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Katarina [22]
3 years ago
11

The graph of g(x) is the graph of f(x)=x+9 reflected across the y-axis.

Mathematics
2 answers:
ICE Princess25 [194]3 years ago
8 0

Answer:

g(x) = -x +9

Step-by-step explanation:

Reflecting across the y-axis involves changing the sign of x, so the reflection of f(x) is ...

... g(x) = f(-x) = (-x) +9

The appropriate choice is ...

... g(x) = -x+9

777dan777 [17]3 years ago
4 0
<h2>Answer:</h2>

The equation which describes the function g is:

               g(x)=-x+9

<h2>Step-by-step explanation:</h2>

The graph of the function f(x) is given by:

                 f(x)=x+9

Now, we know that the transformation of the function f(x) when the graph is shifted across the y-axis then it is given by:

                 f(x) → f(-x)

This means that:

 g(x)=f(-x)

i.e.

g(x)= -x+9

Hence, the equation which will represent the graph of the function g(x) is given by:

                    g(x)=-x+9

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16x^7 - 8x^3 / 4x^3
Sunny_sXe [5.5K]

Answer:

4x^4-2

Step-by-step explanation:

\dfrac{16x^7-8x^3}{4x^3}= \\\\\\\dfrac{4x^3(4x^4-2)}{4x^3}= \\\\\\4x^4-2

Hope this helps!

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3 years ago
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Which expression represent a number that is one fourth as great as 10-2
Nataliya [291]
I think the answer to the expression would be D. (10 - 2) x 4 because (10 - 2) would be multiplied 4 times, which means it would be four times greater. I hope this made sense and also helped you in some way.
5 0
3 years ago
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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
What is the area of an equilateral triangle with apothem 3
Orlov [11]

Answer:

use formula of equilateral triangle = roots 3/4 a square.

6 0
3 years ago
Please helpppppppppppppppppppppppp
salantis [7]

Answer:

252

Step-by-step explanation:

Split up the shape

First part:

12*12=144

Second part:

((12+(12-6))/2)*24-12

((12+6)/2)*12

18/2*12

9*12 =  108

Total:

108+144 = 252

<u>Plz mark brainliest if this was helpful</u>

8 0
3 years ago
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