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Sliva [168]
3 years ago
12

What are the ordered pairs of the

Mathematics
1 answer:
9966 [12]3 years ago
5 0

Start by writing the system down, I will use y to represent f(x)

y=x^2-5x+3\wedge y=-3

Substitute the fact that y=-3 into the first equation to get,

-3=x^2-5x+3

Simplify into a quadratic form (ax^2+bx+c=0),

x^2-5x+6=0

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

x^2+x(m+n)+mn=0

which then must factor into

(x+m)(x+n)=0

And the solutions will be m,n.

Clearly for small coefficients like ours 5,6, this is very easy to figure out. To get 5 and 6 we simply say that m=3, n=2.

This fits the definition as 5=3+2 and 6=2\cdot3.

So as mentioned, solutions will equal to m=3, n=2 but these are just x-values in the solution pairs of a form (x,y).

To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.

So the solution pairs are (2,0) and (3,0).

Hope this helps :)

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I sent help but no one answered :(

Jk, I got you fam

12(t + 2) + 4 ≥ - 8

12t + 24 + 4 ≥ - 8

12t + 28       ≥ - 8

12t               ≥ - 8 - 28

12t               ≥ - 36

t                  ≥ - 36 ÷ 12

t                  ≥ - 3

Solution:

t ≥ -3

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