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3 years ago

PLEASE HELP ASAP

Mathematics

1 answer:

Sophie [7]3 years ago

4 0

I'm pretty sure that the answer is A

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What is the inverse of y = 2x

Nookie1986 [14]

Answer:

........the inverse is 2x = y

4 0

3 years ago

Find the generating function for the sequence 1,1,1,2,3,4,5,6,....

marin [14]

Answer:

$P(x)=\dfrac{1}{1-x}+\dfrac{x^3}{(1-x)^2} \quad \text{for} \mid x \mid < 1$ [/tex]

Step-by-step explanation:

The generating function of a sequence is the power series whose coefficients are the elements of the sequence. For the sequence

$1,1,1,2,3,4,5,6,...$

the generating function would be

$P(x)=1+x+x^2+2x^3+3x^4+4x^5+5x^6+...\\$

we can multiply P(x) by x to get

$xP(x)=x+x^2+x^3+2x^4+3x^5+4x^6+...$

Note that

$P(x)-xP(x)=1+(2x^3-x^3)+(3x^4-2x^4)+(4x^5-3x^5)+(5x^6-4x^6)+...\\ \\=1+x^3+x^4+x^5+x^6+...=1+x^3(1+x+x^2+x^3+x^4+...)$

which for $\mid x \mid < 1$ can be rewritten as

$(1-x)P(x)=1+\dfrac{x^3}{(1-x)} \quad \Rightarrow \\\\P(x)=\dfrac{1}{(1-x)}+\dfrac{x^3}{(1-x)^2}$

8 0

3 years ago

Helpppppp!!! Much needed

Sonbull [250]

29 + (-37) = 29 - 37
answer is C. 29 - 37

5 0

4 years ago

Sixty out of every 100 pieces of candy is red. Which Indicates the

JulsSmile [24]

Answer:

The proportion of red candies is 60/100.

Step-by-step explanation:

Given that sixty out of every 100 pieces of candy is red, to determine which indicates the proportion of red candies, the following calculation must be performed:

60 red candies out of 100 total candies

60/100

Therefore, the ratio of red candies is 60/100.

8 0

3 years ago

4a/3-b/4=6<br> 5a/6+b=13<br> Solve linear system by elimination

kondor19780726 [428]

$\bf \cfrac{4a}{3}-\cfrac{b}{4}=6
\qquad \qquad 
\cfrac{5a}{6}+b=13
\\\\\\
\textit{let us remove the denominators off those folks}\\
\textit{by multiplying the first one by 12, both sides}\\
\textit{and the second one by 6, both sides, thus}
\\\\\\
12\left( \cfrac{4a}{3}-\cfrac{b}{4} \right)=12(6)\implies 16a-4b=72
\\\\\\
6\left( \cfrac{5a}{6}+b \right)=6(13)\implies 5a+6b=78$

$\bf \textit{now, let's do the elimination}
\\\\

\begin{array}{llll}
16a-4b=72&\boxed{\times 3}\implies &48a-\underline{12b}=216\\\\
5a+6b=78&\boxed{\times 2}\implies &10a+\underline{12b}=156\\
&&--------\\
&&58a+0\quad=372
\end{array}$

solve for "a", once you get "a", plug it back into either equation to get "b"

3 0

3 years ago