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kirill115 [55]
2 years ago
14

Remove all perfect squares from inside the square root 25x^4

Mathematics
1 answer:
Serga [27]2 years ago
5 0
The answer:

the main rule:
for all value positive of a real  b,  √b²= b, and   √b*c=   √b * √c, if c is a real positive

let A=√25x^4,  we know that  25x^4=5² * x² * x² 
therefore,  
√25x^4= √5² * x² * x² =√5² √x² √x² = 5* x*x= 5x²

the final answer is A=√25x^4=5x²
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HELP ME PLEASEEEEEEEEEEE
Alborosie

Answer:

2

Step-by-step explanation:

3x2=6

s=2

Plz brainly and like :D

5 0
3 years ago
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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

5 0
3 years ago
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What is the equation of the line that is perpendicular to y=2x+4 and passes through the points (4 6)
Anuta_ua [19.1K]

Answer:

y=-0.5x+8

Step-by-step explanation:

Opposite reciprocal of 2: -0.5

6=-0.5*4+b

6=-2+b

b=8

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3 years ago
Explain how you can tell when a formula represents a length an area or a volume
AlekseyPX
Do you mean how you can tell between an area formula and a volume formula? Well usually like on a reference sheet it says the shape and the letter a or v, a representing area and v representing volume. I hope this helps, if it doesn't please tell me in the comment section
7 0
3 years ago
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More help :3 please and thnx
Mariana [72]

Answer:

a. there's a lot of options but here are a few: 1 and 5, 5 and 6, 2 and 1, 2 and 6

b. also a lot of options but here are a few: 1 and 6, 5 and 2, 3 and 8, 4 and 7

Step-by-step explanation:

supplementary angles are two angles that add up to 180 degrees, so essentially two angles that, combined, are equal to a straight line.

vertical angles are angles that are opposite each other when two lines cross.

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3 years ago
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