Please help me find the function rule, i dont know what to do!! I will give big brain

A line passes through the points R(-1,-5) and S(-3,-1). Write the equation of the line through the point A(4,0) that is parallel

IRISSAK [1]

Sorry I was moving around taking the picture and my handwriting is sloppy. But I hope you understand how I got the answer

4 0

2 years ago

5. Noah is depositing money in his account every week

jeka57 [31]

Answer:

-60/7 or -8 4/7 would be the answer

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

PLEASE HELP URGENTLY!!!!¡!!!!!!¡

DaniilM [7]

<h3>There are 3 answers: Choice A, Choice D, Choice F</h3>

Explanation:

2 & 1/2 = 2.5

10 + (-2.5) can be represented by starting with 10 and then dropping by 2.5, which is what choice A is saying

or we can start with -2.5 and add on 10. This works because we can add numbers in any order, example: 2+3 = 3+2 = 5. So starting with -2.5 and adding on 10 is represented by choice D's and choice F's examples

4 0

2 years ago

Fernanda and Jose were working on an assignment in their geometry class in which they were to write the equation of the graph be

Alborosie

Answer:

yes they both are wrong

Step-by-step explanation:

Center of the Circle is (h,I) which is (4,-3)

so h= 4 and k = -3

radius is 4 units (3+1)

Equation for circle

$(x - h) {}^{2} + (y - k) {}^{2} = r {}^{2}$

substitute all the values in this equation

$(x - 4) { }^{2} + (y - ( - 3)) {}^{2} = 4 {}^{2} \\ (x - 4) {}^{2} + (y + 3) {}^{2} = 16$

4 0

2 years ago