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2 years ago

Find a polynomial with a degree of 3, and the zeros 1,7, & -4 with no other zeros.

Mathematics

1 answer:

jarptica [38.1K]2 years ago

5 0

Answer:

$f(x)=(x+4)(x-7)(x-1)$

Step-by-step explanation:

zeroes are when you plug in that x value and get y=0

therefore if you set x = that number and solve for 0 you get a "function"

multiplying these together will yield the complete 3rd degree polynomial

x = 1

x-1=0

x=7

x-7=0

x=-4

x+4=0

$f(x)=(x+4)(x-7)(x-1)$

If needed you could expand but I'm too lazy

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sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

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The diagonals are perpendicular

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$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

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$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

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The slope of EG is undefined hence it is a vertical line.

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$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

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The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

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3 years ago

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