If I were you, I would make the starting point (3,-6). From there, you will want to use the slope of -1/2 (go down 1 unit and to the right 2 units and draw a point)
Y=mx+b
m=slope
b=yint
given
m=2
b=4/5
easy
y=2x+4/5
A
3/5= 3 fifth size strips
You want to find an equivalent fraction with the denominator of ten, correct? Then you would multiply the fraction in a way that will allow you to have a denominator of ten.
So...
3/5*2/2=6/10
You would need 6 tenth size strips.
Answer:x=1.2
Step-by-step explanation:
3x+2.5=6.1
3x=6.1-2.5
3x=3.6
x=1.2
Answer:
1) 
2) 
3) 
4) 
Step-by-step explanation:
1) 
Solving using exponent rule: 

So, 
2) 
Using the exponent rule: 
We have:

We also know that: 
Using this rule:

So, 
3) 
Solving:

So, 
4) 
We know that: 

So, 