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2 years ago

Octavian became the emperor of Rome in 27 B.C. He ruled until his death in A.D. 14. How long did he rule as emperor?

Mathematics

2 answers:

Archy [21]2 years ago

6 0

He ruled for 40 years (January, 27 B.C until August, AD 14)

Lilit [14]2 years ago

5 0

When counting B.C the years go backward. For example 25 B.C would become 24 B.C. on new years. When counting A.D we're counting forward. Year 1 becomes year 2 and so forth. So knowing that what do you think?

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Joshua randomly selects a vehicle to test drive. There are 12 cars, 9 trucks, and 4 vans on the lot. What is the probability Jos

eimsori [14]

Step-by-step explanation:

The number of vehicles is 25 and the no. of truck is 9 so the probability is 9/25

4 0

2 years ago

Kit bought a soft drink and a sandwich for $9.00. What was the price of each of the sandwich cost 3.5 times as much as the drink

Rainbow [258]

Sandwich cost $7.70 and the soda costs $2.20.

8 0

3 years ago

J. Reexamine the sequence 20, 14, 8, 2, ... from the problem

DENIUS [597]

Answer:

The n th of the given sequence is $t_{n} = 26-6 n$

Step-by-step explanation:

Step 1 :-

Given sequence is 20,14,8,2,.......this sequence in arithmetic progression but this sequence is decreasing sequence.

given first term is 20 and difference is $d = second term- first term = 14-20=-6$

now the nth term of given sequence is

by using formula $t_{n}=a+(n-1)d$

$t_{n}= 20+(n-1)(-6)$

$t_{n}= 20-6 n+6$

final answer:-

$t_{n} = 26-6 n$

verification:-

$t_{n} = 26-6 n$

put n=1 we get first term is 20

put n=2 we get second term is 14

put n=3 we get third term is 8

put n=4 we get fourth term is 2

so the n th term of sequence is

$t_{n} = 26-6 n$

3 0

3 years ago

Fraction less than 1/2. 3/4 1/6 6/12 4/10 5/8 5/2

N76 [4]

Fractions less than 1/2 are 1/6 & 4/10.
Fractions greater than 1 are 5/4 & 15/12.
Fractions closer to 0 than to 1 are 1/8, 1/4, & 3/10.

5 0

3 years ago

5x3 + 14x2 + 9x help

Cerrena [4.2K]

<h3>Answer: x(x+1)(5x+9) </h3>

===================================================

Work Shown:

5x^3 + 14x^2 + 9x

x( 5x^2 + 14x + 9 )

To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.

A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(5)(9)}}{2(5)}\\\\x = \frac{-14\pm\sqrt{16}}{10}\\\\x = \frac{-14\pm4}{10}\\\\x = \frac{-14+4}{10} \ \text{ or } \ x = \frac{-14-4}{10}\\\\x = \frac{-10}{10} \ \text{ or } \ x = \frac{-18}{10}\\\\x = -1 \ \text{ or } \ x = \frac{-9}{5}\\\\$

Then use those two solutions to find the factorization

x = -1 or x = -9/5

x+1 = 0 or 5x = -9

x+1 = 0 or 5x+9 = 0

(x+1)(5x+9) = 0

So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)

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Overall,

5x^3 + 14x^2 + 9x

factors to

x(x+1)(5x+9)

6 0

3 years ago