Simplifying 6(x + 1) + 5 = 13 + -2 + 6x Reorder the terms: 6(1 + x) + 5 = 13 + -2 + 6x (1 * 6 + x * 6) + 5 = 13 + -2 + 6x (6 + 6x) + 5 = 13 + -2 + 6x Reorder the terms: 6 + 5 + 6x = 13 + -2 + 6x Combine like terms: 6 + 5 = 11 11 + 6x = 13 + -2 + 6x Combine like terms: 13 + -2 = 11 11 + 6x = 11 + 6x Add '-11' to each side of the equation. 11 + -11 + 6x = 11 + -11 + 6x Combine like terms: 11 + -11 = 0 0 + 6x = 11 + -11 + 6x 6x = 11 + -11 + 6x Combine like terms: 11 + -11 = 0 6x = 0 + 6x 6x = 6x Add '-6x' to each side of the equation. 6x + -6x = 6x + -6x Combine like terms: 6x + -6x = 0 0 = 6x + -6x Combine like terms: 6x + -6x = 0 0 = 0 Solving 0 = 0 Couldn't find a variable to solve for. This equation is an identity, all real numbers are solutions.

7 0

4 years ago

Vincent bought 25 pound bag of rice. He cooked 6.25 pounds of the rice. He stored the rest of the rice in 3 3/4 pound portions.

Blizzard [7]

The maximum number is infinite but the minimum number of containers is 5
25-6.25=18.75
18.75/3.75=5

7 0

3 years ago

Read 2 more answers

How to estimate the square root of a number that is not a perfect square? Give an example and explain how to find it step by ste

tangare [24]

Answer:

See below

Step-by-step explanation:

There is a formula for estimating square root of non-perfect square numbers
√(A²±B) = A ± B/2A
where A² is the nearest perfect square, B is the difference between the original number and the perfect square

Examples

1) √75 = √(8²+11) = 8 + 11/(2*8) = 8 + 11/16 = 8.6875
2) √114 = √(11² - 7) = 11 - 7/(2*11) = 10.6818

6 0

3 years ago

Now, lets evaluate the same integral using power series. first, find the power series for the function f(x = \frac{32}{x^2+4}. t

BlackZzzverrR [31]

No idea what the previous part of the problem is, but you have

$f(x)=\dfrac{32}{x^2+4}=\dfrac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_{n\ge0}\left(-\frac{x^2}4\right)^n$
$f(x)=\displaystyle8\sum_{n\ge0}\left(-\dfrac14\right)^nx^{2n}$

which is valid for $\left|-\dfrac{x^2}4\right|$ , or $|x|$ . So the integral from 0 to 2 is

$\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_{n\ge0}\left(-\frac14\right)^nx^{2n}\,\mathrm dx$
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\int_0^2x^{2n}\,\mathrm dx$

Note that since the power series only converges on the interval if $x$ is strictly less than 2, which means we have to treat this as an improper integral.

$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\int_0^tx^{2n}\,\mathrm dx$ [/tex]
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}$
$=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}$
$=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}$
$=16-\dfrac{16}3+\dfrac{16}5-\dfrac{16}7+\dfrac{16}9+\cdots$

6 0

3 years ago