What is the value of 80-5(x^2-y)+y when x=4 and y=6
1 answer:
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Answer:
a. From the origin, move back 3 units and down 4 units
Step-by-step explanation:
Given:
The coordinates of a point in a 3-dimensional space is given as:
(-3, -4, 0)
The value of 'x' = -3
The value of 'y' = -4
The value of 'z' = 0
At the origin, the values of 'x', 'y', and 'z' are each equal to 0.
So, the coordinates of origin is (0, 0, 0). In order to reach the point (-3, -4, 0) from the point (0, 0, 0), we don't move along the z-axis as the 'z' value remains the same. We need to move along the 'x' and 'y' directions only.
Now, 'x' is equal to -3. Negative values of 'x' occur backwards from origin. So, we move back by 3 units. to reach the point (-3, 0).
Now, the value of 'y' is -4. Negative values of 'y' occur below the origin. So, we move down by 4 units to reach the point (-3, -4).
Therefore, the correct option is option (a)
A graph in 2 dimensional is shown in support of the answer.
