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3 years ago

11

What is one possible dividend, greater than 1000, if quotient is37 R4

1 answer:

maxonik [38]3 years ago

7 0

Answer:

hello : one possible dividend is 1003

Step-by-step explanation:

let a this dividend : a = 37q +4........( 4 is a remainde)

a > 1000

37q +4 > 1000

q > 996/37

but : 996/7 ≃ 26.91

q = 27 so: a = 37(27)+4 = 1003

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How do you know what angle degree goes on each corner?

Dahasolnce [82]

Let'sSo, we gave 2 parallel lines and 2 transversals, we have to match the angles.

Let's start with angle b,

$\begin{gathered} b+65=180\text{ (angles in a straight line)} \\ b=180-65 \\ b=115^o \end{gathered}$

Let's move on to angle e,

$\begin{gathered} e=b\text{ (Vertically opposite angles)} \\ \text{but b = 115}^o \\ e=115^o^{} \end{gathered}$

Let's move on to angle d,

$d=110^o\text{ (V}ertically\text{ opposite angles)}$

Moving to angle c, we have;

$c=45^o\text{ (Vertically opposite angles)}$

And, angle a;

$\begin{gathered} a=b\text{ (Alternate angles)} \\ a=115^o \end{gathered}$

8 0

1 year ago

A line passes through (4, 5) and (8, 9). Which equation best represents the line?

xxTIMURxx [149]

Answer:

that is the equation of the line, its slope is 1 and the y-intercept is 1 y = x + 1

Step-by-step explanation:

we can calculate the slope of this line passing through (4,5) and (8,9) like this:

m = (y2 - y1)/(x2 - x1)

m = (9 - 5)/(8 - 4)

m = 4/4

m = 1

so the slope is 1, and now we can use the equation of one point and the slope to write the line's equation:

y - y1 = m(x - x1)

y - 5 = 1*(x - 4)

y - 5 = x - 4

y = x + 1

that is the equation of the line, its slope is 1 and the y-intercept is 1

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

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

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

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

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3 years ago

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gayaneshka [121]

Answer:

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Dimas [21]

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Step-by-step explanation:

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2 years ago

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