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il63 [147K]
2 years ago
14

Which Question is equivalent to

Mathematics
2 answers:
iragen [17]2 years ago
6 0

Answer:

\frac{3}{2}a^{2}|b^{3}|c^{4}

Step-by-step explanation:

Setler [38]2 years ago
4 0

Answer:

option. b 3/2a²|b³|c⁴

..............

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The shape is a rhombus if and only if the diagonals are perpendicular and the sides are congruent
harina [27]

Answer:

The statement is True

Step-by-step explanation:

Rhombus is a quadrilateral with the following characteristics;

  • All sides are congruent by definition.
  • The diagonals bisect the angles.
  • The diagonals are perpendicular bisectors of each other.
  • Adjacent angles are supplementary.
  • All the four sides are equal.
5 0
3 years ago
Just need Answers on my late math homework
riadik2000 [5.3K]

Answer:

Step-by-step explanation:

Where's the problem/question?

8 0
3 years ago
Find the slope line (- 6, 1), (4, 8)
AVprozaik [17]

Answer:

The slope is \frac{7}{10}

Step-by-step explanation:

☆Remember:

m =  \frac{y_2 - y_1}{x_2 - x_1}

☆We just plug in now!

m =  \frac{8 - 1}{4 - ( - 6)}  =  \frac{7}{10}

6 0
3 years ago
The graph of f(x) = x2 is translated to form g(x) = (x – 5)2 + 1. Which graph represents g(x)?
levacccp [35]
<span>The graph would be translated 5 units right and 1 unit up, giving an upward facing parabola with a vertex at (5, 1).

Explanation:
Since 5 was subtracted from x before it was squared, this means a horizontal translation 5 units. Since it was subtracted, this means it was translated right 5 units.

The 1 added at the end means it was translated 1 unit up as well.

This is in vertex form, y=a(x-h)^2 + k, where (h, k) is the vertex; h corresponds with 5 and k corresponds with 1, so the vertex is at (5, 1).</span>
6 0
3 years ago
Read 2 more answers
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
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