Answer : 33 horses. Number of Horses (H) + Number of Chickens (C) = 45 therefore C = 45 - H There are 4H horse legs and 2C chicken legs so, 4H + 2C = 156 substituting for C gives :- 4H + 2(45 - H) = 156 4H + 90 - 2H = 156 2H = 156 - 90 = 66 H = 33 There are 33 horses (and 45 - 33 = 12 chickens)
If you mean 4h+6 = 30 then h has a value of 6
To solve for h, first isolate 4h: T=4h+4z T-4z=4h Now divide both sides by 4. T/4-z=h
The like terms are 3hk and 6kh
yea yea yea
4h + 6 = 22 So 4h = 16 and then h = 4
need to be in neutral or park .
If it is the same as my 1974 cj5, then its 1. 4L 2. N 3. 4H 4. 2H My 75 has a pattern of (starting with shifter full forward), which is the same on all Dana 20 transfer cases 4H 2H N 4L
125(2h - p^7)(4h^2 + 2hp^7 + p^14)
t handle shifter has a pattern of 2H, 4H, N, 4L
On hard road surfaces, the normal transfer case range is typically set to 2H or 4H. 2H is used for regular driving on dry, paved surfaces, while 4H can be engaged for added traction in slippery conditions like rain or light snow.
Answer : 33 horses. Number of Horses (H) + Number of Chickens (C) = 45 therefore C = 45 - H There are 4H horse legs and 2C chicken legs so, 4H + 2C = 156 substituting for C gives :- 4H + 2(45 - H) = 156 4H + 90 - 2H = 156 2H = 156 - 90 = 66 H = 33 There are 33 horses (and 45 - 33 = 12 chickens)
-98-4h+8k-40+k-9h = (-98-40)+(-4h-9h)+(8k+k) = -138-13h+9k
If you mean 4h+6 = 30 then h has a value of 6
2g + 5h
Shifting from 4H to 4L (and 4L to 4H) 1. Bring the vehicle to a stop. 2. Depress the brake. 3. Place the gearshift in N (Neutral). 4. Move the 4WD control to the 4H (or 4L) position. Shifting from A4WD to 4H Move the 4WD control from A4WD to 4H at any forward speed. Shifting from 2H to 4H can be done at speeds up to 88 km (55 mph).
I2 + 10 hno3 = 2 hio3 + 10 no2 + 4 h2o