Answer:
(x + 2)² + (y - 1)² = 25
General Formulas and Concepts:
<u>Algebra I</u>
<u>Pre-Calc</u>
Circle Center Formula: (x - h)² + (y - k)² = r²
- <em>(h, k)</em> is center
- <em>r</em> is radius
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
<em>(h, k)</em> = (-2, 1)
<em>r</em> = 5
<u>Step 2: Find Equation</u>
- Substitute in variables [Circle Center Formula]: (x - -2)² + (y - 1)² = 5²
- Simplify: (x + 2)² + (y - 1)² = 25
Topic: Pre-Calculus
Unit: Conics
Book: Pre-Calculus (McGraw Hill)
Answer:
The correct options are;
1) Write tan(x + y) as sin(x + y) over cos(x + y)
2) Use the sum identity for sine to rewrite the numerator
3) Use the sum identity for cosine to rewrite the denominator
4) Divide both the numerator and denominator by cos(x)·cos(y)
5) Simplify fractions by dividing out common factors or using the tangent quotient identity
Step-by-step explanation:
Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;
tan(x + y) = sin(x + y)/(cos(x + y))
sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))
(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))
(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)
∴ tan(x + y) = (tan(x) + tan(y))(1 - tan(x)·tan(y)
Solution:
Given expression is 
To find the inverse of y.
⇒
Do cross multiplication.
⇒ y = –6 × 3
⇒ y = –18
Inverse of y means reverse the function.
We know that inverse of y is 
.
⇒ 
Hence, inverse of y is 
Answer:
A. d ≤ –7 or d > 8.
Step-by-step explanation:
Given : 2d + 3 ≤ –11 or 3d – 9 > 15.
To find : What are the solutions of the compound inequality .
Solution : We have given 2d + 3 ≤ –11 or 3d – 9 > 15.
For 2d + 3 ≤ –11
On subtracting both sides by 3
2d ≤ –11 - 3 .
2d ≤ –14.
On dividing both sides by 2 .
d ≤ –7.
For 3d – 9 > 15.
On adding both sides by 9.
3d > 15 + 9 .
3d > 24 .
On dividing both sides by 3 .
d > 8 .
So, A. d ≤ –7 or d > 8.
Therefore, A. d ≤ –7 or d > 8.
