y=x+14 line 1
y=3x+2 line 2
These are both the equation of lines written in slope intercept form
y=mx+b where m is the slope and the point (0,b) is the y intercept.
The first line has a slope of m=1. The 2nd line has a slope of m=3
Since these lines have different slopes, they are not parallel, thus they will cross at some point. What you have to determine is where the lines cross, which will be a point (x,y) that is on both lines.
We already have y solved in terms of x from either equation so we can use substitution to solve the system.
Since y=x+14 from line 1, put x+14 in place of y in the equation of line 2.
x+14=3x+2
solve for x.
Subtract x from both sides...
14= 3x-x+2
14=2x+2
subtract 2 from both sides
14-2=2x
12=2x
divide both sides by 2
6=x
We now have the x value of the common point. Plug the value 6 in for x in one of the original equations and solve for y.
y=6+14
y=20
These two lines cross at the point (6,20) which is a point the two lines have in common.
Hope I helped (SharkieOwO)
Answer:
1. m=4/9
2.m=-26/7
3.m=0
Step-by-step explanation:
Formula: Y=mx+b
1.y
=
4
/9
x
−
25
/9
2.y
=
−
26
/7
x
+
452
/7
3.y=17
✨Hope this helps!✨
What I did to solve this is finding the equation of the line using slope intercept form y is the cost of the ride which increases as you increase mileage or move along the X axis. My points are (2, 5.25) and (5,10.5)
<span>y=mx+b </span>
<span>m=y2-y1/x2-x1 </span>
<span>m=(10.5-5.25)/(5-2) </span>
<span>m=5.25/3=1.75 or 7/4 </span>
<span>y=7/4x+b enter one point for the values of x and y to find b </span>
<span>10.5=(1.75X5) + b </span>
<span>10.5=8.75 + b </span>
<span>b=1.75 </span>
<span>The equation of the line is y=7/4x+1.75 (I also double checked using the second point) </span>
<span>Then all you have to do is enter the mileage 3.8 for the value of X and solve for y </span>
<span>y=(7/4 X 3.8) + 1.75 </span>
<span>y=6.65+1.75 </span>
<span>y=8.4 The cab ride for the 3.8mile trip is $8.4
I hope this helps</span>
Answer:
28
Step-by-step explanation:
You follow the order of operations - PEMDAS
Answer:
Yes a third plane can be drawn which contains parallel lines.
Step-by-step explanation:
There are two points:
1) a third plane can cab drawn which can have two line parallel as shown in the figure 1.
2) If plane A has a line and plane B has a line such that they are parallel to each other than the third plane cannot be drawn which contains both lines and b parallel.
A third plane can have only two lines parallel if they are not parallel in the in the other two planes.
We see that lines c and d parallel in the third plane are perpendicular in the planes A an d B.