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2 years ago

11

What are the zeros of this function?

1 answer:

blondinia [14]2 years ago

8 0

Answer:

Try D

Step-by-step explanation:

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Line AB passes through points A(−6, 6) and B(12, 3). If the equation of the line is written in slope-intercept form, y=mx+b, the

san4es73 [151]

Answer:

We get value of the value of b = 5

Step-by-step explanation:

Line AB passes through points A(−6, 6) and B(12, 3). If the equation of the line is written in slope-intercept form, y=mx+b, then m=m equals negative StartFraction 1 Over 6 EndFraction.. What is the value of b?

We have slope m: $m=-\frac{1}{6}$

We need to find value of b (y-intercept)

Using the point A(-6,6) and slope $m=-\frac{1}{6}$ we can find b.

Using slope-intercept form, putting values of m and x and y we get the value of b:

$y=mx+b\\6=-\frac{1}{6}(-6)+b\\6=1+b\\b=6-1\\b=5$

So, we get value of the value of b = 5

5 0

2 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

What is the square root of 74 rounded to the nearest tenth

WARRIOR [948]

Answer:

8.6

Step-by-step explanation:

The square root of 74 is 8.602325, which rounded to the nearest tenth, is 8.6

5 0

2 years ago

Create write an expression that simplifies to 2a + 14 using the distributive property

geniusboy [140]

Answer:

2 ( a + 7)

Step-by-step explanation:

2a + 14

both 2a and 14 are divisible by 2

2a/a = a

14/2 = 7

2( a + 7) = 2a + 14

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4 0

3 years ago

Help please I’m trying to do my homework and don’t understand

11Alexandr11 [23.1K]

Answer:

Answer is 9.

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6 0

2 years ago