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3 years ago

What is 12.345 rounded to the nearest tenth

2 answers:

7 0

Rounding to the nearest tenth means rounding to the decimal place directly to the right of the decimal.

The rules of rounding are fairly simple. If the number to the right of the number you are rounding is greater than or equal to 5, you round up. If the number to the right is less than 5, you round down.

Let's take a look at the number 12.345.
First, round to the hundredths place. The number right before the 4 is 5, so we round up. This gives us the number 12.35.
Next, we have to round to the tenths place. The number to the right of the 3 is 5, so we round up as well. This gives us the number 12.4.

Answer: 12.4

horrorfan [7]3 years ago

5 0

The answer to your question 12.3 because it the number the the hundredth place is smaller the 5 you round down

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3 years ago

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3 years ago

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2 years ago

This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

4 years ago

Suppose that the functions r and s are defined for all real numbers x as follows.

mixas84 [53]

ANSWER

Given

$r(x) = x - 1$

and

$s(x) = 3 {x}^{2}$
ANSWER TO QUESTION 1

$(r + s)(x) = r(x) + s(x)$

$(r + s)(x) = x - 1 + 3 {x}^{2}$

$(r + s)(x) = 3 {x}^{2} + x - 1$

ANSWER TO QUESTION 2

$(r \times s)(x) = r(x) \times s(x)$

$(r \times s)(x) = (x - 1) \times 3 {x}^{2}$

$(r \times s)(x) = 3 {x}^{3} - 3 {x}^{2}$

ANSWER TO QUESTION 3

$( r - s)(x) = r(x) - s(x)$

$( r - s)(x) = x - 1 - 3 {x}^{2}$

$( r - s)( - 3) = (- 3) - 1 - 3 ({ - 3}^{2} )$

$( r - s)( - 3) = - 4 - 3 \times 9$

$( r - s)( - 3) = - 4 - 27 = - 31$

4 0

3 years ago