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3 years ago

What is one over six plus three over six?

Mathematics

1 answer:

IrinaK [193]3 years ago

8 0

Answer:4 over 6

Step-by-step explanation:since both denominators are 6, the denominator stays the same. so all you have to do is add 1 and 3. You can also round to 2 over 3.

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Find the value of the expression 2xy^3 if x=3 and y=-2?

Nadusha1986 [10]

So this question basically asks what is 2 x 3 x -2³ and that equals 6 x -8 = -48

6 0

3 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

4 years ago

Is 15 cm greater than or less than 150 mm

Alekssandra [29.7K]

Answer:

Those are the same measurment!

Step-by-step explanation:

Conversion:

Conversion is 1cm=10mm

6 0

2 years ago

Help ok so like help please this is due tonight

Dima020 [189]

A
E
H
all have the same vale of 81

5 0

3 years ago

HELP ASAP!!!!!!!!!!!!!!

gizmo_the_mogwai [7]

Answer: is it a function-yes

why: none of the y values are the same

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers