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1. -4/3, I just graphed it in desmos and fiddled with some stuff.

2. So the distance of the Center, (x,y), to B and A are the same.

We know that y is 5 from tangent rules.

(x,5)

The distance from (x,5) to (9,0) is x, since to B is also x

\(\sqrt{(9-x)^2+25}=x\)

\((9-x)^2+25=x^2\)

\(x^2-18x+106=x^2\)

\(x = \frac{53}{9}\)

\((53/9,5)\)

The points that are 5 units away from the point (2, 7) form a circle, since the definition of a circle is the set of points so many units away from a point.

Remembering our circle formula, the equation of the circle is:

\({(x-2)}^{2}+{(y-7)}^{2}=25\).

Any point that is on the circle and \(y=5x-28\), must satisfy both equations, meaning it is the solution of the system:

\(\begin{cases} {(x-2)}^{2}+{(y-7)}^{2}=25 \\ y=5x-28 \end{cases}\).

Substitute:

\({(x-2)}^{2}+{(5x-35)}^{2}=25\)

\({(x-2)}^{2}+25{(x-7)}^{2}=25\)

\(26{x}^{2}-354x+1204=0\)

\((13x-86)(x-7)=0\)

\(x_{1, 2}= 7, \frac{86}{13}\)

Plugging back in,

\(\begin{cases} x=7, y=7\\x=\frac{86}{13},y=\frac{66}{13}\end{cases}\)

Our two points are \(\bf{(7, 7), (\frac{86}{13}, \frac{66}{13})}\).

If we let 

\(5x+5=3\), which solves to \(x=-\frac{2}{5}\), then if we plug in -2/5, we get

\(\frac{1}{5}*f(3)-5=\frac{4}{5}-5=-\frac{21}{5}\).

So our point is \((-\frac{2}{5}, -\frac{21}{5})\)

By pythagorean theorem looking at a right triangle with the hypotenuse of the side length of the square,

\({2}^{2}+{4}^{2}={c}^{2}\). c² is also the area of the square, so our area is 2²+4²=20.

\(\begin{cases} a + b = 4\\ {a}^{2} b + {b}^{2} a = 6 \end{cases}\)

\(\begin{cases} a + b = 4\\ ab(a+b) = 6 \end{cases}\)

\(\begin{cases} a+b=4 \\ ab=\frac{3}{2}\end{cases}\)

\({a}^{3}+{b}^{3}=(a+b)({(a+b)}^{2}-3ab)\)

\({a}^{3}+{b}^{3}=4(16-3*\frac{3}{2})=46\)

@hairyberry  You need to check \(f(4\sqrt5)\).  It isn't 70.

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.

To simplify the expression (11 - 7i)/(2 + i), we can use the method of multiplying both numerator and denominator by the conjugate of the denominator.

The conjugate of the denominator, 2 + i, is 2 - i. Therefore, we have:

(11 - 7i)/(2 + i) = (11 - 7i)/(2 + i) * (2 - i) / (2 - i)

Expanding the numerator and denominator:

= [(11 - 7i) * (2 - i)] / [(2 + i) * (2 - i)]

= [(11 * 2) + (-7 * 2i) + (11 * -1i) + (-7 * -1 * i^2)] / [(2 * 2) + (2 * -1i) + (1 * -1i) + (1 * -1i^2)]

= [(22 - 14i - 11i + 7 i^2)] / [(4 - 2i - 1i - 1 * i^2)]

= [(22 - 14i - 11i - 7)] / [(4 - 2i - 1i + 1)]

= [15 - 25i] / [5 - 3i]

= (15 - 25i) / 5(1 - 0.6i)

Dividing both numerator and denominator by 5:

= (3 - 5i) / (1 - 0.6i)

Now, multiplying both numerator and denominator by the conjugate of the new denominator (1 + 0.6i):

= (3 - 5i) * (1 + 0.6i) / (1 - 0.6i) * (1 + 0.6i)

Expanding again:

= [(3 * 1) + (-5 * 0.6i) + (3 * 0.6i) + (-5 * 0.6 * i^2)] / [(1 * 1) + (1 * 0.6i) + (-0.6 * 1i) + (-0.6 * 0.6 * i^2)]

= [3 - 3i + 1.8i + 3] / [1 + 0.6i - 0.6i - 0.36]

= [6 - 1.2i] / [0.64]

= (6 - 1.2i) / (0.8 * 0.8)

= (6 - 1.2i) / 0.64

= (15 - 3i) / 4

= 3 - \frac{3}{4}i

Therefore, the simplified form of the complex number is 3 - 3/4*i, which you can write as

3 + (-3/4) * i

 Yay for complex numbers!

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{1 + \cfrac{1}{4}}}}.\\ =\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{4}{5}}}\\ =\cfrac{1}{1 + \cfrac{5}{24 }}\\ \color{blue}=\cfrac{24}{29}\)

 !

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.

Here's how to convert 2231_4 to base 2 without first converting to decimal:

Start with the largest power of 4 that fits into 2231:

4^3 = 64 is the largest power of 4 less than 2231.

Divide 2231 by 64 and get the quotient 34 and the remainder 3.

Write down the quotient 34 as the first digit in the base-2 representation.

Repeat for the remaining digits:

Repeat the process with the remainder 3.

The largest power of 4 that fits into 3 is 4^1 = 4.

Divide 3 by 4 and get the quotient 0 and the remainder 3.

Write down the quotient 0 as the second digit.

Repeat with the remaining 3.

The largest power of 4 that fits into 3 is 4^0 = 1.

Divide 3 by 1 and get the quotient 3 and the remainder 0.

Write down the quotient 3 as the third digit.

Combine the digits:

The digits obtained in the previous steps, in reverse order, give the base-2 representation.

Therefore, 2231_4 in base 2 is 100111_2.

This method directly uses the powers of 4 in the base-4 representation to express the number in the base-2 system, avoiding the need for intermediate decimal conversion.