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Andreyy89
4 years ago
6

What is the value of each number in 1,567.209?

Mathematics
2 answers:
nordsb [41]4 years ago
7 0
1 thousand
5 hundred
6 tens
7 ones
2 tenths
0 hundredths
9 thousandths
Scilla [17]4 years ago
4 0

The value of one (1) is one-thousand (1,000)

The value of five (5) is five-hundred (500)

The value of six (6) is sixty (60)

The value of seven (7) is seven (7)

The value of two (2) is two-tenths (0.2)

The value of zero (0) is zero hundredths (0.00)

The value of nine (9) is nine thousandths (0.009

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pickupchik [31]
No thers not a trend because a trend need to be in a strait line and thers not a strait line in the photo

5 0
3 years ago
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Dale runs 10 miles in 75 minutes. At the same rate, how many miles would he run in 69 minutes?
Tems11 [23]

Answer: 9.2 miles

Step-by-step explanation:

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A theme park had 1099 visitors in 3 days.There were twice as many visitors on saturday than on Friday.There were 234 more visito
Blababa [14]

Answer: there were 692 visitors on Saturday.

Step-by-step explanation:

Let x represent the number of visitors that were there on Friday.

Let y represent the number of visitors that were there on Saturday.

Let z represent the number of visitors that were there on Sunday.

The theme park had 1099 visitors in 3 days. It means that

x + y + z = 1099- - - - - - - - - - - - - -1

There were twice as many visitors on saturday than on Friday. This means that

y = 2x

x = y/2

There were 234 more visitors on Sunday than on Saturday. This means that

z = y + 234

Substituting x = y/2 and z = y + 234 into equation 1, it becomes

y/2 + y + y + 234 = 1099

Cross multiplying by 2, it becomes

y + 2y + 2y + 468 = 2198

5y = 2198 - 468

5y = 1730

x = 1730/5

x = 346

y = 2x = 2 × 346

y = 692

z = y + 234 = 692 + 234

z = 926

3 0
3 years ago
Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.
vitfil [10]

Answer:

(A) e^2

(b) e^{-3}

(c) e^2

(d) e^3

Step-by-step explanation:

We have given expression and we have to simplify the expression using exponent property

(A) (\frac{1}{e})^{-2}

So (\frac{1}{e})^{-2}=\frac{1}{e^{-2}}=e^2

(b) (\frac{e^5}{e^2})^{-1}

So (\frac{e^5}{e^2})^{-1}=(e^{3})^{-1}=e^{-3}

(c) \frac{e^5}{e^3}

So \frac{e^5}{e^3}=e^2

(d) \frac{1}{e^{-3}}=e^3

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3 years ago
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Ghella [55]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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