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3 years ago

Which of the following are integer solutions to the inequality below? -3 < x < 1

Mathematics

1 answer:

iris [78.8K]3 years ago

7 0

Answer:

are you trying to find the place on the number line?

if so, its a open circle starting at -3.0 then going to the right it stops at 1.0

PLEASE BE MORE DEFINED CUZ IM SUPER CONFUSED ILL EDIT MY ANSWER IF THIS IS WRONG BUT JUST TELL ME EXACTLY WHAT I NEEDA DO :)

<em>also i stan your profile picture uwu</em>

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3 years ago

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

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

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3 years ago