With a given parallel line and a given point on the line
we can use the point-line method: y-y0=m(x-x0)
where
y=mx+k is the given line, and
(x0,y0) is the given point.
Here
m=-10, k=-5, (x0,y0)=(-3,5)
=> the required line L is given by:
L: y-5=-10(x-(-3))
on simplification
L: y=-10x-30+5
L: y=-10x-25
Answer:
Step-by-step explanation:
Our inequality is |125-u| ≤ 30. Let's separate this into two. Assuming that (125-u) is positive, we have 125-u ≤ 30, and if we assume that it's negative, we'd have -(125-u)≤30, or u-125≤30.
Therefore, we now have two inequalities to solve for:
125-u ≤ 30
u-125≤30
For the first one, we can subtract 125 and add u to both sides, resulting in
0 ≤ u-95, or 95≤u. Therefore, that is our first inequality.
The second one can be figured out by adding 125 to both sides, so u ≤ 155.
Remember that we took these two inequalities from an absolute value -- as a result, they BOTH must be true in order for the original inequality to be true. Therefore,
u ≥ 95
and
u ≤ 155
combine to be
95 ≤ u ≤ 155, or the 4th option
Answer:
B. 30
Step-by-step explanation:
3x + 2x + x = 180 (angle sum of a triangle is 180)
6x = 180
x = 30
The answer is x=6/11.
Tiger Algebra is a really good tool for algebra if you need some help.
Hope that helped! :)
$117.05
- $ 14.99
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$102.46 left
Divide 102.46 by 0.81, and you get 126, which makes it 126 miles.
Hope this helps!
