Giving brainiliest answer!!

Find the value of the expressions 3x^3-2y^3-6x^2y^2+xy for x=2/3 and y=1/2

fiasKO [112]

Hi!

$3x^3-2y^3-6x^2y^2+xy=\\\\=3\cdot(\frac{2}{3})^3-2\cdot(\frac{1}{2})^3-6\cdot(\frac{2}{3})^2\cdot(\frac{1}{2})^2+\frac{2}{3}\cdot\frac{1}{2}=\\\\=3\cdot\frac{8}{27}-2\cdot\frac{1}{8}-6\cdot\frac{4}{9}\cdot\frac{1}{4}+\frac{1}{3}=\\\\=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}=\frac{8}{9}-\frac{1}{4}-\frac{1}{3}=\frac{32}{36}-\frac{9}{36}-\frac{12}{36}=\boxed{\frac{11}{36}}$

5 0

2 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago

PLEASE TELL ME WHAT I DID WRONG AND TELL ME THE ANSWER I WILL GIVE YOU 30 POINTS AND BRAINLESS

Setler79 [48]

Answer:

1. D. is Brand B

2. 60 mikes per hour

3. is 40 mins

4. is A. 3/20 u are multiplying

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Hey bro's help me!!

Vikentia [17]

Answer:

3x+10

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

For the diagrams, find the following measures. <br> X= <br> DCB= <br> ABC=

Tasya [4]

To set this up you will set the two angles equal to 180° since it is a straight line
This will make the equation x+6x+19=180
Combine like terms. 7x+19=180.
Subtract 19 from both sides. This will make it 7x=161
Divide both sides by 7. That will make it x=23
Next plug in the x to solve for abc. 7(23)+19. This equals 157

X= 23

Hope this helped :)

8 0

3 years ago