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3 years ago

Please help ill give you brainliest

Mathematics

1 answer:

SVEN [57.7K]3 years ago

3 0

The first one is 154

The second one is 12.65^2

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Jack is planning to apply for a new credit card and needs to compare two offers. Complete the table based on the information in

nignag [31]

Answer:

Enjoy, Glad i could help

Step-by-step explanation:

4 0

2 years ago

el precio del kilogramo de jitomate es de 36.50 juana compro 3 1/2 kilogramos cuanto pago por todo el jitomate?

SashulF [63]

El precio total por 3 1/2 kg de jitomate que compró Juana es 127.75

Para saber el precio total de los 3 1/2 kg de jitomate que compró Juana debemos hacer el siguiente procedimiento.

Necesitamos saber cuánto vale medio kg de jitomate, para ello dividimos 36.50 en 2.

36.50/2 = 18.25

Luego multiplicamos 36.50 por 3, para saber cuánto valen 3 kg de jitomate.

36.50 X 3 = 109.5

Por último, sumamos el precio de medio kilo con el precio de tres kilos para tener el total que pagó Juana.

109.5 + 18.25 = 127,75

Aprenda más en: brainly.com/question/16170919

8 0

3 years ago

Read 2 more answers

Triangle FGH with vertices F(1,8), G(5,7), and H(2,3) in the line y=x.

sergiy2304 [10]

Check Hathaway they will

4 0

3 years ago

Read 2 more answers

a triangle has verticals A(-5,-4), B(2,6), C(4,-3) The center of dilation is the origin and (x,y) (3x,3y). what are the vertices

Gnom [1K]

A' (3(-5),3(-4))= (-15,-12)

B' (3(2),3(6))=(6,18)

C'(3(4),3(-3))=(12,-9)

6 0

1 year 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