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castortr0y [4]
3 years ago
6

142 equals 176 minus f

Mathematics
2 answers:
natulia [17]3 years ago
8 0

Answer:

f=34

Step-by-step explanation:

vaieri [72.5K]3 years ago
4 0

Answer:

can you be more clear?

Step-by-step explanation:

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(-4)(5)+(-3)(-2) whats is it???
Vlada [557]

Answer:

-14

Step-by-step explanation:

(-4)(5)+(-3)(-2)

=-20+6

=-14

5 0
3 years ago
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Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

f(x) =  {x}^{5}  -  {x}^{3}  + 6

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

We plug this values into the original function to obtain the y-values of the turning points

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  +750)) \:and \:  (0, - 6) \: and\: (   \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( - 6 \sqrt{15}  +750))

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

f''(x) = 20 {x}^{3}  - 6x

f''( -  \frac{ \sqrt{15} }{5} ) \:   <  \: 0

Hence

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  + 750))

is a maximum point.

f''( \frac{ \sqrt{15} }{5} ) \:    >  \: 0

Hence

(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

is a minimum point.

f''(0) \: =\: 0

Hence (0,-6) is a point of inflexion

4 0
3 years ago
 PLEASE HELP!!!!!!
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The answer is forty five.
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The correct answer is C. 5, 6, -7
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