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2 years ago

9

How much money will be in a bank account after 3 years if $9 is deposited at an interest rate of 5% compounded annually? Round t

o the nearest tenth.

1 answer:

sergij07 [2.7K]2 years ago

4 0

Answer:

Step-by-step explanation:

$13.1

You might be interested in

Type 11/5 in the simplest form

aleksklad [387]

Answer:

$2\frac{1}{5}$

Step-by-step explanation:

11 ÷ 5 = 2 R 1 → $2\frac{1}{5}$

Hope this helps! :)

7 0

3 years ago

2 Nine identical circles are cut from a square sheet of paper whose sides are

Ivanshal [37]

Answer:

278.64cm²

Step-by-step explanation:

Area of the sheet left out = Area of the square - Area of the 9 circles

Area of the square = L^2

L is the side length of the square

A = 36^2

Area of the square = 1296cm^2

Diameter of a circle = 38/3 = 12cm

Area of a circle = πr²

r is the radius = 12/2 = 6cm

Area of a circle = 3.14(6)²

Area of a circle = 3.14 * 36

Area of a circle = 113.04cm²

Area of 9 circles = 9 * 113.04

Area of 9 circles = 1,017.36cm²

Area of the left over = 1296 - 1,017.36

Area of the left over = 278.64cm²

6 0

2 years ago

The equation y=9. 5x 0. 4 represents the line of best fit for the distance (in feet) that Marcella traveled over time (in second

Bas_tet [7]

The average speed of Marcella in feet per seconds is 9.5 feet per second.

<h3>How are time taken for travel, distance traveled, and the average speed are related?</h3>

We have this below shown relation between them

$\text{Average speed} = \dfrac{\text{Distance traveled}}{\text{Total time taken to travel that distance}}\\\\or\\\\\text{Total time taken to travel that distance}= \dfrac{\text{Distance traveled}}{\text{Average speed} }$

For the given case, we have:

The equation $y=9.5x + 0.4$ represents the line of best fit for the distance (in feet) that Marcella traveled over time (in seconds)

The line of best fit can be taken as approximately telling about the data points ( y = distance traveled in feet, and x = time spent )

Let the initial time be 'x' with initial distance traveled 'y', then the time after one second will be (x+1), and the distance traveled be $y_1$

The value of y and $y_1$ are calculated as:

$y = 9.5x + 0.4\\y_1 = 9.5(x+1) + 0.4 = 9.5x + 9.9$

Thus, the average speed is calculated at time 'x' as:

$\text{Average speed} = \dfrac{\text{Distance traveled}}{\text{Total time taken to travel that distance}}\\\\or\\\\\text{Average speed} = \dfrac{y_1 - y}{x + 1 - x} = \dfrac{9.5x + 9.9 - 9.5x - 0.4}{1} = 9.5 \: \rm feet/sec$

We could've also used the fact that in the relation of y to x as $y = mx + c$ , 'm' denotes the rate of y as x changes.

Thus, the average speed of Marcella in feet per seconds is 9.5 feet per second.

Learn more about average rate here:

brainly.com/question/12322912

4 0

1 year ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

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 limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in <em>x</em> = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in <em>x</em> = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

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

Unit: Limits

6 0

2 years ago

Which statement correctly interprets this graph?

Marrrta [24]

D because none of the other statments are correct and it's the only one that lines up with the data.

8 0

3 years ago