Expand (2-x)⁷ using the binomial theorem (a+b)ⁿ = sum_k=0ⁿ nk aⁿ-k b^k.

Here, $a=2$, $b=-x$, and $n=7$. We need terms up to $x^3$ (i.e., for $k=0, 1, 2, 3$).

For $k=0$:

For $k=1$:

For $k=2$:

For $k=3$:

The expansion up to and including the term in $x^3$ is:

b)

We need $2-x = 1.98$.
Solving for $x$:

The first four significant figures are 1, 1, 9, 3. The next digit is 0, so we round down.

The approximate value of $(1.98)^7$ to four significant figures is:

a)

The secants $WQ$ and $YQ$ intersect externally at $Q$.
The theorem states that $QX \cdot QW = QZ \cdot QY$.

From the problem statement and diagram:
$WX = 5$ cm
$XQ = 6$ cm
$WQ = WX + XQ = 5 + 6 = 11$ cm
$ZQ = 4$ cm
Let $YZ = y$. Then $QY = QZ + YZ = 4 + y$.

The length of chord $YZ$ is:

b)

The theorem states that $SQ^2 = QZ \cdot QY$.

From part (a), we found $YZ = 12.5$ cm.
We are given $ZQ = 4$ cm.
So, $QY = QZ + YZ = 4 + 12.5 = 16.5$ cm.

The length of the tangent $SQ$ is:

We need to make $a$ the subject of the formula.

The formula with $a$ as the subject is: