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:
The answer is A
Step-by-step explanation:
Answer:
18.8
14+12+21+26+19+15+22+17+24+18 = 188 188/10
I don't remember how to do stem and leaf plots.
Step-by-step explanation:
Below is an attachment containing the solution.
Answer:
B. AAS
Step-by-step explanation:
The diagram shows two triangles ZVY and WVY.
In these triangles,
- ∠VZY ≅ ∠VWY - given;
- ∠ZVY ≅ ∠WVY - given;
- VY ≅ VY - Reflexive property of Equality
Thus, we have two pairs of congruent angles and a pair of not-included conruent sides.
AAS Postulate states that if two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.
So, we can use AAS Postulate.
