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harkovskaia [24]
3 years ago
12

Look at pic Time in Weeks Which panda was heavier when they were born?

Mathematics
1 answer:
arlik [135]3 years ago
7 0
What the top guy said
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Please help me with number 1, thanks
tekilochka [14]
What don't you get about the problem ?
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Plz help!:&gt; also your so awesome, and I hope you have an amazing day!<br><br>​
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It’s graph 2 because it is. There is a pause for the taking the bath and then the emptiest mhm so it can’t be graph 3
6 0
3 years ago
. How can you determine if two ratios are equivalent. Be sure to include examples to support you response.
BabaBlast [244]

Answer:

By multiplying each ratio by the second number of the other ratio, you can determine if they are equivalent. Multiply both numbers in the first ratio by the second number of the second ratio. For example, if the ratios are 3:5 and 9:15, multiply 3 by 15 and 5 by 15 to get 45:75.

7 0
3 years ago
5 The cost of an adult's ticket into a theme park is $a.
natta225 [31]

Answer:

<u>Total Cost (in dollars) = a + c</u>

Step-by-step explanation:

<u>Algebra</u>

When mathematics quantities are generalized into letters or variables, then we are dealing with algebra.

We are said the cost of an adult's ticket into a theme park is $a and a child's ticket costs $c. Since both quantities are unknown, we must treat them as variables and use the same logic procedure to solve the problem as if they were numbers.

The total cost for an adult and a child is the sum of both individual costs, thus

Total Cost (in dollars) = a + c

4 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
3 years ago
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