Quadrant IV consists of the points with positive \(\displaystyle x\)-coordinates and negative \(\displaystyle y\)-coordinates. Therefore \(\displaystyle (4,-7)\) is the correct choice.

The line of symmetry of the parabola of the equation

\(\displaystyle f(x) = ax^{2}+ bx + c\)

is the vertical line

\(\displaystyle x = -\frac{b}{2a}\)

Substitute \(\displaystyle a = 3, b = -9\):

The line of symmetry is

\(\displaystyle x = -\frac{-9}{2 \cdot 3} = \frac{9}{6} = \frac{3}{2}\)

That is, the line of the equation \(\displaystyle x = \frac{3}{2}\).

Remember that the basic form of the equation of a circle is:

\(\displaystyle (x-c_x)^2+(y-c_y)^2=r^2\)

This means that the center point \(\displaystyle (c_x,c_y)\) is defined by the two values subtracted in the squared terms.  We could rewrite our equation as:

\(\displaystyle (x-(7))^2+(y-(-12))^2=14\)

Therefore, the center is \(\displaystyle (7,-12)\)

For this question, we will need to do three things:

1. Determine the point in question.
2. Use trigonometry to find the area of the angle in question.
3. Use the equation for finding a sector area to finalize our answer.

Let us first solve for the coordinate by substituting into our equation:

\(\displaystyle (6)^2+y^2=100\)

\(\displaystyle 36+y^2=100\)

\(\displaystyle y^2=64\)

\(\displaystyle y=8\)

Our point is, therefore: \(\displaystyle (6,8)\)

Now, we need to calculate the angle formed between the origin and the point that we were given. We can do this using the inverse tangent function. The ratio of \(\displaystyle y\) to \(\displaystyle x\) is here: \(\displaystyle \frac{8}{6}\)

Therefore, the angle is:

\(\displaystyle \tan^{-1}(\frac{8}{6})=53.13010235415592\)

To solve for the sector area, we merely need to use our standard geometry equation. Note that the radius of the circle, based on the equation, is \(\displaystyle 10\).

\(\displaystyle \frac{53.13010235415592}{360}*\pi*10^2 = 46.36476090008\)

This rounds to \(\displaystyle 46.37\).

For this question, we will need to do three things:

1. Determine the point in question.
2. Use trigonometry to find the area of the angle in question.
3. Use the equation for finding a sector area to finalize our answer.

Let us first solve for the coordinate by substituting into our equation:

\(\displaystyle 12.4^2+y^2=271\)

\(\displaystyle 153.76+y^2=271\)

\(\displaystyle y^2=117.24\)

\(\displaystyle y=10.82774214691133\)

Our point is, therefore: \(\displaystyle (12.4,10.82774214691133)\)

Now, we need to calculate the angle formed between the origin and the point that we were given. We can do this using the inverse tangent function. The ratio of \(\displaystyle y\) to \(\displaystyle x\) is here: \(\displaystyle \frac{10.82774214691133}{12.4}\)

Therefore, the angle is:

\(\displaystyle \tan^{-1}(\frac{10.82774214691133}{12.4})=41.1276246899636\)

To solve for the sector area, we merely need to use our standard geometry equation. Note that \(\displaystyle r^2\), based on the equation, is \(\displaystyle 271\).

\(\displaystyle \frac{41.1276246899636}{360}*\pi*271 = 97.2635889213762\)

This rounds to \(\displaystyle 97.26\).

Since you are looking for a point on the \(\displaystyle y-axis\), your \(\displaystyle x\) value will be zero.  

The center of the circle is at the origin and radius is the distance from the center, so that means the point you are looking for must be \(\displaystyle 4\) points away from \(\displaystyle 0\).  

This can be two points on the \(\displaystyle y-axis\) but since you are looking for a positive one, your answer must be \(\displaystyle \left ( 0,4\right )\).

The displacement between negative three and positive one is four.

\(\displaystyle 1-(-3) =4\)

This mean that after flipping the point, it must be symmetrical to its original location.   The new point must also be 4 units to the right of the line.

The new point would be located at:  \(\displaystyle (5,2)\)

The answer is:  \(\displaystyle 5\)

The relation 

\(\displaystyle (x-h)^{2} + (y - k)^{2} = r^{2}\)

is a circle with center \(\displaystyle (h,k)\) and radius \(\displaystyle r\) .

In other words, it is a circle with center at the origin, translated right \(\displaystyle h\) units and up \(\displaystyle k\) units (the radius is irrelevant to the question).

\(\displaystyle x^{2} +\left ( y-2 \right )^{2}= 10\)

or

\(\displaystyle \left (x-0 \right )^{2} +\left ( y-2 \right )^{2}= 10\)

is this circle translated right zero units and up 2 units. The upshot is that the circle moves along the \(\displaystyle y\)-axis only, and therefore is symmetric with respect to the \(\displaystyle y\)-axis, but not the \(\displaystyle x\)-axis. Also, as a consequence, it is not symmetric with respect to the origin.

Reflecting the point across the vertical line \(\displaystyle x=1\) will only change the x-value, but not the y-value.

The point after this reflection is: \(\displaystyle (-1,5)\)

Rotating this point across the origin will swap the x and y-values.

The new point is: \(\displaystyle (5,-1)\)

When a function \(\displaystyle \small f(x)\) is flipped across the x-axis, the new function \(\displaystyle \small g(x)\) is equal to \(\displaystyle \small -f(x)\). Therefore, our function \(\displaystyle \small g(x)\) is equal to:

\(\displaystyle \small g(x)=-f(x)=-(x^4-5x^3+2x^2+9)=-x^4+5x^3-2x^2-9\)

Our final answer is therefore \(\displaystyle \small g(x)=-x^4+5x^3-2x^2-9\)