Please Help Me!!!!!!!!!!!!!!!

Mathematics

2 answers:

nalin [4]2 years ago

8 0

Answer:

s = 11

Step-by-step explanation:

$s-6\geq 5\\\\s-6+6\geq 5+6\\\\\boxed{s\geq 11}$

Any number greater than or equal to 11 would be a solution for the inequality given.

s = 11 is a solution.

Hope this helps.

Semmy [17]2 years ago

3 0

Answer:

s = 11

Step-by-step explanation:

I think this is it because 11 - 6 = 5 this might not be right but i think it is.

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$\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

For the given function:

$f(x)=2x^2-10x-3$ $x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}$ $x=\frac{10\pm\sqrt[]{100+24}}{4}$ $\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}$ $\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}$ $\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Then, the zeros of the given quadratic function are: