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oksano4ka [1.4K]
3 years ago
8

Kenny and johnny went to thr water park. Johnny bought a $1.50 ice cream bar. Write an expression to represent the amount the tw

o friends spent, using p as the price of admission to the park
Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0
2p+1.5=total spent I hope this helps

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Find dy/dx for y= x^3 ln (cot x)
ICE Princess25 [194]
<h3>Answer</h3>

  \dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)

<h3>Explanation</h3>

By the product rule (d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x), we have

  \begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}

By the chain rule:

  \begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}

By the power rule:

  (x^3)' = 3x^2

thus

  \begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}

Nothing to do to simplify any further, other than factoring out x^2.

4 0
3 years ago
How do you use the distributive property to solve for x in the equation -2(x + 5) = 4
lubasha [3.4K]

Answer:

x = - 7

Step-by-step explanation:

Given

- 2(x + 5) = 4 ← distribute the parenthesis by - 2

- 2x - 10 = 4 ( add 10 to both sides )

- 2x = 14 ( divide both sides by - 2 )

x = - 7

3 0
3 years ago
Out of the 20 students in my classroom, 15 wore their pajamas on Pajama Day. In decimal form, represent the portion of students
kvasek [131]
75.0% of students wore pajamas
4 0
3 years ago
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Simplify by commbining the like terms. 8y - 3y
11Alexandr11 [23.1K]
8y-3y=5y
5y is the answer
3 0
3 years ago
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A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
arsen [322]

Answer:

option A

a = 8.65 m/s²

Step-by-step explanation:

Given that,

force applied on a cart (forward direction) = 19N

frictional force experience by cart (backward direction) = 1.7N

mass of the cart = 2 kg

Frictional force always opposes applied force, so the Resultant force on the cart would have to be 19N - 1.7N.

Formula to use

Resultant force = ma

plug values in the formula

19 - 1.7 = 2(a)

17.3 = 2(a)

a = 8.65 m/s²

so the acceleration of the cart is 8.65m/s²

4 0
3 years ago
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