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3 years ago

if 15 is the fifth number in a series of five consecutive odd numbers, what is the third number in the series ?

1 answer:

6 0

The five numbers can be represented by (2n+1), (2n+3), (2n+5), (2n+7), and (2n+9), where n is an integer.
2n+9 = 15 
n = 3 

The third number is 2n+5 = 11

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Help. What’s the correct answer?

vesna_86 [32]

Answer:

f(g(x)) = -9(sqrt(x + 1) ) + 9

-36

Step-by-step explanation:

This is easier to see if you use the expanded notation.

f(g(x)) = ?

But what that means is that the left most function is f(x) and where every you see an x, you put in a g(x) like so

f(g(x)) = -9(g(x)) + 9 Now substitute the right side of g(x) into the equation for f(x)

f(g(x)) = -9(sqrt(x + 1) ) + 9

Edit to get the numerical answer.

This should be your answer or the expression that will give you the answer.

f(g(24)) = -9(sqrt(24+1) ) + 9

f(g(24)) = -9sqrt(25) +9

f(g(24)) = - 9*5 + 9

f(g(24)) = - 45 + 9

f(g(24)) = - 36

7 0

3 years ago

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bixtya [17]

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4 0

2 years ago

Is n+4 equal to 8(n - 4)

weqwewe [10]

No, n+4=4n an 8(n-4)=32n

4 0

3 years ago

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Judy ran the distance s. Monday 0.35, Wednesday 0.2 and Friday 0.25. write the total distance in miles as a fraction in simplest

Alexandra [31]

Answer:

4/5

Step-by-step explanation:

0.35 + 0.2 + 0.25 = 0.80

3 0

2 years ago

Log_c(A)=2 log_c(B)=5, solve log_c(A^5B^3)

vlabodo [156]

We know that $\log_a(bc)=\log_a(b)+\log_a(c)$ .

Using this rule,

$\log_c(A^5B^3)=\log_c(A^5)+\log_c(B^3)$ .

We also know that $\log_c(a^b)=b\log_c(a)$ .

Using this rule,

$\log_c(A^5)+\log_c(B^3)=5\log_c(A)+3\log_c(B)$

Now we know that $\log_c(A)=2,\log_c(B)=5$ so,

$5\cdot2+3\cdot5=10+15=\boxed{25}$ .

Hope this helps :)

3 0

2 years ago