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4 years ago

Predict the next three numbers in the pattern 10,7,4,1

Mathematics

2 answers:

bazaltina [42]4 years ago

7 0

Since the pattern is subtracting 3 each time, the next three numbers would be; -2, -5, and -8.

Anettt [7]4 years ago

6 0

-2, -5, -8. The pattern shows that the numbers decrease by 3.

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Expand:<br><img src="https://tex.z-dn.net/?f=f%28z%29%20%3D%20%20%5Cfrac%7B1%7D%7Bz%28z%20-%202%29%7D%20" id="TexFormula1" title

Monica [59]

Expand f(z) into partial fractions:

$\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)$

Recall that for |z| < 1, we have the power series

$\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n$

Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have

$\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}$

So the series expansion of f(z) for |z| > 2 is

$\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)$

$\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}$

$\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}$

$\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}$

6 0

2 years ago

HELP ASAP IF CORRECT WILL FIVE BRAINLIEST

Nataliya [291]

I’m pretty sure the answer is 160 because the rectangle on the top has a volume of 64(16 x 4) and the rectangle on the bottom has a volume of 96(2 x 4 x 12). Add those together and you get 160 cubed inches. Hope this helped :)

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3 years ago

What is the distance between each pair of points?

Pavlova-9 [17]

Provides the information, the distance between the two points would be (5,1).

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4 years ago

Does a negative plus a negative plus a negative plus a negative plus a negative equal a positive

ra1l [238]

No thats only if your multiplying it gets confusing like that

4 0

4 years ago

Read 2 more answers

Someone show equation form

Oliga [24]

Answer/Step-by-step explanation:

$\begin{gathered}\boxed{\begin{array}{c|c}\boxed{\bf Form} &\boxed{\bf Equation} \\ \sf Slope\: intercept\:form &\sf y=mx+b \\ \sf Intercept\:form &\sf \dfrac{x}{a}+\dfrac{y}{b}=1\\ \sf Normal\:form &\sf xcos\omega+ysin\omega=p \\ \sf Two\: point\:form &\sf y-y_1=\left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x-x_1)\\ \sf Point\:slope\:form &\sf y-y_1=m(x-x_1) \\ \sf Standard\:form &\sf Ax+By+C=0 \end{array}}\end{gathered}$

[RevyBreeze]

3 0

2 years ago