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serg [7]
3 years ago
15

After 4 years, $3,006 deposited in a savings account with simple interest had grown to a total of $3,967.92. What was the intere

st rate?
Mathematics
1 answer:
ella [17]3 years ago
4 0

Answer:

Interest rate is r = 0.33

Step-by-step explanation:

Principal Amount = $3006

Final Amount = $3967.92

Time = 4 years

Interest Rate = ?

The formula used is: Final Amount = Principal \times r \times t

Putting values and finding t

Final\: Amount = Principal \times r \times t \\3967.92=3006\times r \times (4)\\3967.92=12024\times r\\r=\frac{3967.92}{12024}\\r=0.33

So, Interest rate is r= 0.33

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Depends, rise over run could be considered a fraction, or if it's a number that has a whole number like one and 3/4 then you just plot the point to the best of your ability. In other words guess.

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What is the positive solution to this equation? 4x2 + 12x = 135​
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x = 9/2 - 15/2

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3 years ago
A blueprint of a barn is shown.The scale is 5 inches =8 feet. Find the actual length of the wall that is labeled 20 inches.
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Triangle PQR is rotated 180° clockwise about point M. The image of point P is P' and the image of point R is R'. What are the co
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3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

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x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
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