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Masteriza [31]
3 years ago
13

Phyllis solved the compound inequality 16 < 2(3x – 1) < 28. She began by first breaking the inequality into two separate i

nequalities, then she correctly solved each for x. 16 < 2(3x – 1) and 2(3x – 1) < 28 3 < x x < 5 Which graph represents her solution?

Mathematics
2 answers:
poizon [28]3 years ago
3 0
16<2(3x-1)<28  divide all terms by 2

8<3x-1<14  add 1 to all terms

9<3x<15  divide all terms by 3

3<x<5

x=(3,5) in interval notation

On a number line, it would be a line segment from 3 to 5 with open circles at 3 and 5.

On a coordinate system graph, it would be an infinitely high shaded plane between the vertical lines x=3 and x=5
soldi70 [24.7K]3 years ago
3 0
From question ,
16<2(3x-1)=16<6x-2
                 =16+2<6x-2+2
                 =18<6x
                 =18/6<6x/6
                 =3<x
Again,
2(3x-1)<28
=6x-2<28
=6x-2+2<28+2
=6x<30
=6x/6<30/6
=x<5
So both of ther graphs 3<x and x<5 represents her solution 

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\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
(5 points)
Schach [20]

Answer:

The original Slope is y=-1/3x-2

The perpendicular slope would be y=3x+b

Step-by-step explanation:

Original Slope:

You already have the y-interecpet, (0,-2), and you know two points (0,-2) and (-6,0), so you can find the slope with (y1-y2)/(x1-x2) so in this case, it would be (0-(-2))/(-6-0)= -1/3 and you can find y=1/3x-2.

Perpendicular Slope:

The slope of any perpendicular line would be the reciprocal of the original. So the reciprocal of -1/3 is 3. The b can be any real number because no matter the y-intercept the slope would still be perpendicular

3 0
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