Step 1: Expand $(2-x)^7$ using the binomial theorem $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. Here, $a=2$, $b=-x$, and $n=7$. We need terms up to $x^3$ (i.e., for $k=0, 1, 2, 3$). For $k=0$: $$ \binom{7}{0} (2)^{7-0} (-x)^0 = 1 \cdot 2^7 \cdot 1 = 128 $$ For $k=1$: $$ \binom{7}{1} (2)^{7-1} (-x)^1 = 7 \cdot 2^6 \cdot (-x) = 7 \cdot 64 \cdot (-x) = -448x $$ For $k=2$: $$ \binom{7}{2} (2)^{7-2} (-x)^2 = \frac{7 \cdot 6}{2 \cdot 1} \cdot 2^5 \cdot x^2 = 21 \cdot 32 \cdot x^2 = 672x^2 $$ For $k=3$: $$ \binom{7}{3} (2)^{7-3} (-x)^3 = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} \cdot 2^4 \cdot (-x^3) = 35 \cdot 16 \cdot (-x^3) = -560x^3 $$ Step 2: Combine the terms. $$ (2-x)^7 = 128 - 448x + 672x^2 - 560x^3 + \dots $$ The expansion up to and including the term in $x^3$ is: $$ \boxed{128 - 448x + 672x^2 - 560x^3} $$ b) Step 1: Relate $(1.98)^7$ to the expansion of $(2-x)^7$. We need $2-x = 1.98$. Solving for $x$: $$ x = 2 - 1.98 = 0.02 $$ Step 2: Substitute $x=0.02$ into the expansion obtained in part (a). $$ (1.98)^7 \approx 128 - 448(0.02) + 672(0.02)^2 - 560(0.02)^3 $$ Step 3: Calculate each term. $$ 128 $$ $$ -448(0.02) = -8.96 $$ $$ 672(0.02)^2 = 672(0.0004) = 0.2688 $$ $$ -560(0.02)^3 = -560(0.000008) = -0.00448 $$ Step 4: Sum the terms. $$ (1.98)^7 \approx 128 - 8.96 + 0.2688 - 0.00448 = 119.30432 $$ Step 5: Round the result to four significant figures. The first four significant figures are 1, 1, 9, 3. The next digit is 0, so we round down. $$ (1.98)^7 \approx 119.3 $$ The approximate value of $(1.98)^7$ to four significant figures is: $$ \boxed{119.3} $$ a) Step 1: Identify the given lengths and apply the Power of a Point Theorem for intersecting secants. 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 $QY = QZ + YZ = 4 + y$. Step 2: Substitute the values into the theorem equation. $$ QX \cdot QW = QZ \cdot QY $$ $$ 6 \cdot 11 = 4 \cdot (4 + y) $$ $$ 66 = 16 + 4y $$ Step 3: Solve for $y$ (the length of chord $YZ$). $$ 66 - 16 = 4y $$ $$ 50 = 4y $$ $$ y = \frac{50}{4} = \frac{25}{2} = 12.5 $$ The length of chord $YZ$ is: $$ \boxed{12.5 \text{ cm}} $$ b) Step 1: Apply the Power of a Point Theorem for a tangent and a secant from an external point $Q$. 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. Step 2: Substitute the values into the theorem equation. $$ SQ^2 = 4 \cdot 16.5 $$ $$ SQ^2 = 66 $$ Step 3: Solve for $SQ$. $$ SQ = \sqrt{66} $$ The length of the tangent $SQ$ is: $$ \boxed{\sqrt{66} \text{ cm}} $$ Step 1: The given formula is $S = \frac{a(1-r^n)}{1-r}$. We need to make $a$ the subject of the formula. Step 2: Multiply both sides of the equation by $(1-r)$ to clear the denominator. $$ S(1-r) = a(1-r^n) $$ Step 3: Divide both sides by $(1-r^n)$ to isolate $a$. $$ a = \frac{S(1-r)}{1-r^n} $$ The formula with $a$ as the subject is: $$ \boxed{a = \frac{S(1-r)}{1-r^n}} $$ Send me the next one 📸