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3 years ago

I need help please answer correct please

Mathematics

2 answers:

vivado [14]3 years ago

7 0

Answer:

Step-by-step explanation:

1) Perimeter = 2m + 3m +3m + 2 + 2m + 2

= 2m + 3m + 3m + 2m + 2 + 2 {Combine like terms}

= 10m + 4

2) 10m = 2 * 5 * m

4 = 2 * 2

GCF = 2

10m + 4 = 2 *5m + 2*2

= 2*(5m + 2)

eduard3 years ago

5 0

Answer:

I believe the answer is 8m.

Step-by-step explanation:

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Consider the table below that represents a quadratic function.

zlopas [31]

It would be -1/2 because the formula to find slope (rate of change) is (y2-y1)/(x2-x1) and in this case it would be (4.5-2)/(-3--2) which simplifies to (2.5)/(-5) which is -1/2

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3 years ago

On Monday, Takira counted 75 splatters of bird droppings on her car. On Tuesday afternoon, she counted 21 more. What is the perc

Law Incorporation [45]

Answer:

28%

Step-by-step explanation:

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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 $v(x)=\dfrac{y(x)}x$ :

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

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

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

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3 years ago

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Leno4ka [110]

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2 years ago

Marking brainliest!!<br><br> What should I put for the second question? I am confused. HELPP

Levart [38]

if other experts state the same thing in their own paper or document.

8 0

3 years ago

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