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2 years ago

What is the answer to: -|4| - 9 =??

Mathematics

1 answer:

Delvig [45]2 years ago

8 0

$-|4| - 9 =\\
-4-9=\\
-13$

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What is 2.48 as a whole number

lina2011 [118]

ANSWER 2.48 as a whole number would be 2 because you would round to the nearest whole number. GIVE BRAINLIEST PLEASE

3 0

2 years ago

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Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

3 years ago

PLEASE HURRYmy question is in the picture

Andreyy89

The answer t our question is the first one
I just substituted he number one in every x
The original equation is equal to 147 when x equals to 1
And the first option is also equal to 147 when x equals to one

5 0

3 years ago

Read 2 more answers

Pls help asap due tomorrow

Svet_ta [14]

Based on the information given, it should be noted that the least number of marbles that can be in the bag will be 10 marbles.

<h2>Solution to the probability.</h2>

If the bag contains 60 marbles, the number of red marbles will be:

= 1/5 × 60

= 12 marbles.

The number of white marbles will be:

= 3/10 × 60

= 18 marbles.

The number of blue marbles will be:

= 60 - (12 + 18)

= 60 - 30

= 30 marbles.

Furthermore, when the bag has 4 red marbles and 6 white marbles, there'll be 10 blue marbles.

Lastly, to find the probability of choosing a blue marble, we've to multiply 1/2 by the number given.

Learn more about probability on:

brainly.com/question/25870256

4 0

2 years ago

Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?

ivanzaharov [21]

Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.

8 0

3 years ago