Tan θ = y over x tan 11 pi over 6 = the fraction negative 1 over 2 over the fraction square root 3 over 2 tan 11 pi over 6 = negative 1 over 2 times 2 over square root 3 tan 11 pi over 6 = negative 1 over square root 3 tan 11 pi over 6 = negative square root 3 over 3 this is an example on my algebra 2 of the tangent of a point on the unit circle. can anyone tell me how negative 1 over the square root of 3 simplifies to negative square root of 3 over 3?

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

answer. d.289 hope that

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.