The formula that is useful for solving both of these problems is ...
9. Given v₁=15, a=9.8, d=10, find v₂.
... (v₂)² = 15² + 2·9.8·10
... v₂ = √421 ≈ 20.5 . . . . m/s
10. Given d=12 m when a=-9.8 m/s² and v₂=0, find d when a=0.17·(-9.8 m/s²).
The formula tells us that d=(v₁)²/(2a), which is to say that the distance is inversely proportional to the acceleration. If acceleration is 0.17 times that on earth, distance will be 1/0.17 ≈ 5.88 times that on earth.
(12 m)/0.17 ≈ 70.6 m
The range of the following relation R{(3,-2), (1, 2), (-1, -4), (-1, 2)} is O{-1.1,3) -1,-1,1.3 01-4, 2, 2, 2] {-4, -2, 2
maw [93]
Answer:
The range is -2,2,-4
Step-by-step explanation:
hope this helps
Answer:
x = 127 degrees
Step-by-step explanation:
The sum of the interior angles is (n-2)180, so in this case it would be
(6-2)180
(4)180
720
Then you can substitute the given angle measurements to get the equation:
112+133+128+100+120+x = 720
593+x = 720
x = 127
Answer:
A) This is correct. The angles of <em>AEB </em>are the angles of <em>DEC. </em>This means that this is true.
B) This is incorrect. This triangle is NOT an isosceles triangle. The angles of an isosceles triangle are supposed to be about 36° for the top angle, and 72° for the other two angles. (At least this is what I was told)
Hope this helps :)
-wait. .___. they both might be true.
According to G oogle: "All the three angles situated within the isosceles triangle are acute, which signifies that the angles are less than 90°. The sum of three angles of an isosceles triangle is always 180°, which means we can find out the third angle of a triangle if the two angles of an isosceles triangle are known."
