1) 3S = R 2) R+6= 2 (S+6)= 2S+12 ----> R=2S+61,2 )--> 3S = 2S+6 ---><span>S=6 & R= 3 x 6 = 18Ralph is 18 & Sara is 6 .</span>
F, it goes down two for every three spaces, and crosses the y axis at two.
Answer:
So we have 
.
Step-by-step explanation:
Ok so we are in quadrant 2, that means sine is positive while cosine is negative.
We are given 
.
So to find the opposite we will just use the Pythagorean Theorem.
This is the opposite side.
Now to find 
and 
.
.
Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).
So 
.
And now 
.
So we have .
Answer:
(9 - 4 x)/(x (2 x - 4))
Step-by-step explanation:
Simplify the following:
1/(2 x^2 - 4 x) - 2/x
Put each term in 1/(2 x^2 - 4 x) - 2/x over the common denominator x (2 x - 4): 1/(2 x^2 - 4 x) - 2/x = ((x (2 x - 4))/(2 x^2 - 4 x))/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4)):
((x (2 x - 4))/(2 x^2 - 4 x))/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4))
A common factor of 2 x - 4 and 2 x^2 - 4 x is 2 x - 4, so (x (2 x - 4))/(2 x^2 - 4 x) = (x (2 x - 4))/(x (2 x - 4)):
((x (2 x - 4))/(x (2 x - 4)))/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4))
(x (2 x - 4))/(x (2 x - 4)) = 1:
1/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4))
1/(x (2 x - 4)) - (2 (2 x - 4))/(x (2 x - 4)) = (1 - 2 (2 x - 4))/(x (2 x - 4)):
(1 - 2 (2 x - 4))/(x (2 x - 4))
-2 (2 x - 4) = 8 - 4 x:
(8 - 4 x + 1)/(x (2 x - 4))
Add like terms. 1 + 8 = 9:
Answer: (9 - 4 x)/(x (2 x - 4))
Answer:
285,753 cm³/min
Step-by-step explanation:
The rate of change of volume is the product of the water's surface area and its rate of change of depth.
At a depth of 2 m, the water has filled 1/3 of the 6 m depth of the tank. So, the radius at the water's surface will be 1/3 of the tank's radius of 2 m. The water's surface area is ...
A = πr² = π(2/3 m)² = 4π/9 m²
The rate of change of depth is 0.2 m/min, so the volume of water is increasing at the rate ...
dV/dt = (0.20 m/min)(4π/9 m²) = 8π/90 m³/min ≈ 279253 cm³/min
This change in volume is the difference between the rate at which water is being pumped in and the rate at which it is leaking out:
2.8×10⁵ cm³/min = (input rate) - 6500 cm³/min
Adding 6500 cm³/min to the equation, we get ...
input rate ≈ 285,753 cm³/min
