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vovangra [49]
2 years ago
7

Presenting someone else's words or ideas as your own is:

Mathematics
2 answers:
mote1985 [20]2 years ago
7 0

Answer:

<em>B. plagiarizing</em>

Step-by-step explanation:

...........

Mumz [18]2 years ago
6 0
B plagiarizing because it means to copy someone.
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Daniel's store sold 75% fewer pairs of shorts this month than last month. If the store sold 5 pairs of shorts this month, then _
Setler79 [48]

Answer:

EXACT ANSWER: 6.666666666667

ANSWER ROUNDED: 7

Step-by-step explanation:

lol, I can't explain well, but it's correct.

7 0
2 years ago
Compute: 3 .2. -4 plzz help me
Sonja [21]
3x2x(-4)=-24
3x2-4=2
3+2-4=1
not entirely sure what the equation your asking is but i hope this helps
6 0
3 years ago
How would 522 value be classified
In-s [12.5K]

Answer: It will be classified as a whole number!

Step-by-step explanation:

Hope this Helps!

3 0
3 years ago
Read 2 more answers
Aramp is 17 feet long, rises 8 feet above the floor, and covers a horizontal distance of 15 feet, as shown in the figure.
bonufazy [111]

Answer:

The ratio of Tan B is

\tan B= \dfrac{AC}{BC}=\dfrac{8}{15}

OR

\tan A= \dfrac{BC}{AC}=\dfrac{15}{8}

Step-by-step explanation:

In Right Angle Triangle ABC

angle C = 90°

AB = Ramp = 17 feet

BC =Horizontal distance = 15 feet

AC = Height from floor = 8 feet

To Find:

Ratio of Tan B = ?

Solution:

In Right Angle Triangle ABC By Tangent Identity we have

\tan B= \dfrac{\textrm{side opposite to angle B}}{\textrm{side adjacent to angle B}}

Substituting the given values we get

\tan B= \dfrac{AC}{BC}=\dfrac{8}{15}

OR

\tan A= \dfrac{BC}{AC}=\dfrac{15}{8}

5 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
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