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2 years ago

15

What is the constant of the variation for the inverse variation?

1 answer:

Alisiya [41]2 years ago

4 0

Answer:

2nd possible answer is the correct one; the const. of var. is 1/3.

Step-by-step explanation:

xy = 1/3 can be rewritten in inverse variation format: Divide both sides by x to isolate y. Then y = (1/3)(1/x). The constant of (inverse) variation is 1/3.

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Find the area of this figure. Please show work if you can!

Yuri [45]

Here is the answer if you can't read my writing than just ask. also if you have any questions about how I got something feel free to ask

8 0

2 years ago

Help!!! This question is confusing me!!

erica [24]

Answer:

k ≥2

Step-by-step explanation:

k+ 1/3≥7/3

Subtract 1/3 from each side

k+ 1/3-1/3≥7/3-1/3

k ≥6/3

k ≥2

8 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

2 years ago

Solve the system of linear equations by graphing. y = –5/2 x – 7 x + 2y = 4 What is the solution to the system of linear equatio

Anestetic [448]

Keywords

system; linear equations; graphing; solution

we have

$y=-\frac{5}{2}x-7$ -------> equation A

$x+2y=4$

$x=4-2y$ ----> equation B

The <u>system</u> of <u>linear equations</u> is composed of equation A and equation B

Substitute the equation B in equation A

$y=-\frac{5}{2}[4-2y]-7$

$y=-10+5y-7$

$4y=17$

$y=17/4=4.25$

Find the value of x

$x=4-2[17/4]=-4.5$

The<u> solution</u> is the point $(-4.5,4.25)$

Using a <u>graphing</u> tool

see the attached figure

therefore

the answer is the option A

$(-4.5,4.25)$

4 0

2 years ago

Read 2 more answers

Solve.<br><br> y = –4x + 10<br> y = –3x + 5

saw5 [17]

The answers in order are
Y= 10/4
Y=5/3

8 0

1 year ago

Read 2 more answers