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A baby otter was born 3/4 of a month early. At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average

Assoli18 [71]

Answer:

The average weight of new born otter was, $1\frac{31}{40}$

Step-by-step explanation:

Let average weight of new born otter be x.

As per the given statement: At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average weight of a new born otter in the aquarium

" $\frac{9}{10}$ kg less than average weight of a new born otter" means $x-\frac{9}{10}$

As per the given information, we have;

$\frac{7}{8} = x -\frac{9}{10}$

Add $\frac{9}{10}$ both sides, we have;

$\frac{7}{8} + \frac{9}{10} = x$

Take LCM of 8 and 10 is, 40

⇒ $\frac{35+36}{40}=x$

Simplify:

$x = \frac{71}{40} = 1\frac{31}{40}$

Therefore, the average weight of new born otter was, $1\frac{31}{40}$

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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

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