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Harrizon [31]
3 years ago
12

I need help with this

Mathematics
1 answer:
qwelly [4]3 years ago
3 0

Answer:

Step-by-step explanation:

x=√3, y= 3

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Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.
Ira Lisetskai [31]
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i

well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... \bf \begin{cases}
x=-2\implies x+2=0\implies &(x+2)=0\\
x=2i\implies x-2i=0\implies &(x-2i)=0\\
x=-2i\implies x+2i=0\implies &(x+2i)=0
\end{cases}
\\\\\\
(x+2)\underline{(x-2i)(x+2i)}=0\\\\
-----------------------------\\\\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0
\\\\\\
(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0
\\\\\\
x^3+2x^2+4x+8=0

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
6 0
3 years ago
Which two points should the line of best fit go through to best represent the data in the scatterplot?​
Ilya [14]

Given :   A graph with points (1, 3), (2, 7), (3, 8), (4, 11), (5, 12), (6, 13).

To find :  From Which two points the line of best fit goes through

Solution:

(3, 8) &  (5, 12) will be two points

as then line  points will be

( 1 ,  4 ) , ( 2, 6) , ( 3 , 8 ) , ( 4, 10) , (5 . 12)  & ( 6 , 14)

(1, 3)   will be below ( 1 ,  4 )

(2 , 7) will be above ( 2 , 6)

( 3 , 8 ) will be on line

(4 , 11 ) will be above ( 4 , 10)

( 5 , 12  ) will be on line

( 6 , 13)  will be below ( 6 ,  14 )

(1, 3)  & ( 6 , 13)  two points are below line

(2 , 7) & (4 , 11 )  two points above line

Hence points are Equally scattered above  & below line

Hence (3, 8) &  (5, 12) will be two points on   line of best fit

Hope it helpss !!

4 0
2 years ago
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I need help. i dont know where to start......
meriva

Answer:

start at x

Step-by-step explanation:

you are welcome

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3 years ago
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Turner has already taken 6 pictures at home, and he expects to take 3 pictures during every
Helga [31]
Answer:

6+3d=p

Hope this helps ya
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3 years ago
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70% of the days in june were sunny ow many days are sunny?
xxTIMURxx [149]
June has 30 days

so 70% of 30 were sunny

0.70(30) = 21 days <==
7 0
3 years ago
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