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2 years ago

What are the first three terms of a geometric sequence in which (picture below) and the common ratio is 5?

Mathematics

1 answer:

Marta_Voda [28]2 years ago

5 0

Answer:

5, 10, 15

Step-by-step explanation:

the increase in the sequence is 5 so since it starts from 0 which you can find out because 25 is the 5th sequence and that means that +2*5 =25 . x=0

0+ 5 = sequence 1

0+ 5+5 = seq 2

0+5+5+5= seq 3

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Calculate the perimeter of the figure

larisa86 [58]

Perimeter is distance all around the side of the figure, so you would do:

8 + 7 + 19 + 3 + y + x

Plug in y and x that you solved before: 8 + 7 + 19 + 3 + 11 + 4

Add them!: 52

The perimeter is 52 meters.

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4 years ago

Which of the following is equivalent to 10 3/2?

Aleksandr-060686 [28]

Answer:

D) $\sqrt{10^3}$

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4 years ago

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I need help with these two trigonometry problems.

telo118 [61]

Answer:

B= 21° C= 29°

Step-by-step explanation:

Use sin rule.

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2 years ago

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Artyom0805 [142]

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Step-by-step explanation:

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3 years ago

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The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago