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3 years ago

9

EXPERTS/ACE (and people that know it for sure comment) Brainiest Marked to the first correct answer

1 answer:

Goshia [24]3 years ago

6 0

I worked them out separately if I'm correct, then the order is:

x+5y=-2
x+5y=4 These are Inconsistent

y=3x+4
-2x+y=4 These are Consistent independent

3x+y=4
-6x-2y=-8 These are Coincident

Inconsistent lines=No solution
Coincident lines=exactly on top of each other
Consistent Independant= Solution, as they cross each other.

~Hope this helps!

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Consider the function f(x) = x^3 - 7x^2 - 2x + 14

mars1129 [50]

Answer:

There are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14$

To get the zeros of $f(x)$ , set $y$ or $f(x) =0$

so

$0=x^3-7x^2-2x+14$

as

$x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)$

so

$\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0$

Using the zero factor principle:

$\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0$

solving

$x-7=0$

$x=7$

$x+\sqrt{2}=0$

$x=-\sqrt{2}$

$x-\sqrt{2}=0$

$x=\sqrt{2}$

Therefore, there are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

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mash [69]

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3/3=1 and 7/8 is < 1 so 3/3 is bigger

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Oksi-84 [34.3K]

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