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MariettaO [177]
4 years ago
11

numerator of a fraction that has a square root in it (but you don’t have to simplify the denominator with the square root) I alw

ays thought what you do to the top you must do to the bottom
Mathematics
1 answer:
Murrr4er [49]4 years ago
8 0

Answer:

Step-by-step explanation:

Not all operations, only multiplication and division is done on both, numerator and denominator. So that the fraction value doesn't change (like eqivalent fractions)

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Write the equation of the line in fully simplified slope-intercept form.
hoa [83]

Answer:

y = -x+3

Step-by-step explanation:

Slope intercept form =>  y = mx+b

To find 'm', the slope, pick 2 coordinates.

(0,3)

(2,1)

Use this equation to find the slope using these 2 coordinates: (y1 - y2)/(x1 - x2)

(3 - 1)/(0 - 2) = -1

m = slope = -1

'b' is the y-intersept, or the point when a line passes through the y-axis. That's (0,3).

b = y-intercept = 3

<em>So the equation will be y = -1x + 3, or y = -x + 3</em>

8 0
3 years ago
Read 2 more answers
What is 71/72 simplified?
Elina [12.6K]
The fraction can't be simplify since it already is
7 0
3 years ago
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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 are the zeros of (x-2)(x^2-9)
SashulF [63]
So because the factors of the equation are (x-2), (x+3) and (x-3), you set them equal to zero and you get x=2, -3, and 3
4 0
3 years ago
Read 2 more answers
Okay, help me out here!!
luda_lava [24]

Answer:

65.8

Step-by-step explanation:

(2.6*4.7)/2=6.11

11*4.7=51.7

(4.7(17-(2.6+11)))/2=7.99

6.11+51.7+7.99=65.8

3 0
4 years ago
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