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2 years ago

Solve for z: -4 - (z - 6) = 12

Mathematics

2 answers:

Sergeu [11.5K]2 years ago

8 0

Answer:

z = -10

Step-by-step explanation:

-4 -(z+6)=12

2-z=12

-z =12-2

ans= z = -10

aksik [14]2 years ago

7 0

Answer:

z = -10

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Step-by-step explanation:

Step 1: Define equation

-4 - (z - 6) = 12

Step 2: Solve for z

Distribute negative: -4 - z + 6 = 12
Combine like terms: -z + 2 = 12
Subtract 2 on both sides: -z = 10
Divide -1 on both sides: z = -10

Step 3: Check

Plug in z to verify it's a solution.

Substitute: -4 - (-10 - 6) = 12
Subtract: -4 - (-16) = 12
Simplify: -4 + 16 = 12
Add: 12 = 12

Here we see that 12 does indeed equal 12.

∴ z = -10 is a solution of the equation.

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A certain city's population is 120,000 and decreases 1.4% per year for 15 years.

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Answer:

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Step-by-step explanation:

The exponential function for growth/decay is given as:

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In this problem:

The city's initial population is 120,000 and it decreases by 1.4% per year.

Since the population decreases, it is a Decay Problem.

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Therefore, the function is:

$P(t)=120000(1 - 0.014)^t\\P(t)=120000(0.986)^t$

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2 years ago

Suppose that you have $6000 to invest. Which investment yields the greater return over four years: 8.25% compounded quarterly or

WARRIOR [948]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.25\%\to \frac{8.25}{100}\to &0.0825\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.0825}{4}\right)^{4\cdot 4}\implies A=6000(1.020625)^{16}\\\\
-------------------------------\\\\$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.3\%\to \frac{8.3}{100}\to &0.083\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus two}
\end{array}\to &2\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.083}{2}\right)^{2\cdot 4}\implies A=6000(1.0415)^8$

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