Answer:
The determinant is 15.
Step-by-step explanation:
You need to calculate the determinant of the given matrix.
1. Subtract column 3 multiplied by 3 from column 1 (C1=C1−(3)C3):
![\left[\begin{array}{ccc}-25&-23&9\\0&3&1\\-5&5&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%26-23%269%5C%5C0%263%261%5C%5C-5%265%263%5Cend%7Barray%7D%5Cright%5D)
2. Subtract column 3 multiplied by 3 from column 2 (C2=C2−(3)C3):
![\left[\begin{array}{ccc}-25&-23&9\\0&0&1\\-5&-4&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%26-23%269%5C%5C0%260%261%5C%5C-5%26-4%263%5Cend%7Barray%7D%5Cright%5D)
3. Expand along the row 2: (See attached picture).
We get that the answer is 15. The determinant is 15.
-1.5 ------ f(x) = 3(-1.5) - 5
= -4.5 -5
= 9.5
2 -------- f(x) = 3(2) - 5
= 6 -5
= 1
4 -------- f(x) = 3(4) - 5
= 12 - 5
= 7
(9.5, 1 , 7)
9514 1404 393
Answer:
neither; they are the same distance from the line
Step-by-step explanation:
The distance of point (x, y) from line ax+by+c=0 is given by ...
d = |ax +by +c|/(√a² +b²)
Since we're using the same line for both points, we only need to evaluate ...
d = |ax +by +c|
Subtracting y can put the equation in the desired form:
1/2x -y +4 = 0
__
For point A, the expression of interest evaluates to ...
|(1/2)(-2) -(6) +4| = |-1-6+4| = 3
For point B, the expression of interest evaluates to ...
|(1/2)(4) -(3) +4| = |2-3+4| = 3
These values tell us the points are the same distance from the line.
_____
The attached graph shows that circles of the same radius centered on A and B are both tangent to the line. Hence the distances are the same.
Answer:
28
Step-by-step explanation:
The partial circles in each corner mean that all 3 angles are identical, which means this is an equatorial triangle. In an equatorial triangle all three sides are the same.
We can set two of the equations to equal each other and solve for x:
3x -5 = 2x +20
Add 5 to each side:
3x = 2x +25
Subtract 2x from each side:
x = 25