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3 years ago

11020+12 In base 3 pls help

Mathematics

2 answers:

PilotLPTM [1.2K]3 years ago

7 0

It’s 142 because 12020+12 is 142 I did some research

Alinara [238K]3 years ago

4 0

Answer:

I would help but this question is hard lol

Step-by-step explanation:

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Find the area of the regular figure below:

Mice21 [21]

Answer:

1,496.49 cm squared

Step-by-step explanation:

The regular figure below is a hexagon.

The formula for the area of a regular hexagon is $A = \frac{3\sqrt{3} }{2} a^2$ . Here a = 24. Substitute and simplify for the area.

$A = \frac{3\sqrt{3} }{2} a^2\\A = \frac{3\sqrt{3} }{2} 24^2\\A = \frac{3\sqrt{3} }{2} * 576\\A = 1496.49$

6 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

PLEASE HELP ME PLEASEEEEEEEEEEEEEEEEEEE!

mina [271]

Answer:

The most helpful representation would be option C.

Step-by-step explanation:

The horizontal distance traveled corresponds to one of the roots of the quadratic equation.

h(x) = -0.01(x-150)(x+4)

In this case, option c presents the roots of the polynomial, which answers are

x = -4

x = 150

Since we are talking about positive distances, the first option gets discarded.

So the horizontal distance is x = 150 feet

5 0

3 years ago

What is 2.006 in expended form

Verizon [17]

2+ 0.0 + 0.00 + .006

5 0

3 years ago

Can anyone help me with this one

aleksandrvk [35]

50 hours!

825 (paycheck left over) + 25 (the shirt that was subtracted because he bought it) = $850 total
850 (total amount this week) / 17 (the amount he makes per hour) = 50 (total amount of hours worked this week)

8 0

2 years ago