The isosceles triangle ABC has an area of 48 square units if AB=12 what is the perimeter of Abc in units?

What is 1/4x+1+3/4x-2/3-1/2x

Aleks04 [339]

Answer: 1/2x + 1/3

Step-by-step explanation:

Given:

1/4(x) + 3/4(x) - 1/2(x) + 1 - 2/3

Step 1: Combine like terms

1/4(x) and 3/4(x) have a common denominator of 4. This means that you can add them together.

1/4(x) + 3/4(x) = 4/4(x) = x

Step 2: Find the common denominator of x in step 1 and combine like terms

x - 1/2(x) = 2/2(x) - 1/2(x)

Now that we have the common denominator of x, we can combine like terms. Its the same as adding or subtracting fractions without a variable. In this case, you must subtract 1/2(x) from 2/2(x).

2/2(x) - 1/2(x) = 1/2(x)

Step 3: Find the common denominator of the constants and combine like terms

1 - 2/3 = 3/3 - 2/3

Now combine like terms. Simply subtract 2/3 from 3/3.

3/3 - 2/3 = 1/3

Step 4: Write the simplified equation

1/2(x) + 1/3

This is the answer

4 0

2 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

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