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iris [78.8K]
3 years ago
8

Which of the following graphs has the largest average rate of change from -1 to 2?​

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
3 0

Answer:

i dont know either

Step-by-step explanation:

konstantin123 [22]3 years ago
3 0

Answer:

-2 it should be enojy you answer

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Find the solution to the syster<br> y = 2x = 5<br> y = X - 2
lana66690 [7]

Answer:

Both of these equations are in slope-intercept form. The slope-intercept form of a linear equation is: y=mx+b

Where m is the y-intercept value.

y=-2x+5

y=-2x+20

The slope of the two equations are: m1=−2 and m2=−2

Step-by-step explanation:

Because the have the same slope it means the lines represented by these two equations are either parallel or are the same line.

The y-intercepts for the two lines are:b1 = 5 and b1=20

3 0
3 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

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

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

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

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
3 years ago
Put the following equation of a line into slope-intercept form, simplifying all fractions. 4y - 4x -32​
MariettaO [177]

Answer:

y=-4x+4x-32

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Conan had 180$ in the bank. Each week he deposits another 64$ that he earns mowing lawns. Is his account balance proportional to
balu736 [363]
No, because 64 does not go into 180 evenly.
3 0
3 years ago
Under a translation of 5 units up and 2 units to the left, the point (3,4) will become
Vedmedyk [2.9K]
The point would become (8,2)
3+5=8
4-2=2
4 0
3 years ago
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