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3 years ago

What is one solution of the following system?

Mathematics

1 answer:

marta [7]3 years ago

8 0

Answer:

(-6, 0)

Step-by-step explanation:

Check out A): (-6,0). This satisfies 2y-2x=12: 2(0) - 2(-6) = 12, and also satisfies x^2+y^2=36: (-6)^2 + 0^2 = 36. (-6, 0) is therefore a solution of the given system.

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3 years ago

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

<u>System Of Linear Equations </u>

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago