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2 years ago

Does (–3, 17) make the equation y = 6x + –1 true?

Mathematics

1 answer:

telo118 [61]2 years ago

6 0

Answer:

ture

Step-by-step explanation:

Read the article and review the phases of mitosis.

Mitosis

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Plzzz help<br> thank u for those who help

IRISSAK [1]

Answer:

SAS

Step-by-step explanation:

As per the given diagram, the following facts are evident.

(BR) = (CR), Both sides have two small orange lines on them. This shows that these sides are congruent.

(<B) = (<C), this is shown by the box around both angles, indicating that both angles have a measure of (90) degrees.

(AB) = (AC), Both sides have one small orange line on them. This indicates that these sides are congruent to each other.

Therefore, the sides are congruent by the theorem (SAS); side-angle-side, congruence.

8 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

Change this from vertex form to standard form

wolverine [178]

Answer:

y=4x^2-24x+41

hope this helps!

mark brainliest?

3 0

3 years ago

Please help i am haveing alot of trouble with this.

Galina-37 [17]

Answer:

rewrite each sentence change the underline word (s) to a pronoun my friends and i are going to the movies

7 0

3 years ago

Can somebody help pls

Otrada [13]

It’s a ushndu hahaha hahaha Han shxvdbck dhuddj

5 0

3 years ago

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