Answer:
going horizontally, 15, 3, 9, then 1, 3, 6, then 20, then, 40, 60
Step-by-step explanation:
Answer:
If the robot has more mass, then it will take a bigger force to lift up another object. If the robot is going to lift something up, it will use its arms, and its arms have mass, and the object has a mass as well. If the robot's arms have a mass of 10kg, and the object has a mass of 12kg, it will use a certain amount of force to lift up the object using it's arms. If the the robots arms were 5kg instead, it would take less force to lift of the same object using its arms.
9514 1404 393
Answer:
(f/g)(8) = -104/41
Step-by-step explanation:
Fill in the value for x and do the arithmetic.
f(8) = 3 -2·8 = 3 -16 = -13
g(8) = 1/8 +5 = (1+40)/8 = 41/8
Then the ratio is ...
(f/g)(8) = f(8)/g(8) = -13/(41/8) = -13(8)/41
(f/g)(8) = -104/41
Answer:
-20/13 <g
Step-by-step explanation:
-7–5(3g+8)<10g–7+g
Distribute
-7–15g-40<10g–7+g
Combine like terms
-15g - 47 < 11g -7
Add 15 g to each side
-15g+15g -47< 11g+15g -7
-47 < 26g -7
Add 7 to each side
-47+7 < 26g-7+7
-40 < 26g
Divide each side by 26
-40/26 <26g/26
-40/26 <g
Divide top and bottom by 2
-20/13 <g
Answer:
8.75cm²
Step-by-step explanation:
First we need to find the angle subtended by the arc
L = theta/360 * 2pir
3.5 = theta/360 * 2*3.14(5)
3.5 = theta/360 * 31.4
theta/360 = 3.5/31.4
theta/360 = 0.11146
theta = 0.11146 * 360
theta = 40.13 degrees
Area of the sector = theta/360 * pir^2
Area of the sector = 40.13/360 * 3.14*25
Area of the sector = 40.13/360 * 78.5
Area of the sector = 8.75cm
Hence the area of the sector is 8.75cm²
