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2 years ago

Simplify: 3(9 + 2d) + 4d

Mathematics

2 answers:

Inga [223]2 years ago

6 0

Answer:

27 + 10d

Step-by-step explanation:

PilotLPTM [1.2K]2 years ago

4 0

Answer:

10d+27

Step-by-step explanation:

Rearrange terms

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Distribute

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Combine like terms

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Hi please help !!! asap

Taya2010 [7]

Ayurveda is the traditional Hindu system of medicine, which is based on the idea of balance in bodily systems and uses diet, herbal treatment, and yogic breathing

7 0

2 years ago

James puts $3,500 into a savings account that earns 2.5% simple interest. He does not touch that account for 3 years. What would

miskamm [114]

Given :

James puts $3,500 into a savings account that earns 2.5% simple interest.

He does not touch that account for 3 years.

To Find :

The new balance of the account be after 3 years.

Solution :

Interest on $3500 after 3 years is :

$I = \dfrac{P_o\times r\times t}{100}\\\\I = \dfrac{3500\times 2.5\times 3}{100}\\\\I = \$262.5$

So, new balance of the account after 3 years is $( 3500+262.5 ) = $3762.5 .

Hence, this is the required solution.

3 0

3 years ago

Solve h for -9h-15=93

Lerok [7]

The answer to this question:

h=-12

8 0

2 years ago

Fill in the blanks.

Roman55 [17]

Answer:

One way of awarding interest is called simple interest. Before we provide the formula used in calculating simple interest, let’s first define some basic terms.

Balance. The balance is the current amount in an account or the current amount owed on a loan.

Principal. The principal is the initial amount invested or borrowed.

Rate. This is the interest rate, usually given as a percent per year.

Time. This is the time duration of the loan or investment. If the interest rate is per year, then the time must be measured in years.

To calculate the simple interest on an account or loan, use the following formula.

Simple Interest

Simple interest is calculated with the formula

I=Prt,

where I is the interest, P is the principal, r is the interest rate, and t is the time.

Step-by-step explanation:

One way of awarding interest is called simple interest. Before we provide the formula used in calculating simple interest, let’s first define some basic terms.

Balance. The balance is the current amount in an account or the current amount owed on a loan.

Principal. The principal is the initial amount invested or borrowed.

Rate. This is the interest rate, usually given as a percent per year.

Time. This is the time duration of the loan or investment. If the interest rate is per year, then the time must be measured in years.

To calculate the simple interest on an account or loan, use the following formula.

Simple Interest

Simple interest is calculated with the formula

I=Prt,

where I is the interest, P is the principal, r is the interest rate, and t is the time.

5 0

2 years ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

2 years ago