1. Problem a: Find how many times as high Mount Everest is compared to the tallest tree Hyperion. Given: - Height of Hyperion tree = $102$ m - Height of Mount Everest = $104$ m Calculate the ratio: $$\text{Ratio} = \frac{104}{102} \approx 1.02$$ So, Mount Everest is about 1.02 times as high as Hyperion. 2. Problem b: Find how many times as great the diameter of the largest known star is compared to Earth's diameter. Given: - Diameter of Earth = $10^7$ m - Diameter of largest known star = $10^{12}$ m Calculate the ratio: $$\text{Ratio} = \frac{10^{12}}{10^7} = 10^{12-7} = 10^5$$ So, the largest known star's diameter is $10^5$ times Earth's diameter. 3. Problem 12a: Write in standard form: $$(4 \times 10^3) \times (7 \times 10^2) \times (2 \times 10^1) \times (9 \times 10^0)$$ Multiply coefficients: $$4 \times 7 = 28, \quad 28 \times 2 = 56, \quad 56 \times 9 = 504$$ Add exponents: $$3 + 2 + 1 + 0 = 6$$ Standard form: $$504 \times 10^6 = 5.04 \times 10^8$$ 4. Problem 12b: Write in standard form: $$(3 \times 10^5) \times (2 \times 10^2) \times (8 \times 10^0)$$ Multiply coefficients: $$3 \times 2 = 6, \quad 6 \times 8 = 48$$ Add exponents: $$5 + 2 + 0 = 7$$ Standard form: $$48 \times 10^7 = 4.8 \times 10^8$$ 5. Problem 13a: Evaluate $3^4 \times 3^2$ Use exponent rule: $$a^m \times a^n = a^{m+n}$$ So: $$3^{4+2} = 3^6 = 729$$ 6. Problem 13b: Evaluate $( -4)^2 \times ( -4)^3$ Add exponents: $$(-4)^{2+3} = (-4)^5 = -1024$$ 7. Problem 13c: Evaluate $10^3 \div 10^2$ Use exponent rule: $$a^m \div a^n = a^{m-n}$$ So: $$10^{3-2} = 10^1 = 10$$ 8. Problem 13d: Evaluate $( -5)^4 \div ( -5)^2$ Subtract exponents: $$(-5)^{4-2} = (-5)^2 = 25$$ 9. Problem 14a: Evaluate $2^3 \times (5 \times 2)^4$ Calculate inside parentheses: $$5 \times 2 = 10$$ Calculate powers: $$2^3 = 8, \quad 10^4 = 10000$$ Multiply: $$8 \times 10000 = 80000$$ 10. Problem 14b: Evaluate $100 \div 2 \times (4 \times 1)^3$ Calculate inside parentheses: $$4 \times 1 = 4$$ Calculate powers: $$4^3 = 64$$ Calculate division and multiplication: $$100 \div 2 = 50$$ Multiply: $$50 \times 64 = 3200$$ 11. Problem 14c: Evaluate $(62 \times 72)^0 \div (84 \times 24)^0$ Any nonzero number to the zero power is 1: $$(62 \times 72)^0 = 1, \quad (84 \times 24)^0 = 1$$ Divide: $$1 \div 1 = 1$$ 12. Problem 14d: Evaluate $3 \times 2^3 \times 8 \div 4$ Calculate powers: $$2^3 = 8$$ Multiply and divide: $$3 \times 8 = 24$$ $$24 \times 8 = 192$$ $$192 \div 4 = 48$$ 13. Problem 14e: Evaluate $(21 \div 7)^4 \div 2^3$ Calculate inside parentheses: $$21 \div 7 = 3$$ Calculate powers: $$3^4 = 81, \quad 2^3 = 8$$ Divide: $$81 \div 8 = 10.125$$ 14. Problem 14f: Evaluate $[( -4)^0 \times 10]^6 \div (15 \div 10)^2$ Calculate powers and divisions: $$( -4)^0 = 1$$ Inside brackets: $$1 \times 10 = 10$$ Raise to power 6: $$10^6 = 1,000,000$$ Calculate division: $$15 \div 10 = 1.5$$ Square: $$1.5^2 = 2.25$$ Divide: $$1,000,000 \div 2.25 \approx 444,444.44$$ 15. Problem 15a: Number of bacteria doubles every hour starting at noon with 1000 bacteria. At time $t$ hours after noon, number of bacteria: $$N = 1000 \times 2^t$$ Calculate: - Noon ($t=0$): $1000 \times 2^0 = 1000$ - 1:00 P.M. ($t=1$): $1000 \times 2^1 = 2000$ - 2:00 P.M. ($t=2$): $1000 \times 2^2 = 4000$ - 3:00 P.M. ($t=3$): $1000 \times 2^3 = 8000$ 16. Problem 15b: Continue pattern for later times. - 4:00 P.M. ($t=4$): $1000 \times 2^4 = 16000$ - 6:00 P.M. ($t=6$): $1000 \times 2^6 = 64000$ - 9:00 P.M. ($t=9$): $1000 \times 2^9 = 512000$ - Midnight ($t=12$): $1000 \times 2^{12} = 4,096,000$ 17. Problem 16: Evaluate $4^3 \div (2 \times 3)^4 \times 11$ Calculate inside parentheses: $$2 \times 3 = 6$$ Calculate powers: $$4^3 = 64, \quad 6^4 = 1296$$ Calculate expression: $$64 \div 1296 \times 11 = \frac{64}{1296} \times 11 = \frac{704}{1296} \approx 0.5432$$ Changing brackets can produce different results by changing order of operations. 18. Problem 17: Identify errors in student work: Student wrote: $$( -2)^2 \times 2^3 - 3^2 \div ( -3) + ( -4)^2 = ( -2)^5 - 9 \div ( -3) + 16 = -32 - 3 + 16 = -35 + 16 = -19$$ Errors: - Incorrectly combined exponents: $( -2)^2 \times 2^3 \neq ( -2)^5$ - Incorrect arithmetic in division and subtraction steps. Correct approach: $$( -2)^2 = 4, \quad 2^3 = 8$$ Multiply: $$4 \times 8 = 32$$ Calculate: $$3^2 = 9$$ Divide: $$9 \div ( -3) = -3$$ Calculate: $$( -4)^2 = 16$$ Sum all: $$32 - 3 + 16 = 45$$ Final correct answer is 45.