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3 years ago

Determine whether the graph is the graph of a function.

Mathematics

2 answers:

photoshop1234 [79]3 years ago

8 0

Answer:

Yes

Step-by-step explanation:

Unless the line is vertical, the graph of any straight line is the graph of a function.

___

A vertical line does not map x to one single y-value, so it does not represent a function. Any other line does map each x to a single y-value, so does represent a function.

Arada [10]3 years ago

5 0

<h2>Answer:</h2>

Yes, the graph is the graph of a function.

<h2>Step-by-step explanation:</h2>

<u>Function--</u>

A function is a collection of all the ordered pair such that each element of the first set is mapped to a single element of the other set.

i.e. each element has a single image.

i.e. no elements of the first set is mapped to more than one element.

Also, the graph of the function passes the vertical line test.

i.e. any line passing through the domain and parallel to the y-axis should intersect the graph at least once.

Hence, from the graph we observe that the graph passes the vertical line test.

Hence, the graph is a graph of a function.

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In Triangle ABC, the length of Segment AB is equal to the length of segment AC. The length of segment BC is 4 less than twice th

seraphim [82]

Step-by-step explanation:

Let x be the length of segment AB.

Then the length of segment BC is (2x - 4).

The length of segment AC is x.

We know that x + (2x - 4) + x = 52.

Therefore 4x - 4 = 52, 4x = 56, x = 14.

Hence the length of segment AB is 14.

7 0

3 years ago

Which expression is equivalent to 36/(4 + y)?

I am Lyosha [343]

Answer:

B) 36/(y+4)

Step-by-step explanation:

Use the process of elimination.

A cannot work because we are changing the operation. Instead of dividing thirty-six by the sum of four and a number, we are dividing thirty-six by four, then adding the unknown number. This is not the same as the original expression, so we know it is not this.

Next, check C. C doesn't seem right either. Remember that, unlike multiplication and addition, you cannot flip the order in which the operation is being done and still get the same answer. C says four plus a number divided by thirty-six, which is entirely different from our original expression. So the answer is not C.

And finally, check D. D doesn't work either. With D, we are dividing thirty-six by a number, then adding it to thirty-six divided by four. This is not the same as our original expression, so D isn't it.

That leaves us with B. Does it work? Yes, because changing the y and the four still will give the same thing. And yes, because 36 is being divided by the sum of that number and four. So it is our answer.

8 0

1 year ago

Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.

Sladkaya [172]

Answer:

The solution is $\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

Step-by-step explanation:

From the question

The function given is $f(x) = \frac{e^{2x}}{ 25 + e^{4x}} dx$

The indefinite integral is mathematically represented as

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx$

Now let $e^{2x} = u$

=> $\frac{du}{dx} 2e^{2x}$

=> $2 e^{2x}dx = du$

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx = \int\limits {\frac{1}{ 2(25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 5^2 + u^2)} } \, du$

$= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5} + C$

Now substituting for u

$\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

3 0

2 years ago

Complete the table i’ll give brainliest

elena-s [515]

Answer:

1. 4

2. 11

3. 18

4. 25

Step-by-step explanation:

8 0

2 years ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago