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3 years ago

Round 14.561 to two decimal places

Mathematics

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

round 14.561 to two decimal places=15

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N is a positive integer

Murrr4er [49]

Part (1)

n is some positive integer. Let's say for now that n is even. So n = 2k, for some integer k

This means n-1 = 2k-1 is odd since subtracting 1 from an even number leads to an odd number.

Now multiply n with n-1 to get

n(n-1) = 2k(2k-1) = 2m

where m = k(2k-1) is an integer

The result 2m is even showing that n(n-1) is even

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Let's say that n is odd this time. That means n = 2k+1 for some integer k

And also n-1 = 2k+1-1 = 2k showing n-1 is even

Now multiply n and n-1

n(n-1) = (2k+1)(2k) = 2k(2k+1) = 2m

where m = k(2k+1) is an integer

We've shown that n(n-1) is even here as well.

------------

So overall, n(n-1) is even regardless if n is even or if n is odd.

Either n or n-1 will be even. If you multiply an even number with any number, the result will be even.

=======================================================

Part (2)

n is some positive integer

2n is always even since 2 is a factor of 2n

2n+1 is always odd because we're adding 1 to an even number. The sequence of integers goes even,odd,even,odd, etc and it does this forever.

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Another way to see how 2n+1 is odd is to divide 2n+1 over 2 and you'll find that we get (2n+1)/2 = 2n/2+1/2 = n+0.5

The 0.5 at the end is not an integer, so there's no way that (2n+1)/2 is an integer; therefore 2n+1 is odd.

6 0

3 years ago

In a dass of 32 students there are 11 males. Find the ratio of males to females

Levart [38]

$\huge{\boxed{11:21}}$

First, find the number of females in the class. $32-11=21$

Then, just write the ratio. $\text{11 males:21 females}=\boxed{11:21}$

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4 years ago

Which of the following is NOT equivalent to the nxn matrix A being invertible? A. The homogeneous system associated to A has a u

storchak [24]

Answer:

Option C

Step-by-step explanation:

For the matrix A of order $n\times n$ to be invertible, its determinant must not be equal to zero, |A| $\neq$ 0, $A^{- 1}$ exists if- AC = CA = I, where I is identity matrix.

The homogeneous equation with coefficient matrix A has a unique solution:

AB = 0, B = $A^{- 1}.0 = 0$

Thus, B = (0, 0, 0......., 0) is a unique solution

2. The non - homogeneous equation system with coefficient matrix A has a unique solution:

For an equation- AY = D

Y = $A^{- 1}.D$ is a unique solution

3. Every non homogeneous equation with coefficient matrix A is not consistent as:

For an equation- AY = D, has a solution.l Thus coefficient matrix is inconsistent whereas augmented matrix is.

4. Rank of matrix A = n, Thus the column space of A is $R^{n}$

5. Since, column space of A = $R^{n}$ , thus x→xA is one-to-one

5 0

3 years ago

. Solve the given equation <br>3m+7<br>____=13/6<br>5m-4

oee [108]

Answer is X=2
Merry Christmas

5 0

3 years ago

Read 2 more answers

What is the value of the expression 10 − ( fraction 1 over 2 )4 ⋅ 48? (1 point)

OlgaM077 [116]

Answer:

Just want to ask if it is 10 - 1/2 × 4 × 48 right? If the expression is this then the answer is -86.

Hope this helps, thank :) !!

6 0

3 years ago