Please help!

Mathematics

1 answer:

lana [24]3 years ago

6 0

Given Information:

Principle amount = P = $6,000

Interest rate = r = 4% = 0.04

Period in years = t = 5

Required Information:

How much interest will he earn in 5 years = ?

Answer:

Amount of interest = $1,299.92

Step-by-step explanation:

Using the formula given in the question,

$B = P(1 + r )^{t}$

Where B is the final amount, P is the initial amount, r is the interest rate and t is the number of years

$B = P(1 + r )^{t}\\\\B = 6,000(1 + 0.04)^{5}\\\\B = 7,299.92$

The amount of interest earned is

$Amount \:of\: interest = B - P \\\\Amount \:of \:interest = 7,299.92 - 6,000\\\\Amount \:of \:interest = 1,299.92$

Therefore, Quincy has earned $1,299.92 in terms of interest by investing $6,000 in a savings account at the rate of 4% annual interest for a period of 5 years.

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REALLY APPRECIATE YOUR HELP Given log x = 3 and log y = 2, prove that x^2 -7xy =3000y

anyanavicka [17]

Answer:

$if \: log(x ) = 3 \\ log_{10}(x) = 3 \\ {10}^{3} = x \\ therfore \to \: \boxed{x = {10}^{3} } \\ \\ if \: log(y ) = 2 \\ log_{10}(y) = 2 \\ {10}^{2} = y \\ therfore \to \: \boxed{y= {10}^{2} }$

Step-by-step explanation:

$x^2 -7xy =3000y \: then \to \\ x^2 =3000y + 7xy \\ {(10 ^{3} )}^{2} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{5} ) \\ {10}^{6} = 300000 + 700000 \\ \boxed{{10}^{6} = 1000000} \\ hence \: the \: equation \to \: x^2 -7xy =3000y \\ \boxed{ is \: corret}$