Answer:
x = pi/2 + 2 pi n x = pi + 2 pi n where n is an integer
x = 5pi /3 + 2 pi n
Step-by-step explanation:
8 cos^2 x + 4 cos x-4 = 0
Divide by 4
2 cos^2 x + cos x-1 = 0
Let u = cos x
2 u^2 +u -1 =0
Factor
(2u -1) ( u+1) = 0
Using the zero product property
2u-1 =0 u+1 =0
u = 1/2 u = -1
Substitute cosx for u
cos x = 1/2 cos x = -1
Take the inverse cos on each side
cos ^-1(cos x) = cos ^-1(1/2) cos ^-1( cos x) =cos ^-1( -1)
x = pi/2 + 2 pi n x = pi + 2 pi n where n is an integer
x = 5pi /3 + 2 pi n
Answer:
-1/4 , -1
Step-by-step explanation:
I solved it using Factorization method and Quadratic Equation .
Factorization Method
Quadratic Equation
Answer:
We know that triangle ABD and triangle CBD are congruent because of SAS.
Step-by-step explanation:
AB is congruent to BC because of the definition of an isosceles triangle
BD=BD because of the reflexive property
m<ABD=m<CBD because BM is the median of an isosceles triangle
Thus, triangle ABD is congruent to triangle CBD because of SAS
Answer:
Hello! Hope you are having a good day. Your answer is F. I just did this. Good luck!
Step-by-step explanation:
is so that AP:BP=1:3 and point M is the midpoint of segment
CP
. Find the area of △ABC if the area of △BMP is equal to 21m2.
56 m²
A diagram can be helpful.
Triangles with the same altitude will have areas proportional to the length of their bases.
The altitude from B to PC is the same for triangles BMP and BMC, so they have areas that are in the same proportion as MP : MC. Since M is the midpoint of CP, MP = MC and ABMP = ABMC = 21 m². Then ...
... ACPB = 21 m² + 21 m² = 42 m²
The altitude from C to AB is the same for triangles CPA and CPB, so those triangles have areas in the sampe proportion as AP : BP = 1 : 3. Then ...
... ACPA : ACPB = PA : PB = 1 : 3
... ACPA : 42 m² = 1 : 3
So, the area of ∆CPA is 1/3 of 42 m², or 14 m². The area of ABC is the sum of the areas of CPA and CPB, so is ...
... AABC = ACPA + ACPB = 14 m² + 42 m²
... AABC = 56 m²