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3 years ago

6

A) 9x^2-1 B)9x2-6x+1 C)3x^2-1 D) 3x-1

1 answer:

gayaneshka [121]3 years ago

5 0

Answer:

b

Step-by-step explanation:

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Thirty students took a calculus exam. Of the students who passed the exam, 8 students earned an A, 9 earned a B, 6 earned a C, a

Alex73 [517]

Answer:

4/30 or 2/15

Step-by-step explanation:

total of student with either A,B,C, or D equals 26 which means 4 did not pass

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3 years ago

Find an m > 0 such that the the equation x^4−(3m+2)x^2+m^2=0 has four real solutions that form an arithmetic sequence.

Aleonysh [2.5K]

Answer:

The value of m is 6.

Step-by-step explanation:

Here, the given equation,

$x^4-(3m+2)x^2+m^2=0$

$x^4+0x^3-(3m+2)x^2+0x+m^2=0$

Let the roots of the equation are a-3b, a-b, a+b and a + 3b, ( they must be form an AP )

Thus, we can write,

$a-3b+a-b+a+b+a+3b=\frac{\text{coefficient of }x^3}{\text{coefficient of }x^4}$

$=\frac{0}{1}=0$

$\implies a=0----(1)$

$(-3b)(-b)+(-b)(b)+(b)(3b)+(3b)(-3b)+(-b)(3b)+(-3b)(b)=\frac{\text{coefficient of }x^2}{\text{coefficient of }x^4}}$

$=\frac{-3m-2}{1}$

$3b^2-b^2+3b^2-9b^2-3b^2-3b^2=-3m-2$

$-10b^2=-3m-2$

$\implies b^2=\frac{3m+2}{10}-----(2)$

$(-3b)(-b)(b)(3b)=\frac{\text{Constant term}}{\text{coefficient of}x^4}= m^2$

$9b^4=m^2$

$9(\frac{3m+2}{10})^2=m^2$

$9(\frac{9m^2+4+12m}{100})=m^2$

$81m^2+36+108m=100m^2$

$-19m^2+108m+36=0$

$19m^2-108m-36=0$

$19m^2-114m+6m-36=0$

$19m(m-6)+6(m-6)=0$

$(19m+6)(m-6)=0$

$\implies m=-\frac{6}{19}\text{ or }m=6$

But m > 0,

Hence, the value of m is 6.

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The answer to this question is 23

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Step-by-step explanation:

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