Answer:
a
Step-by-step explanation:
Hello AljakeVillena!

You pull horizontally on a 50-kg crate with a force of 450 n and the friction force on the crate is 250 n. the acceleration of the crate is ?

Given,
- mass of crate = 50 kg
- Force you pull it with = 450 N
- Frictional force = 250 N
- Acceleration = ?
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Let's first find the total force/net force acting on the crate. The total force will be :-
- 450 N - 250 N = <u>2</u><u>0</u><u>0</u><u> </u><u>N.</u>
We subtracted the forces because both the forces act on opposite directions.
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Now, let's find the acceleration of the crate. To find the acceleration we must divide the net force by the mass. So,..
- Acceleration = Force/mass
- a = F/m
- a = 200/50
- <u>a = 4 m/s²</u>
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So, the acceleration of the crate is <u>4</u><u> </u><u>m/</u><u>s²</u><u>.</u>
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Hope it'll help you!
ℓu¢αzz ッ
Answer:

General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: ![\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bf%28x%29g%28x%29%5D%3Df%27%28x%29g%28x%29%20%2B%20g%27%28x%29f%28x%29)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>

<u>Step 2: Differentiate</u>
- [Function] Derivative Rule [Product Rule]:
![\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B9x%5E%7B10%7D%5D%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Rewrite [Derivative Property - Multiplied Constant]:
![\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%209%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E%7B10%7D%5D%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Basic Power Rule:
![\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%2090x%5E9%20%5Ctan%5E%7B-1%7D%28x%29%20%2B%209x%5E%7B10%7D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ctan%5E%7B-1%7D%28x%29%5D)
- Arctrig Derivative:

Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Answer:I need so much help with these two and quick if anybody can helpe out thanks.
Step-by-step explanation:
I need so much help with these two and quick if anybody can helpe out thanks.
Answer:
the vertex is:
(2, -1)
Step-by-step explanation:
First solve the equation for the variable y

Add 16y on both sides of the equation


Notice that now the equation has the general form of a parabola

In this case

Add
and subtract
on the right side of the equation

Factor the expression that is inside the parentheses

Divide both sides of the equality between 16


For an equation of the form

the vertex is: (h, k)
In this case

the vertex is:
(2, -1)