Answer:
Hello,
Answer 6
Step-by-step explanation:
The smallest number to be susbtracted is 6
Please give the right answer and show work if possible
2 answers:
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You can solve this by using "similar triangles".
In triangle ABC, we are looking for side AC which is x. Side AC is similar to side DF in triangle EDF.
You can solve for side x by picking two sides in triangle ABC and their corresponding sides in triangle EDF. This is what I mean:
Substitute for the values of AC, BC, DF and EF:
To solve for y, do the same thing. Pick two sides on triangle ABC and their corresponding sides in triangle DEF.
Substitute for the values and solve:
We have the value x to be 5.5 units and y to be 6 units.
In triangle ABC, we are looking for side AC which is x. Side AC is similar to side DF in triangle EDF.
You can solve for side x by picking two sides in triangle ABC and their corresponding sides in triangle EDF. This is what I mean:
Substitute for the values of AC, BC, DF and EF:
To solve for y, do the same thing. Pick two sides on triangle ABC and their corresponding sides in triangle DEF.
Substitute for the values and solve:
We have the value x to be 5.5 units and y to be 6 units.
Answer:
The constant force exerted on the ball by the wall is 119.68 N.
Step-by-step explanation:
Consider the provided information.
It is given that the mass of the ball is m = 2.2 kg
The initial velocity of the ball towards left is 7.4 m/s
So the momentum of the ball when it strikes is =
The final velocity of the ball is -6.2 m/s
So the momentum of the ball when it strikes back is =
Thus change in moment is:
The duration of force exerted on the ball t = 0.25 s
Therefore, the constant force exerted on the ball by the wall is:
Hence, the constant force exerted on the ball by the wall is 119.68 N.