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3 years ago

Rotation 90° clockwise about the origin

Mathematics

1 answer:

sweet [91]3 years ago

5 0

Answer:

point s at (0,5) point t at (5,3) point u at (4,1)

Step-by-step explanation:

happy life

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Completing the sqaure <br> x^2+14x+34

DiKsa [7]

Answer:

-7-√15

Step-by-step explanation:

First move the constant to the right hand side and change it's sign

x^2+14x=-34

Then add (14/2)^2 or 7^2 to both sides

x^2+14x+49=-34+49

Then factor the expression

(x+7)^2=-34+49

Then solve (x+7)^2=15

You should get:

x=-√15-7

x=√15-7

7 0

3 years ago

Pls explain your answer

spayn [35]

Yo sup??

this is of the form called slope and point... hence

y-11=-4(x+2)

therefore the correct answer is option A

Hope this helps

8 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

A quadratic model is fitted to a set of data that has a quadratic trend. What can be

Phoenix [80]

Answer: C & D

Step-by-step explanation:

6 0

2 years ago

Write an equation for each line in slope-intercept form.

Lady bird [3.3K]

Answer:

y = x + 7

Step-by-step explanation:

The slope-intercept form of a line is:

y = mx + b

where m is the slope and b is the y-intercept.

Looking at the graph, we can see that the line intersects the y-axis at y = 7. So 7 would be our y-intercept.

To find the slope, we would divide the rise of the line by the run. Or m = rise/run. From looking at the graph, we can see that for every 1 unit the line moves in the x-direction, the line moves in the y-direction by 1 unit. Therefore, the rise would be 1 and the run would be 1. 1/1 = 1 so the slope of the line would be 1.

Plugging in 7 for b and 1 for m into the equation for the slope-intercept form, we get:

y = x + 7

So that would be the equation for the line in slope-intercept form.

I hope you find my answer and explanation to be helpful. Happy studying.

3 0

3 years ago