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3 years ago

Match the key aspect of a function's graph with its meaning.

Mathematics

1 answer:

Margarita [4]3 years ago

5 0

Answer:

Part 1) Intervals of the domain where the graph is above the x-axis (f(x) > 0)

Part 2) location on graph where input is zero (y-intercept)

Part 3) location on graph where output is zero (x-intercept)

Part 4) Intervals of the domain where the graph is below the x-axis (f(x) < 0)

Step-by-step explanation:

<u><em>Verify each case</em></u>

Part 1) we have

Intervals of the domain where the graph is above the x-axis

we know that

If the graph is above the x-axis, then the value of f(x) is positive

therefore

f(x) > 0

Part 2) we have

location on graph where input is zero

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The y-intercept is the value of f(x) (output value) when the value of x (input value) is zero

therefore

y-intercept

Part 3) we have

location on graph where output is zero

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The x-intercept is the value of x (input value) when the value of the function f(x) (output value) is zero

therefore

x-intercept

Part 4) we have

Intervals of the domain where the graph is below the x-axis

we know that

If the graph is below the x-axis, then the value of f(x) is negative

therefore

f(x) < 0

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I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

$\frac{dT}{dt}=k*(T_r-T)$

If we rearrange and integrate

$\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}$

We know that at time 0, the temperature of the drink was 52°F. Then we have:

$T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20$

We also know that at t=2, T=55°F

$T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08$

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

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3 years ago

Solve the following quadratic equation. Express your answers in exact form.

wlad13 [49]

Answer:

x = (2+√10)/ 3 and x= (2-√10/3)

Step-by-step explanation:

g(x) = -3x² +4x +2

use the quadratic formula x= (-b±√b²-4ac)/2a, when g(x) = ax² +bx +c

x= [-4 ±√4²-4*(-3)*2] / 2*(-3) , solve under the square root

x= (-4 ± √16 +24) / -6 , add under the radical

x= (-4 ±√40) / -6, simplify the square root

x= (-4 ±2√10) /-6, factor -2

x= -2 (2±√10)/-2 *3, simplify -2

x= (2±√10)/3

Therefore,

x = (2+√10)/ 3 and x= (2-√10/3)

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