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The successful ordered outcomes are $TTH$, $THT$ and $HTT$. Each has probability $1/8$ by independence, and these events are mutually exclusive. Adding their probabilities givesEquivalently, the binomial distribution gives $\left(\genfrac{}{}{0ex}{}{3}{2}\right)(1/2{)}^2(1/2)=3/8$.

The three tosses are independent events, each with head probability $1/2$. Consequently

For a constant vector field, the flux integral is its dot product with the vector area. Thus

The prescribed orientation orders the two edges from $A$ as $\mathbf{c}-\mathbf{a}$ followed by $\mathbf{b}-\mathbf{a}$. The vector area is half their cross product:The sign agrees with the right-hand rule for the traversal $A\to C\to B\to A$.

The circles have centres $(1,1)$ and $(-1,1)$. Their Euclidean distance is

Subtracting the two circle equations cancels the common terms and gives $-4x=0$. With $x=0$, either circle equation gives $(y-1{)}^2=0$. Thus the unique common point isThe two circles touch externally there.

The general solution of the first-order linear differential equation is $y=1+C{e}^x$. The initial condition $y(0)=1$ gives $C=0$, so the solution of this initial value problem is the constant function $y(x)=1$. Thus $\boxed{y(1)=1}$.

Writing $v=y-1$ turns the first-order linear differential equation into $v^{\prime }=v$. Its general solution is $v=C{e}^x$, so $y=1+C{e}^x$. The initial condition gives $C=-1$, hence

This geometric series has first term $1/27$ and common ratio $1/3$, whose absolute value is below one. Consequently its sum is

Group consecutive terms in pairs. The $j$th pair, for $j=1,\dots ,25$, isThere are $25$ such pairs in this finite sum, so the answer is $\boxed{25}$.

The derivative of the tangent is ${\sec }^{2}x$, and the order of the endpoints must be retained. Thus the definite integral is

The double-angle identity for the cosine gives ${\cos }^{2}x-1=(\cos 2x-1)/2$. Termwise integration therefore givesHere $C$ is an arbitrary constant in the antiderivative.

For the requested parameter value $a=2$, the function graph has two branches, symmetric about the vertical axis because $f$ is an even function. Its derivatives areThe stationary points are $\boxed{(-2,2)\text{ and }(2,2)}$, both local minima and also the lowest points of their respective branches. The function tends to $+\infty$ as $x\to 0^{\pm }$, so $x=0$ is a vertical asymptote; it also tends to $+\infty$ as $|x|\to \infty$. The negative branch decreases to $(-2,2)$ and then rises towards the vertical asymptote; the positive branch falls from the vertical asymptote to $(2,2)$ and then rises.

The derivative isA stationary point at $x=1$ requires $2/a^2-2a^2=0$, or $a^4=1$. Since $a>0$, $\boxed{a=1}$.

Since $\mathbf{v}=-\mathbf{u}$ and $\mathbf{u}\ne 0$, the vectors point in opposite directions. More explicitly, their dot product givesHence their angle is $\boxed{\theta =\pi \text{ radians}=180^{\circ }}$.

Adding the vectors component by component givesConsequently $\boxed{a=0}$. The requirement that the vectors be nonzero then implies $b\ne 0$.

For $a=b=1$, the stationary points found in parts (a) and (b) have temperatures $0$, $1$, $1/4$ and $1/5$. The unit sphere is compact and $T$ is a continuous function, so the extreme value theorem ensures that a hottest point exists; it must be one of those stationary points. ThusFor a direct bound, put $s=x^2\in [0,1]$. The identity $(y^2-z^2{)}^2\ge 0$ gives $y^2{z}^2\le (y^2+z^2{)}^2/4=(1-s{)}^2/4$, andsince $1-s^2-(1-s{)}^2/4=(1-s)(5s+3)/4\ge 0$. Equality forces $s=1$, giving exactly the two points above.