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2 years ago

What is the leading coefficient of the polynomial function? y=-x^4+4x^2

Mathematics

1 answer:

attashe74 [19]2 years ago

6 0

<h3>Answer: -1</h3>

Explanation:

The given equation is the same as y = -1x^4+4x^2

The leading term is the term with the largest exponent, so it's -1x^4

The leading coefficient is the coefficient of the leading term.

In short, we circle the first coefficient we see. This is assuming that the polynomial is in standard form where the exponents decrease when going from left to right.

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Solve this equation.

Anika [276]

Answer:

Answer: b = 58/5 or 11 3/5 or 11.6 decimal

Step-by-step explanation:

Solve for b:

b - (3 + 2/5) = 8 + 1/5

Put 3 + 2/5 over the common denominator 5. 3 + 2/5 = (5×3)/5 + 2/5:

b - (5×3)/5 + 2/5 = 8 + 1/5

5×3 = 15:

b - (15/5 + 2/5) = 8 + 1/5

15/5 + 2/5 = (15 + 2)/5:

b - (15 + 2)/5 = 8 + 1/5

15 + 2 = 17:

b - 17/5 = 8 + 1/5

Put each term in b - 17/5 over the common denominator 5: b - 17/5 = (5 b)/5 - 17/5:

(5 b)/5 - 17/5 = 8 + 1/5

(5 b)/5 - 17/5 = (5 b - 17)/5:

(5 b - 17)/5 = 8 + 1/5

Put 8 + 1/5 over the common denominator 5. 8 + 1/5 = (5×8)/5 + 1/5:

(5 b - 17)/5 = (5×8)/5 + 1/5

5×8 = 40:

(5 b - 17)/5 = 40/5 + 1/5

40/5 + 1/5 = (40 + 1)/5:

(5 b - 17)/5 = (40 + 1)/5

40 + 1 = 41:

(5 b - 17)/5 = 41/5

Multiply both sides of (5 b - 17)/5 = 41/5 by 5:

(5 b - 17)/(5 1/5) = 1/5×1/(1/5) 41

1/5×1/(1/5) = 1:

5 b - 17 = 1/5×1/(1/5) 41

1/5×1/(1/5) = 1:

5 b - 17 = 41

Add 17 to both sides:

5 b + (17 - 17) = 17 + 41

17 - 17 = 0:

5 b = 41 + 17

41 + 17 = 58:

5 b = 58

Divide both sides of 5 b = 58 by 5:

(5 b)/5 = 58/5

5/5 = 1:

Answer: b = 58/5 or 11 3/5 or 11.6 decimal

5 0

4 years ago

Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$