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16 stepsHere are the solutions to the problems from the image, arranged as requested:
1) Solve for in each of the following linear equations:
- a)
Subtract 5 from both sides.
$$ 2x = 15 - 5 $$ $$ 2x = 10 $$Divide by 2.
$$ x = \frac{10}{2} $$ $$ \boxed{x = \textbf{5}} $$- b)
Add 7 to both sides.
$$ 4x = 9 + 7 $$ $$ 4x = 16 $$Divide by 4.
$$ x = \frac{16}{4} $$ $$ \boxed{x = \textbf{4}} $$- c)
Add 3 to both sides.
$$ 7x = 25 + 3 $$ $$ 7x = 28 $$Divide by 7.
$$ x = \frac{28}{7} $$ $$ \boxed{x = \textbf{4}} $$
2) Calculate the Shaded Area of the region in the figures below:
- Top Right Figure (Rectangle with an inner unshaded rectangle): The outer rectangle has a length of and a width of . The inner unshaded rectangle has a length of and a width of .
Calculate the area of the outer rectangle.
$$ A_{\text{outer}} = \text{length} \times \text{width} = 28 \text{ cm} \times 20 \text{ cm} $$ $$ A_{\text{outer}} = 560 \text{ cm}^2 $$Calculate the area of the inner unshaded rectangle.
$$ A_{\text{inner}} = \text{length} \times \text{width} = 24 \text{ cm} \times 10 \text{ cm} $$ $$ A_{\text{inner}} = 240 \text{ cm}^2 $$Calculate the shaded area by subtracting the inner area from the outer area.
$$ A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}} $$ $$ A_{\text{shaded}} = 560 \text{ cm}^2 - 240 \text{ cm}^2 $$ $$ \boxed{A_{\text{shaded}} = \textbf{320 cm}^2} $$- a) Middle Right Figure (Annulus - two concentric circles): The outer circle has a radius . The inner circle has a radius . We will use .
Calculate the area of the outer circle.
$$ A_{\text{outer}} = \pi R^2 = \frac{22}{7} \times (14 \text{ cm})^2 $$ $$ A_{\text{outer}} = \frac{22}{7} \times 196 \text{ cm}^2 $$ $$ A_{\text{outer}} = 22 \times 28 \text{ cm}^2 = 616 \text{ cm}^2 $$Calculate the area of the inner circle.
$$ A_{\text{inner}} = \pi r^2 = \frac{22}{7} \times (7 \text{ cm})^2 $$ $$ A_{\text{inner}} = \frac{22}{7} \times 49 \text{ cm}^2 $$ $$ A_{\text{inner}} = 22 \times 7 \text{ cm}^2 = 154 \text{ cm}^2 $$Calculate the shaded area by subtracting the inner area from the outer area.
$$ A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}} $$ $$ A_{\text{shaded}} = 616 \text{ cm}^2 - 154 \text{ cm}^2 $$ $$ \boxed{A_{\text{shaded}} = \textbf{462 cm}^2} $$- b) Middle Right Figure (Faint Quarter Circle): Based on the grid, the radius of the quarter circle is . We assume the shaded area is the area of this quarter circle. We will use .
Calculate the area of the quarter circle.
$$ A_{\text{quarter circle}} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (4 \text{ cm})^2 $$ $$ A_{\text{quarter circle}} = \frac{1}{4} \times \frac{22}{7} \times 16 \text{ cm}^2 $$ $$ A_{\text{quarter circle}} = \frac{22 \times 4}{7} \text{ cm}^2 = \frac{88}{7} \text{ cm}^2 $$ $$ \boxed{A_{\text{shaded}} = \textbf{12.57 cm}^2 \text{ (to 2 decimal places)}} $$
Homework 1) Calculate the area of the unshaded region in the figure below:
- Bottom Left Figure (Large rectangle with an inner shaded rectangle): The large rectangle has a length of and a width of (by counting grid squares). The inner shaded rectangle has a length of and a width of .
Calculate the area of the large rectangle.
$$ A_{\text{large}} = \text{length} \times \text{width} = 36 \text{ cm} \times 25 \text{ cm} $$ $$ A_{\text{large}} = 900 \text{ cm}^2 $$Calculate the area of the inner shaded rectangle.
$$ A_{\text{shaded}} = \text{length} \times \text{width} = 20 \text{ cm} \times 18 \text{ cm} $$ $$ A_{\text{shaded}} = 360 \text{ cm}^2 $$Calculate the unshaded area by subtracting the shaded area from the large area.
$$ A_{\text{unshaded}} = A_{\text{large}} - A_{\text{shaded}} $$ $$ A_{\text{unshaded}} = 900 \text{ cm}^2 - 360 \text{ cm}^2 $$ $$ \boxed{A_{\text{unshaded}} = \textbf{540 cm}^2} $$
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