ASAP PLEASE help me with this question! This is really urgent! No nonsense answers please.

Explanation:

Let point A be the center of the circle.

Arc DCF is 316 degrees. Arc FG is 20 degrees. Subtracting the arc measures gives 316-20 = 296 which is the measure of arc DCG, and by extension, it is the measure of reflex angle DAG. The acute angle DAG is 360-296 = 64 degrees

Triangle DAG is isosceles with AD = AG, so the base angles are ADG and AGD. Let's call each of these angles x for now. The sum of the three angles add to 180

x+x+64 = 180

2x+64 = 180

2x = 180-64

2x = 116

x = 116/2

x = 58

Therefore angle ADG is 58 degrees. It adds to angle ADB = 90 to get 90+58 = 148, which is the measure of angle BDG.

Note: angle ADB is 90 because tangent BD forms a right angle with the radius AD.

Another note: angle ADG is half that of the arc it cuts off, so in a sense it's similar to an inscribed angle

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

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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