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love history [14]
3 years ago
9

Find the area of the figure. PLS I NEED HELP FAST I BEG YOU!

Mathematics
1 answer:
Nina [5.8K]3 years ago
6 0
585 is correct!

LxWxH is for area

Mark with crown!
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Domain: (-∞,∞)

Range: (3,∞)

x-intercepts: none

y-intercepts: (0,7)

Interval positive: (3,∞)

Interval negative: none

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Interval decreasing: (-∞,7)

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Read 2 more answers
Why is it necessary to use the absolute value of the difference when finding the distance on a number line, but not necessary wh
padilas [110]
Remember that the distance on a number line is given by the formula: d=|x_{2}-x_{1}|
where 
d is the distance between the points. 
x_{2} is the second point in the number line.  
x_{1} is the first point in the number line.
Now, suppose we are trying to find the distance between the points -1 and -5; our first point is -1, so x_{1}=-1, and our second point is -5, so x_{2}=-5. Suppose we are going to find the distance between the two point without using absolute value:
d=x_{2}-x_{1}
d=-5-(-1)
d=-5+1
d=-4
Look what we have here, a negative distance! Since distances cannot be negative, we must use absolute value to always get postie distances between two point on a umber line:
d=|x_{2}-x_{1}|
d=|-5-(-1)|
d=|-5+1|
d=|-4|
d=4 

Now, the distance between two coordinates on a plane is given by the formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
where
d is the distance between the two coordinates
(x_{1},y_{2}) are the coordinates of the first point
(x_{2},y_{2}) are the coordinates of the second point 
Notice that (x_{2}-x_{1})^2 and (y_{2}-y_{1})^2 are squared, so it doesn't matter if we get a negative distance because a negative number raised to an even power (like 2) is always positive; therefore we don't need absolute value in this case because we won't ever get a negative distance. 


7 0
3 years ago
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