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3 years ago

7

Find all values of x such that

a1" title="\sqrt{4x^2} -\sqrt{x^2}= 6" alt="\sqrt{4x^2} -\sqrt{x^2}= 6" align="absmiddle" class="latex-formula">

1 answer:

tester [92]3 years ago

3 0

There are two answers.
6 and -6

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1) 3(3x - 4) = -2(1 – 4x)

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Let's solve your equation step-by-step.
3
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3
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−
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=
−
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(
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x
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Step 1: Simplify both sides of the equation.
3
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−
4
x
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9
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12
=
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2
+
8
x
9
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−
12
=
8
x
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2
Step 2: Subtract 8x from both sides.
9
x
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12
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8
x
=
8
x
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2
−
8
x
x
−
12
=
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2
Step 3: Add 12 to both sides.
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3 years ago

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. 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)?

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

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