$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{-8}x\stackrel{\stackrel{c}{\downarrow }}{+9} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{-8}{2(2)}~~,~~9-\cfrac{(-8)^2}{4(2)} \right)\implies \left( \cfrac{8}{4}~~,~~9-\cfrac{64}{8} \right) \\\\\\ (2~~,~~9-8)\implies (2~~,~~1)$

5 0

3 years ago

What is the approximate area of the figure?

melisa1 [442]

Option D:

The approximate area of the figure is 109.6 square feet.

Solution:

The figure is splitted into three shapes.

One is rectangle and the other two is semi-circles.

Diameter of the semi-circle = 5 feet

Radius of the semi-circle = 5 ÷ 2 = 2.5 feet

Area of the semi-circle = $\frac{1}{2}\pi r^2$

$$=\frac{1}{2}\times 3.14\times (2.5)^2$

Area of the semi-circle = 9.8 square feet

Area of 2 semi-circles = 2 × 9.8

Area of 2 semi-circles = 19.6 square feet

Length of the rectangle = 18 feet

Width of the rectangle = 5 feet

Area of the rectangle = length × width

= 18 × 5

Area of the rectangle = 90 square feet

Area of the figure = Area of 2 semi-circles + Area of the rectangle

= 19.6 square feet + 90 square feet

= 109.6 square feet

The approximate area of the figure is 109.6 square feet.

Hence Option D is the correct answer.

4 0

3 years ago