3/4 + 1/3 *4 1/3 / 3 1/3

2 answers:

Elena-2011 [213]2 years ago

5 0

$\frac{3}{4}+\frac{1}{3}*4\ \frac{1}{3}/3\ \frac{1}{3}\\\\\frac{3}{4}+\frac{1}{3}*(\frac{4*3}{3}+\frac{1}{3})/ (\frac{3*3}{3}+\frac{1}{3})\\\\\frac{3}{4}+\frac{4}{12}*\frac{13}{3}/\frac{10}{3}\\\frac{3}{4}+\frac{4*13}{12*3}* \frac{3}{10}\\\\\frac{3}{4}+\frac{4*13*3}{12*3*10}\\\\\frac{3}{4}+\frac{2*2*3*13}{2*2*3*3*2*5}\\\\\frac{3}{4}+\frac{13}{3*2*5}\\\\\frac{3}{4}+\frac{13}{30}\\\\\frac{3}{4}*\frac{15}{15}+\frac{13}{30}*\frac{2}{2}\\\\\frac{45}{60}+\frac{26}{60}\\\\\\\frac{71}{60}=1\ \frac{11}{60}=1.833$

Alexeev081 [22]2 years ago

4 0

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given that a1=5 and a2=15 are the first two terms of a geometric sequence, determine the values of a3 and a10. Show the calculat

Alekssandra [29.7K]

a_3 is 45 and a_10 is 98415

Step-by-step explanation:

Given

a_1 = 5

a_2 = 15

First of all we have to calculate the common ratio of the sequence. The common ratio is the ratio between two consecutive terms of a geometric sequence.

So,

$r=\frac{a_2}{a_1} = \frac{15}{5} = 3$