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2 years ago

5

Can somebody help me

1 answer:

Sergeeva-Olga [200]2 years ago

6 0

$cis(x)=\cos x+i\sin x\\\\cis15^o=\cos15^o+i\sin15^o\\\\\cos15^o=\dfrac{\sqrt6+\sqrt2}{4};\ \sin15^o=\dfrac{\sqrt6-\sqrt2}{4}\\\\\text{therefore}\\\\7cis15^o=7\left(\cos15^o+i\sin15^o\right)=7\cos15^o+7i\sin15^o\\\\=7\left(\dfrac{\sqrt6+\sqrt2}{4}\right)+7\left(\dfrac{\sqrt6-\sqrt2}{4}\right)i$

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Pls help me solve this

NISA [10]

Answer:

$50

Step-by-step explanation:

12 x 30 = 360

960 - 360 = 600

600/12 = 50

3 0

2 years ago

Read 2 more answers

Two algebra questions! please help :(

Fiesta28 [93]

The answer is B. h=<span>V<span>2π<span>r<span>2</span></span></span></span>

8 0

2 years ago

Please help me with problem 22a I dont understand how to find the rate of change

Anna [14]

The rate of change is x2. So, since it's forty, it became 80 because 40x2=80!
Good luck for a good grade!

5 0

3 years ago

ANSWER ASAP PLEASE

vazorg [7]

Answer:

C

Step-by-step explanation:

6 0

2 years ago

Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution

Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

$y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\$

for y=e^{6x}:

$y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\$

Now we replace in the differential equation y'' − y' − 30y = 0

for y=e^{-5x}:

$25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0$

for y=e^{6x}:

$36e^{6x}-6e^{6x}-30=0\\36-6+30=0$

Now, to know if both function are linearly independent we calculate the Wronskian

$W(f,g)=fg'-f'g$

$W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0$

I hope this is useful for you

Best regard

7 0

3 years ago