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2 years ago

Does anyone know the answer for his ?

Mathematics

2 answers:

n200080 [17]2 years ago

3 0

The answer is B :) i answered the other one

Tom [10]2 years ago

3 0

The answer is B x=1+2i,1−2i

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The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)

lbvjy [14]

$\bf ~\hspace{10em}\textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 8x^3-125~~ \begin{cases} 8=2\cdot 2\cdot 2\\ \qquad 2^3\\ 125=5\cdot 5\cdot 5\\ \qquad 5^3 \end{cases}\implies 2^3x^3-5^3\implies (2x)^3-5^3 \\[2em] [2x-5][(2x)^2+(2x)(5)+5^2]\implies (2x-5)(\stackrel{\stackrel{j}{\downarrow }}{4}x^2+\stackrel{\stackrel{k}{\downarrow }}{10}x+25)$

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3 years ago

Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in Example 8). [infinity] 6 n2

vekshin1

Answer:

Step-by-step explanation:

given

$\sum\limits^\infty_{n=3}\frac{6}{n^2-1}=\sum\limits^\infty_{n=3}\frac{6}{(n-1)(n+1)}\\\\=\sum\limits^\infty_{n=3}\frac{6}{2}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=\frac{6}{2}\sum\limits^\infty_{n=3}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=3[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+...]\\\\=3[\frac{1}{2}+\frac{1}{3}]=3[\frac{3+2}{6}]=3[\frac{5}{6}]\\\\=\frac{15}{6}=\frac{5}{3}$

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3 years ago

Find the midpoint of the segment between (6,7) and (6,-5)

Norma-Jean [14]

Answer:

(6,1)

Step-by-step explanation:

midpoint = ((x2+x1),(y2+y1)/2)

= (6+6)/2, (-5+7)/2

= 12/2, 2/2

= 6,1

3 0

3 years ago

Read 2 more answers

What is 1/5 x 5 1/3? help please

Law Incorporation [45]

Answer:

1 1/15

Step-by-step explanation:

(I think)

And I am not a very good explainer

(sorry)

8 0

2 years ago

Find the product.<br> 0.18 × 4

ZanzabumX [31]

0.18 x 4 = 0.72
0.72 is the product

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1 year ago