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11 months ago

Can someone help me on these problems and show work

Mathematics

1 answer:

Salsk061 [2.6K]11 months ago

3 0

The simplified expression are as follows;

6mn³ - mn² + 3mn³ + 15mn² = 9mn³ + 14mn²

a⁴b³ + 8a⁴b³ = 9a⁴b³

<h3>How to simplify an expression?</h3>

A simplified expression is express to its lowest term.

Therefore,

6mn³ - mn² + 3mn³ + 15mn²

combine like terms

6mn³ + 3mn³ - mn² + 15mn²

9mn³ + 14mn²

a⁴b³ + 8a⁴b³ = 9a⁴b³

learn more on simplification here: brainly.com/question/403991

#SPJ1

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The first term of a geometric sequence is 15, and the 5th term of the sequence is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B24

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The geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

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Given that the first term of the geometric sequence is 15

The fifth term of the sequence is $\frac{243}{125}$

We need to find the 2nd, 3rd and 4th term of the geometric sequence.

To find these terms, we need to know the common difference.

The common difference can be determined using the formula,

$a_n=a_1(r)^{n-1}$

where $a_1=15$ and $a_5=\frac{243}{125}$

For $n=5$ , we have,

$\frac{243}{125}=15(r)^4$

Simplifying, we have,

$r=\frac{3}{5}$

Thus, the common difference is $r=\frac{3}{5}$

Now, we shall find the 2nd, 3rd and 4th terms by substituting $n=2,3,4$ in the formula $a_n=a_1(r)^{n-1}$

For $n=2$

$a_2=15(\frac{3}{5} )^{1}$

$=9$

Thus, the 2nd term of the sequence is 9

For $n=3$ , we have,

$a_3=15(\frac{3}{5} )^{2}$

$=15(\frac{9}{25} )$

$=\frac{27}{5}$

Thus, the 3rd term of the sequence is $\frac{27}{5}$

For $n=4$ , we have,

$a_4=15(\frac{3}{5} )^{3}$

$=15(\frac{27}{25} )$

$=\frac{81}{25}$

Thus, the 4th term of the sequence is $\frac{81}{25}$

Therefore, the geometric sequence is $15,9,\frac{27}{5},\frac{81}{25}, \frac{243}{125}$

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2 years ago