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3 years ago

I am a trinomial whose leading coefficient is -3 when I'm put in standard form:

Mathematics

1 answer:

Nikolay [14]3 years ago

7 0

Answer:

C. 4x² - 3x⁴ + x

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Standard Form: Largest degree to smallest degree
Leading Coefficient: Constant multiplied to largest degree

Step-by-step explanation:

Step 1: Define

A. 2xy - 3 + 5xy²

B. 3x³ - 3x - 3

C. 4x² - 3x⁴ + x

D. x⁵y - 3x²y - xy

Step 2: Rewrite

Rewrite each trinomials into standard form.

A. 5xy² + 2xy - 3

B. 3x³ - 3x - 3

C. -3x⁴ + 4x² + x

D. x⁵y - 3x²y - xy

Step 3: Identify LC

A. has a leading coefficient of 5

B. has a leading coefficient of 3

C. has a leading coefficient of -3 (our answer)

D. has a leading coefficient of 1

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2 years ago

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nalin [4]

Answer:

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2.2x^2+15x+18

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Step-by-step explanation:

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3 years ago

Can someone help me with this :/

slavikrds [6]

Answer:

2x² + x - 1 = 0

a= 2 b= 1 c= -1

x = [ -b +- sqr rt( b^2 -4ac) ] / 2*a

x = -1 +- sqr rt (1 - 4 *2*-1) / 4

x1 = [ -1 + (sqr rt (1 +8)) / 4

x1 = ( -1 + sqr rt (9)) / 4

x1 = (-1 + 3) / 4

x1 = .5

x2 = (-1 -3) / 4

x2 = -4 / 4

x2 = -1

And to check that answer: http://www.1728.org/quadratc.htm

Step-by-step explanation:

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3 years ago

Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

antiseptic1488 [7]

Answer:

The coordinate of the relative maximum is x=4.

Step-by-step explanation:

Given that the derivative of the function $f(x)$ is $f'(x) = (x-4)(4-2x)$ , the maxima and minima or the critical points can be found where $f'(x)=0,$ that is:

$f'(x) = (x-4)(4-2x)=0.$

The solutions to this equation are $x=4$ and $x=2.$

Now, if the second derivative $f''(x)$ for a function $f(x)$ is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is $f''(x)$ negative. The second derivative is:

$\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).$

Now $f''(4)$ is $-4$ , and $f''(2)$ is $+4$ , therefore we deduce that the relative maxium is located at $x=4$ , because there the second derivative $f''(x)$ is negative.

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3 years ago