Simplify 8x2(2x2+3y), find the value of x if 10 x = 4, find x and y if x+10y=6 and x−10y=4, and make g the subject of T = 2((h2+R2)/(g))

$$= 16x^4 + 24x^2y$$

The simplified expression is $\boxed{\mathbf{16x^4 + 24x^2y}}$.

2. Find the value of $x$ if $\log_{10} x = 4$

$$x = 10 \times 10 \times 10 \times 10$$ $$x = 10000$$

The value of $x$ is $\boxed{\mathbf{10000}}$.

3. Find the value of $x$ and $y$ if $x+10y=6$ and $x-10y=4$
Let the given equations be:
(1) $x+10y=6$
(2) $x-10y=4$

$$(x+10y) + (x-10y) = 6+4$$ $$2x = 10$$

$$x = \frac{10}{2}$$ $$x = 5$$

$$5+10y=6$$ $$10y = 6-5$$ $$10y = 1$$

$$y = \frac{1}{10}$$

The values are $\boxed{\mathbf{x=5, y=\frac{1}{10}}}$.

4. Make $g$ the subject of the formula $T = 2\pi\sqrt{\frac{h^2+R^2}{g}}$

$$\left(\frac{T}{2\pi}\right)^2 = \frac{h^2+R^2}{g}$$ $$\frac{T^2}{4\pi^2} = \frac{h^2+R^2}{g}$$

$$g = (h^2+R^2) \frac{4\pi^2}{T^2}$$ $$g = \frac{4\pi^2(h^2+R^2)}{T^2}$$

The formula with $g$ as the subject is $\boxed{\mathbf{g = \frac{4\pi^2(h^2+R^2)}{T^2}}}$.

5. Factorize $4y^2+15y=3y$

$$4y^2+15y-3y=0$$ $$4y^2+12y=0$$

$$4y(y+3)=0$$

The factored expression is $\boxed{\mathbf{4y(y+3)}}$.

6. Construct angle $75^\circ$
To construct a $75^\circ$ angle:

• Draw a line segment AB.
• At point A, construct a $90^\circ$ angle (e.g., using a compass to draw perpendicular bisector of an arc). Let the perpendicular line be AC.
• At point A, construct a $60^\circ$ angle (e.g., using a compass to mark an arc from A, then another arc from the intersection point on AB with the same radius). Let this line be AD.
• Bisect the angle between the $60^\circ$ line (AD) and the $90^\circ$ line (AC). The angle between them is $30^\circ$. Bisecting it gives $15^\circ$.
• The angle formed by the $60^\circ$ line and the bisector will be $60^\circ + 15^\circ = 75^\circ$.

**7. Construct angle $15