3 years ago

Find the area of a triangle with vertices (1, 5), (1, -2), (3,-2).

Mathematics

1 answer:

ollegr [7]3 years ago

3 0

Answer:

Step-by-step explanation:

A(3, - 2), B(1, 5), C(1, - 2)

ΔABC is right angled Δ

AC and BC are legs (cathetus)

$A_{ABC}$ = (AC)(BC) ÷ 2 = $\frac{2*7}{2}$ = 7 units²

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Give a third degree polynomial that has zeros of 13, 5i, and −5i, and has a value of −680 when x=3. Write the polynomial in stan

Vikki [24]

Answer:

$\dfrac{17x^3-221x^2+425x-5525}{9}\\$

Step-by-step explanation:

hello

we need to find something like, a being real

$a(x-13)(x-5i)(x+5i)$

with a so that

$a(3-13)(3-5i)(3+5i)=-680\\\\a(-10)(3^2-(5i)^2)= -10(9+25)a=-360a=-680\\ a = \dfrac{680}{360}=\dfrac{17*40}{9*40}=\dfrac{17}{9}$

so the third degree polynomial that we are looking for is

$\dfrac{17}{9}(x-13)(x-5i)(x+5i)\\=\dfrac{17}{9}(x-13)(x^2+25)\\\\=\dfrac{17}{9}(x^3+25x-13x^2-13*25)\\\\\\=\dfrac{17}{9}(x^3-13x^2+25x-325)\\\\\\\\=\dfrac{17}{9}x^3-\dfrac{17*13}{9}x^2+\dfrac{17*25}{9}x-\dfrac{17*325}{9}\\=\dfrac{17}{9}x^3-\dfrac{221}{9}x^2+\dfrac{425}{9}x-\dfrac{5525}{9}\\$