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CaHeK987 [17]
3 years ago
13

5(a-2b)-3(a-2b)algebra2 .-.

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
7 0
5(a - 2b) - 3(a - 2b)    |use distributive property: a(b - c) = ab - ac

= 5a - 10b - 3a + 6b

= (5a - 3a) + (-10b + 6b)

= 2a - 4b
Otrada [13]3 years ago
3 0
Formula <span> </span>a(b - c) = ab - ac&#10;

5(a - 2b) - 3(a - 2b)  

5a - 10b - 3a + 6b

(5a - 3a) + (-10b + 6b)

2a - 4b
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If Tanisha has ​$1000 to invest at 7​% per annum compounded semiannually​, how long will it be before she has ​$1600​? If the co
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Answer:

Using continuous interest 6.83 years before she has ​$1600​.

Using continuous compounding, 6.71 years.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

A(t) = P(1 + \frac{r}{n})^{nt}

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

The amount of money earned after t years in continuous interest is given by:

P(t) = P(0)e^{rt}

In which P(0) is the initial investment and r is the interest rate, as a decimal.

If Tanisha has ​$1000 to invest at 7​% per annum compounded semiannually​, how long will it be before she has ​$1600​?

We have to find t for which A(t) = 1600 when P = 1000, r = 0.07, n = 2

A(t) = P(1 + \frac{r}{n})^{nt}

1600 = 1000(1 + \frac{0.07}{2})^{2t}

(1.035)^{2t} = \frac{1600}{1000}

(1.035)^{2t} = 1.6

\log{1.035)^{2t}} = \log{1.6}

2t\log{1.035} = \log{1.6}

t = \frac{\log{1.6}}{2\log{1.035}}

t = 6.83

Using continuous interest 6.83 years before she has ​$1600​

If the compounding is​ continuous, how long will it​ be?

We have that P(0) = 1000, r = 0.07

Then

P(t) = P(0)e^{rt}

1600 = 1000e^{0.07t}

e^{0.07t} = 1.6

\ln{e^{0.07t}} = \ln{1.6}

0.07t = \ln{1.6}

t = \frac{\ln{1.6}}{0.07}

t = 6.71

Using continuous compounding, 6.71 years.

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