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3 years ago

Can someone help me with me with my most recent question please

Mathematics

1 answer:

mojhsa [17]3 years ago

8 0

Answer:

no I can't

Step-by-step explanation:

im a dum dum, sorry.

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PLEASE ANSWER + BRAINLIEST !!!

Alchen [17]

x = 100 cm^2

d = 1.4*(x^1/2)

d = 1.4*(100^1/2)

d = 1.4*10

d = 14 cm

Hope this helps!

5 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

Doug had about $3,215 in savings while his brother had double that amount in savings. What is a reasonable amount that the broth

MaRussiya [10]

Answer: i think you would have to multiply

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud

SSSSS [86.1K]

Answer:

The large sample n = 190.44≅190

The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

$M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }$

The 85% confidence level $z_{\alpha } = 1.44$

$\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}$

$\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}$

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

7 0

3 years ago

Sonia finished the 200-meter dash in 26.12 seconds. Gladys finished the 200-meter dash 1.08 seconds

Schach [20]

Answer:

25.04

Step-by-step explanation:

26.12-1.08

=25.04

7 0

3 years ago

Read 2 more answers