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Phantasy [73]
3 years ago
6

Round 8.923 to the nearest half.

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
8 0

Answer:

answer is : 8.9

Furkat [3]3 years ago
6 0

Answer:

8.923 to the nearest half is 8.9

Step-by-step explanation:

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Could a set of three vectors in ℝ^4 span all ℝ^4? Explain. What about n vectors in ℝ^m when n is less than m?
Elena-2011 [213]

Answer:

Check the ecplanation

Step-by-step explanation:

A set of three vectors in R^{4} represents a matrix of 3 column vectors, and each vector containing 4 entries (that is, a matrix of 4 rows, and 3 columns).

Let A be that 4x 3 matrix. The columns of A span R^{m}. if and only if A has a pivot position in each row. So, there are at most 3 pivot positions in the matrix A, but the number of rows is 4, therefore, there exist at least one row not having a pivot position. If A does not have a pivot position in at least one row, then the columns of A do not span R^{4}. It implies that the set of 3 vectors of A does not span all of R^{4}.

In general, the set of n vectors in R^{m} represents a matrix of in rows, and n columns (an in x matrix). So, there are at most n pivot positions in the matrix A, but n is less than the number of rows. In therefore, there exist at least one row that does not contain a pivot position.

And, hence the set of n vectors of A does not span all of R^{m}. for n < m

8 0
3 years ago
Your subtotal at the store was $58.89. You have a 30% off coupon. If the sales tax rate is 5.5%, what is the total price for you
Mars2501 [29]

Answer:

$43.48

Step-by-step explanation:

58.89x.30=17.68

58.89-17.68=41.21

41.21x.055=2.27

41.21+2.27= $43.48

4 0
3 years ago
Read 2 more answers
Joel has pans of brownies. The length of each brownie pan is 7 cm and the width is 9 cm. Find the area of the brownie pan.
borishaifa [10]

Answer: 63 cm squared

Step-by-step explanation:

9*7=63

3 0
3 years ago
Read 2 more answers
Can someone help me
erik [133]

Answer:

   6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

  (2x -12)/2 = arcsin(16/22×sin(24°))

  x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

  2x -12 = 0

  x -6 = 0 . . . . . divide by 2

  x = 6 . . . . . . . add 6

Then the range of values of x will be ...

  6 < x < 23.206

_____

* One way to do this is to make use of the law of cosines:

  22² = AB² + AC² -2·AB·AC·cos(48°)

  16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

  22² = 2·AB²·(1-cos(48°))

  16² = 2·AB²·(1 -cos(2x-12))

The value of AB drops out of the ratio of these equations, and the result for x is as above.

4 0
3 years ago
Read 2 more answers
Someone please help??
Ksju [112]

Answer:

hey

Step-by-step explanation:

4 0
3 years ago
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