1) The value of 6s in 36,621 are 6,000 and 600.
first 6 is in the thousands place value while the second 6 is in the hundreds place value.
6 x 1000 = 6,000 and 6 x 100 = 600
2) The first 4 is ten times greater than the second 4 is shown in 354,426
the first 4 is in the thousands place value while the second 4 is in the hundreds place value
4 x 1000 = 4000 and 4 x 100 = 400
based on the question the first 4, which is 4000, is ten times greater than the second 4, which is 400.
we can check if this statement is true either by dividing 4000 by ten or multiplying 400 by ten.
4000/10 = 400 ; 400 x 10 = 4000. therefor, out of the 4 numbers given, the correct one is 354,426
3) The relationship of the two 5s in 552 is shown in 500 ÷ 50 = 10
based on its place value, the first five is in the hundreds place and the second five is in the tens place.
500 ÷ 50 = 10. This equation states that 500 is ten times greater than 50.
4) In 2,077 the first 7 is greater than the second 7 by 10 times.
using place value, first 7 is in the tens place while second 7 is in the ones place.
first 7 is 70; second 7 is 7. we divide 70 by 7 to know how many number of times 70 is greater than 7.
70 ÷ 7 = 10 times.
Equation 2 is the only linear one due to the fact that there is no exponent on the x or y values.
-3|x+4|+16=7
-3|x+4|=7-16
-3|x+4|=-9
|x+4|=3
x+4=3
x+4=-3
x=-1
x=-7
X1=-7 X2=-1
Answer:
Verified
Step-by-step explanation:
Let the 2x2 matrix A be in the form of:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
Where det(A) = ad - bc # 0 so A is nonsingular:
Then the transposed version of A is
![A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26c%5C%5Cb%26d%5Cend%7Barray%7D%5Cright%5D)
Then the inverted version of transposed A is
![(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5ET%29%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20cb%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
The inverted version of A is:
![A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-b%5C%5C-c%26d%5Cend%7Barray%7D%5Cright%5D)
The transposed version of inverted A is:
![(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5E%7B-1%7D%29%5ET%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
We can see that
So this theorem is true for 2 x 2 matrices
→ 3 ≠ 1 → A is eliminated.
→ 5 ≠
→ B is eliminated.
→ 2 = 2 → C satisfies the equation.
→ 10 ≠
→ D is eliminated.
→ -1 = -1 → E satisfies the equation.
→ 3 = 3 → F satisfies the equation.