Exponential Functions 10Ce5F

1. **Problem 11:** Find the function $y=7^x$. This is an exponential function with base 7. 2. **Problem 12:** Find the function $y=\left(\frac{1}{9}\right)^x$. This is an exponential decay function with base $\frac{1}{9}$. 3. **Problem 13:** Determine if each graph is a function and if it is one-to-one. - (a) A straight increasing line through the origin is a function and one-to-one because it passes the vertical and horizontal line tests. - (b) Two sideways curves symmetric about the y-axis fail the vertical line test, so not a function. - (c) Two separate branches in quadrants II and IV approaching axes fail vertical line test, so not a function. - (d) An increasing curve rising to the right and leveling leftward is a function and one-to-one. 4. **Problem 14:** Erin has 30 coins worth 4.25, all nickels (5 cents), dimes (10 cents), or quarters (25 cents). Number of quarters equals number of dimes. Let $d$ = number of dimes, $q$ = number of quarters, $n$ = number of nickels. Given: $$q = d$$ $$n + d + q = 30$$ $$5n + 10d + 25q = 425$$ (cents) Substitute $q=d$: $$n + d + d = 30 \Rightarrow n + 2d = 30$$ $$5n + 10d + 25d = 425 \Rightarrow 5n + 35d = 425$$ From first equation: $$n = 30 - 2d$$ Substitute into second: $$5(30 - 2d) + 35d = 425$$ $$150 - 10d + 35d = 425$$ $$150 + 25d = 425$$ $$25d = 275$$ $$d = 11$$ Then: $$n = 30 - 2(11) = 30 - 22 = 8$$ $$q = d = 11$$ **Answer:** 8 nickels, 11 dimes, 11 quarters. 5. **Problem 15:** Given $HP \cong KP$ and $IP \cong JP$, prove $\triangle HPI \cong \triangle KPJ$. **Two-column proof:** | Statements | Reasons | |---|---| | 1. $HP \cong KP$ | Given | | 2. $IP \cong JP$ | Given | | 3. $\angle HPI \cong \angle KPJ$ | Vertical angles are congruent | | 4. $\triangle HPI \cong \triangle KPJ$ | SAS congruence (two sides and included angle) | 6. **Problem 16:** Solve system: $$\begin{cases} x^2 + y^2 = 8 \\ x^2 - y^2 = 4 \end{cases}$$ Add equations: $$ (x^2 + y^2) + (x^2 - y^2) = 8 + 4 $$ $$ 2x^2 = 12 $$ $$ x^2 = 6 $$ $$ x = \pm \sqrt{6} $$ Subtract second from first: $$ (x^2 + y^2) - (x^2 - y^2) = 8 - 4 $$ $$ 2y^2 = 4 $$ $$ y^2 = 2 $$ $$ y = \pm \sqrt{2} $$ Solutions: $$ (\sqrt{6}, \sqrt{2}), (\sqrt{6}, -\sqrt{2}), (-\sqrt{6}, \sqrt{2}), (-\sqrt{6}, -\sqrt{2}) $$ 7. **Problem 17:** Simplify: $$(a - b)(a^{-1}b + ab^{-1})^{-1}$$ Rewrite inside parentheses: $$a^{-1}b + ab^{-1} = \frac{b}{a} + \frac{a}{b} = \frac{b^2 + a^2}{ab}$$ So: $$(a - b) \left( \frac{b^2 + a^2}{ab} \right)^{-1} = (a - b) \frac{ab}{b^2 + a^2}$$ Final simplified form: $$\frac{ab(a - b)}{a^2 + b^2}$$ 8. **Problem 18:** Evaluate $$\frac{\sqrt{3}}{2} \tan 30^\circ + \sqrt{3} \sin 60^\circ - \sqrt{2} \cos 45^\circ$$ Recall: $$\tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}$$ Calculate each term: $$\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} = \frac{1}{2}$$ $$\sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2}$$ $$\sqrt{2} \times \frac{\sqrt{2}}{2} = 1$$ Sum: $$\frac{1}{2} + \frac{3}{2} - 1 = 2 - 1 = 1$$ 9. **Problem 19:** Sphere volume $V = 288\pi$ cubic feet. Volume formula: $$V = \frac{4}{3} \pi r^3$$ Solve for $r$: $$288\pi = \frac{4}{3} \pi r^3$$ $$288 = \frac{4}{3} r^3$$ $$r^3 = 288 \times \frac{3}{4} = 216$$ $$r = \sqrt[3]{216} = 6$$ Surface area formula: $$A = 4 \pi r^2 = 4 \pi (6)^2 = 144 \pi$$ 10. **Problem 20:** Add water to 89 liters of 45% saccharine solution to get 40% solution. Let $x$ = liters of water added. Total saccharine before: $$0.45 \times 89 = 40.05$$ liters After adding water, total volume: $$89 + x$$ liters Concentration after: $$\frac{40.05}{89 + x} = 0.40$$ Solve for $x$: $$40.05 = 0.40 (89 + x)$$ $$40.05 = 35.6 + 0.40x$$ $$40.05 - 35.6 = 0.40x$$ $$4.45 = 0.40x$$ $$x = \frac{4.45}{0.40} = 11.125$$ liters **Answer:** Add 11.125 liters of water.