Answer:
Find the equation of the line perpendicular to x = 0 that passes through (-3,1)
Step-by-step explanation:
Using the equation, y - y1 = m(x - x1)
Since x = 0, the gradient, m = 0
Hence, y - 1 = 0
∴ y = 1
Answer:
<u>Identities used:</u>
- <em>1/cosθ = secθ</em>
- <em>1/sinθ = cosecθ</em>
- <em>sinθ/cosθ = tanθ</em>
- <em>cosθ/sinθ = cotθ</em>
- <em>sin²θ + cos²θ = 1</em>
<h3>Question 1 </h3>
- (1 - sinθ)/(1 + sinθ) =
- (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
- (1 - sinθ)² / (1 - sin²θ) =
- (1 - sinθ)² / cos²θ
<u>Square root of it is:</u>
- (1 - sinθ)/ cosθ =
- 1/cosθ - sinθ / cosθ =
- secθ - tanθ
<h3>Question 2 </h3>
<u>The first part without root:</u>
- (1 + cosθ) / (1 - cosθ) =
- (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
- (1 + cosθ)² / (1 - cos²θ) =
- (1 + cosθ)² / sin²θ
<u>Its square root is:</u>
- (1 + cosθ) / sinθ =
- 1/sinθ + cosθ/sinθ =
- cosecθ + cotθ
<u>The second part without root:</u>
- (1 - cosθ) / (1 + cosθ) =
- (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
- (1 - cosθ)²/ (1 - cos²θ) =
- (1 - cosθ)²/sin²θ
<u>Its square root is:</u>
- (1 - cosθ) / sinθ =
- 1/sinθ - cosθ / sinθ =
- cosecθ - cotθ
<u>Sum of the results:</u>
- cosecθ + cotθ + cosecθ - cotθ =
- 2cosecθ
The area of a square or a rectangle (which is the shape of any room in a house) is:
Area = width x length
So, in this case:
Area = 51 m x 26 m = 1326 Sq meters. Which is pretty big if you ask me.
Anyhow, Ms Melcher needs to buy 1326 sq meters
We are given the functions:
f(x) = 4 x – 5 --->
1
g(x) = 3 x + 7 --->
2
To find for the value of f(x) + g(x), all we have to do is
to add equations 1 and 2:
f(x) + g(x) = 4 x – 5 + 3 x + 7
f(x) + g(x) = 7 x + 2 = y
In this case, for any real number value assign to x, we get
a real number value of y. This is because the function is linear.
Therefore the domain of the function is all real numbers.
