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makvit [3.9K]
3 years ago
13

Classify each statement based on whether a positive or negative rational number represents it. elevation of 3,234 feet above sea

level losing $32.50 depositing $230 in an account withdrawing $2,432 from an account 2°C below 0°C 10 floors above ground level
Mathematics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

elevation of 3,234 feet above sea level   Positive

losing $32.50   Negative  

depositing $230 in an account    Positive

withdrawing $2,432 from an account   Negative

2°C below 0°C  Negative

10 floors above ground level   Positive

Step-by-step explanation:

elevation of 3,234 feet above sea level   Positive   This is above ground

losing $32.50   Negative     it is less than you had

depositing $230 in an account    Positive   putting money in

withdrawing $2,432 from an account   Negative   taking money out

2°C below 0°C    less than 0

10 floors above ground level      above 0

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Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is y=\frac{1}{3} x+2

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3  

And as the line passing through (-3,1) and is  perpendicular to 3x + y = -8, product of slopes of two line will be -1  as lies are perpendicular.

Let required slope = x  

\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

So we need to find the equation of a line whose slope is \frac{1}{3} and passing through (-3,1)

Equation of line passing through (x_1 , y_1) and having lope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

Hence the required equation of line is found using slope intercept form

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