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ASHA 777 [7]
3 years ago
9

HURRY! What are all the possible rational solutions for x^4-3x^2+2x5=0

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer: x = 1

x =(-3-√-15)/4=(-3-i√ 15 )/4= -0.7500-0.9682i

x =(-3+√-15)/4=(-3+i√ 15 )/4= -0.7500+0.9682i

x2 = 0

Step-by-step explanation:

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Find the nonpermissible replacement for 'n' in this expression (-8)/(5n) 
seraphim [82]
In any rational function the denominator should never equal to 0, hence in this case since the denominator is 5n the only answer is n should not equal 0 
7 0
3 years ago
Read 2 more answers
What are the possible rational roots of the polynomial equation?<br><br> 0=2x7+3x5−9x2+6
RoseWind [281]

Answer: \pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0

Where:

- The coefficients are integers.

- a_n is the leading coeffcient (a_n\neq 0)

- a_0 is the constant term a_0\neq 0

Every rational root of the polynomial is in the form:

\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}

For the case of the given polynomial:

2x^7+3x^5-9x^2+6=0

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

 Then, by Rational Roots Test we get the possible rational roots of this polynomial:

\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}

5 0
3 years ago
What is the equation of a line that is parallel to the line 2x + 5y = 10 and passes through the point (–5, 1)? Check all that ap
Ann [662]

Answer:

2x + 5y = -5

Step-by-step explanation:

Since the new line is parallel to the given line 2x + 5y = 10, the equation of the new line has exactly the same form as does 2x + 5y = 10, except that the constant will be different.

Were we to solve this equation (in standard form) for y in slope-intercept form, we'd get:

5y = -2x + 10, or

        -2x + 10

y = ----------------

              5

or

     

y =  (-2/5)x + 2

             

Writing out 2x + 5y = C, we substitute -5 for x and 1 for y, obtaining

2x + 5y = C  =>  2(-5) + 5(1) = -10 + 5 = -5.  Therefore, C = -5 and the equation of the new line is

2x + 5y = -5

6 0
3 years ago
Mr. Jones asked his students to classify the solution(s) to the quadratic equation x^2=24.
bija089 [108]
Qqqqqqqqooejeudmdjdmdndlsbelqbadmd
It’s C
5 0
3 years ago
I need help, Half of the stuff on this test we haven’t learned
iragen [17]

Answers:

1.) 1 and 1/2

2.) 1/2

3 0
3 years ago
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