Ask question

Ask question

All categories

2 years ago

13

11) It's perimeter

1 answer:

natta225 [31]2 years ago

8 0

2.8 m = .0028 km
3.5/.0028 = 1259 revolutions

You might be interested in

Who is correct Julie or Jeremy

Sedaia [141]

Answer:

Julie

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

21×9=9×21 what property is used

kvv77 [185]

Hello There =)

Commutative property is used. The commutative property <span>states that two numbers can be multiplied in either order.</span>

5 0

2 years ago

10. If both equations have the same slope and the same y-intercepts, which

Papessa [141]

The answer is C because it’s the same line

5 0

2 years ago

Which of the following reasons can be used to justify statement #4 in the proof?

Montano1993 [528]

Answer:

D.

Step-by-step explanation:

CPCTC which stands for "corresponding parts of congruent triangles are congruent". Since ΔMQN and ΔPQN have already been proved congruent, So every angle and sides of these two triangle corresponding with each other would be congruent.

3 0

3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

2 years ago