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3 years ago

Dan solves 6p=5-3p in the way shown below. Which properties did Dan use for Step 1 and Step 2?

Mathematics

2 answers:

Vera_Pavlovna [14]3 years ago

8 0

Answer:

Both the Addition Property of Equality and the Additive Identity Property

Step-by-step explanation:

This is Because we know the The Addition Property of Equality means or defines If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. So that would show or represent the Step 1

And for Step two it is The Additive Identity Property because we know that defines : an identity element (such as 0 in the group of whole numbers under the operation of addition) that in a given mathematical system leaves unchanged any element to which it is added.

This is why the correct answer is.... "Both the Addition Property of Equality and the Additive Identity Property"

il63 [147K]3 years ago

4 0

Answer:

Both the addition property of equality and the additive identity property

Step-by-step explanation:

we know that

The addition property of equality states that if the same amount is added to both sides of an equation, then the equality is still true

The additive identity property says that if you add a real number to zero or add zero to a real number, then you get the same real number back

we have

$6p=5-3p$

step 1

Applying the addition property of equality adds 3p both sides

$6p+3p=5-3p+3p$

step 2

Applying the additive identity property

$6p+3p=5+0$

step 3

$9p=5$

step 4

$p=5/9$

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3 years ago

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monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

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Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

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3 years ago

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