It is 7 regions because you can make a triangle with these lines and there can be one middle region then you would have 6 outside regions. Draw out a triangle but the line intersect on each pointy end. If this is right could I possibly have brainliest? Hope this helped!! :)
Answer:
y=1/2x-2
Step-by-step explanation:
RIse / run
find a point then go up one and right two to the next point which gives you 1/2
the y intercept is -2 because thats where the line is interceting
Answer:
Step-by-step explanation:
TYSM
HAVE A GREAT DAY
Answer:
option 3
Step-by-step explanation:
![(4 + 3) \sqrt[5]{x {}^{2}y } = 7 \sqrt[5]{ {x}^{2}y }](https://tex.z-dn.net/?f=%284%20%2B%203%29%20%5Csqrt%5B5%5D%7Bx%20%7B%7D%5E%7B2%7Dy%20%7D%20%3D%207%20%20%5Csqrt%5B5%5D%7B%20%7Bx%7D%5E%7B2%7Dy%20%7D%20)
The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)
[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²
Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N
Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
<span>Units have to be consistent ... so have to convert 75.0 cm to m: </span>
75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
<span>work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J</span>
