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3 years ago

What is the way to write 25 percent

Mathematics

2 answers:

zhenek [66]3 years ago

7 0

There are different ways to write 25% in a
decimal form- .25
fraction form- 1/4

Hope this helps :)

Pani-rosa [81]3 years ago

6 0

1/4 is the right answer.

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Triangle ABC is the pre-image and triangle DFE is the image.

anzhelika [568]

Answer:

Scale factor = ½

Step-by-step explanation:

The original image is the preimage = ∆ACB

The new image is the image = ∆DFE

The scale of factor of dilation = image/preimage = DF/AC

DF = 6 cm (given)

AC = 12 cm (given)

Plug in the values into the equation to find the scale factor of dilation:

Scale factor = 6/12

Scale factor = ½

6 0

3 years ago

HELP PLEASE THIS IS MY 3RD TIME POSTING THIS QUESTION PLEASE HELP

Maurinko [17]

Answer:

11 1/2

Step-by-step explanation:

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8 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

What number goes in the box below to make the statement true?

Schach [20]

Answer:

5000

Step-by-step explanation:

4 0

3 years ago

Find the slope of the following 2 points (2,-5), (9,3)

Rama09 [41]

Answer:

8/7

Step-by-step explanation:

(2,-5) (9,3)

3-(-5)=8 because you apply your keep change change

9-2= 7

so 8/7

4 0

3 years ago