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2 years ago

What is 8.33333 as a fraction?

2 answers:

8 0

The number before the decimal is the whole number and .33333 is equal to one third so the answer would be 8 and one-third

beks73 [17]2 years ago

4 0

$8 \frac{1}{3}$

I only know this because I remember the common fractions, and 1/3 is .3333 repeating.

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Jeremy's password is the combination of letters and numbers shown below. He

wel

Answer:

numbers:letters

5:4 (numbers:letters form)

Step-by-step explanation:

There are 5 numbers in the password and 4 letters

Hopefully this helps :)

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2 years ago

2. Who do you called the number in the expression 2x + 70 A variable B consioni C.bose D exponent

Komok [63]

Answer:

The answer to this question would be constant

Step-by-step explanation:

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2 years ago

Two times a number added to -4, subtracted from triple the sum of eight times the number and five

mr Goodwill [35]

Answer:

2x+(-4)-8x(3)-5

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2 years ago

Read 2 more answers

What digit is in the ten thousands place 1.03098 show how you got this and I will mark brainiest

Luden [163]

The 0 is in the ten thousands

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2 years ago

One Sunday, 120 days before Christmas, Aldsworth store publishes an advertisement saying ‘120 shopping days until Christmas'. Al

Lena [83]

Answer:

(a)18

(b)1089

(c)Sunday

Step-by-step explanation:

The problem presented is an arithmetic sequence where:

First Sunday, a=1
Common Difference (Every subsequent Sunday), d=7

We want to determine the number of Sundays in the 120 days before Christmas.

(a)In an arithmetic sequence:

$\text{The nth term}, T_n=a+(n-1)d\\T_n \leq 120\\$Therefore:$\\1+7(n-1) \leq 120\\1+7n-7\leq 120\\7n-6\leq 120\\7n\leq 120+6\\7n\leq 126\\$Divide both sides by 7$\\n\leq 18$

Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.

Therefore, Aldsworth advertised 18 times.

(b)Next, we want to determine the sum of the first 18 terms of the sequence

1,8,15,...

$\text{Sum of a sequence}, S_n=\frac{n}{2}( 2a+(n-1)d)\\S_{18}=\frac{18}{2}( 2*1+(18-1)*7)\\=9(2+17*7)\\=9(2+119)\\=9*121\\S_{18}=1089$

The sum of the numbers of days published in all the advertisements is 1089.

(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.

6 0

2 years ago