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4 years ago

Kylie starts with $145 in her piggy bank. Each month she adds $20.

Mathematics

1 answer:

djverab [1.8K]4 years ago

4 0

Kylie's first month collection, a1= $ 145

Second month collection , a2= $145+ 20

Third month collection, a3 = $145 +2*20

so for n months collection = an-1+20

We get an= 20 + an-1 and a1=145

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Solve the following questions-

Setler79 [48]

Answer:

2/5

Step-by-step explanation:

To get the ratio as a pure number, it must be expressed as the quotient of two values that have the same units. For the purpose here, it is convenient to convert both values to units of seconds.

<h3>units conversion</h3>

The conversion factor between minutes and seconds is ...

1 minute = 60 seconds

Multiplying this equation by 3 gives ...

3 minutes = 180 seconds

<h3>ratio of interest</h3>

Then the desired ratio is ...

(72 seconds)/(3 minutes) = (72 seconds)/(180 seconds) = 72/180

= (36×2)/(36×5)

= 2/5

The ratio in its simplest form is 2/5.

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2 years ago

Math help please? <3

ipn [44]

A linear equation cannot be used.
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3 years ago

17 dozen cookies<br> are made in 2<br> hours. At this rate,<br> how many are<br> made in 5 hours?

prohojiy [21]

Answer:

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Step-by-step explanation:

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3 years ago

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

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Functions

Function Notation

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

<u>Calculus</u>

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

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Step-by-step explanation:

We are given the following and are trying to find the second derivative at <em>x</em> = 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 <em>x</em> = 2, simply substitute in <em>x</em> = 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

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3 years ago

When you look at the two cylinders with these values of r and h, what do you notice about the proportions of the cylinders?

Elden [556K]

Answer and Explanation:

Volume of a cylinder is given by

V= Pi*r²*h

where v= volume of the cylinder

Pi is constant = 22/7 or 3.14159

r= radius of the circle of the cylinder

h= height of the cylinder

If two cylinders are measured proportionally to the other based on radius and height of each cylinder, we look at the proportional equality of the ratio of the radius and height of one cylinder to the other. If one cylinder for example has ratio of radius and height =2/4 then it is proportional to the other cylinder with ratio 1/2

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3 years ago

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