Because of black Friday, Best Buy had an additional

Mathematics

1 answer:

densk [106]2 years ago

6 0

Answer: $25.20

explanation: 25% + 15% = 40%

42 - 40% = $25.20

hope this helps :)

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Find the rate of discount for a pair of pants that cost 65$ and are on sell for 39

lutik1710 [3]

Answer: 65 - $39 = $26 >> amount you save

26 / 65 = 0.4 = 40% >> discount rate

Step-by-step explanation:

6 0

3 years ago

The radius of one circle is three the time radius of another circle. The sum of their areas is 12pi. Find the radius of each cir

zlopas [31]

$\displaystyle\\
\begin{cases}
R_1 = 3R_2\\
\pi R_1^2 +\pi R_2^2=12\pi ~~~\Big|~~:\pi 
\end{cases} \\ \\ 
\begin{cases} 
R_1 = 3R_2\\
R_1^2 + R_2^2=12 
\end{cases} \\ \\ 
(3R_2)^2+ R_2^2=12 \\ \\ 
9R_2^2 + R_2^2=12 \\ \\ 
10R_2^2=12\\\\
R_2^2 = \frac{12}{10} = \frac{6}{5} \\ \\ 
R_2 = \boxed{\sqrt{\frac{6}{5}} } \\ \\ 
R_1 = \boxed{3\sqrt{\frac{6}{5}} }$

7 0

3 years ago

The formula used to calculate the value of a savings accounty =(1+)120What does theafter t years is A(t)=0.04= 1500 1+120.04frac

yulyashka [42]

We have here the formula for Compound Interest:

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

Where:

• A is the accrued amount.

• P is the Principal (the original amount of money, the starting amount of money).

• r is the interest rate.

• n is the number of times per year compounded.

• t is the time in years.

When we have that n is equal to 12, we are talking here about that the amount of money is being compounded monthly (we have 12 months in a year, 12 periods, n = 12). <em>Therefore, we are dividing the rate, r, by the number of compoundings per year, n, and this is the rate per each new compounding period of time, r/n, and, in this case, n = 12 (monthly interest rate). </em>

Therefore, in few words, the fraction (0.04/12) is the monthly interest rate (option C).

[If we see the other options, we have:

• The daily interest rate would be given by 0.04/365.

• How long the money has been in the account is time, t.

• The starting balance in the account is the Principal, P. ]