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exis [7]
2 years ago
7

The sum of one third of a number and five is equal to six less than the number

Mathematics
1 answer:
abruzzese [7]2 years ago
4 0

Call the number : x

(1/3)x + 5 = x-6

=> (1/3)x + 5 - (x-6) = 0
=> (1/3)x + 5 - x + 6 = 0
Group (1/3)x with -x, 5 with 6

=> [(1/3)x - x] + (5+6) = 0
=> (-2/3)x + 11 = 0
=> (-2/3)x = -11
=> x = -11 : (-2/3) = 33/2.

Recheck : 1/3 x 33/2 + 5 = 33/6 + 5 = 63/6 = 21/2

33/2 - 21/2 = 12/2 = 6 (21/2 is 6 less than 33/2, satisfied.)

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There are 20 entries in the chess tournament. How many ways can they finish in first, second,
Romashka [77]

Answer:

6840

Step-by-step explanation:

Here, the order matters (1st, 2nd, and 3rd), so we need to use permutations.

There are 20 people who can get in first place.  After that, there are 19 people who can get in second place.  Finally, there are 18 people who can get in third place.  So:

20 * 19 * 18 = 6840

3 0
3 years ago
Ryan spent $3.25 on lunch every day, Monday through Friday. If he had $20 at the start of the week, how much money did he have l
shusha [124]

Monday through Friday is 5 days.

Multiply the cost of lunch by number of days:

3.25 x 5 = $16.25

Subtract the total he spent on lunch from what he started with for money:

20 - 16.25 = 3.75

He had $3.75 left.

6 0
3 years ago
Read 2 more answers
Please I need your help on this.
just olya [345]

Answer:

the answer is 7 7/8

Step-by-step explanation:

1.  you will need to change the fractions into an improper fraction.

2. you need to then multiply the numerator and denominator to get a 63/8

3. change 63/8 into a mixed fraction leaving you with 7 7/8

8 0
3 years ago
How do I solve for n? 13=n-8
choli [55]
N in (-oo:+oo)

-13 = n-8 // - n-8

8-n-13 = 0

-n-5 = 0 // + 5

-n = 5 // * -1

n = -5

n = -5
3 0
3 years ago
Read 2 more answers
Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5
Marianna [84]
\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}

\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)

of course, you can always use  (x-1)(x-1)(2x+1)(x²-4x+5)  as well.
7 0
3 years ago
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