2 years ago

Can Someone help me pleases. Marking brainiest

Mathematics

1 answer:

ELEN [110]2 years ago

6 0

Hope these notes help ! https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html

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PLEASE HELP FAST!!!!!!!! Which two values of x are roots of the polynomial below?

worty [1.4K]

Given a quadratic equation

$ax^2+bx+c=0$

you can find its solutions with the following formula:

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

You equation is identified by the coefficients

$x^2 - 3x + 5 \iff a = 1,\ b = -3,\ c = 5$

Which means the solutions are

$\dfrac{3 \pm \sqrt{9-4\cdot 5}}{2} = \dfrac{3 \pm \sqrt{-11}}{2}$

So, the answers are B and D

4 0

2 years ago

If ST bisects RSV, m RST =5x-3 and m VST=2x+15 then choose an equation that could be used to solve for x

Doss [256]

Answer:

x = 6

Step-by-step explanation:

$\because ST \: bisects\: \angle RSV\\\therefore m\angle RST = m\angle VST\\(by\: \angle \: bisector \: property) \\\therefore 5x - 3 = 2x + 15\\\therefore 5x - 2x = 15 +3\\\purple{\bold {\therefore 3x = 18}} \\\therefore x =\frac{18}{3}\\\huge \orange {\boxed {\therefore x = 6}}$