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3 years ago

Find the missing value 3/8 equals y over 24

Mathematics

1 answer:

oee [108]3 years ago

7 0

Answer:

Step-by-step explanation:

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In a geometric sequence, a2 = 2, a3 = 20, & a4 = 200.

babymother [125]

Answer:

the first one

the first one Step-by-step explanation:

the first one Step-by-step explanation:from the formula Un =ar^n-1

where a is 2

where a is 2 r = 20/2=10

where a is 2 r = 20/2=10 so Un= 10×2^n-1

that's all.....have fun

5 0

3 years ago

Can anyone help me with these questions 3? 1-10 pls

Alla [95]

Answer:

Step-by-step explanation:

1) A

2) C

3) B

4) A

5) A

6) C

7) B

8) B

9) B

10) A

3 0

3 years ago

How do I find the dimensions of three different rectangles with the area of 6 cm^2

myrzilka [38]

The area of a rectangle is (length) times (width).
So you have to find a pair of numbers that multiply to produce 6 .

If you only stick to whole numbers, then I don't think there are three
different ones. You're going to need one pair that multiply to 6 and
are not both whole numbers.

After I explain how to solve problems, I hate to give answers. But with
all due respect, I have a feeling that I haven't nudged you enough yet
for you to use my explanation to find the answers on your own.
So here are some answers:

1 and 6
2 and 3

and sets of dimensions that are not both whole numbers, like

0.6 and 10
1.2 and 5
1.25 and 4.8
1.5 and 4
2.4 and 2.5

6 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$