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3 years ago

5

Tom practiced piano 1 and 1/3 hours on Monday. If he spent half as much time practicing on Tuesday, how long did he practice on

Tuesday? Please show/tell me how you got the answer!!

1 answer:

PtichkaEL [24]3 years ago

3 0

Answer: 2/3 hours

Step-by-step explanation: 1 1/3 will equal 4/3 as you have a whole number equaling 3/1 so then you divide by two and get 2/3

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I need 1 more credit to pass my class! Please Help! <br> I will give brainliest If your right.

Ganezh [65]

Answer:

It’s 135

or 67.5

Step-by-step explanation:

7 0

3 years ago

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Rachel used her card o

Amiraneli [1.4K]

Answer: her call lasted for 26 minutes.

Step-by-step explanation:

Let x represent the number of minutes for which her call lasted.

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Converting 6 cents to dollars, it becomes 6/100 = $0.06

This means that the cost of x minutes of long distance call is

0.06 × x = $0.06x.

If the remaining credit on her card is $28.44, it means that

0.06x + 28.44 = 30

0.06x = 30 - 28.44

0.06x = 1.56

x = 1.56/0.06

x = 26

8 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

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

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3 years ago

120 books increased by 55%

mash [69]

55% of 120 is just 66
add that with 120, 186

so... 186 books

8 0

4 years ago

Why do people give other people brainliest for wrong answers?

svet-max [94.6K]

Answer:

because its 2021

Step-by-step explanation:

Yehhh

8 0

3 years ago

Read 2 more answers