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podryga [215]
3 years ago
8

PLZ PLZ PLZ HELP ME SOLVE FOR J AND K!!!I WILL CROWN U!!

Mathematics
1 answer:
Zigmanuir [339]3 years ago
8 0

Step-by-step explanation:

as in the attachment, i have named the lines by letters for easiness

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rewrite the polynomial 2y^2+ 6y^3-11-17y^4+8y^5 in the standard form also find its degree and coefficient of y^4
OleMash [197]

Answer:

The standard form is  8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11

The degree of given polynomial is '5'

the co-efficient of  y⁴  is '-17'

Step-by-step explanation:

Given standard form 2 y²+ 6 y³-11-17 y⁴+8 y⁵

<em>The form ax² + b x + c is called the standard form of the quadratic expression of 'x'.This is second degree standard form of polynomial.</em>

<em>The form ax⁵ + b x⁴ + c x³ +d x² +ex +f  is called the standard form of the quadratic expression of 'x'.This is fifth degree standard form of polynomial</em>

now Given polynomial is  2 y²+ 6 y³-11-17 y⁴+8 y⁵

The standard form is

                                8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11

<u><em>Conclusion</em></u>:-

<em>The degree of given polynomial is '5'</em>

<em>The co-efficient of  y⁴  is '-17'</em>

<em>   </em>

5 0
3 years ago
Please answer this!!​
Firlakuza [10]

Answer:

You can not solve it because Side P does not exist

Step-by-step explanation:

8 0
3 years ago
Jupiter and Saturn have the two largest diameters of all the planets. The diameter of Jupiter is 88,846 miles and the diameter o
andriy [413]
J = 88846
S = 74897
J = 4.6m + S

Plug in what you know.

88846 = 4.6m + 74897
7 0
4 years ago
a point is randomly chosen in the triangle shown below. which statement is true about the likelihood of the location of the poin
storchak [24]
Where is the diagram.
Just Repost it again.
3 0
3 years ago
Find a power series for the function, centered at c. g(x) = 4x x2 2x − 3 , c = 0
BartSMP [9]

The power series for given function g(x)=\frac{4x}{(x-1)(x+3)} is g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)

For given question,

We have been given a function g(x) = 4x / (x² + 2x - 3)

We need to find a power series for the function, centered at c, for c = 0.

First we factorize the denominator of function g(x), we have:

\Rightarrow g(x)=\frac{4x}{(x-1)(x+3)}

We can write g(x) as,

\Rightarrow g(x)=\frac{1}{x-1}+\frac{3}{x+3}\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1+\frac{x}{3} }\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1-(-\frac{x}{3} )}\\

We know that, \frac{1}{1-x}=\sum{_{n=0}^\infty}~{x^n} if |x| < 1

and \frac{1}{1-(-\frac{x}{3} )}=\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n  if |\frac{x}{6}| < 1

\Rightarrow g(x)=-\sum{_{n=0}^\infty}~x^n+\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n\\     if |x| < 1 and  if |\frac{x}{6}| < 1

\Rightarrow g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n) if |x| < 1

Therefore, the power series for given function g(x)=\frac{4x}{(x-1)(x+3)} is g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)

Learn more about the power series here:

brainly.com/question/11606956

#SPJ4

5 0
2 years ago
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