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2 years ago

10

Enter an equation in standard form to model the linear situation. A bathtub that holds 41 gallons of water contains 11 gallons o

f water. You begin filling it, and after 5 minutes, the tub is full.

1 answer:

WITCHER [35]2 years ago

4 0

Answer: In this equation, let "x" equal the rate at which the tub will be filled. The tub holds 32 gallons, which will be one-half of the equation. The tub already contains 12 gallons, so this will be half of the other side of the equation. The rate is "x" gallons per minute, and the time it required was 5 minutes, so this would be "5x". Placing all these parts together gives an equation of (32 = 5x + 12) gallons per minute dispensed.

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3x+2(x+4)=5(x+1)+3. Show work please

Nesterboy [21]

First, we distribute the 2 through the parenthesis.

3x+2x+8=5(x+1)+3

Then we distribute the 5 through the parenthesis

3x+2x+8=5x+5+3

Collect the like terms

5x+8=5x+5+3

Add the numbers

5x+8=5x+8

The statement is true for any value of x, because both sides are identical

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3 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

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3 years ago

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Aleksandr [31]

The bottom answer matches. please give brainlest

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2 years ago

Which number is a perfect cube? 5, 100, 125, 150

KATRIN_1 [288]

Answer:

125

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

|x-1|+2<4 solving for x

Ganezh [65]

Answer:

x < 3 and x > -1

Step-by-step explanation:

Step 1: Subtract 2 from both sides.

$|x-1| + 2 - 2 < 4 - 2$
$|x-1|$

Step 2: Solve absolute value.

We know x - 1 < 2 and x - 1 > -2.

Condition 1:

$x - 1 < 2$
$x - 1 + 1 < 2 + 1$ (Add 1 to both sides)
$x < 3$

Condition 2:

$x - 1 > -2$
$x - 1 + 1 > -2 + 1$ (Add 1 to both sides)
$x>-1$

Therefore, the answer is x < 3 and x > -1.

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2 years ago