answer:Assuming 9x^2 - 24x + c is the square of a binomial, then the binomial can be assumed to be 3x + d (since (3x)^2 = 9x^2). We need to determine d such that the expanded form of (3x + d)^2 matches the given quadratic expression. Expanding (3x + d)^2, we get: [ (3x + d)^2 = 9x^2 + 6dx + d^2. ] We equate the linear term 6dx with the linear term of 9x^2 - 24x + c: [ 6dx = -24x. ] Thus, [ 6d = -24 implies d = -4. ] Substitute d = -4 back into (3x - 4)^2 to find c: [ (3x - 4)^2 = 9x^2 - 24x + 16. ] Therefore, c = 16. Hence, the value of c is boxed{16}.

answer:1. **Arrange equation**: Group all x terms together: [ 3x^2 - 18x - 36y^2 + 27 = 0 ] 2. **Complete the square for x**: [ 3(x^2 - 6x) - 36y^2 + 27 = 0 ] Completing the square inside the parentheses: [ 3((x-3)^2 - 9) - 36y^2 + 27 = 0 ] [ 3(x-3)^2 - 27 - 36y^2 + 27 = 0 ] [ 3(x-3)^2 - 36y^2 = 0 ] 3. **Factor and rearrange to solve for x**: [ 3(x-3)^2 = 36y^2 ] [ (x-3)^2 = 12y^2 ] Taking square roots: [ x-3 = pmsqrt{12}y ] [ x-3 = pm 2sqrt{3}y ] [ x = 3 pm 2sqrt{3}y ] 4. **Interpretation**: This represents two lines: [ x = 3 + 2sqrt{3}y ] [ x = 3 - 2sqrt{3}y ] Conclusion: The graph of the equation is boxed{text{two lines}}.

answer:1. We have a right triangle Rt triangle ABC with the legs AC = 2 and BC = 3. The hypotenuse AB can be found using the Pythagorean theorem: [ AB = sqrt{AC^2 + BC^2} = sqrt{2^2 + 3^2} = sqrt{4 + 9} = sqrt{13} ] 2. Given that AB = sqrt{7}, it seems there is a misinterpretation. Instead, let's follow the reference solution steps assuming AB = sqrt{7} to be additional information. 3. Consider point P on AB. We need to find the dihedral angle between the planes ACP and CPB. 4. Draw perpendiculars from P: - Draw PD perp AC at point D. - Draw PE perp CP intersecting BC at point E. 5. Since ACP perp CPB, then PE perp plane ACP. By the Three Perpendiculars Theorem, since PD perp AC, then DE perp AC. 6. Therefore, angle PDE represents the dihedral angle between the planes ACP and CPB. 7. Let theta = angle BCP, which is the angle between sides BC and CP. We need to find cos(angle ECD). 8. Using trigonometric identities, we have: [ cos(angle ECD) = cos(theta) cdot cosleft(frac{pi}{2} - thetaright) = cos(theta) cdot sin(theta) ] 9. By the Law of Cosines in triangle ACB: [ cos(angle ACB) = frac{AC^2 + BC^2 - AB^2}{2 cdot AC cdot BC} = frac{2^2 + 3^2 - (sqrt{7})^2}{2 cdot 2 cdot 3} = frac{4 + 9 - 7}{12} = frac{6}{12} = frac{1}{2} ] 10. Therefore: [ sin(theta) cdot cos(theta) = frac{1}{2} implies sin(2theta) = 1 ] 11. Given 0 < 2theta < pi, solve: [ 2theta = frac{pi}{2} implies theta = frac{pi}{4} ] 12. Now, let CP = a. Then, based on theta = frac{pi}{4}: [ PD = frac{sqrt{2}}{2}a, quad PE = a ] 13. Finally, compute the tangent of the angle angle PDE: [ tan(angle PDE) = frac{PE}{PD} = frac{a}{frac{sqrt{2}}{2}a} = sqrt{2} ] 14. Therefore: [ angle PDE = arctan(sqrt{2}) ] # Conclusion: [ boxed{arctan sqrt{2}} ]

answer:First, let's calculate the total number of germs in the biology lab: Total germs = 0.037 * 10^5 Now, let's calculate the number of petri dishes: Number of petri dishes = 148000 * 10^(-3) To find out how many germs live in a single dish, we divide the total number of germs by the number of petri dishes: Germs per dish = Total germs / Number of petri dishes Let's do the math: Total germs = 0.037 * 10^5 = 0.037 * 100000 = 3700 germs Number of petri dishes = 148000 * 10^(-3) = 148000 / 1000 = 148 dishes Germs per dish = 3700 germs / 148 dishes = 25 germs/dish So, there are boxed{25} germs living happily in a single dish.