A) 3a + 5 (2a + 6) - 7a

Mathematics

1 answer:

Mrac [35]3 years ago

3 0

Answer:

A) 3a+10a+30-7a

=6a+30

B) 6b-32b+72+3b

=72+23b

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A House needs to some maintenance one sink is dripping water every 10 minutes and the other sink is dripping every 12 minutes if

e-lub [12.9K]

Answer:

60 minutes

Step-by-step explanation:

We solve this question using Lowest Common Factor method

Find and list multiples of each number of minutes( 12 and 10 minutes) until the first common multiple is found. This is the lowest common multiple.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80

Multiples of 12:

12, 24, 36, 48, 60, 72, 84

Therefore,

LCM(10, 12) = 60

The number if minutes until they both drip again is 60 minutes

7 0

3 years ago

. Mrs. Rojas deposited $8000 into an account paying 4% interest annually. After 8 months Mrs. Rojas withdrew the $6000 plus inte

Mashcka [7]

well, keeping in mind that a year has 12 months, that means that 8 months is 8/12 of a year, when Mrs Rojas pull her money out.

$~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\to \frac{8}{12}\dotfill &\frac{2}{3} \end{cases} \\\\\\ A=6000[1+(0.04)(\frac{2}{3})]\implies A=6000\left( \frac{77}{75} \right)\implies A=6160$

well, she put in 6000 bucks, got back 160 extra, that's the interest earned in the 8 months.

what if she had left her money for 1 whole year, then

$~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &1 \end{cases} \\\\\\ A=6000[1+(0.04)(1)]\implies A=6240$

so had she left it in for a year, she'd have gotten 6240, namely 240 in interest, well, what fraction of a year's interest was earned? or worded differently, what fraction is 160(8 months) of 240(1 year)?

$\cfrac{\stackrel{\textit{for 8 months}}{160}}{\underset{\textit{for 12 months}}{240}}\implies \cfrac{2}{3}$