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Yuliya22 [10]
2 years ago
8

Can someone plzzzz help with the last 3 question I really need it !!!

Mathematics
2 answers:
elixir [45]2 years ago
7 0

Answer:

It is 70 % for the first to last three

Step-by-step explanation:

lubasha [3.4K]2 years ago
4 0
Coupon is 345, tax is 138, 2300-345+138. Add that and you’ll get your final answer.
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the lenght of a rectangular storage room is 2 feet longer than its width. What are the dimensions of the room if the perimeter o
Ne4ueva [31]
16.5 x 18.5 is the answer to your question
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What is the probability of rolling a summer five when rolling to six sided number cubes
Vera_Pavlovna [14]
Not sure about the summer part, but  the answer should be 1 / 6.  Maybe your question is wrong.
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3 years ago
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40 POINTS!!
Kaylis [27]

The answer is C 84. Your welcome

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3 years ago
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A group of 2 adults and 4 children spent $38 on tickets to a museum. A group of 3 adults and 3 children spent $40.50 on tickets
mrs_skeptik [129]

Answer:

adult $8.00

child $5.50

Step-by-step explanation:

Let the price of 1 adult ticket = x.

Let the price of 1 child ticket = y.

2x + 4y = 38

3x + 3y = 40.5

Multiply the first equation by 3. Multiply the second equation by -2. Then add them.

         6x + 12y = 114

(+)      -6x - 6y = -81

-----------------------------

                  6y = 33

y = 33/6

y = 5.5

2x + 4y = 38

2x + 4(5.5) = 38

2x + 22 = 38

2x = 16

x = 8

Answer:

adult $8.00

child $5.50

4 0
2 years ago
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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
3 years ago
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