Answer:
I'm going to paint you a picture in words of what this looks like on paper. We have a train leaving from a point on your paper heading straight west. We have another train leaving from the same point on your paper heading straight east. This is the "opposite directions" that your problem gives you.
Now let's make a table:
distance = rate * time
Train 1
Train 2
We will fill in this table from the info in the problem then refer back to our drawing. It says that one train is traveling 12 mph faster than the other train. We don't know how fast "the other train" is going, so let's call that rate r. If the first train is travelin 12 mph faster, that rate is r + 12. Let's put that into the table
distance = rate * time
Train 1 r
Train 2 (r + 12)
Then it says "after 2 hours", so the time for both trains is 2 hours:
distance = rate * time
Train 1 r * 2
Train 2 (r + 12) * 2
Since distance = rate * time, the distance (or length of the arrow pointing straight west) for Train 1 is 2r. The distance (or length of the arrow pointing straight east) for Train 2 is 2(r + 12) which is 2r + 24. The distance between them (which is also the length of the whole entire arrow) is 232. Thus:
2r + 2r + 24 = 232 and
4r = 208 so
r = 52
This means that Train 1 is traveling 52 mph and Train 2 is traveling 12 miles per hour faster than that at 64 mph
Step-by-step explanation:
A(0,-1)
B(3√3/3, 2)
Slope of AB = (y₂ - y₁)/(x₂-x₁)
Slope = (2-1)/(√3/3 , -0)
Slope = 1/√3/3 = 1/√3 = √3/3
OR SLOPE = tan(a°) = √3/3 and a° = 30°
Note:xy means x times y and x(y) means the same thing
so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0
squareroot(y-1)+3=y
isolate the squareroot
subtrac 3 from boht sides
squareroot(y-1)=y-3
square both sides (since they are equal, you should be able to square both sides and still make it true)
(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5
0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0
therefor
y=5 or/and 2 might work
let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work
let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true
y=5
In order for two linear lines to be perpendicular, the product of their gradient must be -1.
Let's take y = x + b and y = -x + b
This is already in the form: y = mx + b, where m is 1 and -1 respectively.
Since the product of their gradient is -1, they are said to be perpendicular.
