1. The problem asks for the leading term of the polynomial $f(x) = 4x^2 - 2x + 16$. The leading term is the term with the highest power of $x$. 2. To find the leading coefficient of $f(x) = 4x^2 - 5x^2 + 2x - 1$, first combine like terms: $4x^2 - 5x^2 = -x^2$. The polynomial simplifies to $-x^2 + 2x - 1$. The leading coefficient is the coefficient of the term with the highest power, which is $-1$. 3. The degree of the polynomial $f(x) = 5x^3 - 3x^5 + 4x^2 - 10x + 1$ is the highest power of $x$, which is $5$ from the term $-3x^5$. 4. To find the zeros of the polynomial $P(x) = x(x+1)(x-2)(x+3)$, set each factor equal to zero: - $x = 0$ - $x + 1 = 0 ightarrow x = -1$ - $x - 2 = 0 ightarrow x = 2$ - $x + 3 = 0 ightarrow x = -3$ Therefore, zeros are $x = 0, -1, 2, -3$. 5. A polynomial with zeros at $x=1$, $x=-1$, and $x=0$ can be written as: $$f(x) = x(x - 1)(x + 1)$$ 6. The graph shows a cubic polynomial because it has one local maximum and one local minimum and crosses the x-axis three times, typical of degree 3. 7. The graph with 4 turning points suggests it is a polynomial of degree 5 because the number of turning points is at most one less than the degree. 8. The end behavior of $$P(x) = -3x^6 + 4x^4 + 2x^3 - 8x + 1$$ is determined by the leading term $-3x^6$. Since degree 6 is even and the coefficient is negative, both ends of the graph go down to $-\, ext{infinity}$: $$ ext{As } x o \\pm \, \\infty, P(x) o -\\infty$$ 9. The maximum number of turning points of $$F(x) = x^4 - 2x^2 + x + 1$$ is one less than degree 4, so at most 3 turning points. By analyzing or graphing, we find it has 3 turning points. 10. The cost function is $$C(x) = 2x^2 + 4x + 100$$. For $x=50$ pairs: $$C(50) = 2(50)^2 + 4(50) + 100 = 2(2500) + 200 + 100 = 5000 + 200 + 100 = 5300$$ 11. For $x=500$ pairs: $$C(500) = 2(500)^2 + 4(500) + 100 = 2(250000) + 2000 + 100 = 500000 + 2000 + 100 = 502100$$ 12. Given total allowance spent: $$A(x) = x^2 + 7x + 10$$ and price per snack: $x + 5$, find number of snacks $n$: $$n(x + 5) = x^2 + 7x + 10$$ Since total cost equals number of snacks times price per snack, divide both sides: $$n = \frac{x^2 + 7x + 10}{x + 5}$$ Factor numerator: $$x^2 + 7x + 10 = (x + 5)(x + 2)$$ Simplify: $$n = x + 2$$ 13. The area of the square garden: $$A(x) = 16x^2 - 40x + 25$$. To find the length of one side, factor $A(x)$: $$A(x) = (4x - 5)^2$$ Thus one side length is: $$s = |4x - 5|$$ 14. The perimeter is $4$ times one side length: $$P = 4|4x - 5|$$ 15. An angle with vertex on the circle is an inscribed angle. 16. An angle with vertex at the center of the circle is a central angle. 17. A line segment with endpoints on the circle is called a chord. 18. Measure of arc $BD$ is twice the central angle $\angle BED$: $$m \overset{\frown}{BD} = 2 \times 60^6 = 120^6$$ 19. Measure of arc $CR$ for inscribed angle $\angle CAR$ is twice the angle: $$m \overset{\frown}{CR} = 2 \times 50^6 = 100^6$$ 20. In circle $\odot \square$, if $\angle PEC$ is a semicircle ($180^6$) and $m\angle PAE = 50^6$, then $\angle MAP$ is the complement inside the semicircle: $$m\angle MAP = 180^6 - 50^6 = 130^6$$ 21. Quadrilateral $MBTP$ inscribed in $\odot C$ with $m\angle ICA=90^6$ and $m\angle AT=45^6$. By properties of angles in circles, $m\angle MCT = 45^6$. 22. In $\odot C$, the measure of $m\angle T$ is $45^6$. 23. Given $m\angle ICA=90^6$ and $m\angle T=45^6$, $m\angle TMP = 45^6$. 24. $m\angle MTO = 45^6$ by angle properties of the circle. 25. $m\angle MPT = 45^6$ similarly. 26. $m\angle AME = 90^6$ since $\angle ICA=90^6$. 27. Given $m\angle T = 64^6$, $m\angle NQ = 19^6$, and vertex outside the circle, the measure of $\angle R$ is: $$m\angle R = \frac{1}{2} |m\overset{\frown}{TQ} - m\overset{\frown}{NR}| = 64^6 - 19^6 = 45^6$$ 28. From the figure, measure of $AV$ is $70^6$ (given central angle). 29. Measure of $RV$ is also $70^6$ (central angle or arc intercept). 30. Area of sector with radius $r=6$ cm and central angle $70^6$: $$\text{Area} = \pi r^2 \times \frac{70}{360} = \pi \times 36 \times \frac{7}{36} = 7\pi$$ 31. Distance between points $M(-2, 3)$ and $H(-2, 0)$: $$d = \sqrt{(-2 + 2)^2 + (3 - 0)^2} = \sqrt{0 + 9} = 3$$ 32. Midpoint between $A(-1,1)$ and $H(-2,0)$: $$\left( \frac{-1 - 2}{2}, \frac{1 + 0}{2} \right) = (-1.5, 0.5)$$ 33. Center of circle given by equation $(x - 2)^2 + (y + 4)^2 = 16$ is at $(2, -4)$. 34. Equation of circle with center $(3, -1)$ and radius $4$: $$(x - 3)^2 + (y + 1)^2 = 16$$ 35. Rewrite circle equation $x^2 + y^2 - 2x + 6y + 6 = 0$ in standard form: Complete the square: $$x^2 - 2x + y^2 + 6y = -6$$ $$x^2 - 2x + 1 + y^2 + 6y + 9 = -6 + 1 + 9$$ $$(x - 1)^2 + (y + 3)^2 = 4$$ Final answers are included in each step above.