Which statement about bq←→ is correct?

bq←→ is a tangent line because m∠abq=90°.

bq←→ is a tangent line because △abq is not a right triangle.

bq←→ is not a tangent line because ∠baq is an acute angle.

bq←→ is not a tangent line because ∠bqa is an acute angle.

We have a circle A and a line segment BQ is passing through the circle A such that it forms a triangle ABQ.

From the diagram, we have some information about the angles of triangle ΔABQ as follows :-

angle QAB = 36 degrees and angle BAQ = 54 degrees.

We know about the Triangle Sum theorem which states "The sum of all interior angles of any triangle is 180 degrees".

We can use this theorem to find the third angle of the triangle ΔABQ.

∠ABQ + ∠BQA + ∠QAB = 180 degrees

∠ABQ + 54° + 36° = 180°

∠ABQ + 90° = 180°

∠ABQ = 180° - 90°

∠ABQ = 90 degrees

If any line makes a right angle with the radius of any circle, then this line must be a tangent to that circle.

So, BQ would be a tangent line because angle ∠ABQ = 90 degrees. Hence, option A is correct.

40 minutes

step-by-step explanation:

add the time and multiply by 2

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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