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Aloiza [94]
3 years ago
10

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
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