Given the trinomial 5x^2 - 2x- 3 predict the type of solutions

Mathematics

1 answer:

Firdavs [7]2 years ago

8 0

Answer:

One rational solution

Step-by-step explanation:

proof

5x2 - 2x - 3=0, the sum of coefficients is 5-2-3=0, so the equation has exactly two solutions, x=1, and x = -3/5 this last is the one rational solution

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A line has a slope of -13. There is a second line that passes through points (-7,-5) and (6,-4) . Would these lines be parallel

Elza [17]

<h3>Answer: The lines are perpendicular</h3>

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Work Shown:

Use the slope formula to find the slope of the line through the two given points.

m = (y2 - y1)/(x2 - x1)

m = (-4 - (-5))/(6 - (-7))

m = (-4 + 5)/(6 + 7)

m = 1/13

The slope of the line through the two given points is 1/13.

This is not equal to -13, which was the slope of the first line, so the two lines are not parallel.

However, the two lines are perpendicular because multiplying the two slope values leads to -1

(slope1)*(slope2) = (-13)*(1/13) = -1

note: It might help to think of -13 as -13/1 to help multiply the fractions.

3 0

2 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1