1. Problem: Solve $\frac{m}{2}+\frac{m}{3}+3=2+\frac{m}{6}$. Combine the fractional $m$ terms on the left to get $\frac{m}{2}+\frac{m}{3}=\frac{3m+2m}{6}=\frac{5m}{6}$. Substitute to obtain $\frac{5m}{6}+3=2+\frac{m}{6}$. Move the $m$ terms: $\frac{5m}{6}-\frac{m}{6}=2-3$. Simplify to $\frac{4m}{6}=-1$ and reduce to $\frac{2m}{3}=-1$. Solve for $m$: $m=-\frac{3}{2}=-1.5$. Final answer: $m=-\tfrac{3}{2}=-1.5$. 2. Problem: Evaluate $\sqrt{\dfrac{(15.62)^2}{29.21\times\sqrt{10.52}}}$. Compute the numerator: $(15.62)^2=243.9844$. Compute the inner square root: $\sqrt{10.52}\approx3.2433009$. Compute the denominator: $29.21\times3.2433009\approx94.736819289$. Form the fraction: $\dfrac{243.9844}{94.736819289}\approx2.57535$. Take the square root: $\sqrt{2.57535}\approx1.6048$. Final answer (approximate): $1.6048$. 3. Problem: Find the intersection points of the line $y=x+2$ and the circle $x^2+y^2=28$, and describe the shapes. Substitute $y=x+2$ into the circle to get $x^2+(x+2)^2=28$. Expand and simplify: $x^2+x^2+4x+4=28$ which gives $2x^2+4x-24=0$. Divide by 2: $x^2+2x-12=0$. Solve the quadratic: $x=\dfrac{-2\pm\sqrt{4+48}}{2}=\dfrac{-2\pm\sqrt{52}}{2}=-1\pm\sqrt{13}$. Compute $y$ from $y=x+2$ to get $y=1\pm\sqrt{13}$. Thus the intersection points are $( -1+\sqrt{13},\;1+\sqrt{13})$ and $( -1-\sqrt{13},\;1-\sqrt{13})$. Decimal approximations are $(2.605551275,\;4.605551275)$ and $(-4.605551275,\;-2.605551275)$. The line $y=x+2$ is a straight line with slope $1$ and $y$-intercept $2$. The circle $x^2+y^2=28$ is centered at $(0,0)$ with radius $\sqrt{28}=2\sqrt{7}\approx5.2915$. 4. Problem: Solve $\sqrt{\dfrac{y+2}{3-y}}=-15+10$. Compute the right-hand side: $-15+10=-5$. The left-hand side is a principal square root and is therefore $\ge 0$, while the right-hand side is $-5<0$. Because a nonnegative quantity cannot equal a negative number, there is no real solution. Final answer: no real solution. 5. Problem: Compute the gravitational force $F=G\dfrac{m_1m_2}{d^2}$ with $G=6.67\times10^{-11}$, $m_1=7.36$, $m_2=15.5$, $d=22.6$. First compute the product $m_1m_2=7.36\times15.5=114.08$. Compute the numerator $Gm_1m_2=6.67\times10^{-11}\times114.08=7.609136\times10^{-9}$. Compute $d^2=22.6^2=510.76$. Compute $F=\dfrac{7.609136\times10^{-9}}{510.76}\approx1.4895\times10^{-11}$. Final answer (approximate): $F\approx1.4895\times10^{-11}$. 6. Problem: Compute the area of a triangle using Heron's formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ with $a=3.60\text{ cm}$, $b=4.00\text{ cm}$, $c=5.20\text{ cm}$. Compute the semiperimeter $s=\dfrac{a+b+c}{2}=\dfrac{3.60+4.00+5.20}{2}=\dfrac{12.80}{2}=6.40$. Compute $s-a=6.40-3.60=2.80$, $s-b=6.40-4.00=2.40$, $s-c=6.40-5.20=1.20$. Compute the product $s(s-a)(s-b)(s-c)=6.40\times2.80\times2.40\times1.20=51.6096$. Take the square root: $A=\sqrt{51.6096}\approx7.1860$. Final answer: $A\approx7.1860\text{ cm}^2$. 7. Problem: Find constants $a,b$ in $F=aL+b$ given $F(8)=5.6$ and $F(2)=4.4$, then compute $F$ when $L=6.5$. Set up the system: $8a+b=5.6$ and $2a+b=4.4$. Subtract the second from the first to get $6a=1.2$, so $a=0.2$. Substitute $a$ into $2a+b=4.4$ to get $0.4+b=4.4$ and hence $b=4.0$. Compute $F$ at $L=6.5$: $F=0.2\times6.5+4.0=1.3+4.0=5.3$. Final answers: $a=0.2$, $b=4.0$, and $F(6.5)=5.3$.