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2 years ago

The graph shown here is the graph of which of the following rational functions?

Mathematics

1 answer:

Iteru [2.4K]2 years ago

7 0

Answer:

Step-by-step explanation:

From the graph you see point (1, -2) is on f(x)

So substitute x = 1 and the resuult should be -2.

The only function that is valid, is function B.

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Her 6. Karen buys a sweatshirt for $30. The sales

Fynjy0 [20]

Answer:

32.1

Step-by-step explanation:

you take 30 times .07 (this was 7% but since you need it to be a decimals you move it over 2 places left). You should get 2.1 and then you add the 2.1 and 30 to get 32.1

7 0

3 years ago

PLEASE HELP MEH WILL ARK BRAINLIST

otez555 [7]

Answer:

a,e,c,b,d

Step-by-step explanation:

i think that is it but im not completly sure

7 0

3 years ago

Read 2 more answers

3 solutions to y=-5x+1

Luda [366]

Y = -5x + 1

x = 1 , y = -4
x = 2 , y = -9
x = 3 , y = -14

so the answers are
(1,-4)
(2,-9)
(3,-14)

7 0

3 years ago

A cup of coffee at 93 degrees Celsius is placed in a room at 23 degrees Celsius. Suppose that the coffee cools at a rate of 1 de

Alinara [238K]

Answer:

Step-by-step explanation:

The rate of change of temperature of the coffee with respect to time is expressed by the differential equation

dT/dt=k(T−A) when k = -1 and A = 23. The equation will become:

dT/dt = -(T-23)

If the coffee cools at the rate of 1°C per minute, then dT/dt = -1 and T = 70

Substituting into the equation:

-1 = k(70-23)

-1 = 47k

k = -1/47

k = -0.0213

Substituting k = -0.0213 into the original equation, the differential equation will be:

dT/dt=k(T−A)

dT/dt = -0.0213(T-23)

dT/dt = -0.0213T+0.489

To get the value of T, we will use variable separable method

dt = dT/-0.0213T+0.489

Integrate both sides

t = -0.0213ln(-0.0213T+0.489)

At t = 1 minute

1 = -0.0213ln(-0.0213T+0.489)

1/-0.0213 = ln(-0.0213T+0.489)

-46.95 = ln(-0.0213T+0.489)

Apply exp to both sides

e^-46.95 = e^ln(-0.0213T+0.489)

4.073×10^-21= -0.0213T+0.489

-0.0213T = 4.073×10^-21-0.489

-0.0213T = -0.489

T = 0.489/0.0213

T = 22.96°C

5 0

3 years ago

Premises: All good students are good readers. Some math students are good students. Conclusion: Some math students are good read

just olya [345]

Answer:

Therefore, the conclusion is valid.

The required diagram is shown below:

Step-by-step explanation:

Consider the provided statement.

Premises: All good students are good readers. Some math students are good students.

Conclusion: Some math students are good readers.

It is given that All good students are good readers, that means all good students are the subset of good readers.

Now, it is given that some math students are good students, that means there exist some math student who are good students as well as good reader.

Therefore, the conclusion is valid.

The required diagram is shown below:

3 0

3 years ago