Which pair of inequalities would overlap if they were graphed?

Mathematics

1 answer:

dsp733 years ago

4 0

Answer:

x ≤ 10, x > 4

Step-by-step explanation:

If x is less than or equal to 10, the line starts on 10 and includes all the numbers to the left . . . .-2-1 0 1 2 3 4 5 6 7 8 9 10,11,12,. . .

If x is greater than 4 it's . -1,0,1 2 3 4,5 6 7 8 9 10 11 12 13 . . .

So there is an overlap of the numbers between 4 and 10 (not including 4, just a little more. . )

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Question is shown below ↓

sladkih [1.3K]

$\bf \stackrel{\textit{using the exponential model}}{N=2^D}~\hspace{7em}\begin{array}{ccll} \stackrel{days}{D}&\stackrel{\$}{N}\\ \cline{1-2} 1&2^1\implies 2\\ 2&2^2\implies 4 \end{array}$

so, using that exponential model, the 1st output value works, but the second value of 2² does not give us 8 as output.

let's check the linear model using slopes to get the equation.

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{8})\qquad \impliedby \begin{array}{|cc|ll} \cline{1-2} D&N\\ \cline{1-2} 1&2\\ 2&8\\ \cline{1-2} \end{array}$

$\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-2}{2-1}\implies \cfrac{6}{1}\implies 6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=6(x-1) \\\\\\ y-2=6x-6\implies y=6x-4$