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3 years ago

Multiply mixed numbers 3 1/2 X 2 1/3

Mathematics

2 answers:

Drupady [299]3 years ago

7 0

Answer:

8 1/6

Step-by-step explanation:

3 1/2 X 2 1/3

7/2x7/3

cricket20 [7]3 years ago

6 0

Answer:

8 1/6

Step-by-step explanation:

that is da answer bbbbb

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5/12-1/9 can you help me?

bezimeni [28]

Answer: 11/6

Step-by-step explanation:

5/12 - 1/9=?

$\frac{5*3}{12*3}-\frac{1*4}{9*4}\\\\\frac{15}{36}-\frac{4}{36}= \\\\15-4=11\\Answer: \frac{11}{36}$

5 0

2 years ago

|x-8|<13<br> What is the solution

maks197457 [2]

Answer:

x = (-5, 21)

Step-by-step explanation:

|x - 8| < 13

1 .

x - 8 < 13
x < 8 + 13
x < 21

x - 8 > -13
x > 8 - 13
x > -5

<u>Combined answer</u>

-5 < x < 21
x = (-5, 21)

5 0

3 years ago

Alison can pay for her gym membership on a monthly basis but if she pays for an entire year of the membership and event shall re

horsena [70]

What do you think the answer is? Tell me and I will help you some more! :)

6 0

3 years ago

Identify the cross section of the solid shown in the diagram

Ivahew [28]

Given:

Image of the pyramid

To identify:

The cross section of the solid.

Solution:

The given image is a pyramid.

The base of the pyramid have 5 edges.

So, it is a pentagonal pyramid.

In the cross section also have 5 edges.

Therefore the cross section of the solid shown in the diagram is pentagon.

6 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