Please help me! this is important.

What is being distributed in the following problem?<br> 5 - 31d + 7)

USPshnik [31]

Answer:

the -3 is being distributed. --> the 5 stays outside till the end. multiply the 3 inside first, then the negative. and then add the 5 to the final answer

5-3(d+21) = -->> 5-(3d+21)=--->>>5-3d-21=3d-16

Step-by-step explanation:

5-3(d+21) =

5-(3d+21)=

5-3d-21=

ans=3d-16

7 0

3 years ago

this isn’t a question but i just wanted to tell everyone you all are doing great. this year has been hard and everyone is workin

BlackZzzverrR [31]

Thank you so much friend.

I think a advantage of this will be that we will be able to tell our grandchildren about how we survived the pandemic, just like our grandparents talk about World War. Lol.

3 0

3 years ago

Read 2 more answers

Help please!!!!!!!!!

In-s [12.5K]

ANSWER

24

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}$

so $C_{11} = M_{11} = 24$

4 0

3 years ago

?????????????????????

valentinak56 [21]

Step-by-step explanation:

$\huge \: 2 \sqrt{x} .4 {x}^{ - \frac{5}{2} } \\ \\ \huge \: = 2.4 {x}^{ \frac{1}{2} } . {x}^{ - \frac{5}{2} } \\ \\ \huge \: = 8 {x}^{ \frac{1}{2} - \frac{5}{2} } \\ \\ \huge \: = 8 {x}^{ \frac{1 - 5}{2} } \\ \\ \huge \: = 8 {x}^{ - \frac{4}{2} } \\ \\ \huge \: = 8. {x}^{ - 2} \\$

4 0

3 years ago

Trina has two brothers. one brother is 7 years older than trina and the other brother is 7 years younger than trina. the product

AlladinOne [14]

(a) 95 = (x+7)*(x-7)

(b) 95 = (x+7)*(x-7)
Expand the brackets:
95 = x² -7x + 7x -49 = x² - 49
Subtract x² from both sides:
95-x² = -49
Subtract 95 from both sides:
-x²=-49-95
-x²= -144
Divide both sides by -1 to get rid of the negative numbers:
-x²/-1=-144/-1
x²=144
Take the square root of 144:
x = ± squareroot of 144
x = 12 or -12
x = 12 because you cannot have a negative age so Trina is 12 years old.

7 0

3 years ago