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Yanka [14]
2 years ago
7

suppose that you put $2,500 into retirement account that grows with an interest rate of 5.25 % compounded once each year after h

ow many years will the balance of the account be at least $15,000?
Mathematics
1 answer:
Jlenok [28]2 years ago
8 0
You will need to use this formula:
<span>Years = {log(total) -log(Principal)} ÷ log(1 + rate)
Years = [log(15,000) - log(2,500)] / log (1.0525)
</span>
<span> <span> <span> 4.1760912591 </span> </span> </span> - <span> <span> <span> 3.3979400087 </span> </span> </span> / <span> <span> <span> 0.0222221045 </span> </span> </span> =
<span> <span> <span> 0.7781512504 </span> </span> </span> / <span> <span> <span> 0.0222221045 =
</span></span></span>
<span> <span> <span> 35.0169917705 </span> </span> </span> years
About 35 years

You'll find the formula here: http://www.1728.org/compint2.htm
And a compound interest calculator here: http://www.1728.org/compint.htm
(You'll find both are helpful for this problem).

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This question is worth 50 points, make sure to provide ALL calculations without using a calculator, though.
Fed [463]

Answer:

<h2>16 cm</h2>

Step-by-step explanation:

Use the formula for the volume of a cone:

V = (1/3)(area of base / round opening)(height of cone)

In this case we're interested in the following:

(1/3)(area of base / round opening)(height of cone) ≤ 535 cm³

First, mult. both sides by 3 to elim. the fraction 1/3:

(area of base)(height) ≤ 1605 cm³

Since the height is 8 cm, we now have:

(area of base) / (8 cm) = 1605 cm³, or

(area of base) = (1605 cm³) · (8 cm) = 200.625 cm²

The formula for the area of a circle is A = 3.14·r², where r is the radius.

If the area of the base is 200.625 cm², as found above, this equals 3.14r², and so

r² = (200.625 cm²) / (3.14) = 63.89 cm²

Then the radius must be +√63.89 cm, or 8 cm

and the diameter must be 2r = 2(8 cm) = 16 cm.  This would be the largest width of the container.

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3 years ago
A quadratic function and an exponential function are graphed below. How do the decay rates of the functions compareover the inte
Kobotan [32]

To check the decay rate, we need to check the variation in y-axis.

Since our interval is

-2We need to evaluate both function at those limits.At x = -2, we have a value of 4 for both of them, at x = 0 we have 1 for the exponential function and 0 to the quadratic function. Let's call the exponential f(x), and the quadratic g(x).[tex]\begin{gathered} f(-2)=g(-2)=4 \\ f(0)=1 \\ g(0)=0 \end{gathered}

To compare the decay rates we need to check the variation on the y-axis of both functions.

\begin{gathered} \Delta y_1=f(-2)-f(0)=4-1=3 \\ \Delta y_2=g(-2)-g(0)=4-0=4 \end{gathered}

Now, we calculate their ratio to find how they compare:

\frac{\Delta y_1}{\Delta y_2}=\frac{3}{4}

This tell us that the exponential function decays at three-fourths the rate of the quadratic function.

And this is the fourth option.

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Answer:

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Step-by-step explanation:

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3 years ago
3x-(2x-1)=7x-(3*5x)+(-x+24)
Paraphin [41]

Answer:

\boxed{\boxed{\sf x=\frac{23}{10}}\: \sf or \:\boxed{x=2.3}}

_________________

\boxed{\sf Step\: By\:Step:- }

\sf 3x-\left(2x-1\right)=7x-\left(3\times \:5x\right)+\left(-x+24\right)

<u>Remove the parentheses:</u>

\to\sf 3x-\left(2x-1\right)=7x-3\times \:5x-x+24

<u>Combine like terms:</u>

\sf ^*7x-x=6x

\to\sf 3x-\left(2x-1\right)=6x-3\times \:5x+24

<u>Multiply 3 and 5x = 15x:-</u>

\to\sf 3x-\left(2x-1\right)=6x-15x+24

<u>Combine like terms:</u>

\sf ^*6x-15x=-9x

\to\sf 3x-\left(2x-1\right)=-9x+24

<u>Expand: 3x-(2x-1)= x+1</u>

\to\sf x+1=-9x+24

<u>Subtract 1 from both sides:</u>

\to\sf x+1-1=-9x+24-1

\to\sf x=-9x+23

<u>Add 9x to both sides:</u>

\to\sf x+9x=-9x+23+9x

\to\sf 10x=23

<u>Divide both sides by 10:</u>

\to\sf \cfrac{10x}{10}=\cfrac{23}{10}

\to\sf x=\cfrac{23}{10}

<u>________________________________</u>

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80% of what how many games is 32 games?
Talja [164]
40 games would be 100% if 32 games is 80%
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