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postnew [5]
3 years ago
15

How can I get the correct answer 2 1/2+1 1/3+1/6

Mathematics
2 answers:
VashaNatasha [74]3 years ago
8 0

Answer:

\frac{43}{3}

Step-by-step explanation:

21/2 + 11/3 + 1/6

Least common multiple of 2 and 3 is 6. Convert \frac{21}{2} and \frac{11}{3} to fractions with denominator 6.

\frac{63}{6} + \frac{22}{6} + \frac{1}{6}

Since \frac{63}{6}  and \frac{22}{6} have the same denominator, add them by adding their numerators.

\frac{63 + 22}{6} + \frac{1}{6}

Add 63 and 22 to get 85.

\frac{85}{6} + \frac{1}{6}

Since \frac{85}{6} and \frac{1}{6} have the same denominator, add them by adding their numerators

\frac{85 + 1}{6}

Add 85 and 1 to get 86.

\frac{86}{6}

Reduce the fraction \frac{86}{6} to lowest terms by extracting and canceling out 2.

\frac{43}{3}

Nataliya [291]3 years ago
3 0

Answer:

2+1 equal to 4 than 3+1 equal to 4 then 21+6 equals to 27

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What is the upper bound of the function f(x)=4x4−2x3+x−5?
inessss [21]

Answer:

(no global maxima found)

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 4 x^4 - 2 x^3 + x - 5

Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.

Find the critical points of f(x):

Compute the critical points of 4 x^4 - 2 x^3 + x - 5

Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.

To find all critical points, first compute f'(x):

d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:

f'(x) = 16 x^3 - 6 x^2 + 1

Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.

Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:

x = -0.303504

Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.

f'(x) exists everywhere:

16 x^3 - 6 x^2 + 1 exists everywhere

Hint: | Collect results.

The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:

x = -0.303504

Hint: | Determine the endpoints of the domain of f(x).

The domain of 4 x^4 - 2 x^3 + x - 5 is R:

The endpoints of R are x = -∞ and ∞

Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.

Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-0.303504 | -5.21365

∞ | ∞

Hint: | Determine the largest and smallest values that f achieves at these points.

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-0.303504 | -5.21365 | global min

∞ | ∞ | global max

Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-0.303504 | -5.21365 | global min

Hint: | Summarize the results.

f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:

Answer: f(x) has a global minimum at x = -0.303504

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12/3=4 
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Answer:

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Answer:

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