The expression 3(x-9) is equivalent to

Mathematics

1 answer:

Inessa05 [86]3 years ago

8 0

The answer to this question would be D. 3(x)-3(9)

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How much should be added to the sum of 25 1/6 and 7 2/5 to get 60?

ehidna [41]

Answer:

3.77 should be added

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4 0

2 years ago

Read 2 more answers

Brandon has a playlist with 83 hip-hop and pop songs. There are 15 more hip-hop songs than pop songs. How many pop songs are on

kogti [31]

Using a system of equations, it is found that there are 34 pop songs are on the playlist.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Variable x: Number of hip-hop songs.

Variable y: Number of pop songs.

Brandon has a playlist with 83 hip-hop and pop songs, hence:

x + y = 83.

There are 15 more hip-hop songs than pop songs, hence:

x = 15 + y.

Replacing the second onto the first equation:

x + y = 83.

15 + y + y = 83.

2y = 68.

y = 34.

There are 34 pop songs are on the playlist.

More can be learned about a system of equations at brainly.com/question/24342899

6 0

2 years ago

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.

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

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