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exis [7]
3 years ago
7

Which solution for the value of x in the figure below is incorrect? explain

Mathematics
1 answer:
ruslelena [56]3 years ago
4 0
B is wrong because they are not supplementary. they a alternate interior angels so they are equal
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Please help me! Thank you!
Mariana [72]
The correct answer is D
4 0
3 years ago
I need help which triangle theorem do I use for these
irina [24]

Answer:

Step-by-step explanation: B. SSS because it shows us three sides ,C.RHS because it has a right angle, d. AAS because it shows us two sides and a non included side,  e. RHS because it has a right angle,  f. AAS because it has two angles and non included side.

4 0
3 years ago
Write 146,000 in scientific notation.
MrMuchimi

Answer:

<h2>1.46 x 10^5</h2>

Step-by-step explanation:

4 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
Read 2 more answers
The row-echelon form of the augmented matrix of a system of equations is given. Find the solution of the system. FYI it’s not b
sleet_krkn [62]

Answer:

c.

Step-by-step explanation:

From the last row in the matrix z = 1.

From the second row:

y + 5z = -4

y = -4 -5(1) = -9.

From the first row:

x + -1 = -3

x = - 3 + 1

x = -2.

8 0
3 years ago
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