Answer and Step-by-step explanation:
The signs didn't really "swap". Instead, the whole function was divided by -1, or we could say the function was divided by -3. That would turn:
-18x² - 15x + 3 = 0
into
(-18 / -3)x² - (15 / -3)x + (3 / -3) = 0
6x² + 5x - 1 = 0
And that gives the "swapped signs".
Answer:
96 square units
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.
My answer:
- The length of the large rectangle is: 5
- The width of the large rectangle is: 7
=> Area of the large rectangle = length × width = 7*5 = 35 square units
- The length of the middle rectangle is: 7
- The width of the middle rectangle is: 4
=> Area of the middle rectangle = length × width = 7*4 = 28 square units
- The length of the small rectangle is: 7
- The width of the small rectangle is: 3
=> Area of the small rectangle = length × width = 7*3 = 21 square units
Area of top and the bottom triangles =2*
Total surface area = 35 + 28 + 21 + 12 = 96 square units
Answer: Option b.
Step-by-step explanation:
First, you need to calculate the ratio of the area. This is:

You know that the area of the smaller trapezoid is 771 m², then you can set up the following proportion, where "x" is the area of the larger trapezoid. Then you have:

Now you must solve for "x". Therefore, you get that the area of the larger trapezoid is:

Answer:
Step-by-step explanation:
A1. C = 104°, b = 16, c = 25
Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°
A = 180-C-B = 37.6°
Law of Sines: a = c·sinA/sinC ≅ 15.7
A2. B = 56°, b = 17, c = 14
Law of Sines: C = arcsin[c·sinB/b] ≅43.1°
A = 180-B-C = 80.9°
Law of Sines: a = b·sinA/sinB ≅ 20.2
B1. B = 116°, a = 11, c = 15
Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2
A = arccos{(b²+c²-a²)/(2bc) ≅26.5°
C = 180-A-B = 37.5°
B2. a=18, b=29, c=30
Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°
Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°
C = 180-A-B = 75.3°