Select the appropriate values of x that are solutions to f (x)=0, where f (x) = -7x^2 +4x+9

1 answer:

7 0

$-7x^2 +4x+9=0\\
\Delta=4^2-4\cdot(-7)\cdot9=16+252=268\\
\sqrt{\Delta}=\sqrt{268}=2\sqrt{67}\\
x_1=\frac{-4-2\sqrt{67}}{2\cdot(-7)}=\frac{-4-2\sqrt{67}}{-14}=\frac{2+\sqrt{67}}{7}\approx1.46\\
x_2=\frac{-4+2\sqrt{67}}{2\cdot(-7)}=\frac{-4+2\sqrt{67}}{-14}=\frac{2-\sqrt{67}}{7}\approx-0.88\\$

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5=1/3(2x+12)<br><br> Solve for X in simplest form

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A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

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Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

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Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$