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3 years ago

The sum of two numbers is 12 and their difference is 20. What is the smaller number?

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

3 0

Answer:

16 and -4

Step-by-step explanation:

Let the 2 numbers be x and y with x > y

x+y=12

x-y=20

Adding the 2 equations, term by term on both sides, will eliminate y leaving an equation in x that we can solve

(x+x)+(-y+y)=(12+20)

2x=32

x=16

16+y=12

y=-4

16+-4=12

12=12

16--4=20

20=20

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

pantera1 [17]

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

(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

ASAP Please answer. Three students sit in the first row. In the second row, 3 students sit behind each of those students, and so

murzikaleks [220]

Answer:

3^2, 9 students.

Step-by-step explanation:

I'm not 100 percent sure about this but I'm pretty sure that because there are 3 rows you can just do 3^2 which is 9.

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3 years ago

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3 years ago

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Lady_Fox [76]

Answer:

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8 0

3 years ago

Can someone please help me with this?

vivado [14]

Answer:

Step-by-step explanation:

As we can see it is a right angled triangle so

hypotenuse (h) = 14 . 7 km

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3 years ago