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vagabundo [1.1K]

3 years ago

14

Ben and irie made a secret code. they wrote some numbers of the code so they could remember the pattern. what should the next nu

mber be?

1 answer:

stiks02 [169]3 years ago

4 0

Well what was the pattern they had

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If you're good at logarithms please help me with question two d and h thanks

Alik [6]

Answer:

d. -0.301

h. 1.38

Step-by-step explanation:

Loga0.5 = loga2^-1

= -1(loga2)

= -1 (0.301)

= -0.301

Loga24 = loga(2^3 x 3)

= loga2^3 + loga3

= 3 (loga2) + loga3

= 3 (0.301) + (0.477)

= 1.38

5 0

3 years ago

In 2014, adele sold 2,500,000 records. In 2015, she sold 90% of the number of records she sold in 2014. In 2016, she sold 80% of

Alla [95]

She sold 70% bc 100% - 90% - 80% -70%

8 0

2 years ago

Read 2 more answers

Point e is located at -5,2 point m is the reflection of point e what is the distance between e and m

nataly862011 [7]

Answer:

the distance is 10

Step-by-step explanation:

-5,2 to 0,0 is 5

and the refection of -5,2 is 5,2

5,2 to 0,0. is also 5

5+5=10

4 0

2 years ago

Whats the Axis of Symmetry? willing to give 15 points !!

nydimaria [60]

The axis of symmetry of a parabola in the form

$y=ax^2+bx+c$

is a vertical line

$x=k$

where k is the x coordinate of the vertex of the parabola.

The x coordinate of the vertex of the parabola is given by

$-\dfrac{b}{2a}$

which in your case is

$-\dfrac{-6}{2\cdot \frac{1}{4}}=\dfrac{6}{\frac{1}{2}}=6\cdot 2=12$

So, the axis of symmetry is the line x=12.

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