What is true about the solution to the system of inequities shown?

Mathematics

1 answer:

SpyIntel [72]2 years ago

5 0

The true solution to the system of inequities shown is B; All values that satisfy $y \leq \dfrac{ 1}{3}x - 3$ are solutions.

<h3>What is a system of inequalities?</h3>

When mathematical expressions are compared, with non-strict equality, then such mathematical statements are called mathematical inequalities.

A collection of inequalities for which we consider a common solution for all inequalities is called a system of inequalities.

If the equation or inequality contains variable terms, then there might be some values of those variables for which that equation or inequality might be true.

A set of such values is called a solution set to the considered equation or inequality.

The true solution to the system of inequities shown is B;

All values that satisfy $y \leq \dfrac{ 1}{3}x - 3$ are solutions.

Learn more about solutions to the system of inequalities here:

brainly.com/question/16339562

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find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. Can someone please show me ho

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For a quadratic of the form $f(x)=ax^2+bx+c$ , we have the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$ ,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given $2x^2-16x+27=0$ , we can get that $a=2, \ b=-16$ , and $c=27$ . We substitute these numbers into the quadratic formula above.

$x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)}$

$x=\dfrac{16 \pm \sqrt{(256 -216)} }{4}$

$x=\dfrac{16 \pm \sqrt{40} }{4}$

$x=\dfrac{16 \pm 2\sqrt{10} }{4}$

$x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}$

This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the $\pm$ symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.