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2 years ago

What is the domain of the function represented by the graph?

Mathematics

2 answers:

kupik [55]2 years ago

6 0

Your answer will be B

Reika [66]2 years ago

3 0

I’m positive the answer is B!

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KIS+KSI=ISK WHAT IS THE NUMBERS?!?!??!?

8090 [49]

Is this a real question ?

5 0

3 years ago

Read 2 more answers

There are 5 very different seats in a car. In how many ways can 5 different people be seated in the car for a road trip if only

avanturin [10]

Answer:

Step-by-step explanation:

Let A and B be the two people who are able to drive. If A is driving, there are 4! ways to arrange the remaining peoplein the car seats. If B is driving, there are also 4! ways to arrange the remaining people. The number of arrangements 'n' is:

$n=2*4!\\n=2*4*3*2*1\\n=48\ ways$

They can be arranged in 48 ways.

6 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

Let f= {(-5,1).( - 4,0).(0,6)} Find f(0) f(0)=

sweet [91]

Given:

It is given that the coordinates of the function are $f=\{(-5,1),(-4,0),(0,6)\}$

We need to determine the value of f(0)

Value of f(0):

The value of f(0) is the value of the function when the input is 0.

We need to determine the value of f(0) when the input is x = 0

Thus, from the coordinates of the function $f=\{(-5,1),(-4,0),(0,6)\}$ , it is obvious that the input value of the function is x = 0 , then the value of the function f(0) = 6.

Thus, when the input value x = 0, the output is f(0) = 6.

Therefore, the value of f(0) = 6.

5 0

2 years ago

How does charge Build Up?

Masteriza [31]

Answer: Charges build up when negative electrons are transferred from one object to another. The object that gives up electrons becomes positively charged, and the object that accepts the electrons becomes negatively charged

7 0

3 years ago