CubeyThePenguin

First, we need to find the slope of the equation, which is \(\frac{6-4}{0+7} =\frac{2}{7}\).

We know the coordinates of one point and the slope, so we can write the equation in point-slope form:

\(y-y_0 = m(x-x_0)\)

\(y - 6 = \frac{2}{7}(x)\)

\(5h>3(3h+4)\)

\(5h > 9h+12\)

\(-4h > 12\)

\(\boxed{h < -3}\)

Simply plug in the coordinates of the point into both equations to see if it works.

6 = (-5) = -11

6 - (-5) = 11

As you can see, the ordered pair satisfies neither of the equations. So, it is not a solution to the system.

We are essentially finding the solutions to \(\frac{x-2}{3x+4} = \frac{x+5}{x}\).

Cross multiply: \((x-2)(x) =(3x+4)(x+5)\)

\(x^2-2x =3x^2 + 19x + 20\)

\(0 = 2x^2 + 21x + 20\)

Using the quadratic formula, we see that the solutions are at \(-\frac{17}{4} \pm \frac{\sqrt{129}}{4}.\)

It is only raining in Seattle if all three friends are truthful. This happens with probability \((\frac{2}{3})^3 =\boxed{\frac{8}{27}}\)

Ducks have two legs and cows have four legs. Let d be the number of ducks and c be the number of cows. We can write a system of equations:

24 in total ==> c + d = 24         . (1)

66 legs ==> 4c + 2d = 66         . (2)

Multiply equation (1) by 2:

2c + 2d = 48

4c + 2d = 66

Subtract to get -2c = -18. So, there are 9 cows and 15 ducks.

\(\log y = \log 16 + \log 49\)

\(\log y = \log (16 \cdot49)\)

\(y = 16 \cdot 49 = \boxed{784}\)

The sides of a 30-60-90 triangle is in the ratio \(1: \sqrt{3} : 2\).

The hypotenuse is 11, so the sides are of lengths \(\frac{11}{2} \) and \(\frac{11\sqrt{3}}{2}\).

The perimeter is \(\frac{11}{2} + \frac{11\sqrt{3}}{2} + 11 \approx \boxed{26.02}\)

\(4 \ln 2 - \ln 3 - 1 \approx \boxed{0.673}\)

\(-\frac{2}{49} + \frac{9}{4} = \frac{2(-4) + 9(49)}{196} = \boxed{\frac{443}{196}}\)