Find the 12th term of the geometric sequence 5, 20, 80, ...

Mathematics

1 answer:

Greeley [361]3 years ago

7 0

20971520

I did 4^11 x 5 (because you could say the first one is 5 x 1 and 4^0 = 1)

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Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How

Scorpion4ik [409]

let's check how much is it after 2 years firstly.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275$

Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}$

4 0

3 years ago

QUESTION 58 Kendra had a choice of 3 breads, 3 meats, and 2 cheeses to make a sandwich. How many different sandwiches can she ma

LUCKY_DIMON [66]

Assume 3 breads are all different, 3 meats are all different and 2 cheeses are all different. A sandwich contains a type of bread, a type of meat and a type of cheese. The number of different types of sandwiches that can be made is

$3*3*2=18$

The correct choice is (C).

7 0

3 years ago

Read 2 more answers

I have a series of 2 questions.

Margarita [4]

1)
I:y=3x-4
II:9x-3y=14

substitute y into II:
9x-3*(3x-4)=14
9x-9x+12=14
12=14

this is obviously not equal so there is no solution, the lines are parallel

2)
I:y=4x+6
II:5x-y=6

substitute y into II:
5x-(4x+6)=6
5x-4x-6=6
x=12
substiute x into II:
5*12-y=6
-y=6-60
-y=-54
y=54

the solution is (12,54)

3 0

3 years ago

Please someone can help me with #13, 15 and 16 Thanks.

ladessa [460]

What's the question though?

8 0

3 years ago

28 < 4v <br> solve for v <br> pls help

DedPeter [7]

Answer:

v = 7

Step-by-step explanation:

Divide both sides by 4 to isolate v

28/4 = 4v/4

v = 7

5 0

3 years ago