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1 year ago

Matthew buys x pens and two more pencils than pens.

Mathematics

1 answer:

nekit [7.7K]1 year ago

4 0

Answer:

Please check the wording of the question. There isn't enough information to select a single value of x.

Step-by-step explanation:

The attached worksheet shows two different combinations of pens and pencils that both sum to a total cost of $10. There are many other combinations that would do the same.

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Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

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3 years ago

Find the volume of a cube with a side length of 12 feet.

konstantin123 [22]

Answer: 1728

Step-by-step explanation: 12x12x12

8 0

3 years ago

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What is the factor expression 5x+30

Masteriza [31]

It is 35x that i think

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3 years ago

Marty's class collected $38.00 in a school-wide penny collection contest. If Tommy's class added $8.00 to the amount they collec

Elanso [62]

Answer: $11

Step-by-step explanation:

To have the same:

11+8=19

19×2=38

4 0

2 years ago

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the vertices of abc are a (-6,-7) b(-3,-10) and c(-5,2) find the vertices of abc given the transition rule (x,y)-->(x,y-3)

bagirrra123 [75]

The vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)

<h3>How to determine the new vertices?</h3>

The vertices are given as:

a (-6,-7)

b(-3,-10)

c(-5,2)

The transition rule is given as

(x,y)-->(x,y-3)

So, we have

a' = (-6, -7 - 3)

a' = (-6, -10)

b' = (-3, -10 - 3)

b' = (-3, -13)

c' = (-5, 2 - 3)

c' = (-5, -1)

Hence, the vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)