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lord [1]
2 years ago
5

Which function results after applying the sequence of transformation in this order to f(x)=x⁵, Stretch vertically by 3; reflect

across the x-axis; shift right 1unit; shift up 2 units?
Mathematics
1 answer:
Nikitich [7]2 years ago
6 0

Using translation concepts, it is found that the resulting function is given by: f(x) = -3(x - 1)^5 + 2.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the function is:

f(x) = x^5

The transformations are as follows:

  • Stretch vertically by a factor of 3, hence f(x) = 3x^5.
  • Reflect across the x-axis, hence f(x) = -3x^5
  • Shift right 1 unit, hence f(x) = -3(x - 1)^5.
  • Shift up 2 units, hence f(x) = -3(x - 1)^5 + 2.

More can be learned about translation concepts at brainly.com/question/4521517

#SPJ1

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Answer: (a) spuriousness relationship

Step-by-step explanation:

Spurious occurs between two variables that are actually caused by a third variable. Examples is like a number of teachers in region and number of people learn from college.

7 0
3 years ago
Can someone help me there's supposed to be two answers but all I got was A
OlgaM077 [116]
D is also correct. if you need an explanation let me know
3 0
3 years ago
Read 2 more answers
Integration using part formula<br> <img src="https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5Enlogx%7D%20%5C%2C%20dx" id="TexFor
liq [111]

Answer:

Integration of I= \int {x^{n}logx \, dx=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

Step-by-step explanation:

Given integral is I= \int {x^{n}logx \, dx

Take logx=t

x=e^{t}

x^{n}=e^{nt}

\frac{1}{x} dx=dt

dx=xdt

dx=e^{t}dt

I= \int (e^{nt})(t)(e^{t})\, dt

I= \int (e^{(n+1)t})(t)\, dt

Using integration by part,

I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]

Writing in terms of x

I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]

I=[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]

I=[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]

I=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

Thus,

Integration of I= \int {x^{n}logx \, dx=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

3 0
3 years ago
What is the annual simple interest rate?<br>I=$17, P=$500, t=2 years​
Ivanshal [37]

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{1.7 \: \%}}}}}

Step-by-step explanation:

Given,

Interest ( I ) = $ 17

Principal ( P ) = $ 500

Time ( T ) = 2 years

Rate ( R ) = ?

<u>Finding </u><u>the</u><u> </u><u>simple </u><u>Interest</u><u> </u><u>rate</u><u> </u><u>:</u>

\boxed{ \bold{ \sf{rate =  \frac{interest \times 100}{principal  \times  time}}}}

\dashrightarrow{ \sf{rate =  \frac{17 \times 100}{500 \times 2} }}

\dashrightarrow{ \sf{rate =  \frac{1700}{1000} }}

\dashrightarrow{ \sf{ rate = 1.7 \: \%}}

Hope I helped!

Best regards! :D

5 0
3 years ago
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EleoNora [17]
Hopefully this is correct and what you wanted. 

6 0
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