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2 years ago

15

PLEASE HELP ITS OVERDUE PLEASE I NEED HELP!!

1 answer:

IgorLugansk [536]2 years ago

5 0

Answer:

First question: Choice A

second question:Choice D

Third question: Choice c

Fourth question: Choise b

5th Question: choice b

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Y=sech(tan‐¹x) find the derivative

Klio2033 [76]

Recall that

d/dx sech(x) = - sech(x) tanh(x)

d/dx tan⁻¹(x) = 1/(1 + x²)

Then by the chain rule,

dy/dx = - sech(x) tanh(x) / (1 + x²)

8 0

2 years ago

A welder works 8 hours a day at a rate of $25.00 per hour. The overtime rate is doubled the normal rate. On a particularday, the

Alborosie

Answer:

C.

Step-by-step explanation:

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5 0

3 years ago

You have a part-time job. You work for 3 hours on Friday and 6 hours on Saturday. You also receive an allowance of $20 per week.

olga55 [171]

Answer:

8 dollars per hour

Step-by-step explanation:

You make 92 a week

20 of that is from your allowance

92-20 = 72

72 is from your part time job

You work for 3 hours on Friday and 6 hours on Saturday, for a total of 3+6 = 9 hours

Take the amount of money earned and divide by the number of hours to determine the money earned per hour

72/9 = 8 dollars per hour

4 0

3 years ago

Read 2 more answers

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

2 2/3 multiplied by 5 4/5

grin007 [14]

$2 \frac{2}{3} = \frac{8}{3} 
$
$5 \frac{4}{5} = \frac{29}{5}$
$\frac{8}{3} * \frac{29}{5} =X$
Solve for X. What do you think the answer is based on that?

8 0

3 years ago