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3 years ago

What does 7/18 estimate to 0,1/2,1

Mathematics

1 answer:

DiKsa [7]3 years ago

7 0

Answer: 11/2

Step-by-step explanation:

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I need help again, as quickly as possible.

mestny [16]

Answer:

${ \tt{ \frac{1}{x} + \frac{1}{3} < \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{x} > - \frac{2}{15} }} \\ \\ { \tt{x > - \frac{15}{2} }} \\ \\ { \bf{ - \frac{15}{2} < x < 0 }}$

5 0

2 years ago

What quadratic formula do I need to use to solve 3x(x+6)=-1

timofeeve [1]

Answer:

$\Large \boxed{x_1=\frac{9+\sqrt{51} }{3} \ \ ; \ \ x_2=\frac{9-\sqrt{51} }{3} }$

Step-by-step explanation:

$\displaystyle \Large \boldsymbol{} 3x(x+6)=-10 \\\\3x^2+18x=-10 \\\\3x^2+18x+10=0 \\\\D=324-120=204 \\\\ x_{1;2}=\frac{18\pm2\sqrt{51} }{6} =\frac{9\pm\sqrt{51} }{3}$

3 0

2 years ago

Frank is trying to factor y^2+6y-27 . He has determined that one factor is (y + 9). What is the other factor? A. y – 27 B. y – 6

beks73 [17]

The other factor is (y - 3). If you multiply the two terms together, you will end up with $y^2+6y-27$ :

$(y - 3)(y + 9) \\ y^{2} +9y-3y-27 \\ y^2+6y-27$

5 0

3 years ago

Read 2 more answers

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

9) EASY POINTS WILL GIVE BRAINLIST TO BEST ANSWERS

ra1l [238]

You bought a magazine for $5 and four erasers. you spent a total of $25.

Magazine cost: 5$ each
Eraser cost: X$ each

Total cost is the number of items (1 magazine) times the cost of each item (5$). We don't know the cost of each eraser, so we represent that using a variable, X.

1(5) + 4(X) = 25

5 + 4X = 25

subtract 5 from both sides

4X = 25 - 5

4X = 20

divide both sides by 4

X = 20/4

X = 5

Each eraser cost 5$

3 0

2 years ago

Read 2 more answers