hairyberry

Since 11 is a factor of 99, we suspect this is a repeating decimal.

We multiply top and bottom by 9,  getting \(\frac{63}{99}\).

Our suspicions are confirmed. In general, \(0.\overline{abc\dots n}\) is represented by \(\frac{\overline{abc \dots n}}{{10}^{n}-1}\), where n is the number of terms.

So, \(\frac{63}{99}=0.\overline{63}\), so our answer is 6363636363.

\({x}^{3}+{y}^{3}=(x+y)({x}^{2}-xy+{y}^{2})\). 

\(5=2({x}^{2}-xy+{y}^{2})\)

\({x}^{2}-xy+{y}^{2}=\frac{5}{2}\)

Because \({x}^{2}-xy+{y}^{2}={(x+y)}^{2}-3xy\)

\({(x+y)}^{2}-3xy=\frac{5}{2}\).

\({2}^{2}-3xy=\frac{5}{2}\)

\(-3xy=-\frac{3}{2}\)

\(xy=\frac{1}{2}\)

\({x}^{2}+{y}^{2}={(x+y)}^{2}-2xy=4-1=3\).

So x²+y² is 3.

Quad 1 is 0 - 90 degrees

Quad 2 is 90 to 180 degrees

Quad 3 is 180-270 degrees

Quad 4 is 270-360 degrees

1) 160 degrees is Q2

2) 104 degrees is Q2

3) 187 degreees is Q3

4) Since 378 degrees, is the same as 18 degrees (this is called coterminal), then it is in Q1.

In general, in a geometric series, three terms, say \(a{r}^{k}, a{r}^{k+1}, a{r}^{k+2}\), then \(a{r}^{k}*a{r}^{k+2}=({a{r}^{k+1}})^{2}\), or, otherwise, in three terms of a geometric series, the middle term squared, equals the product of the two next to it.

Using this, we get the equation

\((60+x)(140+x)={(120+x)}^{2}\), where x is the added number.

\(8400+200x+{x}^{2}=14400+240x+{x}^{2}\). Cancelling out the x²:

\(8400+200x=14400+240x\)

\(40x=-6000\)

\(x=-150\).

The resulting series will be:

\(-90, -30, -10\).

The common ratio is \(\frac{-30}{-90}=\frac{1}{3}\). So, the resulting sequence has a common ratio of 1/3.

\(\begin{cases} a+2b=-3-7a+b\\4a+2b=k-5a-mb \end{cases}\)

\(\begin{cases} 8a+b=-3 \\ 9a+(2+m)b=k \end{cases}\)

To have infinitely, these two equations must be exactly equal, or in other words, one equation is simplifiable to the other. (Another way to visualize it, is that these two lines, when plotted are the same graph).

So, by \(8a*\frac{9}{8}=9a\), the second equation is the first equation, both sides multiplied by \(\frac{9}{8}\).

For the b terms,\(\frac{9}{8}*1=2+m\)

For the constant term, \(-3*\frac{9}{8}=k\)

Solving, we get \(m=\frac{-7}{8}, k=\frac{-27}{8}\). 

\(\sqrt{5}<\pi\), plug into the first case.

\(f(f(f(\sqrt{5})))=f(f(4\sqrt{5}))\)

\(4\sqrt{5}>5.2\), plug into the third case.

\(f(f(4\sqrt{5}))=f(70)\).

\(70>5.2\), plug into third case.

\(f(70)=4893\).

So, \(\underline{\bf{f(f(f(\sqrt{5})))=4893}}\).

The common difference between terms is \(\frac{\frac{1}{4}}{16}=\frac{1}{64}\).

The 30th term is 30-24=6 terms away, meaning 6 common differences from 24.

So, the 30th term is \(\frac{1}{4}*{(\frac{1}{64})}^6\), \(\frac{1}{{2}^{2}}*\frac{1}{{2}^{36}}=\frac{1}{{2}^{38}}\), is our final answer.

\({a}^{3}-\frac{1}{{a}^{3}}=(a-\frac{1}{a})({a}^{2}+1+\frac{1}{{a}^{2}})=(a-\frac{1}{a})({(a-\frac{1}{a})}^{2}+3)\), which is what we want to find.

We see \(a-\frac{1}{a}\) will me good to know.

We have:

\({a}^{2}-a-1=0\), divide both sides by a.

\(a-1-\frac{1}{a}=0\), \(a-\frac{1}{a}=1\).

Plugging in, \({a}^{3}-\frac{1}{{a}^{3}}=(1)({1}^{2}+3)=4\).

So, \(\underline{{a}^{3}-\frac{1}{{a}^{3}}}\) is 4.

\(a_{11}=a_{10}+a_{9}\), so following this logic.

\(4=1+a_{9}\)

\(a_{9}=3\)

Similarly,

\(1=3+a_{8}\)

\(a_{8}=-2\)

\(3=-2+a_{7}\)

\(a_{7}=5\)

\(-2=5+a_{6}\)

\(a_{6}=-7\).

a₆ is -7.