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2 years ago

8

Consider the perimeter of a rectangle being 68 cm and the width being 4 less than the length. Write a linear relation in terms o

f the length. What is the length and the width?

1 answer:

MrMuchimi2 years ago

8 0

Answer:

legth is vetry large sea and width is north america

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Do I have to find a common denominator

antoniya [11.8K]

When it comes to fractions yes you do. But to make it easier just multiply the denominators and that should give u a common denominator, they don't always have to be least common.

5 0

3 years ago

Read 2 more answers

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

HELP PLZ!!!!!!! Is the following graph a function?

Stels [109]

This graph does not represent a function because the x-intercepts repeat themselves several times. In order for a graph or group of coordinates to be a function, there must be only one of a particular x-coordinate.

8 0

2 years ago

Assuming that Switzerland's population is growing exponentially at a continuous rate of 0.19 percent a year and that the 1988 po

Nesterboy [21]

<span>First, the formula would be like this dP / dt= KP.
Secondly, since dP is given which is 6.9. Substitute it from the equation.

P(0) = 6.9 x 10^6
integral 1 divided by pdp which will look like this,

integral 1/ PdP = integral kdt

the answer would be .0019 or
k=.0019</span>

6 0

2 years ago

Read 2 more answers

Michelle bought five packs of crayons for 13.75. what is the cost of a pack ,in dollars, if all the packs cost the same

Schach [20]

Answer:

2.75

Step-by-step explanation:

Divide 13.75 by five.

5 0

2 years ago