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nevsk [136]
2 years ago
12

What is the standard form of the linear function that passes

Mathematics
1 answer:
babunello [35]2 years ago
7 0

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{-2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-2}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{4}}}\implies \cfrac{-3}{-2}\implies \cfrac{3}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{3}{2}}(x-\stackrel{x_1}{4})

\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2(y-1)=2\left( \cfrac{3}{2}(x-4) \right)}\implies 2y-2 = 3(x-4)\implies 2y-2=3x-12 \\\\\\ -3x+2y-2=-12\implies -3x+2y=-10\implies \stackrel{\times -1\textit{ to both sides}}{3x-2y=10}

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Answer:

C) $10,000 invested at 6.7% compounded quarterly over 7 years yields the greater return.

Step-by-step explanation:

-We determine the effective interest rate in both scenarios and use it to calculate the investment's value after 7 years.

#Given n=7yrs, P=$10,000 and i=6.6% compounded monthly:

i_m=(1+i/m)^m-1\\\\=(1+0.066/12)^{12}-1=0.06803\\\\\therefore A=P(1+i_m)^n\\\\=10000(1.06803)^7\\\\\approx \$15,852.00

#Given n=7rs, P=10000, i=6.7%

i_m=(1+i/m)^m-1\\\\=(1+0.067/4)^{4}-1=0.06870\\\\\therefore A=P(1+i_m)^n\\\\=10000(1.06870)^7\\\\\approx \$15,921.75

Hence, the investment has the largest value($15,921.75) when the interest rate is compounded quarterly.

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