Answer:
x = -44/13
y = -65/13
Step-by-step explanation:
Using matrix form means using the crammers rule
The matrix form of the expression is written as;
![\left[\begin{array}{ccc}8&5\\-1&1\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}9\\7\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%265%5C%5C-1%261%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%5C%5C7%5C%5C%5Cend%7Barray%7D%5Cright%5D)
AX = B
taking the determinant of A;
|A| = 8(1) - 5(-1)
|A| = 8 + 5
|A| = 13
After replacing the first row with the column matrix;
![A_x =\left[\begin{array}{ccc}9&5\\7&-1\\\end{array}\right]](https://tex.z-dn.net/?f=A_x%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%265%5C%5C7%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
|Ax| = 9(-1)-5(7)
||Ax| = -9 - 35
|Ax| = -44
x = |Ax|/|A|
x = -44/13
similarly for y
![A_x =\left[\begin{array}{ccc}8&9\\-1&7\\\end{array}\right]](https://tex.z-dn.net/?f=A_x%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%269%5C%5C-1%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
|Ay| = 8(7)+9
|Ay| = 56+9
|Ay| = 65
y = |Ay|/|A|
y = -65/13
Answer:
A; 4,6,8
Step-by-step explanation:
For the first question, simply make a right angle triangle with the following dimensions.
The angle of elevation or the angle from the ground between the first floor and ground is 60 degrees.
The height between the 2 floors is 26 feet.
Solve for the hypotenuse, or the diagonal length between the first and second floors.
Then for the second question, divide the length in feet by the rate to find the time it takes for the object to move.
Answer:
The answer in the procedure
Step-by-step explanation:
we know that
The rule of the reflection of a point over the y-axis is equal to
A(x,y) ----->A'(-x,y)
That means -----> The x-coordinate of the image is equal to the x-coordinate of the pre-image multiplied by -1 and the y-coordinate of both points (pre-image and image) is the same
so
A(3,-1) ------> A'(-3,-1)
The distance from A to the y-axis is equal to the distance from A' to the y-axis (is equidistant)
therefore
To reflect a point over the y-axis
Construct a line from A perpendicular to the y-axis, determine the distance from A to the y-axis along this perpendicular line, find a new point on the other side of the y-axis that is equidistant from the y-axis
Answer:
4h-7
Step-by-step explanation:
Step by Step Solution
STEP1:Equation at the end of step 1 (1 - 2h) + 2 • (3h - 4) STEP2:
Final result :
4h - 7
