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Vikki [24]
3 years ago
11

De sua jarra de suco, Claudete bebeu inicialmente 240 ml. Depois, bebeu 1/4 do que restava e, depois de algum tempo, ela bebeu o

restante que representava 1/3 do volume inicial. A jarra continha inicialmente uma quantidade de suco, em ml, igual a?
Mathematics
1 answer:
docker41 [41]3 years ago
4 0
É uma questão relativamente simples, o problema é cair em uma pegadinha do enunciado.

Se colocarmos que o jarro inicialmente tenha X ml, após Claudete beber 240 ml temos que X-240=Y. Então Claudete bebe 1/4 da jarra, por isso temos 3*Y/4=Z (Se ela bebeu 1/4, reataram 3/4). Depois, ela bebe 1/3 do volume inicial e o suco acaba, logo montamos a equação X/3=Z.

Agora temos um sistema com estas 3 equações:
--> X-240=Y
-->3Y/4=Z
-->Z=X/3

Juntando as 2 ultimas vc terá: X/3=3Y/4. E como Y=X-240.

X/3=3*(X-240)/4

Se resolver isto terá que X=432. logo a resposta é 'E'
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A family has a $93,411, 20-year mortgage at 5.4% compounded monthly. Find the monthly payment. Also find the unpaid balance afte
Mademuasel [1]

Answer:

  • payment: $637.30
  • 5-year balance: $78,505.48
  • 10-year balance: $58,991.59

Step-by-step explanation:

1. The relevant formula for computing the monthly payment A from principal P and interest rate r for loan of t years is ...

  A = P(r/12)/(1 -(1 +r/12)^(-12t))

Filling in the numbers and doing the arithmetic, we get ...

  A = $93,411(0.054/12)/(1 -(1 +0.054.12)^-(12·20)) ≈ $637.30

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2. The relevant formula for computing the remaining balance after n payments of amount p on principal P at interest rate r is ...

  A = P(1 +r/12)^n -p((1 +r/12)^n -1)/(r/12)

Filling in the given values and doing the arithmetic, we get ...

  A = $93,411(1.0045^60) -637.30(1.0045^60 -1)/(0.0045) ≈ $78,505.48

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3. The same formula with n=120 gives ...

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Anton [14]

Answer:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

Step-by-step explanation:

So we have the equation:

\tan(x-y)=\frac{y}{8+x^2}

And we want to find dy/dx.

So, let's take the derivative of both sides:

\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]

Let's do each side individually.

Left Side:

We have:

\frac{d}{dx}[\tan(x-y)]

We can use the chain rule, where:

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Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])

Differentiate x like normally. Implicitly differentiate for y. This yields:

=\sec^2(x-y)(1-y')

Distribute:

=\sec^2(x-y)-y'\sec^2(x-y)

And that is our left side.

Right Side:

We have:

\frac{d}{dx}[\frac{y}{8+x^2}]

We can use the quotient rule, where:

\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}

f is y. g is (8+x²). So:

=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}

Differentiate:

=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

And that is our right side.

So, our entire equation is:

\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)

The right side cancels. Let's distribute the left:

\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2

Move -2xy to the left. So:

\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2

Factor out a y' from the right:

\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)

Divide. Therefore, dy/dx is:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}

We can factor out a (8+x²) from the denominator. So:

\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}

And we're done!

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