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16

I would walk you through it but I'm too tired right now.

Ask me later and I will definitley will walk through it.

1 45/100

1 8/20

1 4/10

1 2/5

This is not what the forum is used for.

If you want to post things like this become a member.

But hey

Morning! I can't guarantee that my idiot friends will be absent today , but thanks, I'll try not to fall asleep!

Have a good day, Coldplay!

CONGRATULATIONS!!

I hope you guys are happy

99.9 as a fraction is 999/1000

Thanks Coldplay!!

You always make me smile.

EVERYDAY

I hope I can do the same.

You have a good day too

Lol!

(sqrt(3)-4+sqrt(5))/((3*sqrt(15))+15-(12*sqrt(5))

\(\begin{array}{rcll} \frac{ \sqrt{3}-4+\sqrt{5} } { 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } &=& \frac{ \sqrt{3}-4+\sqrt{5} } { 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } \cdot \frac{ \sqrt{45} }{ \sqrt{45} } \\\\ &=& \frac{ \sqrt{3}\sqrt{45}-4\sqrt{45}+\sqrt{5}\sqrt{45} } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \quad & | \quad \small{\sqrt{5}\sqrt{45} = \sqrt{225} = 15}\\\\ &=& \frac{ \sqrt{3}\sqrt{45}-4\sqrt{45}+15 } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \quad & | \quad \small{ \sqrt{45} = \sqrt{9\cdot 5} = 3\sqrt{5} }\\\\ &=& \frac{ \sqrt{3}\cdot 3\sqrt{5}-4\cdot3\sqrt{5}+15 } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \\\\ &=& \frac{ 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } { (3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} )\cdot \sqrt{45} } \\\\ &=& \frac{1 } { \sqrt{45} } \\\\ \mathbf{\frac{ \sqrt{3}-4+\sqrt{5} } { 3 \cdot \sqrt{15} +15- 12 \cdot \sqrt{5} } }&\mathbf{=}& \mathbf{\frac{1 } { \sqrt{45} } } \\\\ &\mathbf{=}& \mathbf{0.149071198499986 } \end{array}\)

Unless you are going to eat it.

Then it would be pie.

YES!!!!.

Let the series  \(a_{n+1} =1+\sum{a_n}+a_n\) if \( a_2=7\) what is \(a_5\) ?

\(\begin{array}{rcll} a_{n+1} &=& 1+\sum{a_n}+a_n \\\\ a_2 &=& 1 + a_1 + a_1\\ \mathbf{a_2} &\mathbf{=}& \mathbf{1 + 2a_1} \qquad & | \qquad a_2 = 7\\ 7 &=& 1 + 2a_1 \\ 6 &=& 2a_1 \\ 3 &=& a_1 \\ \mathbf{a_1} &\mathbf{=}& \mathbf{3}\\ \end{array}\)

\(\begin{array}{rcll} a_3 &=& 1 + a_1 + a_2 + a_2\\ a_3 &=& 1 + a_1 + 2a_2 \qquad & | \quad \small{a_2 = 1+2a_1}\\ a_3 &=& 1 + a_1 + 2\cdot(1+2a_1)\\ a_3 &=& 1 + a_1 + 2 + 4a_1\\ \mathbf{a_3} &\mathbf{=}& \mathbf{3 + 5a_1}\qquad & | \qquad a_1 = 3\\ a_3 &=& 3 + 5\cdot 3\\ \mathbf{a_3} &\mathbf{=}& \mathbf{18}\\ \end{array}\)

\(\begin{array}{rcll} a_4 &=& 1 + a_1 + a_2 + a_3 + a_3\\ a_4 &=& 1 + a_1 + a_2 +2a_3 \qquad & | \quad \small{a_2 = 1+2a_1 \quad a_3 = 3 + 5a_1}\\ a_4 &=& 1 + a_1 + 1+2a_1+ 2\cdot( 3 + 5a_1)\\ a_4 &=& 1 + a_1 + 1+2a_1+ 6 + 10a_1\\ \mathbf{a_4} &\mathbf{=}& \mathbf{8 + 13a_1}\qquad & | \qquad a_1 = 3\\ a_4 &=& 8 + 13\cdot 3\\ \mathbf{a_4} &\mathbf{=}& \mathbf{47}\\ \end{array}\)

\(\begin{array}{rcll} a_5 &=& 1 + a_1 + a_2 + a_3 + a_4 + a_4\\ a_5 &=& 1 + a_1 + a_2 + a_3 + 2a_4 \qquad & | \quad \small{a_2 = 1+2a_1 \quad a_3 = 3 + 5a_1 \quad a_4 = 8 + 13a_1}\\ a_5 &=& 1 + a_1 + 1+2a_1+ 3 + 5a_1 +2\cdot (8 + 13a_1)\\ a_5 &=& 1 + a_1 + 1+2a_1+ 3 + 5a_1 + 16 + 26a_1\\ \mathbf{a_5} &\mathbf{=}& \mathbf{21 + 34a_1}\qquad & | \qquad a_1 = 3\\ a_5 &=& 21 + 34\cdot 3\\ \mathbf{a_5} &\mathbf{=}& \mathbf{123}\\ \end{array}\)

Hello!

If I get paid 7.75 pounds an hour, and I work 5.5 hours for 5 days a week and get paid every two weeks how much am I getting paid ?

paid wages = 7.75 pounds * 5.5 h * 5 days * 2 weeks

= 426.25 pounds

Let  sqrt(x) = y, and the first equation becomes

\(\displaystyle y^{3} - 10y - 3 = 0. \)

This clearly has a solution y = -3, so removing the factor (y + 3), the equation can be written

\(\displaystyle (y+3)(y^{2}-3y-1)=0,\)

yielding two further solutions for y,

\(\displaystyle y = (3 \pm\sqrt{13})/2.\)

That suggests three possible values for sqrt(x), but, (and maybe it's implied by the question ?), surely it is reasonable to assume that sqrt(x) is positive ?

If that's the case, then

\(\displaystyle \sqrt{x}=(3+\sqrt{13})/2.\)

Squaring that,

\(\displaystyle x = (11 + 3\sqrt{13})/2\), so that

\(\displaystyle x + \frac{1}{x}=\frac{11+3\sqrt{13}}{2}+\frac{2}{11+3\sqrt{13}}\),

\(\displaystyle =\frac{(11+3\sqrt{13})^{2}+2^{2}}{2(11+3\sqrt{13})}=\frac{121+33\sqrt{13}}{11+3\sqrt{13}} = 11.\)

9(2x-3y)

18x-27y

Just times everything in the brackets by the last thing outside the brackets 

If you expand this you get

9*2x=18x

9*-3y=-27y so the answer is

18x-27y!

\(6(x)\frac{4}{3}\) if x=1

then we substitute x for 1

so we get \(6*\frac{4}{3}\) which is \(8\) so if x=1 the equation equals 8

if x=-1 we just say that instead of plus 8 it's negative 8 so \(-8 \)

As below:

.

Yes!  \(\frac{0.5}{0.5}=1\)

.

is 0.5 over 0.5 1 ?

0.51 is over 0.5

!

f^2+7f+10/f+2

This is certainly a function of  f. 

Looking for the zeros. 

OK?

ƒ(f) = f² + 7f + 10 / f + 2 = 0        [ * f

f³ + 7f² + 10 + 2f = 0

Since the rate of growth is .6% per month, therefore the equivalent annual growth rate is: 1.006^12=

1.0744 -1 X 100=7.44%.

After 20 years, the investment of $10,000 grows to $42,026=~$42,000. So, C.) is your answer.