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Minchanka [31]
3 years ago
7

What is 18k + 30fk in like terms. Will give brainliest to correct answer.

Mathematics
2 answers:
oee [108]3 years ago
4 0

Answer:

6k(5f+3)

Step-by-step explanation:

18k+30fk

Factor out 6.

6(3k+5fk)

Consider 3k+5fk. Factor out k.

k(3+5f)

Rewrite the complete factored expression.

6k(5f+3)

zloy xaker [14]3 years ago
4 0

Answer:

6k(5f+3)

Step-by-step explanation:

18k+30fk

Factor out 6.

6(3k+5fk)

Consider 3k+5fk. Factor out k.

k(3+5f)

Rewrite the complete factored expression.

6k(5f+3)

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(100) points!!!! Help please
inysia [295]

Answer:

\displaystyle a_8=142,11\ mg

Step-by-step explanation:

Geometric sequence

Each term in a geometric sequence can be computed as the previous term by a constant number called the common ratio. The formula to get the term n is

\displaystyle a_n=a_1r^{n-1}

where a_1 is the first term of the sequence

The problem describes Georgie took 275 mg of the medicine for her cold in the first hour and that in each subsequent hour, the amount of medicine in her body is 91% (0.91) of the amount from the previous hour. It can be written as

amount in hour n = amount in hour n-1 * 0.91

a)

This information provides the necessary data to write the general term as

\displaystyle a_n=275\ .\ 0.91^{n-1}

b)

In the 8th hour (n=8), the remaining medicine present is Georgie's body is

\displaystyle a_n=275\ .\ 0.91^{8-1}

\displaystyle a_n=275\ .\ 0.91^{7}

\boxed{\displaystyle a_8=142,11\ mg}

6 0
3 years ago
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Find the sum of the odd integers between 24 and 50
morpeh [17]

Answer:

  481

Step-by-step explanation:

There are several ways you can get there.

1. There are only 13 numbers, so you can write them down and add them up.

  25 + 27 + 29 + ... + 47 + 49 = 481

__

2. You can use the formula for the sum of an arithmetic sequence. This one has a starting value of 25, an ending value of 49, and 13 terms.

  Sum = ((start) + (end))/2 × (number of terms) = (25 +49)/2×13 = 481

__

3. You can use a formula for the terms of the series and evaluate the sum.

  an = 25 +2(n -1) = 2n +23

\sum\limits_{n=1}^{13}{(2n+23)}=2\sum\limits_{n=1}^{13}(n)+\sum\limits_{n=1}^{13}{(23)}=2\dfrac{13\cdot 14}{2}+13\cdot 23=481

4 0
3 years ago
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