Could someone please explain how to solve for the area?

area of equilateral TRAINGLE =√3/4 A²=√3/4×4²=6.9mi² is your answer

9514 1404 393

  6.9 m²

Step-by-step explanation:

All of the sides are the same length. The altitude meets the base at its midpoint, so the altitude (h) of the triangle can be found using the Pythagorean theorem.

  h² + 2² = 4²

  h² = 12

  h = 2√3

Then the area is ...

  A = 1/2bh

  A = (1/2)(4 m)(2√3 m) = 4√3 m² ≈ 6.9 m²

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If you work through the above area formula in generic terms, you find the area of an equilateral triangle to be ...

  A = (s²/4)√3 . . . . . where s is the side length

Another approach is to treat the figure as a 3-gon and use the n-gon area formula that uses the perimeter. That formula is derived here

It is ...

  A = P²/(4n·tan(180°/n))

For a triangle with side length s, this is ...

  A = (3s)²/(4·3·tan(60°) = (s²/4)√3 . . . . same as above

4 + 5 7/9= 9 7/9

step-by-step explanation:

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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