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3 years ago

Solve the inequality 5p -3 < 7 + 3p for p

Mathematics

1 answer:

KiRa [710]3 years ago

5 0

Step-by-step explanation:

p<5 that's the answer for the question

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Robert says he needs 250 plastic sleevs for his basball card collection.He has 3084 baseball cards.Each sleev holds 12 cards do

Iteru [2.4K]

Answer:

Step-by-step explanation:

Given that:

Plastic sleeves needed = 250

Number of baseball cards = 3084

Sleeves held per card = 12

To determine the number of sleeves needed ;

Number of baseball cards / number of sleeves per card

= 3084 / 12

= 257

No, I do not agree with Roberts, from the calculation above, the number of plastic sleeves needed is 257

7 0

3 years ago

Solve the differential equation dy/dx=x/49y. Find an implicit solution and put your answer in the following form: = constant. he

anygoal [31]

Answer:

The general solution of the differential equation is $\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

The equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

The equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

Step-by-step explanation:

This differential equation $\frac{dy}{dx}=\frac{x}{49y}$ is a separable first-order differential equation.

We know this because a first order differential equation (ODE) $y' =f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of x and y

$f(x,y)=p(x)\cdot h(y)$ where p(x) and h(y) are continuous functions. And this ODE is equal to $\frac{dy}{dx}=x\cdot \frac{1}{49y}$

To solve this differential equation we rewrite in this form:

$49y\cdot dy=x \cdot dx$

And next we integrate both sides

$\int\limits {49y} \, dy=\int\limits {x} \, dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\\int\limits {49y} \, dy=\frac{49y^{2} }{2} + c_{1}$

$\int\limits {x} \, dx=\frac{x^{2} }{2} +c_{2}$

$\int\limits {49y} \, dy=\int\limits {x} \, dx\\\frac{49y^{2} }{2} + c_{1} =\frac{x^{2} }{2} +c_{2}$

We can subtract constants $c_{3}=c_{2}-c_{1}$

$\frac{49y^{2} }{2} =\frac{x^{2} }{2} +c_{3}$

An explicit solution is any solution that is given in the form $y=y(t)$ . That means that the only place that y actually shows up is once on the left side and only raised to the first power.

An implicit solution is any solution of the form $f(x,y)=g(x,y)$ which means that y and x are mixed (y is not expressed in terms of x only).

The general solution of this differential equation is:

$\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

To find the equation of the solution through the point (x,y)=(7,1)

We find the value of the $c_{3}$ with the help of the point (x,y)=(7,1)

$\frac{49*1^2\:}{2}-\frac{7^2\:}{2}\:=\:c_3\\c_3 = 0$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:0$

$\mathrm{Add\:}\frac{x^2}{2}\mathrm{\:to\:both\:sides}\\\frac{49y^2}{2}-\frac{x^2}{2}+\frac{x^2}{2}=0+\frac{x^2}{2}\\Simplify\\\frac{49y^2}{2}=\frac{x^2}{2}\\\mathrm{Multiply\:both\:sides\:by\:}2\\\frac{2\cdot \:49y^2}{2}=\frac{2x^2}{2}\\Simplify\\9y^2=x^2\\\mathrm{Divide\:both\:sides\:by\:}49\\\frac{49y^2}{49}=\frac{x^2}{49}\\Simplify\\y^2=\frac{x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$y=\frac{x}{7},\:y=-\frac{x}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{x}{7}=\\1=\frac{7}{7}\\1=1\\$

With the second solution we get:

$\:y=-\frac{x}{7}\\1=-\frac{7}{7}\\1\neq -1$

Therefore the equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

To find the equation of the solution through the point (x,y)=(0,-3)

We find the value of the $c_{3}$ with the help of the point (x,y)=(0,-3)

$\frac{49*-3^2\:}{2}-\frac{0^2\:}{2}\:=\:c_3\\c_3 = \frac{441}{2}$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:\frac{441}{2}$

$y^2=\frac{441+x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\y=\frac{\sqrt{441+x^2}}{7},\:y=-\frac{\sqrt{441+x^2}}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{\sqrt{441+x^2}}{7}\\-3=\frac{\sqrt{441+0^2}}{7}\\-3\neq 3$

With the second solution we get:

$y=-\frac{\sqrt{441+x^2}}{7}\\-3=-\frac{\sqrt{441+0^2}}{7}\\-3=-3$

Therefore the equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

4 0

3 years ago

Which of the following integrals represents the area of the region bounded by x = e and the functions f(x) = ln(x) and g(x) = lo

Neko [114]

Notice that

$\log_{1/e}x=\dfrac{\ln x}{\ln\frac1e}=\dfrac{\ln x}{-\ln e}=-\ln x$

$f(x)=\ln x$ and $g(x)=-\ln x$ intersect when $x=1$ . For all $x>1$ , we have $\ln x>0$ and $-\ln x$ , so $f(x)>g(x)$ . Then the area we want is given by the integral,

$\displaystyle\int_1^e\ln x-(-\ln x)\,\mathrm dx=2\int_1^e\ln x\,\mathrm dx$

or in terms of $\log_{1/e}x$ ,

$\displaystyle\int_1^e\ln x-\log_{1/e}x\,\mathrm dx$

5 0

3 years ago

Each paper clip is 3/4 of an inch long and cost $0.02. Exactly enough paper clips are laid end to end to have a total length of

cupoosta [38]

I believe that the answer is 0.96 cents

6 0

3 years ago

angelo rode his bike around a bike trail that was 1/4 of a mile long.he rode his bike around thetrail 8 times.Angelo says he rod

forsale [732]

Hi there!

From this problem we can pull out some key information and remove any excess.

- The trail is 1/4 of a mile long.
- Angelo rode around the trail 8 times.
- Angelo claims he rode 8/4 miles.
- Teresa claims he rode 2 miles.

From that we know that if Angelo rode around the trail 8 times and the trail is 1/4 a mile, he in total rode 8/4 of a mile.

However, Teresa claims he rode 2 miles!

And the answer is... they're both correct!

Angelo did indeed ride 8/4 miles however that is equivalent to 2 miles so they are both in the right.

7 0

4 years ago