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2 years ago

5

Which of the following statements are true about the equation below?

2 answers:

dangina [55]2 years ago

8 0

The correct answer is 5

seropon [69]2 years ago

5 0

#5 and #1 and #2 are correct

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A mule deer can run one over four of a mile in 25 seconds. At this rate, which expression can be used to determine how fast a mu

kondor19780726 [428]

the answer is D. twenty-five seconds over one-fourth mile = 100 miles per hour.

3 0

2 years ago

A mountain climber starts a climb at an elevation of 453 feet above sea level. At his first rest stop he has climbed 162 feet, a

Sedaia [141]

162
+207
= 369

Okay, so it says he descends 285 feet after he's climbed 369 feet. That means he needs to descend a bit more to reach 453 feet, which we'll just call 0 for now. First, we need to do some subtraction:

369
- 285
= 84

Okay, we know he needs to descend 84 feet to get to his starting point.
So your answer is C

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

Find the nonpermissible replacement for y in<br> this expression.<br> y + 9<br> y+7

BigorU [14]

Answer:

idk

Step-by-step explanation:

3 0

2 years ago

Which value of m makes the statement true? 92-2m ≤7+3m<br><br> A) 12<br> B) 17<br> C) 7<br> D) 9

Mama L [17]

The answer is 17 or B

5 0

2 years ago

Read 2 more answers