Answer:
2375^-3
Step-by-step explanation:
Move the decimal place from 2.375 to the back of the number so it should look like this, 2375.
Than move the decimal place back to its original spot and count the amount of spaces you move. Remember, if you move forward, your exponent should be negative. For example, if the problem is 6.321, you would move the decimal place to the back like this, 6321. And then move it back to where it originally was, 6.321, the amount of spaces I moved was 3 spaces and I moved forward so It would be 6.321^-3
Answer:
DB = CA (Proved)
Step-by-step explanation:
Statement 1.
∠D = ∠C, M is the midpoint of DC and ∠1 = ∠2
Reason 1.
Given
Statement 2.
Between Δ DBM and Δ CAM,
(i) DM = CM,
(ii) ∠D = ∠C and
(iii) ∠DMB = ∠CMA
Reason 2.
(i) given
(ii) given and
(iii) ∠ DMB = ∠1 + ∠AMB and ∠CMA = ∠2 + ∠AMB
Since ∠1 = ∠2, so, ∠DMB = ∠CMA.
Statement 3.
Δ DBM ≅ Δ CAM
Reason 3.
By angle-side-angle rule.
Statement 4.
DB = CA
Reason 4.
Corresponding sides of two congruent triangles. (Answer)
The answer is 333. put the 3 and the 2 in the correct spots. then multiply 8 and 2 which is 16 then 7 and 3 which is 21. then add which is 37. you would have to multiply 37 and 9 and you would get 333.
Answer:
![\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}a+e&b+f\\c+g&d+h\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2Be%26b%2Bf%5C%5Cc%2Bg%26d%2Bh%5Cend%7Barray%7D%5Cright%5D)
We have:
![A=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]\\and\\B=\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D)
And we need to express as a single matrix:
![A+B=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6+(-2)&-3+8\\10+3&-1+(-12)\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6-2&5\\13&-1-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=A%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%2B%28-2%29%26-3%2B8%5C%5C10%2B3%26-1%2B%28-12%29%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6-2%265%5C%5C13%26-1-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
The answer is:
It is expressed as a single matrix.
