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3 years ago

7

The sum of three consecutive odd numbers is one hundred fifty-nine

1 answer:

xeze [42]3 years ago

8 0

First remember that sum means to add. now think about the equation. we need to find 3 numbers that equal 159 when added together. And they are all consecutive odd numbers. so let's call the first number X. the second number would be X+2 since we have to jump over 1 even number. The third number would be X+4 since we have to jump over 2 even numbers. so now an equation to represent this would be X+(X+2)+(X+4)=159. Add like terms, and remember order of operations. if you need help getting the answers leave me a comment.

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5/7x + 15 - 9/14x - 9

muminat

Answer:

1/14x+6

Step-by-step explanation:

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3 years ago

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Can I get some help plz?

Anika [276]

Answer:

X=20 degree

Step-by-step explanation:

∆A+∆D+∆O=180

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3 years ago

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You recently invested $400,000 of your savings in a security issued by a large company. The security agreement pays you 12% per

ANTONII [103]

Answer:

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Step-by-step explanation:

Use Compound Intrest Formula

$a = p(1 + \frac{r}{n} ) {}^{nt}$

where p is the original amount.
R is the amount of percentage compounded
N is amount of times compounded per year.
T is how long the interest last.

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T is 12% or 0.12

N is 4 since it is compounded quarterly

T is 5.

Plug the values in

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6 0

3 years ago

Which expression is equivalent to |1-5| + |3|?<br><br><br> O -2<br><br> O -8<br><br> O 2<br><br> O 8

vfiekz [6]

Answer:

8

Step-by-step explanation:

because you do |1-5| first then do add ur answer to |3|

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3 years ago

An arithmetic sequence is defined by the recursive formula t1 = 11, tn = tn - 1 - 13, where n ∈N and n > 1. Which of these is

galben [10]

coolio

$t_1=11$

$t_n=t_{n-1}-13$

so each term is ound by subtracting 13 from the previous term

an aritmetic sequence can be written as

$t_n=t_1+d(n-1)$ were

$t_n$ is the nth term

$t_1$ is the first term

d is common difference, which can also be found by doing $t_n-t_{n-1}=d$

n=wich term

we know that $t_1=11$ and we can find d

$t_n=t_{n-1}-13$ , $t_n-t_{n-1}=-13=d$

so te general term is $t_n=11-13(n-1)$ which can also be expanded and written as $t_n=-13n+24$

6 0

4 years ago