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2 years ago

Help I’m tired it’s 2:56am it’s due tomorrow morning !

Mathematics

1 answer:

guajiro [1.7K]2 years ago

8 0

Answer:

go to sleep just take the 0 and thinks for the points

Step-by-step explanation:

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Can you help me answer this MATHS question:<br> 4-x<9

aleksley [76]

Answer:

Step-by-step explanation:

4-x<9

-x < 5

x < -5

3 0

2 years ago

Read 2 more answers

In the fifth grade at Lenape Elementary School, there are

Arada [10]

About 48 students are girls

Step - By - Step Explanation:

63 / 4 = 15.75

15.75 x 3 = 47.25

Because you can’t have 0.25 (1/4) of a person, you round to the next number.

Hope this helps!

7 0

2 years ago

A box of 120 crayons contains variations of three colors: blue, red, and purple. The ratio of blue crayons to red crayons is 11:

Kaylis [27]

113 I’m pretty sure or 78 so H or G

4 0

3 years ago

Read 2 more answers

How to this and answer this ASAP i will give 15 points

aliya0001 [1]

Up 3 right 2 ?

left 2 up 2. down 1 right 2

8 0

2 years ago

Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17

V125BC [204]

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

$\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}$

Cramer's rule says that

$x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}$

where $x_i$ is the solution for i-th variable, and $\mathbf A_i$ is a modified version of $\mathbf A$ with its i-th column replaced by $\mathbf b$ .

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

$\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}$

$\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}$

$\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))$

$\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120$

The modified matrices and their determinants are

$\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240$

$\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360$

$\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480$

Then by Cramer's rule, the solution to the system is

$x = \dfrac{-240}{-120} \implies \boxed{x = 2}$

$y = \dfrac{360}{-120} \implies \boxed{y = -3}$

$z = \dfrac{-480}{-120} \implies \boxed{z = 4}$

5 0

1 year ago