Given:
Point B has coordinates (4,1).
The x-coordinate of point A is -4.
The distance between point A and point B is 10 units.
To find:
The possible coordinates of point A.
Solution:
Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).
Distance formula:
The distance between point A and point B is 10 units.
Taking square on both sides, we get
Taking square root on both sides, we get
and 
and 
Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).
Answer:
t ≤ 15
Step-by-step explanation:
One third of 15 is equal to five (1/3 × 15 = 5)
One third of t is lesser than or equal to five (1/3 × 5 ≤ 5)
<em>So, t cannot be more than 15.</em><em> </em><em>It</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>1</em><em>5</em><em>,</em><em> </em><em>though</em><em>.</em>
∴t ≤ 15
Answer: i may be wrong but is it 4?
2x2x5x5 or 2^2*5^5 hope this helps!
Answer:
d = 234.6 m
Step-by-step explanation:
You can consider a system of coordinates with its origin at the beginning of the walk of the student.
When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:
she is at the coordinates (52.97 , 28.16)m.
Next, when she walks 180m to the east, her coordinates are:
(52.97+180 , 28.16)m = (232.97 , 28.16)m
To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:
The displacement of the student on her complete trajectory was of 234.6m
