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3 years ago

Please help me. I need it!

Mathematics

1 answer:

Yanka [14]3 years ago

3 0

Answer:

znd i cant really tell the numbers on the side

Step-by-step explanation:

i like ur shirt XD

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Verizon [17]

The correct answer is B

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3 years ago

Describe the transformation of ƒ(x) = ex which is given by g(x) = −e2x + 6.

Arte-miy333 [17]

Answer:

g(x) is reflected across the x-axis and translated 6 units up compared to ƒ(x).

Step-by-step explanation:

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3 years ago

Read 2 more answers

P is a polynomial of degree 4 with three zeros: -1 and 2 are zeros of multiplicity one, and 3 is a zero with multiplicity two. I

tatyana61 [14]

Answer:

$P(x) = (x+1)(x-2)(x-3)^2 + 108$

Step-by-step explanation:

let me know if you want an explanation

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2 years ago

Humberly has $24 and they want to buy at least 8 containers of yogurt. Yogurt is packaged in both large and small containers. La

yanalaym [24]

Answer:

Hence he will be 4 large containers and 4 small containers

Step-by-step explanation:

Given data

Let the number of small containers be x

and the number of large containers be y

x+y= 8---------1

also

2x+4y= 24-----2

the system of equation to solve the problem is

x+y= 8

2x+4y= 24

from 1

x=8-y

put this in 2

2(8-y)+4y= 24

16-2y+4y= 24

2y= 24-16

2y= 8

y= 8/2

y= 4

put y= 4 in 1

x+4=8

x= 8-4

x= 4

Hence he will be 4 large containers and 4 small containers

6 0

3 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

3 years ago