From sin(180-x) = sin(x)
sin(150) = sin(180-30) = sin(30)
sin(120) = sin(180-60) = sin(60)
from cos(360-x) = cos(x)
cos(300) = cos(360-60) = cos(60)
cos(210) = cos(360-150) = cos(150)
from cos(180-x) = -cos(x)
cos(210) = cos(150) = cos(180-30) = -cos(30)
Answer:
The answer is -22
Step-by-step explanation:
remove the parentheses then you have 11 * -2 = -22
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Answer:
dy/dt = - (1/5) ft/s = - 0.2 ft/s
Step-by-step explanation:
Given
L = 5 ft
Qin = 25 ft³/s
Qout = 30 ft³/s
h = 10 ft
dy/dt = ?
We can apply the relation
ΔQ = Qint - Qout = 25 ft³/s - 30 ft³/s
⇒ ΔQ = - 5 ft³/s
Then we use the formula
Q = v*A
where Q = ΔQ, A = L² is the area of square base and v = dy/dt is the rate of change in the depth of the solution in the tank
⇒ ΔQ = (dy/dt)*L²
⇒ dy/dt = ΔQ/L²
⇒ dy/dt = (- 5 ft³/s)/(5 ft)²
⇒ dy/dt = - (1/5) ft/s = - 0.2 ft/s
Step-by-step explanation:
W=-6, x=1.2, and z=-6/7
(W²x-3)÷10-z
we substitute
((-6²)(1.2)-3)÷10-(-6/7)
((-36)(1.2)-3) ÷10-(-6/7)
(-43.2-3) ÷10(6/7)
(-46.2)÷60/7
-46.2÷60/7
-46.2*7/60
-46.2/1*7/60
-323.4/60
-5.39
