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3 years ago

Graph the solution x

Mathematics

1 answer:

mixas84 [53]3 years ago

8 0

Answer:

I think you need to put more info

Step-by-step explanation:

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Somebody please help

Gnesinka [82]

What do u need help with

5 0

3 years ago

Plz help I need this to be answered

yawa3891 [41]

The correct graph is D.

We can tell this for two reasons. Firstly, you would need a dotted line. Since the equation includes a greater than sign, and not a greater than or equal to sign, it must be dotted.

Then we need to figure out whether to shade above or below the line. The best way to find this is by checking whether the point (0,0) should be shaded.

y + 2 > -3x - 3
0 + 2 > -3(0) - 3
2 > -3 (TRUE)

Since this is a true statement, the side with (0,0) needs to be shaded. Graph D is the only one that has a dotted line and (0,0) shaded.

8 0

3 years ago

In what direction and by how many units is the graph of f(x) = −7 sin(3x + π) − 2 vertically and horizontally shifted?

Nezavi [6.7K]

I think its down 2 left pi/3, but did you get the answer?

7 0

3 years ago

Read 2 more answers

PLEASE HELPPP

posledela

PLEASE HELPPP
The function f is giving by
f(x) = (x - 2)(x +4). What are the x-intercepts of the graph representing f?

6 0

2 years ago

Read 2 more answers

Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

7 0

3 years ago