<h3>
Answer: 110 square miles</h3>
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Explanation:
There are a few ways to do this.
One method involves drawing a horizontal line to form three rectangles as show below. Note the labels A,B,C.
- Rectangle A in the upper left corner has area of base*height = 6*7 = 42 square miles. I'll let A = 42 since we'll use it later.
- Rectangle B is 20 miles across horizontally (base) and 2 miles tall vertically (height). The 2 miles is from 9-7 = 2. The area of rectangle B is 20*2 = 40 square miles. Let B = 40.
- Then finally, C = 28 because rectangle C is 4 miles across and 7 miles tall, so 4*7 = 28.
Add up those three sub areas found: A+B+C = 42+40+28 = 110 square miles
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There are other methods you could do. A second method is to draw two vertical lines to form 3 other rectangles, then add up the sub areas to get 110.
A third method is to draw a horizontal line across the top to form one large rectangle (20 mi by 9 mi) and subtract off the area of the 7 mi by 10 mi inner rectangle (the empty space), so you'd say 20*9 - 7*10 = 180-70 = 110
2y^4 and 5y^4 because same variable (y), same degree (3)
Answer:1.2675x10^37
Step-by-step explanation:
3.9x10^33 x 3.25x10^3
3.9x3.25x10^33x10^3
12.675x10^(33+3)
12.675x10^36
1.2675x10^1x10^36
1.2675x10^(1+36)
1.2675x10^37
Answer:
Carlos is incorrect.
Step-by-step explanation:
We have been given that a line through the origin has a slope of 
. Carlos thinks the slope of a perpendicular line at the origin will be 3.
We know that the slope of a perpendicular line to a given line is always negative reciprocal of the slope of the given line.
The slope of the perpendicular line at the origin will be negative reciprocal of .
Let us find negative reciprocal of as:
Since the slope of a perpendicular line at the origin is 
, therefore, Carlos is incorrect.
NOT NECESSARILY would a triangle be equilateral if one of its angles is 60 degrees. To be an equilateral triangle (a triangle in which all 3 sides have the same length), all 3 angles of the triangle would have to be 60°-angles; however, the triangle could be a 30°-60°-90° right triangle in which the side opposite the 30 degree angle is one-half as long as the hypotenuse, and the length of the side opposite the 60 degree angle is √3/2 as long as the hypotenuse. Another of possibly many examples would be a triangle with angles of 60°, 40°, and 80° which has opposite sides of lengths 2, 1.4845 (rounded to 4 decimal places), and 2.2743 (rounded to 4 decimal places), respectively, the last two of which were determined by using the Law of Sines: "In any triangle ABC, having sides of length a, b, and c, the following relationships are true: a/sin A = b/sin B = c/sin C."¹
