Answer: The answer is 381.85 feet.
Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.
This situation is framed very nicely in the attached figure, where
BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?
From the right-angled triangle WGB, we have

and from the right-angled triangle WAB, we have'

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.
Thus, the height of the building across the street is 381.85 feet.
Answer:
![67.5\text{ [square units]}](https://tex.z-dn.net/?f=67.5%5Ctext%7B%20%5Bsquare%20units%5D%7D)
Step-by-step explanation:
The composite figure consists of one rectangle and two triangles. We can add up the area of these individual shapes to find the total area of the irregular figure.
<u>Formulas</u>:
- Area of rectangle with base
and height
:
- Area of triangle with base
and height
:
By definition, the base and height must intersect at a 90 degree angle.
The rectangle has a base of 10 and a height of 5. Therefore, its area is
.
The smaller triangle to the left of the rectangle has a base of 2 and a height of 5. Therefore, its area is
.
Finally, the larger triangle on top of the rectangle has a base of 5 and a height of 5. Therefore, its area is
.
Thus, the area of the total irregular figure is:
![50+5+12.5=\boxed{67.5\text{ [square units]}}](https://tex.z-dn.net/?f=50%2B5%2B12.5%3D%5Cboxed%7B67.5%5Ctext%7B%20%5Bsquare%20units%5D%7D%7D)
The point
is on the graph of the equation.
Explanation:
The equation is 
We need to determine the point
is on the graph.
To determine the point
is on the graph, we need to substitute the point in the equation and find whether the LHS is equal to RHS.
Thus, substituting the point
in the equation
, we get,

Multiplying the terms within the bracket, we have,

Adding the LHS, we have,

Thus, both sides of the equation are equal.
Hence, the point
is on the graph of the equation 
A , the seal is a carbavore