All categories

3 years ago

Simplify the expression show work

Mathematics

2 answers:

lyudmila [28]3 years ago

5 0

Answer:

4(x−2)

Step-by-step explanation:

4x−24+16

Factor out 4.

4(x−2)

Consider x−6+4. Multiply and combine like terms.

x−2

Rewrite the complete factored expression.

4(x−2)

vladimir1956 [14]3 years ago

4 0

Answer:

4x-8

Step-by-step explanation:

simplify by combining like terms so 4x and then -24+16=-8

You might be interested in

Scientific notation 20,000+3,400,000= an example to would be helpful

Viefleur [7K]

20 000 is as much as 20 * 1 000, so we can write it as: 20 * 10^3
3 400 000 is as much as 34 * 100 000, so we can write it as: 34 * 10^5

The sum in scientific notation would look like this:

20 * 10^3 + 34 * 10^5

I hope that's what you meant :)

4 0

4 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

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

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

A van can travel 18 miles on each gallon of gasoline. at that rate how many miles can the van travel on 15 gallons of gasoline

tensa zangetsu [6.8K]

18 / 1 = x / 15......18 miles to 1 gallon = x miles to 15 gallon
cross multiply
(1)(x) = (18)(15)
x = 270 miles

7 0

3 years ago

Read 2 more answers

What is 5/6 and 3/8 as an equivalent fraction

Veronika [31]

Answer:

5/6 - 10/12, 15/18, 20/24

3/8 - 6/16, 9/24, 12/32

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

How do you solve this logarithmic equation? 3x-5=log(down arrow)2 1024

Marrrta [24]

What does down arrow mean

6 0

3 years ago

Read 2 more answers