Answer:
Multiply vector c by the scalar -1/2.
Step-by-step explanation:
Look at vector c.
It has an x component of 4 and a y component of 4.
You can write vector c as a sum of its components using unit vectors in the x direction (i) and in the y direction (j).
c = 4i + 4j
Now look at vector d, and write it also as a sum of its x and y components.
d = -2i - 2j
Now ask yourself, what operation do I do to 4 to end up with -2?
One answer is to multiply 4 by -1/2.
d = (-1/2)c = (-1/2)(4i) + (-1/2)(4j) = -2i - 2j
That worked. By multiplying vector c by the scalar -1/2, you end up with vector d.
Answer:
16:1
Step-by-step explanation:
The ratio of the surface areas of the similar solids is the square of the lengths.
(4:1)²
4²:1²
⇒ 16:1
<h3>
Answer: c. 8(y-6) = (x-2)^2</h3>
Explanation:
The directrix is horizontal, so the axis of symmetry is vertical. We'll have an x^2 term. The vertical distance from y = 4 to y = 8 is 4 units. Cut this in half to get 2, which is the focal distance p = 2.
The point (2,4) is directly below (2,8), and the point is on the directrix. The midpoint between (2,4) and (2,8) is (2,6). This is the vertex.
(h,k) = (2,6)
4p(y-k) = (x-h)^2
4*2(y-6) = (x-2)^2
8(y-6) = (x-2)^2
Answer: 0.28
Step-by-step explanation:
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min