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2 years ago

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A rectangle with vertices L(3, –1), M(3, –8), N(8, –8), and O(8, –1) is reflected over the y-axis.

1 answer:

Katen [24]2 years ago

7 0

i dont know if you can see it

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Its midpoint i need help

torisob [31]

Answer:

(3.5, 4.8)

Step-by-step explanation:

2.6 - 1.7 = 0.9, 2.6 + 0.9 = 3.5

3 - 1.2 = 1.8, 3 + 1.8 = 4.8

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2 years ago

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What is the area of the triangle 13,13,10

Y_Kistochka [10]

Answer:

depends

Step-by-step explanation:

it depends on the lengths of the base and height since the area of a triangle is A=bh / 2

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3 years ago

Six friends share 4 apples equally. How much apples does each friend friend get?

Zinaida [17]

Half and apple each

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2 years ago

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A tool set is on sale for $424.15.The original price of tthe tool set was $499.00. What percent of the original price is the sal

Jet001 [13]

Answer:

Step-by-step explanation:

The sale price is out of the original price so do 424.15 / 499.00 to find what it is out of in decimal form. Multiply your answer by 100 to find the percentage.

So:

424.15 / 499.00 =0.85

0.85 x 100 = 85

So the answer would 85%.

Hope this helps!

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