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2 years ago

9

The radius of a wheel is 9.0 cm, and the radius of the axle is 0.6 cm. If the effort force is 75 N and the resistance force is 9

00 N, what is the AMA?
5
8
12
15

1 answer:

astraxan [27]2 years ago

7 0

Answer:

Jazmine may I azk you a question?

I'm at NBLC with u btw

Step-by-step explanation:

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Mhanifa please help :( no one has helped me on this yet

VMariaS [17]

Answer:

16)

q / 12 = sin 20
q = 12 sin 20
q = 4.1

p / 12 = cos 20
p = 112 cos 20
p = 11.3

17)

x / 44 = tan 39
x = 44 tan 39
x = 35.6

35.6 / y = sin 39
y = 35.5 / sin 39
y = 56.4

18)

14 / b = sin 57
b = 14 / sin 57
b = 16.7

a / 16.7 = cos 57
a = 16.7 cos 57
a = 9.1

3 0

3 years ago

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

Answer to the equation -6+2c=3c-(6+5)

Phantasy [73]

Answer:

c = 5

Step-by-step explanation:

- 6 + 2c = 3c -(6+5)

-6 + 2c = 3c -11

-6 + 11 = 3c- 2c

5 = c

3 0

2 years ago

What is the area of the figure

mel-nik [20]

The entire picture is needed to solve. the bottom of the pic is cut off

6 0

2 years ago

A copy machine repairman makes $12.50 per hour plus $4 for every service call he performs. Last week he worked 34 hours and made

Anon25 [30]

z=total pay
x=hours
y=service calls

7 0

3 years ago