What point is the projection of Q on RT????? A) R B) S C) T

Mathematics

2 answers:

user100 [1]3 years ago

7 0

Answer:

B) S

Step-by-step explanation:

The projection of Q is the joining point. The answer is B) S

S is the projection of Q on RT.

Hope this will helpful.

Thank you.

goblinko [34]3 years ago

3 0

The answer is S it’s the midpoint

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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: