Atriangle has sides with the following lengths: ab=3, bc=4, ca=5. which angle is the smallest?

A = 36.9°

Step-by-step explanation:

In this triangle we know the three sides:

AB = 3,

BC = 4 and

CA = 5.

Use The Law of Cosines first to find angle A first:

cos A = (BC² + CA² − AB²) / 2BCCA

cos A = (4² + 5² − 3²) / (2×4×5)

cos A = (16 + 25 − 9) / 40

cos A = 0.80

A = cos⁻¹(0.80)

A = 36.86989765°

A = 36.9° to one decimal place.

Next we will find another side. We use The Law of Cosines again, this time for angle B:

cos B = (CA² + AB² − BC²) / 2CAAB

cos B = (5² + 3² − 4²) / (2×5×3)

cos B = (25 + 9 − 16) / 30

cos B = 0.60

B = cos⁻¹(0.60)

B = 53.13010235°

B = 53.1° to one decimal place

Finally, we can find angle C by using 'angles of a triangle add to 180°:

C = 180° − 36.86989765° − 53.13010235°

C = 90°

Now we have completely solved the triangle i.e. we have found all its angles.

So we can analyze from above that the smallest angle in the triangle ABC is A with 36.9°.

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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