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3 years ago

I just need help with 3-6 (explain answers please)

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

8 0

Answer:

3-6=3

Step-by-step explanation:

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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

3 years ago

How do you dilate a figure on the coordinate plane and determine its scale factor?

levacccp [35]

Answer:

All you have to do is scale up the coordinates of the plane by the factor. Multiply the values of each coodinate by the scale factor.

Step-by-step explanation:

3 0

3 years ago

Find the distance between the points J(−8, 0) and K(1, 4).

Tasya [4]

Hope this will help u....

6 0

3 years ago

Read 2 more answers

How many solutions does 3 - 2x = 5 - x + 3 + 4x have?

vitfil [10]

Well let's see.

$3-2x=5-x+3+4x \\3-2x=8+3x \\-5x=5 \\x=-1 \\$

The equation has exactly one solution.

Hope this helps.

5 0

3 years ago

The model includes a new park in the center of the city. If the dimensions of the

11111nata11111 [884]

The question is incomplete. Here is the complete question.

As a part of city building refurbishment project, architects have constructed a scale model of several city builidings to present to the city commission for approval. The scale of the model is 1 inch = 9 feet.

The model includes a new park in the center of the city. If the dimensions of the park in the model are 9 inches by 17 inches, what are the actual dimensions of the park?

Answer: 81 feet by 153 feet

Step-by-step explanation: Unit Scale is a ratio comparing actual dimensions of an object to the dimensions of model representing the actual object.

In the refurbishment project, the unit scale is given by

1 inch = 9 feet

So, the dimensions of the new park in actual dimensions would be

1 inch = 9 feet

9 inches = x

x = 9.9

x = 81 feet

1 inch = 9 feet

17 inches = y

y = 17.9

y = 153 feet

The actual dimensions of the new park are 81 feet by 153 feet.

6 0

3 years ago