The figure consists of three objects, a rectangle, a trapezoid, a triangle
Find the area of the rectangle
The rectangle is 16 in long and 9 in wide
a₁ = l × w
a₁ = 16 × 9
a₁ = 144 in²
Find the area of the trapezoid
The base of trapezoid is 31 in and 16 in, and the height is 35 - 20 = 15 in.
a₂ = 1/2 × (a + b) × h
a₂ = 1/2 × (31 + 16) × 15
a₂ = 1/2 × 47 × 15
a₂ = 352.5 in²
Find the area of the triangle
The base of the triangle is 31 in, the height is 20 in
a₃ = 1/2 × b × h
a₃ = 1/2 × 31 × 20
a₃ = 310 in²
Add the area together
a = a₁ + a₂ + a₃
a = 144 + 352.5 + 310
a = 806.5
The answer is 806.5 in²
Answer:
A) 45=x
B) Yes, since both A and B are 90°
I actually don't have enough information to complete this question but if you have a positive slope the line goes either up and right or down and left but when it is negative it goes down left or up right.
Answer:
x = 75°
Step-by-step explanation:
Sum of angles in a triangle: 180°
So,
50°+55°+x=180°
105°+ x = 180°
Subtract 105° from both sides
x = 75°
Kepler's third law described the relation between semi-major axis (or average distance to the star) and
the orbital period (how long it takes to complete one lap) as follows:
a^3 / p^2 = constant
In the case of our Solar system the constant is 1
This means that, for this problem:
a^3 / p^2 = 1
p^2 = a^3
p = a^(3/2)
The semi major axis is given as 101 million km. We need to convert this into AU where 1 AU is approximately 150 million Km
101 million Km = (101x1) / 150 = 0.67 AU
Now, we substitute in the equation to get the orbital period as follows:
p = (0.67)^(3/2) = 0.548 earth years
