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3 years ago

02.05) simplify i22. (1 point) 1 −1 −i i

Mathematics

1 answer:

olga2289 [7]3 years ago

7 0

I^22 = i^(20 + 2) = i^20 x i^2 = (i^4)^(5) x (i)^2 = 1 x -1 = -1

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3 years ago

Solve the given differential equation by undetermined coefficients. y'' 4y = 7 sin(2x)

kherson [118]

The solution to the given differential equation is yp=−14xcos(2x)

The characteristic equation for this differential equation is:

P(s)=s2+4

The roots of the characteristic equation are:

s=±2i

Therefore, the homogeneous solution is:

yh=c1sin(2x)+c2cos(2x)

Notice that the forcing function has the same angular frequency as the homogeneous solution. In this case, we have resonance. The particular solution will have the form:

yp=Axsin(2x)+Bxcos(2x)

If you take the second derivative of the equation above for yp , and then substitute that result, y′′p , along with equation for yp above, into the left-hand side of the original differential equation, and then simultaneously solve for the values of A and B that make the left-hand side of the differential equation equal to the forcing function on the right-hand side, sin(2x) , you will find:

A=0

B=−14

Therefore,

yp=−14xcos(2x)

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2 years ago

Can someone please help me I really need help please help me thank you

just olya [345]

Answer:

1800 or D.

Step-by-step explanation:

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3 years ago

Read 2 more answers

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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3 years ago

What is the image below simplified as a fraction?

marysya [2.9K]

Answer:

159.8195397043133

Step-by-step explanation:

2²⁰

------------------ = 159.8195397043133

3⁸

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3 years ago