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2 years ago

How long will it take for an investment to double in valve if it earns 12.25% compounded continuously ?

Mathematics

1 answer:

goblinko [34]2 years ago

6 0

Answer:

At the given rate it will take 3 years and 5 months for the investment to be doubled

Step-by-step explanation:

Here, we want to get the amount of time it will take an investment that earns 12.25% to double if it’s compounded continuously

We can have the exponential relation as follows;

P = I( 1 + r)^n

P is the present value, let us call this 2x as it two times the initial

I is the initial value which is x in this case

r is the interest rate = 12.25% which is same as 12.25/100 = 0.1225

n is the number of years which we want to calculate

Thus, we have

2x = x( 1 + 0.1225)^n

2 = 1.225^n

Ln 2 = n ln 1.225

n = ln 2/ln 1.225

n = 3.42

if 3.42 is the number of years, let us have 0.42 in months

That will be 0.42 * 12 = 5.04

So we are looking at the investment being doubled at 3 years and 5 months

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Use the distance formula to find the length of AB. Point A (0 , -8) and Point B (3 , -2). Round your answer to the tenths place.

Lana71 [14]

Answer:

lol you thought i accully knew it

Step-by-step explanation:

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2 years ago

The graph of f(x) = x^2 has been shifted into the form f(x) = (x − h)^2 + k

stepan [7]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see, it went to the right 2 units, and then up 3 units.

that simply means, C = -2, D = 3.

8 0

2 years ago

A 55m and 35m broad park is surrounded by a 2.5m wide path.

daser333 [38]

Step-by-step explanation:

☄ $\underline {\underline{ \text{Given}}} :$

Length of a park ( l ) = 55 m
Breadth of a park ( b ) = 35 m
Width running outside the park ( d ) = 2.5 m
Rate of paving the path with stones = Rs 120 per sq.metre

☄ $\underline{ \underline{ \text{To \: find}}}:$

Area of the park
Cosy of paving the path with stones at Rs 130 per sq.metre

☄ $\underline{ \underline{ \text{Solution}}} :$

Part 1 : $\boxed{\text{Area \:of \:a \:path \: running \:outside = 2d(l + b + 2d)}}$

plug the known values and simplify :

⟹ $\sf{2 \times 2.5(55 + 35 + 2 \times 2.5)}$

⟹ $\sf{5 \times 95}$

⟹ $\sf{475 \: {m}^{2} }$

Part 2 : $\boxed{ \sf{Total \: cost = Area \: of \: paths \: \times Rate}}$

⟹ $\sf{475 \times 120}$

⟹ $\sf{Rs \: 57000}$

$\purple{ \boxed{ \boxed{ \sf{ \tt{⟿ \: Our \: final \: answer : (i) = 475 \: {m}^{2} \: and \: (ii ) = Rs \: 57000}}}}}$

Hope I helped ! ♡

Have a wonderful day /night ツ

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7 0

2 years ago

Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year. Whe

Rama09 [41]

You can see how this works by thinking through what's going on.

In the first year the population declines by 3%. So the population at the end of the first year is the starting population (1200) minus the decline: 1200 minus 3% of 1200. 3% of 1200 is the same as .03 * 1200. So the population at the end of the first year is 1200 - .03 * 1200. That can be written as 1200 * (1 - .03), or 1200 * 0.97

What about the second year? The population starts at 1200 * 0.97. It declines by 3% again. But 3% of what??? The decline is based on the population at the beginning of the year, NOT based no the original population. So the decline in the second year is 0.03 * (1200 * 0.97). And just as in the first year, the population at the end of the second year is the population at the beginning of the second year minus the decline in the second year. So that's 1200 * 0.97 - 0.03 * (1200 * 0.97), which is equal to 1200 * 0.97 (1 - 0.03) = 1200 * 0.97 * 0.97 = 1200 * 0.972.

So there's a pattern. If you worked out the third year, you'd see that the population ends up as 1200 * 0.973, and it would keep going like that.

So the population after x years is 1200 * 0.97x

5 0

3 years ago

How do you graph a solution set of an inequality that only has whole numbers

fomenos

Answer:

Step-by-step explanation:

A solution set is the set of values which satisfy a given inequality. It means, each and every value in the solution set will satisfy the inequality and no other value will satisfy the inequality.

Example:

Solve 2x + 3 ≤ 7, where x is a natural number.

Solution:

2x + 3 ≤ 7

Subtracting 3 from both the sides,

2x ≤ 4

Dividing both sides by 2,

x ≤ 2

Since x is a natural number,

Solution set = {1,2}.

5 0

2 years ago