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3 years ago

What is 2.83 as a fraction

Mathematics

2 answers:

kap26 [50]3 years ago

5 0

Either 283/100 or 2 83/100. They are the same number though

Gre4nikov [31]3 years ago

4 0

Answer:

The improper fraction is 283/100

The mixed number is 2 83/100

Step-by-step explanation:

The whole number remains as a whole number

.83 has a number in the hundredths place so it goes over 100

2 83/100

83 is prime so we cannot simplify the fraction

2 83/100 is a mixed number

To change this to an improper fraction

take 100 *2 then add 83

283 and put it over the denominator of 100

283/100

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What is the perimeter of rhombus WXYZ?

oksian1 [2.3K]

<u>Answer</u>:

The perimeter of rhombus WXYZ is $4 \sqrt{13}$

<u>Step-by-step explanation:</u>

Step 1 :Finding length of XY

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 5

$x_2$ =3

$y_1$ = -1

$y_2$ =2

XY = $\sqrt{(3-5)^2 +(2 -(-1))^2}$

XY = $\sqrt{(3-5)^2 +(2 +1))^2}$

XY = $\sqrt{(-2)^2 +(3))^2}$

XY = $\sqrt{4 +9}$

XY = $\sqrt{(13)}$

Step 2 :Finding length of YZ

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 3

$x_2$ =5

$y_1$ = 2

$y_2$ =5

YZ = $\sqrt{(5-3)^2 +(5-2)^2}$

YZ = $\sqrt{(2)^2 +(3)^2}$

YZ = $\sqrt{4 +9}$

YZ = $\sqrt{(13)}$

Step 3 : :Finding length of ZW

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 5

$x_2$ =7

$y_1$ = 5

$y_2$ =2

ZW = $\sqrt{(7-5)^2 +(5-2)^2}$

ZW = $\sqrt{(2)^2 +(3)^2}$

ZW = $\sqrt{4 +9}$

ZW = $\sqrt{(13)}$

Step 4 :Finding length of WX

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 7

$x_2$ =5

$y_1$ = 2

$y_2$ = -1

WX = $\sqrt{(7-5)^2 +((-1)-2)^2}$

WX = $\sqrt{(2)^2 +(-3)^2}$

WX = $\sqrt{4 +9}$

WX = $\sqrt{(13)}$

Step 5: finding the perimeter of the rhombus

Perimeter= 4 X side

=> $4 \times \sqrt{13}$

=> $4 \sqrt{13}$

5 0

3 years ago

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Click on the value for p that makes this inequality true.

Papessa [141]

Answer:0 is the answer

Step-by-step explanation:

6 0

3 years ago

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grandymaker [24]

Answer:

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Step-by-step explanation:

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3 years ago

The drama club was selling ticketsto the school play. Adult ticketscost $8.00 each, and studenttickets cost $5.00 each. The litt

BlackZzzverrR [31]

We are given a problem that can be solved using a system of linear equations. Let A, be the number of adults, and S the number of students. Since there are in total 142 people and there were two days, this means that the sum of the number of adults and the number of students must be 284, which can be written mathematically as follows:

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This is our first equation. The second equation is found using the total sales of $1948. Since the ticket per adult is $8 and per student is $5, we have the following equations: