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2 years ago

Is (0, 8) a solution of 4x – y = 8? yes no cannot be determined

Mathematics

1 answer:

jolli1 [7]2 years ago

3 0

No, (0-8) is the first point plotted on the graph but not (0,8).

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Hey, can someone solve this for me? 5(x+1)= 4(x-1)+x

Travka [436]

5(x+1)= 4(x-1)+x
5x + 5 = 4x - 4 + x
5x +5 = 5x -4
5 ≠ -4
x ∈ ∅

3 0

3 years ago

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

When the following quadratic equation is written in standard form, what is the value of "c"?

Nikolay [14]

Answer:

Then value is 2

Step-by-step explanation:

standard of quadratic is A $x^{2}$ +Bx+C=0

and 2 is where C is.

sorry for the bad explanation but, trust me I did this type of question in class yesterday

7 0

2 years ago

Solve for x 14 -2x = \frac{3 + x}{2}

Stels [109]

I think x = 2 is the answer

6 0

2 years ago

A new grocery store is expected to have a fair market value of $2,856,000. It will be assessed at 40%, and the tax rate is $5.24

olganol [36]

Assessed Value = (Fair Market Value) x (0.40) Where 0.40 is the decimal equivalent of 40%. Tax Rate = $5.24/$100 of assessed value = $0.0524 per dollar of assessed value. Taxes = (Assessed Value) x (Tax Rate) = (Assessed Value) x ($0.0524) Hope this helps!

4 0

3 years ago

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