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2 years ago

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(1/4)(5/6)^x=75/432. Solve the Equation

1 answer:

Ksivusya [100]2 years ago

4 0

This one I believe is easier than it looks. First you want to start by isolating the part with the exponent. To do this start by dividing both sides by (1/4) and you get….

(5/6)^x= (25/36)

Immediately, it’s noticeable that 25 is 5^2 and 36 is 6^2

So you can rewrite this equation as

(5/6)^x=(5/6)^2

Because the bases are the same now, all you need to do is solve for the exponents. In which you get x=2.

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Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

creativ13 [48]

Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

$a_1=a_{0}+3-1$

$a_1=0+2$

$a_1=2$

For n=2,

$a_2=a_{2-1}+3(2)-1$

$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

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Answer:

x = 4

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Step-by-step explanation:

You can make the equations 3x - 1 = 2x + 3

x = 4

y = 11

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