Answer:
Verified
Step-by-step explanation:
Let A matrix 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)
Then det(A) = ad - bc
Matrix A transposed would be in the form of:
![\left[\begin{array}{cc}a&c\\b&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26c%5C%5Cb%26d%5Cend%7Barray%7D%5Cright%5D)
Where we can also calculate its determinant:
det(AT) = ad - bc = det(A)
So the determinant of the nxn matrix is the same as its transposed version for 2x2 matrices
>
absolute 17 would be greater because the absolute signs cancel out the negatives
Let the faculties be X and the number of students be Y.
X/Y = 17/3
3X= 17Y
X=17Y/3
Let that be equation 1
We also know that X+Y = 740. Let it be equation 2
Substitute equation 1 in equation 2
(17Y/3)+Y= 740
20Y/3 = 740
Y=111
Since the total is 740, then X equal 740-111 =629.
The number of faculties is 111 and the number of students is 629.
Answer:
x = -3
Step-by-step explanation:
You can try the answers to see which works. Or, you can actually solve the equation for x.
... 7 - log2(x+5) = 6
... 1 - log2(x+5) = 0 . . . . subtract 6
... 1 = log2(x+5) . . . . . . . add log2(x+5)
... 2 = x +5 . . . . . . . . . . . take the antilog
... -3 = x . . . . . . . . . . . . . subtract 5
Answer:
∠ WZX = 50°
XW is not an altitude.
Step-by-step explanation:
16. See the attached figure.
XW is the angle bisector of ∠ YXZ, hence, ∠ WXY = ∠ WXZ
Now, given that ∠ YXZ = 8x + 34 and ∠ WXY = 10x - 13
Hence, ∠ YXZ = 2 ∠ WXY
⇒ 8x + 34 = 2(10x - 13)
⇒ 8x + 34 = 20x - 26
⇒ 12x = 60
⇒ x = 5.
Hence, ∠ XZY = ∠ WZX = 10x = 50° (Answer)
Now, ∠ WXZ = ∠ WXY = 10x - 13 = 37°
Hence, from Δ WXZ,
∠ WZX + ∠ WXZ + ∠ XWZ = 180°
⇒ 50° + 37° + ∠ XWZ = 180°
⇒ ∠ XWZ = 93° ≠ 90°
Hence, XW is not an altitude. (Answer)
