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Maslowich
3 years ago
7

Find the area of the figure. Round your answer to the nearest tenth

Mathematics
1 answer:
expeople1 [14]3 years ago
5 0

Answer:

4 in

Step-by-step explanation:

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The measures of four interior angles of a pentagon are 63º, 119º, 125º, and 116º. which is the measure of the fifth interior ang
san4es73 [151]
The sum of interior angles of a pentagon is given by:
(n-2)*180
where
n=number of sides
thus the sum of angles in a pentagon will be:
(5-2)*180
=540
hence to evaluate the missing angles we shall proceed as follows:
sum of the given angles is:
63+119+125+116
=423
thus the missing angle, x, will be:
x+423=540
x=540-423
x=117°

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3 years ago
SIMPLIFY<br> 3(4x+4)-4=?
kykrilka [37]
12x+8 is the answer
3 0
2 years ago
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Baseball player wants to hit a baseball over the fence the fence is 25 feet tall baseball players 52 feet away from the fence at
Igoryamba

Answer:

75 degree angle

heart if this helped :3

7 0
3 years ago
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

5 0
3 years ago
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Multiply or divide as indicated 10x^5/2x^2
shepuryov [24]
To perform the following operation, make sure to do the division parts separately and place them back together in the final answer.

10/2 and x^5/x^2

5 and using the exponential laws of division X ^m/X^n = X ^ m-n

The second part evaluates to X^3.

Putting the terms together we get the final answer, which is 5x^3.
3 0
3 years ago
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