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Nata [24]
3 years ago
10

1 3/10 x 18 please show work and answer

Mathematics
1 answer:
klasskru [66]3 years ago
7 0

Answer:

Step-by-step explanation:

1 3/10 x 18

(1 x 10 +3)/10 = 13/10

13/10 x 18 = (13 x 18) / 10 = 234/10 = (234/2) / (10/2) = 117/5 = 23 2/5 (a fraction is a division)

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For the function, find all critical numbers and then use the second-derivative test to determine whether the function has a rela
Serhud [2]

Answer:

relative maximum: x = 1

relative minimum: x = 7

Step-by-step explanation:

Critical points:

Values of x for which f'(x) = 0.

Second derivative test:

For a critical point, if f''(x) > 0, the critical point is a relative minimum.

Otherwise, if f''(x) < 0, the critical point is a relative maximum.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

f(x) = x^{3} - 12x^{2} + 21x - 8

Finding the critical points:

f'(x) = 3x^{2} - 24x + 21

3x^{2} - 24x + 21 = 0

Simplifying by 3

x^{2} - 8x + 7 = 0

So a = 1, b = -8, c = 7

\bigtriangleup = (-8)^{2} - 4*1*7 = 36

x_{1} = \frac{-(-8) + \sqrt{36}}{2} = 7

x_{2} = \frac{-(-8) - \sqrt{36}}{2} = 1

Second derivative test:

The critical points are x = 1 and x = 7.

The second derivative is:

f''(x) = 6x - 24

f''(1) = 6*1 - 24 = -18

Since f''(1) < 0, at x = 1 there is a relative maximum.

f''(7) = 6*7 - 24 = 18

Since f''(x) > 0, at x = 7 there is a relative minumum.

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MATH HELP!!!!!!!!!!!!!!!!!!!!!!!!1
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