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2 years ago

Question in picture

Mathematics

2 answers:

Bumek [7]2 years ago

7 0

A sounds corrects to me Zhen spends 6 1/2

Sidana [21]2 years ago

6 0

Answer A. is correct I’m pretty sure

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Which two expressions are equivelent?!! Please help me with it!!☺

arsen [322]

J

You get the same answer, you just start with (x + 3) in the second equation...

That moment when you realize you're dumb and the correct answer is F

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2 years ago

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The sum of 4 consecutive intergers is 50, what is the first interger in this relationship

Vesnalui [34]

Answer:

x = 11

Step-by-step explanation:

Consecutive means in a row

Let x be the first integer

x+1 is the second

x+2 is the third

x+3 is the 4th

The sum is 50

x+ x+1 + x+2 + x+3 = 50

Combine like terms

4x+6 = 50

Subtract 6 from each side

4x+6-6 = 50-6

4x = 44

Divide each side by 4

4x/4 = 44/4

x = 11

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2 years ago

HELP QUICK! You must find all intervals where f(x) is negative. f(x)=2(x+3)(x-7)

beks73 [17]

Answer:

any # where -2<=x<7

Step-by-step explanation:

(x-7) would be negative and (x+3) positive, resulting in f(x) being negative.

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1 year ago

Please help me with this question

MariettaO [177]

Answer:

It would be A,B

Step-by-step explanation:

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2 years ago

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

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2 years ago