Answer:
i think its A but not %100
Step-by-step explanation:
Answer:
what idea and experience do you get after being lockdown for long three or four months due to corona virus. write your experience in the form of article that is published in school magazine on the topic lack of physical activity.
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
Hello!
This expression can be broken down into 3 separate parts:
1. "the quotient of a number and (-7)"
2. "decreased by 2"
3. "is 10"
We'll begin with the first part. The word "quotient" implies the result of a
division problem. Therefore, this part can be represented by the following expression (let "x" represent the unknown number):

Now we'll move on to the second part. If something is decreased by a value of 2, we know that 2 is being
subtracted. Therefore, this part can be represented by the following expression:
– 2
Now we'll move on to the third and final part. The phrase "is 10" implies that the operations preceding this part are
equal to 10. Therefore, this part can be represented by the following expression:
= 10
Finally, put the three parts together to create the following equation:

– 2 = 10
The equation has now been fully translated. If, however, you are required to find the unknown value (x), begin by adding 2 to both sides of the equation:

= 12
Now multiply both sides by (-7):
x = (-84)
We have now proven that
the unknown number is equal to (-84).I hope this helps!
Answer:
Height of the fighter plane =1.5km=1500 m
Speed of the fighter plane, v=720km/h=200 m/s
Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, u=600 m/s
Time taken by the shell to hit the plane =t
Horizontal distance travelled by the shell =u
x
t
Distance travelled by the plane =vt
The shell hits the plane. Hence, these two distances must be equal.
u
x
t=vt
u Sin θ=v
Sin θ=v/u
=200/600=1/3=0.33
θ=Sin
−1
(0.33)=19.50
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.
H
max
=u
2
sin
2
(90−θ)/2g=600
2
/(2×10)=16km