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Mamont248 [21]
3 years ago
6

Ms. Moran has started an investment club at BSS. $8000 is invested, some at 10%

Mathematics
1 answer:
labwork [276]3 years ago
6 0
The answer can be readily calculated using a single variable, x:

Let x = the amount being invested at an annual rate of 10%
Let (8000 - x) = the amount being invested at an annual rate of 12%

The problem is then stated as:

(x * 0.10) + ((8000 - x) * 0.12) = 900
0.10(x) + ((8000 * 0.12) - 0.12(x)) = 900
0.10(x) + 960 - 0.12(x) = 900
0.10(x) - 0.12(x) = 900 - 960
-0.02(x) = -60
-0.02(x) * -100/2 = -60 * -100/2
x = 6000 / 2
x = 3000

Thus, $3,000 is invested at 10% = $300 annually; and $8,000 - $3,000 = $5,000 invested at 12% = $600 annually, which sum to $900 annual investment.
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4 0
3 years ago
3^x= 3*2^x solve this equation​
kompoz [17]

In the equation

3^x = 3\cdot 2^x

divide both sides by 2^x to get

\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3

Take the base-3/2 logarithm of both sides:

\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}

Alternatively, you can divide both sides by 3^x:

\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13

Then take the base-2/3 logarith of both sides to get

\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}

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8 0
3 years ago
The population of Centerville increases each year. The function C(t) = P(1 +r)^t represents the population of centerville at yea
strojnjashka [21]

Answer:

P represents the population in year 0 ⇒ D

Step-by-step explanation:

* Lets explain the exponential growth function

- The exponential growth function is f(x) = a (1 + r)^t, where a is the initial

  amount (at t = 0), (1 + r) is the factor of growth , r is the rate of growth

  in decimal ant is the time of growth

* Lets solve the problem

∵ The function C(t) = P(1 + r)^t represents the population of

  centerville at year t, where P is the initial population and r is the

  rate of increase

- Ex: If your investment is increased 10% annually, then that means

 each year, your total has multiplied itself by 110% (the growth factor

 is 1 + 10/100 = 1.1)

∴ (1 + r) is the factor grows each year

∵ C(t) = P(1 + r)^t

∴ C depends on P(starting population) , r(the increasing rate and

   t(the time in year)

∵ r is the rate of increase means the percentage of increasing , then

  0 < r < 1

∴ r is not less than 0

∵ P is the initial amount when t = 0

∴ P represents the population in year 0

4 0
4 years ago
Read 2 more answers
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