Answer:
(-3, -5) This is my answer I hope it helps
Answer:
Width = 2x²
Length = 7x² + 3
Step-by-step explanation:
∵ The area of a rectangle is 
∵ Its width is the greatest common monomial factor of
and 6x²
- Let us find the greatest common factor of 14 , 6 and
, x²
∵ The factors of 14 are 1, 2, 7, 14
∵ The factors of 6 are 1, 2, 3, 6
∵ The common factors of 14 and 6 are 1, 2
∵ The greatest one is 2
∴ The greatest common factor of 14 and 6 is 2
- The greatest common factor of monomials is the variable with
the smallest power
∴ The greatest common factor of
and x² is x²
∴ The greatest common monomial factor of
and 6x² is 2x²
∴ The width of the rectangle is 2x²
To find the length divide the area by the width
∵ The area = 
∵ The width = 2x²
∴ The length = (
) ÷ (2x²)
∵
÷ 2x² = 7x²
∵ 6x² ÷ 2x² = 3
∴ (
) ÷ (2x²) = 7x² + 3
∴ The length of the rectangle is 7x² + 3

By the quadratic formula,

Then


Multiply numerator and denominator by the denominator's conjugate:

Answer:
4cotα=tanα
4(1/tanα)=tanα
(4/tanα)=tanα
cross multiply
=> 4=tan²α
√4=√tan²α
±2=tanα
α=arc( tan) |2|
α=63.4° ( in first quadrant)
and
α=180+63.4=243.4 in the third quadrant
since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants)
α=180–63.4=116.6° in the second quadrant
α=360–63.4=296.6 in the fourth quadrant
therefore theta( in my case, alpha) lies in all four quadrants and is equal to:
α=63.4°,243.4°,116.6°and 296.6°
Answer:
Step-by-step explanation:
The sample space is shown in the attached photo.
Probability = number of favourable outcomes/total number of outcomes.
Looking at the diagram, the total possible outcomes is 36.
A) from the diagram, the total number of events in which the sum of the numbers is odd is 18. Therefore, the probability that the sum of the numbers is odd is
18/36 = 1/2
B) the total number of events in which the sum of the numbers is 10 or more is 6. Therefore, the probability that the sum of the numbers is 10 or more is
6/36 = 1/6
C) the total number of events in which 3 appears on each of the two dice is 1. Therefore, the probability that 3 appears on each of the two dice is
1/36