9

Since we can write 9 as 81/9 we have

x = 81/9 - 11/9 = (81 - 11)/9 = 70/9

If this is ⁵√(-13.7) then the only real number answer is

$${\mathtt{\,-\,}}\left({{\mathtt{13.7}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)}\right) = -{\mathtt{1.687\: \!889\: \!896\: \!014\: \!528\: \!1}}$$

Check

$${\left(-{\mathtt{1.687\: \!889\: \!896\: \!014\: \!528\: \!1}}\right)}^{{\mathtt{5}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{137}}}{{\mathtt{10}}}} = -{\mathtt{13.700\: \!000\: \!000\: \!000\: \!000\: \!5}}$$

Ok (apart from small numerical rounding error in the decimal representation)

He must have bought 60/4 = 15 tickets.

The total regular price he would have paid for these is $60 + $37.50 = $97.50, so the regular price per ticket is $97.50/15 = $6.50

It's not coming to me Bros   

Just let me think!

My answer is L8; though I think there should probably be a 68 between 06 and 88 !!

y = 2sin^-1(x-1)

Divide both sides by 2 and take sin of both sides of the result

sin(y/2) = x - 1      ...(1)

d(sin(y/2))/dx = d(x-1)/dx

d(sin(y/2))/dy*dy/dx = d(x-1)/dx    using the chain rule on the LHS

(1/2)*cos(y/2)*dy/dx = 1     performing the differentiation on each side

Rearrange:

dy/dx = 2/cos(y/2)

dy/dx = 2/√(1-sin²(y/2))

Make use of (1)
dy/dx = 2/√(1 - (x-1)²)

Rewrite these with the x's and y's on the same side of the equations

10x + 15y = 17       ...(1)

15x - 25y  = -3       ...(2)

Multiply all terms in (1) by 3 and all terms in (2) by 2

30x + 45y = 51     ...(3)

30x - 50y  = -6     ...(4)

Subtract (4) from (3) term by term

45y - (-50y) = 51 - (-6)

or 95y = 57

y = 57/95 or y = 3/5

Substitute this into (1)

10x +15*3/5 =17

10x + 9 = 17

10x = 8

x = 8/10 or x = 4/5

My logic leads me to answer 33.

Let z = (b^x - 1)/x

Multiply both sides by x and add 1 to both sides of the result:

1 + zx = b^x

Take (natural) logs of both sides

ln(1+zx) = xln(b)   using the property of logs that ln(mⁿ) = n ln(m)

Divide both sides by x 

(1/x)*ln(1 + zx) = ln(b)

Use the same property of logs as we used above, but in reverse:

ln( (1 + zx)^1/x ) = ln(b)

By definition, e^z = limit as x goes to zero of  (1 + zx)^1/x

So in the limit as x goes to zero we have

ln(e^z) = ln(b)

or just z = ln(b)

or, using the definition of z

(b^x - 1)/x = ln(b)

Thanks. I got that when I did it or close. I probahbly lost a sign somewhere in my calculations. cheers.

Hey, neither answer works, nor the step before the solution with the theta ^2 in side or outside the brackets.

-7x - 3y = -6    ...(1)

-3x + 7y = 14   ...(2)

Multiply through (1) by 7 and (2) by 3

-49x - 21y = -42    ...(3)

-9x + 21y  = 42     ...(4)

Add (3) and (4) term by term

-58x = 0

x = 0

Put this in (1)

-7*0 - 3y = -6

-3y = -6

y = 2

sine is opposite over hypotenuse, so

sin(angle) = 2.7/6

angle = sin^-1(2.7/6)

$${\mathtt{angle}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.7}}}{{\mathtt{6}}}}\right)} \Rightarrow {\mathtt{angle}} = {\mathtt{26.743\: \!683\: \!950\: \!403^{\circ}}}$$

This graph might help.

for (U sub n) Un=3n-1, list U1+U2+U3+....+U20 and find its sum

$$U_n=3n-1\\ \\
\begin{array}{|c|c|}
\hline
n & U_n \\
\hline
1 & 2 \\
2 & 5 \\
3 & 8 \\
4 & 11 \\
5 & 14 \\
6 & 17 \\
7 & 20 \\
8 & 23 \\
9 & 26 \\
10 & 29 \\
11 & 32 \\
12 & 35 \\
13 & 38 \\
14 & 41 \\
15 & 44 \\
16 & 47 \\
17 & 50 \\
18 & 53 \\
19 & 56 \\
20 & 59 \\
\hline
\end{array}$$

sum:

$$\begin{array} {*{20}{c}}
&
2
&+&
5
&+&
8
&+&
11
&+&
14
&...&
47
&+&
50
&+&
53
&+&
56
&+&
59
\\
&
59
&+&
56
&+&
53
&+&
50
&+&
47
&...&
14
&+&
11
&+&
8
&+&
5
&+&
2
\\
\hline
sum
&
61
&+&
61
&+&
61
&+&
61
&+&
61
&...&
61
&+&
61
&+&
61
&+&
61
&+&
61
\end{array}
\\ \\
sum = \frac{1}{2}*20 * 61 = 610$$

it was a casio fx-82MS calculator. 

Anyways though i figured it out. I just did e^184.5*e^184.5  (369/2 - 184.5) which is (1.34x10^80)*(1.34x10^80) which then became (1.34*1.34)x10^(80+80) = 1.7956x10^160 which was confirmed to be right by the calculator in the homepage.

Riddle help