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📰 Solution: Rewrite the equation as $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Possible integer values for $y$ are $-2, -1, 0, 1, 2$. For each $y$, solve for $x$: 📰 $y = 0$: $x^2 = 16 \Rightarrow x = \pm4$ (2 points). 📰 $y = \pm1$: $x^2 + 4(1) = 16 \Rightarrow x^2 = 12$ (no integer solutions). 📰 Solution The Area A Of A Regular Hexagon With Side Length S Is Given By The Formula A Frac3Sqrt32S2 Setting This Equal To 54Sqrt3 We Have 📰 Solution The Area A Of The Triangle Is Given By 📰 Solution The Arithmetic Mean Is Calculated By Summing The Values And Dividing By The Count Frac10 14 163 Frac403 Approx 13Overline3 The Exact Average Is Boxeddfrac403 📰 Solution The Cosine Function Oscillates Between 1 And 1 So 5Cosleftfracpi6Xright Ranges From 5 To 5 Adding 10 Shifts The Range To 5 15 The Range Is Boxed5 15 📰 Solution The Expression Under The Square Root Must Satisfy 4 X2 Geq 0 So X In 2 2 The Maximum Value Of Sqrt4 X2 Is 2 At X 0 And The Minimum Is 0 At X Pm 2 The Range Is Boxed0 2 Think 📰 Solution The Minimum Depth Where Both Measurements Coincide Is The Lcm Of 5 And 7 Since 5 And 7 Are Prime Lcm5 7 5 Cdot 7 35 The Answer Is Boxed35 📰 Solution The Slope Of The Line From 2 3 To 5 11 Is Frac11 35 2 Frac83 Since Tan Theta Equals The Slope Tan Theta Frac83 Boxeddfrac83 📰 Solution The Volume V Of A Cone Is Given By 📰 Solution To Divide Both 48 And 72 Regions Into Equal Sized Groups With No Regions Left Out We Must Find The Greatest Common Divisor Gcd Of 48 And 72 📰 Solution To Find The Least Common Multiple Lcm Of 12 And 18 Factorize Both 12 22 Cdot 3 And 18 2 Cdot 32 The Lcm Takes The Highest Powers Of All Primes 22 Cdot 32 4 Cdot 9 36 Thus The Lcm Is Boxed36 📰 Solution To Find The Smallest Number Of Samples Per Column Such That The Number Of Columns Is A Multiple Of Both 6 And 10 We Compute The Least Common Multiple Lcm Of 6 And 10 📰 Solution To Find The Smallest Positive Integer Divisible By Both 7 And 5 We Compute The Least Common Multiple Lcm Of 7 And 5 Since 7 And 5 Are Coprime Their Lcm Is Simply Their Product 📰 Solution We Analyze The Inequality X Y X Y Leq 8 This Expression Is Symmetric And Represents A Diamond Rhombus Centered At The Origin 📰 Solution We Are Given A B C 1 With A B C 0 So 1 A B C 1 B A C And 1 C A B The Expression Becomes 📰 Solution We Are Given 180 Dots And Must Arrange Them Into Rows With The Same Number Of Dots Where The Number Of Rows Is A Prime Number Greater Than 3 To Maximize The Number Of Dots Per Row We Must Minimize The Number Of Rows Among Valid Prime Numbers