What is a complementary angle versus a supplementary angle?
2 answers:
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Please refer to my attachments for visual guidelines.
We are going to solve your problem by using the pythagorean theorem, a^2+b^2 = c^2, where a and b are the legs of the triangle, and c is the hypotenuse (the longest side).
The length of the ladder is equal to 70ft (hypotenuse); one leg is the distance between the wall and the bottom of the ladder - 40 ft, the other leg is unknown for it is the distance between 10 ft above the ground and the top of the ladder-represented by "x". Using pythagorean theorem, a^2+b^=c^2, we have x^2+40^2 = 70^2. Solving the exponents, we have x^2 + 1600 = 4900.
Isolating the variable x, we have x^2 = 4900-1600. Futher simplying, x^2 = 3300. Thus, x = √3300 or 57.4456264654 ft.
Adding 10 ft to x, therefore, the top of the leadder is 67.4456264654 ft off the ground.
We are going to solve your problem by using the pythagorean theorem, a^2+b^2 = c^2, where a and b are the legs of the triangle, and c is the hypotenuse (the longest side).
The length of the ladder is equal to 70ft (hypotenuse); one leg is the distance between the wall and the bottom of the ladder - 40 ft, the other leg is unknown for it is the distance between 10 ft above the ground and the top of the ladder-represented by "x". Using pythagorean theorem, a^2+b^=c^2, we have x^2+40^2 = 70^2. Solving the exponents, we have x^2 + 1600 = 4900.
Isolating the variable x, we have x^2 = 4900-1600. Futher simplying, x^2 = 3300. Thus, x = √3300 or 57.4456264654 ft.
Adding 10 ft to x, therefore, the top of the leadder is 67.4456264654 ft off the ground.
Answer: Choice C
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Reason:
The portion on the left represents because of the open circle at -2 and shading pointing to the left. This represents all numbers smaller than -2.
The portion on the right represents all values that are 4 or larger; hence . Note the use of a closed circle at 4 to include this value.
We connect those two inequalities with the keyword "or" because x can only pick one of those options, but not both at the same time.