40 oz= 2.5 lb
One pound (lb) =16 Ounces
40/16=2.5
( f ∘ g ) ( x ) is equivalent to f ( g ( x ) ) . We solve this problem just as we solve f ( x ) . But since it asks us to find out f ( g ( x ) ) , in f ( x ) , each time we encounter x, we replace it with g ( x ) . In the above problem, f ( x ) = x + 3 . Therefore, f ( g ( x ) ) = g ( x ) + 3 . ⇒ ( f ∘ g ) ( x ) = 2 x − 7 + 3 ⇒ ( f ∘ g ) ( x ) = 2 x − 4 Basically, write the g ( x ) equation where you see the x in the f ( x ) equation. f ∘ g ( x ) = ( g ( x ) ) + 3 Replace g ( x ) with the equation f ∘ g ( x ) = ( 2 x − 7 ) + 3 f ∘ g ( x ) = 2 x − 7 + 3 we just took away the parentheses f ∘ g ( x ) = 2 x − 4 Because the − 7 + 3 = 4 This is it g ∘ f ( x ) would be the other way around g ∘ f ( x ) = 2 ( x + 3 ) − 7 now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them. g ∘ f ( x ) = 2 x + 6 − 7 Next, + 6 − 7 = − 1 g ∘ f ( x ) = 2 x − 1
15 = 3 parts
15 divided by 3 = 5
5 = 1 part
A calls for 5 parts
5 • 5 = 25
Answer: 25 Milliliters
Answer: if you have infinity points, which i will asume are the events, they cant have the same probability because then the probability will not be normalized, because in graph of prob vs variable, you will se infinite area under the curve if the probability is constant.
And yes, can all points have positive probability of occurring, but besides you medium value (the bell for example) you will see an asintotic decrease to the zero.
Answer:
2758 Nm
Step-by-step explanation
Work done usually depends on two things force applied and distance travelled due to applied force. In this current scenario, the force is being applied at an angle so we will have to find a component of force in the direction of the movement.
We usually find component using cos θ.
Here θ is 40°
Now, the modified equation becomes,
Work Done = Force * Distance * Component of force along the direction of distance
∴ Work = 30 N * 120 m * Cos 40°
⇒Work = 30 * 120 * 0.766
⇒Work = 2757.6 Nm
Rounding to the nearest whole number.
∴ The work done by force is 2758 Nm which is option B
