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poizon [28]
3 years ago
13

When rounding to.the nearest hundred are you to add or subtract?

Mathematics
2 answers:
Rashid [163]3 years ago
6 0
Basically you don't really do either, mostly because when your rounding you mostly just look at the number your rounding to. okay so say your number is like 127 you would round to 100 because your number would've been below 150. And we know that five is the middle of every thing. 10 and the whole number system is made up of tens. So when you have to choose to round you have to go up to the next ten so say your number is 27 since it's above 5 you would round to 30. So if you have 127 it's below the 5 mark (which would be 150) so you would round to 100 because it's below that mark. 
kicyunya [14]3 years ago
5 0

Answer:

When you round to the nearest hundred:

150 or higher turns into 200, so you'd do add.

But when it's 149 or lower, you simply subtract it by 49.

Step-by-step explanation:


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What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S
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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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3 years ago
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