PLEASE Help mE wiTh the two QUESTIONS
1 answer:
You might be interested in
Answer:
Area of the walkway border is 736 ft²
Step-by-step explanation:
A rectangle is a quadrilateral (has four sides and four angles) with two pairs of opposite and parallel sides. All angles in a rectangle are 90 degrees each. Also, opposite sides are equal.
The entire pool area has a length of 60 feet and width of 40 feet. The area of the entire pool area is:
Area of entire pool = length * width = 60 feet * 40 feet = 2400 ft²
The area of entire pool
A walkaway border of 4 ft is made around the pool area. There is a decrease of 4 ft round the sides of the pool area.
Therefore length of pool = length of entire pool area - 4 feet border at left - 4 feet border at right = 60 ft - 4 ft - 4 ft = 52 ft
width of pool = width of entire pool area - 4 feet border at top - 4 feet border at bottom = 40 ft - 4 ft - 4 ft = 32 ft
Therefore the pool is 52 ft × 32 ft
Area of pool = length * width = 52 ft * 32 ft = 1664 ft²
Area of the walkway border = Entire pool area - Area of pool = 2400 ft² - 1664 ft² = 736 ft²
You have the right idea and you are close to the correct answer. However that's not what your teacher is looking for in terms of steps.
The starting inequality is
She starts off selling 8 buckets. Then she sells b more to get a total of b+8
This total must be 20 or larger which is why I set b+8 greater than or equal to 20
Solve for b by subtracting 8 from both sides
So she needs to sell at least 12 more buckets to reach her goal
If you graphed this on a number line, then you would draw a closed circle at 12. Then shade to the right of the close circle.
The starting inequality is
She starts off selling 8 buckets. Then she sells b more to get a total of b+8
This total must be 20 or larger which is why I set b+8 greater than or equal to 20
Solve for b by subtracting 8 from both sides
So she needs to sell at least 12 more buckets to reach her goal
If you graphed this on a number line, then you would draw a closed circle at 12. Then shade to the right of the close circle.