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Marizza181 [45]
3 years ago
11

Mario transferred a balance of $6050 to a new credit card at the beginning of

Mathematics
2 answers:
Kryger [21]3 years ago
7 0

Answer:

the answer is (6050){1+0.031/12}^3{1+0.206/12}^9 for APEX

Step-by-step explanation:

Maru [420]3 years ago
5 0

Answer:

6050(1+\frac{0.031}{12})^{3}(1+\frac{0.206}{12})^{9}

Step-by-step explanation:

p = $6050

The card offered an introductory APR of 3.1% for the first 3 months

r = 3.1% or 0.031

t = 3

n = 12

So, compound interest formula for this period is :

6050(1+\frac{0.031}{12})^{3}

And a standard APR of 20.6% thereafter,

r = 20.6% or 0.206

t = 9

n = 12

So, compound interest formula for this period is :

6050(1+\frac{0.206}{12})^{9}

Now as the p is common, we can re write the expressions as :

6050(1+\frac{0.031}{12})^{3}(1+\frac{0.206}{12})^{9}

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REALLY APPRECIATE YOUR HELP<br><br>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}

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