8x^2-3 = sqrt(16x+9)
[8x^2-3]^2 = [sqrt(16x+9)]^2 ... square both sides
64x^4-48x^2+9 = 16x+9
64x^4-48x^2+9-16x-9 = 0
64x^4-48x^2-16x = 0
16x(4x^3-3x-1) = 0
16x(x-1)(2x+1)^2 = 0
16x=0 or x-1=0 or (2x+1)^2 = 0
x=0 or x=1 or x = -1/2
The possible solutions are x=0 or x=1 or x = -1/2
We need to check all the possible solutions
Checking x=0
8x^2-3 = sqrt(16x+9)
8(0)^2-3 = sqrt(16*0+9)
-3 = 3
The equation is false so x=0 is extraneous (not a real solution)
Checking x=1
8x^2-3 = sqrt(16x+9)
8(1)^2-3 = sqrt(16*1+9)
5 = 5
Equation is true. The value x=1 is a solution
Checking x=-1/2
8x^2-3 = sqrt(16x+9)
8(-1/2)^2-3 = sqrt(16(-1/2)+9)
-1 = 1
The equation is false so x=-1/2 is extraneous (not a real solution)
Therefore, the only answer is choice A) 1
For -7pi/6 is an angle in second quadrant, then sine and cosecant must be positive; and cosine, secant, tangent and cotangent must me negative.
The reference angle is:
7pi/6-pi=7pi/6-6pi/6=(7pi-6pi)/6=pi/6
Then
sin(-7pi/6)=sin(pi/6)→sin(-7pi/6)=1/2
cos(-7pi/6)=-cos(pi/6)→cos(-7pi/6)=-sqrt(3)/2
csc(-7pi/6)=1/sin(-7pi/6)=1/(1/2)=1(2/1)=2/1→csc(-7pi/6)=2
sec(-7pi/6)=1/cos(-7pi/6)=1/(-sqrt(3)/2)=-1(2/sqrt(3))=-2/sqrt(3)→
sec(-7pi/6)=-[2/sqrt(3)]*sqrt(3)/sqrt(3)=-2sqrt(3)/[sqrt(3)]^2→
sec(-7pi/6)=-2sqrt(3)/3
tan(-7pi/6)=sin(-7pi/6)/cos(-7pi/6)=(1/2)/(-sqrt(3)/2)=-(1/2)*(2/sqrt(3))→
tan(-7pi/6)=-2/[2sqrt(3)]=-1/sqrt(3)=-[1/sqrt(3)]*[sqrt(3)/sqrt(3)]→
tan(-7pi/6)=-sqrt(3)/[sqrt(3)]^2→tan(-7pi/6)=-sqrt(3)/3
cot(-7pi/6)=cos(-7pi/6)/sin(-7pi/6)=[-sqrt(3)/2]/(1/2)=-sqrt(3)/2*(2/1)→
cot(-7pi/6)=-2sqrt(3)/2→cot(-7pi/6)=-sqrt(3)
Answers:
sin(-7pi/6) = 1/2
cos(-7pi/6) = - sqrt(3)/2
tan(-7pi/6) = - sqrt(3)/3
csc(-7pi/6) = 2
sec(-7pi/6) = - 2*sqrt(3)/2
cot(-7pi/6) = - sqrt(3)
The answer is d the last one
Answer:
650
Step-by-step explanation:
822-72=650
Answer:
the answer is 3/8
Step-by-step explanation:
multiply 9/10 by 5/12 then you get the answer 3/8
