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3 years ago

Write 4% as a reduced fraction and as a decimal

Mathematics

1 answer:

grandymaker [24]3 years ago

5 0

The answer is 0.04
All you have to do is go to a calculator and type 4 then put %

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Write the ratio as a fraction in lowest terms. Be sure to make necessary conversions:

agasfer [191]

Answer:12533

Step-by-step explanation:

4 0

2 years ago

Use crayons to color to shows at least two of it's parts<br><br><br>

Luden [163]

Answer:

im not exactly sure what you mean by that, could you please rephrase and expand on the pregunta

Step-by-step explanation:

5 0

1 year ago

What is the quotient?

yuradex [85]

Answer:

-1.58208955224

Step-by-step explanation:

DO THE MATH

5 0

2 years ago

Plz help meeeeeeeeeee

PSYCHO15rus [73]

Answer:

Please check the explanation.

Step-by-step explanation:

We know the equation

y = kx

k = y/x

where k is the proportionality constant

so,

First ratio can be calculated as

Miles = 45

Hours = 3/4

The ratio = 45 / (3/4)

= (45 × 4) / 3

= 180/3

= 60

Thus, the proportionality constant = 60

Thus, the remaining ratios can be calculated as:

The Pair (135, Hours)

135/Hours = 60

Hours = 135 / 60

= 27/12

= 9/4

= 2.25

Thus, the pair is (135, 2.25)

The Pair (Miles, 3)

Miles/3 = 60

Miles = 60×3

= 180

The Pair (90, Hours)

90/Hours = 60

Hours = 90 / 60

= 3/2

= 1.5

Thus, the complete table:

Miles 45 135 180 90

Hours 3/4 9/4 3 3/2

Verification:

Miles / Hour = 45/(3/4) = 60

Miles / Hour = 135 / 2.25 = 60

Miles / Hour = 180 / 3 = 60

Miles / Hour = 90 / 1.5 = 60

8 0

2 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