<em>Answer:</em>
<em>x = 3</em>
<em>Step-by-step explanation:</em>
<em>Hi there ! </em>
<em>0 = - 2x + 6</em>
<em>2x = 6</em>
<em>x = 6 : 2</em>
<em>x = 3</em>
<em>Good luck !</em>
Answer:
D) 5
Step-by-step explanation:
Isolate the variable, y. Note the equal sign, what you do to one side, you do to the other. Divide 2 from all terms within the equation:
(2y)/2 = (10x - 8)/2
y = (10x)/2 - (8)/2
y = 5x - 4
Note the slope-intercept form:
y = mx + b
y = y
x = x
m = slope
b = y-intercept.
In this case, 5 is in place of m, or your slope.
D) 5 is your answer.
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If you had the number 4 1/4 you would divide the numerator (1) by the denominator (4) and then take the decimal number (0.25) you get and add the decimal to the whole number on your mixed number (4+0.25) and that would give you the decimal.
Answer:
Graph A
Step-by-step explanation:

Its mean graph should intersect x axis at x = 7 and x = - 1. This everything is satisfied by graph A.
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